MATHEMATICS, BASIC MATH AND ALGEBRA

MATHEMATICS, VOLUME 1

results of counting, it is too cumbersome for purposes of calculation. Before arithmetic could develop as we know it today, the following two important concepts were needed as addi- tions to the counting process: 1. The idea of 0 as a number. 2. Positional notation (place value). Positional notation is a form of coding in which the value of each digit of a number de- pends upon its position in relation to the other digits of the number. The convention used in our number system is that each digit has a higher place value than those digits to the right of it. The place value which corresponds to a given position in a number is determined by the BASE of the number system. The base which is most commonly used is ten, and the system with ten as a base is called the decimal system (decem is the Latin word for ten). Any number is as- sumed to be a base-ten number, unless some other base is indicated. One exception to this rule occurs when the subject of an entire dis- cussion is some base other than ten. For ex- ample, in the discussion of binary (base two) numbers later in this chapter, all numbers are assumed to be binary numbers unless some other base is indicated.

DECIMAL SYSTEM

In the decimal system, each digit position in a number has ten times the value of the position adjacent to it on the right. For example, in the number 11, the 1 on the left is said to be in the "tens place," and its value is 10 times as great as that of the 1 on the right. The 1 on the right is said to be in the "units place," with the un- derstanding that the term "unit" in our system refers to the numeral 1. Thus the number 11 is actually a coded symbol which means "one ten plus one unit." Since ten plus one is eleven, the symbol 11 represents the number eleven. Figure 1-1 shows the names of several digit positions in the decimal system. If we apply this nomenclature to the digits of the integer 235, then this number symbol means "two hun- dreds plus three tens plus five units." This number may be expressed in mathematical symbols as follows:

2 x 10³ + 3 x 10 + 5 x 1

Notice that this bears out our earlier statement: each digit position has 10 times the value of the position adjacent to it on the right.

1999

UNIT'S TENS HUNDREDS THOUSANDS

Figure 1-1.—Names of digit positions.

The integer 4,372 is a number symbol whose meaning is "four thousands plus three hundreds plus seven tens plus two units." Expressed in mathematical symbols, this number is as fol- lows:

4 x 1000 + 3 x 100 + 7 x 10 + 2 x 1

This presentation may be broken down further, in order to show that each digit position as 10 times the place value of the position on its right, as follows:

4 x 10 x 100 + 3 x 10 x 10 + 7 x 10 x 1 + 2 x 1

The comma which appears in a number sym- bol such as 4,372 is used for "pointing off" the digits into groups of three beginning at the right-hand side. The first group of three digits on the right is the units group; the second group is the thousands group; the third group is the millions group; etc. Some of these groups are shown in table 1-1.

Table 1-1.—Place values and grouping.

| Billions group | Millions group | Thousands group | Units group | |---|---|---|---| | Hundred billions Tens billions Billions | Hundred millions Tens millions Millions | Hundred thousands Tens thousands Thousands | Hundreds Tens Units |

By reference to table 1-1, we can verify that 5,432,786 is read as follows: five million, four

Chapter 1—NUMBER SYSTEMS AND SETS

hundred thirty-two thousand, seven hundred eighty-six. Notice that the word "and" is not necessary when reading numbers of this kind.

Practice problems: 1. Write the number symbol for seven thousand two hundred eighty-one. 2. Write the meaning, in words, of the symbol 23,469. 3. If a number is in the millions, it must have at least how many digits? 4. If a number has 10 digits, to what number group (thousands, millions, etc.) does it belong?

Answers: 1. 7,281 2. Twenty-three thousand, four hundred sixty- nine. 3. 7 4. Billions

BINARY SYSTEM

The binary number system is constructed in the same manner as the decimal system. How- ever, since the base in this system is two, only two digit symbols are needed for writing num- bers. These two digits are 1 and 0. In order to understand why only two digit symbols are needed in the binary system, we may make some observations about the decimal system and then generalize from these. One of the most striking observations about number systems which utilize the concept of place value is that there is no single-digit sym- bol for the base. For example, in the decimal system the symbol for ten, the base, is 10. This symbol is compounded from two digit symbols, and its meaning may be interpreted as "one base plus no units." Notice the implication of this where other bases are concerned: Every system uses the same symbol for the base, namely 10. Furthermore, the symbol 10 is not called "ten" except in the decimal system. Suppose that a number system were con- structed with five as a base. Then the only digit symbols needed would be 0, 1, 2, 3, and 4. No single-digit symbol for five is needed, since the everyday 10 in a base-five system with place value means "one five plus no units." In gen- eral, in a number system using base N, the largest number for which a single-digit symbol is needed is N minus 1. Therefore, when the base is two the only digit symbols needed are 1 and 0.

An example of a binary number is the sym- bol 101. We can discover the meaning of this symbol by relating it to the decimal system. Figure 1-2 shows that the place value of each digit position in the binary system is two times the place value of the position adjacent to it on the right. Compare this with figure 1-1, in which the base is ten rather than two.

0101

UNIT'S TWO'S FOUR'S EIGHT'S

Figure 1-2.—Digit positions in the binary system.

Placing the digits of the number 101 in their respective blocks on figure 1-2, we find that 101 means "one four plus no twos plus one unit." Thus 101 is the binary equivalent of decimal 5. If we wish to convert a decimal number, such as 7, to its binary equivalent, we must break it into parts which are multiples of 2. Since 7 is equal to 4 plus 2 plus 1, we say that it "con- tains" one 4, one 2, and one unit. Therefore the binary symbol for decimal 7 is 111. The most common use of the binary number system is in electronic digital computers. All data fed to a typical electronic digital computer is converted to binary form and the computer performs its calculations using binary arith- metic rather than decimal arithmetic. One of the reasons for this is the fact that electrical and electronic equipment utilizes many switch- ing circuits in which there are only two operat- ing conditions. Either the circuit is "on" or it is "off," and a two-digit number system is ideally suited for symbolizing such a situation. Details concerning binary arithmetic are be- yond the scope of this volume, but are available in Mathematics, Volume 3, NavPers 10013, and in Basic Electronics, NavPers 10087-A.

Practice problems: 1. Write the decimal equivalents of the binary numbers 1101, 1010, 1001, and 1111. 2. Write the binary equivalents of the decimal numbers 12, 7, 14, and 3.

Answers: 1. 13, 10, 9, and 15 2. 1100, 111, 1110, and 11

MATHEMATICS, VOLUME 1

SETS

Any serious study of mathematics leads the student to investigate more than one text and more than one way of approaching each new topic. At the time of printing of this course, much emphasis is being placed on so-called modern math in the public schools. Consequently, the trainee who uses this course is likely to find considerable material in his parallel reading, which uses the ideas and terminology of the "new" math.

In the following paragraphs, a very brief introduction to some of the set theory of modern math is presented. Although the remainder of this course is not based on set theory, this brief introduction should help in making the transition from traditional methods to newer, experimental methods.

DEFINITIONS AND SYMBOLS

The word "set" implies a collection or grouping of similar objects or symbols. The objects in a set have at least one characteristic in common, such as similarity of appearance or purpose. A set of tools would be an example of a group of objects not necessarily similar in appearance but similar in purpose. The objects or symbols in a set are called members or ELEMENTS OF THE SET.

The elements of a mathematical set are usually symbols, such as numerals, lines, or points. For example, the integers greater than zero and less than 6 form a set, as follows:

{1, 2, 3, 4}

Notice that braces are used to indicate sets. This is often done where the elements of the set are not too numerous.

Since the elements of the set {2, 4, 6} are the same as the elements of {4, 2, 6}, these two sets are said to be equal. In other words, equality between sets has nothing to do with the order in which the elements are arranged. Furthermore, repeated elements are not necessary. That is, the elements of {2, 2, 3, 4} and {2, 3, and 4. Therefore the sets {2, 3, 4} and {2, 2, 3, 4} are equal.

Practice problems:

1. Use the correct symbols to designate the set of odd positive integers greater than 0 and less than 10.

2. Use the correct symbols to designate the set of names of days of the week which do not contain the letter "s".

3. List the elements of the set of natural numbers greater than 15 and less than 20.

4. Suppose that we have sets as follows:

A = {1, 2, 3} C = {1, 2, 3, 4} B = {1, 2, 2, 3} D = {1, 1, 2, 3}

Which of these sets are equal?

Answers: 1. {1, 3, 5, 7, 9} 2. {Monday, Friday} 3. 16, 17, 18, and 19 4. A = B = D

SUBSETS

Since it is inconvenient to enumerate all of the elements of a set each time it is mentioned, sets are often designated by a letter. For example, we could let S represent the set of all integers greater than 0 and less than 10. In symbols, this relationship could be stated as follows:

S = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Now suppose that we have another set, T, which comprises all positive even integers less than 10. This set is then defined as follows:

T = {2, 4, 6, 8}

Notice that every element of T is also an element of S. This establishes the SUBSET relationship; T is said to be a subset of S.

POSITIVE INTEGERS

The most fundamental set of numbers is the set of positive integers. This set comprises the counting numbers (natural numbers) and includes, as subsets, all of the sets of numbers which we have discussed. The set of natural numbers has an outstanding characteristic: it is infinite. This means that the successive elements of the set continue to increase in size without limit, each number being larger by 1 than the number preceding it. Therefore there is no "largest" number; any number that we might choose as larger than all others could be

Chapter 1—NUMBER SYSTEMS AND SETS

increased to a larger number simply by adding 1 to it.

One way to represent the set of natural numbers symbolically would be as follows:

{1, 2, 3, 4, 5, 6, ...}

The three dots, called ellipsis, indicate that the pattern established by the numbers shown continues without limit. In other words, the next number in the set is understood to be 7, the next after that is 8, etc.

POINTS AND LINES

In addition to the many sets which can be formed with number symbols, we frequently find it necessary in mathematics to work with sets composed of points or lines.

A point is an idea, rather than a tangible object, just as a number is. The mark which is made on a piece of paper is merely a symbol representing the point. In strict mathematical terms, a point has no dimensions (physical size) at all. Thus a pencil dot is only a rough picture of a point, useful for indicating the location of the point but certainly not to be confused with the idea.

Now suppose that a large number of points are placed side by side to form a "string." Picturing this arrangement by drawing dots on paper, we would have a "dotted line." If more dots were placed between the dots already in the string, with the number of dots increasing until we could not see between them, we would have a rough picture of a line. Once again, it is important to emphasize that the picture is only a symbol which represents an ideal line. The ideal line would have length but no width or thickness.

The foregoing discussion leads to the conclusion that a line is actually a set of points. The number of elements in the set is infinite, since the line extends in both directions without limit.

The idea of arranging points together to form a line may be extended to the formation of

planes (flat surfaces). A mathematical plane is determined by three points which do not lie on the same line. It is also determined by two intersecting lines.

Line Segments and Rays

When we draw a "line," label its end points A and B, and call it "line AB," we really mean LINE SEGMENT AB. A line segment is a subset of the set of points comprising a line.

When a line is considered to have a starting point but no stopping point (that is, it extends without limit in one direction), it is called a RAY. A ray is not a line segment, because it does not terminate at both ends; it may be appropriate to refer to a ray as a "half-line."

THE NUMBER LINE

As in the case of a line segment, a ray is a subset of the set of points comprising a line. All three—lines, line segments, and rays—are subsets of the set of points comprising a plane.

Among the many devices used for representing a set of numbers, one of the most useful is the number line. To illustrate the construction of a number line, let us place the elements of the set of natural numbers in one-to-one correspondence with points on a line. Since the natural numbers are equally spaced, we select points such that the distances between them are equal. The starting point is labeled 0, the next point is labeled 1, the next 2, etc., using the natural numbers in normal counting order. (See fig. 1-3.) Such an arrangement is often referred to as a scale, a familiar example being the scale on a thermometer.

Thus far in our discussion, we have not mentioned any numbers other than integers. The number line is an ideal device for picturing the

| | | | | | | | 0 1 2 3 4 5 6 7

Figure 1-3.—A number line.

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MATHEMATICS, VOLUME 1

relationship between integers and other num- bers such as fractions and decimals. It is clear that many points, other than those representing integers, exist on the number line. Examples are the points representing the numbers 1/2 (located halfway between 0 and 1) and 2.5 (lo- cated halfway between 2 and 3).

An interesting question arises, concerning the "in-between" points on the number line: How many points (numbers) exist between any two integers? To answer this question, suppose that we first locate the point halfway between 0 and 1, which corresponds to the number 1/2. Then let us locate the point halfway between 0 and 1/2, which corresponds to the number 1/4. The result of the next such halving operation would be 1/8, the next 1/16, etc. If we need more and more space to continue our halving operations on the number line, we can enlarge our "picture" and then continue.

It soon becomes apparent that the halving process could continue indefinitely; that is, without limit. In other words, the number of points between 0 and 1 is infinite. The same is

true of any other interval on the number line. Thus, between any two integers there is an infi- nite set of numbers other than integers. It this seems physically impossible, considering that even the sharpest pencil has some width, remember that we are working with ideal points, which have no physical dimensions whatsoever. Although it is beyond the scope of this course to discuss such topics as orders of infinity, it is interesting to note that the set of integers contains many subsets which are themselves infinite. Not only are the many subsets of num- bers other than integers infinite, but such subsets as the set of all odd integers and the set of all even integers. By intuition we see that these two subsets are infinite, as follows: If we select a particular odd or even integer which we think is the largest possible, a larger one can be formed immediately by merely adding 2.

Perhaps the most practical use for the num- ber line is in explaining the meaning of nega- tive numbers. Negative numbers are discussed in detail in chapter 3 of this course.

6

CHAPTER 2 POSITIVE INTEGERS

The purpose of this chapter is to review the methods of combining integers. We have al- ready used one combination process in our dis- cussion of counting. We will extend the idea of counting, which is nothing more than simple ad- dition, to develop a systematic method for add- ing numbers of any size. We will also learn the meaning of subtraction, multiplication, and division.

ADDITION AND SUBTRACTION

In the following discussion, it is assumed that the reader knows the basic addition and subtraction tables, which present such sums as the following: 2 + 3 = 5, 9 + 8 = 17, 8 - 3 = 5, etc.

The operation of addition is indicated by a plus sign (+) as in 8 + 4 = 12. The numbers 8 and 4 are ADDENDS and the answer (12) is their SUM. The operation of subtraction is indicated by a minus sign (-) as in 9 - 3 = 6. The number 9 is the MINUEND, 3 is the SUBTRAHEND, and the answer (6) is their DIFFERENCE.

REGROUPING

Addition may be performed with the addends arranged horizontally, if they are small enough and not too numerous. However, the most com- mon method of arranging the addends is to place them in vertical columns. In this arrangement, the units digits of all the addends are aligned vertically, as are the tens digits, the hundreds digits, etc. The following example shows three addends arranged properly for addition:

357 1,845 22

It is customary to draw a line below the last addend, placing the answer below this line. Sub- traction problems are arranged in columns in the same manner as for addition, with a line at the bottom and the answer below this line.

Carry and Borrow

Problems involving several addends, with two or more digits each, usually produce sums in one or more of the columns which are greater than 9. For example, suppose that we perform the following addition:

357 845 22

1,224

The answer was found by a process called "carrying." In this process extra digits, gen- erated when a column sum exceeds 9, are car- ried to the next column to the left and treated as addends in that column. Carrying may be explained by grouping the original addenda. For example, 357 actually means 3 hundreds plus 5 tens plus 7 units. Rewriting the problem with each addend grouped in terms of units, tens, etc., we would have the following:

300 + 50 + 7 800 + 40 + 5 20 + 2

1,100 + 110 + 14

The "extra" digit in the units column of the answer represents 1 ten. We regroup the col- umns of the answer so that the units column has no digits representing tens, the tens column has no digits representing hundreds, etc., as follows:

1,100 + 110 + 14 = 1,100 + 110 + 10 + 4 = 1,100 + 120 + 4 = 1,100 + 100 + 20 + 4 = 1,200 + 20 + 4 = 1,000 + 200 + 20 + 4 = 1,224

When we carry the 10 from the expression 10 + 4 to the tens column and place it with the 110 to make 120, the result is the same as if

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MATHEMATICS, VOLUME 1

we had added 1 to the digits 5, 4, and 2 in the tens column of the original problem. There- fore, the thought process in addition as is as follows: Add the 7, 5, and 2 in the units column, getting a sum of 14. Write down the 4 in the units column of the answer and carry the 1 to the tens column. Mentally add the 1 along with the other digits in the tens column, getting a sum of 12. Write down the 2 in the tens column of the answer and carry the 1 to the hundreds column. Mentally add the 1 along with the other digits in the hundreds column, getting a sum of 12. Write down the 2 in the hundreds column of the answer and carry the 1 to the thousands column. If there were other digits in the thou- sands column to which the 1 could be added, the process would continue as before. Since there are no digits in the thousands column of the original problem, this final 1 is not added to anything, but is simply written in the thousands place in the answer.

The borrow process is the reverse of carry- ing and is used in subtraction. Borrowing is not necessary in such problems as 46 - 5 and 58 - 53. In the first problem, the thought proc- ess may be "6 from 6 is 1 and bring down the 4 to get the difference, 41." In the second prob- lem, the thought process is "3 from 8 is 5" and "5 from 5 is zero," and the answer is 5. More explicitly, the subtraction process in these ex- amples is as follows:

40 + 6 50 + 8 5 50 + 3 40 + 1 = 41 0 + 5 = 5

This illustrates that we are subtracting units from units and tens from tens.

Now consider the following problem where borrowing is involved:

43 _8

If the student uses the borrowing method, he may think "8 from 13 is 5 and bring down the 4 to get the difference, 35." In this case what actu- ally was done is as follows:

30 + 13 8 30 + 5 = 35

A 10 has been borrowed from the tens column and combined with the 3 in the units column to make a number large enough for subtraction of the 8. Notice that borrowing to increase the value of the digit in the units column reduces the value of the digit in the tens column by 1.

Sometimes it is necessary to borrow in more than one column. For example, suppose that we wish to subtract 2,345 from 5,234. Grouping the minuend and subtrahend in units, tens, hun- dreds, etc., we have the following:

5,000 + 200 + 30 + 4 2,000 + 300 + 40 + 5

Borrowing a 10 from the 30 in the tens column, we regroup as follows:

5,000 + 200 + 20 + 14 2,000 + 300 + 40 + 5

The units column is now ready for subtrac- tion. By borrowing from the hundreds column, we can regroup so that subtraction is possible in the tens column, as follows:

5,000 + 100 + 120 + 14 2,000 + 300 + 40 + 5

In the final regrouping, we borrow from the thousands column to make subtraction possible in the hundreds column, with the following result:

4,000 + 1,100 + 120 + 14 2,000 + 300 + 40 + 5 2,000 + 800 + 80 + 9 = 2,889

In actual practice, the borrowing and re- grouping are done mentally. The numbers are written in the normal manner, as follows:

5,234 -2,345

2,889

The following thought process is used: Borrow from the tens column, making the 4 become 14. Subtracting in the units column, 5 from 14 is 9. In the tens column, we now have a 2 in the min- uend as a result of the first borrowing operation. Some students find it helpful at first to cancel any digits that are reduced as a result of borrowing, jotting down the digit of next lower

Chapter 2—POSITIVE INTEGERS

value just above the canceled digit. This has been done in the following example:

4 12 5,2̶3̶4̶ -2,345

2,889

After canceling the 3, we proceed with the subtraction, one column at a time. We borrow from the hundreds column to change the 2 that we now have in the tens column into 12. Sub- tracting in the tens column, 4 from 12 is 8. Proceeding in the same way for the hundreds column, 3 from 11 is 8. Finally, in the thou- sands column, 2 from 4 is 2.

Practice problems. In problems 1 through 4, add the indicated numbers. In problems 5 through 8, subtract the lower number from the upper.

1. Add 23, 468, 7, and 9,045.

2. 129 3. 9,497 4. 67,856 958 6,364 22,851 787 4,269 44,238 436 9,785 97,156

5. 709 6. 8,700 7. 7,928 8. 75,168 594 5,008 5,349 28,089

Answers:

1. 9,543 2. 2,310 3. 39,915 4. 232,101 5. 115 6. 3,692 7. 2,579 8. 47,079

Denominate Numbers

Numbers that have a unit of measure associ- ated with them, such as yard, kilowatt, pound, pint, etc., are called DENOMINATE NUMBERS. The word "denominate" means the numbers have been given a name; they are not just ab- stract symbols. To add denominate numbers, add all units of the same kind. Simplify the re- sult, if possible. The following example illus- trates the addition of 6 ft 8 in. to 4 ft 5 in.:

6 ft 8 in. 4 ft 5 in.

10 ft 13 in.

Since 13 in. is the equivalent of 1 ft 1 in., we regroup the answer as 11 ft 1 in.

A similar problem would be to add 20 de- grees 44 minutes 6 seconds to 13 degrees 22 minutes 5 seconds. This is illustrated as fol- lows:

20 deg 44 min 6 sec 13 deg 22 min 5 sec

33 deg 66 min 11 sec

This answer is regrouped as 34 deg 6 min 11 sec.

Numbers must be expressed in units of the same kind. In order to be combined. For in- stance, the sum of 6 kilowatts plus 1 watt is not 7 kilowatts nor is it 7 watts. The sum can only be indicated (rather than performing the opera- tion) unless some method is used to write these numbers in units of the same value.

Subtraction of denominate numbers also in- volves the regrouping idea. If we wish to sub- tract 16 deg 8 min 2 sec from 28 deg 4 min 3 sec, for example, we would have the following arrangement:

28 deg 4 min 3 sec -16 deg 8 min 2 sec

In order to subtract 8 min from 4 min we re- group as follows:

27 deg 64 min 3 sec -16 deg 8 min 2 sec

11 deg 56 min 1 sec

Practice problems. In problem 1, 2, and 3 add. In problem 4, 5, and 6 subtract the lower number from the upper.

1. 6 yd 2 ft 7 in. 4. 15 hr 25 min 10 sec 1 yd 9 in. 6 hr 50 min 35 sec 2 yd 10 in.

2. 9 hr 47 min 51 sec 5. 125 deg 3 hr 36 min 23 sec 47 deg 9 min 14 sec 5 hr 15 min 23 sec

3. 10 wks 5 days 7 hrs 6. 20 wks 2 days 10 hrs 22 wks 3 days 10 hrs 7 wks 6 days 15 hrs 3 wks 4 days 12 hrs

MATHEMATICS, VOLUME 1

Answers:

1. 9 yd 2 ft 2 in. 2. 18 hr 39 min 37 sec 3. 36 wks 6 days 5 hr 4. 8 hr 34 min 35 sec 5. 77 deg 50 min 46 sec 6. 12 wks 2 days 19 hr

Mental Calculation

Mental regrouping can be used to avoid the necessity of writing down some of the steps, or of rewriting in columns, when groups of one-digit or two-digit numbers are to be added or subtracted.

One of the most common devices for rapid addition is recognition of groups of digits whose sum is 10. For example, in the following problem two "ten groups" have been marked with braces:

7

8 | 10 4 |

5

1 | 10 9 |

To add this column as grouped, you would say to yourself, "7, 17, 22, 32." The thought should be just the successive totals as shown above and not such cumbersome steps as "7 + 10, 17, +5, 22, +10, 32."

When successive digits appear in a column and their sum is less than 10, it is often convenient to think of them, too, as a unit rather than separately. Thus, if adding a column in which the sum of two successive digits is 10 or less, group them as follows:

3 | 5 1 |

8 | 9 1 |

4 | 10 6 |

The thought process here might be, as shown by the grouping, "5, 14, 24."

Practice problems. Add the following columns from the top down, as in the preceding example:

1. 2 2. 4 3. 88 4. 57 7 6 36 32 3 7 59 64 6 8 82 97 4 1 28 79 1 8 57 44

Answers, showing successive mental steps:

1. 2, 12, 22, 23 - - Final answer, 23 2. 10, 17, 26, 34 - - Final answer, 34 3. Units column: 14, 23, 33, 40 - - Write down 0, carry 4. Tens column: 12, 20, 30, 35 - - Final answer, 350. 4. Units column: 9, 20, 29, 33 - - Write down 3, carry 3. Tens column: 8, 17, 26, 37 - - Final answer, 373.

SUBTRACTION.--In an example such as 73 - 46, the conventional approach is to place 46 under 73 and subtract units from units and tens from tens, and write only the difference without the intermediate steps. To do this, the best method is to begin at the left. Thus, in the example 73 - 46, we take 40 from 73 and then take 6 from the result. This is done mentally, however, and the thought would be "73, 33, 27." In the example 84 - 21 the thought is "64, 63" and in the example 64 - 39 the thought is "34, 25."

Practice problems. Mentally subtract and write only the difference:

1. 47 - 24 4. 86 - 73 2. 69 - 38 5. 82 - 41 3. 87 - 58 6. 30 - 12

Answers, showing successive mental steps:

1. 27, 23 - - Final answer, 23 2. 39, 31 - - Final answer, 31 3. 37, 29 - - Final answer, 29 4. 16, 13 - - Final answer, 13 5. 42, 41 - - Final answer, 41 6. 20, 18 - - Final answer, 18

MULTIPLICATION AND DIVISION

Multiplication may be indicated by a multiplication sign (x) between two numbers, a dot

Chapter 2--POSITIVE INTEGERS

between two numbers, or parentheses around one or both of the numbers to be multiplied. The following examples illustrate these methods:

6 x 8 = 48 6 · 8 = 48 6(8) = 48 (6)(8) = 48

Notice that when a dot is used to indicate multiplication, it is distinguished from a decimal point or a period by being placed above the line of writing, as in example 2, whereas a period or decimal point appears on the line. Notice also that when parentheses are used to indicate multiplication, the numbers to be multiplied are spaced closer together than they are when the dot or x is used.

In each of the examples just given, 6 is the MULTIPLIER and 8 is the MULTIPLICAND. Both the 6 and the 8 are FACTORS, and the more modern texts refer to them this way. The "answer" in a multiplication problem is the PRODUCT; in the examples just given, the product is 48.

Division usually is indicated either by a division sign (÷) or by placing one number over another number with a line between the numbers, as in the following examples:

1. 8 ÷ 4 = 2

2. 8/4 = 2

The number 8 is the DIVIDEND, 4 is the DIVISOR, and 2 is the QUOTIENT.

MULTIPLICATION METHODS

The multiplication of whole numbers may be thought of as a short process of adding equal numbers. For example, 6(5) is 5 read as six 5's. Of course we would write 5 six times and add, but if we learn that the result is 30 we can save time. Although the concept of adding equal numbers is quite adequate in explaining multiplication of whole numbers, it is only special case of a more general definition, which will be explained later in multiplication involving fractions.

Grouping

Let us examine the process involved in multiplying 6 times 27 to get the product 162. We first arrange the factors in the following manner:

27 x6 162

The thought process is as follows:

1. 6 times 7 is 42. Write down the 2 and carry the 4. 2. 6 times 2 is 12. Add the 4 that was carried over from step 1 and write the result, 16, beside the 2 that was written in step 1. 3. The final answer is 162.

Table 2-1 shows that the factors were grouped in units, tens, etc. The multiplication was done in three steps: Six times 7 units is 42 units (or 4 tens and 2 units) and six times 2 tens is 12 tens (or 1 hundred and 2 tens). Then the tens were added and the product was written as 162.

Table 2-1.--Multiplying by a one-digit number.

[Table with columns: Hundreds, Tens, Units] 6(27) = 162 2 7 6 4 2 1 6 2

In preparing numbers for multiplication as in table 2-1, it is important to place the digits of the factors in the proper columns; that is, units must be placed in the units column, tens in tens column, and hundreds in hundreds column. Notice that it is not necessary to write the zero in the case of 12 tens (120) since the 1 and 2 are written in the proper columns. In practice, the addition is done mentally, and just the product is written without the intervening steps.

Multiplying a number with more than two digits by a one-digit number, as shown in table 2-2, involves no new ideas. Three times 6 units is 18 units (1 ten and 8 units), 3 times 0 tens is 0, and 3 times 4 hundreds is 12 hundreds (1

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MATHEMATICS, VOLUME 1

under 5, take the difference, and bring down 6); and "4 is contained in 16 four times" (write 4 in units place over the 6). After a little prac- tice, many people can do the work shown under the dividend mentally and write only the quo- tient, if the divisor has only 1 digit. The divisor is sometimes too large to be contained in the first digit of the dividend. The following example illustrates a problem of this kind:

36 7)252 21

42 42

Since 2 is not large enough to contain 7, we divide 7 into the number formed by the first two digits. 25. Seven is contained 3 times in 25; we write 3 above the 5 of the dividend. Multiplying, 3 times 7 is 21; we write 21 below the first two digits of the dividend. Subtracting, 25 minus 21 is 4; we write down the 4 and bring down the 2 in the units place of the dividend. We have now formed a new dividend, 42. Seven is contained 6 times in 42; we write 6 above the 2 of the dividend. Multiplying as before, 6 times 7 is 42; we write this product below the dividend 42. Subtracting, we have nothing left and the divi- sion is complete.

Estimation

When there are two or more digits in the divisor, it is not always easy to determine the first digit of the quotient. An estimate must be made, and the resulting trial quotient may be too large or too small. For example, if 1,862 is to be divided by 38, we might estimate that 38 is contained 5 times in 186 and the first digit of our trial divisor would be 5. However, mul- tiplication reveals that the product of 5 and 38 is larger than 186. Thus we would change the 5 in our quotient to 4, and the problem would then appear as follows:

49 38)1862 152

342 342

On the other hand, suppose that we had esti- mated that 38 is contained in 186 only 3 times. We would then have the following:

3 38/1862 114 72

Now, before we make any further moves in the division process, it should be obvious that some- thing is wrong. If our new dividend is large enough to contain the divisor before bringing down a digit from the original dividend, then the trial quotient should have been larger. In other words, our estimate is too small. Proficiency in estimating trial quotients is gained through practice and familiarity with number combinations. For example, after a little experience we realize that a close esti- mate can be made in the foregoing problem by thinking of 38 as "almost 40." It is easy to see that 40 is contained 4 times in 186, since 4 times 40 is 160. Also, since 5 times 40 is 200, we are reasonably certain that 5 is too large for our trial divisor.

Uneven Division

In some division problems such as 7 ÷ 3, there is no other whole number that, when mul- tiplied by the divisor, will give the dividend. We use the distributive idea to show how divi- sion is done in such a case. For example, 7 ÷ 3 could be written as follows:

(6 + 1) 6 1 1 ------- = --- + --- = 2 --- 3 3 3 3

Thus, we see that the quotient also carries one unit that is to be divided by 3. It should now be clear that 3/37 = 3/30 + 7, and that this can be further reduced as follows:

30 6 1 1 1 --- + --- + --- = 10 + 2 + --- = 12 --- 3 3 3 3 3

In elementary arithmetic the part of the divi- dend that cannot be divided evenly by the divisor is often called a REMAINDER and is placed next to the quotient with the prefix R. Thus, in the foregoing example where the quotient was 1 12 ---, the quotient could be written 12 R 1. This 3

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Chapter 2—POSITIVE INTEGERS

method of indicating uneven division is useful in examples such as the following: Suppose that $18 is available for the pur- chase of spare parts, and the parts needed cost $3 each. Four parts can be bought with the available money, and $1 will be left over. Since it is not possible to buy 1/3 of a part, express- ing the result as 4 R 1 gives a more meaningful answer than 4 1/3.

Placement of Decimal Points

In division, as in multiplication, the place- ment of the decimal point is important. Deter- mining the location of the decimal point and the number of places in the quotient can be rela- tively simple if the work is kept in the proper columns. For example, notice the vertical alinement in the following problem:

311 31)9,641 9 3

34 31

31 31

We notice that the first two places in the divi- dend are used to obtain the first place in the quotient. Since 3 is in the hundreds column there are two more places in the quotient (tens place and units place). The decimal point in the quotient is understood to be directly above the position of the decimal point in the dividend. In the example shown here, the decimal point is not shown but is understood to be immediately after the second 1.

Checking Accuracy

The accuracy of a division of numbers can be checked by multiplying the quotient by the divisor and adding the remainder, if any. The result should equal the dividend. Consider the following example:

5203 42)218541 Check: 5203 210 x 42 --- ------ 85 10406 84 20812 --- 141 218526 126 + 15 --- ------ 15 218541

DENOMINATE NUMBERS

We have learned that denominate numbers are not difficult to add and subtract, provided that units, tens, hundreds, etc., are retained in their respective columns. Multiplication and division of denominate numbers may also be performed with comparative ease, by using the experience gained in addition and subtraction.

Multiplication

In multiplying denominate numbers by inte- gers, no new ideas are needed. If in the prob- lem 3(5 yd 2 ft 6 in.) we remember that we can multiply each part separately to get the correct product (as in the example, 6(8) = 6(3) + 6(5)), we can easily find the product, as follows:

5 yd 2 ft 6 in. x 3 ________________ 15 yd 6 ft 18 in.

Simplifying, this is

17 yd 1 ft 6 in.

When one denominate number is multiplied by another, a question arises concerning the products of the units of measurement. The product of one unit times another of the same kind is one square unit. For example, 1 ft times 1 ft is 1 square foot, abbreviated sq ft; 2 in. times 3 in. is 6 sq in.; etc. If it becomes necessary to multiply such numbers as 2 yd 1 ft times 6 yd 2 ft, the foot units may be converted to fractions of a yard, as follows:

(2 yd 1 ft)(6 yd 2 ft) = (2 1/3 yd)(6 2/3 yd)

In order to complete the multiplication, a knowledge of fractions is needed. Fractions are discussed in chapter 4 of this training course.

Division

The division of denominate numbers requires division of the highest units first; and if there is a remainder, conversion to the next lower unit, and repeated division until all units have been divided. In the example (24 gal 1 qt 1 pt) ÷ 5, we per- form the following steps:

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MATHEMATICS, VOLUME 1

Step 1: 4 gal multiplications may be performed in any order. 5/24 gal Thus, in 20 4 + 2 + 7 + 5 = 18 4 gal (left over) or Step 2: Convert the 4 gal left over to 16 qt and add to the 1 qt. 100 - 20 - 10 - 3 = 67

Step 3: 3 qt or 5/17 qt 15 4 x 2 x 7 x 5 = 280

2 qt (left over) the numbers may be combined in any order de- sired. For example, they may be grouped easily Step 4: Convert the 2 qt left over to 4 pt and add to the 1 pt. to give

Step 5: 1 pt 6 + 12 = 18 5/5 pt and Therefore, 24 gal 1 qt 1 pt divided by 5 is 4 gal 3 qt 1 pt. 97 - 30 = 67

Practice problems. In problems 1 through 4, divide as indicated. In problems 5 through 8, and multiply or divide as indicated. 40 x 7 = 280 1. 540 ÷ 9 5. 4 hr 26 min 16 sec x 5 A series of divisions should be taken in the 2. 470/63 order written. 6. 3(4 gal 3 qt 1 pt) 3. 25/2,300 Thus, 7. 67 deg 43 min 12 sec 4. 84/74,518 2 100 ÷ 10 ÷ 2 = 10 ÷ 2 = 5 8. 5/63 lb 11 oz In a series of mixed operations, perform multi- Answers: plications and divisions in order from left to right, then perform additions and subtractions in order from left 1. 61 5. 22 hr 11 min 20 sec to right. 2. 7 R 20 6. 14 gal 2 qt 1 pt 3. 92 7. 33 deg 51 min 36 sec For example 4. 1,159 8. 12 lb 11 4/5 oz 100 ÷ 4 x 5 = 25 x 5 = 125 ORDER OF OPERATIONS and When a series of operations involving addi- tion, subtraction, multiplication, or division is 60 - 25 ÷ 5 = 60 - 5 = 55 indicated, the order in which the operations are performed is important only if division is in- Now consider volved or if the operations are mixed. A se- ries of individual additions, subtractions, or 60 - 25 ÷ 5 + 15 - 100 + 4 x 10 = 60 - 5 + 15 - 100 + 40 = 115 - 105 = 10

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Chapter 2– POSITIVE INTEGERS

Practice problems. Evaluate each of the ZERO AS A FACTOR following expressions: If any number is multiplied by zero, the 1. 9 - 3 + 2 product is zero. For example, 5 times zero 2. 18 - 2 x 5 + 4 equals zero and may be written 5(0) = 0. The 3. 90 ÷ 2 - 9 zero factor law tells us that, if the product of 4. 75 ÷ 5 x 3 + 5 two or more factors is zero, at least one of the 5. 7 + 1 - 8 x 4 ÷ 16 factors must be zero.

Answers: PRIME FACTORS

1. 5 4. 9 A number that has factors other than itself 2. 12 5. 6 and 1 is a COMPOSITE NUMBER. For exam- 3. 5 ple, the number 15 is composite. It has the factors 5 and 3. MULTIPLES AND FACTORS A number that has no factors except itself and 1 is a PRIME NUMBER. Since it is some- Any number that is exactly divisible by a times advantageous to separate a composite given number is a MULTIPLE of the given number into prime factors, it is helpful to be number. For example, 24 is a multiple of 2, 3, able to recognize a few prime numbers quickly. 4, 6, 8, and 12, since it is divisible by each of The following series shows all the prime num- these numbers. Saying that 24 is a multiple of bers up to 60: 9, for instance, is equivalent to saying that 3 multiplied by some whole number will give 24. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, Any number is a multiple of itself and also of 1. 43, 47, 53, 59.

Any number that is a multiple of 2 is an Notice that 2 is the only even prime number. EVEN NUMBER. The even numbers begin with All other even numbers are divisible by 2. 2 and progress by 2's as follows: Notice also that 51, for example, does not ap- pear in the series, since it is a composite num- 2, 4, 6, 8, 10, 12, . . . ber equal to 3 x 17. If a factor of a number is prime, it is called Any number that is not a multiple of 2 is an a PRIME FACTOR. To separate a number into ODD NUMBER. The odd numbers begin with 1 prime factors, begin by taking out the smallest and progress by 2's, as follows: factor. If the number is even, take out all the 2's first, then try 3 as a factor, etc. Thus, we 1, 3, 5, 7, 9, 11, 13, . . . have the following example:

Any number that can be divided into a given 540 = 2 · 270 number without a remainder is a FACTOR of = 2 · 2 · 135 the given number. The given number is a mul- = 2 · 2 · 3 · 45 tiple of any number that is one of its factors. = 2 · 2 · 3 · 3 · 15 For example, 2, 3, 4, 6, 8, and 12 are factors = 2 · 2 · 3 · 3 · 5 of 24. The following four equalities show vari- ous combinations of the factors of 24: Since 1 is an understood factor of every num- ber, we do not waste space recording it as one 24 = 24 · 1 24 = 8 · 3 of the factors in a presentation of this kind. 24 = 12 · 2 24 = 6 · 4 A convenient way of keeping track of the prime factors is in the short division process If the number 24 is factored as completely as as follows: possible, it assumes the form 2/540 24 = 2 · 2 · 2 · 3 2/270 3/135 3/45 3/15 5/5 1

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MATHEMATICS, VOLUME 1

If a number is odd, its factors will be odd numbers. To separate an odd number into prime factors, take out the 3's first, it there are any. Then try 5 as a factor, etc. As an example,

5,775 = 3 · 1,925 = 3 · 5 · 385 = 3 · 5 · 5 · 77 = 3 · 5 · 5 · 7 · 11

Practice problems:

1. Which of the following are prime numbers and which are composite numbers?

25, 7, 18, 29, 51

2. What prime numbers are factors of 36?

3. Which of the following are multiples of 3?

45, 53, 51, 39, 47

4. Find the prime factors of 27.

Answers:

1. Prime: 7, 29 Composite: 25, 18, 51 2. 36 = 2 · 2 · 3 · 3 3. 45, 51, 39 4. 27 = 3 · 3 · 3

Tests for Divisibility

It is often useful to be able to tell by inspection whether a number is exactly divisible by one or more of the digits from 2 through 9. An expression which is frequently used, although it is sometimes misleading, is "evenly divisible." This expression has nothing to do with the concept of even and odd numbers, and it probably should be avoided in favor of the more descriptive expression, "exactly divisible." For the remainder of this discussion, the word "divisible"

has the same meaning as "exactly divisible." Several tests for divisibility are listed in the following paragraphs:

1. A number is divisible by 2 if its right-hand digit is even. 2. A number is divisible by 3 if the sum of its digits is divisible by 3. For example, the digits of the number 6,561 add up to produce the sum 18. Since 18 is divisible by 3, we know that 6,561 is divisible by 3. 3. A number is divisible by 4 if the number formed by the two right-hand digits is divisible by 4. For example, the two right-hand digits of the number 3,524 form the number 24. Since 24 is divisible by 4, we know that 3,524 is divisible by 4. 4. A number is divisible by 5 if its right-hand digit is 0 or 5. 5. A number is divisible by 6 if it is even and the sum of its digits is divisible by 3. For example, the sum of the digits of 64,236 is 21, which is divisible by 3. Since 64,236 is also an even number, we know that it is divisible by 6. 6. No short method has been found for determining whether a number is divisible by 7. 7. A number is divisible by 8 if the number formed by the three right-hand digits is divisible by 8. For example, the three right-hand digits of the number 54,272 form the number 272, which is divisible by 8. Therefore, we know that 54,272 is divisible by 8. 8. A number is divisible by 9 if the sum of its digits is divisible by 9. For example, the sum of the digits of 546,372 is 27, which is divisible by 9. Therefore we know that 546,372 is divisible by 9. Practice problems. Check each of the following numbers for divisibility by all of the digits except 7:

1. 242,431,231,320 2. 344,624,221,840 3. 988,446,662,640 4. 207,634,542,480

Answer: All of these numbers are divisible by 2, 3, 4, 5, 6, 8, and 9.

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CHAPTER 3 SIGNED NUMBERS

The positive numbers with which we have worked in previous chapters are not sufficient for every situation which may arise. For example, a negative number results in the operation of subtraction when the subtrahend is larger than the minuend.

NEGATIVE NUMBERS

When the subtrahend happens to be larger than the minuend, this fact is indicated by placing a minus sign in front of the difference, as in the following:

12 - 20 = -8

The difference, -8, is said to be NEGATIVE. A number preceded by a minus sign is a NEGATIVE NUMBER. The number -8 is read "minus eight." Such a number might arise when we speak of temperature changes. If the temperature was 12 degrees yesterday and dropped 20 degrees today, the reading today would be 12 - 20 or -8 degrees.

Numbers that show either a plus or minus sign are called SIGNED NUMBERS. An unsigned number is understood to be positive and is treated as though there were a plus sign preceding it.

If it is desired to emphasize the fact that a number is positive, a plus sign is placed in front of the number, as in +5, which is read "plus five." Therefore, either +5 or 5 indicates that the number 5 is positive. If a number is negative, a minus sign must appear in front of it, as in -9.

In dealing with signed numbers it should be emphasized that the plus and minus signs have two separate and distinct functions. They may indicate whether a number is positive or negative, or they may indicate the operation of addition or subtraction.

When operating entirely with positive numbers, it is not necessary to be concerned with this distinction since plus or minus signs indicate only addition or subtraction. However, when negative numbers are also involved in a computation, it is important to distinguish between a sign of operation and the sign of a number.

DIRECTION OF MEASUREMENT

Signed numbers provide a convenient way of indicating opposite directions with a minimum of words. For example, an altitude of 20 ft above sea level could be designated as +20 ft. The same distance below sea level would then be designated as -20 ft. One of the most common devices utilizing signed numbers to indicate direction of measurement is the thermometer.

Thermometer

The Celsius (centigrade) thermometer shown in figure 3-1 illustrates the use of positive and negative numbers to indicate direction of travel above and below 0. The 0 mark is the changeover point, at which the signs of the scale numbers change from + to +.

When the thermometer is heated by the surrounding air or by a hot liquid in which it is placed, the mercury expands and travels up the tube. After the expanding mercury passes 0, the mark at which it comes to rest is read as a positive temperature. If the thermometer is allowed to cool, the mercury contracts. After passing 0 in its downward movement, any mark at which it comes to rest is read as a negative temperature.

Rectangular Coordinate System

As a matter of convenience, mathematicians have agreed to follow certain conventions as to the use of signed numbers in directional measurement. For example, in figure 3-2, a direction to the right along the horizontal line is positive, while the opposite direction (toward the left) is negative. On the vertical line, direction upward is positive, while direction downward is negative. A distance of -3 units along the horizontal line indicates a measurement of 3 units to the left of starting point 0. A distance of -3 units on the vertical line indicates

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MATHEMATICS, VOLUME 1

130°C 120 110 WATER BEGINS BOILING STEAM BEGINS CONDENSING 100 90 80 70 60 100 DEGREES 50 40 30 20 +10 0 -10 -20 ICE BEGINS MELTING WATER BEGINS FREEZING MERCURY -30 -40 BOILING WATER 0

Figure 3-1.—Celsius (centigrade) temperature scale.

3 2 1 STARTING POINT -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 -1 -2 -3

Figure 3-2.—Rectangular coordinate system.

The Number Line

Sometimes it is important to know the rela- tive greatness (magnitude) of positive and nega- tive numbers. To determine whether a particu- lar number is greater or less than another number, think of all the numbers both positive and negative as being arranged along a hori- zontal line. (See fig. 3-3.)

-5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5

Figure 3-3.—Number line showing both positive and negative numbers.

Place zero at the middle of the line. Let the positive numbers extend from zero toward the right. Let the negative numbers extend from zero toward the left. With this arrangement, positive and negative numbers are so located that they progress from smaller to larger num- bers as we move from left to right along the line. Any number that lies to the right of a given number is greater than the given number. A number that lies to the left of a given number is less than the given number. This arrange- ment shows that any negative number is smaller than any positive number. The symbol for "greater than" is >. The symbol for "less than" is <. It is easy to dis- tinguish between these symbols because the symbol used always opens toward the larger number. For example, "7 is greater than 4" can be written 7 > 4 and "-5 is less than -1" can be written -5 < -1.

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Chapter 3—SIGNED NUMBERS

Absolute Value

The ABSOLUTE VALUE of a number is its numerical value when the sign is dropped. The absolute value of either +5 or -5 is 5. Thus, two numbers that differ only in sign have the same absolute value. The symbol for absolute value consists of two vertical bars placed one on each side of the number, as in |-5| = 5. Consider also the following:

|4 - 20| = 16 |-7| = |-7| = 7

The expression |-7| is read "absolute value of minus seven." When positive and negative numbers are used to indicate direction of measurement, we are concerned only with absolute value. If we wish to know only the distance covered. For example, in figure 3-2, if an object moves to the left from the starting point to the point in- dicated by -2, the actual distance covered is 2 units. We are concerned only with the fact that |-2| = 2, if our only interest is in the distance and not the direction.

OPERATING WITH SIGNED NUMBERS

The number line can be used to demonstrate addition of signed numbers. Two cases must be considered; namely, adding numbers with like signs and adding numbers with unlike signs.

ADDING WITH LIKE SIGNS

As an example of addition with like signs, suppose that we use the number line (fig. 3-4) to add 2 + 3. Since these are signed numbers, we indicate this addition as (+2) + (+3). This emphasizes that, among the three + signs shown, two are number signs and one is a sign of

operation. Line a (fig. 3-4) above the number line shows this addition. Find 2 on the number line. To add 3 to it, go three units more in a positive direction and get 5. To add two negative numbers on the number line, such as -2 and -3, find -2 on the number line and then go three more in the nega- tive direction to get -5, as in b (fig. 3-4) above the number line. Observation of the results of the foregoing operations on the number line leads us to the following conclusion, which may be stated as a law: To add numbers with like signs, add the absolute values and prefix the common sign.

ADDING WITH UNLIKE SIGNS

To add a positive and a negative number, such as (-4) + (+5), find +5 on the number line and go four units in a negative direction, as in line c above the number line in figure 3-4. Notice that this addition could be performed in the other direction. That is, we could start at -4 and move 5 units in the positive direction. (See line d, fig. 3-4.) The result of our operations with mixed signs on the number line lead to the following conclusion, which may be stated as a law: To add numbers with unlike signs, find the difference between their absolute values and prefix the sign of the numerically greater number. The following examples show the addition of the numbers 3 and 5 with the four possible com- binations of signs:

3 -3 3 -3 5 -5 -5 5

8 -8 -2 2

In the first example, 3 and 5 have like signs and the common sign is understood to be posi- tive. The sum of the absolute values is 8 and this sign is prefixed to this sum, thus signifying that the sign of the 8 is understood to be positive. In the second example, the 3 and 5 again have like signs, but their common sign is negative. The sum of the absolute values is 8, and this time the common sign is prefixed to the sum. The answer is thus -8. In the third example, the 3 and 5 have unlike signs. The difference between their absolute values is 2, and the sign of the larger addend is negative. Therefore, the answer is -2. In the fourth example, the 3 and 5 again have unlike signs. The difference of the absolute

Figure 3-4.—Using the number line to add.

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MATHEMATICS, VOLUME 1

values is still 2, but this time the sign of the larger addend is positive. Therefore, the sign prefixed to the 2 is positive (understood) and the final answer is simply 2.

These four examples could be written in a different form, emphasizing the distinction be- tween the sign of a number and an operational sign, as follows:

(+3) + (+5) = +8 (-3) + (-5) = -8 (+3) + (-5) = -2 (-3) + (+5) = +2

Practice problems. Add as indicated:

1. -10 + 5 = (-10) + (+5) = ? 2. Add -9, -16, and 25 3. - 7 - 1 - 3 = (-7) + (-1) + (-3) = ? 4. Add -22 and -13

Answers:

1. -5 3. -11 2. 0 4. -35

SUBTRACTION

Subtraction is the inverse of addition. When subtraction is performed, we "take away" the subtrahend. This means that whatever the value of the subtrahend, its effect is to be reversed when subtraction is indicated. In addition, the sum of 5 and -2 is 3. In subtraction, however, to take away the effect of the -2, the quantity +2 must be added. Thus the difference between +5 and -2 is +7.

Keeping this idea in mind, we may now pro- ceed to examine the various combinations of subtraction involving signed numbers. Let us first consider the four possibilities where the

minuend is numerically greater than the sub- trahend, as in the following examples:

8 8 -8 -8 5 -5 5 -5 ___ ___ ___ ___ 3 13 -13 -3

We may show how each of these results is obtained by use of the number line, as shown in figure 3-5. In the first example, we find +8 on the num- ber line, then subtract 5 by making a movement that reverses its sign. Thus, we move to the left 5 units. The result (difference) is +3. (See line a, fig. 3-5.) In the second example, we find +8 on the number line, then subtract (-5) by making a movement that will reverse its sign. Thus we move to the right 5 units. The result in this case is +13. (See line b, fig. 3-5.) In the third example, we find -8 on the num- ber line, then subtract 5 by making a movement that reverses its sign. Thus we move to the left 5 units. The result is -13. (See line c, fig. 3-5.) In the fourth example, we find -8 on the number line, then reverse the sign of -5 by moving 5 units to the right. The result is -3. (See line d, fig. 3-5.) Next, let us consider the four possibilities that arise when the subtrahend is numerically greater than the minuend, as in the following examples:

5 5 -5 -5 8 -8 8 -8 ___ ___ ___ ___ -3 13 -13 3

In the first example, we find +5 on the num- ber line, then subtract 8 by making a movement

h ┐ ┐ ├─ ─┤ F ←─────────→ │ │ ├────────┤e d ├────→ │ │ ├─ ─┤c │ │ ├──── ├─────────┤ │ b ├──→ ├──────→ a -13-12-11-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13

Figure 3-5.—Subtraction by use of the number line.

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Chapter 3—SIGNED NUMBERS

that reverses its sign. Thus we move to the left 8 units. The result is -3. (See line e, fig. 3-5.) In the second example, we find +5 on the number line, then subtract -8 by making a move- ment to the right that reverses its sign. The result is 13. (See line f, fig. 3-5.) In the third example, we find -5 on the num- ber line, then reverse the sign of 8 by a move- ment to the left. The result is -13. (See line g, fig. 3-5.) In the fourth example, we find -5 on the num- ber line, then reverse the sign of -8 by a move- ment to the right. The result is 3. (See line h, fig. 3-5.) Careful study of the preceding examples leads to the following conclusion, which is stated as a law for subtraction of signed num- bers: In any subtraction problem, mentally change the sign of the subtrahend and proceed as in addition.

Practice problems. In problems 1 through 4, subtract the lower number from the upper. In 5 through 8, subtract as indicated.

1. 17 2. -12 3. -9 4. 7 -10 8 -13 -16

5. 1 -(-5) = ? 6. -6 -(-8) = ? 7. 14 - 7 -(-3) = ? 8. -9 - 2 = ?

Answers:

1. 27 2. -20 3. 4 4. -9 5. 6 6. 2 7. 10 8. -11

MULTIPLICATION

To explain the rules for multiplication of signed numbers, we recall that multiplication of whole numbers may be thought of as short- ened addition. Two types of multiplication problems must be examined; the first type in- volves numbers with unlike signs, and the sec- ond involves numbers with like signs.

Unlike Signs

Consider the example 3(-4), in which the multiplicand is negative. This means we are to add -4 three times; that is, 3(-4) is equal to (-4) + (-4) + (-4), which is equal to -12. For example, if we have three -4-dollar debts, we owe 12 dollars in all.

When the multiplier is negative, as in -3(7), we are to take away 7 three times. Thus, -3(7) is equal to -(7) - (7) - (7) which is equal to -21. For example, if 7 shells were expended in one firing, 7 the next, and 7 the next, there would be a loss of 21 shells in all. Thus, the rule is as follows: The product of two numbers with unlike signs is negative. The law of signs for unlike signs is some- times stated as follows: Minus times plus is minus; plus times minus is minus. Thus a problem such as 3(-4) can be reduced to the following two steps: 1. Multiply the signs and write down the sign of the answer before working with the numbers themselves. 2. Multiply the numbers as if they were un- signed numbers. Using the suggested procedure, the sign of the answer for 3(-4) is found to be minus. The product of 3 and 4 is 12, and the final answer is -12. When there are more than two numbers to be multiplied, the signs are taken in pairs until the final sign is determined.

Like Signs

When both factors are positive, as in 4(5), the sign of the product is positive. We are to add +5 four times, as follows:

4(5) = 5 + 5 + 5 + 5 = 20

When both factors are negative, as in -4(-5), the sign of the product is positive. We are to take away -5 four times.

-4(-5) = -(-5) - (-5) - (-5) - (-5) = +5 +5 +5 +5 = 20

Remember that taking away a negative 5 is the same as adding a positive 5. For example, suppose someone owes a man 20 dollars and pays him back (or diminishes the debt) 5 dollars at a time. He takes away a debt of 20 dollars by giving him four positive 5-dollar bills, or a total of 20 positive dollars in all. The rule developed by the foregoing example is as follows: The product of two numbers with like signs is positive. Knowing that the product of two positive num- bers or two negative numbers is positive, we can conclude that the product of any even num- ber of negative numbers is positive. Similarly,

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MATHEMATICS, VOLUME 1

the product of any odd number of negative numbers is negative. The laws of signs may be combined as follows: Minus times plus is minus; minus is minus; minus times minus is plus; plus times plus is plus. Use of this combined rule may be illustrated as follows:

4(-2) · (-5) · (6) · (-3) = -720

Taking the signs in pairs, the understood plus on the 4 times the minus on the 2 produces a minus. This minus times the minus on the 5 produces a plus. This plus times the understood plus on the 6 produces a plus. This plus times the minus on the 3 produces a minus, so we know that the final answer is negative. The product of the numbers, disregarding their signs, is 720; therefore, the final answer is -720.

Practice problems. Multiply as indicated:

1. 5(-8) = ? 2. -7(3) (2) = ? 3. 6(-1) (-4) = ? 4. -2(3) (-4) (5) (-6) = ?

Answers:

1. -40 3. 24 2. -42 4. -720

DIVISION

Because division is the inverse of multiplication, we can quickly develop the rules for division of signed numbers by comparison with the corresponding multiplication rules, as in the following examples:

1. Division involving two numbers with unlike signs is related to multiplication with unlike signs, as follows:

3(-4) = -12

Therefore, -12/3 = -4

Thus the rule for division with unlike signs is: The quotient of two numbers with unlike signs is negative.

2. Division involving two numbers with like signs is related to multiplication with like signs, as follows:

3(-4) = -12

Therefore,

-12/-4 = 3

Thus the rule for division with like signs is: The quotient of two numbers with like signs is positive.

The following examples show the application of the rules for dividing signed numbers:

12/3 = 4 -12/3 = -4

-12/-3 = 4 12/-3 = -4

Practice problems. Multiply and divide as indicated:

1. 15 ÷ -5 3. (-3)(4)/-6

2. -2(-3)/-6 4. -81/9

Answers:

1. -3 3. 2 2. -1 4. -9

SPECIAL CASES

Two special cases arise frequently in which the laws of signs may be used to advantage. The first such usage is in simplifying subtraction; the second is in changing the signs of the numerator and denominator when division is indicated in the form of a fraction.

Subtraction

The rules for subtraction may be simplified by use of the laws of signs, if each expression to be subtracted is considered as being multiplied by a negative sign. For example, 4 -(-5) is the same as 4 + 5, since minus times minus is plus. This result also establishes a basis for the rule governing removal of parentheses.

The parentheses rule, as usually stated, is: Parentheses preceded by a minus sign may be removed, if the signs of all terms within the parentheses are changed. This is illustrated as follows:

12 -(3 - 2 + 4) = 12 - 3 + 2 - 4

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Chapter 3—SIGNED NUMBERS

The reason for the changes of sign is clear when the negative sign preceding the parentheses is considered to be a multiplier for the whole mathematical expression.

Division in Fractional Form

Division is often indicated by writing the dividend as the numerator, and the divisor as the denominator, of a fraction. In algebra, every fraction is considered to have three signs. The numerator has a sign, the denominator has a sign, and the fraction itself, taken as a whole, has a sign. In many cases, one or more of these signs will be positive, and thus will not be shown. For example, in the following fraction the sign of the numerator and the sign of the denominator are both positive (understood) and the sign of the fraction itself is negative:

-4/5

Fractions with more than one negative sign are always reducible to a simpler form with at most one negative sign. For example, the sign of the numerator and the sign of the denominator may be both negative. We note that minus divided by minus gives the same result as plus divided by plus. Therefore, we may change to the less complicated form having plus signs (understood) for both numerator and denominator, as follows:

-15/-5 = +15/+5 = 15/5

Since -15 divided by -5 is 3, and 15 divided by 5 is also 3, we conclude that the change of sign does not alter the final answer. The same reasoning may be applied in the following example, in which the sign of the fraction itself is negative:

-15/-5 = -+15/+5 = -15/5

When the fraction itself has a negative sign, as in this example, the fraction may be enclosed in parentheses temporarily, for the purpose of working with the numerator and denominator only. Then the sign of the fraction is applied separately to the result, as follows:

-15/-5 = -(−15/−5) = -(-3) = -3

All of this can be done mentally.

If a fraction has a negative sign in one of the three sign positions, this sign may be moved to another position. Such an adjustment is an advantage in some types of complicated expressions involving fractions. Examples of this type of sign change follow:

-15/5 = -15/5 = 15/-5

In the first expression of the foregoing example, the sign of the numerator is positive (understood) and the sign of the fraction is negative. Changing these signs, we obtain the second expression. To obtain the third expression from the second, we change the sign of the numerator and the sign of the denominator. Observe that the sign changes in each case involve a pair of signs. This leads to the law of signs for fractions: Any two of the three signs of a fraction may be changed without altering the value of the fraction.

AXIOMS AND LAWS

An axiom is a self-evident truth. It is a truth that is so universally accepted that it does not require proof. For example, the statement that "a straight line is the shortest distance between two points" is an axiom from plane geometry. One tends to accept the truth of an axiom without proof, because anything which is axiomatic is, by its very nature, obviously true.

On the other hand, a law (in the mathematical sense) is the result of defining certain quantities and relationships and then developing logical conclusions from the definitions.

AXIOMS OF EQUALITY

The four axioms of equality with which we are concerned in arithmetic and algebra are stated as follows:

1. If the same quantity is added to each of two equal quantities, the resulting quantities are equal. This is sometimes stated as follows: If equals are added to equals, the result are equal. For example, by adding the same quantity (3) to both sides of the following equation, we obtain two sums which are equal:

-2 = -3 + 1 -2 + 3 = -3 + 1 + 3 1 = 1

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MATHEMATICS, VOLUME 1

2. If the same quantity is subtracted from each of two equal quantities, the resulting quantities are equal. This is sometimes stated as follows: If equals are subtracted from equals, the results are equal. For example, by sub-tracting 2 from both sides of the following equation we obtain results which are equal:

5 = 2 + 3 5 - 2 = 2 + 3 - 2 3 = 3

3. If two equal quantities are multiplied by the same quantity, the resulting products are equal. This is sometimes stated as follows: If equals are multiplied by equals, the products are equal. For example, both sides of the following equation are multiplied by -3 and equal results are obtained:

5 = 2 + 3 (-3)(5) = (-3)(2 + 3) -15 = -15

4. If two equal quantities are divided by the same quantity, the resulting quotients are equal. This is sometimes stated as follows: If equals are divided by equals, the results are equal. For example, both sides of the following equation are divided by 3, and the resulting quotients are equal:

12 + 3 = 15 12 + 3 = 15 3 3 4 + 1 = 5

These axioms are especially useful when letters are used to represent numbers. If we know that 5x = -30, for instance, then dividing both 5x and -30 by 5 leads to the conclusion that x = -6.

LAWS FOR COMBINING NUMBERS

Numbers are combined in accordance with the following basic laws: 1. The associative laws of addition and multiplication. 2. The commutative laws of addition and multiplication. 3. The distributive law.

Associative Law of Addition

The word "associative" suggests association or grouping. This law states that the sum of three or more addends is the same regardless of the manner in which they are grouped. For example, 6 + 3 + 1 is the same as 6 + (3 + 1) or (6 +3) +1. This law can be applied to subtraction by changing signs in such a way that all negative signs are treated as number signs rather than operational signs. That is, some of the addends can be negative numbers. For example, 6 - 4 - 2 can be rewritten as 6 + (-4) + (-2). By the associative law, this is the same as

6 + [(-4) + (-2)] or [6 + (-4)] + (-2).

However, 6 - 4 - 2 is not the same as 6 - (4 - 2); the terms must be expressed as addends before applying the associative law of addition.

Associative Law of Multiplication

This law states that the product of three or more factors is the same regardless of the manner in which they are grouped. For example, 6 · 3 · 2 is the same as (6 · 3) · 2 or 6 · (3 · 2). Negative signs require no special treatment in the application of this law. For example, 6 · (-4) · (-2) is the same as [6 · (-4)] · (-2) or 6 · [(-4) · (-2)].

Commutative Law of Addition

The word "commute" means to change, substitute or move from place to place. The commutative law of addition states that the sum of two or more addends is the same regardless of the order in which they are arranged. For example, 4 + 3 + 2 is the same as 4 + 2 + 3 or 2 + 4 + 3. This law can be applied to subtraction by changing signs so that all negative signs become number signs and all signs of operation are positive. For example, 5 - 3 - 2 is changed to 5 + (-3) + (-2), which is the same as 5 + (-2) + (-3) or (-3) + 5 + (-2).

Commutative Law of Multiplication

This law states that the product of two or more factors is the same regardless of the order in which the factors are arranged. For example, 3 · 4 · 5 is the same as 5 · 3 · 4 or

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Chapter 3—SIGNED NUMBERS

4 · 3 · 5. Negative signs require no special treatment in the application of this law. For example, 2 · (-4) · (-3) is the same as (-3) · 2 or (-3) · 2 · (-4).

Distributive Law

This law combines the operations of addition and multiplication. The word "distributive" refers to the distribution of a common multiplier among the terms of an additive expression. For example,

2(3 + 4 + 5) = 2 · 3 + 2 · 4 + 2 · 5 = 6 + 8 + 10

To verify the distributive law, we note that 2(3 + 4 + 5) is the same as 2(12) or 24. Also, 6 + 8 + 10 is 24. For application of the distributive law where negative signs appear, the following procedure is recommended:

3(4 - 2) = 3 [4 + (-2)] = 3(4) + 3(-2) = 12 - 6 = 6

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MATHEMATICS, VOLUME 1

For example, if 1 in the form 2/7 is multiplied by 3/5, the product will still have a value of 6/35 but will be in a different form, as follows:

2/7 · 3/5 = 2·3/7·5 = 6/35

Figure 4-4 shows that 3/5 of line a is equal to 6/10 of line b where line a equals line b. Line a is marked off in fifths and line b is marked off in tenths. It can readily be seen that 6/10 and 3/5 measure distances of equal length.

[FIGURE: Two horizontal lines labeled a and b. Line a is divided into sections marked 0, 1/5, 2/5, 3/5, 4/5, 5/5, with a bracket above showing 3/5. Line b is divided into tenths marked 0, 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10, 10/10, with a bracket above showing 6/10.]

Figure 4-4.—Equivalent fractions.

The markings on a ruler show equivalent fractions. The major division of an inch divides it into two equal parts. One of these parts represents 1/2. The next smaller markings divide the inch into four equal parts. It will be noted that two of these parts represent the same distance as 1/2, that is, 2/4 equals 1/2. Also, the next smaller markings break the inch into 8 equal parts. How many of these parts are equivalent to 1/2 inch?

The answer is found by noting that 4/8 equals 1/2.

Practice problems. Using the divisions on a ruler for reference, complete the following exercise: 1. 1/4 = ?/8 3. 3/4 = ?/16 2. 1/8 = ?/16 4. 1/4 = ?/16

Answers: 1. 2 3. 12 2. 2 4. 4

A review of the foregoing exercise will reveal that in each case the right-hand fraction could be formed by multiplying both the numerator and the denominator of the left-hand fraction by the same number. In each case the number may be determined by dividing the denominator of the right-hand fraction by the denominator of the left-hand fraction. Thus in problem 1, both terms of 1/4 were multiplied by 2. In problem 3, both terms were multiplied by 4. It is seen that multiplying both terms of a fraction by the same number does not change the value of the fraction.

Since 1/2 equals 2/4, the reverse must also be true; that is 2/4 must be equal to 1/2. This can likewise be verified on a ruler. We have already seen that 3/8 is the same as 1/2; 12/16 equals 3/4, and 2/8 equals 1/4. We see that dividing both terms of a fraction by the same number does not change the value of the fraction.

FUNDAMENTAL RULE OF FRACTIONS

The foregoing results are combined to form the fundamental rule of fractions, which is stated as follows: Multiplying or dividing both terms of a fraction by the same number does not change the value of the fraction. This is one of the most important rules used in dealing with fractions.

The following examples show how the fundamental rule is used:

1. Change 1/4 to twelfths. This problem is set up as follows:

1/4 = ?/12

The first step is to determine how many 4's are contained in 12. The answer is 3, so we know that the multiplier for both terms of the fraction is 3, as follows:

3 · 1/3 · 1/4 = 3/12

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Chapter 4—COMMON FRACTIONS

2. What fraction with a numerator of 6 is equal to 3/4?

SOLUTION: 6/8 = 3/4

We note that 6 contains 3 twice; therefore we need to double the numerator of the right-hand fraction to make it equivalent to the numerator of the fraction we seek. We multiply both terms of 3/4 by 2, obtaining 8 as the denominator of the new fraction, as follows:

6/8 = 3 · 2/4 · 2

3. Change 6/16 to eighths.

SOLUTION: 6/16 = ?/8

We note that the denominator of the fraction which we seek is 1/2 as large as the denominator of the original fraction. Therefore the new fraction may be formed by dividing both terms of the original fraction by 2, as follows:

6/16 ÷ 2/2 = 3/8

Practice problems. Supply the missing number in each of the following:

1. 3/8 = 30/? 3. ?/90 = 3/10 5. 1/7 = ?/12 2. 44/48 = ?/12 4. 1/6 = 6/7 6. 3/5 = ?/25

Answers: 1. 80 3. 27 5. 6 2. 11 4. 36 6. 15

REDUCTION TO LOWEST TERMS

It is frequently desirable to change a fraction to an equivalent fraction with the smallest possible terms; that is, with the smallest possible numerator and denominator. This process is called REDUCTION. Thus, 6/30 reduced to lowest terms is 1/5. Reduction can be accomplished by finding the largest factor that is common to both the numerator and denominator and dividing both of these terms by it. Dividing

both terms of the preceding example by 6 reduces the fraction to lowest terms. In computation, fractions should usually be reduced to lowest terms where possible.

If the greatest common factor cannot readily be found, any common factor may be removed and the process repeated until the fraction is in lowest terms: Thus, 18/48 could first be divided by 2 and then by 3.

18 ÷ 2/48 ÷ 2 = 9/24 9 ÷ 3/24 ÷ 3 = 3/8

Practice problems. Reduce the following fractions to lowest terms:

1. 18/48 2. 15/20 3. 35/56 4. 12/60 5. 18/24 6. 9/144

Answers: 1. 3/8 2. 3/4 3. 5/8 4. 1/5 5. 3/4 6. 1/10

IMPROPER FRACTIONS

Although the "improper" fraction is really quite "proper" mathematically, it is usually customary to change it to a mixed number. A recipe may call for 1 2/3 cups of milk, but would not call for 5/3 cups of milk.

Since a fraction is an indicated division, a method is already known for reduction of improper fractions to mixed numbers. The improper fraction 8/3 may be considered as the division of 8 by 3. This division is carried out as follows:

2 R 2 = 2 2/3 3/8 6 2

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MATHEMATICS, VOLUME 1

The truth of this can be verified another way: If 1 equals 3/3, then 2 equals 6/3. Thus,

2 2/3 = 2 + 2/3 = 6/3 + 2/3 = 8/3

These examples lead to the following conclusion, which is stated as a rule: To change an improper fraction to a mixed number, divide the numerator by the denominator and write the fractional part of the quotient in lowest terms.

Practice problems. Change the following fractions to mixed numbers:

1. 31/20 3. 65/20 2. 33/9 4. 45/8

Answers:

1. 1 11/20 3. 3 1/4

2. 3 2/3 4. 5 5/8

OPERATING WITH MIXED NUMBERS

In computation, mixed numbers are often unwieldy. As it is possible to change any improper fraction to a mixed number, it is likewise possible to change any mixed number to an improper fraction. The problem can be reduced to the finding of an equivalent fraction and a simple addition.

EXAMPLE: Change 2 1/5 to an improper fraction.

SOLUTION:

Step 1: Write 2 1/5 as a whole number plus a fraction, 2 + 1/5

Step 2: Change 2 to an equivalent fraction with a denominator of 5, as follows:

2/1 = ?/5

2(5)/1(5) = 10/5

Step 3: Add 10/5 + 1/5 = 11/5

Thus, 2 1/5 = 11/5

EXAMPLE: Write 5 2/9 as an improper fraction.

SOLUTION: 5 2/9 = 5 + 2/9

5/1 = ?/9

5(9)/1(9) = 45/9

45/9 + 2/9 = 47/9

Thus, 5 2/9 = 47/9

In each of these examples, notice that the multiplier used in step 2 is the same number as the denominator of the fractional part of the original mixed number. This leads to the following conclusion, which is stated as a rule: To change a mixed number to an improper fraction, multiply the whole-number part by the denominator of the fractional part and add the numerator to this product. The result is the numerator of the improper fraction; its denominator is the same as the denominator of the fractional part of the original mixed number.

Practice problems. Change the following mixed numbers to improper fractions:

1. 1 1/5 3. 3 2/7

2. 2 11/20 4. 4 9/10

Answers:

1. 6/5 3. 23/7

2. 51/20 4. 43/10

NEGATIVE FRACTIONS

A fraction preceded by a minus sign is negative. Any negative fraction is equivalent to a positive fraction multiplied by -1. For example,

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Chapter 4—COMMON FRACTIONS

- 2/5 = -1(2/5)

The number - 2/5 is read "minus two-fifths."

We know that the quotient of two numbers with unlike signs is negative. Therefore,

-2/5 = - 2/5 and -2/5 = - 2/5

This indicates that a negative fraction is equivalent to a fraction with either a negative numerator or a negative denominator.

The fraction -2/5 is read "two over minus five." The fraction -2/5 is read "minus two over five." A minus sign in a fraction can be moved about at will. It can be placed before the numerator, before the denominator, or before the fraction itself. Thus,

-2/5 = 2/-5 = - 2/5

Moving the minus sign from numerator to denominator, or vice versa, is equivalent to multiplying the terms of the fraction by -1. This is shown in the following examples:

-2(-1)/5(-1) = 2/-5 and 2(-1)/-5(-1) = -2/5

A fraction may be regarded as having three signs associated with it—the sign of the numerator, the sign of the denominator, and the sign preceding the fraction. Any two of these signs may be changed without changing the value of the fraction. Thus,

- 3/4 = -3/4 = 3/-4 = - -3/-4

OPERATIONS WITH FRACTIONS

It will be recalled from the discussion of denominate numbers that numbers must be of the same denomination to be added. We can add pounds to pounds, pints to pints, but not ounces to pints. If we think of fractions loosely as denominate numbers, it will be seen that the rule

of likeness applies also to fractions. We can add eighths to eighths, fourths to fourths, but not eighths to fourths. To add 1/8 inch to 2/5 inch we simply add the numerators and retain the denominator unchanged. The denomination is fifths; as with denominate numbers, we add 1 fifth to 2 fifths to get 3 fifths, or 3/5.

LIKE AND UNLIKE FRACTIONS

We have shown that like fractions are added by simply adding the numerators and keeping the denominator. Thus,

3/8 + 2/8 = 3 + 2/8 = 5/8

or

5/16 + 2/16 = 7/16

Similarly we can subtract like fractions by subtracting the numerators.

7/8 - 2/8 = 7 - 2/8 = 5/8

The following examples will show that like fractions may be divided by dividing the numerator of the dividend by the numerator of the divisor.

3/8 ÷ 1/8 = ?

SOLUTION: We may state the problem as a question: "How many times does 1/8 appear in 3/8, or how many times may 1/8 be taken from 2/8"

3/8 - 1/8 = 2/8 (1) 2/8 - 1/8 = 1/8 (2) 1/8 - 1/8 = 0/8 = 0 (3)

We see that 1/8 can be subtracted from 3/8 three times. Therefore,

3/8 ÷ 1/8 = 3

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MATHEMATICS, VOLUME 1

When the denominators of fractions are un- equal, the fractions are said to be unlike. Ad- dition, subtraction, or division cannot be per- formed directly on unlike fractions. The proper application of the fundamental rule, however, can change their form so that they become like fractions; then all the rules for like fractions apply.

LOWEST COMMON DENOMINATOR

To change unlike fractions to like fractions, it is necessary to find a COMMON DENOMINA- TOR and it is usually advantageous to find the LOWEST COMMON DENOMINATOR (L.C.D). This is nothing more than the least common multiple of the denominators.

Least Common Multiple

If a number is a multiple of two or more different numbers, it is called a COMMON MULTIPLE. Thus, 24 is a common multiple of 6 and 2. There are many common multiples of these numbers. The numbers 36, 48, and 54, to name a few, are also common multiples of 6 and 2.

The smallest of the common multiples of a set of numbers is called the LEAST COMMON MULTIPLE. It is abbreviated LCM. The least common multiple of 6 and 2 is 6. To find the least common multiple of a set of numbers, first separate each of the numbers into prime factors.

Suppose that we wish to find the LCM of 14, 24, and 30. Separating these numbers into prime factors we have:

14 = 2 · 7 24 = 2³ · 3 30 = 2 · 3 · 5

The LCM will contain each of the various prime factors shown. Each prime factor is used the greatest number of times that it occurs in any one of the numbers. Notice that 3, 5, and 7 each occur only once in any one number. On the other hand, 2 occurs three times in one number. We get the following result:

LCM = 2³ · 3 · 5 · 7 = 840

Thus, 840 is the least common multiple of 14, 24, and 30.

Greatest Common Divisor

The largest number that can be divided into each of two or more given numbers without a remainder is called the GREATEST COMMON DIVISOR of the given numbers. It is abbreviated GCD. It is also sometimes called the HIGHEST COMMON FACTOR.

In finding the GCD of a set of numbers, se- parate the numbers into prime factors just as for LCM. The GCD is the product of only those factors that appear in all of the numbers. Notice in the example of the previous section that 2 is the greatest common divisor of 14, 24, and 30.

Find the GCD of 650, 900, and 700. The pro- cedure is as follows:

650 = 2 · 5² · 13 900 = 2² · 3² · 5² 700 = 2² · 5² · 7 GCD = 2 · 5² = 50

Notice that 2 and 5² are factors of each num- ber. The greatest common divisor is 2 x 25 = 50.

USING THE LCD

Consider the example

1/2 + 1/3

The numbers 2 and 3 are both prime; so the LCD is 6.

Therefore 1/2 = 3/6

and 1/3 = 2/6

Thus, the addition of 1/2 and 1/3 is performed as follows:

1/2 + 1/3 = 3/6 + 2/6 = 5/6

In the example

1/5 + 3/10

10 is the LCD.

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Chapter 4—COMMON FRACTIONS

Therefore, 1/5 + 3/10 = 2/10 + 3/10 = 5/10 = 1/2

Practice problems. Change the fractions in each of the following groups to like fractions with least common denominators:

1. 1/3, 1/6 3. 1/2, 1/4, 2/3

2. 5/12, 2/3 4. 1/6, 3/10, 1/5

Answers:

1. 2/6, 1/6 3. 6/12, 3/12, 8/12

2. 5/12, 8/12 4. 5/30, 9/30, 6/30

ADDITION

It has been shown that in adding like frac- tions we add the numerators. In adding unlike fractions, the fractions must first be changed so that they have common denominators. We apply these same rules in adding mixed numbers. It will be remembered that a mixed number is an indicated sum. Thus, 2 1/3 is really 2 + 1/3. Add- ing can be done in any order. The following examples will show the application of these rules:

EXAMPLE:

2 1/3 3 1/3 5 2/3

This could have been written as follows:

2 + 1/3 3 + 1/3 5 + 2/3

EXAMPLE:

4 5/7 6 3/7 10 8/7

Here we change 8/7 to the mixed number 1 1/7. Then

10 8/7 = 10 + 1 + 1/7 = 11 1/7

EXAMPLE:

Add 1/4 2 2/3

We first change the fractions so that they are like and have the least common denominator and then proceed as before.

1/4 = 3/12 2 2/3 = 2 8/12 2 11/12

EXAMPLE:

Add 4 5/8 = 4 5/8 2 1/2 = 2 4/8 1/4 = 2/8 6 11/8

Since 11/8 equals 1 3/8, the final answer is found as follows:

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MATHEMATICS, VOLUME 1

6 11/8 = 6 + 1 + 3/8 (A)

= 7 3/8

Practice problems. Add, and reduce the sums to simplest terms:

1. 1 1/7 2. 3/4 3. 6 2/5 4. 5/8 5. 4 1/2

2 1/4 1 1/2 3 1/4 2 1/20 1 1/8

Answers:

1. 3 25/28 2. 2 1/4 3. 9 13/20 4. 2 31/40 5. 5 5/8

The following example demonstrates a practical application of addition of fractions:

EXAMPLE: Find the total length of the piece of metal shown in figure 4-5 (A).

SOLUTION: First indicate the sum as follows:

9/16 + 3/4 + 7/8 + 3/4 + 9/16 = ?

Changing to like fractions and adding numerators,

9/16 + 12/16 + 14/16 + 12/16 + 9/16 = 56/16

= 3 8/16

= 3 1/2

The total length is 3 1/2 inches.

Practice problem. Find the distance from the center of the first hole to the center of the last hole in the metal plate shown in figure 4-5 (B).

Answer: 2 7/16 inches

SUBTRACTION

The rule of likeness applies in the subtraction of fractions as well as in addition. Some examples will show that cases likely to arise may be solved by use of ideas previously developed.

Figure 4-5.—Adding fractions to obtain total length or spacing.

EXAMPLE: Subtract 1 1/3 from 5 2/3

5 2/3

1 1/3

4 1/3

We see that whole numbers are subtracted from whole numbers; fractions from fractions.

EXAMPLE: Subtract 1/8 from 4/5

4/5 1/8

Changing to like fractions with an LCD, we have

32/40 5/40

27/40

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Chapter 4—COMMON FRACTIONS

EXAMPLE: Subtract 11/12 from 3 2/3

3 2/3 = 3 8/12

11/12 = 11/12

Regrouping 3 8/12 we have

2 + 1 + 8/12 = 2 + 12/12 + 8/12

Then

3 2/3 = 2 20/12

11/12 = 11/12

2 9/12 = 2 3/4

Practice problems. Subtract the lower number from the upper number and reduce the difference to simplest terms:

1. 7/9 2. 2/3 3. 5 5/12 4. 5 5. 2 3/8

1/6 1/3 2 7/12 2 2/3 5/8

Answers:

1. 11/18 2. 1/3 3. 2 5/6 4. 2 1/3 5. 1 3/4

The following problem demonstrates subtraction of fractions in a practical situation.

EXAMPLE: What is the length of the dimension marked X on the machine bolt shown in figure 4-6 (A)?

SOLUTION: Total the lengths of the known parts.

1/4 + 1/64 + 1/2 = 16/64 + 1/64 + 32/64 = 49/64

Subtract this sum from the overall length.

2 = 1 64/64

49/64 = 49/64

1 15/64

The answer is 1 15/64 inch.

(A)

(B)

Figure 4-6.—Finding unknown dimensions by subtracting fractions.

Practice problem. Find the length of the dimension marked Y on the machine bolt in figure 4-6 (B).

Answer: 2 3/32 inches

MULTIPLICATION

The fact that multiplication by a fraction does not increase the value of the product may confuse those who remember the definition of multiplication presented earlier for whole numbers. It was stated that 4(5) means 5 is taken as an addend 4 times. How is it then that 2/3(4) is 2, a number less than 4? Obviously our idea of multiplication must be broadened. Consider the following products: 4(4) = 16 3(4) = 12 2(4) = 8 1(4) = 4

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MATHEMATICS, VOLUME 1

1/2(4) = 2

From these examples a general rule is developed: To find the product of two or more fractions multiply their numerators together and write the result as the numerator of the product; multiply their denominators and write the result as the denominator of the product; reduce the answer to lowest terms.

1/4(4) = 1

In using this rule with whole numbers, write each whole number as a fraction with 1 as the denominator. For example, multiply 4 times 1/12 as follows:

Notice that as the multiplier decreases, the product decreases, until, when the multiplier is a fraction, the product is less than 4 and continues to decrease as the fraction decreases. The fraction introduces the "part of" idea: 1/2(4) means 1/2 of 4; 1/4(4) means 1/4 of 4.

4 x 1/12 = 1/1 x 1/12

The definition of multiplication stated for whole numbers may be extended to include fractions. Since A(5) means 5 is to be used 4 times as an addend, we can say that with fractions the numerator of the multiplier tells how many times the numerator of the multiplicand is to be used as an addend. By the same reasoning, the denominator of the multiplier tells how many times the denominator of the multiplicand is to be used as an addend. The following examples illustrate the use of this idea:

= 1/12 + 1/12 + 1/12 + 1/12

= 4/12 = 1/3

This example shows that 4 (1/12) is the same as 4(1)/12

1. The fraction 1/12 is multiplied by the whole number 4 as follows:

Another way of thinking about the multiplication of 1/12 by 4 is as follows:

4 x 1/12 = 1/12 + 1/12 + 1/12 + 1/12

= 4/12 = 1/3

2. The fraction 2/3 is multiplied by 1/2 as follows:

1/2 x 2/3 = 2/6

= 1/3

In using this rule with mixed numbers, rewrite all mixed numbers as improper fractions before applying the rule, as follows:

2 1/3 x 1/2 = 7/3 x 1/2

= 7/6

A second method of multiplying mixed numbers makes use of the distributive law. This law states that a multiplier applied to a two-part expression is distributed over both parts. For example, to multiply 6 1/3 by 4 we may rewrite

6 1/3 as 6 + 1/3. Then the problem can be written as 4(6 + 1/3) and the multiplication proceeds as follows:

4(6 + 1/3) = 24 + 4/3

= 25 + 1/3

= 25 1/3

Cancellation

Computation can be considerably reduced by dividing out (CANCELLING) factors common to both the numerator and the denominator. We recognize a fraction as an indicated division.

Thinking of 6/8 as an indicated division, we remember that we can simplify division by showing both dividend and divisor as the indicated

38

Chapter 4—COMMON FRACTIONS

products of their factors and then dividing like factors, or canceling. Thus,

6/9 = (2 x 3)/(3 x 3)

Dividing the factor 3 in the numerator by 3 in the denominator gives the following simplified result:

(2 x 1)/(3 x 1) = 2/3

This method is most advantageous when done before any other computation. Consider the example,

1/3 x 3/2 x 2/5

The product in factored form is

(1 x 3 x 2)/(3 x 2 x 5)

Rather than doing the multiplying and then reducing the result 6/30, it is simpler to cancel like factors first, as follows:

(1 x 3 x 2)/(3 x 2 x 5) = 1/5

Likewise,

(2 x 4 x 5)/(2 x 4 x 9) = 5/9

Here we mentally factor 6 to the form 3 x 2, and 4 to the form 2 x 2. Cancellation is a valuable tool in shortening operations with fractions.

The general rule may be applied to mixed numbers by simply changing them to improper fractions.

Thus,

2 1/4 x 3 2/3 = ?

(9/4) x (11/3) = (9 x 11)/(4 x 3) = (3 x 11)/4 = 33/4

= 7 1/2

Practice problems. Determine the following products, using the general rule and canceling where possible:

1. 5/6 x 12 2. 1/2 x 1/3 x 2/5 3. 5 x 4/9 4. 1/4 x 6 5. 1/3 x 2/3 6. 2/3 x 1/6

Answers:

1. 7 1/2 2. 1/15 3. 2 2/9 4. 4 1/2 5. 2/9 6. 2/9

The following problem illustrates the multiplication of fractions in a practical situation.

EXAMPLE: Find the distance between the center lines of the first and fifth rivets connecting the two metal plates shown in figure 4-7 (A).

SOLUTION: The distance between two adjacent rivets, centerline to centerline, is 4 1/2 times the diameter of one of them.

Thus,

1 space = 4 1/2 x 5/8

= 9/2 x 5/8

= 45/16

There are 4 such spaces between the first and fifth rivets. Therefore, the total distance, D, is found as follows:

D = 4 x 45/16 = 45/4 = 11 1/4

39

MATHEMATICS, VOLUME 1

(A) ──────────────────────────────────────────────────────────── ←RIVET SPACING=4½ DIAMETERS

5⁄8=DIAMETER ?

(B) ─────┬─────────────┬───── ⊕ ⊕ ─────┴─────────────┴───── ? 5⁄8=DIAMETER ←RIVET SPACING=5½ DIAMETERS→

Figure 4-7.— Application of multiplication of fractions in determining rivet spacing.

The distance is 11 1/2 inches = 12 + 4 ___ __ 1 1

Practice problem. Find the distance between = 3 the centers of the two rivets shown in figure 4-7 (B). If the dividend and divisor are both fractions, as in 1/3 divided by 1/4, we proceed as follows: Answer: 4 13/16 inches 1 1 4 3 — ÷ — = —— ÷ —— 3 4 12 12 DIVISION 4 ÷ 3 —— ÷ —— There are two methods commonly used for 12 ÷ 12 performing division with fractions. One is the common denominator method and the other is 4 ÷ 3 the reciprocal method. = ——— 1

Common Denominator Method = 4 ÷ 3 = 1 1/3

The common denominator method is an adap- tation of the method of like fractions. The rule Reciprocal Method is as follows: Change the dividend and divisor to like fractions and divide the numerator of The word "reciprocal" denotes an inter- the dividend by the numerator of the divisor. changeable relationship. It is used in mathe- This method can be demonstrated with whole matics to describe a specific relationship be- numbers, first changing them to fractions with tween two numbers. We say that two numbers 1 as the denominator. For example, 12 ÷ 4 can are reciprocals of each other if their product be written as follows: is one. In the example 4 × 1/4 = 1, the fractions 4/1 and 1/4 are reciprocals. Notice the interchange- 12 4 12 ÷ 4 = —— ÷ —— ability: 4 is the reciprocal of 1/4 and 1/4 is the re- 1 1 ciprocal of 4. 12 ÷ 4 = ———— 1 ÷ 1

40

Chapter 4—COMMON FRACTIONS

What is the reciprocal of 3/7? It must be a SOLUTION: 1 ÷ 5/2 = 2/5 × 2/5 number which, when multiplied by 3/7, produces the product, 1. Therefore, = 2 × 5

3/7 × ? = 1 = 2/5

1 1 —— × —— = 1 Check: 5/2 × 2/5 = 1 7 7 ÷ ÷ = 1 1 1

4. What is the reciprocal of 3 5/6? We see that 7/3 is the only number that could ful- fill the requirement. Notice that the numerator SOLUTION: 1 ÷ 3 5/6 = 8/6 ÷ 25/6 and denominator of 3/7 were simply interchanged to get its reciprocal. If we know a number, we = 8 ÷ 25 can always find its reciprocal by dividing 1 by the number. Notice this principle in the follow- = 8/25 ing examples:

1. What is the reciprocal of 7? Check: 25/6 × 8/25 = 1

1 ÷ 7 = 1/7 The foregoing examples lead to the rule for Check: finding the reciprocal of any number: The re- ciprocal of a number is the fraction formed 7/1 × 1/7 = 1 when 1 is divided by the number. (If the final result is a whole number, it can be considered Notice that the cancellation process in this ex- as a fraction whose denominator is 1.) A short- ample does not show the usual 1's which result cut rule which is purely mechanical and does when dividing a number into itself. For ex- not involve reasoning may be stated as follows: ample, when 7 cancels 7, the quotient 1 could be To find the reciprocal of a number, express shown beside each of the 7's. However, since a 1 the number as a fraction and then invert the as a factor has the same effect whether it is fraction. written in or simply understood, the 1's need When the numerator of a fraction is 1, the not be written. reciprocal is a whole number. The smaller the fraction, the greater is the reciprocal. For ex- 2. What is the reciprocal of 3/8? ample, the reciprocal of ___1___ is 1,000. 1,000 1 ÷ 3/8 = 8/8 ÷ 3/8 Also, the reciprocal of any whole number is a = 8 ÷ 3, or 8/3 proper fraction. Thus the reciprocal of 50 is ____1____. 50 Check: Practice problems. Write the reciprocal of 3/8 × 8/3 = 1. each of the following numbers:

3. What is the reciprocal of 5/2? 1. 4 2. 1/3 3. 2 1/2 4. 17 5. 3/2 6. 5/1

Answers:

1. 1/4 2. 3 3. 2/5 4. 1/17 5. 2/3 6. 1/5

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MATHEMATICS, VOLUME 1 Figure 5-1.–Place values including decimals. Figure 5-2.–Conversion of a decimal fraction to shortened form. are zeros in the denominator of the fractional form. Figure 5-3 shows the fraction 24358 and100000 what is meant when it is changed to the short- ened form. This figure is presented to show further that each digit of a decimal fraction holds a certain position in the digit sequence and has a particular value. By the fundamental rule of fractions, it should be clear that 5 50 500= = . Writing10 100 1000 the same values in the shortened way, we have 0.5 = 0.50 = 0.500. In other words, the value of a decimal is not changed by annexing zeros at the right-hand end of the number. This is not 46 Figure 5-3.–Steps in the conversion of a decimal fraction to shortened form. true of whole numbers. Thus, 0.3, 0.30, and 0.300 are equal but 3, 30, and 300 are not equal. Also notice that zeros directly after the deci- mal point do change values. Thus 0.3 is not equal to either 0.03 or 0.003. Decimals such as 0.125 are frequently seen. Although the 0 on the left of the decimal point is not required, it is often helpful. This is par- ticularlytrue in an expression such as 32 ÷ 0.1. In this expression, the lower dot of the division symbol must not be crowded against the decimal point; the 0 serves as an effective spacer. If any doubt exists concerning the clarity of an expression such as .125, it should be written as 0.125.

Chapter 5 – DECIMALS

Practice problems. In problems 1 through 4, change the fractions to decimals. In problems 5 through 8, write the given numbers as deci- mals:

1. 8/100 5. Four hundredths 2. 5/1000 6. Four thousandths 3. 43/1000 7. Five hundred one ten- 4. 32/10000 thousandths 8. Ninety-seven thousandths

Answers:

1. 0.08 5. 0.04 2. 0.005 6. 0.004 3. 0.043 7. 0.0501 4. 0.0032 8. 0.097

READING DECIMALS

To read a decimal fraction in full, we read both its numerator and denominator, as if reading common fractions. To read 0.305, we read "three hundred five thousandths." The denomi- nator is always 1 with as many zeros as deci- mal places. Thus the denominator for 0.14 is 1 with two zeros, or 100. For 0.03 it is 1,000; for 0.101 it is 1,000; and for 0.3 is 10. The denominator may also be determined by count- ing off place values of the decimal. For 0.13 we may think "tenths, hundredths" and the frac- tion is in hundredths. In the example 0.1276 we may think "tenths, hundredths, thousandths, ten-thousandths." We see that the denominator is 10,000 and we read the fraction "one thou- sand two hundred seventy-six ten-thousandths." A whole number with a fraction in the form of a decimal is called a MIXED DECIMAL. Mixed decimals are read in the same manner as mixed numbers. We read the whole number in the usual way followed by the word "and" and then read the decimal. Thus, 160.32 is read "one hundred sixty and thirty-two hundredths." The word "and" in this case, as with mixed numbers, means plus. The number 3.2 means three plus two tenths.

It is also possible to have a complex deci- mal. A COMPLEX DECIMAL contains a com- mon fraction. The number 0.3¾ is a complex

decimal and is read "three and one-third tenths."

The number 0.87½ means 87½ hundredths. The

common fraction in each case forms a part of the last or right-hand place.

In actual practice when numbers are called out for recording, the above procedure is not used. Instead, the digits are merely called out in order with the proper placing of the decimal point. For example, the number 216.003 is read, "two one six point zero zero three." The number 0.05 is read, "zero point zero five."

EQUIVALENT DECIMALS

Decimal fractions may be changed to equiv- alent fractions of higher or lower terms, as is the case with common fractions. If each deci- mal fraction is rewritten in its common frac- tion form, changing to higher terms is accom- plished by multiplying both numerator and denominator by 10, or 100, or some higher power of 10. For example, if we desire to change 5/10 to hundredths, we may do so by mul- tiplying both numerator and denominator by 10. Thus,

5/10 = 50/100

In the decimal form, the same thing may be ac- complished by simply annexing a zero. Thus,

0.5 = 0.50

Annexing a 0 on a decimal has the same ef- fect as multiplying the common fraction form of the decimal by 10/10. This is an application of the fundamental rule of fractions. Annexing two 0's has the same effect as multiplying the common fraction form of the decimal by 100/100; annexing three 0's has the same effect as mul- tiplying by 1000/1000; etc.

REDUCTION TO LOWER TERMS

Reducing to lower terms is known as ROUND- OFF, or simply ROUNDING, when dealing with decimal fractions. If it is desired to reduce 6.3000 to lower terms, we may simply drop as many end zeros as necessary since this is equivalent to dividing both terms of the fraction by some power of ten. Thus, we see that 6.3000 is the same as 6.300, 6.30, or 6.3.

It is frequently necessary to reduce a num- ber such as 6.427 to some lesser degree of precision. For example, suppose that 6.427 is to be rounded to the nearest hundredth. The question to be decided is whether 6.427 is closer

47

MATHEMATICS, VOLUME 1

to 6.42 or 6.43. The best way to decide this question is to compare the fractions 420/1000, 427/1000, and 430/1000. It is obvious that 427/1000 is closer to 430/1000 and 430/1000 is equivalent to 43/100; therefore we say that 6.427, correct to the nearest hundredth, is 6.43.

A mechanical rule for rounding off can be developed from the foregoing analysis. Since the digit in the tenths place is not affected when we round off .427 to hundredths, we may limit our attention to the digits in the hundredths and thousandths places. Thus the decision reduces to the question whether 27 is closer to 20 or 30. Noting that 25 is halfway between 20 and 30, it is clear that anything greater than 25 is closer to 30 than it is to 20.

In any number between 20 and 30, if the digit in the thousandths place is greater than 5, then the number formed by the hundredths and thou- sandths digits is greater than 25. Thus we would round the 27 in our original problem to 30, as far as the hundredths and thousands digits are concerned. This result could be summarized as follows: When rounding to hun- dredths, if the digit in the thousandths place is greater than 5, increase the digit in the hun- dredths place by 1 and drop the digit in the thousandths place.

The digit in the thousandths place may be any one of the ten digits, 0 through 9. If these ten digits are split into two groups, one com- posed of the five smaller digits (0 through 4) and the other composed of the five larger digits, then 5 is counted as one of the larger digits. Therefore, the general rule for rounding off is stated as follows: If the digit in the decimal place to be eliminated is 5 or greater, increase the digit in the next decimal place to the left by 1. If the digit to be eliminated is less than 5, leave the retained digits unchanged.

The following examples illustrate the rule for rounding off: 1. 0.141 rounded to thousandths is 0.141. 2. 3.147 rounded to tenths is 3.1. 3. 475 rounded to the nearest hundred is 500.

Observe carefully that the answer to exam- ple 2 is not 3.2. Some trainees make the error of treating the rounding process as a kind of chain reaction, in which one first rounds 3.147 to 3.15 and then rounds 3.15 to 3.2. The error of this method is apparent when we note that 147/1000 is closer to 100/1000 than it is to 200/1000.

Problems of the following type are some- times confusing: Reduce 2.998 to the nearest

hundredth. To drop the end figure we must in- crease the next figure by 1. The final result is 3.00. We retain the zeroes to show that the an- swer is carried to the nearest hundredth.

Practice problems. Round off as indicated:

1. 0.5862 to hundredths 2. 0.345 to tenths 3. 2346 to hundreds 4. 3.999 to hundredths

Answers:

1. 0.59 3. 2300 2. 0.3 4. 4.00

CHANGING DECIMALS TO COMMON FRACTIONS

Any decimal may be reduced to a common fraction. To do this we simply write out the numerator and denominator in full and reduce to lowest terms. For example, to change 0.12 to a common fraction, we simply write out the fraction in full,

12/100

and reduce to lowest terms,

3/24 = 3/24

100/4 = 25 25

Likewise, 0.77 is written

77/100

but this is in lowest terms so the fraction can- not be further reduced.

One way of checking to see if a decimal frac- tion can be reduced to lower terms is to con- sider the makeup of the decimal denominator. The denominator is always 10 or a power of 10. Inspection shows that the prime factors of 10 are 5 and 2. Thus, the numerator must be di- visible by 5 or 2 or both, or the fraction cannot be reduced.

EXAMPLE: Change the decimal 0.0625 to a common fraction and reduce to lowest terms.

48

Chapter 5—DECIMALS

SOLUTION: 0.0625 = 625/10000

= (625 ÷ 25)/(10000 ÷ 25) = 25/400

= 1/16

Complex decimals are changed to common fractions by first writing out the numerator and denominator in full and then reducing the resulting complex fraction in the usual way. For example, to reduce 0.12½, we first write

12½/2 ——— 100

Writing the numerator as an improper fraction we have

25/2 ——— 100

and applying the reciprocal method of division, we have

25/2 × 1/100 = 1/8

Practice problems. Change the following decimals to common fractions in lowest terms:

1. 0.25 3. 0.6⅔ 2. 0.375 4. 0.03⅕

Answers: 1. 1/4 3. 5/8 2. 3/8 4. 4/125

CHANGING COMMON FRACTIONS TO DECIMALS

The only difference between a decimal fraction and a common fraction is that the decimal fraction has 1 with a certain number of zeros (in other words, a power of 10) for a denominator. Thus, a common fraction can be changed to a decimal if it can be reduced to a fraction having a power of 10 for a denominator.

If the denominator of the common fraction in its lowest terms is made up of the prime factors 2 or 5 or both, the fraction can be converted to an exact decimal. If some other prime factor is present, the fraction cannot be converted exactly. The truth of this is evident when we consider the denominator of the new fraction. It must always be 10 or a power of 10, and we know the factors of such a number are always 2's and 5's.

The method of converting a common fraction to a decimal is illustrated as follows:

EXAMPLE: Convert 3/4 to a decimal.

SOLUTION: 3/4 = 300/400

= (300/4) × (1/100)

= 75 × 1/100

= 0.75

Notice that the original fraction could have been rewritten as 3000/4000, in which case the result would have been 0.750. On the other hand, if the original fraction had been rewritten as 30/40, the resulting division of 4 into 30 would not have been possible without a remainder. When the denominator in the original fraction has only 2's and 5's as factors, so that no remainder is not necessary, the fraction should be rewritten with enough 0's to complete the division with no remainder.

Observation of the results in the foregoing example leads to a shortcut in the conversion method. Noting that the factor 1/100 ultimately enters the answer in the form of a decimal, we could introduce the decimal point as the final step without ever writing the fraction 1/100. Thus the rule for changing fractions to decimals is as follows:

1. Annex enough 0's to the numerator of the original fraction so that the division will be exact (no remainder). 2. Divide the original denominator into the new numerator formed by annexing the 0's. 3. Place the decimal point in the answer so that the number of decimal places in the answer is the same as the number of 0's annexed to the original numerator.

49

MATHEMATICS, VOLUME 1

If a mixed number in common fraction form is to be converted, convert only the fractional part and then write the two parts together. This is illustrated as follows:

2¾ = 2 + 3/4 = 2 + .75 = 2.75

Practice problems. Convert the following common fractions and mixed numbers to decimal form:

1. 1/4 2. 3/8 3. 5/32 4. 2 5/16

Answers:

1. 0.25 2. 0.375 3. 0.15625 4. 2.3125

Nonterminating Decimals

As stated previously, if the denominator of a common fraction contains some prime factor other than 2 or 5, the fraction cannot be converted completely to a decimal. When such fractions are converted according to the foregoing rule, the decimal resulting will never terminate. Consider the fraction 1/3. Applying the rule, we have

.333 ... 3)1.0000 9 10 9 10 9 1

The division will continue indefinitely. Any common fraction that cannot be converted exactly yields a decimal that will never terminate and in which the digits sooner or later recur. In the previous example, the recurring digit was 3. In the fraction 5/11, we have

.4545 11)5.0000 44 60 55 50 44 60 55

The recurring digits are 4 and 5.

When a common fraction generates such a repeating decimal, it becomes necessary to arbitrarily select a point at which to cease the repetition. This may be done in two ways. We may write the decimal fraction by writing of at the desired point. For example, to round off the decimal generated by 1/3 to hundredths, we carry the division to thousandths, see that this figure is less than 5, and drop it. Thus, 1/3 rounded to hundredths is 0.33. The other method is to carry the division to the desired number of decimal places and carry the remaining incomplete division as a common fraction—that is, we write the result of a complex decimal. For example, 1/3 carried to thousandths would be

.333⅓ 1/3 = 3/1.000 9 10 9 10 9 1

Practice problems. Change the following common fractions to decimals with three places and carry the incomplete division as a common fraction:

1. 7/13 2. 5/9 3. 4/15 4. 5/12

Answers:

1. 0.538⁶/₁₃ 3. 0.266⅔ 2. 0.555⅔ 4. 0.41⅔

OPERATION WITH DECIMALS

In the study of addition of whole numbers, it was established that units must be added to units, tens to tens, hundreds to hundreds, etc. For convenience, in adding several numbers, units were written under units, tens under tens, etc. The addition of decimals is accomplished in the same manner.

50

Chapter 5—DECIMALS

ADDITION

In adding decimals, tenths are written under tenths, hundredths under hundredths, etc. When this is done, the decimal points fall in a straight line. The addition is the same as in adding whole numbers. Consider the following example:

2.18 34.35 0.14 4.90 41.57

Adding the first column on the right gives 17 hundredths or 1 tenth and 7 hundredths. As with whole numbers, we write the 7 under the hundredths column and add the 1 tenth in the tenths column—that is, the column of the next higher order. The sum of the tenths column is 15 tenths or 1 unit and 5 tenths. The 5 is written under the tenths column and the 1 is added in the units column.

It is evident that if the decimal points are kept in a straight line—that is, if the place values are kept in the proper columns—addition with decimals may be accomplished in the ordinary manner of addition of whole numbers. It should also be noted that the decimal point of the sum falls directly under the decimal points of the addends.

SUBTRACTION

Subtraction of decimals likewise involves no new principles. Notice that the place values of the subtrahend in the following example are fixed directly under the corresponding place values in the minuend. Notice also that this causes the decimal points to be aligned and that the figures in the difference (answer) also retain the correct columnar alignment.

45.76 -31.87 13.89

We subtract column by column, as with whole numbers, beginning at the right. Practice problems. Add or subtract as indicated:

1. 12.3 + 2.13 + 4 + 1.234 2. 0.5 + 0.04 + 12.001 + 10 3. 237.5 - 217.9 4. 9.04 - 7.156

Answers:

1. 19.664 3. 19.6 2. 22.541 4. 1.884

MULTIPLICATION

Multiplication of a decimal by a whole number may be explained by expressing the decimal as a fraction.

EXAMPLE: Multiply 6.12 by 4.

SOLUTION: 4/1 × 612/100 = 2448/100 = 24.48

When we perform the multiplication keeping the decimal form, we have

6.12 4 24.48

By common sense, it is apparent that the whole number 4 times the whole number 6, with some fraction, will yield a number in the neighborhood of 24. Hence, the placing of the decimal point is reasonable.

An examination of several examples will reveal that the product of a decimal and a whole number has the same number of decimal places as the factor containing the decimal. Zeros, if any, at the end of the decimal should be rejected.

Multiplication of Two Decimals

To show the rule for multiplying two decimals together, we multiply the decimal in fractional form first and then in the conventional way, as in the following example:

0.4 × 0.37

Writing these decimals as common fractions, we have

4/10 × 37/100 = 4 × 37/(10 × 100) = 148/1000 = 0.148

51

MATHEMATICS, VOLUME 1

In decimal form the problem is

0.37 0.4 0.148

The placing of the decimal point is reasonable, since 4 tenths of 37 hundredths is a little less than half of 37 hundredths, or about 15 hundredths.

Consider the following example:

4.316 × 3.4

In the common fraction form, we have

4316/1000 × 34/10 = 4316 × 34/(1000 × 10) = 146744/10000 = 14.6744

In the decimal form the problem is

4.316 3.4 17264 pd48 14.6744

We note that 4 and a fraction times 3 and a fraction yields a product in the neighborhood of 12. Thus, the decimal point is in the logical place.

In the above examples it should be noted in each case that when we multiply the decimals together we are multiplying the numerators. When we place the decimal point by adding the number of decimal places in the multiplier and multiplicand, we are in effect multiplying the denominators.

When the numbers multiplied together are thought of as the numerators, the decimal points may be temporarily disregarded and the numbers may be considered whole. This justifies the apparent disregard for place value in the multiplication of decimals. We see that the rule for multiplying decimals is only a modification of the rule for multiplying fractions.

To multiply numbers in which one or more of the factors contain a decimal, multiply as though the numbers were whole numbers. Mark off as many decimal places in the product as there are decimal places in the factors together.

Practice problems. Multiply as indicated:

1. 3.7 × 0.02 2. 0.45 × 0.7 3. 6.5 4. 0.0073 x0.01 x5.4

Answers:

1. 0.074 2. 0.315 3. 0.065 4. 0.03942

Multiplying by Powers of 10

Multiplying by a power of 10 (10, 100, 1,000, etc.) is done mechanically by simply moving the decimal point to the right as many places as there are zeros in the multiplier. For example, 0.00687 is multiplied by 1,000 by moving the decimal point three places to the right as follows:

1,000 × 0.00687 = 6.87

Multiplying a number by 0.1, 0.01, 0.001, etc., is done mechanically by simply moving the decimal point to the left as many places as there are decimal places in the multiplier. For example, 348.2 is multiplied by 0.001 by moving the decimal point three places to the left as follows:

348.2 × 0.001 = 0.3482

DIVISION

When the dividend is a whole number, we recognize the problem of division as that of converting a common fraction to a decimal. Thus in the example 5 ÷ 8, we recall that the problem could be written

5000 ÷ 8 = 5000 ÷ 8 1000 1000 = 625 1000 = .625

This same problem may be worked by the following, more direct method:

625 5 = 8/5.000 8 4 8 20 16 40 40

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Chapter 5—DECIMALS

Since not all decimals generated by division terminate so early as that in the above example, if at all, it should be predetermined as to how many decimal places it is desired to carry the quotient. If it is decided to terminate a quotient at the third decimal place, the division should be carried to the fourth place so that the correct rounding off of the third place may be determined.

When the dividend contains a decimal, the same procedure applies as when the dividend is whole. Notice the following examples (rounded to three decimal places):

1. 6.31 ÷ 8

.7887 = .789 8/6.3100 5 6 71 64 70 64 60 56 4

2. 0.0288 ÷ 32

0.0009 = .001 32/0.0288 288

Observe in each case (including the case where the dividend is whole), that the quotient contains the same number of decimal places as the number used in the dividend. Notice also that the place values are rigid; that is, tenths in the quotient appear as tenths in the dividend, hundredths over hundredths, etc.

Practice problems. In the following division problems, round off each quotient correct to three decimal places.

1. 10 ÷ 6 3. 2.743 ÷ 77 2. 23.5 ÷ 16 4. 1.00 ÷ 3

Answers:

1. 1.667 3. 0.036 2. 1.469 4. 0.333

Decimal Divisors

In the foregoing examples, the divisor in each case was an integer. Division with divisors which are decimals may be accomplished by changing the divisor and dividend so that the divisor becomes a whole number.

Recalling that every division expression may be written in fraction form, we use the fundamental rule of fractions as follows: Rewrite the division problem as a fraction. Multiply the numerator (dividend) and denominator (divisor) by 10, 100, or some higher power of 10; the power of 10 must be large enough to change the divisor to a whole number. This rule is illustrated as follows:

2.568 ÷ 0.24 = 2.568/0.24

= 2.568/0.24 × 100/100

= 256.8/24

Thus 2.568 divided by 0.24 is the same as 256.8 divided by 24.

From the mechanical standpoint, the foregoing rule has the effect of moving the decimal point to the right, as many places as necessary to change the divisor to an integer. Therefore the rule is sometimes stated as follows: When the divisor is a decimal, change it to a whole number by moving the decimal point to the right. Balance the change in the divisor by moving the decimal point in the dividend an equal number of places to the right.

The following example illustrates this version of the rule:

9.1.1 0.9÷/81.9.9

The inverted v, called a caret, is used as a marker to indicate the new position of the decimal point. Notice that the decimal point in the quotient is placed immediately above the caret in the dividend. Alignment of the first quotient digit immediately above the 1 in the dividend, and the second quotient digit above the 9, assures that these digits are placed properly with respect to the decimal point.

Practice problems. In the following division problems, round off each quotient to three decimal places:

1. 0.02958 ÷ 0.12 3. 4610 ÷ 0.875 2. 30.625 ÷ 3.5 4. 0.000576 ÷ 0.008

Answers:

1. 0.247 3. 5268.571 2. 8.750 4. 0.072

53

MATHEMATICS, VOLUME 1

Dividing by Powers of 10

Division of any number by 10, 100, 1,000, etc., is really just an exercise in placing the decimal point of a decimal fraction. Thus, 5,031 ÷ 100 may be thought of as the decimal fraction 5031/100; to remove the denominator, we simply count off two places from the right. Thus,

5031/100 = 50.31

The following three examples serve to illustrate this procedure further:

401 ÷ 10 = 40.1 2 ÷ 1,000 = .002 11,431 ÷ 100 = 114.31

If the dividend already contains a decimal point, begin counting with the first number to the left of the decimal point. Thus, 243.6 ÷ 100 = 2.436. When the decimal point is not shown in a number, it is always considered to be to the right of the right-hand digit.

Dividing by 0.1, 0.01, 0.001, etc., may also be accomplished by a simple mechanical rule. We simply begin at the position of the decimal point in the dividend and count off as many places to the right as there are decimal places in the divisor. The decimal point is then placed to the right of the last digit counted. If there are not enough digits, zeros may be added.

The foregoing rule is based on the fact that 0.1 is really 1/10, 0.01 is 1/100, 0.001 is 1/1000, etc.

For example,

23 ÷ 0.1 = 23 ÷ 1/10 = 23 × 10/1 = 230

Notice that dividing by 0.1 is the same as multiplying by 10. Likewise,

234.1 ÷ 0.001 = 234.1 ÷ 1/1000 = 234.1 × 1000/1 = 234,100

and

24 ÷ 0.01 = 24 ÷ 1/100 = 24 × 100/1 = 2,400

Practice problems. Divide by relocation of the decimal point.

1. 276 ÷ 100 3. 276 ÷ 0.01 2. 2,845 ÷ 1,000 4. 2,845 ÷ 0.001

Answers:

1. 2.76 3. 27,600 2. 2.845 4. 2,845,000

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CHAPTER 6 PERCENTAGE AND MEASUREMENT

In the discussion of decimal fractions, it was shown that for convenience in writing fractions whose denominators are 10 or some power of 10, the decimal point could be employed and the denominators could be dropped. Thus, this special group of fractions could be written in a much simpler way. As early as the 15th century, businessmen made use of certain decimal fractions so much that they gave them the special designation PERCENT.

MEANING OF PERCENT

The word "percent" is derived from Latin. It was originally "per centum," which means "by the hundred." Thus the statement is often made that "percent means hundredths." Percentage deals with the group of decimal fractions whose denominators are 100—that is, fractions of two decimal places. Since hundredths were used so frequently, the decimal point was dropped and the symbol % was placed after the number and read "percent" (per 100). Thus, 0.15 and 15% represent the same value, 15/100. The first is read "15 hundredths," and the second is read "15 percent." Both mean 15 parts out of 100.

Ordinarily, percent is used in discussing relative values. For example, 25 percent may convey an idea of relative value or relationship. To say "25 percent of the crew is ashore" gives an idea of what part of the crew is gone, but it does not tell how many. For example, 25 percent of the crew would represent vastly different numbers if the comparison were made between an LSM and a cruiser. When it is necessary to use a percent in computation, the number is written in its decimal form, to avoid confusion. By converting all decimal fractions so that they had the common denominator 100, men found that they could mentally visualize the relative size of the part of the whole that was being considered.

CHANGING DECIMALS TO PERCENT

Since percent means hundredths, any decimal may be changed to percent by first expressing it as a fraction with 100 as the denominator. The numerator of the fraction thus indicated indicates how many hundredths we have, and therefore it indicates "how many percent" we have. For example, 0.36 is the same as 36/100. Therefore, 0.36 expressed as a percentage would be 36 percent. By the same reasoning, since 0.052 is equal to 5.2/100, 0.052 is the same as 5.2 percent.

In actual practice, the step in which the denominator 100 occurs is seldom written down. The expression in terms of hundredths is converted mentally to percent. This results in the following rule: To change a decimal to percent, multiply the decimal by 100 and annex the percent sign (%). Since multiplying by 100 has the effect of moving the decimal point two places to the right, the rule is sometimes stated as follows: To change a decimal to percent, move the decimal point two places to the right and annex the percent sign.

Changing Common Fractions and Whole Numbers To Percent

Common fractions are changed to percent by first expressing them as decimals. For example, the fraction 1/4 is equivalent to the decimal 0.25. Thus 1/4 is the same as 25 percent. Whole numbers may be considered as special types of decimals (for example, 4 may be written as 4.00) and thus may be expressed in terms of percentage. The meaning of an expression such as 400 percent is vague unless we keep in mind that percentage is a form of comparison. For example, a question which often arises is "How can I have more than 100 percent of something, if 100 percent means all of it?" This question seems reasonable, if we limit our attention to such quantities as test scores. However, it is also reasonable to use percentage in comparing a current set of data with a previous set. For example, if the amount of electrical power used by a Navy facility this year is double the amount used last year, then this year's power usage is 200 percent of last year's usage.

MATHEMATICS, VOLUME 1

The meaning of a phrase such as "200 percent of last year's usage" is often misinterpreted. A total amount that is 200 percent of the previous amount is not the same as an increase of 200 percent. The increase in this case is only 100 percent, for a total of 200. If the increase had been 200 percent, then the new usage figure would be 300 percent of the previous figure.

Baseball batting averages comprise a special case in which percentage is used with only occasional reference to the word "percent." The percentages in batting averages are expressed in their decimal form, with the figure 1.000 representing 100 percent. Although a batting average of 0.300 is referred to as "batting 300," this is actually erroneous nomenclature from the strictly mathematical standpoint. The correct statement, mathematically, would be "batting point three zero zero" or "batting 30 percent."

Practice problems. Change each of the following numbers to percent:

1. 0.0065 3. 0.363 5. 7 2. 1.25 4. 3/4 6. 1/2

Answers:

1. 0.65% 3. 36.3% 5. 700% 2. 125% 4. 75% 6. 50%

CHANGING A PERCENT TO A DECIMAL

Since we do not compute with numbers in the percent form, it is often necessary to change a percent back to the decimal form. The procedure is just opposite to that used in changing decimals to percents: To change a percent to a decimal, drop the percent sign and divide the number by 100. Mechanically, the decimal point is simply shifted two places to the left and the percent sign is dropped. For example, 25 percent is the same as the decimal 0.25. Percents larger than 100 percent are changed to decimals by the same procedure as ordinary percents. For example, 125 percent is equivalent to 1.25.

Practice problems. Change the following percents to decimals:

1. 2.5% 3. 125% 5. 5¾% 2. 0.63% 4. 25% 6. 9½%

Answers:

1. 0.025 3. 1.25 5. 5.75% = 0.0575 2. 0.0063 4. 0.25 6. 9.50% = 0.095

THE THREE PERCENTAGE CASES

To explain the cases that arise in problems involving percents, it is necessary to define the terms that will be used. Rate (r) is the number of hundredths parts taken. This is the number followed by the percent sign. The base (b) is the whole on which the rate operates. Percentage (p) is the part of the base determined by the rate. In the example

5% of 40 = 2

5% is the rate, 40 is the base, and 2 is the percentage.

There are three cases that usually arise in dealing with percentage, as follows:

Case I—To find the percentage when the base and rate are known.

EXAMPLE: What number is 6% of 50?

Case II—To find the rate when the base and percentage are known.

EXAMPLE: 20 is what percent of 60?

Case III—To find the base when the percentage and rate are known.

EXAMPLE: The number 5 is 25% of what number?

Case I

In the example

6% of 50 = ?

the "of" has the same meaning as it does in fractional examples, such as

1/4 of 16 = ?

In other words, "of" means to multiply. Thus, to find the percentage, multiply the base by the rate. Of course the rate must be changed from a percent to a decimal before multiplying can

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Chapter 6—PERCENTAGE AND MEASUREMENT

least precise addend. It can also be shown that the precision of a difference is no greater than the less precise number compared. To add or subtract numbers of different or- ders, all numbers should first be rounded off to the order of the least precise number. In the foregoing example, 1.426 should be rounded to tenths—that is, 1.4.

This rule also applies to repeating decimals. Since it is possible to round off a repeating decimal at any desired point, the degree of pre- cision desired should be determined and all re- peating decimals to be added should be rounded to this level. Thus, to add the decimals gener- ated by 1 2/3, 2/3, and 5/12 to correct to thousandths, first round off each decimal to thousandths, and then add, as follows:

.333 .667 .417 ______ 1.417

When a common fraction is used in recording the results of measurement, the denominator of the fraction indicates the degree of precision. For example, a pipe marked in sixty-fourths of an inch is smaller divisions than one marked in sixteenths of an inch. Therefore a

measurement of 3⁵⁄₆₄ inches is more precise

than a measure of 3—inches, even though the 16

two fractions are numerically equal. Remember that a measurement of 3⁵⁄₆₄ inches contains a probable error of only one-half of one sixty- fourth of an inch. On the other hand, if the smallest division on the ruler is one-sixteenth of an inch, then a measurement of 3—inches con- 16 tains a probable error of one thirty-second of an inch.

ACCURACY

Even though a number may be very precise, which indicates that it was measured with an instrument having closely spaced divisions, it may not be very accurate. The accuracy of a measurement depends upon the relative size of the probable error when compared with the quantity being measured. For example, a dis- tance of 25 yards on a pistol range may be

measured carefully enough to be correct to the nearest inch. Since there are 900 inches in 25 yards, this measurement is between 899.5 inches and 900.5 inches. When compared with the total of 900 inches, the 0.5-inch probable error is not very great.

On the other hand, a length of pipe may be measured rather precisely and found to be 3.2 inches long. The probable error here is 0.05 inch, and this measurement is thus more pre- cise than that of the pistol range mentioned be- fore. To compare the accuracy of the two meas- urements, we note that 0.05 inch out of a total of 3.2 inches is the same as 0.5 inch out of 32 inches. Comparing this with the figure obtained in the other example (0.5 inch out of 900), we conclude that the more precise measurement is actually the less accurate of the two measure- ments considered.

It is important to realize that the location of the decimal point has no bearing on the accu- racy of the number. For example, 1.25 dollars represents exactly the same amount of money as 125 cents. These are equally accurate ways of representing the same quantity, despite the fact that the decimal point is placed differently.

Practice problems. In each of the following problems, determine which number of each pair is more accurate and which is more precise:

1. 3.72 inches or 2,417 feet 2. 2.5 inches or 17.5 inches 3. 5²⁄₃ inches or 12⁵⁄₈ inches 4. 34.2 seconds or 13 seconds

Answers:

1. 3.72 inches is more precise. 2,417 feet is more accurate. 2. The numbers are equally precise. 17.5 inches is more accurate. 3. 12⁵⁄₈ inches is more precise and more accu- rate. 4. 34.2 seconds is more precise and more ac- curate.

Percent of Error

The accuracy of a measurement is deter- mined by the RELATIVE ERROR. The relative

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MATHEMATICS, VOLUME 1

error is the ratio between the probable error and the quantity being measured. This ratio is simply the fraction formed by using the prob- able error as the numerator and the measure- ment itself as the denominator. For example, suppose that a metal plate is found to be 5.4 inches long, correct to the nearest tenth of an inch. The maximum probable error is five hundredths of an inch (one-half of one tenth of an inch) and the relative error is found as follows:

probable error 0.05 ─────────────── = ───── measured value 5√4

= 5 ───── 540

Thus the relative error is 5 parts out of 540. Relative error is usually expressed as PER- CENT OF ERROR. When the denominator of the fraction expressing the error is divided into the numerator, a decimal is obtained. This decimal, converted to percent, gives the percent of error. For example, the error in the foregoing problem could be stated as 0.93 percent, since the 5÷540 reduces to 0.0093 (rounded off) in decimal form.

Significant Digits

The accuracy of a measurement is often de- scribed in terms of the number of significant digits used in expressing it. If the digits of a number resulting from measurement are exam- ined one by one, beginning with the left-hand digit, the first digit that is not 0 is the first significant digit. For example, 2345 has four significant digits and 0.023 has only two sig- nificant digits.

The digits 2 and 3 in a measurement such as 0.023 inch signify how many thousandths of an inch comprise the measurement; their presence is required only as "place holders" in placing the decimal point.

A rule often used states that the sig- nificant digits in a number begin with the first nonzero digit (counting from left to right) and end with the last digit. This implies that 0 can be a significant digit if it is not the first digit in the number. For example, 0.205 inch is a measurement having three significant digits. The 0 between the 2 and 5 is significant

because it is a part of the number specifying how many hundredths are in the measurement.

The rule stated in the foregoing paragraph fails to classify final 0's on the right. For ex- ample, in a number such as 4,700, the number of significant digits might be two, three, or four. If the 0's merely locate the decimal point (that is, if they show the number to be approxi- mately forty-seven hundred rather than forty seven), then the number of significant digits is two. However, if the number 4,700 represents a number such as 4,730 rounded off to the near- est hundred, there are three significant digits. The last 0 merely locates the decimal point. If the number 4,700 represents a number such as 4,700.4 rounded off, then the number of signifi- cant digits is four.

Unless we know how a particular number was measured, it is sometimes impossible to determine whether right-hand 0's are the result of rounding off. However, in a practical situa- tion it is normally possible to obtain informa- tion concerning the instruments used and the degree of precision of the original data before any rounding was done.

In a number such as 49.30 inches, it is rea- sonable to assume that the 0 in the hundredths place would not have been recorded at all if it were not significant. In other words, the in- strument used for the measurement can be read to the nearest hundredth of an inch. The 0 on the right is thus significant. This conclusion can be reached another way by observing that the 0 in 49.30 is not needed as a holder in placing the decimal point. Therefore its pres- ence must have some other significance.

The facts concerning significant digits may be summarized as follows:

1. Digits other than 0 are always significant. 2. Zero is significant when it falls between significant digits. 3. Any final 0 to the right of the decimal point is significant. 4. When a 0 is present only as a place holder for locating the decimal point, it is not significant. 5. The following categories comprise the significant digits of any measurement number: a. The first nonzero left-hand digit is significant. b. The digit which indicates the precision of the number is significant. This is the digit farthest to the right, except when the right-hand digit is 0. If it is 0, it may be only a place holder when the number is an integer.

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Chapter 6—PERCENTAGE AND MEASUREMENT

c. All digits between significant digits are significant.

Practice problems. Determine the percent of error and the number of significant digits in each of the following measurements:

1. 5.4 feet 3. 4.17 sec 2. 0.00042 inch 4. 147.50 miles

Answers:

1. Percent of error: 0.93% Significant digits: 2

2. Percent of error: 1.19% Significant digits: 2

3. Percent of error: 0.12% Significant digits: 3

4. Percent of error: 0.0034% Significant digits: 5

CALCULATING WITH APPROXIMATE NUMBERS

The concepts of precision and accuracy form the basis for the rules which govern calculation with approximate numbers (numbers resulting from measurement).

Addition and Subtraction

A sum or difference can never be more precise than the least precise number in the calculation. Therefore, before adding or subtracting approximate numbers, they should be rounded to the same degree of precision. The more precise numbers are all rounded to the precision of the least precise number in the group to be combined. For example, the numbers 2.95, 32.7, and 1.414 would be rounded to tenths before adding as follows:

3.0 32.7 1.4

Multiplication and Division

When two numbers are multiplied, the result often has several more digits than either of the original factors. Division also frequently produces more digits in the quotient than the original data possessed, if the division is "carried out" to several decimal places. Results such

as these appear to have more significant digits than the original measurements from which they came, giving the false impression of greater accuracy than is justified. In order to correct this situation, the following rule is used:

In order to multiply or divide two approximate numbers having an equal number of significant digits, round the answer to the same number of significant digits as are shown in one of the original numbers. If one of the original factors has more significant digits than the other, round the more accurate number before multiplying. It should be rounded to one more significant digit than appears in the less accurate number; the extra digit protects the answer from the effects of multiple rounding. After performing the multiplication or division, round the result to the same number of significant digits as are shown in the less accurate of the original factors.

Practice problems:

1. Find the sum of the sides of a triangle in which the lengths of the three sides are as follows: 2.5 inches, 3.72 inches, and 4.996 inches.

2. Find the product of the length and width of a rectangle which is 2.95 feet long and 0.9046 foot wide.

Answers:

1. 11.2 inches 2. 2.67 square feet

MICROMETERS AND VERNIERS

Closely associated with the study of decimals is a measuring instrument known as a micrometer. The ordinary micrometer is capable of measuring accurately to one-thousandth of an inch. One-thousandth of an inch is about the thickness of a human hair or a thin sheet of paper. The parts of a micrometer are shown in figure 6-1.

MICROMETER SCALES

The spindle and the thimble move together. The end of the spindle (hidden from view in figure 6-1) is a screw with 40 threads per inch. Consequently, one complete turn of the thimble moves the spindle one-fortieth of an inch or

MATHEMATICS. VOLUME 1 0.200 0.025 0.002 0.227 Figure 6-1.–(A) Parts of a micrometer; (B) micrometer scales. of an inch. Multiplying times 0.025 inch, we find that each marking on the thimble repre- sents 0.001 inch. 0.025 inch since 1 is equal to 0.025. The40 sleeve has 40 markings to the inch. Thus each space between the markings on the sleeve is also 0.025 inch. Since 4 such spaces are 0.1 inch (that is, 4 x 0.025), every fourth mark is labeled in tenths of an inch for convenience in reading. Thus, 4 marks equal 0.1 inch, 8 marks equal 0.2 inch, 12 marks equal 0.3 inch, etc. To enable measurement of a partial turn, the beveled edge of the thimble is divided into 25 equal parts. Thus each marking on the thimble is 1 of a complete turn, or 1 of 1 25 25 40 1 25 READING THE MICROMETER It is sometimes convenient when learning to read a micrometer to writedown the component 62 parts of the measurement as read on the scales and then to add them. For example, in figure 6-1 (B) there are two major divisions visible (0.2 inch). One minor division is showing clearly (0.025 inch). The marking on the thimble nearest the horizontal or index line of the sleeve is the second marking (0.002 inch). Adding these parts, we have Thus, the reading is 0.227 inch. As explained previously, this is read verbally as "two hun- dred twenty-seven thousandths." A more skill- ful method of reading the scales is to read all digits as thousandths directly and to do any adding mentally. Thus, we read the major divi- sion on the scale as “two hundred thousandths” and the minor division is added on mentally. The mental process for the above setting then would be “two hundred twenty-five; two hundred twenty-seven thousandths.” Practice problems: 1. Read each of the micrometer settings shown in figure 6-2. Figure 6-2.–Micrometer settings.

Chapter 6—PERCENTAGE AND MEASUREMENT

Answers:

1. (A) 0.750 (F) 0.009 (B) 0.201 (G) 0.662 (C) 0.655 (H) 0.048 (D) 0.075 (I) 0.526 (E) 0.527

VERNIER

Sometimes the marking on the thimble of the micrometer does not fall directly on the index line of the sleeve. To make possible readings even smaller than thousandths, an ingenious device is introduced in the form of an additional scale. This scale, called a VERNIER, was named after its inventor, Pierre Vernier. The vernier makes possible accurate readings to the ten-thousandth of an inch.

Principle of the Vernier

Suppose a ruler has markings every tenth of an inch but it is desired to read accurately to hundredths. A separate, freely sliding vernier scale (fig. 6-3) is added to the ruler. It has 10 markings on it that take up the same distance as 9 markings on the ruler scale. Thus, each space on the vernier is (9/10) of (1/10) inch, or (9/100) inch. How much smaller is a space on the vernier than a space on the ruler? The ruler space is (1/10) inch, and the vernier space is (9/100) inch. The vernier space is smaller by the difference between these two numbers, as follows:

10/100 - 9/100 = 1/100

[DECIMAL RULER (ENLARGED) - diagram showing ruler markings from 3 to 4 with 1/10 and 9/100 measurements, and a VERNIER scale below]

Figure 6-3.—Vernier scale.

Each vernier space is (1/100) inch smaller than a ruler space.

As an example of the use of the vernier scale, suppose that we are measuring the steel bar shown in figure 6-4. The end of the bar almost reaches the 3-inch mark on the ruler, and we estimate that it is about halfway between 2.9 inches and 3.0 inches. The vernier marks help us to decide whether the exact measurement is 2.94 inches, 2.95 inches, or 2.96 inches.

[DECIMAL RULER (ENLARGED) - diagram showing ruler with vernier scale and a steel bar being measured]

Figure 6-4.—Measuring with a vernier.

The 0 on the vernier scale is spaced the distance of exactly one ruler mark (in this case, one tenth of an inch) from the left hand end of the vernier. Therefore the 0 is at a position between ruler marks which is comparable to the position of the end of the bar. In other words, the 0 on the vernier is about halfway between two adjacent marks on the ruler, just as the end of the bar is about halfway between two adjacent marks. The 1 on the vernier scale is a little closer to alignment with an adjacent ruler mark; in fact, it is one hundredth of an inch closer to alignment than the 0. This is because each space on the vernier is one hundredth of an inch shorter than each space on the ruler.

Each successive mark on the vernier scale is one hundredth of an inch closer to alignement than the preceding mark, until finally alignement is achieved at the 5 mark. This means that the 0 on the vernier must be five hundredths of an inch from the nearest ruler mark, since five increments, each one hundredth of an inch in size, were used before a mark found in alignement.

We conclude that the end of the bar is five hundredths of an inch from the 2.9 mark on the ruler, since its position between marks is exactly comparable to that of the 0 on the vernier scale. Thus the value of our measurement is 2.95 inches.

MATHEMATICS, VOLUME 1 The foregoing example could be followed through for any distance between markings. Suppose the 0 mark fell seven tenths of the dis- tance between ruler markings. It would take seven vernier markings, a loss of one-hundredth of an inch each time, to bring the marks in line at 7 on the vernier. The vernier principle may be used to get fine linear readings, angular readings, etc. The principle is always the same. The vernier has one more marking than the number of mark- ings on an equal space of the conventional scale of the measuring instrument. For example, the vernier caliper (fig. 6-5) has 25 markings on the vernier for 24 on the caliper scale. The caliper is marked off to read to fortieths (0.025) of an inch, and the vernier extends the accuracy to a thousandth of an inch. 0.3000 0.0750 0.0080 0.0004 0.3834 Figure 6-5.–A vernier caliper. Vernier Micrometer By adding a vernier to the micrometer, it is possible to read accurately to one ten-thousandth of an inch. The vernier markings are on the sleeve of the micrometer and are parallel to the thimble markings. There are 10 divisions on the vernier that occupy the same space as 9 divisions on the thimble. Since a thimble space is one thousandth of an inch, a vernier space is 1 of 9 inch, or 9 inch. It is 1 inch10 1000 10000 10000 less than a thimble space. Thus, as in the pre- ceding explanation of verniers, it is possible to read the nearest ten-thousandth of an inch by reading the vernier digit whose marking coin- cides with a thimble marking. In figure 6-6 (A), the last major division showing fully on the sleeve index is 3. The third minor division is the last mark clearly 64 showing (0.075). The thimble division nearest and below the index is the 8 (0.008). The ver- nier marking that matches a thimble marking is the fourth (0.0004). Adding them all together, we have, The reading is 0.3834 inch. With practice these readings can be made directly from the microm- eter, without writing the partial readings. Figure 6-6.–Vernier micrometer settings. Practice problems: 1. Read the micrometer settings in figure 6-6. Answers: 1. (A) See the foregoing example. (B) 0.1539 (E) 0.4690 (C) 0.2507 (F) 0.0552 (D) 0.2500

CHAPTER 7 EXPONENTS AND RADICALS

The operation of raising a number to a power is a special case of multiplication in which the factors are all equal. For examples such as 4² = 4 x 4 = 16 and 5³ = 5 x 5 x 5 = 125, the number 16 is the second power of 4 and the number 125 is the third power of 5. The expression "5³" means that the five 5's are to be multiplied together. Similarly, 4³ means 4 x 4. The first power of any number is the number itself. The power is the number of times the number itself is to be taken as a factor.

The process of finding a root is the inverse of raising a number to a power. A root is a special factor of a number, such as 4 in the expression 4² = 16. When a number is taken as a factor two times, as in the expression 4 x 4 = 16, it is called a square root. Thus, 4 is a square root of 16. By the same reasoning, 2 is a cube root of 8, since 2 x 2 x 2 is equal to 8. This relationship is usually written as 2³ = 8.

POWERS AND ROOTS

A power of a number is indicated by an EXPONENT, which is a number in small print placed to the right and toward the top of the number. Thus, in 4³ = 64, the number 3 is the EXPONENT of the number 4. The exponent 3 indicates that the number 4, called the BASE, is to be raised to its third power. The expression "64 to the third power (or 4 cubed) equals 64." Similarly, 5² = 25 is read "5 to the second power (or 5 squared) equals 25." Higher powers are read according to the degree indicated; for example, "fourth power," "fifth power," etc.

When an exponent occurs, it must always be written unless its value is 1. The exponent 1 usually is not written, but is understood. For example, the number 5 is actually 5¹. When we work with exponents, it is important to remember that any number that has no written exponent really has an exponent equal to 1.

A root of a number can be indicated by placing a radical sign, √, over the number and showing the root by placing a small number within the notch of the radical sign. Thus, ³√64 indicates the cube root of 64, and ⁵√32 indicates the fifth root of 32. The number that indicates the root is called the INDEX of the root. In the case of the square root, the index is usually is not shown. When a radical has no index, the square root is understood to be the one desired. For example, √36 indicates the square root of 36. The line above the number whose root is to be found is a symbol of grouping called the vinculum. When the radical symbol is used, a vinculum, long enough to extend over the entire expression whose root is to be found, should be attached.

Practice problems. Raise to the indicated power or find the root indicated.

1. 2³ 2. 6² 3. 4³ 4. 25³ 5. √16 6. ³√8 7. ⁵√125 8. ⁴√32

Answers:

1. 8 2. 36 3. 64 4. 15,625 5. 4 6. 2 7. 5 8. 2

NEGATIVE INTEGERS

Raising to a power is multiplication in which all the numbers being multiplied together are equal. The sign of the product is determined, as in ordinary multiplication, by the number of minus signs. The number of minus signs is odd or even, depending on whether the exponent of the base is odd or even. For example, in the problem

(-2)³ = (-2)(-2)(-2) = -8

there are three minus signs. The result is negative. In

(-2)⁶ = 64

there are six minus signs. The result is positive.

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MATHEMATICS, VOLUME 1

Thus, when the exponent of a negative number is odd, the power is negative; when the exponent is even, the power is positive.

As other examples, consider the following:

(-3)⁴ = 81 (-2/5)³ = -8/125 (-2)⁸ = 256 (-1)⁵ = -1

Positive and negative numbers belong to the class called REAL NUMBERS. The square of a real number is positive. For example, (-7)² = 49 and 7² = 49. The expression (-7)² is read "minus seven squared." Note that either seven squared or minus seven squared gives us +49. We cannot obtain -49 or any other negative number by squaring any real number, positive or negative.

Since there is no real number whose square is a negative number, it is sometimes said that the square root of a negative number does not exist. However, an expression under a square root sign may take on negative values. While the square root of a negative number cannot actually be found, it can be indicated.

The indicated square root of a negative number is called an IMAGINARY NUMBER. The number √-1, for example, is said to be imaginary. It is read "square root of minus seven." Imaginary numbers are discussed in chapter 15 of this course.

FRACTIONS

We recall that the exponent of a number tells the number of times that the number is to be taken as a factor. A fraction is raised to a power by raising the numerator and the denominator separately to the indicated power. The expression (3/7)² means 3/7 is used twice as a factor. Thus,

(3/7)² = 3/7 × 3/7 = 3²/7² = 9/49

Similarly,

(-1/5)² = 1/25

Since a minus sign can occupy any one of three locations in a fraction, notice that evaluating (-1/5)² is equivalent to

(-1)²(1/5)² or (-1)²/5² or 1/(-5)²

The process of taking a root of a number is the inverse of the process of raising the number to a power, and the method of taking the root of a fraction is similar. We may simply take the root of each term separately and write the result as a fraction. Consider the following examples:

1. ⁴√(36/49) = √36/√49 = 6/7

2. ³√(8/125) = ³√8/³√125 = 2/5

Practice problems. Find the values for the indicated operations:

1. (1/3)² 2. (3/4)² 3. (6/5)² 4. (2/3)³ 5. ⁴√(16/36) 6. ⁴√(16/25) 7. ³√(8/27) 8. ⁴√(81/49)

Answers:

1. 1/9 2. 9/16 3. 36/25 4. 8/27 5. 4/6 6. 4/5 7. 2/3 8. 3/7

DECIMALS

When a decimal is raised to a power, the number of decimal places in the result is equal to the number of places in the decimal multiplied by the exponent. For example, consider (0.12)³. There are two decimal places in 0.12 and 3 is the exponent. Therefore, the number of places in the power will be 3(2) = 6. The result is as follows:

(0.12)³ = 0.001728

The truth of this rule is evident when we recall the rule for multiplying decimals. Part of the rule states: Mark off as many decimal places in the product as there are decimal places in the factors together. If we carry out

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Chapter 7—EXPONENTS AND RADICALS

the multiplication, (0.12) x (0.12) x (0.12), it is obvious that there are six decimal places in the three factors together. The rule can be shown for any decimal raised to any power by simply carrying out the multiplication indicated by the exponent. Consider these examples:

(1.4)² = 1.96 (0.12)² = 0.0144 (0.4)³ = 0.064 (0.02)² = 0.0004 (0.2)² = 0.04

Finding a root of a number is the inverse of raising a number to a power. To determine the number of decimal places in the root of a per- fect power, we divide the number of decimal places in the radicand by the index of the root. Notice that this is just the opposite of what was done in raising a number to a power. Consider √0.0625. The square root of 625 is 25. There are four decimal places in the radicand, 0.0625, and the index of the root is 2. Therefore, 4 ÷ 2 = 2 is the number of decimal places in the root. We have

√0.0625 = 0.25

Similarly, √1.69 = 1.3 ³√0.027 = 0.3 ⁴√1.728 = 1.2 ⁶√0.0001 = 0.1

LAWS OF EXPONENTS

All of the laws of exponents may be devel- oped directly from the definition of exponents. Separate laws are stated for the following five cases: 1. Multiplication. 2. Division. 3. Power of a power. 4. Power of a product. 5. Power of quotient.

MULTIPLICATION

To illustrate the law of multiplication, we examine the following problem:

4³ x 4² = ?

Recalling that 4³ means 4 x 4 x 4 and 4² means 4 x 4, we see that 4 is used as a factor five times. Therefore 4³ x 4² is the same as 4⁵. This result could be written as follows: 4³ x 4² = 4 x 4 x 4 x 4 x 4 = 4⁵

Notice that three of the five 4's came from the expression 4³, and the other two 4's came from the expression 4². Thus we may rewrite the problem as follows: 4³ x 4² = 4⁽³⁺²⁾ = 4⁵

The law of exponents for multiplication may be stated as follows: To multiply two or more powers having the same base, add the exponents and raise the common base to the sum of the exponents. This law is further illustrated by the following examples:

2³ x 2⁴ = 2⁷ 3 x 3² = 3³ 15⁴ x 15² = 15⁶ 10² x 10⁰·⁵ = 10²·⁵

Common Errors

It is important to realize that the base must be the same for each factor in order to apply the laws of exponents. For example, 2³ x 3² is neither 2⁵ nor 3⁵. There is no way to apply the law of exponents to a problem of this kind. An- other common mistake is to multiply the bases together. For example, this kind of error in the foregoing problem would imply that 2³ x 3² is equivalent to 6⁵, or 7776. The error of this may be proved as follows: 2³ x 3² = 8 x 9 = 72

DIVISION

The law of exponents for division may be developed from the following example:

6⁷ ÷ 6⁵ = 6 x 6 x 6 x 6 x 6 x 6 x 6 6 x 6 x 6 x 6 x 6 = 6²

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MATHEMATICS, VOLUME 1

Cancellation of the five 6's in the divisor with five of the 6's in the dividend leaves only two 6's, the product of which is 6². This result can be reached directly by noting that 6⁵ is equivalent to 6⁽⁷⁻⁵⁾. In other words, we have the following:

6⁷ ÷ 6⁵ = 6⁽⁷⁻⁵⁾ = 6²

Therefore the law of exponents for division is as follows: To divide one power into another having the same base, subtract the exponent of the divisor from the exponent of the dividend. Use the number resulting from this subtraction as the exponent of the base in the quotient. Use of this rule sometimes produces a nega- tive exponent or an exponent whose value is 0. These two special types of exponents are dis- cussed later in this chapter.

POWER OF A POWER

Consider the example (3²)⁴. Remembering that an exponent shows the number of times the base is to be taken as a factor and noting in this case that 3² is considered the base, we have

(3²)⁴ = 3² · 3² · 3² · 3²

Also in multiplication we add exponents. Thus,

3² · 3² · 3² · 3² = 3⁽²⁺²⁺²⁺²⁾ = 3⁸

Therefore,

(3²)⁴ = 3⁽⁴ˣ²⁾ = 3⁸

The laws of exponents for the power of a power may be stated as follows: To find the power of a power, multiply the exponents. It should be noted that this case is the only one in which multiplication of exponents is performed.

POWER OF A PRODUCT

Consider the example (3 · 2 · 5)³. We know that

(3 · 2 · 5)³ = (3 · 2 · 5)(3 · 2 · 5)(3 · 2 · 5)

Thus 3, 2, and 5 appear three times each as factors, and we can show this with exponents as 3³, 2³, and 5³. Therefore,

(3 · 2 · 5)³ = 3³ · 2³ · 5³

The law of exponents for the power of a product is as follows: The power of a product is equal to the product obtained when each of the original factors is raised to the indicated power and the resulting powers are multiplied together.

POWER OF A QUOTIENT

The law of exponents for a power of an indi- cated quotient may be developed from the fol- lowing example:

(2/3)³ = 2/3 · 2/3 · 2/3 = 2 · 2 · 2 / 3 · 3 · 3 = 2³/3³

Therefore,

(2/3)³ = 2³/3³

The law is stated as follows: The power of a quotient is equal to the quotient obtained when the dividend and divisor are each raised to the indicated power separately, before the division is performed. Practice problems. Raise each of the fol- lowing expressions to the indicated power:

1. (3² · 2³)² 3. (3/5 · 2/6)³ 5. 5³/5 2. 3⁵ + 3² 4. (-3²)³ 6. (3 · 2 · 7)²

Answers:

1. 3⁴ x 2⁶ = 5,184 2. 27 3. 1/125 4. [(-3)²]³ = 729 5. 25 6. 9 · 4 · 49 = 1,764

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Chapter 7—EXPONENTS AND RADICALS

SPECIAL EXPONENTS ONE AS AN EXPONENT

Thus far in this discussion of exponents, the The number 1 arises as an exponent some- emphasis has been on exponents which are posi- times as a result of division. In the example tive integers. There are two types of exponents 5³ 5² we subtract the exponents to get which are not positive integers, and two which are treated as special cases even though they 5³⁻² = 5¹ may be considered as positive integers.

ZERO AS AN EXPONENT This problem may be worked another way as follows: Zero occurs as an exponent in the answer to a problem such as 4³ ÷ 4³. The law of expo- 5³ 5 · 5 · 5 nents for division states that the exponents are 5² = ―――――――― = 5 5 · 5 to be subtracted. This is illustrated as follows: Therefore, 4³ ―― = 4⁽³⁻³⁾ = 4⁰ 4³ 5¹ = 5

Another way of expressing the result of We conclude that any number raised to the dividing 4³ by 4³ is to use the fundamental first power is the number itself. The exponent axiom which states that any number divided by 1 usually is not written but is understood to itself is 1. In order for the law of exponents to exist. hold true in all cases, this must also be true when any number raised to a power is divided NEGATIVE EXPONENTS by itself. Thus, 4³/4³ must equal 1. Since 4³/4³ has been shown to be equal to If the law of exponents for division is ex- 4⁰ and 1, we are forced to the conclusion tended to include cases where the exponent of that 4⁰ = 1. the denominator is larger, negative exponents By the same reasoning, arise. Thus,

5 ― = 5¹⁻¹ = 5⁰ 3² 5 ―― = 3²⁻⁵ = 3⁻³ 3⁵ Also, Another way of expressing this problem is as 5 follows: ― = 1 5 3² 5 · 5 ―― = ――――――――――――――― = ――― Therefore, 3⁵ 3 · 3 · 3 · 3 · 3 3³

5⁰ = 1 Therefore,

Thus we see that any number divided by itself results in a 0 exponent and has a value of 1. 3⁻³ = ――― By definition then, any number (other than zero) 1 raised to the zero power equals 1. This is fur- ――― 3³ ther illustrated in the following examples:

3⁰ = 1 We conclude that a number N with a negative 400⁰ = 1 exponent is equivalent to a fraction having the 0.02⁰ = 1 following form: Its numerator is 1; its denomi- (⅙)⁰ = 1 nator is N with a positive exponent whose abso- (√3)⁰ = 1 lute value is the same as the absolute value of the original exponent. In symbols, this rule may be stated as follows:

N⁻ᵃ = ――― 1 ――― Nᵃ

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MATHEMATICS, VOLUME 1

Also, Another way of expressing this would be

1 4¹/² × 4¹/² = (4¹/²)² ――― = N ᵃ N⁻ᵃ = 4⁽¹/² ˣ ²⁾

The following examples further illustrate = 4¹ = 4 the rule: Observe that the number 4¹/² , when squared 5⁻¹ = ― 1 in the foregoing example, produced the number 5 4 as an answer. Recalling that a square root of a number N is a number x such that x² = N, we 6⁻² = ――― conclude that 4¹/² is equivalent to √4. Thus 1 we have a definition, as follows: A fractional ―― exponent of the form 1/r indicates a root, the 6² index of which is r. This is further illustrated in the following examples: 4⁻¹² = ――― 1 2¹/² = √2 ―――― 4¹² 4¹/³ = ³√4

―――³/⁻² = 3² 6²/³ = (6¹/³)² = (³√6)² 3 Also, Notice that the sign of an exponent may be changed by merely moving the expression which 6²/³ = (6²)¹/³ = ³√36 contains the exponent to the other position in the fraction. The sign of the exponent is changed Notice that in an expression such as 8²/³ we as this move is made. For example, can either find the cube root of 8 first or square 8 first, as shown by the following example: ―――― = 1 ÷ ――― 1 1 10⁻² 10² (8¹/³)² = 2² = 4 and (8²)¹/³ = ³√64 = 4

10² All the numbers in the evaluation of 8²/³ = 1 × ――― remain small if the cube root is found before 1 raising the number to the second power. This order of operation is particularly desirable in Therefore, evaluating a number like 64⁵/⁶. If 64 were first raised to the fifth power, a large number would ―――― = ―――― result. It would require a great deal of unnec- 1 10² essary effort to find the sixth root of 64⁵. The 10⁻² 1 result is obtained easily, if we write

By using the foregoing relationship, a prob- 64⁵/⁶ = (64¹/⁶)⁵ = 2⁵ = 32 lem such as 3 ÷ 5⁻⁴ may be simplified as fol- lows: If an improper fraction occurs in an expo- ――― = 3 × 5⁻⁽⁻⁴⁾ nent, such as 7/3 in the expression 2⁷/³, it is 3 5⁻⁴ customary to keep the fraction in that form rather than express it as a mixed number. In 5⁴ fraction form an exponent shows immediately = 3 × ――― what power is intended and what root is in- 1 tended. However, 2⁷/³ can be expressed in another form and simplified by changing the = 3 × 5⁴ improper fraction to a mixed number and writ- ing the fractional part in the radical form as FRACTIONAL EXPONENTS follows:

Fractional exponents obey the same laws as 2⁷/³ = 2²⁺¹/³ = 2² · 2¹/³ = 4 ³√2 do integral exponents. For example,

4¹/² × 4¹/² = 4⁽¹/² ⁺ ¹/²⁾

= 4²/²

= 4¹ = 4

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Chapter 7—EXPONENTS AND RADICALS

The law of exponents for multiplication may be combined with the rule for fractional expo- ents to solve problems of the following type:

PROBLEM: Evaluate the expression 4^2.5.

SOLUTION: 4^2.5 = 4^2 x 4^0.5 = 16 x 4^1/2 = 16 x 2 = 32

Practice problems:

1. Perform the indicated division: 2/(2^1/3)

2. Find the product: 7^2/5 x 7^1/10 x 7^3/10

3. Rewrite with a positive exponent and sim- plify: 9^-1/2

4. Evaluate 100^3/2

5. Evaluate (8^9)^5

Answers:

1. 2^3/3 ÷ 2^1/3 = ^3√4 2. 7^8/10 3. 1/(9^1/2) = 1/3 4. 1,000 5. 1

SCIENTIFIC NOTATION AND POWERS OF 10

Technicians, engineers, and others engaged in scientific work are often required to solve problems involving very large and very small numbers. Problems such as

22,684 x 0.00189 0.0713 x 83 x 7

are not uncommon. Solving such problems by the rules of ordinary arithmetic is laborious and time consuming. Moreover, the tedious arithmetic process lends itself to operational errors. Also there is difficulty in locating the decimal point in the result. These difficulties can be greatly reduced by a knowledge of the powers of 10 and their use.

The laws of exponents form the basis for calculation using powers of 10. The following list includes several decimals and whole num- bers expressed as powers of 10:

10,000 = 10^4 1,000 = 10^3 100 = 10^2 10 = 10^1 1 = 10^0 0.1 = 10^-1 0.01 = 10^-2 0.001 = 10^-3 0.0001 = 10^-4

The concept of scientific notation may be demonstrated as follows:

60,000 = 6.0000 x 10,000 = 6 x 10^4

538 = 5.38 x 100 = 5.38 x 10^2

Notice that the final expression in each of the foregoing examples involves a number be- tween 1 and 10, multiplied by a power of 10. Furthermore, in each case the exponent of the power of 10 is a number equal to the number of digits between the new position of the decimal point and the original position (iterated) of the decimal point.

We apply this reasoning to write any number in scientific notation; that is, as a number be- tween 1 and 10 multiplied by the appropriate power of 10. The appropriate power of 10 is found by the following mechanical steps:

1. Shift the decimal point to standard posi- tion, which is the position immediately to the right of the first nonzero digit.

2. Count the number of digits between the new position of the decimal point and its origi- nal position. This number indicates the value of the exponent for the power of 10.

3. If the decimal point is shifted to the left, the sign of the exponent of 10 is positive; if the decimal point is shifted to the right, the sign of the exponent is negative.

The validity of this rule, for those cases in which the exponent of 10 is negative, is demon- strated as follows:

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MATHEMATICS, VOLUME 1

0.00657 = 6.57 x 0.001 = 6.57 x 10^-3

0.348 = 3.48 x 0.1 = 3.48 x 10^-1

Further examples of the use of scientific notation are given as follows:

543,000,000 = 5.43 x 10^8

186 = 1.86 x 10^2

243.01 = 2.4301 x 10^2

0.0000007 = 7 x 10^-7

0.00023 = 2.3 x 10^-4

Multiplication Using Powers of 10

From the law of exponents for multiplication we recall that to multiply two or more powers to the same base we add their exponents. Thus, 10^6 x 10^2 = 10^8

We see that multiplying powers of 10 together is an application of the general rule. This is demonstrated in the following examples:

1. 10,000 x 100 = 10^4 x 10^2 = 10^4 + 2 = 10^6

2. 0.0000001 x 0.001 = 10^-7 x 10^-3 = 10^-7 + (-3) = 10^-10

3. 10,000 x 0.001 = 10^4 x 10^-3 = 10^4 - 3 = 10

4. 23,000 x 500 = ? 23,000 = 2.3 x 10^4 500 = 5 x 10^2 Therefore, 23,000 x 500 = 2.3 x 10^4 x 5 x 10^2 = 2.3 x 5 x 10^4 x 10^2 = 11.5 x 10^6 = 1.15 x 10^7

5. 62,000 x 0.0003 x 4,600 = ? 62,000 = 6.2 x 10^4 0.0003 = 3 x 10^-4 4,600 = 4.6 x 10^3

Therefore, 62,000 x 0.0003 x 4,600 = 6.2 x 3 x 4.6 x 10^4 x 10^-4 x 10^3 = 85.56 x 10^3 = 8.556 x 10^4

Practice problems. Multiply, using powers of 10. For the purposes of this exercise, treat all numbers as exact numbers:

1. 10,000 x 0.001 x 100

2. 0.000350 x 5,000,000 x 0.0004

3. 3,875 x 0.000032 x 3,000,000

4. 7,000 x 0.015 x 1.78

Answers:

1. 1.0 x 10^3

2. 7.0 x 10^-1

3. 3.72 x 10^5

4. 1.869 x 10^2

Division Using Powers of 10

The rule of exponents for division states that, for powers of the same base, the exponent of the denominator is subtracted from the ex- ponent of the numerator. Thus,

10^7/10^3 = 10^7 - 3 = 10^4

It should be remembered that powers may be transferred from numerator to denominator or from denominator to numerator by simply changing the sign of the exponent. The follow- ing examples illustrate the use of this rule for powers of 10:

1. 72,000/0.0012 = 7.2 x 10^4 / 1.2 x 10^-3 = 7.2/1.2 x 10^4 x 10^3 = 6 x 10^7

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Chapter 7—EXPONENTS AND RADICALS

2. 44 x 10⁻⁴ 44 -4 5 ———————— = —— x 10 x 10 11 x 10⁻⁵ 11 = 4 x 10

Combined Multiplication and Division

Using the rules already shown, multiplication and division involving powers of 10 may be combined. The usual method of solving such problems is to multiply and divide alternately until the problem is completed. For example,

36,000 x 1.1 x 0.06 —————————————————— 0.012 x 2,200

Rewriting this problem in scientific notation, we have

3.6 x 10⁴ x 1.1 x 6 x 10⁻² 3.6 x 1.1 x 6 ———————————————————————— = ——————————— x 10 1.2 x 10⁻² x 2.2 x 10³ 1.2 x 2.2

= 9 x 10 = 90

Notice that the elimination of 0's, wherever possible, simplifies the computation and makes it an easy matter to place the decimal point. SIGNIFICANT DIGITS.—One of the most important advantages of scientific notation is the fact that it simplifies the task of determining the number of significant digits in a number. For example, the fact that the number 0.00045 has two significant digits is sometimes obscured by the presence of the 0's. The confusion can be avoided by writing the number in scientific notation, as follows:

0.00045 = 4.5 x 10⁻⁴

Practice problems. Express the numbers in the following problems in scientific notation and round off before performing the calculation. In each problem, round off calculation numbers to one more digit than the number of significant digits in the least accurate number; round the answer to the number of significant digits in the least accurate number:

0.000063 x 50.4 x 0.007213 1. ———————————————————————————— 780 x 0.682 x 0.018

0.015 x 216 x 1.78 2. ———————————————————————— 72 x 0.0624 x 0.0353

0.000079 x 0.00036 3. ———————————————————————— 29 x 10⁻⁸

Answers:

1. 2.4 x 10⁻⁶ 2. 3.6 x 10 3. 9.8 x 10⁻²

Other Applications

The applications of powers of 10 may be broadened to include problems involving reciprocals and powers of products. RECIPROCALS.—The following example illustrates the use of powers of 10 in the formation of a reciprocal:

1 ————————————————— 250,000 x 300 x 0.02

1 = —————————————————————————— 2.5 x 10⁵ x 3 x 10² x 2 x 10⁻²

10⁻⁵ = ———————— 2.5 x 3 x 2

10⁻⁵ = —— 15

Rather than write the numerator as 0.00001, write it as the product of two factors, one of which may be easily divided, as follows:

10⁻⁵ 10² x 10⁻⁷ —— = —————————— 15 15

100 -7 = —— x 10 15

= 6.67 x 10⁻⁷ = 0.000000667

POWER OF A PRODUCT.—The following example illustrates the use of powers of 10 in finding the power of a product:

(80,000 x 2 x 10⁵)² = (8 x 10⁴ x 2 x 10⁵)² = 8² x 2² x (10⁴⁺⁵)² = 64 x 4 x 10¹⁸ = 256 x 10¹⁸ = 2.56 x 10²⁰

RADICALS

An expression such as √z, ³√5, or √a + b that exhibits a radical sign, is referred to as a

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MATHEMATICS, VOLUME 1

RADICAL. We have already worked with radicals in the form of fractional exponents, but it is also frequently necessary to work with them in the radical form. The word "radical" is denominate number from the Latin word "radix," which means "root." The word "radix" itself is more often used in modern mathematics to refer to the base of a number system, such as the base 2 in the binary system. However, the word "radical" is retained with its original meaning of "root." The radical symbol (√) appears to be a distortion of the initial letter "r" from the word "radix." With long usage, the r gradually lost its significance as a letter and became distorted into the symbol as we use it. The vinculum helps to specify exactly which of the letters and numbers following the radical sign actually belong to the radical expression. The number under a radical sign is the RADICAND. The index of the root (except in the case of a square root) appears in the trough of the radical sign. The index tells what root of the radicand is intended. For example, in ⁵√32, the radicand is 32 and the index of the root is 5. The fifth root of 32 is intended. In √50, the square root of 50 is intended. When the index is 2, it is not written, but is understood. If we can find one square root of a number we can always find two of them. Remember (3)² is 9 and (-3)² is also 9. (-4)² both equal 16 and (5)² and (-5)² both equal 25. Conversely, √9 is +3 or -3; √16 is +4 or -4, and √25 is +5 or -5. When we wish to show a number that may be either positive or negative, we may use the symbol ± which is read "plus or minus." Thus ± 3 means "plus or minus 3." Usually when a number is placed under the radical sign, only its positive root is desired and, unless otherwise specified, it is the only root that need be found.

COMBINING RADICALS

A number written in front of another number and intended as a multiplier is called a COEFFICIENT. The expression 5x means 5 times x; ay means a times y; and 7 √2 means 7 times √2. In these examples, 5 is the coefficient of x, a is the coefficient of y, and 7 is the coefficient of √2. Radicals having the same index and the same radicand are SIMILAR. Similar radicals may have different coefficients in front of the radical sign. For example, 3 √2, √2, and ⅕ √2

are similar radicals. When a coefficient is not written, it is understood to be 1. Thus, the coefficient of √2 is 1. The rule for adding radicals is the same as that stated for adding decimals: Add only units of the same kind. For example, we could add 2 √3 and 4 √3 because the "unit" in each of these numbers is the same (√3). The same reasoning, we could not add 2 √3 and 4 √5 because these are not similar radicals.

Addition and Subtraction

When addition or subtraction of similar radicals is indicated, the radicals are combined by adding or subtracting their coefficients and placing the result in front of the radical. Adding 3 √2 and 5 √2 is similar to adding 3 bolts and 5 bolts. The following examples illustrate the addition and subtraction of similar radical expressions:

1. 3 √2 + 5 √2 = 8 √2

2. 1/2 (³√3) + 1/3 (³√3) = 5/6 (³√3)

3. √5 - 6 √5 + 2 √5 = -3 √5

4. -5 ⁷√7 -2 ⁷√7 + 7 ⁷√7 = 0

Example 4 illustrates a case that is sometimes troublesome. The sum of the coefficients, -5, -2, and 7, is 0. Therefore, the coefficient of the answer would be 0, as follows:

0(³√7) = 0 x ³√7

Thus the final answer is 0, since 0 multiplied by any quantity is still 0. Practice problems. Perform the indicated operations:

1. 4 √3 - √3 + 5 √3

2. ½ √6 + √6

3. ³√5 - 6 ³√5

4. -2 √10 - 7 √10

Answers

1. 8 √3 3. -5 ³√5

2. ³⁄₂ √6 4. -9 √10

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Chapter 7—EXPONENTS AND RADICALS

Multiplication and Division If a radical is written immediately after another radical, multiplication is intended. Sometimes a dot is placed between the radicals, but not always. Thus, either √ · √11 or √ √11 means multiplication. When multiplication or division of radicals is indicated, several radicals having the same index can be combined into one radical, if desired. Radicals having the same index are said to be of the SAME ORDER. For example, √2 is a radical of the second order. The radicals √2 and √5 are of the same order. If radicals are of the same order, the radicands can be multiplied or divided and placed under one radical symbol. For example, √5 multiplied by √3 is the same as √5 · √3. Also, √6 divided by √3 is the same as √(6/3) = 3. If coefficients appear before the radicals, they must also be included in the multiplication or division. This is illustrated in the following examples:

1. 2 √2 · 3 √5 = 2 · √2 · 3 · √5 = 2 · 3 √2 · √5 = 2 · 3 √2·5 = 6 √10

2. (15 √6)/(3 √3) = (15/3) × √(6/3) = 5 × √2 = 5 √2

It is important to note that what we have said about multiplication and division does not apply to addition. A typical error is to treat the expression √9 + 4 as if it were equivalent to √9 + √4. These expressions cannot be equivalent, since 3 + 2 is not equivalent to √13. FACTORING RADICALS.—A radical can be split into two or more radicals of the same order if the radicand can be factored. This is illustrated in the following examples:

1. √20 = √4 · √5 = 2 √5 2. ∛54 = ∛27·2 = ∛27 · ∛2 = 3 ∛2 3. (√20)/√5 = (√4 · √5)/√5 = √4 = 2

SIMPLIFYING RADICALS Some radicals may be changed to an equivalent form that is easier to use. A radical is in its simplest form when no factor can be removed from the radical, when there is no fraction under the radical sign, and when the index of the root cannot be reduced. A factor can be removed from the radical if it occurs a number of times equal to the index of the root. The following examples illustrate this:

1. √28 = √2² · 7 = 2 √7 2. ∛54 = ∛3³ · 2 = 3(∛2) 3. ⁴√160 = ⁴√2⁵ · 5 = 2(⁴√5)

Removing a factor that occurs a number of times equal to the index of the root is equivalent to separating a radical into two radicals so that one radicand is a perfect power. The radical sign can be removed if the radicand is a perfect square, cube, fourth power, etc. The root taken becomes the coefficient of the remaining radical. In order to simplify radicals easily, it is convenient to know the squares of whole numbers up to about 25 and a few of the smaller powers of the numbers 2, 3, 4, 5, and 6. Table 7-1 shows some frequently used powers of numbers.

Table 7-1.—Powers of numbers.

1² = 1 14² = 196 2² = 4 15² = 225 3² = 9 16² = 256 4² = 16 17² = 289 5² = 25 18² = 324 6² = 36 19² = 361 7² = 49 20² = 400 8² = 64 21² = 441 9² = 81 22² = 484 10² = 100 23² = 529 11² = 121 24² = 576 12² = 144 25² = 625 13² = 169

(A)

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MATHEMATICS, VOLUME 1

Table 7-1.—Powers of numbers—Continued.

2¹ = 2 3¹ = 3 2² = 4 3² = 9 2³ = 8 3³ = 27 2⁴ = 16 3⁴ = 81 2⁵ = 32 3⁵ = 243 2⁶ = 64 (C) 2⁷ = 128 2⁸ = 256

(B)

4¹ = 4 5¹ = 5 4² = 16 5² = 25 4³ = 64 5³ = 125 4⁴ = 256 5⁴ = 625

(D) (E)

6¹ = 6 6² = 36 6³ = 216

(F)

Referring to table 7-1 (A), we see that the series of numbers

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

comprises all the perfect squares from 1 to 100 inclusive. If any one of these numbers appears under a square root symbol, the radical sign can be removed immediately. This is illustrated as follows:

√25 = 5 √81 = 9

A radicand such as 75, which has a perfect square (25) as a factor, can be simplified as follows:

√75 = √25 · 3 = √25 · √3 = 5 √3

This procedure is further illustrated in the following problems:

1. √8 = √4 · 2 = √4 · √2 = 2 √2

2. √72 = √36 · 2 = √36 · √2 = 6 √2

By reference to the perfect fourth powers in table 7-1, we may simplify a radical such as ⁴√405. Noting that 405 has the perfect fourth power 81 as a factor, we have the following: ⁴√405 = ⁴√81 · 5 = ⁴√81 · ⁴√5 = 3 (⁴√5)

As was shown with fractional exponents, taking a root is equivalent to dividing the exponent of a power by the index of the root. If a factor of the radicand has an exponent that is not a multiple of the index of the root, the factor may be separated so that one exponent is divisible by the index, as in

√37 = √3⁵· 3 = 3⁶/² · 3¹/² = 3³ · √3 = 27 √3

Consider also

√2³ · 3⁷ · 5 = √2² · 2 · 3⁶ · 3 · 5 = 2 · 3³(√2 · 3 · 5) = 54 √30

If the radicand is a large number, the perfect powers that are factors are not always obvious. In such a case the radicand can be separated into prime factors. For example,

√8,820 = √2² · 3² · 5 · 7² = 2 · 3 · 7 √5 = 42 √5

Practice problems. Simplify the radicals and reduce to lowest terms:

1. √3 · √15 3. 18(√30) √5 3(√10)

2. ∛81 4. √8,820 ∛27 √160

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Chapter 7—EXPONENTS AND RADICALS

Answers:

1. 3 2. ∛3

3. 6(∛3) 4. 7

RATIOONAL AND IRRATIONAL NUMBERS

Real and imaginary numbers make up the number system of algebra. Imaginary numbers are discussed in chapter 15 of this course. Real numbers are either rational or irrational. The word RATIONAL comes from the word "ratio." A number is rational if it can be expressed as the quotient, or ratio, of two whole numbers. Rational numbers include fractions like 2/7, whole numbers, and radicals if the radical sign is removable.

Any whole number is rational. Its denominator is 1. For instance, 8 equals 8/1, which is the quotient of two integers. A number like √16 is rational, since it can be expressed as the quotient of two integers in the form 4/1. The following are also examples of rational numbers:

1. √(25/9), which equals 5/3

2. -6, which equals -6/1

3. 5²/7, which equals 27/7

Any rational number can be expressed as the quotient of two integers in many ways. For example,

7 = 7/1 = 14/2 = 21/3 ...

An IRRATIONAL number is a real number that cannot be expressed as the ratio of two integers. The numbers √3, 5√2, √7, 3/5√20, and -2/√5 are examples of irrational numbers.

Rationalizing Denominators

Expressions such as 7/√2 and √3/(5√3) have irrational numbers in the denominator. If the

denominators are changed immediately to decimals, as in

7/√2 = 7/1.4142

the process of evaluating a fraction becomes an exercise in long division. Such a fraction can be evaluated quickly by first changing the denominator to a rational number. Converting a fraction with an irrational number in the denominator to an equivalent fraction with a rational number in the denominator is called RATIONALIZING THE DENOMINATOR.

Multiplying a fraction by 1 leaves the value of the fraction unchanged. Since any number divided by itself equals 1, it follows, for example, that

√2/√2 = 1

If the numerator and denominator of 7/√2 are each multiplied by √2, another fraction having the same value is obtained. The result is

7/√2 · √2/√2 = 7√2/2

The denominator of the new equivalent fraction is 2, which is rational. The decimal value of the fraction is

7√2/2 = 7(1.4142)/2 = 7(0.7071) = 4.9497

To rationalize the denominator in √3/(5√3) we multiply the numerator and denominator by √3. We get

√3/(5√3) · √3/√3 = √3·√3/(5(3)) = 3/(15) = 1/5√6

Practice problems. Rationalize the denominator in each of the following:

1. 6/√2 3. 2/√6

2. √3/√3 4. 6/√y

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MATHEMATICS, VOLUME 1

Answers:

1. 3√2 2. √15/3

3. √6/3 4. 6√y/y

EVALUATING RADICALS

Any radical expression has a decimal equivalent which may be exact if the radicand is a rational number. If the radicand is not rational, the root may be expressed as a decimal approximation, but it can never be exact. A procedure similar to long division may be used for calculating square root and cube root, and higher roots may be calculated by means of methods based on logarithms and higher mathematics. Tables of powers and roots have been calculated for use in those scientific fields in which it is frequently necessary to work with roots.

SQUARE ROOT PROCESS

The arithmetic process for calculation of square root is outlined in the following paragraphs:

1. Begin at the decimal point and mark the number off into groups of two digits each, moving both to the right and to the left from the decimal point. This may leave an odd digit at the right-hand or left-hand end of the number, or both. For example, suppose that the number whose square root we seek is 9025. The number marked off as specified would be as follows:

√90'25.

2. Find the greatest number whose square is contained in the left-hand group (90). This number is 9, since the square of 9 is 81. Write 9 above the first group. Square this number (9), place its square below the left-hand group, and subtract, as follows:

9 √90'25. 81 9 25

Bring down the next group (25) and place it beside the 9, as shown. This is the new dividend (925).

3. Multiply the first digit in the root (9) by 20, obtaining 180 as a trial divisor. This trial

divisor is contained in the new dividend (925) five times; thus the second digit of the root appears to be 5. However, this number must be added to the trial divisor to obtain a "true divisor." If the true divisor is then too large to use with the second quotient digit, this digit must be reduced by 1. The procedure for step 3 is illustrated as follows:

9 5. √90'25. 81 180 9 25 185 9 25 0 00

The number 180, resulting from the multiplication of 9 by 20, is written as a trial divisor beside the new dividend (925), as shown. The quotient digit (5) is then recorded and the trial divisor is adjusted, becoming 185. The trial quotient (180) is crossed out.

4. The true divisor (185) is multiplied by the second digit (5) and the product is placed below the new dividend (925). This step is shown in the illustration for step 3. When the product in step 4 is subtracted from the new dividend, the difference is 0. Thus, in the example, the root is exact.

5. In some problems, the difference is not 0 after all of the digits of the original number have been used to form new dividends. Such problems may be carried further by adding 0's on the right-hand end of the original number, just as in normal long division. However, in the square root process the 0's must be added and used in groups of 2.

Practice problems. Find the square root of each of the following numbers:

1. 9.61 2. 123.21 3. 0.0025

Answers:

1. 3.1 2. 11.1 3. 0.05

TABLES OF ROOTS

The decimal values of square roots and cube roots of numbers with as many as 3 or 4 digits can be found from tables. The table in appendix I of this course gives the square roots and cube roots of numbers from 1 to 100. Most of the values given in such tables are approximate numbers which have been rounded off.

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Chapter 7—EXPONENTS AND RADICALS

For example, the fourth column in appendix I shows that √72 = 8.4853, to 4 decimal places. By shifting the decimal point we can obtain other square roots. A shift of two places in the decimal point in the radicand corresponds to a shift of one place in the same direction in the square root.

The following examples show the effect, as reflected in the square root, of shifting the location of the decimal point in the number whose square root we seek:

√72 = 8.4853 √0.72 = 0.84853 √0.0072 = 0.084853 √7,200 = 84.853

Cube Root

The fifth column in appendix I shows that the cube root of 72 is 4.1602. By shifting the deci- mal point we immediately have the cube roots of certain other numbers involving the same digits. A shift of three places in the decimal point in the radicand corresponds to a shift of one place in the same direction in the cube root.

Compare the following examples:

∛72 = 4.1602 ∛0.072 = 0.41602 ∛72,000 = 41.602

Many irrational numbers in their simplified forms involve √2 and √3. Since these radicals occur often, it is convenient to remember their decimal equivalents as follows:

√2 = 1.4142 and √3 = 1.7321

Thus any irrational numbers that do not contain any radicals other than √2 or √3 can be con- verted to decimal forms quickly without re- ferring to tables.

For example consider

√72 = 6 √2 = 6(1.4142) = 8.485 √27 = 3 √3 = 3(1.7321) = 5.196

Keep in mind that the decimal equivalents of √2 and √3 as used in the foregoing examples are not exact numbers and the results obtained with them are approximate in the fourth deci- mal place.

79

CHAPTER 8 LOGARITHMS AND THE SLIDE RULE

Logarithms represent a specialized use of exponents. By means of logarithms, computa- tion with large masses of data can be greatly simplified. For example, when logarithms are used, the process of multiplication is replaced by simple addition and division is replaced by subtraction. Raising to a power by means of logarithms is done in a single multiplication, and extracting a root reduces to simple division.

DEFINITIONS

In the expression 2³ = 8, the number 2 is the base (not to be confused with the base of the number system), and 3 is the exponent which must be used with the base to produce the num- ber 8. The exponent 3 is the logarithm of 8 when the base is 2. This relationship is usually stated as follows: The logarithm of 8 to the base 2 is 3. In general, the logarithm of a number N with respect to a given base is the exponent which must be used with the base to produce N. Table 8-1 illustrates this.

Table 8-1.—Logarithms with various bases.

Exponential form | Logarithmic form 2³ = 8 | log₂ 8 = 3 4² = 16 | log₄ 16 = 2 5⁰ = 1 | log₅ 1 = 0 27^(2/3) = 9 | log₂₇ 9 = 2/3

Table 8-1 shows that the logarithmic rela- tionship may be expressed equally well in either of two forms; these are the exponential form and the logarithmic form. Observe, in table 8-1, that the base of a logarithmic expression is indicated by placing a subscript just below and to the right of the abbreviation "log." Ob- serve also that the word "logarithm" is abbrevi- ated without using a period.

The equivalency of the logarithmic and ex- ponential forms may be used to relate the fun- damental definition of logarithms in its most useful form, as follows:

b^x = N implies that log_b N = x

In words, this definition is stated as follows: If the base b raised to the x power equals N, then x is the logarithm of the number N to the base b.

One of the many uses of logarithms may be shown by an example in which the base is 2. Table 8-2 shows the powers of 2 from 0 through 20. Suppose that we wish to use logarithms to multiply the numbers 512 and 256, as follows:

From table 8-2, 512 = 2⁹ 256 = 2⁸

Then 512 × 256 = 2⁹ × 2⁸ = 2¹⁷

and from the table again 2¹⁷ = 131072

It is seen that the problem of multiplication is reduced to the simple addition of the expo- nents 9 and 8 and finding the corresponding power in the table.

Table 8-2 (A) shows the base 2 in the expo- ential form with its corresponding powers. The actual computation in logarithmic work does not require that we record the exponential form. All that is required is that we add the appropriate exponents and have available a table in which we can look up the number cor- responding to the new exponent after adding. Therefore, table 8-2 (B) is adequate for our purpose. Solving the foregoing example by this table, we have the following:

log₂ 512 = 9 log₂ 256 = 8

log₂ of the product = 17

Therefore, the number we seek is the one in the table whose logarithm is 17. This number is 131,072. In this example, we found the expo- nents directly, added them since this was a

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Chapter 8--LOGARITHMS AND THE SLIDE RULE

Table 8-2.--Exponential and logarithmic tables for the base 2.

(A) Powers of 2 from | (B) Logarithms for the 0 through 20 | base 2 and corresponding powers

| Log | Number 2⁰ = 1 | 0 | 1 2¹ = 2 | 1 | 2 2² = 4 | 2 | 4 2³ = 8 | 3 | 8 2⁴ = 16 | 4 | 16 2⁵ = 32 | 5 | 32 2⁶ = 64 | 6 | 64 2⁷ = 128 | 7 | 128 2⁸ = 256 | 8 | 256 2⁹ = 512 | 9 | 512 2¹⁰ = 1024 | 10 | 1024 2¹¹ = 2048 | 11 | 2048 2¹² = 4096 | 12 | 4096 2¹³ = 8192 | 13 | 8192 2¹⁴ = 16384 | 14 | 16384 2¹⁵ = 32768 | 15 | 32768 2¹⁶ = 65536 | 16 | 65536 2¹⁷ = 131072 | 17 | 131072 2¹⁸ = 262144 | 18 | 262144 2¹⁹ = 524288 | 19 | 524288 2²⁰ = 1048576 | 20 | 1048576

NATURAL AND COMMON LOGARITHMS

Many natural phenomena, such as rates of growth and decay, are most easily described in terms of logarithmic or exponential formulas. Furthermore, the geometric patterns in which certain seeds grow (for example, sunflower seeds) is a logarithmic spiral. These facts explain the name "natural logarithms." Natural logarithms use the base e, which is an irrational number approximately equal to 2.71828. This system is sometimes called the Napierian system of logarithms, in honor of John Napier, who is credited with the invention of logarithms.

To distinguish natural logarithms from other logarithmic systems the abbreviation, ln, is sometimes used. When ln appears, the base is understood to be e and need not be shown. For example, either log_e 45 or ln 45 signifies the natural logarithm of 45.

COMMON LOGARITHMS

As has been shown in preceding paragraphs, any number may be used as a base for a system of logarithms. The selection of a base is a matter of convenience. Briggs in 1617 found that base 10 possessed many advantages not obtainable in ordinary calculations with other bases. The selection of 10 as a base proved so satisfactory that today it is used almost exclusively for ordinary calculations. Logarithms with 10 as a base are therefore called COMMON LOGARITHMS.

When 10 is used as a base, it is not necessary to indicate it in writing logarithms. For example,

log 100 = 2

is understood to mean the same as

log₁₀ 100 = 2

If the base is other than 10, it must be specified by the use of a subscript to the right and below the abbreviation "log". As noted in the foregoing discussion of natural logarithms, the use of the distinctive abbreviation "ln" eliminates the need for a subscript when the base is e.

It is relatively easy to convert common logarithms to natural logarithms or vice versa, if necessary. It should be noted further that each system has its peculiar advantages, but for most everyday work, the common system is

multiplication problem, and located the corresponding power. This avoided the unnecessary step of writing the base 2 each time.

Practice problems. Use the logarithms in table 8-2 to perform the following multiplication:

1. 64 x 128 3. 128 x 4,096 2. 1,024 x 256 4. 512 x 2,048

Answers: 1. 8,192 3. 524,288 2. 262,144 4. 1,048,576

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MATHEMATICS, VOLUME 1

more often used. A simple relation connects the two systems. If the common logarithm of a number can be found, multiplying by 2.3026 gives the natural logarithm of the number. For example,

log 1.60 = 0.2041 ln 1.60 = 2.3026 x 0.2041 = 0.4700

Thus the natural logarithm of 1.60 is 0.4700, correct to four significant digits.

Conversely, multiplying the natural logarithm by 0.4343 gives the common logarithm of a number. As might be expected, the conversion factor 0.4343 is the reciprocal of 2.3026. This is shown as follows:

1/2.3026 = 0.4343

Positive Integral Logarithms

The derivation of positive whole logarithms is readily apparent. For example, we see in table 8-3 (B) that the logarithm of 10 is 1. The number 1 is simply the exponent of the base 10 which yields 10. This is shown in table 8-3 (A), opposite the logarithmic equation. Similarly,

10⁰ = 1 ........... log 1 = 0 10² = 100 ......... log 100 = 2 10³ = 1,000 ......log 1,000 = 3 10⁴ = 10,000 ..... log 10,000 = 4

Table 8-3.--Exponential and corresponding logarithmic notations using base 10.

A. B.

10⁻⁴ = 1/10⁴ = 0.0001 log 0.0001 = -4

10⁻³ = 1/10³ = 0.001 log 0.001 = -3

10⁻² = 1/10² = 0.01 log 0.01 = -2

10⁻¹ = 1/10 = 0.1 log 0.1 = -1

10⁻¹/² = 1/√10 = √10/10 = 0.31623 log 0.31623 = -0.5 = 0.5 -1

10⁰ = 1 log 1 = 0

10¹/² = √10 = 3.1623 log 3.1623 = 0.5

10¹ = 10 log 10 = 1

10³/² = 10·√10 = 31.623 log 31.623 = 1.5

10² = 100 log 100 = 2

10⁵/² = 10²(√10) = 316.23 log 316.23 = 2.5

10³ = 1,000 log 1,000 = 3

10⁷/² = 10³(√10) = 3162.3 log 3162.3 = 3.5

10⁴ = 10,000 log 10,000 = 4

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Chapter 8—LOGARITHMS AND THE SLIDE RULE

Positive Fractional Logarithms

Referring to table 8-3, notice that the logarithm of 1 is 0 and the logarithm of 10 is 1. Therefore, the logarithm of a number between 1 and 10 is between 0 and 1. An easy way to verify this is to consider some numbers between 1 and 10 which are powers of 10; the exponent in each case will then be the logarithm we seek. Of course, the only powers of 10 which produce numbers between 1 and 10 are fractional powers.

EXAMPLE: 10^1/2 = 3.1623 (approximately) 10^0.5 = 3.1623

Therefore, log 3.1623 = 0.5

Other examples are shown in the table for 10^3/2, 10^5/2, and 10^7/2. Notice that the number that represents 10^3/2, 31.623, logically lies between the numbers 10^1 and 10^2—that is, 10 and 100. Notice also that 10^5/2 appears between 10^2 and 10^3, and 10^7/2 lies between 10^3 and 10^4.

Negative Logarithms

Table 8-3 shows that negative powers of 10 may be fitted into the system of logarithms. We recall that 10^-1 means 1/10, or the decimal fraction, 0.1. What is the logarithm of 0.1?

SOLUTION: 10^-1 = 0.1; log 0.1 = -1

Likewise 10^-2 = 0.01; log 0.01 = -2

Negative Fractional Logarithms

Notice in table 8-3 that negative fractional exponents present no new problem in logarithmic notation. For example, 10^-1/2 means 1/√10

1/√10 = √10/10 = 0.31623

What is the logarithm of 0.31623?

SOLUTION: 10^-1/2 = 0.31623; log 0.31623 = -1/2 = -0.5

Table 8-3 shows logarithms for numbers ranging from 0.0001 to 10,000. Notice that there are only 8 integral logarithms in the entire range. Excluding zero logarithms, the logarithms for all other numbers in the range are fractional or contain a fractional part. By the year 1628, logarithms for all integers from 1 to 100,000 had been computed. Practically all of these logarithms contain a fractional part. It should be remembered that finding the logarithm of a number is nothing more than expressing the number as a power of 10. Table 8-4 shows the numbers 1 through 10 expressed as powers of 10. Most of the exponents which comprise logarithms are found by methods beyond the scope of this text. However, it is not necessary to know the process used to obtain logarithms in order to make use of them.

Table 8-4.—The numbers 1 through 10 expressed as powers of 10.

1 = 10^0 6 = 10^0.77815 2 = 10^0.30103 7 = 10^0.84510 3 = 10^0.47712 8 = 10^0.90309 4 = 10^0.60206 9 = 10^0.95424 5 = 10^0.69897 10 = 10^1

COMPONENTS OF LOGARITHMS

The fractional part of a logarithm is usually written as a decimal. The whole number part of a logarithm and the decimal part have been given separate names because each plays a special part in relation to the number which the logarithm represents. The whole number part of a logarithm is called the CHARACTERISTIC. This part of the logarithm shows the position of the decimal point in the associated number.

The decimal part of a logarithm is called the MANTISSA.

For a particular sequence of digits making up a number, the mantissa of a common logarithm is always the same regardless of the position of the decimal point in that number.

For example, log 5270 = 3.72181; the mantissa is 0.72181 and the characteristic is 3.

CHARACTERISTIC

The characteristic of a common logarithm shows the position of the decimal point in the

MATHEMATICS, VOLUME 1

associated number. The characteristic for a given number may be determined by inspection. It will be remembered that a common logarithm is simply an exponent of the base 10. It is the power of 10 when a number is written in scientific notation.

When we write log 360 = 2.55630, we understand this to mean 10^2.55630 = 360. We know that the number is 360 and not 36 or 3,600 because the characteristic is 2. We know 10^1 is 10, 10^2 is 100, and 10^3 is 1,000. Therefore, the number whose value is 10^2.55630 must lie between 100 and 1,000 and of course any number in that range has 3 digits. Notice how the position of the decimal point changes with the value of the characteristic in the following examples:

log 36,000 = 4.55630 log 3,600 = 3.55630 log 360 = 2.55630 log 36 = 1.55630 log 3.6 = 0.55630

Note that it is only the characteristic that changes when the decimal point is moved. An advantage of using the base 10 is thus revealed: If the characteristic is known, the decimal point may easily be placed. If the number is known, the characteristic may be determined by inspection: that is, by observing the location of the decimal point.

Although an understanding of the relation of the characteristic to the powers of 10 is necessary for thorough comprehension of logarithms, the characteristic may be determined mechanically by application of the following rules:

1. For a number greater than 1, the characteristic is positive and is one less than the number of digits to the left of the decimal point in the number.

2. For a positive number less than 1, the characteristic is negative and has an absolute value one more than the number of zeros between the decimal point and the first nonzero digit of the number.

Table 8-5 contains examples of each type of characteristic.

Practice problems. In problems 1 through 4, write the characteristic of the logarithm for each number. In 5 through 8, place the decimal point.

Table 8-5.—Positive and negative characteristics.

| Number | Power of 10 Between: | Digits in number to the left of decimal point | Characteristic | |--------|---------------------|----------------------------------------------|----------------| | 134 | 10² and 10³ | 3 | 2 | | 13.4 | 10¹ and 10² | 2 | 1 | | 1.34 | 10⁰ and 10¹ | 1 | 0 | | | | Zeros between decimal point and first nonzero digit | | | 0.134 | 10⁻¹ and 10⁰ | 0 | -1 | | 0.0134 | 10⁻² and 10⁻¹ | 1 | -2 | | 0.00134 | 10⁻³ and 10⁻² | 2 | -3 |

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Chapter 8—LOGARITHMS AND THE SLIDE RULE

point in each number as indicated by the char- acteristic (c) given for each.

1. 4,321 2. 1,23 3. 0.05 4. 12 5. 123; c = 4 6. 8,210; c = 0 7. 8; c = -1 8. 321; c = -2

Answers:

1. 3 2. 0 3. -2 4. 1 5. 12,300 6. 8,210 7. 0.8 8. 0.0321

Negative Characteristics

When a characteristic is negative, such as -2, we do not carry out the subtraction, since this would involve a negative mantissa. There are several ways of indicating a negative char- acteristic. Mantissas as presented in the table in the appendix are always positive and the sign of the characteristic is indicated separately. For example, where log 0.023 = 2.36173, the bar over the 2 indicates that only the charac- teristic is negative—that is, the logarithm is -2 + 0.36173.

Another way to show the negative character- istic is to place it after the mantissa. In this case we write 0.36173-2.

A third method, which is used where possi- ble throughout this chapter, is to add a certain quantity to the characteristic and to subtract the same quantity to the right of the mantissa. In the case of the example, we may write:

2.36173 10 -10 8.36173-10

In this way the value of the logarithm remains the same but we now have a positive character- istic as well as a positive mantissa.

MANTISSA

The mantissa is the decimal part of a loga- rithm. Tables of logarithms usually contain only mantissas since the characteristic can be readily determined as explained previously. Table 8-6 shows the characteristic, mantissa, and logarithm for several positions of the deci- mal point using the sequence of digits 4, 5, 6. It will be noted that the mantissa remains the same for that particular sequence of digits, re- gardless of the position of the decimal point.

Table 8-6.—Effect of changes in the location of the decimal point.

Number | Charac- | Mantissa | Logarithm | teristic| | 45,600 | 4 | 0.6590 | 4.6590 4,560 | 3 | 0.6590 | 3.6590 456 | 2 | 0.6590 | 2.6590 45.6 | 1 | 0.6590 | 1.6590 4.56 | 0 | 0.6590 | 0.6590 0.456 | -1 | 0.6590 | 0.6590-1 0.0456 | -2 | 0.6590 | 0.6590-2 0.00456 | -3 | 0.6590 | 0.6590-3

Appendix I of this training course is a table which includes the logarithms of numbers from 1 to 100. For our present purpose in using this table, we are concerned only with the first and sixth columns.

The first column contains the number and the sixth column contains the logarithm. For example, if it is desired to find the logarithm of 45, we would find the number 45 in the first column, look horizontally across the page to column 6 and read the logarithm, 1.65321. A glance down the logarithm column will reveal that the logarithms increase in value as the numbers increase in value.

It must be noted in this particular table that both the mantissa and the characteristic are given for the number in the first column. This is simply an additional aid, since the charac- teristic can easily be determined by inspection. Suppose that we wish to use the table of Appendix I to find the logarithm of a number not shown in the "number" column. By recall- ing that the mantissa does not change when the decimal point moves, we may be able to deter- mine the desired logarithm. For example, the number 450 does not appear in the number col- umn of the table. However, the number 45 has the same mantissa as 450; the only difference between the two logs is in their characteristics. Thus the logarithm of 450 is 2.65321.

Practice problems. Find the logarithms of the following numbers:

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MATHEMATICS, VOLUME 1

1. 64 2. 98 3. 6400 4. 9.8

Answers:

1. 1.80618 2. 1.99123 3. 3.80618 4. 0.99123

THE SLIDE RULE

In 1620, not long after the invention of loga- rithms, Edmund Gunter showed how logarithmic calculations could be carried out mechanically. This is done by laying off lengths on a rule, representing the logarithms of numbers, and by combining these lengths in various ways. The idea was developed with the contributions of Mannheim in 1851 the slide rule came into being as we know it today.

The slide rule is a mechanical device by which we can carry out any arithmetic calcula- tion with the exception of addition and subtrac- tion. The most common operations with the slide rule are multiplication, division, finding the square or cube of a number, and finding the square root or cube root of a number. Also trigonometric operations are frequently per- formed. The advantage of the slide rule is that it can be used with relative ease to solve com- plicated problems. One limitation is that it will give results with a maximum of only three accurate significant digits. This is sufficient in most calculations, however, since most phys- ical constants are only correct to two or three significant digits. When greater accuracy is required, other methods must be used.

A simplified diagram of a slide rule is pic- tured in Figure 8-1. The sliding, central part of the rule is called the SLIDE. The movable glass or plastic runner with a hairline imprinted on it is called the INDICATOR. There is a C scale printed on the slide, and a D scale exactly the same as the C scale printed on the BODY or STOCK of the slide rule. The mark that is associated with the primary number 1 on any slide rule scale is called the INDEX. There is

an index at the extreme left and at the extreme right on both the C and D scales. There are other scales, each having a particular use. Some of these will be mentioned later.

SLIDE RULE THEORY

We have mentioned that the slide rule is based on logarithms. Recall that, to multiply two numbers, we simply add their logarithms. Previously we found these logarithms in tables, but if the logarithms are laid out on scales such as the C and D scale of the slide rule, we can add the lengths representing these loga- rithms. To make such a scale we could mark off mantissas ranging from 0 to 1 on a rule as in figure 8-2. We then find in the tables the logarithms for numbers ranging from 1 to 10 and write the number opposite its correspond- ing logarithm on the scale.

Figure 8-2.—Logarithms and corresponding numbers on a scale.

Table 8-7 lists the numbers 1 through 10 and their corresponding logarithms to three places. These numbers are written opposite their logarithms on the scale shown in figure 8-2. If we have two such scales, exactly alike, arranged so that one of them is free to slide along the other, we can perform the operation of multiplication, for example, by ADDING LENGTHS; that is, by adding logarithms. For example, if we wish to multiply 2 × 3, we find the logarithm of 2 on the stationary scale and move the sliding scale so that its index is over that mark. We then add the logarithm of 3 by finding that logarithm on the sliding scale and by reading below it, on the stationary scale, the logarithm that is the sum of the two.

Since we are not interested in the logarithms themselves, but rather in the numbers they represent, it is possible to remove the loga- rithmic notation on the scale in figure 8-2 and leave only the logarithmically spaced number scale. The C and D scales of the ordinary slide rule are made up in this manner. Figure 8-3 shows the multiplication of 2 × 3. Although the logarithm scales have been removed, the num- bers 2 and 3 in reality signify the logarithms of

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Chapter 8—LOGARITHMS AND THE SLIDE RULE

Table 8-7.—Numbers and their corresponding logarithms.

Number | Logarithm | Number | Logarithm 1 | 0.000 | 6 | 0.778 2 | 0.301 | 7 | 0.845 3 | 0.477 | 8 | 0.903 4 | 0.602 | 9 | 0.954 5 | 0.699 | 10 | 1.000

EXAMPLE: Use logs (positions on the slide rule) to multiply 20 times 30.

SOLUTION:

log 20 = 1.301 (2 on the slide rule)

log 30 = 1.477 (3 on the slide rule)

log of answer = 2.778 (6 on the slide rule)

Since the 2 in the log of the answer is merely the indicator of the position of the deci- mal point in the answer itself, we do not expect to find it on the slide rule scale. As in the foregoing example, we find the digit 6 opposite the multiplier 3. This time, however, the 6 represents 600, because the characteristic of the log represented by 6 in this problem is 2.

READING THE SCALES

Reading a slide rule is no more complicated than reading a yard stick or ruler. If the dif- ferences in its markings are understood. Between the two indices of the C or D scales (the large digit 1 at the extreme left and right of the scales) are divisions numbered 2, 3, 4, ,5, 6, 7, 8, and 9. Each length between two con- secutive divisions is divided into 10 sections, and each section is divided into spaces. (See fig. 8-4.)

DIVISION

SECTION SPACE

Figure 8-4.—Division, section, and space of a slide rule scale.

Notice that the division between 1 and 2 occupies about one-third of the length of the rule. This is sufficient space in which to write a number for each of the section marks. The sections in the remaining divisions are not numbered, because the space is more limited. Notice also that in the division between 1 and 2, the sections are each divided into 10 spaces. The sections of the divisions from 2 to 4 are

MATHEMATICS, VOLUME 1

subdivided into only 5 spaces, and those from 4 to the right index are subdivided into only 2 spaces. These subdivisions are so arranged because of the limits of space.

Only the sequence of significant digits is read on the slide rule. The position of the dec- imal point is determined separately. For ex- ample, if the hairline of the indicator is in the left-hand position shown in figure 8-5, the sig- nificant digits are read as follows:

103 194

Figure 8-5.—Readings in the first division of a slide rule.

1. Any time the hairline falls in the first division, the first significant digit is 1. 2. Since the hairline lies between the index and the first section mark, we know the number lies between 1.0 and 1.1, or 10 and 11, or 100 and 110, etc. The second significant digit is 0. 3. We next find how far from the index the hairline is located. It lies on the marking for the third space. 4. The three significant digits are 103.

In the second example shown in figure 8-5, the hairline is located in the first division, the ninth section, and on the fourth space mark of that section. Therefore, the significant digits are 194.

Thus, we see that any number falling in the first division of the slide rule will always have 1 as its first significant digit. It can have any

number from 0 through 9 as its second digit, and any number from 0 through 9 as its third digit. Sometimes a fourth digit can be roughly approx- imated in this first division, but the number is really accurate to only three significant digits.

In the second and third divisions, each sec- tion is divided into only 5 spaces. (See fig. 8-6.) Thus, each space is equal to 0.2 of the section. Suppose, for example, that the hairline lies on the third space after the large 2 indicating the second division. The first significant digit is 2. Since the hairline lies between 2 and the first section mark, the second digit is 0. The hairline lies on the third space mark or 0.6 of the way between the division mark and the first section mark, so the third digit is 6. Thus, the significant digits are 206. Notice that if the hairline lies on a space mark the third digit can be written accurately; otherwise it must be approximated.

From the fourth division to the right index, each section is divided into only two spaces. Thus, if the hairline is in the fourth division and lies on the space mark between the sixth and seventh sections, we would read 465. If the hairline did not fall on a space mark, the third digit would have to be approximated.

OPERATIONS WITH THE SLIDE RULE

There are two parts in solving problems with a slide rule. In the first part the slide rule is used to find the digit sequence of the final result. The second part is concerned with the placing of the decimal point in the result. Let us consider first the digit sequence in mul- tiplication and division.

Multiplication

Multiplication is performed on the C and D scales of the slide rule. The following proce- dure is used:

TEN SPACES IN EACH ONLY FIVE SPACES IN EACH SECTION SECTION

Figure 8-6.—Reading in the second division of a slide rule.

Chapter 8—LOGARITHMS AND THE SLIDE RULE

1. Locate one of the factors to be multiplied on the D scale, disregarding the decimal point. 2. Place the index of the C scale opposite that number. 3. Locate the other factor on the C scale and move the hairline of the indicator to cover this factor. 4. The product is on the D scale under the hairline.

Sometimes in multiplying numbers, such as 25 x 6, the number on the C scale extends to the right of the stock and the product cannot be read. In such a case, we simply shift indices. Instead of the left-hand index of the C scale, the right-hand index is placed opposite the fac- tor on the D scale. The rest of the problem remains the same. By shifting indices, we are simply multiplying or dividing by 10, but this plays no part in reading the significant digits. Shifting indices affects the characteristic only.

EXAMPLE: 252 x 3 = 756

1. Place the left index of the C scale over 252. 2. Locate 3 on the C scale and set the hair- line of the indicator over it. 3. Under the hairline on the D scale read the product, 756.

EXAMPLE: 4 x 64 = 256

1. Place the right index of the C scale over 4. 2. Locate 64 on the C scale and set the hairline of the indicator over it. 3. Under the hairline on the D scale read the product, 256.

Practice problems. Determine the following products by slide rule to three significant digits:

1. 2.8 x 16 3. 6 x 85

2. 7 x 1.3 4. 2.56 x 3.5

Answers:

1. 44.8 3. 510

2. 9.10 4. 8.96

Division

Division being the inverse of multiplication, the process of multiplication is reversed to

perform division on a slide rule. We subtract the length representing the logarithm of the divisor from the length representing the loga- rithm of the dividend to get the logarithm of the quotient.

The procedure is as follows:

1. Locate the dividend on the D scale and place the hairline of the indicator over it. 2. Move the slide until the divisor (on the C scale) lies under the hairline. 3. Read the quotient on the D scale opposite the C scale index.

If the divisor is greater numerically than the dividend, the slide will extend to the left. If the divisor is less, the slide will extend to the right. In either case, the quotient is the number on the D scale that lies opposite the C scale in- dex, falling within the limits of the D scale.

EXAMPLE: 6 ÷ 3 = 2

1. Locate 6 on the D scale and place the hairline of the indicator over it. 2. Move the slide until 3 on the C scale is under the hairline. 3. Opposite the left C scale index, read the quotient, 2, on the D scale.

EXAMPLE: 378 ÷ 63 = 6

1. Locate 378 on the D scale and move the hairline of the indicator over it. 2. Move the slide to the left until 63 on the C scale is under the hairline. 3. Opposite the right-hand index of the C scale, read the quotient, 6, on the D scale.

Practice problems. Determine the following quotients by slide rule.

1. 126 ÷ 3 3. 142 ÷ 71

2. 960 ÷ 15 4. 459 ÷ 17

Answers:

1. 42 3. 2

2. 64 4. 27

PLACING THE DECIMAL POINT

Various methods have been advanced regard- ing the placement of the decimal point in num- bers derived from slide rule computations. Probably the most universal and most easily remembered method is that of approximation.

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MATHEMATICS, VOLUME 1

The method of approximation means simply the rounding off of numbers and the mechanical shifting of decimal points in the numbers of the problem so that the approximate size of the solution and the exact position of the decimal point will be seen for inspection. The slide rule may then be used to derive the correct se- quence of significant digits. The method may best be demonstrated by a few examples. Re- member, shifting the decimal point in a number one place to the left is the same as dividing by 10. Shifting it one place to the right is the same as multiplying by 10. Every shift must be compensated for in order for the solution to be correct.

EXAMPLE: 0.573 x 1.45

SOLUTION: No shifting of decimals is neces- sary here. We see that 0.6 is to be multiplied by approximately 1 1/2. Immedi- ately, we see that the solution is in the neigh- borhood of 0.9. By slide rule we find that the significant digit sequence of the product is 832. From our approximation we know that the deci- mal point is to the immediate left of the first significant digit, 8. Thus,

0.573 x 1.45 = 0.832

EXAMPLE: 239 x 52.3

SOLUTION: For ease in multiplying, we shift the decimal point in 52.3 one place to the left, making it 5.23. To compensate, the decimal point is shifted to the right one place in the other factor. The new position of the decimal point is indicated by the presence of the caret symbol.

239.0 x 5.2.3 ∧ ∧

Our problem is approximately the same as

2,400 x 5 = 12,000

By slide rule the digit sequence is 125. Thus,

239 x 52.3 = 12,500

EXAMPLE: 0.000134 x 0.092

SOLUTION:

Shifting decimal points, we have 0.00.000134 x 0.09.2 ∧∧ ∧∧

Approximation: 9 x 0.0000013 = 0.0000117. By slide rule the digit sequence is 123. From approximation the decimal point is located as follows:

0.0000123

Thus,

0.000134 x 0.092 = 0.0000123

EXAMPLE: 53.1 42.4

SOLUTION: The decimal points are shifted so that the divisor becomes a number between 1 and 10. The method employed is cancellation. Shifting decimal points, we have

5.3.1 4.2.4

Approximation: 5 = 1.2 4

Digit sequence by slide rule:

1255

Placing the decimal point from the approxi- mation:

1.255

Thus,

53.1 = 1.255 42.4

EXAMPLE: 0.00645 0.0935

SOLUTION:

Shifting decimal points

0.00.645 0.09.35 ∧∧

Approximation:

0.6 = 0.7 9

Digit sequence by slide rule: 690

Chapter 8—LOGARITHMS AND THE SLIDE RULE

Placing the decimal point from the approximation:

0.0690

Thus,

0.00645/0.0935 = 0.0690

Practice problems. Solve the following problems with the slide rule and use the method of approximation to determine the position of the decimal point:

1. 0.00453 x 0.1645 3. 0.0362 x 1.21 2. 53.1 ÷ 1.255 4. 67 ÷ 316

Answers:

1. 0.000745 3. 0.0438 2. 42.4 4. 0.212

MULTIPLICATION AND DIVISION COMBINED

In problems such as

0.644 x 330 161 x 12

It is generally best to determine the position of the decimal point by means of the method of approximation and to determine the significant digit sequence from the slide rule. Such problems are usually solved by dividing and multiplying alternately throughout the problem. That is, we divide 0.644 by 161, multiply the quotient by 330, and divide that product by 12.

Shifting decimal points, we have

0.0644 x 3.30 1.61 x 1.2

Since there is a combined shift of three places to the left in the divisor, there must also be a combined shift of three places to the left in the dividend.

2 Approximation: 0.06 x 2 = 0.06 x 2 = 0.12 1.2

The step-by-step process of determining the significant digit sequence of this problem is as follows:

1. Place the hairline over 644 on the D scale. 2. Draw the slide so that 161 of the C scale lies under the hairline opposite 644. 3. Opposite the C scale index (on the D scale) is the quotient of 644 ÷ 161. This is to be multiplied by 330, but 330 projects beyond the rule so the C scale indices must be shifted. 4. After shifting the indices, find 330 on the C scale and place the hairline over it. Opposite 330 under the hairline on the D scale is the product of 644 x 330/161. 5. Next, move the C scale until 12 is under the hairline. Opposite the C scale index (on the D scale) is the final quotient. The digit sequence is 110.

The decimal point is then placed according to our approximation: 0.11. Thus,

0.644 x 330/161 x 12 = 0.11

Practice problems. Solve the following problems, using a slide rule:

1. 22 x 18.5 x 157/(17 x 18.3 x 85) 2. 432 x 9,600/(25,600 x 198) 3. 2.77 x 0.064/(0.17 x 1.97)

Answers:

1. 10.2 2. 0.817 3. 0.529

SQUARES

Squares of numbers are found by reference to the A scale. The numbers on the A scale are the squares of those on the D Scale. The A scale is really a double scale, each division being one-half as large as the corresponding division on the D scale. The use of a double scale for squaring is based upon the fact that the logarithm of the square of a number is twice the logarithm of the number itself. In other words,

log N² = 2 log N

This is reasonable, since

log N² = log (N x N) = log N + log N

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MATHEMATICS, VOLUME 1

For a numerical example, suppose that we seek to square 2 by means of logarithms.

log 2 = 0.301 log 2² = 2 log 2 = 2 x 0.301 = 0.602

Since each part of the A scale is half as large as the corresponding part of the D scale, the logarithm 0.602 on the A scale will be the same length as the logarithm 0.301 on the D scale. That is, these logarithms will be opposite on the A and D scales. On the A scale as on the D scale, the numbers are written rather than their logarithms. Select several numbers on the D scale, such as 2, 4, 8, 11, and read their squares on the A scale, namely 4, 16, 64, 121.

Notice also that the same relation exists for the B and C scales as for the A and D scales. Of interest, also, is the fact that since the A and B scales are made up as are the C and D scales, they too could be used for multiplying or dividing.

Placing the Decimal Point

Usually the decimal may be placed by the method of approximation. However, close observation will reveal certain facts that eliminate the need for approximations in squaring numbers. Two rules suffice for squaring whole or mixed numbers, as follows:

1. When the square of a number is read on the left half of the A scale, that number will contain twice the number of digits to the left of the decimal point in the original number, less 1. 2. When the square of a number is read on the right half of the A scale, that number will contain twice the number of digits to the left of the decimal point in the original number.

EXAMPLE: Square 2.5.

SOLUTION: Place the hairline over 25 on the D scale. Read the digit sequence, 625, under the hairline in the left half of the A scale. By rule 1: (2 x number of digits)-1 = 2(1)-1=1. There is one digit to the left of the decimal point. Thus,

(2.5)² = 6.25

EXAMPLE: Square 6,340.

SOLUTION:

Digit sequence, right half A scale: 402. By rule 2: 2 x number of digits = 2 x 4 = 8 (digits in answer). Thus,

(6,340)² = 40,200,000

Positive Numbers Less Than One

If positive numbers less than one are to be squared, a slightly different version of the preceding rules must be employed. Count the zeros between the decimal point and the first nonzero digit. Consider this count negative. Then the number of zeros between the decimal point and the first significant digit of the squared number may be found as follows:

1. Left half A scale: Multiply the zeros counted by 2 and subtract 1. 2. Right half A scale: Multiply the zeros counted by 2.

EXAMPLE: Square 0.0045

SOLUTION:

Digit sequence, right half A scale: 2025. By rule 2: 2(-2) = -4. (Thus, 4 zeros between the decimal point and the first digit.)

(0.0045)² = 0.00002025

EXAMPLE: Square 0.0215

SOLUTION:

Digit sequence, left half A scale: 462. By rule 1: 2(-1) -1 = -3

(0.0215)² = 0.000462

SQUARE ROOTS

Taking the square root of a number with the slide rule is the inverse process of squaring a number. We find the number on the A scale, set the hairline of the indicator over it, and read the square root on the D scale under the hairline.

Positioning Numbers on the A Scale

Since there are two parts of the A scale exactly alike and the digit sequence could be

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Chapter 8–LOGARITHMS AND THE SLIDE RULE

found on either part, a question arises as to which section to use. Generally, we think of the left half of the rule as being numbered from 1 to 10 and the right half as being numbered from 10 to 100. The numbering continues—left half 100 to 1,000, right half 1,000 to 10,000, and so forth.

A simple process provides a check of the location of the number from which the root is to be taken. For whole or mixed numbers, be- gin at the decimal point of the number and mark off the digits to the left (including end zeros) in groups of two. This is illustrated in the follow- ing two examples:

1. √40,300.21 √4'03'00.21

2. √2,034.1 √20'34.1

Look at the left-hand group. If it is a 1-digit number, use the left half of the A scale. If it is a 2-digit number, use the right half of the A scale. The number in example 1 is thus located in the left half of the A scale and the number in example 2 is located in the right half.

Numbers Less Than One

For positive numbers less than one, begin at the decimal point and mark off groups of two to the right. This is illustrated as follows:

1. √0.000245 √0.00'02'45

2. √0.00402 √0.00'40'2

Looking from left to right, locate the first group that contains a digit other than zero. If the first figure in this group is zero, locate the number in the left half of the A scale. If the first figure is other than zero, locate the num- ber in the right half of the A scale. Thus,

√0.00'02'45 is located left

and

√0.00'40'2 is located right

Powers of 10

When the square root of 10, 1,000, 100,000, and so forth, is desired, the center index is used. That is, when the number's digits in a power of 10 is even, use the center index.

The slide rule uses only the first three significant digits of a number. Thus, if the rule is used, √23451.6 must be considered as √23500.0. Likewise, 1.43567 would be consid- ered 1.43000, and so forth. For greater accu- racy, other methods must be used.

Practice problems. State which half of the A scale should be used for each of the following:

1. √432 5. √4,320 2. √0.014 6. √0.00301 3. √241.67 7. √0.0640 4. √0.00045 8. √9.41

Answers:

1. Left 5. Right 2. Left 6. Right 3. Left 7. Left 4. Left 8. Left

Placing the Decimal Point

To place the decimal point in the square root of a number, mark off the original number in groups of two as explained previously.

For whole or mixed numbers, the number of groups marked off is the number of digits in- cluding end zeros to the left of the decimal point in the root. The following problems il- lustrate this:

1. √23,415 √2'34'15 Three digits to left of dec- imal point in square root

2. √421,562.4 √42'15'62.4 Three digits to left of dec- imal point in square root

3. √231.321 √2'31.321 Two digits to left of deci- mal point in square root

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MATHEMATICS, VOLUME 1

For positive numbers less than one, there will be one zero in the square root between the decimal point and the first significant digit for every pair of zeros counted between the deci- mal point and the first significant digit of the original number. This is illustrated as follows:

1. √0.0004 √0.00'04 One zero before first digit in square root

2. √0.00008 √0.00'00'8 Two zeros before first digit in square root

3. √0.087 No zeros before first digit in square root

EXAMPLE: √4,521 √45'21

(Two digits in left-hand group)

Place the hairline over 452 on the right half of the A scale. Read the digit sequence of the root, 672, on the D scale under the hairline. Since there are two groups in the original num- ber, there are two digits to the left of the deci- mal point in the root. Thus,

√4,521 = 67.2

EXAMPLE: √0.000741 √0.00'01'41

(First figure is zero in this group)

Place the hairline over 741 on the left half of the A scale. Read the digit sequence of the root, 272, under the hairline on the D scale. Since there is one pair of zeros to the left of the group containing the first digit, there is one zero between the decimal point and the first significant digit of the root. Thus,

√0.000741 = 0.0272

Practice problems. Evaluate each of the following by means of a slide rule:

1. (17.75)² 3. √8.42 2. (0.65)² 4. √0.074

Answers:

1. 315 3. 3.07 2. 0.422 4. 0.272

CUBES AND CUBE ROOTS

Cubes and cube roots are read on the K and D scales of the slide rule. On the K scale are compressed three complete logarithmic scales in the same space as that of the D scale. Thus, any logarithm on the K scale is three times the logarithm opposite it on the D scale. To cube a number by logarithms, we multiply its loga- rithm by three. Therefore, the logarithms of cubed numbers will lie on the K scale opposite the numbers on the D scale.

As with the other slide rule scales men- tioned, the numbers the logarithms represent, rather than the logarithmic notations, are printed on the rule. In the left-hand third of the K scale, the numbers range from 1 to 10; in the middle third they range from 10 to 100; and in the right-hand third, they range from 100 to 1,000.

To cube a number, find the number on the D scale, place the hairline over it, and read the digit sequence of the cubed number on the K scale under the hairline.

Placing the Decimal Point

The decimal point of a cubed whole or mixed number may be easily placed by application of the following rules:

1. If the cubed number is located in the left third of the K scale, its number of digits to the left of the decimal point is 3 times the number of digits to the left of the decimal point in the original number, less 2.

2. If the cubed number is located in the middle third of the K scale, its number of digits is 3 times the number of digits of the original number, less 1.

3. If the cubed number is located in the right third of the K scale, its number of digits is 3 times the number of digits of the original number.

EXAMPLE: (1.6)³

SOLUTION: Place the hairline over 16 on D scale. Read the digit sequence, 409, on the K scale under the hairline.

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Chapter 8—LOGARITHMS AND THE SLIDE RULE

Number of digits to left of decimal point in the number 1.6 is 1 and the cubed number is in the left-hand third of the K scale.

3 x (No. of digits)-2 = (3 x 1)-2 = 1

Therefore,

(1.6)³ = 4.09 EXAMPLE: (4.1)³

Digit sequence = 689.

SOLUTION: Number of digits to left of decimal point in the number 4.1 is 1, and the cubed number is in the middle third of the K scale.

3 x (No. of digits)-1 = (3 x 1)-1 = 2

Therefore,

(4.1)³ = 68.9

EXAMPLE: (52)³

SOLUTION: Digit sequence = 141.

Number of digits to left of decimal point in the number 52 is 2, and the cubed number is in the right-hand third of the K scale.

3 x No. of digits = 3 x 2 = 6

Therefore,

(52)³ = 141,000

Positive Numbers Less Than One

If positive numbers less than one are to be cubed, count the zeros between the decimal point and the first nonzero digit. Consider the count negative. Then the number of zeros between the decimal point and the first significant digit of the cubed number may be found as follows:

1. Left third of K scale: Multiply the zeros counted by 3 and subtract 2. 2. Middle third of K scale: Multiply the zeros counted by 3 and subtract 1. 3. Right third of K scale: Multiply the zeros counted by 3.

EXAMPLE: Cube 0.034

SOLUTION: Digit sequence = 393

Zero count of 0.034 = -1, and 393 is in the middle third of the K scale.

3 x (No. of zeros)-1 = (3 x -1)-1 = -4

Therefore,

(0.034)³ = 0.0000393

Practice problems. Cube the following numbers using the slide rule.

1. 21 2. 0.7 3. 0.0128 4. 40²

Answers:

1. 9260 3. 0.0000021 2. 0.342 4. 66,000,000

Cube Roots

Taking the cube root of a number on the slide rule is the inverse process of cubing a number. To take the cube root of a number, find the number on the K scale, set the hairline over it, and read the cube root on the D scale under the hairline.

POSITIONING NUMBERS ON THE K SCALE.— Since a given number can be located in three positions on the K scale, the question arises as to which third of the K scale to use when locating a number. Generally, the left index, the left middle index, the right middle index, and the right index are considered to be numbered as shown in figure 8-7.

┌───────┬───────┬───────┬───────┐ │ 1 │ 1 │ 1 │ 1 │ ────┤ │ │ │ ├──── │ .000 │ 10,000│100,000│1,000,000 │ │ │ │ │ └───────┴───────┴───────┴───────┘

Figure 8-7.— Powers of 10 associated with K-scale indices.

A system similar to that used with square roots may be used to locate the position of a number on the K scale. Groups of three are used rather than groups of two. The grouping for cube root is illustrated as follows:

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MATHEMATICS, VOLUME 1

1. ³√40,531.6 ³√40'531.6

2. ³√4,561.43 ³√4'561.43

3. ³√0.000043 ³√0.000'043

For whole or mixed numbers the following rules apply:

1. If the left-hand group contains one digit, locate the number in the left third of the K scale. 2. If the left group contains two digits, locate the number in the middle third of the K scale. 3. If the left group contains three digits, locate the number in the right third of the K scale.

The following examples illustrate the foregoing:

1. ³√4'561.43 (One digit)—left third K scale.

2. ³√40'531.6 (Two digits)—middle third K scale.

3. ³√453'561 (Three digits)—right third of K scale.

For positive numbers less than one, look from left to right and find the first group that contains a digit other than a zero.

1. If the first two figures in this group are zeros, locate the number in the left third of the K scale. 2. If only the first figure in this group is zero, locate the number in the middle third of the K scale. 3. If the first figure of the group is not zero, locate the number in the right third of the K scale.

The following examples illustrate these rules:

1. ³√0.000'004'53 (Two zeros)—left third K scale.

2. ³√0.000'050'43 (One zero)—middle third K scale.

3. ³√0.000'000'430 (No zero)—right third K scale.

PLACING THE DECIMAL POINT.— To place the decimal point in the cube root of a number, we use the system of marking off in groups of three as shown above.

For whole or mixed numbers, there is one digit in the root to the left of the decimal point for every group marked in the original number. Thus,

³√4'531.6

(Two digits in root to left of decimal point.)

For positive numbers less than one, there will be one zero in the root between the decimal point and the first significant digit for every three zeros counted between the decimal point and the first significant digit of the original number. Thus,

³√0.000'000'004

(Two zeros between decimal point and first significant digit of root.)

EXAMPLE: ³√216000.4 ³√216'000.4

(Three digits in left group)

Place the hairline over 216 in the right third of the K scale. Read the digit sequence, 6, under the hairline on the D scale. Since there are two groups in the original number, there are two digits to the left of the decimal point in the root. Thus,

³√216000.4 = 60

EXAMPLE: ³√0.0000451 ³√0.000'045'1

(Only first figure is zero in this group)

Place the hairline over 451 in the middle third of the K scale. Read the digit sequence, 357, under the hairline on the D scale. Since there is one group of three zeros, there is one zero between the decimal point and the first significant digit of the root. Thus,

³√0.0000451 = 0.0357

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Chapter 8—LOGARITHMS AND THE SLIDE RULE

POWERS OF 10.—To take the cube root of a power of 10, mark it off as explained in the preceding paragraphs. The number in the left group will then be 1, 10, or 100. We know that the cube root of 10 is a number between 2 and 3. Thus, for the cube root of any number whose left group is 10, use the K scale index which lies between 2 and 3 on the D scale. The cube root of 100 lies between 4 and 5. Therefore, for a number whose left group is 100, use the K scale index between 4 and 5 on the D scale.

Practice problems. Following are some problems and the digit sequence (d. s.) of the roots. Locate the decimal point for each root.

1. ³√0.000023 d. s. 2844 2. ³√0.051 d. s. 371 3. ³√127 d. s. 5026

4. ³√204,000 d. s. 589 5. ³√734,000,000 d. s. 902 6. ³√4,913 d. s. 17

Answers:

1. 0.02844 2. 0.371 3. 5.026 4. 58.9 5. 902 6. 17

CHAPTER 9

FUNDAMENTALS OF ALGEBRA

The numbers and operating rules of arithmetic form a part of a very important branch of mathematics called ALGEBRA.

Algebra extends the concepts of arithmetic so that it is possible to generalize the rules of operating with numbers and use these rules in manipulating symbols other than numbers. It does not involve an abrupt change into a distinctly new field, but rather provides a smooth transition into many branches of mathematics with a continuation of knowledge already gained in basic arithmetic.

The idea of expressing quantities in a general way, rather than in the specific terms of arithmetic, is fairly common. A typical example is the formula for the perimeter of a rectangle, P = 2L + 2W, in which the letter P represents perimeter, L represents length, and W represents width. It should be understood that 2L = 2(L) and 2W = 2(W). If the L and the W were numbers, parentheses or some other multiplication sign would be necessary, but the meaning of a term such as 2L is clear without additional signs or symbols.

All formulas are algebraic expressions, although they are not always identified as such. The letters used in algebraic expressions are often referred to as LITERAL NUMBERS (literal implies "letterai").

Another typical use of literal numbers is in the statement of mathematical laws of operation. For example, the commutative, associative, and distributive laws, introduced in chapter 3 with respect to arithmetic, may be restated in general terms by the use of algebraic symbols.

COMMUTATIVE LAWS

The word "commutative" is defined in chapter 3. Remember that the commutative laws refer to those situations in which the factors and terms of an expression are rearranged in a different order.

ADDITION

The algebraic form of the commutative law for addition is as follows:

a + b = b + a

From this law, it follows that

a + (b + c) = a + (c + b) = (c + b) + a

In words, this law states that the sum of two or more addends is the same regardless of the order in which the addends are arranged.

The arithmetic example in chapter 3 shows only one specific numerical combination in which the law holds true. In the algebraic example, a, b, and c represent any numbers we choose, thus giving a broad inclusive example of the rule. (Note that once a value is selected for a literal number, that value remains the same wherever the letter appears in that particular example or problem. Thus, if we give a the value of 12, in the example just given, a's value is 12 wherever it appears.)

MULTIPLICATION

The algebraic form of the commutative law for multiplication is as follows:

ab = ba

In words, this law states that the product of two or more factors is the same regardless of the order in which the factors are arranged.

ASSOCIATIVE LAWS

The associative laws of addition and multiplication refer to the grouping (association) of terms and factors in a mathematical expression.

ADDITION

The algebraic form of the associative law for addition is as follows:

a + b + c = (a + b) + c = a + (b + c)

In words, this law states that the sum of three or more addends is the same regardless of the manner in which the addends are grouped.

MULTIPLICATION

The algebraic form of the associative law for multiplication is as follows:

a · b · c = (a · b) · c = a · (b · c)

Chapter 9—FUNDAMENTALS OF ALGEBRA

In words, this law states that the product of three or more factors is the same regardless of the manner in which the factors are grouped.

DISTRIBUTIVE LAW

The distributive law refers to the distribu- tion of factors among the terms of an additive expression. The algebraic form of this law is as follows:

a(b + c) = ab + ac

From this law, it follows that: If the sum of two or more quantities is multiplied by a third quantity, the product is found by applying the multiplier to each of the original quantities separately and summing the resulting expres- sions.

ALGEBRAIC SUMS

The word "sum" has been used several times in this discussion, and it is important to realize the full implication where algebra is concerned. Since a literal number may represent either a positive or a negative quantity, a sum of sev- eral literal numbers is always understood to be an ALGEBRAIC SUM. That is, it is the sum that results when the algebraic signs of all the addends are taken into consideration. The following problems illustrate the proce- dure for finding an algebraic sum:

Let a = 3, b = -2, and c = 4.

Then a + b + c = (3) + (-2) + (4)

= 5

Also, a - b - c = a + (-b) + (-c)

= 3 + (+2) + (-4)

= 1

The second problem shows that every expres- sion containing two or more terms to be com- bined by addition and subtraction may be re- written as an algebraic sum, all negative signs being considered as belonging to specific terms and all operational signs being positive. It should be noted, in relation to this subject, that the laws of signs for algebra are the same as those for arithmetic.

ALGEBRAIC EXPRESSIONS

An algebraic expression is made up of the signs and symbols of algebra. These symbols

include the Arabic numerals, literal numbers, the signs of operation, and so forth. Such an expression represents one number or one quan- tity. Thus, just as the sum of 4 and 2 is one quantity, that is, 6, the sum of c and d is one quantity, that is, c + d. Likewise, a/b, √b, ab, a - b, and so forth, are algebraic expres- sions each of which represents one quantity or number. Longer expressions may be formed by com- binations of the various signs of operation and the other algebraic symbols, but no matter how complex such expressions are they still repre- sent one number. Thus the algebraic expres-

sion -a + √(2a + b)/c - c is one number

The arithmetic value of any algebraic ex- pression depends on the values assigned to the literal numbers. For example, in the expres- sion 2x² - 3ay, if x = -3, a = 5, and y = 1, then we have the following:

2x² - 3ay = 2(-3)² -3(5)(1)

= 2(9) - 15 = 18 - 15 = 3

Notice that the exponent is an expression such as 2x² applies only to the x. If it is de- sired to indicate the square of 2x, rather than 2 times the square of x, then parentheses are used and the expression becomes (2x)². Practice problems. Evaluate the following algebraic expressions when x = 4, b = 2, c = 3, x = 7, and y = 5. Remember, the order of op- eration is multiplication, division, addition, and subtraction.

1. 3x + 7y - c 3. ax/b + y

2. xy - 4a² 4. c + ay²/b

Answers:

1. 53 3. 19

2. -29 4. 53

TERMS AND COEFFICIENTS

The terms of an algebraic expression are the parts of the expression that are connected by plus and minus signs. In the expression 3abx + cy - k, for example, 3abx, cy, and k are the terms of the expression.

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MATHEMATICS, VOLUME 1

An expression containing only one term, such as 3ab, is called a monomial (mono means one). A binomial contains two terms; for example, 2r + by. A trinomial consists of three terms. Any expression containing two or more terms may also be called by the general name, poly- nomial (poly means many). Usually special names are not given to polynomials of more than three times. The expression x³ - 3x² + 7x + 1 is a polynomial of four terms. The trinomial x² + 2x + 1 is an example of a polynomial which has a special name. Practice problems. Identify each of the fol- lowing expressions as a monomial, binomial, trinomial, or polynomial. (Some expressions may have two names.)

1. x 3. abx 5. 3y² + 4

2. 3y + a + b 4. 4 + 2b + y + z 6. 2y/6 + 1

Answers:

1. Monomial 2. Trinomial (also polynomial)

3. Monomial 4. Polynomial

5. Binomial 6. Binomial (also polynomial) (also polynomial)

In general, a COEFFICIENT of a term is any factor or group of factors of a term by which the remainder of the term is to be multi- plied. Thus in the term 2axy, 2ax is the coeffi- cient of y, 2a is the coefficient of xy, and 2 is the coefficient of axy. The word "coefficient" is usually used in reference to that factor which is expressed in Arabic numerals. This factor is sometimes called the NUMERICAL COEF- FICIENT. The numerical coefficient is cus- tomarily written as the first factor of the term. In 4x, 4 is the numerical coefficient, or simply the coefficient, of x. Likewise, in 24xy², 24 is the coefficient of xy² and in 16(a + b), 16 is the coefficient of (a + b). When no numerical coef- ficient is written it is understood to be 1. Thus in the term xy, the coefficient is 1.

COMBINING TERMS

When arithmetic numbers are connected by plus and minus signs, they can always be com- bined into one number. Thus,

5 - 7½ + 8 = 5½

Here three numbers are added algebraically (with due regard for signs) to give one number. The terms have been combined into one term. Terms containing literal numbers can be combined only if their literal parts are the same. Terms containing literal factors in which the same letters are raised to the same power are called like terms. For example, 3y and 2y are like terms since the literal parts are the same. Like terms are added by adding the coefficients of the like parts. Thus, 3y + 2y = 5y just as 3 bolts + 2 bolts = 5 bolts. Also 3a²b and a²b are like; 3a²b + a²b = 4a²b and 3a²b - a²b = 2a²b. The numbers ay and by are like terms with respect to y. Their sum could be indicated in two ways: ay + by or (a + by. The latter may be explained by comparing terms to denominate numbers. For instance, a bolts - b bolts = (a + b) bolts. Like terms are added or subtracted by add- ing or subtracting the numerical coefficients and placing the result in front of the literal factor, as in the following examples:

7x² - 5x² = (7 - 5)x² = 2x²

5b²x - 3ay² - 8b²x + 10ay² = -3b²x + 7ay²

Dissimilar or unlike terms in an algebraic expression cannot be combined when numerical values have not been assigned to the literal factors. For example, -5x² + 3xy - 3y² con- tains three dissimilar terms. This expression cannot be further simplified by combining terms through addition or subtraction. The expres- sion may be rearranged as x(3y - 5x) - 8y² or y(3x - 8y) - 5x², but such a rearrangement is not actually a simplification. Practice problems. Combine like terms in the following expression:

1. 2a + 4a 4. 2ay² - ay²

2. y + y² + 2y 5. txc² + 2bx²

3. 4 ay/c - ay/c 6. 2y + y²

Answers:

1. 6a 4. ay²

2. y² + 3y 5. 3bx²

3. 3ay/c 6. 2y + y²

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Chapter 9 – FUNDAMENTALS OF ALGEBRA

SYMBOLS OF GROUPING

Often it is desired to group two or more terms to indicate that they are to be considered and treated as though they were one term even though there may be plus and minus signs between them. The symbols of grouping are parentheses ( ) (which we have already used), brackets [ ], braces { }, and the vinculum ___. The vinculum is sometimes called the "over-score." The fact that -7 + 2 - 5 is to be subtracted from 15, for example, could be indicated in any one of the following ways:

15 - (-7 + 2 - 5) 15 - [-7 + 2 - 5] 15 - {-7 + 2 - 5} 15 - -7 + 2 - 5

Actually the vinculum is seldom used except in connection with a radical sign, such as in √a + b, or in a Boolean algebra expression. Boolean algebra is a specialized kind of symbolic notation which is discussed in Mathematics, 3, NavPers 10073. Parentheses are the most frequently used symbols of grouping. When several symbols are needed to avoid confusion in grouping, parentheses usually comprise the innermost symbols, followed by brackets, and then by braces as the outermost symbols. This arrangement of grouping symbols is illustrated as follows:

2x - {3y + [- 8 - 5y - (x - 4)]}

REMOVING AND INSERTING GROUPING SYMBOLS

Discussed in the following paragraphs are various rules governing the removal and insertion of parentheses, brackets, braces, and the vinculum. Since the rules are the same for all grouping symbols, the discussion in terms of parentheses will serve as a basis for all.

Removing Parentheses

If parentheses are preceded by a minus sign, the entire quantity enclosed must be regarded as a subtrahend. This means that each term of the quantity in parentheses is subtracted from the expression preceding the minus sign. Accordingly, parentheses preceded by a minus sign can be removed, if the signs of all terms within the parentheses are changed. This may be explained with an arithmetic example. We recall that to subtract one number from another, we change the sign of the subtrahend and proceed as in addition. To subtract -7 from 16, we change the sign of -7 and proceed as in addition, as follows:

16 - (-7) = 16 + 7 = 23

It is sometimes easier to see the result of changing signs in the subtrahend if the minus sign preceding the parentheses is regarded as a multiplier. Thus, the thought process in removing parentheses from an expression such as - (4 - 3 + 2) would be as follows: Minus times plus is minus, so the first term of the expression with parentheses removed is - 4. (Remember that the 4 in the original expression is understood to be a +4, since it has no sign showing.) Minus times minus is plus, so the second term is +3. Minus times plus is minus, so the third term is -2. The result is -4 + 3 - 2, which reduces to -3. This same result can be reached just as easily, in an arithmetic expression, by combining the numbers within the parentheses before applying the negative sign which precedes the parentheses. However, in an algebraic expression with no like terms such combination is not possible. The following example shows how the rule for removal of parentheses is applied to algebraic expressions:

2a - (-4x + 3by) = 2a + 4x - 3by

Parentheses preceded by a plus sign can be removed without any other changes, as the following example shows:

2b + (a - b) = 2b + a - b = a + b

Many expressions contain more than one set of parentheses, brackets, and other symbols of grouping. In removing symbols of grouping, it is possible to proceed from the outside inward or from the inside outward. For the beginner, it is simpler to start on the inside and work toward the outside, collecting terms and simplifying as one proceeds. In the following example the inner grouping symbols are removed first:

101

MATHEMATICS, VOLUME 1

2a - [x + (x - 3a) - (9a - 5x)] = 2a - [x + x - 3a - 9a + 5x] = 2a - [7x - 12a] = 2a - 7x + 12a = 14a - 7x

Enclosing Terms in Parentheses

When it is desired to enclose a group of terms in parentheses, the group of terms remains unchanged if the sign preceding the parentheses is positive. This is illustrated as follows:

3x - 2y + 7x - y = (3x - 2y) + (7x - y)

Note that this agrees with the rule for removing parentheses preceded by a plus sign. If terms are enclosed within parentheses preceded by a minus sign, the signs of all the terms enclosed must be changed as in the following example:

3x - 2y + 7x - y = 3x - (2y - 7x + y)

Practice problems. In problems 1 through 4, remove the symbols of grouping and combine like terms. In problems 5 through 8, enclose the first two terms in parentheses preceded by a plus sign (understood) and the last two in parentheses preceded by a minus sign.

1. 6a - (4a - 3) 2. 3x + [2x - 4y(6 - 4x)] + 2y - (3 - x + 3y) 3. -a + [-a - (2a + 3)] + 3 4. (7x - 3ay) - (4a - b) + 16 5. 4a - 3b - 2c + 4d 6. -2 -3x +4y - z 7. x + 4y + 3z + 7 8. -4 + 2a - 6c + 3d

Answers:

1. 2a + 3 2. 8x + 16xy - 25y - 3 3. -4a 4. 7x - 3ay - 4a + b + 16

5. (4a - 3b) - (2c - 4d) 6. (-2 -3x) - (-4y + z) 7. (x + 4y) - (-3z - 7) 8. (-4 + 2a) - (6c - 3d)

EXPONENTS AND RADICALS

Exponents and radicals have the same meaning in algebra as they do in arithmetic. Thus, if n represents any number then n² = n·n, n³ = n·n·n, etc. By the same reasoning, nᵐ means that n is to be taken as a factor m times. That is, nᵐ is equal to n · n · n · . . . . , with n appearing m times. The series of dots, called ellipsis (not to be confused with the geometric figure having a similar name, ellipse), represents continuation of the same pattern or the same symbol. The rules of operation with exponents are also the same in algebra as in arithmetic. For example, n² · n³ = n²⁺³ = n⁵. Some care is necessary to avoid confusion over an expression such as 3² · 3. In this example, n = 3 and the product desired is 3⁵, not 9⁵. In general, aᵇ · aᶜ = aᵇ⁺ᶜ, and a similar result is reached whether the factor which acts as a base for the exponents is a number or a letter. Thus the general form can be expressed as follows:

nᵃ · nᵇ = nᵃ⁺ᵇ

In words, the general rule for multiplication involving exponents is as follows: When multiplying terms whose literal factors are like, the exponents are added. This rule may be applied to problems involving division, if all expressions containing exponents in denominators are rewritten as expressions with negative exponents. For example, the fraction x²y/x²y² can be rewritten as (x²y)(x⁻¹y⁻²), which is equal to (x²⁻¹)(y¹⁻²). This reduces to x²⁻¹y⁻¹ = x/y. Notice that the result is the same as it would have been if we had simply subtracted the exponents of literal factors in the denominator from the exponents of the same literal factors in the numerator. The algebraic rules for radicals also remain the same as those of arithmetic. In arithmetic, √4 = 4^(1/2) = 2. Likewise, in algebra ⁿ√a = a^(1/n) and ⁿ√aᵐ = aᵐ/ⁿ.

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Chapter 9—FUNDAMENTALS OF ALGEBRA

MULTIPLYING MONOMIALS

If a monomial such as 3abc is to be multiplied by a numerical multiplier, for example 5, the coefficient alone is multiplied, as in the following example:

5 x 3abc = 15abc

When the numerical factor is not the initial factor of the expression, as in x(2a), the result of the multiplication is not written as x2a. Instead, the numerical factor is interchanged with literal factors by use of the commutative law of multiplication. The literal factors are usually interchanged to place them in alphabetical order, and the final result is as follows:

x(2a) = 2ax

The rule for multiplication of monomials may be stated as follows: Multiply the numerical coefficients to form the coefficient of the product. Multiply the literal factors, combining exponents in like factors, to form the literal part of the product. The complete process is illustrated in the following example:

(2ab)(3a²)(2b³) = 12a³b⁴

Practice problems. Perform the indicated operations:

1. (2x²)(5x³) 4. (2ᵃ)(2ᵇ) 2. (-5ab²)(2a²b) 5. (-4a³)² 3. (-4x⁴y)(-3xy⁴) 6. (3a²b)²

Answers:

1. 10x⁷ 4. 2ᵃ⁺ᵇ 2. -10a³b³ 5. 16a⁶ 3. 12x⁵y⁵ 6. 9a⁴b²

DIVIDING MONOMIALS

As may be expected, the process of dividing is the inverse of multiplying. Because 3 x 2a = 6a, 6a ÷ 3 = 2a, or 6a ÷ 2 = 3a. Thus, when the divisor is numerical, divide the coefficient of the dividend by the divisor.

When the divisor contains literal parts that are also in the dividend, cancellation may be

performed as in arithmetic. For example, 6ab - 3a may be written as follows:

(2)(3a)(b) 3a

Cancellation of the common literal factor, 3a, from the numerator and denominator leaves 2b as the answer for this division problem.

When the same literal factors appear in both the divisor and the dividend, but with different exponents, cancellation may still be used, as follows:

14a³b³x (7)(2)a²ab³x -21a⁴b³x = (7)(-3)a²b³b³x

= 2a = - 2a -3b² 3b²

This same problem may be solved without thinking in terms of cancellation, by rewriting with negative exponents as follows:

14a³b³x 2a³⁻²b³⁻³x¹⁻¹ -21a⁴b³x = -3

= 2ab⁻² = 2a -3 -3b²

= - 2a 3b²

Practice problems. Perform the indicated operations:

1. x⁵/x⁶ 6. √x⁴ᵃy²ˢ 2. a⁹b⁴/b³ 7. 5a⁴b/10a⁺b³ 3. a²bc²/abc 8. 10x²y³z⁴/-5xy²z³ 4. a²b/ab² 9. √100a³b⁴ 5. √16x⁴y⁶ 10. √a⁵b⁶ⁿ

Answers:

1. x⁻¹ 6. ± x²yᵃ 2. a³b 7. a²/2b² 3. ac 8. - 2xyz 4. a/b 9. ± 10a⁺b² 5. ± 4x²y³ 10. ± a²b³ⁿ

MATHEMATICS, VOLUME 1

OPERATIONS WITH POLYNOMIALS

Adding and subtracting polynomials is simply the adding and subtracting of their like terms. There is a great similarity between operations with polynomials and denominate numbers. Compare the following examples:

1. Add 5 qt and 1 pt to 3 qt and 2 pt.

3 qt + 2 pt 5 qt + 1 pt 8 qt + 3 pt

2. Add 5x + y to 3x + 2y.

3x + 2y 5x + y 8x + 3y

One method of adding polynomials (shown in the above examples) is to place like terms in columns and to find the algebraic sum of the like terms. For example, to add 3a + b - 3c, 3b + c - d, and 2a + 4d, we would arrange the polynomials as follows:

3a + b - 3c 3b + c - d 2a + 4d 5a + 4b - 2c + 3d

Subtraction may be performed by using the same arrangement—that is, by placing terms of the subtrahend under the like terms of the minuend and carrying out the subtraction with due regard for sign. Remember, in subtraction the signs of all the terms of the subtrahend must first be mentally changed and then the process completed as in addition. For example, subtract 10a + b from 8a - 2b, as follows:

8a - 2b 10a + b -2a - 3b

Again, note the similarity between this type of subtraction and the subtraction of denominate numbers.

Addition and subtraction of polynomials also can be indicated with the aid of symbols of grouping. The rule regarding changes of sign when removing parentheses preceded by a minus sign automatically takes care of subtraction. For example, to subtract 10a + b from 8a - 2b, we can use the following arrangement:

(8a - 2b) - (10a + b) = 8a - 2b - 10a - b = -2a - 3b

Similarly, to add -3x + 2y to -4x - 5y, we can write

(-3x + 2y) + (-4x - 5y) = -3x + 2y - 4x - 5y = -7x - 3y

Practice problems. Add as indicated, in each of the following problems:

1. 3a + b 2a + 5b 2. (6s⁴t + 3s²t + st + 5) + (a³t - 5) 3. 4a + b + c, a + c - d, and 3a + 2b + 2c 4. 4x + 2y 3x - y + z x - z

In problems 5 through 8, perform the indicated operations and combine like terms.

5. (4a + b) - (3a + 5b) 6. (5x²y + 3x²y) - (x²y) 7. (x + 6) + (3x + 7) 8. (4a² - b) - (2a² + b)

Answers:

1. 5a + 6b 5. -(a + 4b) 2. 7s⁴t + 3s²t + st 6. 4x²y + 3x²y 3. 8a + 3b + 4c - d 7. 4x + 13 4. 8x + y 8. 2(a² - b)

MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL

We can explain the multiplication of a polynomial by a monomial by using an arithmetic example. Let it be required to multiply the binomial expression, 7 - 2, by 4. We may write this 4 x (7 - 2) or simply 4(7 - 2). Now 7 - 2 = 5. Therefore, 4(7 - 2) = 4(5) = 20. Now, let us solve the problem a different way. Instead of

Chapter 9—FUNDAMENTALS OF ALGEBRA

subtracting first and then multiplying, let us multiply each term of the expression by 4 and then subtract. Thus, 4(7 - 2) = (4 x 7) - (4 x 2) = 20. Both methods give the same result. The second method makes use of the distributive law of multiplication.

When there are literal parts in the expres- sion to be multiplied, the first method cannot be used and the distributive method must be employed. This is illustrated in the following examples:

4(5 + a) = 20 + 4a 3(a + b) = 3a + 3b ab(x + y - z) = abx + aby - abz

Thus, to multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial.

Practice problems. Multiply as indicated:

1. 2a(a - b) 2. 4a²(a² + 5a + 2) 3. -4x(-y - 3z) 4. 2a³(a² - ab)

Answers:

1. 2a² - 2ab 2. 4a⁴ + 20a³ + 8a² 3. 4xy + 12xz 4. 2a⁵ - 2a⁴b

MULTIPLICATION OF A POLY- NOMIAL BY A POLYNOMIAL

As with the monomial multiplier, we explain the multiplication of a polynomial by a poly- nomial by use of an arithmetic example. To multiply (3 + 2)(6 - 4), we could do the opera- tion within the parentheses first and then mul- tiply, as follows:

(3 + 2)(6 - 4) = (5)(2) = 10

However, thinking of the quantity (3 + 2) as one term, we can use the method described for a monomial multiplier. That is, we can multiply each term of the multiplicand by the multiplier, (3 + 2), with the following result:

(3 + 2)(6 - 4) = [(3 + 2) x 6 - (3 + 2) x 4]

Now considering each of the two resulting products separately, we note that each is a bi- nomial multiplied by a monomial.

The first is

(3 + 2)6 = (3 x 6) + (2 x 6)

and the second is

-(3 + 2)4 = - [(3 x 4) + (2 x 4)] = -(3 x 4) - (2 x 4)

Thus we have the following result:

(3 + 2)(6 - 4) = (3 x 6) + (2 x 6) - (3 x 4) - (2 x 4) = 18 + 12 - 12 - 8 = 10

The complete product is formed by multiplying each term of the multiplicand separately by each term of the multiplier and combining the results with due regard to signs.

Now let us apply this method in two exam- ples involving literal numbers.

1. (a + b)(m + n) = am + an + bm + bn

2. (2b + c)(r + s + 3t - u) = 2br + 2bs + 6bt - 2bu + cr + cs + 3ct - cu

The rule governing these examples is stated as follows: The product of any two polynomials is found by multiplying each term of one by each term of the other and adding the results alge- braically.

It is often convenient, especially when either of the expressions contains more than two terms, to place the polynomial with the fewer terms beneath the other polynomial and multi- ply term by term beginning at the left. Like terms of the partial products are placed one beneath the other to facilitate addition.

Suppose we wish to find the product of 3x² - 7x - 9 and 2x - 3. The procedure is

3x² - 7x - 9 2x - 3 _____________ 6x³ - 14x² - 18x - 9x² + 21x + 27 _____________ 6x³ - 23x² + 3x + 27

Practice problems. In the following prob- lems, multiply and combine like terms:

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MATHEMATICS, VOLUME 1

1. (2a - 3)(a + 2) 2. (ax + b)(ax - b) 3. x³ + 5x² - x + 2 2x + 3 4. 2a² + 5ab - b² a + b

Answers:

1. 2a² + a - 6 2. a²x² - b² 3. 2x⁴ + 13x³ + 13x² + x + 6 4. 2a³ + 7a²b + 4ab² - b³

SPECIAL PRODUCTS

The products of certain binomials occur fre- quently. It is convenient to remember the form of these products so that they can be written immediately without performing the complete multiplication process. We present four such special products as follows, and then show how each is derived:

1. Product of the sum and difference of two numbers.

EXAMPLE: (x - y)(x + y) = x² - y²

2. Square the sum of two numbers.

EXAMPLE: (x + y)² = x² + 2xy + y²

3. Square of the difference of two numbers.

EXAMPLE: (x - y)² = x² - 2xy + y²

4. Product of two binomials having a com- mon term.

EXAMPLE: (x + a)(x + b) = x² + (a + b)x + ab

Product of Sum and Difference

The product of the sum and difference of two numbers is equal to the square of the first number minus the square of the second number. If, for example, x - y is multiplied by x + y, the middle terms cancel one another. The result is the square of x minus the square of y, as shown in the following illustration:

x - y x + y _______ x² - xy + xy - y² _______ x² - y²

By keeping this rule in mind, the product of the sum and difference of two numbers can be written down immediately by writing the differ- ence of the squares of the numbers. For ex- ample, consider the following three problems:

(x + 3)(x - 3) = x² - 3² = x² - 9

(5a + 2b)(5a - 2b) = (5a)² - (2b)² = 25a² - 4b²

(7x + 4y)(7x - 4y) = 49x² - 16y²

RATIONALIZING DENOMINATORS.—The product of the sum and difference of two num- bers is useful in rationalizing a denominator that is a binomial. For example, in a fraction such as

2 √2 - 6

the denominator can be altered so that no radi- cal terms appear in it. (This process is called rationalizing.) The denominator must be mul- tiplied by √2 + 6, which is called the conjugate of √2 - 6. Since the value of the original frac- tion would be changed if we multiplied only the denominator, our multiplier must be applied to both the numerator and the denominator. Mul- tiplying the original fraction by

√2 + 6 √2 + 6

is, in effect, the same as multiplying it by 1. The result of rationalizing the denominator of this fraction is as follows:

2 √2 + 6 2(√2 + 6) ―――― · ――――――― = ―――――――――――――― √2 - 6 √2 + 6 (√2)² - 6²

2(√2 + 6) = ――――――――――――― 2 - 36

2(√2 + 6) = ――――――――――――― 2(1 - 18)

2(√2 + 6) = ――――――――――――― 2(-17)

√2 + 6 = ―――――――― -17

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Chapter 9—FUNDAMENTALS OF ALGEBRA

MENTAL MULTIPLICATION.—The product of the sum and difference can be utilized to mentally multiply two numbers that differ from a multiple of 10 by the same amount, one greater and the other less. For example, 67 is 3 less than 70 while 73 is 3 more than 70. The product of 67 and 73 is then found as follows:

67(73) = (70 - 3)(70 + 3) = 70² - 3² = 4,900 - 9 = 4,891

Square of Sum or Difference

The square of the SUM of two numbers is equal to the square of the first number plus twice the product of the numbers plus the square of the second number. The square of the DIF- FERENCE of the same two numbers has the same form, except that the sign of the middle term is negative.

These results are evident from multiplication. When x and y represent the two numbers, we obtain

x + y x - y x + y x - y x² + xy x² - xy + xy + y² - xy + y² x² + 2xy + y² x² - 2xy + y²

Applying this rule to the squares of the binomials 3a + 2b and 3a - 2b, we have the following two cases:

1. (3a + 2b)² = (3a)² + 2(3a)(2b) + (2b)² = 9a² + 12ab + 4b²

2. (3a - 2b)² = 9a² - 12ab + 4b²

The square of the sum or difference of two numbers is applicable to squaring a binomial that contains one or two irrational terms, as in the following examples:

1. (√3 + 8)² = (√3)² + 2(8)(√3) + 64 = 3 + 16 √3 + 64 = 67 + 16 √3

2. (√3 - 8)² = (√3)² - 2(8)(√3) + 64 = 3 - 16 √3 + 64 = 67 - 16 √3

3. (√5 + √7)² = (√5)² + 2 √5 √7 + (√7)² = 5 + 2 √35 + 7 = 12 + 2√35

4. (√5 - √7)² = 12 - 2 √35

The square of the sum or difference of two numbers can be applied to the process of mentally squaring certain numbers. For example, 82² can be expressed as (80 + 2)² while 67² can be expressed as (70 - 3)². We find that

(80 + 2)² = 80² + 2(80)(2) + 2² = 6,400 + 320 + 4 = 6,724

(70 - 3)² = 70² - 2(70)(3) + 3² = 4,900 - 420 + 9 = 4,489

Binomials Having a Common Term

The binomials x + 2 and x - 3 have a common term, x. They have two unlike terms, +2 and -3. The product of these binomials is

x + 2 x - 3 x² + 2x - 3x - 6 x² - x - 6

Inspection of this product shows that it is obtained by squaring the common term, adding the sum of the unlike terms multiplied by the common term, and finally adding the product of the unlike terms.

Apply this rule to the product of 3y - 5 and 3y + 4. The common term is 3y; its square is 9y². The sum of the unlike terms is -5 + 4 = -1; the sum of the unlike terms multiplied by the common term is -3y; and the product of the unlike terms is -5(4) = -20. The product of the two binomials is

(3y - 5)(3y + 4) = 9y² - 3y - 20

The product of two binomials having a common term is applicable to the multiplication of numbers like √3 + 7 and √3 - 2 which contain irrational terms. For example,

(√3 + 7)(√3 - 2) = (√3)² + 5 √3 - 14

= 3 + 5 √3 - 14 = -11 + 5 √3

Practice problems. In problems 1 through 4, multiply and combine terms. In 5 through 8, simplify by using special products.

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MATHEMATICS, VOLUME 1

1. (x + 4)(x + 2) 2. (√a - b)² 3. (7a + 4b)(7a - 4b) 4. (ax + y)²

5. 2 √2 - 2

6. 48(52) 7. (√3 + 7)² 8. (73)²

Answers: 1. x² + 6x + 8 2. a - 2b√a + b² 3. 49a² - 16b² 4. a²x² + 2axy + y²

5. -(√2 + 2) 6. (50 - 2)(50 + 2) = 2496 7. 52 + 14√3 8. (70 + 3)(70 + 3) = 5329

DIVISION OF A POLYNOMIAL BY A MONOMIAL

Division, like multiplication, may be distributive. Consider, for example, the problem (4 + 6 - 2) - 2, which may be solved by adding the numbers within the parentheses and then dividing the total by 2. Thus,

4 + 6 - 2 = 8 = 4 2 2

Now notice that the problem may also be solved distributively.

4 + 6 - 2 = 4 + 6 - 2 2 2 2 2

= 2 + 3 - 1 = 4

CAUTION: Do not confuse problems of the type just described with another type which is similar in appearance but not in final result. For example, in a problem such as 2 - (4 + 6 - 2), the beginner is tempted to divide 2 successively by 4, then 6, and then -2, as follows:

2 ÷ 2 ÷ 2 - 2 ≠ 2 + 2 - 2 4 + 6 - 2 4 6 2

Notice that we have cancelled the "equals" sign, because 2 - 8 is obviously not equal to 1/2 + 2/6 - 1. The distributive method applies only in those cases in which several different numerators are to be used with the same denominator.

When literal numbers are present in an expression, the distributive method must be used, as in the following two problems:

1. 2ax + aby + a = 2ax + aby + a a a a a

= 2x + by + 1

2. 18ab² - 12bc = 18ab² - 12bc 6b 6b 6b

= 3ab - 2c

Quite often this division may be done mentally, and the intermediate steps need not be written out.

DIVISION OF A POLYNOMIAL BY A POLYNOMIAL

Division of one polynomial by another proceeds as follows:

1. Arrange both the dividend and the divisor in either descending or ascending powers of the same letter. 2. Divide the first term of the dividend by the first term of the divisor and write the result as the first term of the quotient. 3. Multiply the complete divisor by the quotient just obtained, write the terms of the product under the like terms of the dividend, and subtract this expression from the dividend. 4. Consider the remainder as a new dividend and repeat steps 1, 2, and 3.

EXAMPLE:

(10x³ - 7x²y - 16xy² + 12y³) ÷ (5x - 6y)

SOLUTION:

2x² + xy - 2y² 5x - 6y 10x³ - 7x²y - 16xy² + 12y³ 10x³ - 12x²y 5x²y - 16xy² 5x²y - 6xy² -10xy² + 12y³ -10xy² + 12y³

In the example just shown, we began by dividing the first term, 10x³, of the dividend by the first term, 5x, of the divisor. The result is 2x². This is the first term of the quotient.

Next, we multiply the divisor by 2x² and subtract this product from the dividend. Use

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Chapter 9—FUNDAMENTALS OF ALGEBRA

the remainder as a new dividend. Get the sec- ond term, xy, in the quotient by dividing the first term, 5x²y, of the new dividend by the first term, 5x, of the divisor. Multiply the divi- sor by xy and again subtract from the dividend. Continue the process until the remainder is zero or is of a degree lower than the divisor. In the example being considered, the remainder is zero (indicated by double line at the bot- tom). The quotient is 2x² + xy - 2y². The following long division problem is an example in which a remainder is produced:

x² - x + 3 x + 3 | x³ + 2x² + 5 x³ + 3x² - x² - x² - 3x 3x + 5 3x + 9 - 4

The remainder is -4. Notice that the term -3x in the second step of this problem is subtracted from zero, since there is no term containing x in the dividend. When writing down a dividend for long division, leave spaces for missing terms which may en- ter during the long division process. In arithmetic, division problems are often arranged as follows, in order to emphasize the relationship between the remainder and the divisor:

5/2 = 2 + 1/2

This same type of arrangement is used in alge- bra. For example, in the problem just shown, the results could be written as follows:

x³ + 2x² + 5 = x² - x + 3 - 4/(x + 3) x + 3

Remember, before dividing polynomials ar- range the terms in the dividend and divisor according to either descending or ascending powers of one of the literal numbers. When only one literal number occurs, the terms are usually arranged in order of descending powers. For example, in the polynomial 2x² + 4x³ + 5 - 7x the highest power among the literal terms

is x³. If the terms are arranged according to de- scending powers of x, the term in x² should ap- pear first. The x³ term should be followed by the x² term, the x term, and finally the constant term. The polynomial arranged according to descending powers of x is 4x³ + 2x² - 7x + 5. Suppose that 4ab + b² + 15a² is to be divided by 3a + 2b. Since 3a can be divided evenly into 15a², arrange the terms according to descend- ing powers of a. The division takes the form

15a² + 4ab + b²

Synthetic Division

Synthetic division is a shorthand method of dividing a polynomial by a binomial of the form x - a. For example, if 3x³ + 2x² + 2x² - x - 6 is to be divided by x - 1, the long form would be as follows:

3x³ + 5x² + 7x + 6 x - 1 | 3x³ + 2x² + 2x² - x - 6 3x³ - 3x²

+ 5x² + 2x² + 5x² - 5x²

+ 7x² - x + 7x² - 7x

+ 6x - 6 + 6x - 6

Notice that every alternate line of work in this example contains a term which duplicates the one above it. Furthermore, when the sub- traction is completed in each step, these dupli- cated terms cancel each other and thus have no effect on the final result. Another unnecessary duplication results when terms from the divi- dend are brought down and rewritten prior to subtraction. By omitting these duplications, the work may be condensed as follows:

3x³ +5x² +7x +6 x - 1 | 3x³ +2x² +2x² -x -6 -3x³ -5x² -7x -6

+5x³ +7x² +6x 0

The coefficients of the dividend and the con- stant term of the divisor determine the results of each successive step of multiplication and subtraction. Therefore, we may condense still further by writing only the nonliteral factors, as follows:

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MATHEMATICS, VOLUME 1

1 |3 +5 +7 +6 |3 +2 +2 -1 -6 |-3 -5 -7 -6 | |3 +5 +7 +6 0

Notice that if the coefficient of the first term in the dividend is brought down to the last line, then the numbers in the last line are the same as the coefficients of the terms in the quotient. Thus we do not really need to write a separate line of coefficients to represent the quotient. Instead, we bring down the first coefficient of the dividend and make the subtraction "subtracts" serve as coefficients for the rest of the quotient, as follows:

x - 1 |3 3 2 -1 -6 | 3 -5 -7 -6 | |3 5 7 6 0

The unnecessary writing of plus signs is also eliminated here. The use of synthetic division is limited to divisions of the form x - a, in which the degree of x is 1. Thus the degree of each term in the quotient is 1 less than the degree of the corre- sponding term in the dividend. The quotient in this example is as follows:

3x³ + 5x² + 7x + 6

The sequence of operations in synthetic di- vision may be summarized as follows, using as an example the division of 3x - 4x² + x³ - 3 by x + 2:

2 |1 0 -4 3 -3 | 2 -4 0 6 | |1 -2 0 3 -9

First, rearrange the terms of the dividend in descending powers of x. The dividend then becomes x⁴ - 4x² + 3x - 3, with 1 understood as the coefficient of the first term. No x³ term appears in the polynomial, we supply a zero as a place holder for the x³ position. Second, bring down the 1 and multiply it by the +2 of the divisor. Place the result under the zero, and subtract. Multiply the result (-2) by the +2 of the divisor, place the product under the -4 of the dividend, and subtract. Continue this process, finally obtaining 3x - 2x² + 3 as the quotient. The remainder is -9. Practice problems. In the following prob- lems, perform the indicated operations. In 4, 5, and 6, first use synthetic division and then check your work by long division:

1. (a² - 3a² + a) ÷ a

2. x⁶ - 7x⁵ + 4x⁴ x⁴

3. (10x³ - 7x²y - 16xy² + 12y³) ÷ (2x² + xy - 2y²)

4. (x² + 11x + 30) ÷ (x + 6)

5. (12 + x² - 7x) ÷ (x - 3)

6. (a² - 11a + 30) ÷ (a - 5)

Answers:

1. a² - 3a + 1 4. x + 5

2. x⁴ - 7x³ + 4x² 5. x - 4

3. 5x - 6y 6. a - 6

110

CHAPTER 10 FACTORING POLYNOMIALS

A factor of a quantity N, as defined in chap- ter 2 of this course, is any expression which can be divided into N without producing a re- mainder. Thus 2 and 3 are factors of 6, and the factors of 5x are 5 and x. Conversely, when all of the factors of N are multiplied together, the product is N. This definition is extended to include polynomials. The factors of a polynomial are two or more expressions which, when multiplied together, give the polynomial as a product. For example, 3, x, and x2 - 4 are factors of 3x3 - 12x, as the following equation shows:

(3)(x)(x² - 4) = 3x³ - 12x

The factors 3 and x, which are common to both terms of the polynomial 3x³ - 12x, are called COMMON FACTORS. The distributive principle, mentioned in chapters 3 and 9 of this course, is an important part of the concept of factoring. It may be stated as follows:

If the sum of two or more quantities is multi- plied by a third quantity, the product is found by applying the multiplier to each of the origi- nal quantities separately and summing the re- sulting expressions. It is this principle which allows us to separate common factors from the terms of a polynomial. Just as with numbers, an algebraic expres- sion is a prime factor if it has no other factors except itself and 1. The factor x² - 1 is not prime, since it can be separated into x - 2 and x + 2. The factors x - 2 and x + 2 are both prime factors, since they cannot be separated into other factors. The process of finding the factors of a poly- nomial is called FACTORING. An expression is said to be factored completely when it has been separated into its prime factors. The polynomial 3x³ - 12x is factored completely as follows:

3x³ - 12x = 3x(x - 2)(x + 2)

COMMON FACTORS

Factoring any polynomial begins with the removal of common factors. Notice that "re- moval" of a factor does not mean discarding it. To remove a factor is to insert parentheses and move the factor outside the parentheses as a common multiplier. The removal of common factors proceeds as follows: 1. Inspect the polynomial and find the fac- tors which are common to all terms. These common factors, multiplied together, comprise the "largest common factor." 2. Mentally divide each term of the poly- nomial by the largest common factor and write the quotients within a set of parentheses. 3. Write the largest common factor outside the parentheses as a common multiplier. For example, the expression x²y - xy² con- tains xy as a factor of each term. Therefore, it is factored as follows:

x²y - xy² = xy(x - y)

Other examples of factoring by the removal of common factors are found in the following expressions:

6m⁴n + 3m²n² - 3m²n³ = 3m²n(2m² + mn - n²) -5z² - 15z = -5z(z + 3) 7x - 7y + 7z = 7(x - y + z)

In selecting common factors, always remove as many factors as possible from each term in order to factor completely. For example, if a factor of 3ax² - 3ax, so that 3ax² - 3ax is equal to x(3ax - 3a). However, 3 and a are also factors. Thus the largest common factor is 3ax. When factored completely, the expression is as follows:

3ax² - 3ax = 3ax(x - 1)

Practice problems: Remove the common factors:

MATHEMATICS, VOLUME 1

1. y² - y 4. 6mn² + 30m²n Substituting +w(y - z) for -w(z - y) in the origi- nal expression, we may now factor as follows: 2. a³b² - a²b² 5. ⅔x - ⅓y + ⅓ 3. 2b³ - 8b² - 6b

Answers: x(y - z) - w(z - y) = x(y - z) + w(y - z) 1. y(y - 1) 4. 6mn(n + 5m) = (y - z)(x + w)

2. a²b²(a - 1) In factoring an expression such as ax + bx + ay + by, common monomial factors are re- 3. 2b(b² - 4b - 3) 5. ⅓(2x - y + 1) moved first, as follows:

LITERAL EXPONENTS ax + bx + ay + by = x(a + b) + y(a + b)

It is frequently necessary to remove com- Having removed the common monomial factors, mon factors involving literal exponents; that is, we then remove the common binomial factor to exponents composed of letters rather than num- obtain (a + b)(x + y). bers. A typical expression involving literal Notice that we could have rewritten the ex- exponents is x^(a+1) + x, in which x^a is a common pression as ax + ay + bx + by, based on the factor. The factored form is x^a(x + 1). An- commutative law of addition, which states that other example of this type is a^(m+n) + 2a^n. Re- the sum of two or more terms is the same re- member that a^(m+n) is equivalent to a^m · a^n. Thus gardless of the order in which they are ar- the factored form is as follows: ranged. The first step in factoring would then produce a(x + y) + b(x + y) and the final form a^(m+n) + 2a^n = a^n · a^m + 2a^n would be (x + y)(a + b). This is equivalent to = a^n(a^m + 2) (a + b)(x + y), by the commutative law of multi- plication, which states that the product of two BINOMIAL FORM or more factors is the same regardless of the order in which they are arranged. The distinctions between monomial, bino- Practice problems. Factor each of the fol- mial, and trinomial factors are discussed in lowing: detail in chapter 9 of this course. An expres- sion such as a(x + y) + b(x + y) has a common 1. x³ + 3x² factor in binomial form. The factor (x + y) can 2. xy² + y + x²y + x be removed from both terms, with the following 3. e^x + 4e^(4x) result: 4. 7(x² + y²) - 3z(x² + y²) a(x + y) + b(x + y) = (x + y)(a + b) 5. a² + ab - ac - cb Sometimes it is easier to see this if a single letter is substituted temporarily for the bino- 6. ⅓e²r - ⅓er² mial. Thus, if (x + y) = n, so that a(x + y) + b(x + y) reduces to (an + bn). The factored 7. a^(x+2) + a² form is n(a + b), which becomes (x + y)(a + b) 8. xy - 3x - 2y + 6 when n is replaced by its equal, (x + y). Another form of this type is x(y - z) - w(z - y). Answers: Notice that this expression could be factored easily if the binomial in the second term were 1. x³(x² + 3) (y - z). We can show that -w(z - y) is equiva- 2. (xy + 1)(x + y) lent to +w(y - z), as follows: 3. e^x(1 + 4e^(3x)) -w(z - y) = -w [(-1) · (-1) · z + (-1) · y] = -w [(-1) {(-1) z + y}] 4. (x² + y²)(7 - 3z) = (-w)(-1) [-z + y] 5. (a + b)(a - c) = +w(y - z) 6. ⅓er(e - ⅓r)

7. a²(a^x + 1)

8. (y - 3)(x - 2)

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Chapter 10–FACTORING POLYNOMIALS

BINOMIAL FACTORS 1. a² - b² 5. x² - y² 2. b² - 9 6. y² - 36 After any common factor has been removed 3. a²b² - 1 7. 1 - 4y² from a polynomial, the remaining polynomial 4. a² - 144 8. 9a² - 16 factor must be examined further for other fac- tors. Skill in factoring is principally the ability Answers: to recognize certain types of products such as the square of a sum or difference. Therefore, 1. (a + b)(a - b) 5. (x + y)(x - y) it is important to be familiar with the special 2. (b + 3)(b - 3) 6. (y + 6)(y - 6) products discussed in chapter 9. 3. (ab + 1)(ab - 1) 7. (1 + 2y)(1 - 2y) 4. (a + 12)(a - 12) 8. (3a + 4)(3a - 4) DIFFERENCE OF TWO SQUARES SPECIAL BINOMIAL FORMS In chapter 9 we learned that the product of the sum and difference of two numbers is the Special cases involving binomial expressions difference of their squares. Thus, (a + b)(a - b) are frequently encountered. All such expres- = a² - b². Conversely, if a binomial is the dif- sions may be factored by reference to general ference of two squares, its factors are the sum formulas, but these formulas are beyond the and difference of the square roots. For exam- scope of this course. For our purposes, analy- ple, in 9a² - 4b² both 9a² and 4b² are perfect sis of some special cases will be sufficient. squares. The square roots are 3a and 2b, re- spectively. Connect these square roots with a Even Exponents plus sign to get one factor of 9a² - 4b² and with a minus sign to get the other factor. The two When the exponents on both terms of the bi- binomial factors are 3a - 2b and 3a + 2b. There- nomial are even, the expression may be treated fore, factored completely, the binomial can be as the sum or difference of two squares. For written as follows: example, x⁶ - y⁶ can be rewritten as (x³)² - (y³)² which results in the following factored 9a² - 4b² = (3a - 2b)(3a + 2b) form:

We may check to see if these factors are x⁶ - y⁶ = (x³ - y³)(x³ + y³) correct by multiplying them together to see if their product is the original binomial. In general, a binomial with even exponents The expression 20x³y - 5xy³ reduces to the has the form x²ⁿ - y²ⁿ, since all even numbers difference of two squares after the common have 2 as a factor. If the connecting sign is factor 5xy is removed. Completely factored, positive, the expression may not be factorable; this expression produces the following: for example, x² + y², x⁴ + y⁴, and x⁸ + y⁸ are all nonfactorable binomials. If the connecting 20x³y - 5xy³ = 5xy(4x² - y²) sign is negative, a binomial with even exponents = 5xy(2x - y)(2x + y) is factorable as follows:

Other examples that show the difference of x²ᵐ - y²ⁿ = (xᵐ - yⁿ)(xᵐ + yⁿ) two squares in factored form are as follows: A special case which is particularly impor- 49 - 16 = (7 + 4)(7 - 4) tant because it occurs so often is the binomial which has the numeral 1 as one of its terms. 16a² - 4x² = 4(4a² - x²) For example, x⁴ - 1 is factorable as the differ- = 4(2a + x)(2a - x) ence of two squares, as follows:

4x²y - 9y = y(4x² - 9) x⁴ - 1 = (x² - 1)(x² + 1) = y(2x + 3)(2x - 3) = (x - 1)(x + 1)(x² + 1)

Practice problems: Factor each of the fol- lowing:

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MATHEMATICS, VOLUME 1

Odd Exponents Answers:

Two special cases involving odd exponents 1. (x + y)(x - y)(x² + y²) are of particular importance. These are the 2. (m + n)(m² - mn + n²) sum of two cubes and the difference of two 3. (x + y)(x - y)(x² + xy + y²)(x² - xy + y²) cubes. Examples of the sum and difference of 4. (x - y)(x² + xy + y²) two cubes, showing their factored forms, are 5. (a - b)(a² + ab + b²)(a⁶ + a³b³ + b⁶) as follows: 6. (xᵐ - yⁿ)(xᵐ + yⁿ) 7. (1 + x²)(1 - x)(1 + x) x³ + y³ = (x + y)(x² - xy + y²) 8. (x² + 1)(x⁴ - x² + 1) x³ - y³ = (x - y)(x² + xy + y²) 9. (1 - x)(1 + x + x²)

Notice that each of these factored forms in- volves a first degree binomial factor ((x + y) TRINOMIAL SQUARES in the first case and (x - y) in the second). The connecting sign in the first degree binomial A trinomial that is the square of a binomial factor corresponds to the connecting sign in the is called a TRINOMIAL SQUARE. Trinomials original unfactored binomial. that are perfect squares factor into either the We are now in a position to give the com- square of a sum or the square of a difference. pletely factored form of x⁶ - y⁶, as follows: Recalling that (x + y)² = x² + 2xy + y² and (x - y)² = x² - 2xy + y², the form of a trinomial x⁶ - y⁶ = (x³ - y³)(x³ + y³) square is apparent. The first term and the last = (x - y)(x² + xy + y²) term are perfect squares and their signs are (x + y)(x² - xy + y²) positive. The middle term is twice the product of the square roots of these two numbers. The In general, (x + y) is a factor of (xⁿ + yⁿ) if sign of the middle term is plus if a sum has n is odd. If n is even, (xⁿ + yⁿ) is not factor- been squared; it is minus if a difference has able unless it can be expressed as the sum been squared. of two cubes. When the connecting sign is nega- The polynomial 16x² - 8xy + y² is a trino- tive, the binomial is always factorable if n is mial in which the first term, 16x², and the last a whole number greater than 1. Thus (x - y) term, y², are perfect squares with positive is a factor of (xⁿ - yⁿ) for both odd and even signs. The square roots are 4x and y. Twice values of n. the product of these square roots is 2(4x)(y) = The special case in which one of the terms 8xy. The middle term is preceded by a minus of the binomial is the numeral 1 occurs fre- sign indicating that a difference has been quently. An example of this is x³ + 1, which is squared. In factored form this trinomial is factorable as the sum of two cubes, as follows: as follows:

x³ + 1 = (x + 1)(x² - x + 1) 16x² - 8xy + y² = (4x - y)²

In a similar manner, 1 + x⁶ can be treated To factor the trinomial, we simply take the as the sum of two cubes and factored as follows: square roots of the end terms and join them with a plus sign if the middle term is preceded 1 + x⁶ = 1 + (x²)³ by a plus or with a minus if the middle term is = (1 + x²)(1 - x² + x⁴) preceded by a minus. The terms of a trinomial may appear in any Practice problems. In each of the following order. Thus, 8xy + y² + 16x² is a trinomial problems, factor completely: square and may be factored as follows:

1. x⁴ - y⁴ 4. x³ - y³ 7. 1 - x⁴ 2. m³ + n³ 5. a⁹ - b⁹ 8. x⁶ + 1 3. x⁶ - y⁶ 6. x²ᵃ - y²ᵇ 9. 1 - x³

8xy + y² + 16x² = 16x² + 8xy + y² = (4x + y)²

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Practice problems. Among the following expressions, factor those which are trinomial squares:

1. y² - 8y + 16 5. 12y + 9y² - 4 2. 16y² + 30x + 9 6. 4x² + y² + 4xy 3. 36 + 12x + x² 7. 9 - 6cd + c²d² 4. a² + 2ab + b² 8. x⁴ + 4x² + 4

Answers: 1. (y - 4)² 5. Not a trinomial 2. Not a trinomial square square 6. (2x + y)² 3. (6 + x)² 7. (3 - cd)² 4. (a + b)² 8. (x² + 2)²

SUPPLYING THE MISSING TERM

Skill in recognizing trinomial squares may be improved by practicing the solution of prob- lems which require supplying a missing term. For example, the expression y² + (?) + 16 can be made to form a perfect trinomial square by supplying the correct term to fill the paren- theses.

The middle term must be twice the product of the square roots of the two perfect square terms; that is, (2)(4)(y), or 8y. Check: y² + 8y + 16 = (y + 4)². The missing term is 8y.

Suppose that we wish to supply the missing term in 16x² + 24xy + (?) so that the three terms will form a perfect trinomial square. The square root of the first term is 4x. One- half the middle term is 12xy. Dividing 12xy by 4x. The result is 3y which is the square root of the last term. Thus, our missing term is 9y². Checking, we find that (4x + 3y)² = 16x² + 24xy + 9y².

Practice problems. In each of the following problems, supply the missing term to form a perfect trinomial square:

1. x² + (?) + y² 4. 4m² + 16m + (?) 2. t² + (?) + 25 5. x² + 4x + (?) 3. 9a² - (?) + 25b² 6. c² - 6cd + (?)

Answers: 1. 2xy 4. 16 2. 10t 5. 4 3. 30ab 6. 9d²

OTHER TRINOMIALS

It is sometimes possible to factor trinomials that are not perfect squares. Following are some examples of such trinomials, and the ex- pressions of which they are products:

1. (x + 3)(x + 4) = x² + 7x + 12 2. (x - 3)(x - 4) = x² - 7x + 12 3. (x - 3)(x + 4) = x² + x - 12 4. (x + 3)(x - 4) = x² - x - 12

It is apparent that trinomials like these may be factored into binomials as shown. Notice how the trinomial in each of the preceding ex- amples is formed. The first term is the square of the common term of the binomial factors. The second term is the algebraic sum of their unlike terms times their common term. The third term is the product of their unlike terms.

Such trinomials may be factored as the prod- uct of two binomials if there are two numbers such that their algebraic sum is the coefficient of the middle term and their product is the last term.

For example, let us factor the expression x² - 12x + 32. If the expression is factorable, there will be a common term, x, in each of the binomial factors. We begin factoring by placing this term within each set of parentheses, as follows:

(x )(x )

Next, we must find the other terms that are to go in the parentheses. They will be two num- bers such that their algebraic sum is -12 and their product is -32. We see that -8 and -4 satisfy the conditions. Thus, the following ex- pression results:

x² - 12x + 32 = (x - 8)(x - 4)

It is of value in factoring to note some use- ful facts about trinomials. If both the second and third terms of the trinomial are positive, the signs of the terms to be found are positive as in example 1 of this section. If the second term is negative and the last is positive, both terms to be found will be negative as in exam- ple 2. If the third term of the trinomial is neg- ative, one of the terms to be found is positive and the other is negative as in examples 3 and 4. Concerning this last case, if the second term is

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MATHEMATICS, VOLUME 1

positive as in example 3, the positive term in the factors has the greater numerical value. If the second term is negative as in example 4, the negative term in the factors has the greater numerical value.

It should be remembered that not all trino- mials are factorable. For example, x² + 4x + 2 cannot be factored since there are no two ra- tional numbers whose product is 2 and whose sum is 4.

Practice problems. Factor completely, in the following problems:

1. y² + 15y + 50 5. x² - 12x - 45 2. y² - 2y - 24 6. x² - 15x + 56 3. x² + 8x - 48 7. x² + 2x - 48 4. x² - 4x - 60 8. x² + 14x + 24

Answers: 1. (y + 5)(y + 10) 5. (x - 15)(x + 3) 2. (y - 6)(y + 4) 6. (x - 7)(x - 8) 3. (x + 12)(x - 4) 7. (x - 6)(x + 8) 4. (x - 10)(x + 6) 8. (x + 12)(x + 2)

Thus far we have considered only those ex- pressions in which the coefficient of the first term is 1. When the coefficient of the first term is other than 1, the expression can be fac- tored as shown in the following example:

6x² - x - 2 = (2x + 1)(3x - 2)

Although this result can be obtained by the trial and error method, the following procedure saves time and effort. First, find two numbers whose sum is the coefficient of the second term (-1 in this example) and whose product is equal to the product of the third term and the coeffi- cient of the first term (in this example, (6)(-2) or -12). By inspection, the desired numbers are found to be -4 and +3. Using these two numbers as coefficients for x, we can rewrite the original expression as 6x² - 4x + 3x - 2 and factor as follows:

6x² - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x + 1)(3x - 2)

Practice problems. Factor completely, in the following problems:

1. 2x² + 13x + 21 3. 15x² - 16x - 7 2. 16x² + 26x + 3 4. 12x² - 8x - 15

Answers: 1. (2x + 7)(x + 3) 3. (3x + 1)(5x - 7) 2. (2x + 3)(8x + 1) 4. (6x + 5)(2x - 3)

REDUCING FRACTIONS TO LOWEST TERMS

There are many useful applications of fac- toring. One of the most important is that of simplifying algebraic fractions. Fractions that contain algebraic expressions in the numerator or denominator, or both, can be reduced to lower terms, if there are factors common to numerator and denominator. If the terms of a fraction are monomials, common factors are immediately apparent, as in the following ex- pression:

3x²y 3xy(x) x —— = ——— = — 6xy 3xy(2) 2

If the terms of a fraction are polynomials, the polynomials must be factored in order to recognize the existence of common factors, as in the following two examples:

a - b a - b 1. ————— = ————— = —— a² - 2ab + b² (a - b)(a - b) (a - b)

4x² - 9 (2x + 3)(2x - 3) (2x + 3) 2. ————— = —————— = —— 6x² - 9x 3x(2x - 3) 3x

Notice that without the valuable process of fac- toring, we would be forced to use the fractions in their more complicated form. When there are factors common to both numerator and de- nominator, it is obviously more practical to cancel them (first using the factoring process) before proceeding.

Practice problems. Reduce to lowest terms in each of the following:

12 x² - 25 1. ————— 4. ————— 6x + 12 x² - 8y - 15

a² - b² a² - 5a - 24 2. ————— 5. ————— a² - 2ab + b² a² - 64

y² - 14y + 45 4x²y - 9y 3. ————— 6. ————— y² - 8y - 9 4x² + 12x + 9

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Chapter 10–FACTORING POLYNOMIALS

Answers:

1. 2/(x + 2)

2. (a + b)/(a - b)

3. (y - 5)/(y + 1)

4. (y + 5)/(y - 3)

5. (a + 3)/(a + 6)

6. y(2x - 3)/(2x + 3)

OPERATIONS INVOLVING FRACTIONS

Addition, subtraction, multiplication, and division operations involving algebraic fractions are often simplified by means of factoring, whereas they would be quite complicated without the use of factoring.

MULTIPLYING FRACTIONS

Multiplication of fractions that contain polynomials is similar to multiplication of fractions that contain only arithmetic numbers. If this fact is kept in mind, the student will have little difficulty in mastering multiplication in algebra. For instance, we recall that to multiply a fraction by a whole number, we simply multiply the numerator by the whole number. This is illustrated in the following example:

Arithmetic: 4 x 3/17 = 12/17

Algebra: (x - 4) · x/(x² - 5) = 3x - 12/(x² - 5)

Sometimes the work may be simplified by factoring and canceling before carrying out the multiplication. The following example illustrates this:

(2a - 8) · 3/(a² - 8a + 16) = 2(a - 4)/(a - 4)(a - 4) · 3/(a - 4)(a - 4)

= 2(3)/(a - 4) = 6/(a - 4)

When the multiplier is a fraction, the rules of arithmetic remain applicable—that is, multiply numerators together and denominators together. This is illustrated as follows:

Arithmetic: 4/5 x 2/3 = 8/15

Algebra: (a + b)/(a - b) · a/(a - b) = a(a + b)/(a - b)²

Where possible, the work may be considerably reduced by factoring, canceling, and then carrying out the multiplication, as in the following example:

(x² - 2x + 1)/(x² - 9) · (x² + x - 6)/(x² - 1)

= (x - 1)(x - 1)/(x - 3)(x + 3) · (x - 3)(x + 2)/(x + 1)(x - 1)

= (x - 1)(x - 2)/(x - 3)(x + 1) = (x² - 3x + 2)/(x² - 2x - 3)

Although the factors may be multiplied to form two trinomials as shown, it is usually sufficient to leave the answer in factored form.

Practice problems. In the following problems, multiply as indicated:

1. 5a² · 3b/(a + b)

2. (x + y)/x² · (x - y)/(x - 1)

3. (a² + 2ab + b²)/(a² - b²) · 6a/(3a + 3b)

4. (a - 1)/(2a² + 4a + 2) · (a + 1)²/(a - 1)

Answers:

1. 15a²b/(a + b)

2. (x² - y²)/(x³ - x²)

3. 2a/(a - b)

4. 1/2

DIVIDING FRACTIONS

The rules of arithmetic apply to the division of algebraic fractions; as in arithmetic, simply invert the divisor and multiply, as follows:

Arithmetic: 3/8 ÷ 9/16 = 3/8 × 16/9

= 2/8 · {8}(2)/{8}(3) = 2/3

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MATHEMATICS, VOLUME 1

Algebra: (x - 3y)/(x + 3y) ÷ (x² - 6xy + 9y²)/(x² + 7xy + 12y²)

= (x - 3y)/(x + 3y) · (x² + 7xy + 12y²)/(x² - 6xy + 9y²)

= (x - 3y)/[(x + 3y)] · [(x + 3y)(x + 4y)]/[(x - 3y)(x - 3y)]

= (x + 4y)/(x - 3y)

Practice problems. In the following problems, divide and reduce to lowest terms:

1. (x - 2)/(x² + 4x + 4) ÷ 1/(x² - 4)

2. (2a - 1)/(a³ + 3a) ÷ (a + 1)/(a² + 3)

3. (a³ - 4a² + 3a)/(a + 2) ÷ (a - 3)

4. (6t + 12)/(9t² + 6t - 24) ÷ (8t - 12)/(15t - 20)

Answers:

1. (x - 2)²/(x + 2)

2. (2a - 1)/(a² + a)

3. a(a - 1)/(a + 2)

4. 5/(4t - 6)

ADDING AND SUBTRACTING FRACTIONS

The rules of arithmetic for adding and subtracting fractions are applicable to algebraic fractions. Fractions that are to be combined by addition or subtraction must have the same denominator. The numerators are then combined according to the operation indicated and the result is placed over the denominator. For example, in the expression

(x - 4)/(x - 2) + (2 - 11x)/(2 - x)

the second denominator will be the same as the first, if its sign is changed. The value of the fraction will remain the same if the sign of the numerator is also changed. Thus, we have the following simplification:

(x - 4)/(x - 2) + (2 - 11x)/(2 - x) = (x - 4)/(x - 2) + (-2 - 11x)/(-2 - x)

= (x - 4)/(x - 2) + (11x - 2)/(x - 2)

= (x - 4 + 11x - 2)/(x - 2)

= (12x - 6)/(x - 2)

= 6(2x - 1)/(x - 2)

When the denominators are not the same, we must reduce all fractions to be added or subtracted to a common denominator and then proceed.

Consider, for example,

4/(x² - 4) + 3/(x² - 4x - 12)

We first must find the least common denominator (LCD). Remember this is the least number that is exactly divisible by each of the denominators. To find such a number, as in arithmetic, we first separate each of the denominators into prime factors. The LCD will contain all the various prime factors, each one as many times as it occurs in any of the denominators.

Factoring, we have

4/[(x + 2)(x - 2)] + 3/[(x - 6)(x + 2)]

and the LCD is (x + 2)(x - 2)(x - 6). Rewriting the fractions with this denominator and adding numerators, we have the following expression:

4(x - 6)/[(x + 2)(x - 2)(x - 6)] + 3(x - 2)/[(x + 2)(x - 2)(x - 6)]

= [4(x - 6) + 3(x - 2)]/LCD

= (4x - 24 + 3x - 6)/LCD

= (7x - 30)/[(x + 2)(x - 2)(x - 6)]

As another example, consider

4/(x + 3) - (x + 2)/(x² + 4x + 3)

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Chapter 10—FACTORING POLYNOMIALS

Factoring the denominator of the second fraction, we find that the LCD is (x + 3)(x + 1). Rewriting the original fractions with the LCD as denominator, we may now combine the fractions as follows:

4(x + 1) (x + 2) ————————— - ————————— (x + 3)(x + 1) (x + 3)(x + 1)

4x + 4 - x - 2 = —————————————— (x + 3)(x + 1)

3x + 2 = ———————————— (x + 3)(x + 1)

Practice problems. Perform the indicated operations in each of the following problems:

3x - 4 x - 2 1. ————————— - ——————— x² + x - 2 x - 1

3a 3 2. ———— - ——— a² - 9 3 - a

x - 3 x + 2 3. ——— + ——— 3x 2x

1 1 4. ———— - ——— a² - 1 a + 1

3 2 1 5. ———— - ———— + ——— (a + 4)² a(a + 4) 6(a + 4)

Answers:

3x - x² 1. ———————————— (x + 2)(x - 1)

6a + 9 2. —————————— (a + 3)(a - 3)

5 3. ——— 6

2 - a³ + a² - a 4. ——————————————— (a² + 1)(a + 1)(a - 1)

a² + 10a + 48 5. ——————————— 6a(a + 4)²

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CHAPTER 11 LINEAR EQUATIONS IN ONE VARIABLE

One of the principal reasons for an intensive study of polynomials, grouping symbols, factoring, and fractions is to prepare for solving equations. The equation is perhaps the most important tool in algebra, and the more skillful the student becomes in working with equations, the greater will be his ease in solving problems. Before learning to solve equations, it is necessary to become familiar with the words used in the discussion of them. An EQUATION is a statement that two expressions are equal in value. Thus,

4 + 5 = 9

and

A = lw

(Area of a rectangle = length x width)

are equations. The part to the left of the equality sign is called the LEFT MEMBER, or first member, of the equation. The part to the right is the RIGHT MEMBER, or second member, of the equation.

The members of an equation are sometimes thought of as a corresponding two weights that balance a scale. (See Fig. 11-1.) This comparison is often helpful to students who are learning to solve equations. It is obvious, in

the case of the scale, that any change made in one pan must be accompanied by an equal change in the other pan. Otherwise the scale will not balance. Operations on equations are based on the same principle. The members must be kept balanced or the equality is lost.

CONSTANTS AND VARIABLES

Expressions in algebra consist of constants and variables. A CONSTANT is a quantity whose value remains the same throughout a particular problem. A VARIABLE is a quantity whose value is free to vary.

There are two kinds of constants—fixed and arbitrary. Numbers such as 7, -3, 1/2, and π are examples of FIXED constants. Their values never change. In 5x + 7 = 0, the numbers 0, 5, and 7, are fixed constants.

ARBITRARY constants can be assigned different values for different problems. Arbitrary constants are indicated by letters—quite often letters at the beginning of the alphabet such as a, b, c, and d. In

ax + b = 0 ,

the letters a and b represent arbitrary constants. The form ax + b = 0 represent many linear equations. If we give a and b particular values, say a = 5 and b = 7, then these constants become fixed, for this particular problem, and the equation becomes

5x + 7 = 0

A variable may have one value or it may have many values in a discussion. The letters at the end of the alphabet, such as x, y, z, and w, usually are used to represent variables. In 5x + 7, the letter x is the variable. If x = 1, then

5x + 7 = 5 + 7 = 12

If x = 2, then

5x + 7 = 5(2) + 7 = 10 + 7 = 17

[Figure 11-1. Equation compared to a balance scale.]

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Chapter 11— LINEAR EQUATIONS IN ONE VARIABLE

and so on for as many values of x as we desire to select. If the expression 5x + 7 is set equal to some particular number, say -23, then the resulting equality

5x + 7 = -23

holds true for just one value of x. The value is -6, since

5(-6) + 7 = -23

In an algebraic expression, terms that con- tain a variable are called VARIABLE TERMS. Terms that do not contain a variable are CON- STANT TERMS. The expression 5x + 7 con- tains one variable term and one constant term. The variable term is 5x, while 7 is the constant term. In ax + b, ax is the variable term and b is the constant term. A variable term often is designated by nam- ing the variable it contains. In 5x + 7, 5x is the x-term. In ax + by, ax is the x-term, while by is the y-term.

DEGREE OF AN EQUATION

The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. The equation

3x - 17 = 0

is a FIRST-DEGREE equation, since x is raised only to the first power. An example of a SECOND-DEGREE equa- tion is

5x² - 2 x + 1 = 0.

The equation,

4x³ - 7x² = 0,

is of the THIRD DEGREE . The equation,

3x - 2y = 5

is of the first degree in two variables, x and y. When more than one variable appears in a term, as in xy = 5, it is necessary to add the expo- nents of the variables within a term to get the

degree of the equation. Since 1 + 1 = 2, the equation xy = 5 is of the second degree.

LINEAR EQUATIONS

Graphs are used in many different forms to give visual pictures of certain related facts. For example, they are used to show business trends, production output, continued individual attainment, and so forth. We find bar graphs, line graphs, circle graphs, and many other types, each of which is used for a particular need. In algebra, graphs are also used to give a visual picture containing a great deal of in- formation about equations. Sometimes many numerical values, when substituted for the variables of an equation, will satisfy the conditions of the equation. On a particular type of graph (which will be explained fully in chapter 12) several of these values are plotted (located), and when enough are plotted, a line is drawn through these points. For each particular equation a certain type of curve re- sults. For equations in the first degree in one or two variables, the resulting shape of the "curve" is a straight line. . Thus, the name LINEAR EQUATION is derived. Equations of a higher degree form other shapes. The name "linear equation" now applies to equations of the first degree, regardless of the number of variables they contain. Chapter 12 shows how an equation may be pictured on a graph. The purpose and value of graphing an equation will also be developed.

IDENTITIES

If a statement of equality involves one or more variables, it may be either an IDENTITY (identical equation) or a CONDITIONAL EQUA- TION. An identity is an equality that states a fact, such as the following examples:

1. 9 + 5 = 14

2. 2n + 5n = 7n

3. 6(x - 3) = 6x - 18

Notice that equation 3 merely shows the fac- tored form of 6x - 18 and holds true when any value of x is substituted. For example, if x = 5, it becomes

6(5-3) = 6(5) - 18

6(2) = 30 - 18

12 = 12

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MATHEMATICS, VOLUME 1

If x assumes the negative value -10, this iden- tity becomes

6(-10-3) = 6(-10)-18

6(-13) = -60-18

-78 = -78

An identity is established when both sides of the equality have been reduced to the same number or the same expression. When 5 is substituted for x, the value of either side of 6(x-3) = 6x - 18 is 12. When -10 is substituted for x, the value on either side is -78. The fact that this equality is an identity can be shown also by factoring the right side so that the equality becomes

6(x-3) = 6(x-3)

The expressions on the two sides of the equality are identical.

CONDITIONAL EQUATIONS

A statement such as 2x-1 = 0 is an equality only when x has one particular value. Such a statement is called a CONDITIONAL EQUA- TION, since it is true only under the condition that x = 1/2. Likewise, the equation y - 7 = 8 holds true only if y = 15. The value of the variable for which an equa- tion in one variable holds true is a ROOT, or SOLUTION, of the equation. When we speak of solving equations in algebra, we refer to condi- tional equations. The solution of a conditional equation can be verified by substituting for the variable its value, as determined by the solution. The solution is correct if the equality re- duces to an identity. For example, if 1/2 is substituted for x in 2x - 1 = 0, the result is

2(1/2) - 1 = 0

1 - 1 = 0

0 = 0 (an identity)

The identity is established for x = 1/2, since the value of each side of the equality reduces to -ero.

SOLVING LINEAR EQUATIONS

Solving a linear equation in one variable means finding the value of the variable that makes the equation true. For example, 11 is the SOLUTION of x - 7 = 4, since 11 - 7 = 4. The number 11 is said to SATISFY the equation. Basically, the operation used in solving equa- tions is to manipulate both members, by adi- tion, subtraction, multiplication, or division until the value of the variable becomes apparent. This manipulation may be accomplished in a straightforward manner by use of the axioms outlined in chapter 3 of this course. These axioms may be summed up in the following rule: If both members of an equation are in- creased, decreased, multiplied, or divided by the same number, or by equal numbers, the re- sults will be equal. (Division by zero is ex- cluded.) As mentioned earlier, an equation may be compared to a balance. What is done to one member must also be done to the other to main- tain a balance. An equation must always be kept in balance or the equality is lost. We use the above rule to remove or adjust terms and coefficients until the value of the variable is discovered. Some examples of equations solved by means of the four operations mentioned in the rule are given in the following paragraphs.

ADDITION

Find the value of x in the equation

x - 3 = 12

As in any equation, we must isolate the variable on either the right or left side. In this prob- lem, we leave the variable on the left and per- form the following steps:

1. Add 3 to both members of the equation, as follows:

x - 3 + 3 = 12 + 3

In effect, we are "undoing" the subtraction indi- cated by the expression x - 3, for the purpose of isolating x in the left member.

2. Combining terms, we have

x = 15

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Chapter 11—LINEAR EQUATIONS IN ONE VARIABLE

SUBTRACTION

Find the value of x in the equation

x + 14 = 24

1. Subtract 14 from each member. In effect, this undoes the addition indicated in the expression x + 14.

x + 14 - 14 = 24 - 14

2. Combining terms, we have

x = 10

MULTIPLICATION

Find the value of y in the equation

y/5 = 10

1. The only way to remove the 5 so that the y can be isolated is to undo the indicated division. Thus we use the inverse of division, which is multiplication. Multiplying both members by 5, we have the following:

5(y/5) = 5(10)

2. Performing the indicated multiplications, we have

y = 50

DIVISION

Find the value of x in the equation

3x = 15

1. The multiplier 3 may be removed from the x by dividing the left member by 3. This must be balanced by dividing the right member by 3 also, as follows:

3x/3 = 15/3

2. Performing the indicated divisions, we have

x = 5

Practice problems. Solve the following equations:

1. m + 2 = 8 4. x/14 = 2

2. x - 5 = 11 5. 2n = 5

3. 6x = -48 6. 1/6y = 6

Answers:

1. m = 6 4. x = 28

2. x = 16 5. n = 2 1/2

3. x = -8 6. y = 36

SOLUTIONS REQUIRING MORE THAN ONE OPERATION

Most equations involve more steps in their solutions than the simple equations already described, but the basic operations remain unchanged. If the basic axioms are kept well in mind, these more complicated equations will not become too difficult. Equations may require one or all of the basic operations before a solution can be obtained.

Subtraction and Division

Find the value of x in the following equation:

2x + 4 = 16

1. The term containing x is isolated on the left by subtracting 4 from the left member. This operation must be balanced by also subtracting 4 from the right member, as follows:

2x + 4 - 4 = 16 - 4

2. Performing the indicated operations, we have

2x = 12

3. The multiplier 2 is removed from the x by dividing both sides of the equation by 2, as follows:

2x/2 = 12/2

x = 6

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MATHEMATICS, VOLUME 1

Addition, Multiplication, and Division

Find the value of y in the following equation:

3y/2 - 4 = 11

1. Isolate the term containing y on the left by adding 4 to both sides, as follows:

3y/2 - 4 + 4 = 11 + 4

3y/2 = 15

2. Since the 2 will not divide the 3 exactly, multiply the left member by 2 in order to eliminate the fraction. This operation must be balanced by multiplying the right member by 2, as follows:

2(3y/2) = 2(15)

3y = 30

3. Divide both members by 3, in order to isolate the y in the left member, as follows:

3y/3 = 30/3

y = 10

Equations Having the Variable in More Than One Term

Find the value of x in the following equation:

3x/4 + x = 12 - x

1. Rewrite the equation with no terms containing the variable in the right member. This requires adding x to the right member to eliminate the -x term, and balance requires that we also add x to the left member, as follows:

3x/4 + x + x = 12 - x + x

3x/4 + 2x = 12

2. Since the 4 will not divide the 3 exactly, it is necessary to multiply the first term by 4

to eliminate the fraction. However, notice that this multiplication cannot be performed on the first term only; any multiplier which is introduced for simplification purposes must be applied to the entire equation. Thus each term in the equation is multiplied by 4, as follows:

4(3x/4) + 4(2x) = 4(12)

3x + 8x = 48

3. Add the terms containing x and then divide both sides by 11 to isolate the x in the left member, as follows:

11x = 48

x = 48/11

= 4 7/11

Practice problems. Solve each of the following equations:

1. x - 1 = 1/2 4. 4 - 7x = 9 - 8x

2. x/3 + y = 8 5. y/2 + 6y = 13

3. x/4 + 3x = 7 6. 1/2x - 2x = 25 + x

Answers:

1. x = 3/2 4. x = 5

2. y = 6 5. y = 2

3. x = 28/13 6. x = -10

EQUATIONS WITH LITERAL COEFFICIENTS

As stated earlier, the first letters of the alphabet usually represent known quantities (constants), and the last letters represent unknown quantities (variables). Thus, we usually solve for x, y, or z. An equation such as

ax - 8 = bx - 5

has letters as coefficients. Equations with literal coefficients are solved in the same way as

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equations with numerical coefficients, except that when an operation cannot actually be per- formed, it merely is indicated. In solving for x in the equation

ax - 8 = bx - 5

subtract bx from both members and add 8 to both members. The result is

ax - bx = 8 - 5

Since the subtraction on the left side cannot actually be performed, it is indicated. The quantity, a - b, is the coefficient of x when terms are collected. The equation takes the form

(a-b) x = 3

Now divide both sides of the equation by a-b. Again the result can be only indicated. The solution of the equation is

x = 3 a-b

In solving for y in the equation

ay + b = 4

subtract b from both members as follows:

ay = 4 - b

Dividing both members by a, the solution is

y = 4-b a

Practice problems. Solve for x in each of the following:

1. 3 + x = b 3. 3x + 6m = 7m

2. 4x = 8 + t 4. ax - 2(x + b) = 3a

Answers:

1. x = b - 3 3. x = m 3

2. x = 8 + t 4. x = 3a + 2b 4 a - 2

REMOVING SIGNS OF GROUPING

If signs of grouping appear in an equation they should be removed in the manner indicated in chapter 9 of this course. For example, solve the equation

5 = 24 - [x-12(x-2) - 6(x-2)]

Notice that the same expression, x-2, occurs in both parentheses. By combining the terms con- taining (x-2), the equation becomes

5 = 24 - [x-18(x-2)]

Next, remove the parentheses and then the bracket, obtaining

5 = 24 - [x-18x + 36] = 24 - [36 - 17x] = 24 - 36 + 17x = -12 + 17x

Subtracting 17x from both members and then subtracting 5 from both members, we have

-17x = - 12 - 5 -17x = - 17

Divide both members by -17. The solution is

x = 1

EQUATIONS CONTAINING FRACTIONS

To solve for x in an equation such as

2x + x - 1 = x + x 3 12 4 2

first clear the equation of fractions. To do this, find the least common denominator of the fractions. Then multiply both sides of the equa- tion by the LCD. The least common denominator of 3, 12, 4, and 2 is 12. Multiply both sides of the equation by 12. The resulting equation is

8x + x - 12 = 3 + 6x

Subtract 6x from both members, add 12 to both members, and collect like terms as follows:

6x - 6x = 12 + 3

3x = 15

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MATHEMATICS, VOLUME 1

The solution is

x = 5

To prove that x = 5 is the correct solution, substitute 5 for x in the original equation and show that both sides of the equation reduce to the same value. The result of substitution is

2(5) + 5 - 1 = 1 + 5 3 12 4 2

In establishing an identity, the two sides of the equality are treated separately, and the op- erations are performed as indicated. Some- times, as here, fractions occur on both sides of the equality, and it is desirable to find the least common denominator for more than one set of fractions. The same denominator could be used on both sides of the equality, but this might make some of the terms of the fractions larger than necessary.

Proceeding in establishing the identity for x = 5 in the foregoing equation we obtain

10 + 5 - 3 = 1 + 10 3 12 3 4 4

7 + 5 = 11 3 12 4

28 + 5 = 11 12 12 4

33 = 11 12 4

11 = 11 4 4

Each member of the equality has the value 11/4 when x = 5. The fact that the equation be- comes an identity when x is replaced by 5 proves that x = 5 is the solution.

Practice problems. Solve each of the fol- lowing equations:

1. x - 2 = x 4 6

2. 1 - 1 = 1 2 v 3

3. x - y = 5 2 3

4. 3 = 6 4x

Answers:

1. x = 24 3. y = 30

2. v = 6 4. x = 1/8

GENERAL FORM OF A LINEAR EQUATION

The expression GENERAL FORM, in mathe- matics, implies a form to which all expressions or equations of a certain type can be reduced. The only possible terms in a linear equation in one variable are the first-degree term and the constant term. Therefore, the general form of a linear equation in one variable is

ax + b = 0

By selecting various values for a and b, this form can represent any linear equation in one variable after such an equation has been simpli- fied. For example, if a = 7 and b = 5, ax + b = 0 represents the numerical equation

7x + 5 = 0

If a = 2m - n and b= p - q, then ax + b = 0 rep- resents the literal equation

(2m-n)x + p - q = 0

This equation is solved as follows:

(2m-n)x + (p-q) - (p-q) = 0 - (p-q) (2m-n)x = 0 - (p-q)

x = -(p-q) = q-p 2m-n 2m-n

USING EQUATIONS TO SOLVE PROBLEMS

To solve a problem, we first translate the numerical sense of the problem into an equa- tion. To see how this is accomplished, con- sider the following examples and their solutions.

EXAMPLE 1: Together Smith and Jones have $120. Jones has 5 times as much as Smith. How much has Smith?

SOLUTION:

Step 1. Get the problem clearly in mind. There are two parts to each problem—what is given (the facts) and what we want to know (the question). In this problem we know that Jones has 5 times as much as Smith and together they have $120. We want to know how much Smith has.

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Chapter 11—LINEAR EQUATIONS IN ONE VARIABLE

Step 2. Express the unknown as a letter. Solving the equation, Usually we express the unknown or number we know the least about as a letter (conventionally we use x). Here we know least about Smith's money. Let x represent the number of dollars Smith has.

Step 3. Express the other facts in terms of the unknown. If x is the number of dollars Smith has and Jones has 5 times as much, then 5x is the number of dollars Jones has.

Step 4. Express the facts as an equation. The problem will express or imply a relation between the expressions in steps 2 and 3. Smith's dollars plus Jones' dollars equal $120. Translating this statement into algebraic sym- bols, we have

x + 5x = 120

Solving the equation for x,

6x = 120 x = 20

Thus Smith has $20.

Step 5. Check: See if the solution satisfies the original statement of the problem. Smith and Jones have $120.

$20 + $100 = $120 (Smith's money) (Jones' money)

EXAMPLE 2: Brown can do a piece of work in 5 hr. If Olsen can do it in 4 hr how long will it take them to do the work together?

SOLUTION:

Step 1. Given: Brown could do the work in 5 hr. Olsen could do it in 4 hours. Unknown: How long it takes them to do the work together.

Step 2. Let x represent the time it takes them to do the work together.

Step 3. Then 1/x is the amount they do to- gether in 1 hr. Also, in 1 hour Brown does 1/5 of the work and Olsen does 1/4 of the work.

Step 4. The amount done in 1 hr is equal to the part of the work done by Brown in 1 hr plus that done by Olsen in 1 hr.

1/x = 1/5 + 1/4

20x(1/x) = 20x(1/5) + 20x(1/4)

20 = 4x + 5x 20 = 9x 20/9 = x, or x = 2 2/9 hours

They complete the work together in 2 2/9 hours.

Step 5. Check: 2 2/9 × 1/5 = amount Brown does

2 2/9 × 1/4 = amount Olsen does

(20/9 × 1/5) + (20/9 × 1/4) = 4/9 + 5/9 = 9/9

Practice problems. Use a linear equation in one variable to solve each of the following problems:

1. Find three numbers such that the second is twice the first and the third is three times as large as the first. Their sum is 180.

2. A seaman drew $75.00 pay in dollar bills and five-dollar bills. The number of dollar bills was three more than the number of five- dollar bills. How many of each kind did he draw? (Hint: If x is the number of five-dollar bills, then 5x is the number of dollars they represent.)

3. Airman A can complete a maintenance task in 4 hr. Airman B requires only 3 hr to do the same work. If they work together, how long should it take them to complete the job?

Answers:

1. First number is 30. Second number is 60. Third number is 90.

2. Number of five-dollar bills is 12. Number of one-dollar bills is 15.

3. 1 5/7 hr.

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MATHEMATICS, VOLUME 1

INEQUALITIES Addition

Modern mathematical thought gives consid- erable emphasis to the concept of inequality. A meaningful comparison between two quantities can be set up if they are related in some way, even though the relationship may not be one of equality.

The expression "number sentence" is often used to describe a general relationship which may be either an equality or an inequality. If the number sentence states an equality, it is an EQUATION; if it states an inequality, it is an INEQUALITY.

ORDER PROPERTIES OF REAL NUMBERS

The idea of order, or relative rank accord- ing to size, is based upon two intuitive concepts: "greater than" and "less than." Mathematicians use the symbol > to represent "greater than" and the symbol < to represent "less than." For example, the inequation stating that 7 is greater than 5 is written in symbols as follows:

7 > 5

The inequation stating that x is less than 10 is written as follows:

x < 10

A "solution" of an inequation involving a variable is any number which may be substi- tuted for the variable without changing the re- lationship between the left member and the right member. For example, the inequation x < 10 has many solutions. All negative num- bers zero, and all positive numbers less than 10, may be substituted for x successfully. These solutions comprise a set of numbers, called the SOLUTION SET.

The SENSE of an inequality refers to the direction in which the inequality symbol points. For example, the following two inequalities have opposite sense:

7 > 5 10 < 12

PROPERTIES OF INEQUALITIES

Inequations may be manipulated in accord- ance with specific operational rules, in a man- ner similar to that used with equations.

The rule for addition is as follows: If the same quantity is added to both members of an inequation, the result is an inequation having the same sense as the original inequation. The following examples illustrate this:

1. 5 < 8 5 + 2 < 8 + 2 7 < 10

The addition of 2 to both members does not change the sense of the inequation.

2. 5 < 8 5 + (-3) < 8 + (-3) 2 < 5

The addition of -3 to both members does not change the sense of the inequation.

Addition of the same quantity to both mem- bers is a useful method for solving inequations. In the following example, 2 is added to both members in order to isolate the x term on the left:

x - 2 > 6 x - 2 + 2 > 6 + 2 x > 8

Multiplication

The rule for multiplication is as follows: If both members of an inequation are multiplied by the same positive quantity, the sense of the resulting inequation is the same as that of the original inequation. This is illustrated as follows:

1. -3 < -2 2(-3) < 2(-2) -6 < -4

Multiplication of both members by 2 does not change the sense of th inequation.

2. 10 < 12 1/2(10) < 1/2(12) 5 < 6

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Chapter 11—LINEAR EQUATIONS IN ONE VARIABLE

Multiplication of both members by 1/2 does not change the sense of the inequality. Notice that example 2 illustrates division of both members by 2. Since any division can be rewritten as multiplication by a fraction, the multiplication rule is applicable to both multi- plication and division.

Multiplication is used to simplify the solu- tion of inequalities such as the following:

x/3 > 2

Multiply both members by 3:

3(x/3) > 3(2)

x > 6

Sense Reversal

If both sides of an inequality are multiplied or divided by the same negative number, the sense of the resulting inequality is reversed. This is illustrated as follows:

1. -4 < -2 (-3)(-4) > (-3)(-2) 12 > 6

2. 7 > 5 (-2)(7) < (-2)(5) -14 < -10

Sense reversal is useful in the solution of an inequality in which the variable is preceded by a negative sign, as follows:

2 - x < 4

Add -2 to both members to isolate the x term: 2 - x - 2 < 4 - 2 - x < 2

Multiply both members by -1:

x > -2

Practice problems. Solve each of the fol- lowing inequalities:

1. x + 2 > 3 3. 3 - x < 6

2. y/3 - 1 < 2 4. 4y > 8

Answers:

1. x > 1 3. x > -3

2. y < 9 4. y > 2

GRAPHING INEQUALITIES

An inequality such as x > 2 can be graphed on a number line, as shown in figure 11-2. The heavy line in figure 11-2 contains all values of x which comprise the solution set. Notice that this line continues indefinitely in the positive direction, as indicated by the arrow head. Notice also that the point representing x = 2 is designated by a circle. This signifies that the solution set does not contain the num- ber 2.

Figure 11-3 is a graph of the inequality x² > 4. Since the square of any number greater than 2 is greater than 4, the solution set con- tains all values of x greater than 2. Further- more, the solution set contains all values of x less than -2. This is because the square of any negative number smaller than -2 is a positive number greater than 4.

[Number line from -4 to 4 with circle at 2, heavy line and arrow to the right]

Figure 11-2.—Graph of the inequation x > 2.

[Number line from -4 to 4 with circles at -2 and 2, heavy lines to left and right]

Figure 11-3.—Graph of x² > 4.

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CHAPTER 12

LINEAR EQUATIONS IN TWO VARIABLES

Thus far in this course, discussions of equa- tions have been limited to linear equations in one variable. Linear equations which have two variables are common, and their solution in- volves extending some of the procedures which have already been introduced.

RECTANGULAR COORDINATES

An outstanding characteristic of equations in two variables is their adaptability to graphical analysis. The rectangular coordinate system, which was introduced in chapter 3 of this course, is used in analyzing equations graphically. Systems of vertical and horizontal lines, meeting each other at right angles and thus forming a rectangular grid, is often called the Cartesian coordinate system. It is named after the French philosopher and mathematician, Rene Descartes, who invented it.

COORDINATE AXES

The rectangular coordinate system is devel- oped on a framework of reference similar to figure 3-2 in chapter 3 of this course. On a piece of graph paper, two lines are drawn inter- secting each other at right angles, as in figure 12-1. The vertical line is usually labeled with the capital letter Y and called the Y axis. The horizontal line is usually labeled with the capital letter X and called the X axis. The point where the X and Y axes intersect is called the ORIGIN and is labeled with the letter o.

Above the origin, numbers measured along or parallel to the Y axis are positive; below the origin they are negative. To the right of the origin, numbers measured along or parallel to the X axis are positive; to the left they are negative.

COORDINATES

A point anywhere on the graph may be lo- cated by two numbers, one showing the distance of the point from the Y axis, and the other show- ing the distance of the point from the X axis.

[Rectangular coordinate system diagram with labeled points P, Q, R, S]

Figure 12-1.—Rectangular coordinate system.

Point P (fig. 12-1) is 6 units to the right of the Y axis and 3 units above the X axis. We call the numbers that indicate the position of a point COORDINATES. The number indicating the distance of the point measured horizontally from the origin is the X coordinate (6 in this example), and the number indicating the dis- tance of the point measured vertically from the origin (3 in this example) is the Y coordinate.

In describing the location of a point by means of rectangular coordinates, it is customary to place the coordinates within parentheses and separate them with a comma. The X coordinate is always written first. The coordinates of point P (fig. 12-1) are written (6, 3). The co- ordi nates for point Q are (4, -5); for point R, they are (-5, -2); and for point S, they are (-8, 5).

Usually when we indicate a point on a graph, we write a letter and the coordinates of the point. Thus, in figure 12-1, for point S, we write S(-8, 5). The other points would ordinarily

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Chapter 12—LINEAR EQUATIONS IN TWO VARIABLES

be written, P(6, 3), Q(4, -5), and R(-5, -2). The Y coordinate of a point is often called its ORDI- NATE and the X coordinate is often called its ABSCISSA.

QUADRANTS

The X and Y axes divide the graph into four parts called QUADRANTS. In figure 12-1, point P is in quadrant I, point S is in quadrant II, R is in quadrant III, and Q is in quadrant IV. In the first and fourth quadrants, the X coordinate is positive, because it is to the right of the origin. In the second and third quadrant it is negative, because it is to the left of the origin. Likewise, the Y coordinate is positive in the first and second quadrants, being above the origin; it is negative in the third and fourth quadrants, being below the origin. Thus, we know in advance the signs of the coordinates of a point by knowing the quadrant in which the point appears. The signs of the coordinates in the four quadrants are shown in figure 12-1. Locating points with respect to axes is called PLOTTING. As shown with point P (fig. 12-1), plotting a point is equivalent to completing a rectangle that has segments of the axes as two of its sides with lines dropped perpendicularly to the axes forming the other two sides. This is the reason for the name "rectangular co- ordinates."

PLOTTING A LINEAR EQUATION

A linear equation in two variables may have many solutions. For example, in solving the equation 2x - y = 5, we can find an unlimited number of values of x for which there will be a corresponding value of y. When x is 3, y is 3, since (2 x 4) - 3 = 5. When x is 6, y is 7. When we graph an equa- tion, these pairs of values are considered co- ordinates of points on the graph. The graph of an equation is nothing more than a line joining the points located by the various pairs of num- bers that satisfy the equation. To picture an equation, we first find several pairs of values that satisfy the equation. For example, for the equation 2x - y = 5, we assign several values to x and solve for y. A conven- ient way to find values is to first solve the equation for either variable, as follows:

2x - y = 5

-y = -2x + 5

y = 2x - 5

Once this is accomplished, the value of y is readily apparent when values are substituted for x. The information derived may be re- corded in a table such as table 12-1. We then lay off X and Y axes on graph paper, select some convenient unit distance for measurement along the axes, and then plot the pairs of values found for x and y as coordinates of points on the graph. Thus, we locate the pairs of values shown in table 12-1 on a graph, as shown in figure 12-2 (A).

Table 12-1.—Values of x and y in the equation 2x - y = 5.

If x = ------ | -2 | 1 | 3 | 5 | 6 | 7 | 8 | Then y = --- | -9 | -3 | 1 | 5 | 7 | 9 | 11 |

[Two coordinate plane graphs showing plotted points and a line]

Figure 12-2.—Graph of 2x - y = 5.

Finally, we draw a line joining these points, as in figure 12-2 (B). It is seen that this is a straight line; hence the name "linear equation." Once the graph is drawn, it is customary to write the equation it represents along the line, as shown in figure 12-2 (B). It can be shown that the graph of an equation is the geometric representation of all the points whose coordinates satisfy the conditions of the equation. The line represents an infinite num- ber of pairs of coordinates for this equation. For example, selecting at random the point on the line where x is 2½ and y is 0 and substitut- ing these values in the equation, we find that they satisfy it. Thus,

2(2½) - 0 = 5

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MATHEMATICS, VOLUME 1

If two points that lie on a straight line can be located, the position of the line is known. The mathematical language for this is "Two points DETERMINE a straight line." We now know that the graph of a linear equation in two variables is a straight line. Since two points are sufficient to determine a straight line, a linear equation can be graphed by plotting two points and drawing a straight line through these points. Very often pairs of whole numbers which satisfy the equation can be found by in- spection. Such points are easily plotted. After the line is drawn through two points, it is well to plot a third point as a check. If this third point whose coordinates satisfy the equa- tion lies on the line the graph is accurately drawn.

X AND Y INTERCEPTS

Any straight line which is not parallel to one of the axes has an X intercept and a Y inter- cept. These are the points at which the line crosses the X and Y axes. At the X intercept, the graph line is touching the X axis, and thus the Y value at that point is 0. At the Y inter- cept, the graph line is touching the Y axis; the X value at that point is 0. In order to find the X intercept, we simply let y = 0 and find the corresponding value of x. The Y intercept is found by letting x = 0 and finding the corresponding value of y. For ex- ample, the line

5x + 3y = 15

crosses the Y axis at (0,5). This may be veri- fied by letting x = 0 in the equation. The X in- tercept is (3,0), since x is 3 when y is 0. Fig- ure 12-3 shows the line

5x + 3y = 15

graphed by means of the X and Y intercepts.

EQUATIONS IN ONE VARIABLE

An equation containing only one variable is easily graphed, since the line it represents lies parallel to an axis. For example, in

2y = 9

the value of y is

9/2, or 4½

[A coordinate grid with a line showing 5x + 3y = 15]

Figure 12-3.—Graph of 5x + 3y = 15.

The line 2y = 9 lies parallel to the X axis at a distance of 4½ units above it. (See fig. 12-4.) Notice that each small division on the graph paper in figure 12-4 represents one-half unit. The line 4x + 15 = 0 lies parallel to the Y axis. The value of x is -15/4. Since this value is negative, the line lies to the left of the Y axis at a distance of 3¾ units. (See fig. 12-4.)

[A coordinate grid with two lines: 2y = 9 and 4x + 15 = 0]

Figure 12-4.—Graphs of 2y = 9 and 4x + 15 = 0.

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Chapter 12—LINEAR EQUATIONS IN TWO VARIABLES

From the foregoing discussion, we arrive at two important conclusions: 1. A pair of numbers that satisfy an equa- tion are the coordinates of a point on the graph of the equation. 2. The coordinates of any point on the graph of an equation will satisfy that equation.

SOLVING EQUATIONS IN TWO VARIABLES

A solution of a linear equation in two vari- ables consists of a pair of numbers that satisfy the equation. For example, x = 2 and y = 1 constitute a solution of

3x - 5y = 1

When 2 is substituted for x and 1 is substituted for y, we have

3(2) - 5(1) = 1

The numbers x = -3 and y = -2 also form a solution. This is true because substituting -3 for x and -2 for y reduces the equation to an identity:

3(-3) -5(-2) = 1 -9 + 10 = 1 1 = 1

Each pair of numbers (x, y) such as (2, 1) or (-3, -2) locates a point on the line 3x - 5y = 1. Many more solutions could be found. Any two numbers that constitute a solution of the equa- tion are the coordinates of a point on the line represented by the equation.

Suppose we were asked to solve a problem such as: Find two numbers such that their sum is 33 and their difference is 5. We could indi- cate the problem algebraically by letting x rep- resent one number and y the other. Thus, the problem may be indicated by the two equations

x + y = 33 x - y = 5

Considered separately, each of these equations represents a straight line on a graph. There are many pairs of values for x and y which sat- isfy the first equation, and many other pairs which satisfy the second equation. Our problem

is to find ONE pair of values that will satisfy BOTH equations. Such a pair of values is said to satisfy both equations at the same time, or simultaneously. Hence, two equations for which we seek a common solution are called SIMUL- TANEOUS EQUATIONS. The two equations, taken together, comprise a SYSTEM of equa- tions.

Graphical Solution

If there is a pair of numbers that can be substituted for x and y in two different equations, the pair form the coordinates of a point which lies on the graph of each equation. The only way in which a point can lie on two lines simultaneously is for the point to be at the intersection of the lines. Therefore, the graphical solution of two simultaneous equations involves drawing their graphs and locating the point at which the graph lines intersect.

For example, when we graph the equations x + y = 33 and x - y = 5, as in figure 12-5, we see that they intersect in a single point. There is one pair of values comprising coordinates of that point (19, 14), and that pair of values sat- isfies both equations, as follows:

x + y = 33 x - y = 5 19 + 14 = 33 19 - 14 = 5

[GRAPH SHOWING TWO INTERSECTING LINES WITH GRID, WITH POINT OF INTERSECTION AT (19, 14) MARKED]

Figure 12-5.—Graph of x + y = 33 and x - y = 5.

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MATHEMATICS, VOLUME 1

This pair of numbers satisfies each equation. It is the only pair of numbers that satisfies the two equations simultaneously.

The graphical method is a quick and simple means of finding an approximate solution of two simultaneous equations. Each equation is graphed, and the point of intersection of the two lines is read as accurately as possible. A high degree of accuracy can be obtained but this, of course, is dependent on the precision with which the lines are graphed and the amount of accu- racy possible in reading the graph. Sometimes the graphical method is quite adequate for the purpose of the problem.

Figure 12-6 shows the graphs of x + y = 11 and x - y = -3. The intersection appears to be the point (4, 7). Substituting x = 4 and y = 7 into the equations shows that this is the actual point of intersection, and this pair of num- bers satisfies both equations.

[GRAPH SHOWING TWO INTERSECTING LINES WITH GRID]

Figure 12-6.—Graph of x + y = 11 and x - y = -3.

The equations 7x - 8y = 2 and 4x + 3y = 5 are graphed in figure 12-7. The lines intersect where y is approximately 1/2 and x is approxi- mately 5/6.

Practice problems. Solve the following si- multaneous systems graphically:

1. x + y = 8 2. 3x + 2y = 12 x - y = 2 4x + 5y = 2

Answers:

1. x = 5 2. x = 8 y = 3 y = -6

Addition Method

The addition method of solving systems of equations is illustrated in the following ex- ample:

x - y = 2 x + y = 8 2x + 0 = 10 x = 5

The result in the foregoing example is obtained by adding the left member of the first equation to the left member of the second, and adding the right member of the first equation to the right member of the second.

Having found the value of x, we substitute this value in either of the original equations to find the value of y, as follows:

x - y = 2 (5) - y = 2 -y = 2 - 5 -y = -3 y = 3

Notice that the primary goal in the addition method is the elimination (temporarily) of one of the variables. If the coefficient of y is the same in both equations, except for its sign, adding the equations eliminates y as in the foregoing example. On the other hand, suppose that the coefficient of the variable which we de- sire to eliminate is exactly the same in both equations.

In the following example, the coefficient of x is the same in both equations, including its sign:

x + 2y = 4 x - 3y = -1

Adding the equations would not eliminate either x or y. However, if we multiply both members of the second equation by -1, then addition will eliminate x, as follows:

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Chapter 12—LINEAR EQUATIONS IN TWO VARIABLES

[FIGURE: A coordinate grid showing two intersecting lines with points labeled (-4, 3), (2, -1), and (6, 5)]

Figure 12-7.—Graph of 7x - 8y = 2 and 4x + 3y = 5.

x + 2y = 4 are multiplied by 4 and both members of the -x + 3y = 1 second equation by -3. Then addition will elim- 5y = 5 inate x. y = 1 Following this procedure to get the value of y, we multiply the first equation by 4 and the The value of x is found by substituting 1 for y second equation by -3, as follows: in either of the original equations, as follows: 12x + 8y = 48 x + 2(1) = 4 -12x - 15y = -6 x + 2 = 4 -7y = 42 x = 2 y = -6

As a second example of the addition method, Substituting for y in the first equation to get the find the solution of the simultaneous equations value of x, we have

3x + 2y = 12 3x + 2(-6) = 12 4x + 5y = 2 x + 2(-2) = 4 x - 4 = 4 Here both x and y have unlike coefficients. The coefficients of one of the variables must be x = 8 made the same, except for their signs. The coefficients of x will be the same except for signs, if both members of the first equation

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This solution is checked algebraically by Since equals may be substituted for equals, substituting 8 for x and -6 for y in each of the we may substitute this value of y wherever y original equations, as follows: appears in the second equation. Thus,

1. 3x + 2y = 12 x + 2(11 - 4x) = 8 3(8) + 2(-6) = 12 24 - 12 = 12 We now have one equation that is linear in x; that is, the equation contains only the variable x. 2. 4x + 5y = 2 Removing the parentheses and solving for x, 4(8) + 5(-6) = 2 we find that 32 - 30 = 2 x + 22 - 8x = 8 Practice problems. Use the addition method to solve the following problems: -7x = 8 - 22

1. x + y = 24 3. x - 2y = -1 -7x = -14 x - y = 12 2x + 3y = 12 x = 2

2. 5t + 2v = 9 4. 2x + 7y = 3 To get the corresponding value of y, we sub- 3t - 2v = -5 3x - 5y = 51 stitute x = 2 in y = 11 - 4x. The result is

Answers: y = 11 -4(2) = 11 - 8 1. x = 18 3. x = 3 = 3 y = 6 y = 2 Thus, the solution for the two original equa- 2. t = 1/2 4. x = 12 tions is x = 2 and y = 3. v = 13/4 y = -3 Practice problems. Solve the following sys- tems by the substitution method: Substitution Method 1. 2x - 9y = 1 3. 5r + 2s = 23 In some cases it is more convenient to use x - 4y = 1 4r + s = 19 the substitution method of solving problems. In this method we solve one equation for one of 2. 2x + y = 0 4. t - 4v = 1 the variables and substitute the value obtained 2x - y = 1 2t - 9v = 3 into the other equation. This eliminates one of the variables, leaving an equation in one un- Answers: known. For example, find the solution of the following system: 1. x = 5 3. r = 5 y = 1 s = -1 4x + y = 11 x + 2y = 8 2. x = 1/4 4. t = -3 y = -1/2 v = -1 It is easy to solve for either y in the first equa- tion or x in the second equation. Let us solve Literal Coefficients for y in the first equation. The result is Simultaneous equations with literal coeffi- y = 11 - 4x cients and literal constants may be solved for the value of the variables just as the other equations discussed in this chapter, with the exception that the solution will contain literal

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Chapter 12 - LINEAR EQUATIONS IN TWO VARIABLES

numbers. For example, find the solution of the system:

3x + 4y = a 4x + 3y = b

We proceed as with any other simultaneous linear equation. Using the addition method, we may proceed as follows: To eliminate the y term we multiply the first equation by 3 and the second equation by -4. The equations then become

9x + 12y = 3a -16x - 12y = -4b _________________ -7x = 3a - 4b

x = 3a - 4b 7

x = 4b - 3a 7

To eliminate x, we multiply the first equa- tion by 4 and the second equation by -3. The equations then become

12x + 16y = 4a -12x - 9y = -3b ________________ 7y = 4a - 3b

y = 4a - 3b 7

We may check in the same manner as that used for other equations, by substituting these values in the original equations.

INTERPRETING EQUATIONS

Recall that the general form for an equation in the first degree in one variable is ax + b = 0. The general form for first-degree equations in two variables is

ax + by + c = 0.

It is interesting and often useful to note what happens graphically when equations differ, in certain ways, from the general form. With this information, we know in advance certain facts concerning the equation in question.

LINES PARALLEL TO THE AXES

If in a linear equation the y term is miss- ing, as in

2x - 15 = 0

the equation represents a line parallel to the Y axis and 7½ units from it. Similarly, an equa- tion such as

4y - 9 = 0

which has no x term, represents a line paral- lel to the X axis and 2¼ units from it. (See fig. 12-8.) The fact that one of the two variables does not appear in an equation means that there are no limitations on the values the missing vari- able can assume. When a variable does not ap- pear, it can assume any value from zero to plus or minus infinity. This can happen only if the line represented by the equation lies paral- lel to the axis of the missing variable.

Lines Passing Through the Origin

A linear equation, such as

4x + 3y = 0

that has no constant term, represents a line passing through the origin. This fact is obvi- ous since x = 0, y = 0 satisfies any equation not having a constant term. (See fig. 12-8.)

Lines Parallel to Each Other

An equation such as

3x - 2y = 6

has all possible terms present. It represents a line that is not parallel to an axis and does not pass through the origin. Equations that are exactly alike, except for the constant terms, represent parallel lines. As shown in figure 12-8, the lines represented by the equations

3x - 2y = -18 and 3x - 2y = 6

are parallel.

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MATHEMATICS, VOLUME 1

Figure 12-8.—Interpreting equations.

Parallel lines have the same slope. Chang- ing the constant term moves a line away from or toward the origin while its various positions remain parallel to one another. Notice in fig- ure 12-8 that the line 3x - 2y = 6 lies closer to the origin than 3x - 2y = -18. This is revealed at sight by comparing their constant terms. That one which has the constant term of greater absolute value will lie farther from the origin. In this case 3x - 2y = -18 will be farther from the origin since |-18| > |6|. The fact that lines are parallel is indicated by the result when we try to solve two equations such as 3x - 2y = -18 and 3x - 2y = 6 simultane- ously. Subtraction eliminates both x and y im- mediately. If both variables disappear, we can- not find values for them such that both equations

are satisfied at the same time. This means that there is no solution. No solution implies that there is no point of intersection for the straight lines represented by the equations. Lines that do not intersect in the finite plane are parallel.

USING TWO VARIABLES IN SOLVING WORD PROBLEMS

Many problems can be solved quickly and easily using one equation with one variable. Other problems that might be rather difficult to solve in terms of one variable can easily be solved using two equations and two variables. The difference in the two methods is shown in the following example, solved first by using one variable and then using two.

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Chapter 12—LINEAR EQUATIONS IN TWO VARIABLES

EXAMPLE: Find the two numbers such that half the first equals a third of the second and twice their sum exceeds three times the second by 4.

SOLUTION USING ONE VARIABLE:

1. Let x = the first number.

2. Then x/2 = 1/3 of the second number.

3. Thus 3x/2 = the second number.

From the statement of the problem, we then have

2(x + 3x/2) = 3(3x/2) + 4

2x + 3x = 9x/2 + 4

10x = 9x + 8

x = 8 (first number)

3x/2 = 12 (second number)

SOLUTION USING TWO VARIABLES:

If we let x and y be the first and second numbers, respectively, we can write two equations almost directly from the statement of the problem. Thus,

1. x/2 = y/3

2. 2(x + y) = 3y + 4

Solving for x in the first equation and substituting this value in the second, we have

x = 2y/3

2(2y/3 + y) = 3y + 4

4y/3 + 2y = 3y + 4

4y + 6y = 9y + 12

y = 12 (second number)

x/2 = 12/3

x = 8 (first number)

Thus, we see that the solution using two variables is more direct and simple. Often it would require a great deal of skill to manipulate a problem so that it might be solved using one variable; whereas the solution using two variables might be very simple. The test of two variables, of course, involves the fact that the student must be able to form two equations from the information given in the problem.

Practice problems. Solve the following problems using two variables:

1. A Navy tug averages 12 miles per hour downstream and 9 miles per hour upstream. How fast is the stream flowing?

2. The sum of the ages of two boys is 18. If 4 times the younger boy's age is subtracted from 3 times the older boy's age, the difference is 12. What are the ages of the two boys?

Answers:

1. 1½ mph.

2. 6 years and 12 years.

INEQUALITIES IN TWO VARIABLES

Inequalities in two variables are of the following form:

x + y > 2

Many solutions of such an inequation are apparent immediately. For example, x could have the value 2 and y could have the value 3, since 2 + 3 is greater than 2.

The existence of a large number of solutions suggests that a graph of the inequation would contain many points. The graph of an inequation in two unknowns is, in fact, an entire area rather than just a line.

PLOTTING ON THE COORDINATE SYSTEM

It would be extremely laborious to plot enough points at random to define an entire area of the coordinate system. Therefore our method consists of plotting a boundary line and shading the area, on one side of this line, wherein the solution points lie.

The equation of the boundary line is formed by changing the inequation to an equation. For

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MATHEMATICS, VOLUME 1

example, the equation of the boundary line for the graph of

x + y > 2

is the equation

x + y = 2

Figure 12-9 is a graph of x + y > 2. Notice that the boundary line x + y = 2 is not solid. This is intended to indicate that points on the boundary line are not members of the solution set. Every point lying above and to the right of the boundary line is a member of the solution set. Any solution point may be verified by substituting its X and Y coordinates for x and y in the original inequation.

[FIGURE 12-9: Graph of x + y > 2 showing shaded region above and to the right of dashed line]

Figure 12-9.—Graph of x + y > 2.

SIMULTANEOUS INEQUALITIES

The areas representing the solutions of two different inequations may overlap. If such an overlap occurs, the area of the overlap includes all points whose coordinates satisfy both inequations simultaneously. An example of this is shown in figure 12-10, in which the following two inequations are graphed:

x + y > 2 x - y > 2

[FIGURE 12-10: Graph showing two shaded regions with cross-hatching indicating intersection of x + y > 2 and x - y > 2]

Figure 12-10.—Graph of x + y > 2 and x - y > 2.

The double crosshatched area in figure 12-10 contains all points which comprise the solution set for the system.

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CHAPTER 13 RATIO, PROPORTION, AND VARIATION

The solution of problems based on ratio, proportion, and variation involves no new prin-ciples. However, familiarity with these topics will often lead to quick and simple solutions to problems that would otherwise be more com-plicated.

RATIO

The results of observation or measurement often must be compared with some standard in order to have any meaning. For ex-ample, to say that a man can read 400 words per minute has little meaning as it stands. However, when his rate is compared to the 250 words per minute of the average reader, one can see that he reads considerably faster than the average reader. How much faster? To find out, his rate is divided by the average rate, as follows:

400/250 = 8/5

Thus, for every 5 words read by the average reader, this man reads 8. Another way of mak-ing this comparison is to say that he reads 8/5 times as fast as the average reader.

When the relationship between two numbers is shown in this way, they are compared as a RATIO. A ratio is a comparison of two like quantities. It is the quotient obtained by divid-ing the first number of a comparison by the second.

Comparisons may be stated in more than one way. For example, if one person has 40 teeth and another has 10, one way of stating the com-parison would be 40 teeth to 10 teeth. This comparison could be shown as a ratio in four ways as follows:

1. 40:10 2. 40 ÷ 10 3. 40/10 4. The ratio of 40 to 10.

When the emphasis is on "ratio," all of these expressions would be read, "the ratio of 40 to 10." The form 40 ÷ 10 may also be read "40 divided by 10." The form 40/10 may also be read "40 over 10."

Comparison by means of a ratio is limited to quantities of the same kind. For example, in order to express the ratio between 6 ft and 3 yd, both quantities must be written in terms of the same unit. Thus the proper form of this ratio is 2 yd : 3 yd, not 6 ft : 3 yd. When the parts of the ratio are expressed in terms of the same unit, the units cancel and the ratio consists simply of two numbers. In this exam-ple, the final form of the ratio is 2 : 3.

Since a ratio is also a fraction, all the rules that govern fractions may be used in working with ratios. Thus, the terms may be reduced, increased, simplified, and so forth, according to the rules for fractions. To reduce the ratio 15:20 to lowest terms, write the ratio as a fraction and then proceed as for fractions. Thus, 15:20 becomes

15/20 = 3/4

Hence the ratio of 15 to 20 is the same as the ratio of 3 to 4.

Notice the distinction in thought between 3/4 as a fraction and 3/4 as a ratio. As a fraction we think of 3/4 as the single quantity "three-fourths."

As a ratio, we think of 3/4 as a comparison be-tween the two numbers, 3 and 4. For example, the lengths of two sides of a triangle are 1 9/16 ft and 2 ft. To compare these lengths by means of a ratio, divide one number by the other and reduce to lowest terms, as follows:

1 9/16 ÷ 2 = 25/16 ÷ 2 = 25/32

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The two sides of the triangle compare as 25 to 32.

INVERSE RATIO

It is often desirable to compare the numbers of a ratio in the inverse order. To do this, we simply interchange the numerator and the de-nominator. Thus, the inverse of 15:20 is 20:15. When the terms of a ratio are interchanged, the INVERSE RATIO results.

Practice problems. In problems 1 through 6, write the ratio as a fraction and reduce to low-est terms. In problems 7 through 10, write the inverse of the given ratio.

1. The ratio of 5 lb to 15 lb 2. $16 : $12 3. 16 - 4 4. One quart to one gallon 5. 5x to 10x 6. 3 3/4 : 4 1/2 7. The ratio of 6 ft to 18 ft 8. 4/8 9. 5 : 8 10. 15 to 21

Answers:

1. 1/3 2. 4/3 3. 4/1 4. 1/4 5. 1/2 6. 20/27 7. 3/1 8. 2/1 9. 8/5 10. 7/5

PROPORTION

Closely allied with the study of ratio is the subject of proportion. A PROPORTION is nothing more than an equation in which the members are ratios. In other words when two ratios are set equal to each other, a proportion is formed. The proportion may be written in three different ways as in the following ex-amples:

15:20 : : 3:4 15:20 = 3:4 15/20 = 3/4

The last two forms are the most common. All these forms are read, "15 is to 20 as 3 is to 4." In other words, 15 has the same ratio to 20 as 3 has to 4.

One reason for the extreme importance of proportions is that if any three of the terms are given, the fourth may be found by solving a simple equation. In science many chemical and physical relations are expressed as propor-tions. Consequently, a familiarity with propor-tions will provide one method for solving many applied problems. It is evident from the last form shown, 15/20 = 3/4, that a proportion is really a fractional equation. Therefore, all the rules for fraction equations apply.

TERMS OF A PROPORTION

Certain names have been given to the terms of the two ratios that make up a proportion. In a proportion such as 3:8 = 9:24, the first and the last terms (the outside terms) are called the EXTREMES. In other words, the numerator of the first ratio and the denominator of the second are called the extremes. The second and third terms (the inside terms) are called the MEANS. The means are the denominator of the first ratio and the numerator of the second. In the example just given, the extremes are 3 and 24; the means are 8 and 9.

Four numbers, such as 5, 8, 15, and 24, form a proportion if the ratio of the first two in the order named equals the ratio of the second two. When these numbers are set up as ratios with the equality sign between them, the members will reduce to an identity if a true proportion exists. For example, consider the following proportion:

5/8 = 15/24

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Chapter 13—RATIO, PROPORTION, AND VARIATION

In this proportion, 15/24 must reduce to 5/8 for the proportion to be true. Removing the same factor from both members of 15/24 we have

5/8 = 3(5)/3(8)

The number 3 is the common factor that must be removed from both the numerator and the denominator of one fraction in order to show that the expression

5/8 = 15/24

is a true proportion. To say this another way, it is the factor by which both terms of the ratio 5/8 must be multiplied in order to show that this

ratio is the same as 15/24.

Practice problems. For each of the following proportions, write the means, the extremes, and the factor of proportionality.

1. 3/16 = 15/80 3. 25/75 = 1/3

2. 4:5 = 12:15 4. 12:3 :: 4:1

Answers:

1. Means: 16 and 15 Extremes: 3 and 80 Factor of proportionality: 5

2. M: 5 and 12 E: 4 and 15 FP: 3

3. M: 75 and 1 E: 25 and 3 FP: 25

4. M: 3 and 4 E: 12 and 1 FP: 3

OPERATIONS OF PROPORTIONS

It is often advantageous to change the form of a proportion. There are rules for changing or combining the terms of a proportion without altering the equality between the numbers. These rules are simplifications of fundamental rules for equations; they are not new, but are simply adaptations of laws or equations presented earlier in this course.

Rule 1. In any proportion, the product of the means equals the product of the extremes. This is perhaps the most commonly used rule of proportions. It provides a simple way to rearrange a proportion so that no fractions are present. In algebraic language the rule is illustrated as follows:

a/b = c/d

bc = ad

To prove this rule, we note that the LCD of the two ratios b and d is bd. Multiplying both members of the equation in its original form by this LCD, we have

bd · a/b = bd · c/d

ad = bc

The following numerical example illustrates the simplicity of rule 1:

3/8 = 9/24

8(9) = 3(24)

If one of the terms of a proportion is a variable to the first power as in

7:5 = x:6

the proportion is really a linear equation in one variable. Such an equation can be solved for the unknown.

Equating the products of the means and extremes produces the following:

5x = 42

x = 8⅖

Mean Proportional

When the two means of a proportion are the same quantity, that quantity is called the MEAN

MATHEMATICS, VOLUME 1

PROPORTIONAL between the other two terms. In the proportion

a/x = x/c

x is the mean proportional between a and c.

Rule 2. The mean proportional between two quantities is the square root of their product. This rule is stated algebraically as follows:

a/x = x/c

x = ± √ac

To prove rule 2, we restate the proportion and apply rule 1, as follows:

a/x = x/c

x² = ac

x = ± √ac

Rule 2 is illustrated by the following numerical example:

2/8 = 8/32

8 = √2(32)

8 = √64

OTHER FORMS FOR PROPORTIONS

If four numbers, for example, a, b, c, and d, form a proportion, such as

a/b = c/d

they also form a proportion according to other arrangements.

Inversion

The four selected numbers are in proportion by INVERSION in the form

b/a = d/c

The inversion relationship is proved as follows, by first multiplying both members of the original proportion by bd/ac:

(bd/ac)(a/b) = (bd/ac)(c/d)

d/c = b/a

Note that the product of the means and the product of the extremes still yield the same equality as in the original proportion.

The inversion relationship may be illustrated by the following numerical example:

5/8 = 10/16

Therefore,

8/5 = 16/10

Alteration

The four selected numbers (a, b, c, and d) are in proportion by ALTERNATION in the following form:

a/c = b/d

To prove the alternation relationship, first multiply both sides of the original proportion by c/b, as follows:

a/b = c/d

c(a/b) = c(c/d)

a/c = b/d

The following numerical example illustrates alternation:

5/8 = 10/16

Therefore,

5/10 = 8/16

SOLVING PROBLEMS BY MEANS OF PROPORTION

One of the most common types of problems based on proportions involves triangles with

Chapter 13—RATIO, PROPORTION, AND VARIATION

proportional sides. Suppose that the corre- sponding sides of two triangles are known to be proportional. (See fig. 13-1.) The lengths of the sides of one triangle are 8, 9, and 11. The length of the side of the second triangle corre- sponding to side 8 in the first triangle is 10. We wish to find the lengths of the remaining sides, b and c.

with the following result:

8c = 110

c = 13¾

The sides of the second triangle are 10, 11¼, and 13¾. The result can also be obtained by using the factor of proportionality. Since 8 and 10 are lengths of corresponding sides, we can write

8k = 10

k = 10/8 = 5/4

The factor of proportionality is thus found to be 5/4.

Multiplying any side of the first triangle by 5/4 gives the corresponding side of the second triangle, as follows:

b = 9(5/4) = 45/4 = 11¼

c = 11(5/4) = 55/4 = 13¾

Proportional sides of similar triangles may be used to determine the height of an object by measuring its shadow. (See fig. 13-2.)

[FIGURE 13-1 showing two triangles with sides labeled: First triangle has sides 8, 9, 11. Second triangle has sides 10, b, c]

Figure 13-1.—Triangles with corresponding sides proportional.

Since the corresponding sides are propor- tional, the pairs of corresponding sides may be used to form proportions as follows:

8/10 = 9/b

9/b = 11/c

8/10 = 11/c

To solve for b, we use the proportion

8/10 = 9/b

and obtain the following result:

8b = 90

4b = 45

b = 11¼

The solution for c is similar to that for b, using the proportion

8/10 = 11/c

[FIGURE 13-2 showing two right triangles: Triangle ABC with base AB = 20 FT., and triangle DEF with base DE = 16 FT. and height DF = 12 FT.]

Figure 13-2.—Measuring height by shadow length.

In figure 13-2, mast AC casts a shadow 20 ft. long (AB). At the same time, DF (12 ft. long) casts a shadow of 16 ft. long (DE). Assuming that both masts are vertical and on level ground, triangle ABC is similar to triangle DEF and their corresponding sides are therefore proportional. Thus the height of AC may be found as follows:

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MATHEMATICS, VOLUME 1

AC/12 = 20/16

AC = (12)(20)/16 = 15

Practice problems. In each of the following problems, set up a proportion and then solve for the unknown quantity:

1. Referring to figure 13-1, if the shortest side of the larger triangle is 16 units long, rather than 10, how long is side c ?

2. If a mast 8 ft high casts a shadow 10 ft long, how high is a mast that casts a shadow 40 ft long ?

Answers:

1. 8/16 = 11/c

8c = (11)(16)

c = (11)(16)/8

c = 22

2. 8/10 = h/40

(8)(40)/10 = h

h = 32

Word Problems

A knowledge of proportions often provides a quick method of solving word problems. The following problem is a typical example of the types that lend themselves to solution by means of proportion.

If an automobile runs 36 mi on 2 gal of gas, how many miles will it run on 12 gal? Com- paring miles to miles and gallons to gallons, we have

36:x = 2:12

Rewriting this in fraction form, the solution is as follows:

36/x = 2/12

2x = 12(36)

2x = 432

x = 216 mi

Practice problems. In each of the following problems, first set up a proportion and then solve for the unknown quantity:

1. The ratio of the speed of one aircraft to that of another is 2 to 5. If the slower aircraft has a speed of 300 knots, what is the speed of the faster aircraft ?

2. If 6 seamen can empty 2 cargo spaces in 1 day, how many spaces can 150 seamen empty in 1 day ?

3. On a map having a scale of 1 in. to 50 mi, how many inches represent 540 mi ?

Answers:

1. 750 kt 2. 50 3. 10.8 in.

VARIATION

When two quantities are interdependent, changes in the value of one may have a predictable effect on the value of the other. Variation is the name given to the study of the effects of changes among related quantities. The three types of variation which occur frequently in the study of scientific phenomena are DIRECT, INVERSE, and JOINT.

DIRECT VARIATION

An example of direct variation is found in the following statement: The perimeter (sum of the lengths of the sides) of a square increases if the length of a side increases. In everyday language, this statement might become: The longer the side, the bigger the square. In mathematical symbols, using p for perimeter and s for the length of the side, the relationship is stated as follows:

p = 4s

Since the number 4 is constant, any variations which occur are the results of changes in p and s. Any increase or decrease in the size of s results in a corresponding increase or decrease in the size of p. Thus p varies in the same way (increasing or decreasing) as s. This explains the terminology which is frequently used: p varies directly as s.

In general, if a quantity can be expressed in terms of a second quantity multiplied by a constant, it is said to VARY DIRECTLY AS THE SECOND QUANTITY. For example if x and y are variables and k is a constant, x varies directly as y, if x=ky. Thus, as y increases x increases,

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Chapter 13—RATIO, PROPORTION, AND VARIATION

and as y decreases, x decreases. There is a direct effect on x caused by any change in y. The fact that x varies as y is sometimes indicated by x = y, or x ~ y. however, it is usually written in the form x = ky. The relationship x = ky is equivalent to x/y = k. If one quantity varies directly as a second quantity, the ratio of the first quantity to the second quantity is a constant. Thus, whatever the value of x, where it is divided by y, the result will always be the same value, k. A quantity that varies directly as another quantity is also said to be DIRECTLY PROPORTIONAL to the second quantity. In x = ky, the coefficient of x is 1. The relationship x = ky can be written in proportion form as

x/k = y/1

or

k/x = 1/y

Notice that the variables, x and y, appear either in the numerators or in the denominators of the equal ratios. This implies that x and y are directly proportional. The constant, k, is the CONSTANT OF PROPORTIONALITY. Practice problems. Write an equation showing the stated relationship, in each of the following problems:

1. The cost, C of a dozen wrenches varies directly as the price, p, of one wrench.

2. X is directly proportional to Y (use k as the constant of proportionality).

3. The circumference, C, of a circle varies directly as its diameter, d (use π as the constant of proportionality).

In the following problems, based on the formula p = 4s, find the appropriate word or symbol to fill the blank.

4. When s is doubled, p will be___________.

5. When s is halved, p will be______________.

6. ______________is directly proportional to s.

Answers:

1. C = 12p 2. X = kY 3. C = πd

4. doubled 5. halved 6. p

Variation as the Power of a Quantity

Another form of direct variation occurs when a quantity varies as some power of another. For example, consider the formula

A = πr²

Table 13-1 shows the values of r and the corresponding values of A.

Table 13-1.—Relation between values of radius and area in a circle.

When r = ---- | 1 | 2 | 3 | 4 | 5 | 7 | 9 Then A = ---- | π | 4π | 9π | 16π | 25π | 49π | 81π

Notice how A changes as a result of a change in r. When r changes from 1 to 2, A changes from π to 4 times π or 2² times π. Likewise when r changes from 3 to 4, A changes not as p, but as the SQUARE of r. In general, when one quantity varies as the power of another if it is equal to a constant times that quantity raised to the power. Thus, in an equation such as x = ky^n, x varies directly as the n^th power of y. As y increases, x increases but more rapidly than y, and as y decreases, x decreases, but again more rapidly.

Practice problems.

1. In the formula V = e³, how does V vary? 2. In the formula A = s², if s is doubled how much is A increased? 3. In the formula s = gt²/2, g is a constant. If t is halved, what is the resulting change in s?

Answers:

1. Directly as the cube of e. 2. It is multiplied by 4. 3. It is multiplied by 1/4.

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INVERSE VARIATION

A quantity VARIES INVERSELY as another quantity if the product of the two quantities is a constant. For example, if x and y are variables and k is a constant, the fact that x varies inversely as y is expressed by

xy = k

or

x = k/y

If values are substituted for x and y, we see that as one increases, the other must decrease, and vice versa. Otherwise, their product will not equal the same constant at each time. If a quantity varies inversely as a second quantity, it is INVERSELY PROPORTIONAL to the second quantity. In xy = k, the coefficient of k is 1. The equality xy = k can be written in the form

x/k = 1/y

or

k/x = y/1

Notice that when one of the variables, x or y, occurs in the numerator of a ratio, the other variable occurs in the denominator of the second ratio. This implies that x and y are inversely proportional.

Inverse variation may be illustrated by means of the formula for the area of a rectangle. If A stands for area, L for length, and W for width, the expression for the area of a rectangle in terms of the length and width is

A = LW

Suppose that several rectangles, all having the same area but varying lengths and widths, are to be compared. Then LW = A has the same form as xy = k, where A and k are constants. Thus L is inversely proportional to W, and W is inversely proportional to L. If the constant area is 12 sq ft, this relationship becomes

LW = 12

If the length is 4 ft, the width is found as follows:

W = 12/L = 12/4 = 3 ft

If the length increases to 6 ft, the width decreases as follows:

W = 12/6 = 2 ft

If a constant area is 12, the width of a rectangle decreases from 3 to 2 as the length increases from 4 to 6. When two inversely proportional quantities vary, one decreases as the other increases.

Another example of inverse variation is found in the study of electricity. The current flowing in an electrical circuit at a constant potential varies inversely as the resistance of the circuit. Suppose that the current, I, is 10 amperes when the resistance, R, is 11 ohms and it is desired to find the current when the resistance is 5 ohms.

Since I and R vary inversely, the equation for the relationship is IR = k, where k is the constant voltage. Therefore, (10)(11) = k. Also, when the resistance changes to 5 ohms, (5)(I) = k. Quantities equal to the same quantity are equal to each other, so we have the following equation:

5I = (10)(11)

I = 110/5 = 22

The current is 22 amperes when the resistance is 5 ohms. As the resistance decreases from 11 to 5 ohms, the current increases from 10 to 22 amperes.

One type of variation problem which tends to be confusing to the beginner involves rates of speed or rates of doing work. For example, if 7 men can complete a job in 20 days, how long will 50 men require to complete the same job? The strictly mechanical approach to this problem might result in the following false solution, relating men to men and days to days:

7 men/50 men = 20 days/T

However, a little thought brings out the fact that we are dealing with an INVERSE relationship rather than a direct one. In other words,

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CHAPTER 14 DEPENDENCE, FUNCTIONS, AND FORMULAS

In chapter 13 of this course, use is made of several formulas, such as A = LW, P = lR, etc. It is the purpose of this chapter to explain the function and dependency relationships which make formulas so useful.

DEPENDENCE AND FUNCTIONS

Dependence may be defined as any relationship between two variables which allows the prediction of change in one of them as a result of change in the other. For example, the cost of 200 bolts depends upon the price per hundred. If C represents cost and p represents the price of 100 bolts, then the cost of 200 bolts may be expressed as follows:

C = 2p

In the example just given, C is called the DEPENDENT VARIABLE because its value depends upon the changing values of p. The INDEPENDENT VARIABLE is p. It is standard practice to isolate the dependent variable on the left side of an equation, as in the example. Consider the formula for the area of a rectangle, A = LW. Here we have two independent variables, L and W. Figure 14-1 (A) shows what happens if we double the length. Figure 14-1 (B) shows the result of doubling the width. Figure 14-1 (C) shows the effect of doubling both length and width. Notice that when the length or width alone is doubled the area is doubled, but when both length and width are doubled the area is four times as great. In any equation showing a dependency relationship, the dependent variable is said to be a FUNCTION of the independent variable. Another use of the term "function" in describing an equation such as C = 2p is to refer to the whole expression as "the function C = 2p." This terminology is especially useful when the right-hand expression has several terms. For example, consider the equation y = 2x² + 3x - 4. Mathematicians frequently use a shorthand notation and rewrite the equation as y = f(x). The

expression f(x) is understood to mean "a function of x" and reference to the function by calling it f(x) saves the space and time that would otherwise be required to write out all three terms. Practice problems. Answer the following questions concerning the function r = d/t:

1. When t increases and d remains the same, does r increase, decrease, or remain the same?

2. When d increases and t remains the same, does r increase, decrease, or remain the same?

3. When t decreases and d remains the same, does r increase, decrease, or remain the same?

4. When d decreases and t remains the same, does r increase, decrease, or remain the same?

5. When d is doubled·and t remains the same, is r doubled or halved?

6. When t is doubled and d remains the same, is r doubled or halved?

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Answers:

1. Decreases. 4. Decreases. 2. Increases. 5. Doubled. 3. Increases. 6. Halved.

FORMULAS

One of the most common uses of algebra is in the solution of formulas. Formulas have a wide and varied use throughout the Navy. It is important to know how formulas are derived, how to translate them into words, how to make them from word statements, and how to use them to solve problems.

A formula is a general fact, rule, or principle expressed in algebraic symbols. It is a shorthand expression of a rule in which letters and signs of operation take the place of words. The formula always indicates the mathematical operations involved. For example, the formula P = 2L + 2W indicates that the perimeter (sum of the lengths of the sides) of a rectangle is equal to twice its length plus twice its width. (See fig. 14-2.)

┌──────────────────┐ │ P = 2L + 2W │ W └──────────────────┘ L

Figure 14-2.—Perimeter of a rectangle.

A formula obtained by logical or mathematical reasoning is called a mathematical formula. A formula whose reliability is based on a limited number of observations, or on immediate experience, and not necessarily on established theories or laws is called an EMPIRICAL formula. Empirical formulas are found frequently in engineering and physical sciences. They sometimes are valid for only a limited number of values.

SUBJECT OF A FORMULA

Usually a formula is taken almost directly from the verbal rule or law. For instance, the perimeter of a rectangle is equal to twice the length plus twice the width. Where possible, letters are used as symbols for the words.

Thus, P = 2L + 2W. A simple formula such as this is like a declarative sentence. The left half is the SUBJECT and all the rest is the predicate. The subject is P. It corresponds to the part of the verbal rule that reads "the perimeter of a rectangle." This subject is usually a single letter followed by the equals sign.

All formulas are equations, but not all equations are formulas. Some distinctions between a formula and an ordinary equation are worthy of note. The equation may not have a subject, while the formula typically does. In the formula, the unknown quantity stands alone in the left-hand member. No computation is performed upon it, and it does not appear more than once. In the equation, on the other hand, the unknown quantity may appear once or more in either or both members, and computation may be performed with it or on it. We evaluate a formula by substituting for the literal numbers in the right member. An equation is solved by computation in either or both members until all that remains is an unknown in one member and a known quantity in the other. The solution of an equation usually requires a knowledge of algebraic principles, while the evaluation of a formula may ordinarily be accomplished with only a knowledge of arithmetic.

SYMBOLS

Letters that represent words have been standardized in many cases so that certain formulas may be written the same in various texts and reference books. However, to avoid any misunderstanding a short explanation often accompanies formulas as follows:

A = hw,

where

A = area in square units h = height w = width

Subscripts and Primes

In a formula in which two or more of the same kind of letters are being compared, it is desirable to make a distinction between them. In electronics, for example, a distinction between resistances may be indicated by Ra and Rb or R₁ and R₂. These small numbers or

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letters written to the right and below the R's are called subscripts. Those shown here are read: R sub a, R sub b, R sub one, and R sub two. Primes are also used in the same manner to distinguish between quantities of the same kind. Primes are written to the right and above the letters, as in S', S'', and S'''. They are read: S prime, S double prime, and S triple prime.

CHANGING THE SUBJECT OF A FORMULA

If values are given for all but one of its variables, a formula can be solved to obtain the value of that variable. The first step is usually the rearranging of the formula so that the unknown value is the subject—that is, a new formula is derived from the original. For example, the formula for linear motion—distance equals rate times time—is usually written

d = rt

Suppose that instead of the distance we wish to know the rate, r, or the time, t. We simply change the subject of the formula by the algebraic means developed in earlier chapters. Thus, in solving the formula for r, we divide both sides by t, with the following result:

d/t = rt/t

d/t = r, or r = d/t

In words, this formula states that rate equals distance divided by time. Likewise, in solving for t, we have the following:

d/r = rt/r

d/r = t, or t = d/r

In words, this formula states that time equals distance divided by rate.

We have in effect two new formulas, the subject of one being rate and the subject of the other being time. They are related to the original formula because they were derived from it, but they are different in that they have different subjects.

Practice problems. Derive new formulas from the following expressions with subjects as indicated:

1. A = ½bh, subject h

2. P = 2L + 2W, subject L

3. i = prt, subject r

4. p = br, subject b

5. E = IR, subject I

6. The modern formula for converting Fahrenheit temperatures to Celsius (centigrade) is C = (F + 40)(5/9) - 40. Express the formula for converting Celsius (centigrade) temperatures to Fahrenheit.

Answers:

1. h = 2A/b 4. b = p/r

2. L = (P - 2W)/2 5. I = E/R

3. r = 1/pt 6. F = (C + 40)(9/5) - 40

EVALUATING FORMULAS

The first step in finding the value of the unknown variable of a formula is usually the derivation of a formula that has the unknown as its subject. Once this is accomplished, the evaluation of a formula consists of nothing more than substituting numerical values for the letters representing known quantities and performing the indicated operations.

For example, suppose we wish to find the time required to fly 1,250 nautical miles at the rate of 250 knots. The formula is d = rt. We can change the subject by dividing both sides of the equation by r, as follows:

d/r = rt/r

d/r = t

t = 1250/250 = 5 hr

Formulas can be solved for an unknown by substituting directly in the original formula

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MATHEMATICS, VOLUME 1

even though that unknown is not the subject. Generally, however, it is simpler to first make the unknown the subject.

Formulas vary widely, from the simple type such as we have been considering to some that are very complex. All formulas have certain characteristics in common. There is always a subject, the quantity whose value is sought as a final answer. This subject usually stands alone, being placed equal to at least one and possibly several literal numbers, which are combined according to certain indicated operations. The formula can always be evaluated for a specific case when numerical values are known for all these literal quantities.

Evaluating formulas may be facilitated by developing a routine order of doing the work. If someone else can read the work and clearly understand what has been done, the work is in good order. The original formula should be written first, then the derived formula that is going to be used in solving the problem, and finally the actual substitutions. The indicated operations may then be carried out. Care should be taken to label answers with correct units; that is, miles per hour, foot-pounds, square feet, etc.

Practice problems.

1. E = IR. Solve for R in ohms if E is 110 volts and I is 5 amperes.

2. d = rt. Solve for t in hours if d is 840 nautical miles and r is 25 knots.

3. F = (C + 40)(9/5) - 40. Solve for C if F is 32°.

Answers:

1. 22 ohms 2. 33.6 hr 3. 0°

DEVELOPING FORMULAS

Developing a formula from a verbal statement is nothing more than reducing the statement to a shorthand form and showing the mathematical relationships between the elements of the statement.

For example, suppose that we wish to develop a formula showing the distance, D, traveled at the rate of 20 knots for t hours. If the distance traveled in 1 hr is 20 nautical miles,

then the distance traveled in t hours is 20t. Therefore, the formula is

D = 20t

Practice problems.

1. Write a formula for the cost, C of p pounds of sugar at 15 cents per pound.

2. Write the formula for the cost, C, of one article when the total cost, T, of n similar articles is known.

3. Write a formula for the number of days, d, in w weeks.

4. Write a formula for the number of ounces, n, in p pounds.

Answers:

1. C = 15p 3. d = 7w

2. C = T/n 4. n = 16p

Developing Formulas from Tables

In technical work, instrument readings and other data are often recorded in a tabular arrangement. By careful observation of such tables of data, it is frequently possible to find values that are related in a definite pattern. The table can thus be used in developing a formula showing the relationship between the related quantities.

For example, table 14-1 shows the results of time trials on a ship, with the data rounded to the nearest whole hour and the nearest whole mile.

Table 14-1.—Time trials.

| Nautical miles (d) | 20 | 40 | 60 | 80 | 100 | | Hours (t) | 1 | 2 | 3 | 4 | 5 |

By inspection of the table, it soon becomes clear that the number of miles traveled is always 20 times the corresponding number of hours. Therefore the formula developed from this table is as follows:

d = 20t

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Chapter 14—DEPENDENCE, FUNCTIONS, AND FORMULAS

A second example of the derivation of a formula from a table is shown in figure 14-3. Figure 14-3 (A) shows several polygons (many-sided plane figures), each with one or more diagonals. A diagonal is a straight line joining one vertex (point where two sides meet) with another.

[FIGURE: Four polygons showing increasing number of diagonals]

(A)

DIAGONALS (d) | 1 | 2 | 3 | 4 SIDES (n) | 4 | 5 | 6 | 7

(B)

Figure 14-3.—Diagonals of plane figures.

The table in figure 14-3 (B) compares the number of sides of each polygon with the number of diagonals that can be drawn from any one vertex. Using this table, we make a formula for the number, d, of diagonals that can be drawn from one vertex of a polygon of n sides. In the table we note that the number of diagonals is always 3 less than the number of sides. Therefore the formula is d = n - 3.

Practice problems. Complete the following tables and write formulas to show the relationship between the numbers.

1. L | 2 | 5 | 8 | 11 | 14 | 17 | 20 P | 12 | 30 | 48 | 66 | 84 | |

2. a | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 b | 4 | 5 | 6 | 7 | | | |

3. x | 0 | 1 | 2 | 3 | 4 | 5 | 6 y | 0 | 3 | 6 | 9 | 12| | |

4. n | 1 | 2 | 3 | 4 | 5 | 6 | 7 s | 3 | 6 | 11| 18| 27| 38| |

Answers:

1. P = 6L 2. b = a + 4 3. y = 3x 4. s = n² + 2

TRANSLATING FORMULAS

Thus far, we have been concerned primarily with reducing verbal rules to formula form. It is also necessary to be able to do the reverse, and translate a formula into words. Technical publications frequently take advantage of the fact that it is more convenient to write formulas than longhand rules. Understanding is hampered if we are not able to translate these formulas into words. As an example of translation, we may translate the formula V = lwh into words, with the literal factors representing words as follows:

V = volume of a rectangular solid l = length w = width h = height

This produces the following translation: The volume of a rectangular solid equals the length times the width times the height.

As a second example, we translate the algebraic expression 2 √x - 4 into words as follows: Twice the square root of a certain number, minus 4.

Practice problems. Translate each of the following expressions into words.

1. PV = k, where P represents pressure of a gas and V represents volume. (Assume constant temperature.)

2. x = y + 4, where x and y are numbers.

3. A = LW, where A is the area of a rectangle, L is its length, and W is its width.

4. d = rt, where d is distance, r is rate, and t is time.

Answers:

1. The pressure of a gas multiplied by its volume is constant, if the temperature is constant.

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MATHEMATICS, VOLUME 1

2. A certain number, x, is equal to the sum of another number, y, and 4.

3. The area of a rectangle is equal to the product of its length times its width.

4. Distance is equal to rate multiplied by time.

GRAPHING FORMULAS

We have seen that the formula is an equation. Since all formulas are equations they may be graphed. Graphs of formulas have wide use in the Navy in such fields as electronics and engineering. In practical applications it is often convenient to derive information from graphs of formulas rather than from formulas directly.

As an example, suppose that a fuel costs 30 cents per gallon. The formula for the cost in dollars of n gallons is

C = 0.30n

We see that this is a linear equation, the result of which passes through the origin (no constant term). Since we are interested only in positive values, we can eliminate three quadrants of the graph and use only the first quadrant. We already know one point on the graph is (0,0). We need plot only one other point to graph the formula. The result is shown in figure 14-4.

[FIGURE: Graph labeled "Figure 14-4.—Graph for the formula C = 0.30n." with x-axis "GALLONS (n)" ranging from 0-11 and y-axis "COST (DOLLARS)" ranging from 0.30-3.00. A linear line is plotted starting from origin.]

We may read the cost directly from the graph when the number of gallons is known, or the number of gallons when the cost is known. For instance, if 5-1/2 gal are sold, find 5-1/2 on the gallons scale and follow the vertical line from that point to the point where it intersects the graph of the formula. From this point, follow the horizontal line to the cost scale. The horizontal line intersects the cost scale at 1.65. Therefore the cost of 5-1/2 gal is $1.65.

Likewise, to answer the question, "How many gallons may be bought for $1.27," we would enlarge the graph enough to estimate to the exact cent. Then we would follow a horizontal line from 1.27 on the cost scale to the formula graph and follow a vertical line from that point to the gallons scale. Thus, 4-1/4 gal may be bought for $1.27.

Plotting two formulas on the same graph may help to solve certain kinds of problems. For example, suppose that two ships leave port at the same time. One averages 10 knots and the other averages 15 knots. How far has each traveled at the end of 3 hr and at the end of 5 hr? A graph to relate the two ships' movement at any time can be made as follows: Let the vertical scale be in nautical miles and the

[FIGURE: Graph labeled "Figure 14-5.—Graph of the formulas d = 10t and d = 15t." with x-axis "HOURS (t)" ranging from 0-10 and y-axis "NAUTICAL MILES (d)" ranging from 0-110. Two linear lines are plotted, one labeled "d = 10t SHIP 1" and another labeled "d = 15t SHIP 1".]

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Chapter 14—DEPENDENCE, FUNCTIONS, AND FORMULAS

horizontal scale be in hours. The formula for the first ship's distance related to time is

d = 10t

The formula for the second ship's distance re- lated to time is

d = 15t

We see that these formulas are linear and their curves pass through the origin. They are graphed in figure 14-5. With this graph we can now answer the ques- tions originally posed, at a glance. Thus in 3 hr the first ship traveled 30 mi and the second

traveled 45 mi. In 5 hr the first ship traveled 50 mi and the second traveled 75 mi. We could also answer such questions as: When the second ship has traveled 100 mi, how far has the other traveled? We first find the point on the graph of d = 15t where the ship has traveled 100 mi. We then follow the vertical line from that point to the point where it inter- sects the graph of the other formula. From the point of intersection we follow a horizontal line to the distance axis and see that the first ship has traveled about 67 mi when the second has traveled 100 mi. The foregoing examples serves to illustrate the wide variety of applications in which graphs of formulas are useful.

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CHAPTER 15 COMPLEX NUMBERS

In certain calculations in mathematics and related sciences, it is necessary to perform operations with numbers unlike any mentioned thus far in this course. These numbers, unfor- tunately called "imaginary" numbers by early mathematicians, are quite useful and have a very real meaning in the physical sense. The number system, which consists of ordinary numbers and imaginary numbers, is called the COMPLEX NUMBER system. Complex num- bers are composed of a "real" part and an "imaginary" part. This chapter is designed to explain imagi- nary numbers and to show how they can be com- bined with the numbers we already know.

REAL NUMBERS

The concept of number, as has been noted in previous chapters, has developed gradually. At one time the idea of number was limited to positive whole numbers. The concept was broadened to include posi- tive fractions; numbers that lie between the whole numbers. At first, fractions included only those numbers which could be expressed with terms that were integers. Since any frac- tion may be considered as a ratio, this gave rise to the term RATIONAL NUMBER, which is defined as any number which can be ex- pressed as the ratio of two integers. (Remem- ber that any whole number is an integer.) It soon became apparent that these numbers were not enough to complete the positive num- ber range. The ratio, π, of the circumference of a circle to its diameter, did not fit the con- cept of number thus far advanced, nor did such

numbers as √2 and √3. Although decimal values are often assigned to these numbers, they are only approximations. That is, π is not exactly equal to 22/7 or to 3.142. Such num- bers are called IRRATIONAL numbers to distinguish them from the other numbers of the system. With rational and irrational numbers, the posi- tive number system included all the numbers from zero to infinity in a positive direction. Since the number system was not complete with only positive numbers, the system was ex- panded to include negative numbers. The idea of negative rational and irrational numbers to minus infinity was an easy extension of the system. Rational and irrational numbers, positive and negative to ± infinity as they have been presented in this course, comprise the REAL NUMBER system. The real number system is pictured in figure 15-1.

OPERATORS

As shown in a previous chapter, the plus sign in an expression such as 5 + 3 can stand for either of two separate things: It indicates the positive number +3, or it indicates that 3 is to be added to 5; that is, it indicates the op- eration to be performed on +3. Likewise, in the problem 5 - 3, the minus sign may indicate the negative number -3, in which case the operation would be addition; that is, 5 + (-3). On the other hand, it may indicate the sign of operation, in which case +3 is to be subtracted from 5; that is, 5 - (+3). Thus, plus and minus signs may indicate positive and negative numbers, or they may in- dicate operations to be performed.

-∞←─────────────────────────────+∞ -4 -3 -2 -1 0 +1 +2 +3 +4 -π -2½ -√2 +½ +√2 +π

Figure 15-1.—The real number system.

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Chapter 15—COMPLEX NUMBERS

IMAGINARY NUMBERS

The number line pictured in figure 15-1 rep- resents all positive and negative numbers from plus infinity to minus infinity. However, there is a type of number which does not fit into the picture. Such a number occurs when we try to solve the following equation:

x² + 4 = 0 x² = -4 x = ± √-4

Notice the distinction between this use of the radical sign and the manner in which it was used in chapter 7. There the symbol is in- cluded with the radical sign to emphasize the fact that two values of x exist. Although both roots exist, only the positive one usually given. This is in accordance with usual mathe- matical convention.

The equation

x = ± √-4

raises an interesting question: What number multiplied by itself yields -4? The square of -2 is +4. Likewise, the square of +2 is +4. There is no number in the system of real numbers that is the square root of a negative number. The square root of a nega- tive number came to be called an IMAGINARY NUMBER. When this name was assigned the square roots of negative numbers, it was natu- ral to refer to the other known numbers as the REAL numbers.

IMAGINARY UNIT

To reduce the problem of imaginary num- bers to its simplest terms, we proceed as far as possible using ordinary numbers in the so- lution. Thus, we may write √-4 as a product

√-1·4 = √4·√-1 = ±2 √-1

Likewise,

√-5 = √5 √-1

Also,

3 √-7 = 3 √7 √-1

Thus, the problem of giving meaning to the square root of any negative number reduces to that of finding a meaning for √-1. The square root of minus 1 is designated i by mathematicians. When it appears with a co- efficient, the symbol i is written last unless the coefficient is in radical form. This con- vention is illustrated in the following examples:

±2 √-1 = ±2i √5 √-1 = i √5 3 √7 √-1 = 3i √7

The symbol i stands for the imaginary unit √-1. An imaginary number is any real multi- ple, positive or negative, of i. For example, -7i, +7i, i √5, and 0i are all imaginary num- bers.

In electrical formulas the letter i denotes current. To avoid confusion, electronic techni- cians use the letter j to indicate √-1 and call it "operator j." The name "imaginary" should be thought of as a technical mathematical term of convenience. Such numbers have a very real purpose in the physical sense. Also it can be shown that ordinary mathematical operations such as addition, multiplication, and so forth, may be performed in exactly the same way as for the so-called real numbers.

Practice problem. Express each of the following as some real number times i:

1. √-16 3. √-5 5. √-25 2. 2 √-1 4. 4/i √-i² 6. ∛√(3/16)

Answers:

1. 4i 3. i √5 5. 5i 2. 2i 4. di 6. 3i/4

Powers of the Imaginary Unit

The following examples illustrate the re- sults of raising the imaginary unit to various powers:

i = √-1 i² = √-1 √-1, or -1 i³ = i²i = -1i, or -i i⁴ = i²i² = -1 · -1 = +1 i⁻¹ = 1/i = i/i² = i/-1 = -i

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MATHEMATICS, VOLUME 1

We see from these examples that an even power of i is a real number equal to +1 or -1. Every odd power of i is imaginary and equal to i or -i. Thus, all powers of i reduce to one of the following four quantities: √-1, -1, -√-1, or +1.

GRAPHICAL REPRESENTATION

Figure 15-1 shows the real numbers repre- sented along a straight line, the positive num- bers extending from zero to the right for an infinite distance, and the negative numbers ex- tending to the left of zero for an infinite dis- tance. Every point on this line corresponds to a real number, and there are no gaps between them. It follows that there is no possibility of representing imaginary numbers on this line.

Earlier, we noted that certain signs could be used as operators. The plus sign could stand for the operation of addition. The minus sign could stand for the operation of subtraction. Likewise, it is easy to explain the imaginary number i graphically as an operator indicating a certain operation to be performed on the number of which it is the coefficient.

If we graphically represent the length, n, on the number line pictured in figure 15-2 (A), we start at the point O and measure to the right (positive direction) a distance representing n units. If we multiply n by -1, we may repre- sent the result -n by measuring from O in a negative direction a distance equal to n units.

Graphically, multiplying a real number by -1 is equivalent to rotating the line that repre- sents the number about the point O through 180° so that the new position of n is in the opposite direction and a distance n units from O. In this case we may think of -1 as the operator that rotates n through two right angles to its new position (fig. 15-2 (B)).

As we have shown, i² = -1. Therefore, we have really multiplied n by i twice in succession, or i x i. In other words, multiplying by -1 is the same as multi- plying by i twice in succession. Logically, if we multiplied n by i once, the line would be rotated only half as much as before—that is, through only one right angle, or 90°. The new segment ni would be measured in a direction 90° from the line n. Thus, i is an operator that rotates a number through one right angle. (See fig. 15-3.)

We have shown previously that a positive number may have two real square roots, one positive and one negative. For example, √9 = ±3.

[FIGURE: Diagram showing (A) number line with -n, 0, n marked, and (B) semicircle showing 90° rotations]

Figure 15-2.—Graphical multiplication by -1 and by operator i².

[FIGURE: Graph showing coordinate system with ni and n marked, with 90° angle]

Figure 15-3.—Graphical multipli- plication by operator i.

We also saw that an imaginary number may have two roots. For example, √-i² is equal to ±2i. When the operator i graphically rotates a number, it may do so in a counterclockwise or a clockwise direction. Likewise, the opera- tor i may graphically rotate a number in either direction. This gives meaning to numbers such as ±2i. It has been agreed that a number multiplied by +i is to be rotated 90° in a coun- terclockwise direction. A number multiplied by -i is to be rotated 90° in a clockwise di- rection.

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Chapter 15—COMPLEX NUMBERS

In figure 15-4, +2i is represented by rotating the line that represents the positive real num- ber 2 through 90° in a counterclockwise direc- tion. It follows that -2i is represented by rotating the line that represents the positive real number 2 through 90° in a clockwise direction.

+2i

-2 -1 0 1 2

-2i

Figure 15-4.—Graphical representation of ±2i.

In figure 15-5, notice that the idea of i as an operator agrees with the concept advanced con- cerning the powers of i. Thus, i rotates a num- ber through 90°; i² or -1 rotates the number through 180°, and the number is real and nega- tive; i³ rotates the number through 270°, which has the same effect as -i; and i⁴ rotates the number through 360°, and the number is once again positive and real.

THE COMPLEX PLANE

All imaginary numbers may be represented graphically along a line extending through zero and perpendicular to the line representing the real numbers. This line may be considered in- finite in both the positive and negative direc- tions, and all multiples of i may be represented on it. This graph is similar to the rectangular coordinate system studied earlier.

In this system, the vertical or y axis is called the axis of imaginaries, and the horizon- tal or x axis is called the axis of reals. In the rectangular coordinate system, real numbers are laid off on both the x and y axes and the plane on which the axes lie is called the real plane. When the y axis is the axis of imagi- naries, the plane determined by the x and y axes is called the COMPLEX PLANE (fig. 15-6).

In any system of numbers a unit is necessary for counting. Along the real axis, the unit is the number 1. As shown in figure 15-6, along the imaginary axis the unit is i. Numbers that lie along the imaginary axis are called PURE IMAGINARIES. They will always be some multiple of i, the imaginary unit. The numbers 5i, 3i, √2 i, and √-7 are examples of pure imaginaries.

NUMBERS IN THE COMPLEX PLANE

All numbers in the complex plane are complex numbers, including reals and pure imaginaries. However, since the reals and imaginaries have

bi = b√-1

bi² = b√-1√-1 = -b

i² = -i

bi⁴ = b√-1√-1√-1√-1 = b(-1)(-1) = b

bi³ = b√-1√-1√-1 = -bi

Figure 15-5.—Operation with powers of i.

161

MATHEMATICS, VOLUME 1

AXIS OF IMAGINARIES Y 4i 3i 2i i AXIS OF REALS -4 -3 -2 -1 0 1 2 3 4 X -2i -3i -4i

Figure 15-6.—The complex plane.

the special property of being located on the axes, they are usually identified by their dis- tinguishing names.

The term complex number has been defined as the indicated sum or difference of a real number and an imaginary number.

For example, 3 + 5 √-1 or 3 + 5i, 2 - 6i, and -2 + √-5 are complex numbers. In the complex number 7 - i √2, 7 is the real part and -i √2 is the imaginary part.

All complex numbers correspond to the gen- eral form a + bi, where a and b are real num- bers. When a has the value 0, the real term disappears and the complex number becomes a pure imaginary. When b has the value of 0, the imaginary term disappears and the complex number becomes a real number. Thus, 4 may be thought of as 4 + 0i, and 3i may be consid- ered 0 + 3i. From this we may reason that the real number and the pure imaginary number are special cases of the complex number. Con- sequently, the complex number may be thought of as the most general form of a number and can be constructed to include all the numbers of algebra as shown in the chart in figure 15-7.

Plotting Complex Numbers

Complex numbers may easily be plotted in the complex plane. Pure imaginaries are plotted along the vertical axis, the axis of imag- inaries, and real numbers are plotted along the horizontal axis, the axis of reals. It follows that other points in the complex plane must represent numbers that are part real and part imaginary; in other words, complex numbers. If we wish to plot the point 3 + 2i, we note that the number is made up of the real number 3 and the imaginary number 2i. Thus, as in fig- ure 15-8, we measure along the real axis in a

COMPLEX NUMBERS (a + bi)

REAL NUMBERS PURE IMAGINARIES (FORM IS a WHERE (FORM IS bi WHERE b IS o) a IS o)

RATIONAL IRRATIONAL

NEGATIVE POSITIVE √2 √2 INTEGERS INTEGERS √5 -√3

NEGATIVE POSITIVE Π -Π FRACTIONS FRACTIONS etc. etc

Figure 15-7.—The complex number system.

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Chapter 15–COMPLEX NUMBERS

Y AXIS OF IMAGINARIES Y 5i P 4i 3i 3+2i -3+2i 2i 8 1i -5 -4 -3 -2 -1 0 1 2 3 4 5 X O 6 A -1i 53.1° AXIS OF REALS X -2i 3-2i -3-2i -3i -4i -5i

Figure 15-8.–Plotting complex numbers. Figure 15-9.–A complex number shown as a vector. positive direction. At point (3, 0) on the real axis we turn through one right angle and meas- ure 2 units up and parallel to the imaginary the legs of the triangle, which in this case are axis. Likewise, the number -3 + 2i is 3 units the real numbers, 6 and 8, thus, to the left and up 2 units; the number 3 - 2i is 3 units to the right and down 2 units; and the OP = √6² + 8² number -3 -2i is 3 units to the left and down = √100 2 units. = 10

Complex Numbers as Vectors However, since a vector has direction as well as magnitude, we must also show the di- A vector is a directed line segment. A com- rection of the segment; otherwise the seg- plex number represents a vector expressed in ment OP could radiate in any direction on the the RECTANGULAR FORM. For example, the complex plane from point O. The expression complex number 6 + 8i in figure 15-9 may be 10/53.1° indicates that the vector OP has been considered as representing either the point P rotated counterclockwise from the initial posi- or the line OP. The real parts of the complex tion through angle of 53.1°. (The initial po- number (6 and 8) are the rectangular compo- sition in a line extending from the origin to the nents of the vector. The real parts are the legs right along OX.) This method of expressing the of the right triangle (sides adjacent to the right vector quantity is called the POLAR FORM. angle), and the vector OP is its hypotenuse (side The number represents the magnitude of the opposite the right angle). If we merely wish to quantity, and the angle represents the position indicate the vector OP, we may do so by writ- of the vector with respect to the horizontal ref- ing the complex number that represents it along erence, OX. Positive angles represent coun- the segment as in figure 15-9. This method not terclockwise rotation of the vector, and nega- only fixes the position of point P, but also shows tive angles represent clockwise rotation. The what part of the vector is imaginary (PA) and polar form is generally simpler for multiplica- what part is real (OA). tion and division, but its use requires a knowl- If we wish to indicate a number that shows edge of trigonometry. the actual length of the vector OP, it is neces- sary to solve the right triangle OAP for its hypotenuse. This may be accomplished by tak- ing the square root of the sum of the squares of 163

MATHEMATICS, VOLUME 1

ADDITION AND SUBTRACTION OF 5. (-5 + 3i) - (4 - 2i) COMPLEX NUMBERS 6. (a + bi) - (-c + di) Pure imaginaries are added and subtracted in the same way as any other algebraic quanti- Answers: ties. The coefficients of similar terms are added or subtracted algebraically, as follows: 1. 3a + 2i 4. -1 2. 5i 5. -9 + 5i 4i + 3i = 7i 3. a + c + (b + d)i 6. a + c + (b - d)i 4i - (-3i) = 7i

4i - (-3i) = 7i MULTIPLICATION OF COMPLEX NUMBERS Likewise, complex numbers in the rectangular form are combined like any other algebraic Generally, the rules for the multiplication of polynomials. Add or subtract the coefficients complex numbers and pure imaginaries are the of similar terms algebraically. If parentheses same as for other algebraic quantities. How- enclose the numbers, first remove the paren- ever, there is one exception that should be theses. Next, place the real parts together and noted: The rule for multiplying numbers under the imaginary parts together. Collect terms. radical signs does not apply to TWO NEGA- As examples, consider the following: TIVE numbers. When at least one of two radi- cands is positive, the radicands can be multi- 1. (2 - 3i) + (5 + 4i) = 2 - 3i + 5 + 4i plied immediately, as in the following example: = 2 + 5 - 3i + 4i = 7 + i √2 √3 = √6 √2 √-3 = √-6 2. (2 - j3) - (5 + j4) = 2 - j3 - 5 - j4 = 2 - 5 - j3 - j4 When both radicands are negative, however, = -3 - j7 as in √-2 √-3, an inconsistent result is ob- tained if we multiply both numbers under the In example 2, notice that the convention for radical signs immediately. To get the correct writing operator j (the electronics form of the result, express the imaginary numbers first in imaginary unit) with numerical coefficients is terms of i, as follows: to place it first. If the complex numbers are placed one un- √-2 √-3 = i √2 · i √3 der the other, the results of addition and sub- = i² √2 √3 traction appear as follows: = i² √6 ADDITION SUBTRACTION = (-1) √6 = -√6 3 + 4 √-1 a + jb 2 - 7 √-1 ← -c + jd Multiplying complex numbers is equivalent 5 - 3 √-1 (a - c) + j(b - d) to multiplying binomials in the manner ex- plained previously. After the multiplication is Practice problems. Add or subtract as in- performed, simplify the powers of i as in the dicated, in the following problems: following examples:

1. (3a + 4i) + (0 - 2i) 1. 4 - i 2. (3 + 2i) + (-3 + 3i) 3 + i 3. (a + bi) + (c + di) 12 - 3i 4. (1 + 2 √-1) + (-2 - 2 √-1) + 4i - i² 12 + i - i² = 12 + i - (-1) = 13 + i

164

Chapter 15—COMPLEX NUMBERS

2. (-6 + 5 √-7) (8 - 2 √-7)

= (-6 + 5i √7) (8 - 2i √7) = -48 - 12i√7 + 40i√7 - 10(7)i² = -48 + 28i √7 + 70 = 22 + 28i √7

Practice problems. Perform the indicated operations:

1. √-3 √-16 2. √-2 √18 3. √-9 √-4 4. a √-ba · √-b 5. (2 + 5i) (3 - 2i) 6. (a + √-b) (a - √-b) 7. (-2 + √-4) (-1 + √-4) 8. (6 - √-7) (6 + √-7)

Answers:

1. -12 5. 16 + 11i 2. 6i 6. a² + b 3. -6 7. -2 - 6i 4. -ab √a 8. 55 + 2i √7

CONJUGATES AND SPECIAL PRODUCTS

Two complex numbers that are alike except for the sign of their imaginary parts are called CONJUGATE COMPLEX NUMBERS. For example, 3 + 5i and 3 - 5i are conjugates. Either number is the conjugate of the other.

If one complex number is known, the conjugate can be obtained immediately by changing the sign of the imaginary part. The conjugate of -8 + √-10 is -8 - √-10. The conjugate of √-6 is -√-6.

The sum of two conjugate complex numbers is a real number, as illustrated by the following:

1. (3 + j5) + (3 - j5) = 2(3) = 6

2. (-1/2 + √-3/2) + (-1/2 - √-3/2) = -1/2 + √3/2 i - 1/2 - √3/2 i = -1/2 + (-1/2) = -1

Product of Two Conjugates

The product of two conjugate complex numbers is a real number. Multiplying two conjugates is equivalent to finding the product of the sum and difference of two numbers.

Consider the following examples:

1. (3 + j5) ( 3 - j5) = 3² - (j5)² = 9 - 25(-1) = 9 + 25 = 34

2. (-1/2 + √3/2 i)(-1/2 - √3/2 i) = (-1/2)² - (√3/2 i)² = 1/4 - [3/4(-1)] = 1/4 + 3/4 = 1

Squaring a Complex Number

Squaring a complex number is equivalent to raising a binomial to the second power. For example:

(-6 - √-25)² = (-6 - j5)² = [(-1) · (6 + j5)]² = (-1)² · (6² + j60 + j²25) = 36 + j60 - 25 = 11 + j60

DIVISION OF COMPLEX NUMBERS

When dividing by a pure imaginary, the denominator may be rationalized and the problem thus simplified by multiplying both numerator and denominator by the denominator. Thus,

12/(√-2) = 12/(i√2) · (i√2)/(i√2) = 12i√2/2i² = 6i√2/-1 = -6i√2

165

MATHEMATICS, VOLUME 1

Division of complex numbers can be accomplished by multiplying the numerator and denominator by the number that is the conjugate of the denominator. This process is similar to the process of rationalizing a denominator in the case of real numbers that are irrational.

As an example, consider

(5 - 2i)/(3 + i)

The denominator is 3 + i. Its conjugate is 3 - i. Multiplying numerator and denominator by 3 - i gives

(5 - 2i)/(3 + i) · (3 - i)/(3 - i) = (15 - 11i + 2i²)/(9 - i²)

= (15 - 11i - 2)/(9 + 1)

= (13 - 11i)/10

= 13/10 - 11/10 i

Practice problems. Rationalize the denominators and simplify:

1. (2 √-1)/(4 + 2 √-1)

2. (-2 + 4i)/(-1 + 4i)

3. (3 + √-2)/(3 - √-2)

4. 3/(1 - i √3)

5. (1 - i)/(2 - i)

6. 8/(2 + √-2)

Answers:

1. (2i + 1)/5

2. (18 + 4i)/17

3. (7 + 6i √2)/11

4. 3/4 + 3/4 i √3

5. (3 - i)/5

6. (8 - 4i √2)/3

166

CHAPTER 16 QUADRATIC EQUATIONS IN ONE VARIABLE

The degree of an equation in one variable is the exponent of the highest power to which the variable is raised in that equation. A second-degree equation in one variable is one in which the variable is raised to the second power. A second-degree equation is often called a QUADRATIC EQUATION. The word quadratic is derived from the Latin word quadratus, which means "squared." In a quadratic equation the term of highest degree is the squared term. For example, the following are quadratic equations:

x² + 3x + 4 = 0 3m + 4m² = 6

The terms of degree lower than the second may or may not be present. The possible terms of lower degree than the squared term in a quadratic equation are the first-degree term and the constant term. In the equation

3x² - 8x - 5 = 0

-5 is the coefficient of x⁰. If we wished to emphasize the powers of x in this equation, we could write the equation in the form

3x² - 8x¹ - 5x⁰ = 0

Examples of quadratic equations in which either the first-degree term or the constant term is missing are:

1. 4x² = 16 2. y² + 16y = 0 3. e² + 12 = 0

GENERAL FORM OF A QUADRATIC EQUATION

Any quadratic equation can be arranged in the general form:

ax² + bx + c = 0

If it has more than three terms, some of them will be alike and can be combined, after which the final form will have at most three terms. For example,

2x² + 3 + 5x - 1 + x² = 4 - x² - 2x - 3

reduces to the simpler form

4x² + 7x + 1 = 0

In this form, it is easy to see that a, the coefficient of x², is 4; b, the coefficient of x, is 7; and c, the constant term, is 1.

Sometimes the coefficients of the terms of a quadratic appear as negative numbers, as follows:

2x² - 3x - 5 = 0

This equation can be rewritten in such a way that the connecting signs are all positive, as in the general form. This is illustrated as follows:

2x² + (-3)x + (-5) = 0

In this form, the value of a is seen to be 2, b is -3, and c is -5.

An equation of the form

x² + 2 = 0

has no x term. This can be considered as a case in which a is 1 (coefficient of x² understood to be 1), b is 0, and c is 2. For the purpose of emphasizing the values of a, b, and c with reference to the general form, this equation can be written

x² + 0x + 2 = 0

The coefficient of x² can never be 0; if it were 0, the equation would not be a quadratic. If the coefficients of x and x⁰ are 0, then those terms do not normally appear. To say that the coefficient of x⁰ is 0 is the same as saying that the constant term is 0 or is missing.

167

MATHEMATICS, VOLUME 1

A ROOT of an equation in one variable is a value of the variable that satisfied the equation. Every equation in one variable, with constants as coefficients and positive integers as exponents, has as many roots as the exponent of the highest power. In other words, the number of roots is the same as the degree of the equation. A fourth-degree equation has four roots, a cubic (third-degree) equation has three roots, a quadratic equation has two roots, and a linear equation has one root.

As an example, 6 and -1 are roots of the quadratic equation

x² - 5x - 6 = 0

This can be verified by substituting these values into the equation and noting that an identity results in each case.

Substituting x = 6 gives 6² - 5(6) - 6 = 0 36 - 36 = 0 0 = 0

Substituting x = -1 gives (-1)² - 5(-1) - 6 = 0 1 + 5 - 6 = 0 6 - 6 = 0 0 = 0

Several methods of finding the roots of quadratic equations (SOLVING) are possible. The most common methods are solution by FACTORING and solution by the QUADRATIC FORMULA. Less commonly used methods of solution are accomplished by completing the square and by graphing.

SOLUTION BY FACTORING

The equation x² - 36 = 0 is a pure quadratic equation. There are two numbers which, when substituted for x, will satisfy the equation as follows:

(+6)² - 36 = 0 36 - 36 = 0

also

(-6)² - 36 = 0 36 - 36 = 0

Thus, +6 and -6 are roots of the equation

x² - 36 = 0

The most direct way to solve a pure quadratic (one in which no x term appears and the constant term is a perfect square) involves rewriting with the constant term in the right member, as follows:

x² = 36

Taking square roots on both sides, we have

x = ±6

The reason for expressing the solution as both plus and minus 6 is found in the fact that both +6 and -6, when squared, produce 36.

The equation

x² - 36 = 0

can also be solved by factoring, as follows:

x² - 36 = 0 (x + 6)(x - 6) = 0

We now have the product of two factors equal to zero. According to the zero factor law, if a product is zero, then one or more of its factors is zero. Therefore, at least one of the factors must be zero, and it makes no difference which one. We are free to set first one factor and then the other factor equal to zero. In so doing we derive two solutions or roots of the equation.

If x + 6 is the factor whose value is 0, then we have

x + 6 = 0 x = -6

If x - 6 is the zero factor, we have

x - 6 = 0 x = 6

When a three-term quadratic is put into simplest form, it is customary to place all terms on the left side of the equality sign with the squared term first, the first-degree term next, and the constant term last, as in

9x² - 2x + 7 = 0

168

Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

If the trinomial in the left member is readily factorable, the equation can be solved quickly by separating the trinomial into factors. Con- sider the equation

3x² - x - 2 = 0

By factoring the trinomial, the equation be- comes

(3x + 2)(x - 1) = 0

Once again we have two factors, the product of which is 0. This means that one or the other of them (or both) must have the value 0. If the zero factor is 3x + 2, we have

3x + 2 = 0

3x = -2

x = - 2/3

If the zero factor is x - 1, we have

x - 1 = 0

x = 1

Substituting first x = 1 and then x = -2/3 in the original equation, we see that both roots satisfy it. Thus,

3(1)² - (1) - 2 = 0

3 - 1 - 2 = 0

0 = 0

3(-2/3)² - (-2/3) - 2 = 0

4/3 + 2/3 - 2 = 0

0 = 0

In summation, when a quadratic may be readily factored, the process for finding its roots is as follows:

1. Arrange the equation in the order of the descending powers of the variable so that all the terms appear in the left member and zero appears in the right. 2. Factor the left member of the equation. 3. Set each factor containing the variable equal to zero and solve the resulting equations. 4. Check by substituting each of the derived roots in the original equation.

EXAMPLE: Solve the equation x² - 4x = 12 for x.

1. x² - 4x - 12 = 0

2. (x - 6)(x + 2) = 0

3. x - 6 = 0 x + 2 = 0 x = 6 x = -2

4. (6)² - 4(6) = 12 (x = 6) 36 - 24 = 12 12 = 12

(-2)² - 4(-2) = 12 (x = -2) 4 + 8 = 12 12 = 12

Practice problems. Solve the following equa- tions by factoring:

1. x² + 10x - 24 = 0 4. 7y² - 19y - 6 = 0 2. a² - a - 56 = 0 5. m² - 4m = 96 3. y² - 2y = 63

Answers:

1. x = -12 4. y = 3 x = 2 y = -2/7

2. a = 8 5. m = -8 a = -7 m = 12

3. y = -7 y = 9

SOLUTION BY COMPLETING THE SQUARE

When a quadratic cannot be solved by fac- toring, or the factors are not readily seen, an- other method of finding the roots is needed. A method that may always be used for quadratics in one variable involves perfect square trino- mials. These, we recall, are trinomials whose factors are identical. For example,

x² - 10x + 25 = (x - 5)(x - 5) = (x - 5)²

Recall that in squaring a binomial, the third term of the resulting perfect square trinomial

MATHEMATICS, VOLUME 1

is always the square of the second term of the binomial. The coefficient of the middle term of the trinomial is always twice the second term of the binomial. For example, when (x + 4) is squared, we have

x + 4 x + 4 ___________ x² + 4x + 4x + 16 ___________ x² + 8x + 16

Hence if both the second- and first-degree terms of a perfect square trinomial are known, the third term may be written by squaring one-half the coefficient of the first-degree term. Essentially, in completing the square, cer- tain quantities are added to one member and subtracted from the other, and the equation is so arranged that the left member is a perfect square trinomial. The square roots of both members may then be taken, and the subsequent equalities may be solved for the variable. For example,

x² + 5x - 11/4 = 0

cannot be readily factored. To solve for x by completing the square, we proceed as follows:

1. Leave only the second- and first-degree terms in the left member.

x² + 5x = 11/4

(If the coefficient of x² is not 1, divide through by the coefficient of x².) 2. Complete the square by adding to both members the square of half the coefficient of the x term. In this example, one-half of the

coefficient of the x term is 5/2, and the square

of 5/2 is 25/4. Thus,

x² + 5x + 25/4 = 11/4 + 25/4

3. Factor the left member and simplify the right member.

(x + 5/2)² = 9

4. Take the square root of both members.

√(x + 5/2)² = ±√9

x + 5/2 = ±3

Remember that, in taking square roots on both sides of an equation, we must allow for the fact that two roots exist in every second-degree equation. Thus we designate both the plus and the minus root of 9 in this example. 5. Solve the resulting equations.

x + 5/2 = 3 x + 5/2 = -3

x = 6/2 - 5/2 x = -6/2 - 5/2

x = 1/2 x = -11/2

6. Check the results.

(1/2)² + 5/2 - 11/4 = 0

5/2 - 10/4 = 0

0 = 0

(-11/2)² + (5)(-11/2) - 11/4 = 0

121/4 - 55/2 - 11/4 = 0

110/4 - 55/2 = 0

0 = 0

The process of completing the square may always be used to solve a quadratic equation. However, since this process may become com- plicated in more complex equations, a formula based on completing the square has been devel- oped in which known quantities may be substi- tuted in order to derive the roots of the quad- ratic equation. This formula is explained in the following paragraphs.

SOLUTION BY THE QUADRATIC FORMULA

The quadratic formula is derived by apply- ing the process of completing the square to

Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

solve for x in the general form of the quadratic equation, ax² + bx + c = 0. Remember that the general form represents every possible quad- ratic equation. Thus, if we can solve this equa- tion for x, the solution will be in terms of a, b, and c. To solve this equation for x by complet- ing the square, we proceed as follows:

1. Subtract the constant term, c, from both members.

ax² + bx + c = 0 ax² + bx = -c

2. Divide all terms by a so that the coeffi- cient of the x² term becomes unity.

x² + (b/a)x = -c/a

3. Add the square of one-half the coefficient of the x term, b/a, to both members.

Square b/2a: (b/2a)² = b²/4a²

Add: x² + (b/a)x + b²/4a² = b²/4a² - c/a

4. Factor the left member and simplify the right member.

(x + b/2a)² = b²/4a² - 4ac/4a²

5. Take the square root of both members.

x + b/2a = ± √(b² - 4ac)/2a

6. Solve for x.

x = -b/2a ± √(b² - 4ac)/2a

= (-b ± √(b² - 4ac))/2a

Thus, we have solved the equation repre- senting every quadratic for its unknown in terms of its constants a, b, and c. Hence, in a given quadratic we need only substitute in the ex- pression

(-b ± √(b² - 4ac))/2a

the values of a, b, and c, as they appear in the particular equation, to derive the roots of that equation. This expression is called the QUAD- RATIC FORMULA. The general quadratic equation, ax² + bx + c = 0, and the quadratic formula should be memorized. Then, when a quadratic cannot be solved quickly by factoring, it may be solved at once by the formula.

EXAMPLE: Use the quadratic formula to solve the equation

x² + 30 - 11x = 0.

SOLUTION:

1. Set up the equation in standard form.

x² - 11x + 30 = 0

Then a (coefficient of x²) = 1 b (coefficient of x) = -11 c (the constant term) = 30

2. Substituting,

x = (-b ± √(b² - 4ac))/2a

= -(-11) ± √((-11)² - 4(1)(30))/2(1)

= (11 ± √(121 - 120))/2

= (11 ± 1)/2 = 6 or 5

3. Checking:

When When

x = 6, x = 5, (6)² - 11(6) + 30 = 0 (5)² - 11(5) + 30 = 0 36 - 66 + 30 = 0 25 - 55 + 30 = 0 0 = 0 0 = 0

EXAMPLE: Find the roots of

2x² - 3x - 1 = 0

Here, a = 2, b = -3, and c = -1.

MATHEMATICS, VOLUME 1

Substituting into the quadratic formula gives

x = (-(-3) ± √((-3)² - 4(2)(-1)))/(2(2))

= (3 ± √(9 + 8))/4

= (3 ± √17)/4

The two roots are

x = (3/4) + (1/4)√17 and x = (3/4) - (1/4)√17

These roots are irrational numbers, since the radicals cannot be removed. If the decimal values of the roots are de- sired, the value of the square root of 17 can be taken from appendix I of this course. Substi- tuting √17 = 4.1231 and simplifying gives

x₁ = (3 + 4.1231)/4 and x₂ = (3 - 4.1231)/4

x₁ = 7.1231/4 x₂ = -1.1231/4

x₁ = 1.781 x₂ = -0.281

In decimal form, the roots of 2x² - 3x - 1 = 0 to the nearest tenth are 1.8 and -0.3.

Notice that the subscripts, 1 and 2, are used to distinguish between the two roots of the equa- tion. The three roots of a cubic equation in x might be designated x₁, x₂, and x₃. Sometimes the letter r is used for root. Using r, the roots of a cubic equation could be labeled r₁, r₂, and r₃.

Checking:

When x₁ = (3 + √17)/4

2x² - 3x - 1 = 0

then

2((3 + √17)/4)² - 3((3 + √17)/4) - 1 = 0

(3 + √17)²/8 - 3(3 + √17)/4 - 1 = 0

(9 + 6√17 + 17)/8 - (18 - 6√17)/8 - 8 = 0

0 = 0

When

x₂ = (3 - √17)/4

then

2((3 - √17)/4)² - 3((3 - √17)/4) - 1 = 0

(9 - 6√17 + 17)/8 - (9 - 3√17)/4 - 1 = 0

Multiplying both members of the equation by 8, the LCD, we have

8((9 - 6√17 + 17)/8) - 8((9 - 3√17)/4) - 8(1) = 0

9 - 6√17 + 17 - 2(9 - 3√17) - 8 = 0

9 - 6√17 + 17 - 18 + 6√17 - 8 = 0

0 = 0

Practice problems. Use the quadratic for- mula to find the roots of the following equations:

1. 3x² - 20 - 7x = 0 3. 15x² - 22x - 5 = 0

2. 4x² - 3x - 5 = 0 4. x² + 7x = 8

Answers:

1. x₁ = 4 3. x₁ = 5/3

x₂ = -5/3 x₂ = -1/5

2. x₁ = (3 + √89)/8 4. x₁ = 1

x₂ = (3 - √89)/8 x₂ = -8

GRAPHICAL SOLUTION

A fourth method of solving a quadratic equa- tion is by means of graphing. In graphing lin- ear equations using both axes as reference, we recall that an independent variable, x, and a dependent variable, y, were needed. The co- ordi nates of points on the graph of the equation were designated (x, y).

Since the quadratics we are considering con- tain only one variable, as in the equation

x² - 8x + 12 = 0

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Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

we cannot plot values for the equations in the present form using both x and y axes. A dependent variable, y, is necessary. If we think of the expression

x² - 8x + 12

as a function, then this function can be considered to have many possible numerical values, depending on what value we assign to x. The particular value or values of x which cause the value of the function to be 0 are solutions for the equation

x² - 8x + 12 = 0

For convenience, we may choose to let y represent the function

x² - 8x + 12

If numerical values are now assigned to x, the corresponding values of y may be calculated. When these pairs of corresponding values of x and y are tabulated, the resulting table provides the information necessary for plotting a graph of the function.

EXAMPLE: Graph the equation

x² + 2x - 8 = 0

and from the graph write the roots of the equation.

SOLUTION:

1. Let y = x² + 2x - 8.

2. Make a table of the y values corresponding to the value assigned x, as shown in table 16-1.

Table 16-1.—Tabulation of x and y values for the function y = x² + 2x - 8.

if x = ----- | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 then y --- | 7 | 0 | -5 | -8 | -9 | -8 | -5 | 0 | 7

3. Plot the pairs of x and y values that appear in the table as coordinates of points on a rectangular coordinate system as in figure 16-1 (A).

4. Draw a smooth curve through these points, as shown in figure 16-1 (B). Notice that this curve crosses the X axis in two places. We also recall that, for any point on the X axis, the y coordinate is zero. Thus, in the figure we see that when y is zero, x is -4 or +2. When y is zero, furthermore, we have the original equation,

[Two coordinate system diagrams labeled (A) and (B) showing plotted points and a parabola]

Figure 16-1.—Graph of the equation y = x² + 2x - 8. (A) Points plotted; (B) curve drawn through plotted points.

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MATHEMATICS, VOLUME 1

x² + 2x - 8 = 0

Thus, the values of x at these points where the graph of the equation crosses the X axis (x = -4 or +2) are solutions to the original equation. We may check these results by solving the equation algebraically. Thus,

x² + 2x - 8 = 0 (x + 4)(x - 2) = 0

x₁ + 4 = 0 x₂ - 2 = 0 x₁ = -4 x₂ = 2

Check: (-4)² + 2(-4) - 8 = 0 (2)² + 2(2) - 8 = 0 16 - 8 - 8 = 0 4 + 4 - 8 = 0 0 = 0 0 = 0

The curve in figure 16-1 (B) is called a PARABOLA. Every quadratic of the form ax² + bx + c = y will have a graph of this general shape. The curve will open downward if a is negative, and upward if a is positive. Graphing provides a fourth method of finding the roots of a quadratic in one variable. When the equation is graphed, the roots will be the X intercepts (those values of x where the curve crosses the X axis). The X intercepts are the points at which y is 0. Practice problems. Graph the following quadratic equations and read the roots of each equation from its graph

1. x² - 4x - 8 = 0

2. 6x - 5 - x² = 0

Answers: 1. See figure 16-2. x = 5.5; x = -1.5 2. See figure 16-3. x = 1; x = 5

MAXIMUM AND MINIMUM POINTS

It will be seen from the graphs of quadratics in one variable that a parabola has a maximum or minimum value, depending on whether the curve opens upward or downward. Thus, when a is negative the curve passes through a maximum value; and when a is positive, the curve passes through a minimum value. Often these maximum or minimum values comprise the only information needed for a particular problem.

In higher mathematics it can be shown that the X coordinate, or abscissa, of the maximum or minimum value is

x = -b/2a

In other words, if we divide minus the coefficient of the x term by twice the coefficient of the x² term, we have the X coordinate of the maximum or minimum point. If we substitute this value for x in the original equation, the result is the Y value or ordinate, which corresponds to the X value. For example, we know that the graph of the equation

x² + 2x - 8 = y

passes through a minimum value because a is positive. To find the coordinates of the point where the parabola has its minimum value, we note that a = 1, b = 2, c = -8. From the rule given above, the X value of the minimum point is

x = -b/2a

x = -2/2(1)

x = -1

Substituting this value for x in the original equation, we have the value of the Y coordinate of the minimum point. Thus,

(-1)² + 2(-1) - 8 = y 1 - 2 - 8 = y -9 = y

The minimum point is (-1, -9). From the graph in figure 16-1 (B), we see that these coordinates are correct. Thus, we can quickly and easily find the coordinates of the minimum or maximum point for any quadratic of the form ax² + bx + c = 0. Practice problems. Without graphing, find the coordinates of the maximum or minimum points for the following equations and state whether they are maximum or minimum.

1. 2x² - 5x + 2 = 0 2. 6s - 3x - x² = 0 3. 3 + 7x - 6x² = 0 4. 24x² - 14x = 3

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Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

Figure 16-2.—Graph of x² - 4x - 8 = 0.

Answers:

1. x = 5/4 Minimum

y = -9/8

2. x = -3/2 Maximum

y = 281/4

3. x = 7/12 Maximum

y = 121/24

4. x = 7/24 Minimum

y = -121/24

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MATHEMATICS, VOLUME 1

Figure 16-3.—Graph of 6x - 5 - x² = 0.

THE DISCRIMINANT

The roots of a quadratic equation may be classified in accordance with the following criteria:

1. Real or imaginary. 2. Rational or irrational. 3. Equal or unequal.

The task of discriminating among these possi- ble characteristics to find the nature of the roots is best accomplished with the aid of the quadratic formula. The part of the quadratic formula which is used is called the DISCRIMI- NANT.

If the roots of a quadratic are denoted by the symbols r₁ and r₂, then the following relations may be stated:

r₁ = -b + √(b² - 4ac)/2a

r₂ = -b - √(b² - 4ac)/2a

We can show that the character of the roots is dependent upon the form taken by the expres- sion

b² - 4ac

which is the quantity under the radical in the formula. This expression is the DISCRIMI- NANT of a quadratic equation.

IMAGINARY ROOTS

Since there is a radical in each root, there is a possibility that the roots could be imaginary. They are imaginary when the number under the radical in the quadratic formula is negative (less than 0). In other words, when the value of the discriminant is less than 0, the roots are imaginary.

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Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

EXAMPLE: In this case the roots are equal; such roots are sometimes called double roots. x² + x + 1 = 0 Consider the equation

a = 1, b = 1, c = 1 9x² + 12x + 4 = 0

b² - 4ac = (1)² - 4(1)(1) = 1 - 4 Comparing with the general quadratic, we no- = -3 tice that

Thus, without further work, we know that the a = 9, b = 12, and c = 4 roots are imaginary. The discriminant is CHECK: The roots are b² - 4ac = 12² - 4(9)(4) r₁ = (-1 + √-3)/2 r₂ = (-1 - √-3)/2 = 144 - 144 = 0 r₁ = -½ + i(√3/2) r₂ = -½ - i(√3/2) Therefore, the roots are equal. We recognize both of these numbers as being imaginary. CHECK: From the formula We may also conclude that when one root is imaginary the other will also be imaginary. r₁ = (-12 ± 0)/(2(9)) r₂ = (-12 - 0)/(2(9)) This is because the pairs of imaginary roots are always conjugate complex numbers. If one r₁ = -2/3 r₂ = -2/3 root is of the form a + ib, then a - ib is also a root. Knowing that imaginary roots always The equality of the roots is thus verified. occur in pairs, we can conclude that a quad- The roots can be equal only if the trinomial ratic equation always has either two imaginary is a perfect square. Its factors are equal. roots or two real roots. Factoring the trinomial in Practice problems. Using the discriminant, state whether the roots of the following equa- 9x² + 12x + 4 = 0 tions are real or imaginary: we see that 1. x² - 6x - 16 = 0 2. x² - 6x = -12 (3x + 2)² = 0 3. 3x² - 10x + 50 = 0 4. 6x² + x = 1 Since the factor 3x + 2 is squared, we actu- ally have Answers:

1. Real 3x + 2 = 0 2. Imaginary 3. Imaginary twice, and we have 4. Real

EQUAL OR DOUBLE ROOTS x = -2/3

If the discriminant b² - 4ac equals zero, the radical in the quadratic formula becomes zero. twice. The fact that the same root must be counted twice explains the use of the term "double root." A double root of a quadratic equation is always rational because a double root can oc- cur only when the radical vanishes.

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MATHEMATICS, VOLUME 1

REAL AND UNEQUAL ROOTS Consider the equation

When the discriminant is positive, the roots 2x² - 4x + 1 = 0 must be real. Also they must be unequal since equal roots occur only when the discriminant in which is zero. a = 2, b = -4, and c = 1. Rational Roots If the discriminant is a perfect square, the The discriminant is roots are rational. For example, consider the equation b² - 4ac = (-4)² - 4(2)(1) = 16 - 8 3x² - x - 2 = 0 = 8

in which This discriminant is positive and not a perfect square. Thus the roots are real, unequal, and a = 3, b = -1, and c = -2 irrational. To check the correctness of this information, The discriminant is we derive the roots by means of the formula. Thus, b² - 4ac = (-1)² - 4(3)(-2) = 1 + 24 x = (-b ± √(b² - 4ac))/(2a) = 25 = (4 ± √8)/4 We see that the discriminant, 25, is a per- fect square. The perfect square indicates that = (2 ± √2)/2 the radical in the quadratic formula can be re- moved, that the roots of the equation are ra- x = 1 + (√2/2) or x = 1 - (√2/2) tional, and that the trinomial can be factored. In other words, when we evaluate the discrim- This verifies the conclusions reached in inant and find it to be a perfect square, we know evaluating the discriminant. When the dis- that the trinomial can be factored. criminant is a positive number, not a perfect square, it is useless to attempt to factor the Thus, trinomial. The formula is needed to find the roots. They will be real, unequal, and irrational. 3x² - x - 2 = 0 (3x + 2)(x - 1) = 0 SUMMARY

from which The foregoing information concerning the discriminant may be summed up in the follow- 3x + 2 = 0 x - 1 = 0 ing four rules: x = -2/3 x = 1 1. If b² - 4ac is a perfect square or zero, the roots are rational; otherwise they are We see that the information derived from the irrational. discriminant is correct. The roots are real, 2. If b² - 4ac is negative (less than zero), unequal, and rational. the roots are imaginary. 3. If b² - 4ac is zero, the roots are real, Irrational Roots equal, and rational. If the discriminant is not a perfect square, 4. If b² - 4ac is greater than zero, the roots the radical cannot be removed and the roots are real and unequal. are irrational.

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Chapter 16—QUADRATIC EQUATIONS IN ONE VARIABLE

Practice problems. Determine the character of the roots of each of the following equations:

1. x² - 7x + 12 = 0 2. 9x² - 6x + 1 = 0 3. 2x² - x + 1 = 0 4. 2x - 2x² + 6 = 0

Answers:

1. Real, unequal, rational 2. Real, equal, rational 3. Imaginary 4. Real, unequal, irrational

GRAPHICAL INTERPRETATION OF ROOTS

When a quadratic is set equal to y and the resulting equation is graphed, the graph will reveal the character of the roots, but it may not reveal whether the roots are rational or irrational. Consider the following equations:

1. x² + 6x - 3 = y 2. x² + 6x + 9 = y 3. x² + 6x + 13 = y

The graphs representing these equations are shown in figure 16-4. We recall that the roots of the equation are the values of x at those points where y is zero. Y is zero on the graph anywhere along the X axis. Thus, the roots of the equation are the positions where the graph crosses the X axis. In parabola No. 1 (fig. 16-4) we see immedi- ately that there are two roots to the equation and that they are unequal. These roots appear to be -6.5 and 0.5. Algebraically, we find them to be the irrational numbers

-3 + 2 √3 and -3 -2 √3.

For equation No. 2 (fig. 16-4), the parabola just touches the X axis at x = -3. This means that both roots of the equation are the same— that is, the root is a double root. At the point where the parabola touches the X axis, the two roots of the quadratic equation have moved

together and the two points of intersection of the parabola and the X axis are coincident. The quantity -3 as a double root agrees with the algebraic solution. When the equation No. 3 (fig. 16-4) is solved algebraically, we see that the roots are -3 + 2i and -3 - 2i. Thus they are imaginary. Para- bola No. 3 does not cross the X axis. When this situation occurs, imaginary roots are implied. Only equations having real roots will have graphs that cross or touch the X axis. Thus we may determine from the graph of an equation whether the roots are real or imaginary.

VERBAL PROBLEMS INVOLVING QUADRATIC EQUATIONS

Many practical problems give rise to quad- ratic equations. In such problems it often hap- pens that one of the roots will have no meaning. We must select the root that satisfies the con- ditions of the problem. Consider the following example: The length of a plot of ground exceeds its width by 7 ft and the area of the plot is 120 sq ft. What are the dimensions?

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MATHEMATICS, VOLUME 1

SOLUTION:

Let x = length y = width

then

x - y = 7 (1)

and

xy = 120 (2)

Solving (1) for y, y = x - 7

Substituting (x - 7) for y in (2)

x(x - 7) = 120

Therefore

x² - 7x - 120 = 0

(x - 15)(x + 8) = 0

x = 15, x = -8

Thus, length = +15 or -8.

But the length obviously cannot be a negative value. Therefore, we reject -8 as a value for x and use only the positive value, +15. Then from equation (1),

15 - y = 7

y = 8

Length = 15, Width = 8

Practice problems. Solve the following problems by forming quadratic equations:

1. A rectangular plot is 8 yd by 24 yd. If the length and width are increased by the same amount, the area is increased by 144 sq yd. How much is each dimension increased?

2. Two cars travel at uniform rates of speed over the same route a distance of 180 mi. One goes 5 mph slower than the other and takes 1/2 hr longer to make the run. How fast does each car travel?

Answers:

1. Length and width are each increased by 4 yd.

2. Faster car: 45 mph. Slower car: 40 mph.

180

CHAPTER 17 PLANE FIGURES

The discussion of lines and planes in chap- ter 1 of this course was limited to their con- sideration as examples of sets. The present chapter is concerned with lines, angles, and areas as found in various plane (flat) geometric figures.

LINES

In the strictly mathematical sense, the term "line segment" should be used whenever we re- fer to the straight line joining some point A to some other point B. However, since the straight lines comprising geometric figures have clearly designated end points, we may simplify our terminology. Throughout the remaining chap- ters of this course, the general term "line" is used to designate straight line segments, unless stated otherwise.

TYPES OF LINES

The two basic types of lines in geometry are straight lines and curved lines. A curved line joining points A and B is designated as "curve AB." (See fig. 17-1.) If curve AB is an arc of a circle, it may be designated as "arc AB."

Figure 17-1.—Straight and curved lines.

The term "broken line" in mathematics means a series of two or more straight seg- ments connected end-to-end but not running in the same direction. In mathematics, a series of short, straight segments with breaks be- tween them, which would form a single straight line if joined end-to-end, is a DASHED LINE. (See fig. 17-2.)

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BROKEN LINE DASHED LINE

Figure 17-2.—Broken and dashed lines.

ORIENTATION

Straight lines may be classified in terms of their orientation to the observer's horizon or in terms of their orientation to each other. For example, lines in the same plane which run be- side each other without meeting at any point, no matter how far extended, are PARALLEL. (See fig. 17-3 (A).) Lines in the same plane which are not parallel are OBLIQUE. Oblique lines meet to form angles (discussed in the following section). If two oblique lines cross or meet in such a way as to form four equal angles, as in figure 17-3 (B), the lines are PERPENDICULAR. This definition includes the case in which only one angle is formed, such as angle AEC in figure 17-3 (C). By ex- tending line AE to form line AD, and extending CE to form CB, four equal angles (AEC, CED, DEB, and BEA) are formed.

Figure 17-3.—(A) Parallel lines; (B) and (C) perpendicular lines.

Lines parallel to the horizon are HORIZON- TAL. Lines perpendicular to the horizon are VERTICAL.

MATHEMATICS, VOLUME 1

ANGLES

Lines which meet or cross each other are said to INTERSECT. Angles are formed when two straight lines intersect. The two lines which form an angle are its SIDES, and the point where the sides intersect is the VERTEX. In figure 17-4, the sides of the angles are AV and BV, and the vertex is V in each case. Figure 17-4 (A) is an ACUTE angle; (B) is an OBTUSE angle.

Figure 17-4.—(A) Acute angle; (B) obtuse angle.

CLASSIFICATION BY SIZE

When the sides of an angle are perpendicular to each other, the angle is a RIGHT angle. This term is related to the Latin word "rectus," which may be translated "erect" or "upright." Thus, if one side of a right angle is horizontal, the other side is erect or upright.

The size of an angle refers to the amount of separation between its sides, and the unit of angular size is the angular DEGREE. A right angle contains 90 degrees, abbreviated 90°. An angle smaller than a right angle is acute; an angle larger than a right angle is obtuse. There- fore, acute angles are angles of less than 90°, and obtuse angles are angles between 90° and 180°.

If side AV in figure 17-5 (A) is moved down- ward, the size of the obtuse angle AVB is in- creased. If side AV is moved so far that it alines with line CB (as in figure 17-5 (B), an angle is formed which is equal to the sum of two right angles. The special angle thus formed (AVB) is a straight angle, so called because it is visually indistinguishable from a straight line.

GEOMETRIC RELATIONSHIPS

Angles are often classified by their relation- ship to other angles or to other parts of a geo- metric figure. For example, angles 1 and 3 in

182

Figure 17-5.—(A) Large obtuse angle; (B) straight angle.

figure 17-6 are VERTICAL angles, so called because they share a common vertex. Angles 2 and 4 are opposite each other and are also vertical angles. Lines which cross, as in figure 17-6, always form two pairs of vertical angles, and the vertical angles thus formed are equal in pairs; that is, angle 1 equals angle 3, and angle 2 equals angle 4.

Figure 17-6.—Vertical angles.

Angles 1 and 2 in figure 17-6 are ADJACENT angles. Other pairs of adjacent angles in fig- ure 17-6 are 2 and 3, 3 and 4, and 1 and 4. In the sense used here, adjacent means side by side, not merely close together or touching. For example, angles 1 and 3 are not adjacent angles even though they touch each other.

COMPLEMENTS AND SUPPLEMENTS

Two angles whose sum is 90° are comple- mentary. For example, a 60° angle is the com- plement of a 30° angle, and conversely. "Con- versely" is a mathematical word meaning "vice versa." Two angles whose sum is 180° are supplementary. For example, a 100° angle is the supplement of an 80° angle, and conversely.

Practice problems.

1. Describe the angle which is the complement of an acute angle.

2. Describe the angle which is the supplement of a right angle.

Chapter 17—PLANE FIGURES

3. If two equal angles are complementary, each contains how many degrees?

4. Find the size of an angle which is twice as large as its own complement.

(Hint: If x is the angle, then 90° - x is its complement.)

Answers:

1. Acute 2. Right 3. 45° 4. 60°

GEOMETRIC FIGURES

The discussion of geometric figures in this chapter is limited to polygons and circles. A POLYGON is a plane closed figure, the sides of which are all straight lines. Among the poly- gons discussed are triangles, parallelograms, and trapezoids.

TRIANGLES

A triangle is a polygon which has three sides and three angles. In general, any polygon has as many angles as it has sides, and conversely.

Parts of a Triangle

Each of the three angles of a triangle is a VERTEX; therefore, every triangle has three vertices. The three straight lines joining the vertices are the SIDES (sometimes called legs), and the side upon which the triangle rests is its BASE, often designated by the letter b. This

definition assumes that the standard position of a triangle drawn for general discussion is as shown in figure 17-7, in which the triangle is lying on one of its sides. The vertex opposite the base is the highest point of a triangle in standard position, and is thus called the APEX.

APEX

BASE

Figure 17-7.—Triangle in standard position.

A straight line perpendicular to the base of a triangle, joining the base to the apex, is the ALTITUDE, often designated by the letter a. The altitude is sometimes referred to as the height, and is then designated by the letter h. Figure 17-8 (B) shows that the apex may not be situated directly above the base. In this case, the base must be extended, as shown by the dashed line, in order to drop a perpendicular from the apex to the base. Mathematicians often use the term "drop a perpendicular." The meaning is the same as "draw a straight, perpendicular line." In general, the geometrical term "distance from a point to a line" means the length of a perpendicular dropped from the point to the line. Many straight lines could be drawn from a line to a point not on the line, but the shortest of these is the one we use in measuring the

C

C

a

a

A B

A B

b=AB b b

(A) (B)

b=AB

Figure 17-8.—(A) Interior altitude line; (B) exterior altitude line.

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MATHEMATICS, VOLUME 1

distance from the point to the line. The short- est one is perpendicular to the line.

Perimeter and Area

The PERIMETER of a triangle is the sum of the lengths of its sides. In less precise terms, this is sometimes stated as "the distance around the triangle." If the three sides are labeled a, b, and c, the perimeter P can be found by the following formula:

P = a + b + c

The area of a triangle is the space bounded (enclosed) by its sides. The formula for the area can be found by using a triangle which is part of a rectangle. In Figure 17-9, triangle ABC is one-half of the rectangle. Since the area of the rectangle is a times b (that is, ab), the area of the triangle is given by the follow- ing formula:

Area = 1/2 ab

Written in terms of h, representing height, the formula is:

A = 1/2 bh

This formula is valid for every triangle, in- cluding those with no two sides perpendicular.

5 3 5 4 5

4 6 (1) (2)

10

8

3 2.4 4 6 5 (3) (4)

Figure 17-10.—Perimeters and areas of triangles.

Answers: 1. P = 12 units 3. P = 12 units A = 6 square units A = 6 square units 2. P = 16 units 4. P = 24 units A = 12 square units A = 24 square units

CAUTION: The concept of area is meaning- less if the units of the multiplied dimensions are not the same. For example, if the base of a triangle is 2 feet long and the altitude is 6 inches long, the area might be carelessly stated as 1/2 (6)(2). However, the units must be con- sidered in order to decide whether the answer is in square feet or square inches. When the units are considered, we realize that the cor- rect answer is

1/2 (6 in.) (24 in.) = 72 sq in.

1/2 (1/2 ft) (2 ft) = 1/2 sq ft

C

A B

Figure 17-9.—Area of a triangle.

Practice problems. Find the perimeter and area of each of the triangles in figure 17-10.

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Chapter 17—PLANE FIGURES

Special Triangles

The classification of triangles depends upon their special characteristics, if any. For ex-ample, a triangle may have all three of its sides equal in length; it may have two equal sides and a third side which is longer or shorter than the other two; it may contain a right angle or an obtuse angle. If .it has none of these special characteristics, it is a SCALENE triangle. A scalene triangle has no two of its sides equal and no two of its angles equal.

RIGHT TRIANGLE.—If one of the angles of a triangle is a right angle, the figure is a right triangle. The sides which form the right angle are the LEGS of the triangle, and the third side (opposite the right angle) is the HYPOTENUSE. The area of a right triangle is always easy to determine. If the base of the triangle is one of its legs, as in figure 17-10 (4), the other leg is the altitude. If the hypotenuse is acting as the base, as in figure 17-10 (3), the triangle can be turned until one of its legs is the base, as in figure 17-10 (1). If the triangle is not known to be a right triangle, then the altitude must be given, as in figure 17-10 (2), in order to calculate the area.

Any triangle whose sides are in the ratio of 3:4:5 is a right triangle. Thus, triangles with sides as follows are right triangles:

Side 1 Side 2 Side 3

3 4 5 6 8 10 12 16 20 3x 4x 5x (x is any positive number)

In addition to the 3-4-5 triangle, two other types of right triangles occur frequently. Any triangle having one 30° angle and one 60° angle is a right triangle; that is, the third angle is 90°. Any triangle having two 45° angles is a right triangle.

ISOSCELES TRIANGLE.—A triangle having two of its sides equal in length is an ISOSCELES triangle. Since the length of the side opposite an angle is determined by the size of the angle, the isosceles triangle has two equal angles. In figure 17-11 (A), triangle ABC is isosceles. Sides AC and BC are equal in length, and angles A and B are equal.

Figure 17-11.—(A) isosceles triangle; (B) equilateral triangle.

Figure 17-11 (B) illustrates an EQUILATERAL triangle, which is a special case of an isosceles triangle. An equilateral triangle has all three of its sides equal in length. Since the lengths of the sides are directly related to the size of the angles opposite them, an equilateral triangle is also equiangular; that is, all three of its angles are equal.

OBLIQUE TRIANGLES.—Any triangle con-taining no right angle is an OBLIQUE triangle. Figure 17-12 illustrates two possible configurations, both of which are oblique triangles. An oblique triangle which contains an obtuse angle is often called an OBTUSE triangle.

Figure 17-12.—Oblique triangles. (A) Acute; (B) obtuse.

Sum of the Angles

The sum of the angles in any triangle is 180°. For example, if one of the angles is 40° and another is 20°, the third angle is 120°. It is this relationship that justifies the statements made in the preceding section concerning 45° triangles and 30°-60°-90° triangles. If two of the angles are 45° each, then the third angle is 180° - (45°+45°) and the figure is a right triangle. If one angle is 60° and another is 30°, the third angle is 90° and the figure is a right triangle.

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MATHEMATICS, VOLUME 1

QUADRILATERALS

A QUADRILATERAL is a polygon with four sides. The parts of a quadrilateral are its sides, its four angles, and its two DIAGONALS. A diagonal is a straight line joining two alternate vertices of a polygon. Figure 17-13 illustrates the parts of a quadrilateral, in which AC and DB are the diagonals.

Figure 17-13.—Parts of a quadrilateral.

Perimeter and Area

The perimeter of a quadrilateral is the sum of the lengths of its sides. For example, the perimeter of the quadrilateral in figure 17-13 is 30 units.

The area of a quadrilateral can be found by dividing it into triangles and summing the areas of the triangles. However, the altitudes of the triangles are usually difficult to calculate unless the quadrilateral has at least one pair of parallel sides.

Parallelograms

A PARALLELOGRAM is a quadrilateral in which the opposite sides are parallel. For example, in the parallelogram in figure 17-14, side AB is parallel to side CD. Furthermore, side BC is parallel to side AD.

Figure 17-14.—A parallelogram.

Since lines AB and CD are parallel, lines DE and CF (both perpendicular to line AF in figure 17-14) are equal. Angles DAE and CBF in figure 17-14 are equal, because a straight line cutting two parallel lines, such as AD and BC, forms equal angles with the parallel lines. Thus, triangles AED and BFC are equal, and line AD equals line BC. Therefore we have proved that the opposite sides of a parallelogram are equal. If all four of the sides are the same length, the parallelogram is a RHOMBUS.

In addition to the equality of the opposite sides, the opposite angles of a parallelogram are also equal. For example, angle DAB equals angle BCD in figure 17-14, and angle ADC equals angle ABC.

RECTANGLES AND SQUARES.—When all of the angles of a parallelogram are right angles, it is a RECTANGLE. A rectangle with all four of its sides the same length is a SQUARE. Thus a square is a rhombus having 90° angles. Every square is a rectangle, and every rectangle is a parallelogram. Notice that the reverse of this statement is not true.

The area of a rectangle is found by multiplying its length times its width. Therefore, if each side of a square has length s, the area of the square is s².

Written as formulas, these areas are as follows:

Rectangle: A = lw or A = bh, where b = base, h = height

Square: A = s²

AREA.—The area of a parallelogram can be found by dividing it into rectangles and triangles. For example, in figure 17-14 the area of the parallelogram is the sum of the areas of

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triangle AED and figure EBCD. Since triangle AED is equal to triangle BFC, the sum of AED and EBCD is equal to the sum of BFC and EBCD. Thus the area of parallelogram ABCD is the same as the area of rectangle EFCD. Since the area of EFCD is DC multiplied by DE, and DC has the same length as AB, we conclude that the area of a parallelogram is the product of its base times its altitude. Written as a formula, this is

A = ba

or

A = bh, where h is height

Trapezoids

A TRAPEZOID is a quadrilateral in which two sides are parallel and the other two sides are not parallel. By orienting a trapezoid so that its parallel sides are horizontal, we may call the parallel sides bases. Observe that the bases of a trapezoid are not equal in length. (See fig. 17-15.)

(A) (B)

Figure 17-15.—Typical trapezoids.

The area of a trapezoid may be found by separating it into two triangles and a rectangle, as in figure 17-16. The total area A of the trapezoid is the sum of A₁ plus A₂ plus A₃, and is calculated as follows:

A = A₁ + A₂ + A₃

= ½ha + hb₁ + ½hc

= ½h (a + 2b₁ + c)

= ½h [(a + b₁ + c) + b₁]

= ½h (b + b₁)

Thus the area of a trapezoid is equal to one- half the altitude times the sum of the bases.

b₁

A₁ h A₂ h A₃

a b₁ c ←————————— b ——————————→

Figure 17-16.—Area of a trapezoid.

Practice problems. Find the area of each of the following figures:

1. Rhombus; base 4 in., altitude 3 in.

2. Rectangle; base 6 ft, altitude 4 ft

3. Parallelogram; base 10 yd, altitude 12 ft

4. Trapezoid; bases 6 ft and 4 ft, altitude 2 yd.

Answers:

1. 12 sq in. 3. 40 sq yd

2. 24 sq ft 4. 30 sq ft

CIRCLES

The mathematical definition of a circle states that it is a plane figure bounded by a curved line, every point of which is equally distant from the center of the figure. The parts of a circle are its circumference, its radius, and its diameter.

Parts of a Circle

The CIRCUMFERENCE of a circle is the line that forms its outer boundary. Circumfer- ence is the special term used in referring to the "perimeter" of a circle. (See fig. 17-17.) A RADIUS of a circle is a line joining the center to a point on the circumference, as shown in figure 17-17. A straight line joining two points on the circumference of a circle, and passing through the center, is a DIAMETER. A straight line which touches the circle at just one point is a TANGENT. A tangent is perpendicular to a radius at the point of tangency.

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Figure 17-17.—Parts of a circle.

An ARC is a portion of the circumference of a circle. A CHORD is a straight line joining the end points of any arc. The portion of the area of a circle cut off by a chord is a SEG- MENT of the circle, and the portion of the circle's area cut off by two radii (radius lines) is a SECTOR. (See fig. 17-18.)

Formulas for Circumference and Area

The formula for the circumference of a circle is based on the relationship between the circumference and the diameter. This com- parison can be made experimentally by mark- ing the edge of a circular object, such as a coin, and rolling it (without slippage) along a flat surface. (See fig. 17-19.) The distance from the initial position to the final position of the disk in figure 17-19 is ap- proximately 3.14 times as long as the diameter of the disk. With any circle, this is always found to be the case; but it is not possible to give the value of C/d (circumference divided by diameter) exactly. The ratio C/d is repre- sented by the symbol π, which is the Greek letter pi. Thus we have the following equations:

C/d = π

C = πd

Figure 17-18.—Arc, chord, segment, and sector.

D=1 INCH

←———— C= 3.14 INCHES ————→

INITIAL POSITION FINAL POSITION

Figure 17-19.—Measuring the circumference of a circle.

This formula states that the circumference of a circle is π times the diameter. Notice that it could be written as

C = 2r · π or C = 2πr

since the diameter d is the same as 2r (twice the radius). Although the value of π is not exactly equal to any of the numerical expressions which are sometimes used for it, the ratio is very close to 3.14. If extreme accuracy is required, 3.1416 is used as an approximate value of π. Many calculations involving π are satisfactory if the fraction 22/7 is used as the value of π.

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Practice problems. Calculate the circum-ference of each of the following circles, using 22/7 as the value of π:

1. Radius = 21 in. 3. Radius = 14 ft

2. Diameter = 7.28 in. 4. Diameter = 2.8 yd

Answers:

1. 132 in. 3. 88 ft

2. 22.88 in. 4. 8.8 yd

AREA.–The area of a circle is found by multiplying the square of its radius by π. The formula is written as follows:

A = πr²

EXAMPLE: Find the area of a circle whose diameter is 4 ft, using 3.14 as the value of π.

SOLUTION: The radius is one-half the diameter. Therefore,

r = 1/2(4 ft)

= 2 ft

A = πr² = π(2 ft)²

A = 3.14 (4 sq ft)

= 12.56 sq ft

Practice problems. Find the area of each of the following circles, using 3.14 as the value of π.

1. Radius = 7 in. 3. Diameter = 2.8 ft 2. Diameter = 42 mi 4. Radius = 14 yd

Answers:

1. A = 154 sq in. 3. 6.15 sq ft

2. A = 1,385 sq mi 4. 615 sq yd

Concentric Circles

Circles which have a common center are said to be CONCENTRIC. (See fig. 17-20.) The area of the ring between the concentric circles in figure 17-20 is calculated as follows:

Figure 17–20.–Concentric circles.

Let R = radius of large circle

r = radius of small circle

AR = area of large circle

Ar = area of small circle

A = area of ring

Then A = AR - Ar,

= πR² - πr²

= π(R² - r²)

Notice that the last expression is the difference of two squares. Factoring, we have

A = π(R + r)(R - r)

Therefore, the area of a ring between two circles is found by multiplying π times the product of the sum and difference of their radii.

Practice problems. Find the areas of the rings between the following concentric circles:

1. R = 4 in. 2. R = 6 ft

r = 3 in. r = 2 ft

Answers:

1. 22 sq in. 2. 100.5 sq ft

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GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES

Many ratings in the Navy involve work which requires the construction or subdivision of geometric figures. For example, materials must be cut into desired shapes, perpendicular lines must be drawn, etc. In addition to these skills, some Navy ratings require the ability to recognize various solid figures and calculate their volumes and surface areas.

CONSTRUCTIONS

From the standpoint of geometry, a CONSTRUCTION may be either the process of building up a figure or that of breaking down a figure into smaller parts. Some typical constructions are listed as follows:

1. Dividing a line into equal segments. 2. Erecting the perpendicular bisector of a line. 3. Erecting a perpendicular at any point on a line. 4. Bisecting an angle. 5. Constructing an angle. 6. Finding the center of a circle. 7. Constructing an ellipse.

EQUAL DIVISIONS ON A LINE

A line may be divided into any desired number of equal segments by the method shown in figure 18-1.

Suppose that line AB (fig. 18-1) is to be divided into seven equal segments. Draw line AC at any convenient angle with AB and mark

off seven spaces of some convenient length, say 1/2 inch, on it. Extend AC, if necessary, in order to get seven intervals of the chosen length on it. This produces the points a, b, c, d, e, f, and g, as shown in figure 18-1. Draw a line from g to B, and then draw lines parallel to gB, starting at each of the points a, b, c, d, e, and f. The segments of AB cut off by these lines are equal in length.

It is frequently necessary to rule a predetermined number of lines on a blank sheet of material. This may be done by a method based on the foregoing discussion. For example, suppose that the sheet of typing paper in figure 18-2 is to be divided into 24 equal spaces.

The 12-inch ruler is laid across the paper at an angle, in such a way that the ends of the

Figure 18-1.—Dividing a line into equal segments.

Figure 18-2.—Ruling equal spaces on a sheet of paper.

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ruler coincide with the top and bottom edges of the paper. There are 24 spaces, each 1/2 inch wide, on a 12-inch ruler. Therefore, we mark the paper beside each 1/2-inch division marker on the ruler. After removing the ruler, we draw a line through each of the marks on the paper, parallel to the top and bottom edges of the paper.

PERPENDICULAR BISECTOR OF A LINE

To bisect a line or an angle means to divide it into two equal parts. A line may be bisected satisfactorily by measurement, or by a geo- metric method. If the measuring instrument does not reach the full length of the line, pro- ceed as follows:

1. Starting at one end, measure about half the length of the line and make a mark. 2. Starting at the other end, measure exactly the same distance as before and make a second mark. 3. The bisector of the line lies halfway be- tween these two marks.

The geometric method of bisecting a line is not dependent on measurement. It is based upon the fact that all points equally distant from the ends of a straight line lie on the perpendi- cular bisector of the line.

Bisecting a line geometrically requires the use of a mathematical compass, which is an in- strument for drawing circles and comparing distances. If a line AB is to be bisected as in figure 18-3, the compass is opened until for the distance between its points is more than half as long as AB. Then a short arc is drawn above the approximate center of the line and another below, using A as the center of the arcs' circle. (See fig. 18-3.)

Two more short arcs are drawn, one above and one below the approximate center of line AB, this time using B as the center of the arcs' circle.

The two arcs above line AB are extended until they intersect, forming point C, and the two arcs below line AB intersect to form point D. The line joining point C and point D is the perpendicular bisector of line AB.

PERPENDICULAR AT ANY POINT ON A LINE

Figure 18-4 shows a line AB with point C between A and B. A perpendicular to AB is erected at C as follows:

C ✕ A B ✕ D

Figure 18-3.—Bisecting a line geometrically.

E / \ / \ / \ / \ / O \ / \ / \ A C D B \ / \ / \ / \ / \ / \ /

Figure 18-4.—Erecting a perpendicular at a point.

1. Using any convenient point above the line (such as O) as a center, draw a circle with ra- dius OC. This circle cuts AB at C and at D. 2. Draw line DO and extend it to intersect the circle at E. 3. Draw line EC. This line is perpendicular to AB at C.

BISECTING AN ANGLE

Let angle AOB in figure 18-5 be an angle which is to be bisected. Using O as a center and any convenient radius, draw an arc inter- secting OA and a second arc intersecting OB. Label these intersections C and D.

Using C and D as centers, and any conven- ient radius, draw two arcs intersecting halfway between lines OA and OB. A line from O through the intersection of these two arcs is the bisector of angle AOB.

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MATHEMATICS, VOLUME 1

centers, draw arcs intersecting at C. When A and B are connected to C by straight lines, all three angles of triangle ABC are 60° angles. The special angles already discussed are used in constructing 45° and 30° angles. A 90° angle is bisected to form two 45° angles, and a 60° angle is bisected to form two 30° angles.

A / / C / O ──────────────── X \ \ \ D \ B

Figure 18-5.—Bisecting an angle.

SPECIAL ANGLES

Several special angles may be constructed by geometric methods, so that an instrument for measuring angles is not necessary in these special cases.

Figure 18-4 illustrates a method of con- structing a right triangle in a semicircle. By inscribing a right triangle in a semicircle, DCE. But an alternate method is needed for those situations in which drawing circles is inconvenient. The method described herein makes use of a right triangle having its sides in the ratio of 3 to 4 to 5. It is often used in laying out the foundations of build- ings. The procedure is as follows:

1. A string is stretched as shown in figure 18-6, forming line AC. The length of AC is 1 foot. 2. A second string is stretched, crossing line AC at A, directly above the point intended as the corner of the foundation. Point D on this line is 4 feet from A. 3. Attach a third string, 5 feet long, at C and D. When AC and AD are spread so that line CD is taut, angle DAC is a right angle. A 60° angle is constructed as shown in fig- ure 18-7. With AB as a radius and A and B as

FINDING THE CENTER OF A CIRCLE

It is sometimes necessary to find the center of a circle of which only an arc or a segment is given. (See fig. 18-8.)

From any point on the arc, such as A, draw two chords intersecting the arc in any two points, such as B and C. With the points A, B, and C as centers, use any convenient radius and draw short intersecting arcs to form the perpendicular bisectors of chords AC and AB. Join the intersecting arcs on each side of AC, obtaining line MP, and join the arcs on each side of AB, obtaining line NQ. The intersection of MP and NQ is point O, the center of the circle.

ELLIPSES

An ellipse of specified length and width is constructed as follows:

1. Draw the major axis, AB, and the minor axis, CD, as shown in figure 18-9. 2. On a straightedge or ruler, mark a point (labeled a in the figure) and from this point measure one-half the length of the minor axis and make a second mark (b in figure 18-9). From point a, measure one-half the length of

5 4 3 2 ! C D ┌─────┬─────┬─────┬─────┐ ┌─┐ │ 1 │ 2 │ 3 │ │2│ 3 ├─────┼─────┼─────┼─────┤ ├─┤ │ │ │ │ │ │ │ │ │ │ │ │

Figure 18-6.—Constructing a right angle by the 3-4-5 method.

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SOLID FIGURES

The plane figures discussed in chapter 17 of this course are combined to form solid figures. For example, three rectangles and two triangles may be combined as shown in figure 18-10. The flat surfaces of the solid figure are its FACES; the top and bottom faces are the BASES, and the faces forming the sides are the LAT- ERAL FACES.

Figure 18-7.—Constructing 60° angles.

EDGE

UPPER BASE

LATERAL FACE

LATERAL EDGE

Figure 18-10.—Parts of a solid figure.

LOWER BASE

Figure 18-8.—Finding the center of a circle.

Some solidfigures do not have any flat faces, and some have a combination of curved surfaces and flat surfaces. Examples of solids with curved surfaces include cylinders, cones, and spheres.

Figure 18-9.—Constructing an ellipse.

PRISMS

The solid shown in figure 18-10 is a PRISM. A prism is a solid with three or more lateral faces which intersect in parallel lines.

Types of Prisms

The name of a prism depends upon its base polygons. If the bases are triangles, as in fig- ure 18-10, the figure is a TRIANGULAR prism. A RECTANGULAR prism has bases which are rectangles.

If the bases of a prism are perpendicular to the planes forming its lateral faces, the prism is a RIGHT prism.

A PARALLELEPIPED is a prism with paral- lelograms for bases. Since the bases are par- allel to each other, this means that they cut the lateral faces to form parallelograms. There- fore, in a parallelepiped, all of the faces are parallelograms. If a parallelepiped is a right

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MATHEMATICS, VOLUME 1

prism, and if its bases are rectangles, it is a rectangular solid. A CUBE is a rectangular solid in which all of the six rectangular faces are squares.

Parts of a Prism

The parts of a prism are shown in figure 18-10. The line formed by the joining of two faces of a prism is an EDGE. If the two faces forming an edge are lateral faces, the edge thus formed is a LATERAL EDGE.

Surface Area and Volume

The SURFACE AREA of a prism is the sum of the areas of all of its faces, including the bases. The VOLUME of a prism may be con- sidered as the sum of the volumes of many thin wafers, each having a thickness of one unit and a shape that duplicates the shape of the base. (See fig. 18-11.)

Figure 18-12.—Height of a prism which is not a right prism.

CIRCULAR CYLINDERS

Any surface may be considered as the result of moving a straight line in a direction at right angles to its length. For example, suppose that the stick of charcoal in figure 18-13 is moved from position CD to position AB by dragging it across the paper. The broad mark made by the charcoal represents a plane surface. The sur- face is said to be "generated" by moving line AB.

Figure 18-11.—Volume of a prism.

The wafers which comprise the prism in figure 18-11 all have the same area, which is the area of the base. Therefore, the volume of the prism is found by multiplying the area of the base times the number of wafers. Since each wafer is 1 inch thick, the number of wafers is the same as the height of the prism in inches. The resulting formula for the volume of a prism, using B to represent the area of the base and h to represent the height, is as follows:

V = Bh

When a prism has lateral edges which are not perpendicular to the bases, the height of the prism is the perpendicular distance between the bases. (See fig. 18-12.) The formula for the volume remains the same, even though the prism is no longer a right prism.

Figure 18-13.—Surface generated by a moving line.

The movement of the line in figure 18-13 may be controlled by requiring that its lower end trace a particular path. For example, if line AB moves so as to trace an ellipse as in figure 18-14 (A), a cylindrical surface is gen- erated by the line. This surface, shown in fig- ure 18-14 (B), is an elliptical cylinder.

Any line in the surface, parallel to the generating line, such as CD or EF in figure 18-14 (B), is an ELEMENT of the cylinder. If the elements are perpendicular to the bases, the cylinder is a RIGHT CYLINDER. If the

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Chapter 18—GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES

(A) Line generating a cylinder; (B) elliptical cylinder; (C) circular cylinder.

Figure 18-14.—(A) Line generating a cylinder; (B) elliptical cylinder; (C) circular cylinder.

ases are circles, the cylinder is a CIRCULAR CYLINDER. Figure 18-14 (C) illustrates a right circular cylinder. Line O-O', joining the centers of the bases of a right circular cylinder, is the AXIS of the cylinder.

Surface Area and Volume

The lateral area of a cylinder is the area of its curved surface, excluding the area of its bases. Figure 18-15 illustrates an experimental method of determining the lateral area of a right circular cylinder.

The card of length L and width W in figure 18-15 is rolled into a cylinder. The height of the cylinder is W and the circumference is L. The lateral area is the same as the original area of the card, LW. Therefore, the lateral area of the cylinder is found by multiplying its height by the circumference of its base. Written as a formula, this is

A = Ch

EXAMPLE: Find the lateral area of a right circular cylinder whose base has a radius of 4 inches and whose height is 6 inches.

(A) CARD

(B) CARD STARTING TO ROLL

(C) CARD HALF ROLLED

(D) CARD ROLLED INTO CYLINDER

Figure 18-15.—Lateral area of a cylinder.

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MATHEMATICS, VOLUME 1

SOLUTION: The circumference of the base is

C = πd C = 3.14 x 8 in. = 25.12 in.

Therefore,

A = 25.12 in. x 6 in. = 151 sq in. (approximately)

The formula for the volume of a cylinder is obtained by the same reasoning process that was used for prisms. The cylinder is considered to be composed of many circular wafers, or disks, each one unit thick. The area of each disk, multiplied by the number of disks, is the volume of the cylinder. With V representing volume, A d representing the area of each disk, and n representing the number of disks, the formula is as follows:

V = A d n

Since the number of disks is the same as the height of the cylinder, the formula for the volume of a cylinder is normally written

V = Bh

In this formula, B is the area of the base and h is the height of the cylinder.

EXAMPLE: Determine the volume of a circular cylinder with a base of radius 5 inches and a height of 14 inches.

SOLUTION:

V = Bh = (π x 5²) x 14 = 3.14 x 25 x 14 = 78.5 x 14 = 1099 cu in. = 1,100 cu in. (approximately)

Practice problems:

1. Determine the lateral area of a right circular cylinder with a base of diameter 7 inches and a height of 4 inches.

2. Determine the volume of the cylinder in problem 1.

Answers:

1. 88 sq in. 2. 154 cu in.

REGULAR PYRAMIDS AND RIGHT CIRCULAR CONES

A PYRAMID is a solid figure, the lateral faces of which are triangles. (See fig. 18-16.) A REGULAR PYRAMID has all of its lateral faces equal.

Figure 18-16.—(A) Irregular pyramid; (B) regular pyramid.

A regular pyramid with a very large number of lateral faces would have a base polygon with many sides. If the number of sides is sufficiently large, the base polygon is indistinguishable from a circle and the surface formed by the many lateral faces becomes a smoothly curved surface. The solid figure thus formed is a RIGHT CIRCULAR CONE. (See fig. 18-17.)

Figure 18-17.—Right circular cone.

Slant Height

The slant height of a regular pyramid is the perpendicular distance from the vertex to the center of any side of the base. For example,

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Chapter 18—GEOMETRIC CONSTRUCTIONS AND SOLID FIGURES

the length of line AV in figure 18-18 (A) is the slant height. The slant height of a right circular cone is the length of any straight line joining the vertex to the circumference line of the base. Such a line is perpendicular to a line tangent to the base at the point where the slant height intersects the base. (See fig. 18-18 (B).) Lines AV, BV, and CV in figure 18-18 (B) are all slant heights.

Volume

The volume of a pyramid is determined by its base and its altitude, as is the case with other solid figures. Experiments show that the volume of any pyramid is one-third of the product of its base and its altitude. This may be stated as a formula with V representing volume, B representing the area of the base, and h representing height (altitude), as follows:

V = 1/3 Bh

The formula for the volume of a pyramid does not depend in any way upon the number of faces. Therefore, we use the same formula for the volume of a right circular cone. Since the base is a circle, we replace B with πr² (where r is the radius of the base). The formula for the volume of a right circular cone is then

V = 1/3 Bh

= 1/3 πr²h

Practice problems:

1. Find the lateral area of a regular pyramid with a 5-sided base measuring 3 inches on each side, if the slant height is 12 inches.

2. Find the lateral area of a right circular cone whose base has a diameter of 6 cm and whose slant height is 14 cm.

3. Find the volume of a regular pyramid with a square base measuring 4 cm on each side, if the vertex is 9 cm above the base.

4. Find the volume of a right circular cone whose base has a diameter of 14 inches, if the altitude is 21 inches.

Answers:

1. 90 sq in. 3. 48 cu cm 2. 132 sq cm 4. 1077 cu in.

Lateral Area

The lateral area of a pyramid is the sum of the areas of its lateral faces. If the pyramid is regular, its lateral faces have equal bases; furthermore, the slant height is the altitude of each face. Therefore, the area of each lateral face is one-half the slant height multiplied by the length of one side of the base polygon. Since the sum of these sides is the perimeter of the base, the total lateral area of the pyramid is the product of one-half its slant height multiplied by the perimeter of its base. Using s to represent slant height and P to represent the perimeter of the base, the formula is as follows:

Lateral Area = 1/2 sP

A right circular cone can be considered as a regular pyramid with an infinite number of faces. Therefore, using C to represent the circumference of the base, the formula for the lateral area of a right circular cone is

Lateral Area = 1/2 sC

SPHERES

A SPHERE is a solid figure with all points on its surface equally distant from its center.

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MATHEMATICS, VOLUME 1

The formula for the surface area of a sphere may be rewritten as follows:

A = (2πr)(2r)

When the formula is factored in this way, it is easy to see that the surface area of a sphere is simply its circumference times its diameter.

Volume

The volume of a sphere whose radius is r is given by the formula

V = 4/3 πr³

EXAMPLE: Find the volume of a sphere whose diameter is 42 inches.

SOLUTION:

V = 4/3 πr³

= 4/3 x 3.14 x (21 in.)³

= 4/3 x 3.14 x 21 x 21 x 21 cu in.

= 4.187 x 21 x 21 x 21 cu in.

= 38,776 cu in. (approximately)

Practice problems. Calculate the surface area and the volume of the sphere in each of the following problems:

1. Radius = 7 inches 2. Radius = 14 cm

Answers:

1. Area = 615 sq in. (approx.) Volume = 1436 cu in. (approx.)

2. Area = 2462 sq cm (approx.) Volume = 11,489 cu cm (approx.)

[Figure 18-19 shows a sphere with center O and various radii and diameters marked, with points labeled A, B, C, D, E, and F]

Figure 18-19.—Parts of a sphere.

In figure 18-19, the center of the sphere is point O. A RADIUS of a sphere is a straight line segment joining the center of the sphere to a point on the surface. Lines OA, OB, OC, OD, OE, and OF in figure 18-19 are radii. A DIAMETER of a sphere is a straight line segment joining two points on the surface and passing through the center of the sphere. Lines AB, CD, and EF in figure 18-19 are diameters. A HEMISPHERE is half of a sphere.

Circles of various sizes may be drawn on the surface of a sphere. The largest circle that may be so drawn is one with a radius equal to the radius of the sphere. Such a circle is a GREAT CIRCLE. In figure 18-19, circles AEBF, ACBD, and CEDF are great circles.

On the surface of a sphere, the shortest distance between two points is an arc of a great circle drawn so that it passes through the two points. This explains the importance of great circles in the science of navigation, since the earth is approximately a sphere.

Surface Area

The surface area of a sphere may be calculated by multiplying 4 times π times the square of the radius. Written as a formula, this is

A = 4πr²

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CHAPTER 19 NUMERICAL TRIGONOMETRY

The word "trigonometry" means "measure- ment by triangles." As it is presented in many textbooks, trigonometry includes topics other than triangles and measurement. However, this chapter is intended only as an introduction to the numerical aspects of trigonometry as they relate to measurement of lengths and angles.

SPECIAL PROPERTIES OF RIGHT TRIANGLES

A RIGHT TRIANGLE has been defined as any triangle containing a right angle. The side opposite the right angle in a right triangle is a HYPOTENUSE. (See fig. 19-1.) In figure 19-1, side AC is the hypotenuse.

C

HYPOTENUSE

A 90° B

Figure 19-1.—A right triangle.

An important property of all right triangles, which relates the lengths of the three sides, was discovered by the Greek philosopher Pythagoras.

PYTHAGOREAN THEOREM

The rule of Pythagoras, or PYTHAGOREAN THEOREM, states that the square of the length of the hypotenuse (in any right triangle) is equal to the sum of the squares of the lengths of the other two sides. For example, if the sides are

x² + y² = r² 4² + 3² = r²

y 3 r r

x 4 (A) (B)

Figure 19-2.—The Pythagorean Theorem. (A) General triangle; (B) triangle with sides of specific lengths.

labeled as in figure 19-2 (A), the Pythagorean Theorem is stated in symbols as follows:

x² + y² = r²

An example of the use of the Pythagorean Theorem in a problem follows:

EXAMPLE: Find the length of the hypotenuse in the triangle shown in figure 19-2 (B).

SOLUTION: r² = 3² + 4² r = √9 + 16 = √25 = 5

EXAMPLE: An observer on a ship at point A, figure 19-3, knows that his distance from point C is 1,200 yards and that the length of BC is 1,300 yards. He measures angle A and finds that it is 90°. Calculate the distance from A to B.

SOLUTION: By the rule of Pythagoras, (BC)² = (AB)² + (AC)² (1,300)² = (AB)² + (1,200)² (1,300)² - (1,200)² = (AB)² (13 x 10²)² - (12 x 10²)² = (AB)² (169 x 10⁴) - (144 x 10⁴) = (AB)² 25 x 10⁴ = (AB)² 500 yd = AB

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C angle of the second have all of their corre- sponding angles equal. Thus the two triangles are similar. B Practical situations frequently occur in which similar right triangles are used to solve prob- 90° lems. For example, the height of a tree can be determined by comparing the length of its A shadow with that of a nearby flagpole, as shown in figure 19-5. Figure 19-3.—Using the Pythagorean Theorem. B' SIMILAR RIGHT TRIANGLES B FLAGPOLE Two right triangles are SIMILAR if one of the acute angles of the first is equal to one of A C A' C' SHADOW SHADOW the acute angles of the second. This conclusion is supported by the following reasons: Figure 19-5.—Calculation of height by 1. The right angle in the first triangle is comparison of shadows. equal to the right angle in the second, since all right angles are equal. Assume that the rays of the sun are parallel 2. The sum of the angles of any triangle is and that the tree and flagpole both form 90° equal to 180°. Therefore, the sum of the two acute angles with the ground. Then triangles ABC angles in a right triangle is 90°. and A'B'C' are right triangles and angle B is 3. Let the equal acute angles in the two tri- equal to angle B'. Therefore, the triangles are angles be represented by A and A' respectively. similar and their corresponding sides are pro- (See fig. 19-4.) Then the other acute angles, portional, with the following result: B and B', are as follows:

B = 90° - A BC B'C' B' = 90° - A' AC = A'C'

(AC) x (B'C') B' BC = A'C' B Suppose that the flagpole is known to be 30 feet high, the shadow of the tree is 12 feet long, A C A' C' and the shadow of the flagpole is 24 feet long. Then Figure 19-4.—Similar right triangles.

BC = 12 x 30 = 15 feet 4. Since angles A and A' are equal, angles 24 B and B' are also equal. 5. We conclude that right triangles with one acute angle of the first equal to one acute angle of the second have all of their corresponding Practice problems. angles equal.

1. A mast at the top of a building casts a shadow whose tip is 48 feet from the base of the build- ing. If the building is 12 feet high and its shadow is 32 feet long, what is the length of the mast? (NOTE: If the length of the mast is x, then the height of the mast above the ground is x + 12.)

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Chapter 19—NUMERICAL TRIGONOMETRY

2. Figure 19-6 represents an L-shaped build- ing with dimensions as shown. On the line of sight from A to D, a stake is driven at C, a point 8 feet from the building and 10 feet from A. If ABC is a right angle, find the length of AB and the length of AD. Notice that AE is 18 feet and ED is 24 feet.

24 FT

16 FT

8FT 10 FT

Figure 19-6.—Using similar triangles.

Answers: 1. 6 feet 2. AB = 6 feet AD = 30 feet

TRIGONOMETRIC RATIOS

The relationships between the angles and the sides of a right triangle are expressed in terms of TRIGONOMETRIC RATIOS. For example, in figure 19-7, the sides of the triangle are named in accordance with their relationship to angle θ. In trigonometry, angles are usually named by means of a Greek letters. The Greek name of the symbol θ is theta. The six trigonometric ratios for the angle θ are listed in table 19-1. The ratios are defined as follows:

1. sin θ = side opposite θ/hypotenuse = y/r

2. cos θ = side adjacent to θ/hypotenuse = x/r

3. tan θ = side opposite θ/side adjacent to θ = y/x

4. cot θ = side adjacent to θ/side opposite θ = x/y

SIDE ADJACENT TO ANGLE θ (A)

(B)

Figure 19-7.—Relationship of sides and angles in a right triangle. (A) Names of the sides; (B) symbols used to designate the sides.

Table 19-1.—Trigonometric ratios.

Name of ratio | Abbreviation ---|--- sine of θ | sin θ cosine of θ | cos θ tangent of θ | tan θ cotangent of θ | cot θ secant of θ | sec θ cosecant of θ | csc θ

5. sec θ = hypotenuse/side adjacent to θ = r/x

6. csc θ = hypotenuse/side opposite to θ = r/y

The other acute angle in figure 19-7 (B) is labeled α (Greek alpha). The side opposite α is x and the side adjacent to α is y. Therefore the six ratios for α are as follows:

1. sin α = x/r 4. cot α = y/x

2. cos α = y/r 5. sec α = r/y

3. tan α = x/y 6. csc α = r/x

Suppose that the sides of triangle (B) in fig- ure 19-7 are as follows: x = 3, y = 4, r = 5. Then each of the ratios for angles θ and α may

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be expressed as a common fraction or as a decimal. For example,

sin θ = 4/5 = 0.800

sin α = 3/5 = 0.600

Decimal values have been computed for ratios of angles between 0° and 90°, and values for angles above 90° can be expressed in terms of these angles by means of conversion formulas. Appendix II of this training course gives the sine, cosine, and tangent of angles from 0° to 90°. The secant, cosecant, and cotangent are calculated, when needed, by using their relationships to the three principal ratios. These relationships are as follows:

secant θ = 1/cosine θ

cosecant θ = 1/sine θ

cotangent θ = 1/tangent θ

TABLES

Tables of decimal values for the trigono- metric ratios may be constructed in a variety of ways. Some give the angles in degrees, min- utes, and seconds; others in degrees and tenths of a degree. The latter method is more com- pact and is the method used for appendix II. The "headings" at the bottom of each page in appendix II provide a convenient reference showing the minute equivalents for the decimal fractions of a degree. For example, 12' (12 minutes) is the equivalent of 0.2°.

Finding the Function Value

The trigonometric ratios are sometimes called FUNCTIONS, because the value of the ratio depends upon (is a function of) the angle size. Finding the function value in appendix II is easily accomplished. For example, the sine 35° is found by looking in the "sin" row oppo- site the large number 35, which is located in the extreme left-hand column. Since our angle in this example is exactly 35°, we look for the decimal value of the sine in the column with the 0.0° heading. This col- umn contains decimal values for functions of

the angle plus 0.0°; in our example, 35° plus 0.0°, or simply 35.0°. Thus we find that the sine of 35.0° is 0.5736. By the same reasoning, the sine of 42.7° is 0.6782, and the tangent of 32.3° is 0.6322.

A typical problem in trigonometry is to find the value of an unknown side in a right triangle when only one side and one acute angle are known. EXAMPLE: In triangle ABC (fig. 19-8), find the length of AC if AB is 13 units long and angle CAB is 34.7°.

B

13

34.7° A C

Figure 19-8.—Using the trigonometric ratios to evaluate the sides.

SOLUTION:

AC/13 = cos 34.7°

AC = 13 cos 34.7°

= 13 × 0.8221

= 10.69 (approx.)

The angles of a triangle are frequently stated in degrees and minutes, rather than degrees and tenths. For example, in the foregoing prob- lem, the angle might have been stated as 34°42'. When the stated number of minutes is an exact multiple of 6 minutes, the minute entries at the bottom of each page in appendix II may be used.

Finding the Angle

Problems are frequently encountered in which two sides are known, in a right triangle, but neither of the acute angles is known. For ex- ample, by applying the Pythagorean Theorem we can verify that the triangle in figure 19-9 is

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The following arrangement of numbers is recommended for interpolation:

ANGLE TANGENT

22°36' 0.4163 } 6' θ 0.4167 .0004 - .0020 } 22°42' 0.4183

The spread between 22°36' and 22°42' is 6', and we use the comparison of the tangent values to determine how much of this 6' spread is in- cluded in θ, the angle whose value is sought. Notice that the tangent of θ is different from tan 22°36' by only 0.0004, and the total spread in the tangent values is 0.0020. Therefore, the tangent of θ is 0.0004/0.0020 of the way between the tangents of the two angles given in the table. This is 1/5 of the total spread, since

0.0004/0.0020 = 4/20 = 1/5

Another way of arriving at this result is to observe that the total spread is 20 ten- thouandths, and that the partial spread cor- responding to angle θ is 4 ten-thousandths. Since 4 out of 20 is the same as 1 out of 5, angle θ is 1/5 of the way between 22°36' and 22°42'. Taking 1/5 of the 6' spread between the angles, we have the following calculation:

1/5 x 6' = 1/5 x 5'60"

= 1'12" (1 minute and 12 seconds)

The 12" obtained in this calculation causes our answer to appear to have greater accuracy than the tables from which it is derived. This appar- ent increase in accuracy is a normal result of interpolation. Final answers based on inter- polated data should be rounded off to the same degree of accuracy as that of the original data. The value of 1 minute and 12 seconds found in the foregoing problem is added to 22°36', as follows:

θ = 22°36' + 1'12" = 22°37'12"

Therefore θ is 22°37', approximately. The foregoing problem could have been solved in terms of tenths and hundredths of a degree, rather than minutes, as follows:

13 12 5

Figure 19-9.—Using trigonometric ratios to evaluate angles.

a right triangle. The only information given, concerning angle θ, is the ratio of sides in the triangle. The size of θ is calculated as follows:

tan θ = 5/12 = 0.4167

θ = the angle whose tangent is 0.4167

Assuming that the sides and angles in figure 19-9 are in approximately the correct propor- tions, we estimate that angle θ is about 20°. The table entries for the tangent in the vicinity of 20° are slightly too small, since we need a number near 0.4167. However, the tangent of 23°36' is 0.4163 and the tangent of 23°42' is 0.4183. Therefore, θ is- between 22°36' and 22°42'.

Interpolation

It is frequently necessary to estimate the value of an angle to a closer approximation than is available in the table. This is equiva- lent to estimating between table entries, and the process is called INTERPOLATION. For example, in the foregoing problem it was deter- mined that the angle value was between 22°36' and 22°42'. The following paragraphs describe the procedure for interpolating to find a closer approximation to the value of the angle.

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ANGLE TANGENT 2. Find the angle which corresponds to each of the following decimal values in appendix II: 22.60° 0.4163 } a. sin θ = 0.2790 c. tan θ = 0.7604 0.1° < θ 0.4167 0.0004 > 0.0020 } b. cos θ = 0.9018 d. sin θ = 0.8142 22.70° 0.4183 Answers:

In this example, we are concerned with an 1. a. 1 d. 0.6051 angular spread of 0.10° and θ is located 1/5 of b. 0.8660 e. 0.6225 the way through this spread. Thus we have c. 0.7420 f. 0.2447

θ = 22.60° + (1/5 x 0.10°) 2. a. θ = 16.2° c. θ = 37°15' θ = 22.60° + 0.02° b. θ = 25°36' d. θ = 54°30' θ = 22.62°

Interpolation must be approached with com- RIGHT TRIANGLES WITH mon sense, in order to avoid applying correc- SPECIAL ANGLES AND SIDE RATIOS tions in the wrong direction. For example, the cosine of an angle decreases in value as the Three types of right triangles are especially angle increases from 0° to 90°. If we need the significant because of their frequent occur- value of the cosine of an angle such as 22°39', rence. These are the 30°-60°-90° triangle, the the calculation is as follows: 45°-90° triangle, and the 3-4-5 triangle.

ANGLE COSINE THE 30°-60°-90° TRIANGLE

22°36' 0.9232 The 30°-60°-90° triangle is so named be- } cause these are the sizes of its three angles. 6' < 22°39' } 3' < 0.0007 The sides of this triangle are in the ratio of } 1 to √3 to 2, as shown in figure 19-10. 22°42' 0.9225

In this example, it is easy to see that 22°39' is halfway between 22°36' and 22°42'. There- fore the cosine of 22°39' is halfway between the cosine of 22°36' and that of 22°42'. Taking B one-half of the spread between these cosines, 60° we then SUBTRACT from 0.9232 to find the 2 cosine of 22°39', as follows:

cos 22°39' = 0.9232 - (1/2 x 0.0007) 1

= 0.9232 - 0.00035 30° 90° = 0.92285 A C = 0.9229 (approximately) √3

Practice problems: Figure 19-10.—30°-60°-90° triangle.

1. Use the table in appendix II to find the deci- mal value of each of the following ratios: The sine ratio for the 30° angle in figure 19-10 establishes the proportionate values of a. tan 45° d. sin 37°14' the sides. For example, we know that the sine b. sin 60° e. cos 51.5° of 30° is 1/2; therefore side AB must be twice c. cos 42°6' f. tan 13.75° as long as BC. If side BC is 1 unit long, then

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Chapter 19—NUMERICAL TRIGONOMETRY

side AB is 2 units long and, by the rule of Pythagoras, AC is found as follows:

AC = √(AB)² - (BC)² = √4 - 1 = √3

Regardless of the size of the unit, a 30°- 60°-90° triangle has a hypotenuse which is 2 times as long as the shortest side. The short- est side is opposite the 30° angle. The side op- posite the 60° angle is √3 times as long as the shortest side. For example, suppose that the hypotenuse of a 30°-60°-90° triangle is 30 units long; then the shortest side is 15 units long, and the length of the side opposite the 60° angle is 15√3 units.

Practice problems. Without reference to tables or to the rule of Pythagoras, find the following lengths and angles in figure 19-11:

1. Length of AC. 4. Length of RT. 2. Size of angle A. 5. Length of RS. 3. Size of angle B. 6. Size of angle T.

[DIAGRAM: Triangle ABC with angle C = 90°, angle A = (unmarked), side AB = 2, and a right angle at C marked as 1]

[DIAGRAM: Triangle TRS with angle R = 60°, angle S = 90°, and side TS = 2√3]

Figure 19-11.—Finding parts of 30°-60°-90° triangles.

Answers:

1. √3 4. 4 2. 30° 5. 2 3. 60° 6. 30°

THE 45°-90° TRIANGLE

Figure 19-12 illustrates a triangle in which two angles measure 45° and the third angle

measures 90°. Since angles A and B are equal, the sides opposite them are also equal. There- fore, AC equals CB. Suppose that CB is 1 unit long; then AC is also 1 unit long, and the length of AB is calculated as follows:

(AB)² = 1² + 1² = 2

AB = √2

Regardless of the size of the triangle, if it has two 45° angles and one 90° angle, its sides are in the ratio 1 to 1 to √2. For example, if sides AC and CB are 3 units long, AB is 3√2 units long.

Practice problems. Without reference to tables or to the rule of Pythagoras, find the following lengths and angles in figure 19-13:

1. AB 2. BC 3. Angle B

Answers:

1. 2√2 2. 2 3. 45°

THE 3-4-5 TRIANGLE

The triangle shown in figure 19-14 has its sides in the ratio 3 to 4 to 5. Any triangle with its sides in this ratio is a right triangle.

It is a common error to assume that a tri- angle is a 3-4-5 type because two sides are known to be in the ratio 3 to 4, or perhaps 4 to 5. Figure 19-15 shows two examples of tri- angles which happen to have two of their sides in the stated ratio, but not the third side. This

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MATHEMATICS, VOLUME 1

It is interesting to note that the third side in figure 19-15 (B) is √7. This is a very unusual coincidence, in which one side of a right tri- angle is the square root of the sum of the other two sides.

Related to the basic 3-4-5 triangle are all triangles whose sides are in the ratio 3 to 4 to 5 but are longer (proportionately) than these basic lengths. For example, the triangle pic- tured in figure 19-6 is a 3-4-5 triangle.

[DIAGRAM: Triangle ABC with angle A = 45°, angle C = 90°, side AC = 2]

Figure 19-13.—Finding unknown parts in a 45°-90° triangle.

[DIAGRAM: Triangle with sides 10, 6, and 8]

Figure 19-16.—Triangle with sides which are multiples of 3, 4, and 5.

The 3-4-5 triangle is very useful in calcula- tions of distance. If the data can be adapted to fit a 3-4-5 configuration, no tables or calcula- tion of square root (Pythagorean Theorem) are needed.

EXAMPLE: An observer at the top of a 40-foot vertical tower knows that the base of the tower is 30 feet from a target on the ground. How does he calculate his slant range (direct line of sight) from the target?

SOLUTION: Figure 19-17 shows that the de- sired length, AB, is the hypotenuse of a right triangle whose shorter sides are 30 feet and 40 feet long. Since these sides are in the ratio 3 to 4 and angle C is 90°, the triangle is a 3-4-5 triangle. Therefore, side AB represents the 5-unit side of the triangle. The ratio 30 to 40 to 50 is equivalent to 3-4-5, and thus side AB is 50 units long.

Practice problems. Without reference to tables or to the rule of Pythagoras, solve the following problems:

1. An observer is at the top of a 30-foot verti- cal tower. Calculate his slant range from a target on the ground which is 40 feet from the base of the tower.

[DIAGRAM: Triangle with hypotenuse 5, legs 3 and 4, marked as (A)]

[DIAGRAM: Triangle with hypotenuse √7, legs 4 and 3, angle = 110°, marked as (B)]

Figure 19-15.—Triangles which may be mistaken for 3-4-5 triangles.

can be because the triangle is not a right tri- angle, as in figure 19-15 (A). On the other hand, even though the triangle is a right tri- angle its longest side may be the 4-unit side, in which case the third side cannot be 5 units long. (See fig. 19-15 (B).)

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Chapter 18—NUMERICAL TRIGONOMETRY

B B

35 40

A 30 C A 40° D 75° C

Figure 19-17.—Solving problems with a Figure 19-18.—Finding the unknown parts 3-4-5 triangle. of an oblique triangle.

2. A guy wire 15 feet long is stretched from CAUTION: A careless appraisal of this prob- the top of a pole to a point on the ground 9 feet lem may lead the unwary trainee to represent from the base of the pole. Calculate the height the ratio AC/AB as the cosine of 40°. This of the pole. error is avoided only by the realization that the trigonometric ratios are based on RIGHT tri- Answers: angles.

1. 50 feet 2. 12 feet 2. In order to find the length of DC, first calculate BD. OBLIQUE TRIANGLES BD Oblique triangles were defined in chapter 17 ---- = sin 40° of this training course as triangles which con- 35 tain no right angles. A natural approach to the solution of problems involving oblique triangles BD = 35 sin 40° is to construct perpendicular lines and form right triangles which subdivide the original tri- = 35 (0.6428) angle. Then the problem is solved by the usual methods for right triangles. = 22.4 (approximately)

DIVISION INTO RIGHT TRIANGLES 3. Find the length of DC

The oblique triangle ABC in figure 19-18 22.4 has been divided into two right triangles by -------- = tan 75° drawing line BD perpendicular to AC. The DC length of AC is found as follows: 22.4 22.4 1. Find the length of AD. DC = ------- = ------- tan 75° 3.732 AD ---- = cos 40° DC = 6.01 (approximately) 35 4. Add AD and DC to find AC. AD = 35 cos 40°

= 35 (0.7660) 26.8 + 6.01 = 32.81

= 26.8 (approximately) AC = 32.8 (approximately)

SOLUTION BY SIMULTANEOUS EQUATIONS

A typical problem in trigonometry is the determination of the height of a point such as B in figure 19-19.

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x tan 70° = (50 + x) tan 30° B x (2.748) = 50 (0.5774) + x(0.5774) h x (2.748) - x (0.5774) = 28.8

x (2.171) = 28.8

28.8 x = ------- = 13.3 feet A 50 FT C x D 2.171

Figure 19-19.—Calculation of unknown Knowing the value of x, it is now possible to quantities by means of oblique triangles. compute h as follows:

Suppose that point B is the top of a hill, and h = x tan 70° point D is inaccessible. Then the only meas- urements possible on the ground are those = 13.3 (2.748) shown in figure 19-19. If we let h represent BD and x represent CD, we can set up the fol- = 36.5 feet (approximately) lowing system of simultaneous equations: Practice problems: h ---- = tan 70° x 1. Find the length of side BC in figure 19-20 (A).

h 2. Find the height of point B above line AD in -------- = tan 30° 50 + x figure 19-20 (B).

Solving these two equations for h in terms of x, we have Answers:

h = x tan 70° 1. 21.3 feet 2. 41.7 feet

and LAW OF SINES h = (50 + x) tan 30°

The law of sines provides a direct approach Since the two quantities which are both equal to the solution of oblique triangles, avoiding the to h must be equal to each other, we have necessity of subdividing into right triangles. Let the triangle in figure 19-21 (A) represent any oblique triangle with all of its angles acute. The labels used in figure 19-21 are stand- ardized. The small letter a is used for the side opposite angle A; small b is opposite angle B; small c is opposite angle C.

B B

40 FT 55° 70 FT 65° 32° 85° 55° A C A C D

(A) (B)

Figure 19-20.—(A) Oblique triangle with all angles acute; (B) obtuse triangle.

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Chapter 19—NUMERICAL TRIGONOMETRY

B B /| / / |a / |a / | / | / | / | A/____C A/____C b b (A) (B)

Figure 19-21.—(A) Acute oblique triangle with standard labels; (B) obtuse triangle with standard labels.

The law of sines states that in any triangle, SOLUTION: By the law of sines, whether it is acute as in figure 19-21 (A) or obstuse as in figure 19-21 (B), the following is 20 = c true: sin 15° = sin 85°

a b c 20 sin 85° ——— = ——— = ——— c = ——————————— sin A sin B sin C sin 15°

EXAMPLE: In figure 19-21 (A), let angle A be 20 (0.9962) 15° and let angle C be 85°. If BC is 20 units, c = ————————— = 77.0 find the length of AB. 0.2588

APPENDIX I

SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS, LOGARITHMS, AND RECIPROCALS OF NUMBERS

┌────┬────────┬────────┬────────┬────────┬────────┬──────────┬─────────┬──────────┐ │ No.│ Square │ Cube │Square │ Cube │ Log. │1000 │No.= Dia.│ │ │ │ │ │ Root │ Root │ │÷ Recip. │ Circum. │ Area │ ├────┼────────┼────────┼────────┼────────┼────────┼──────────┼─────────┼──────────┤ │ 1 │ 1 │ 1 │ 1.0000 │ 1.0000 │ 0.00000│1000.000 │ 3.142 │ 0.7854 │ │ 2 │ 4 │ 8 │ 1.4142 │ 1.2599 │ 0.30103│ 500.000 │ 6.283 │ 3.1416 │ │ 3 │ 9 │ 27 │ 1.7321 │ 1.4422 │ 0.47712│ 333.333 │ 9.425 │ 7.0686 │ │ 4 │ 16 │ 64 │ 2.0000 │ 1.5874 │ 0.60206│ 250.000 │12.566 │12.5664 │ │ 5 │ 25 │ 125 │ 2.2361 │ 1.7100 │ 0.69897│ 200.000 │15.708 │19.6350 │ │ │ │ │ │ │ │ │ │ │ │ 6 │ 36 │ 216 │ 2.4495 │ 1.8171 │ 0.77815│ 166.667 │18.850 │28.2743 │ │ 7 │ 49 │ 343 │ 2.6458 │ 1.9129 │ 0.84510│ 142.857 │21.991 │38.4845 │ │ 8 │ 64 │ 512 │ 2.8284 │ 2.0000 │ 0.90309│ 125.000 │25.133 │50.2655 │ │ 9 │ 81 │ 729 │ 3.0000 │ 2.0801 │ 0.95424│ 111.111 │28.274 │63.6173 │ │ 10 │ 100 │ 1000 │ 3.1623 │ 2.1544 │ 1.00000│ 100.000 │31.416 │78.5398 │ │ │ │ │ │ │ │ │ │ │ │ 11 │ 121 │ 1331 │ 3.3166 │ 2.2240 │ 1.04139│ 90.9091 │34.558 │ 95.0332 │ │ 12 │ 144 │ 1728 │ 3.4641 │ 2.2894 │ 1.07918│ 83.3333 │37.699 │113.097 │ │ 13 │ 169 │ 2197 │ 3.6056 │ 2.3513 │ 1.11394│ 76.9231 │40.841 │132.732 │ │ 14 │ 196 │ 2744 │ 3.7417 │ 2.4101 │ 1.14613│ 71.4286 │43.982 │153.938 │ │ 15 │ 225 │ 3375 │ 3.8730 │ 2.4662 │ 1.17609│ 66.6667 │47.124 │176.715 │ │ │ │ │ │ │ │ │ │ │ │ 16 │ 256 │ 4096 │ 4.0000 │ 2.5198 │ 1.20412│ 62.5000 │50.265 │201.062 │ │ 17 │ 289 │ 4913 │ 4.1231 │ 2.5713 │ 1.23045│ 58.8235 │53.407 │226.980 │ │ 18 │ 324 │ 5832 │ 4.2426 │ 2.6207 │ 1.25527│ 55.5556 │56.549 │254.469 │ │ 19 │ 361 │ 6859 │ 4.3589 │ 2.6684 │ 1.27875│ 52.6316 │59.690 │283.529 │ │ 20 │ 400 │ 8000 │ 4.4721 │ 2.7144 │ 1.30103│ 50.0000 │62.832 │314.159 │ │ │ │ │ │ │ │ │ │ │ │ 21 │ 441 │ 9261 │ 4.5826 │ 2.7589 │ 1.32222│ 47.6190 │65.973 │346.361 │ │ 22 │ 484 │ 10648 │ 4.6904 │ 2.8020 │ 1.34242│ 45.4545 │69.115 │380.133 │ │ 23 │ 529 │ 12167 │ 4.7958 │ 2.8439 │ 1.36173│ 43.4783 │72.257 │415.476 │ │ 24 │ 576 │ 13824 │ 4.8990 │ 2.8845 │ 1.38021│ 41.6667 │75.398 │452.389 │ │ 25 │ 625 │ 15625 │ 5.0000 │ 2.9240 │ 1.39794│ 40.0000 │78.540 │490.874 │ │ │ │ │ │ │ │ │ │ │ │ 26 │ 676 │ 17576 │ 5.0990 │ 2.9625 │ 1.41497│ 38.4615 │81.681 │530.929 │ │ 27 │ 729 │ 19683 │ 5.1962 │ 3.0000 │ 1.43136│ 37.0370 │84.823 │572.555 │ │ 28 │ 784 │ 21952 │ 5.2915 │ 3.0366 │ 1.44716│ 35.7143 │87.965 │615.752 │ │ 29 │ 841 │ 24389 │ 5.3852 │ 3.0723 │ 1.46240│ 34.4828 │91.106 │660.520 │ │ 30 │ 900 │ 27000 │ 5.4772 │ 3.1072 │ 1.47712│ 33.3333 │94.248 │706.858 │ │ │ │ │ │ │ │ │ │ │ │ 31 │ 961 │ 29791 │ 5.5678 │ 3.1414 │ 1.49136│ 32.2581 │97.389 │754.768 │ │ 32 │ 1024 │ 32768 │ 5.6569 │ 3.1748 │ 1.50515│ 31.2500 │100.531 │804.248 │ │ 33 │ 1089 │ 35937 │ 5.7446 │ 3.2075 │ 1.51851│ 30.3030 │103.673 │855.299 │ │ 34 │ 1156 │ 39304 │ 5.8310 │ 3.2396 │ 1.53148│ 29.4118 │106.814 │907.920 │ │ 35 │ 1225 │ 42875 │ 5.9161 │ 3.2711 │ 1.54407│ 28.5714 │109.956 │962.113 │ │ │ │ │ │ │ │ │ │ │ │ 36 │ 1296 │ 46656 │ 6.0000 │ 3.3019 │ 1.55630│ 27.7778 │113.097 │1017.88 │ │ 37 │ 1369 │ 50653 │ 6.0828 │ 3.3322 │ 1.56820│ 27.0270 │116.239 │1075.21 │ │ 38 │ 1444 │ 54872 │ 6.1644 │ 3.3619 │ 1.57978│ 26.3158 │119.381 │1134.11 │ │ 39 │ 1521 │ 59319 │ 6.2450 │ 3.3912 │ 1.59106│ 25.6410 │122.522 │1194.59 │ │ 40 │ 1600 │ 64000 │ 6.3246 │ 3.4200 │ 1.60206│ 25.0000 │125.66 │1256.64 │ │ │ │ │ │ │ │ │ │ │ │ 41 │ 1681 │ 68921 │ 6.4031 │ 3.4482 │ 1.61278│ 24.3902 │128.81 │1320.25 │ │ 42 │ 1764 │ 74088 │ 6.4807 │ 3.4760 │ 1.62325│ 23.8095 │131.95 │1385.44 │ │ 43 │ 1849 │ 79507 │ 6.5574 │ 3.5034 │ 1.63347│ 23.2558 │135.09 │1452.20 │ │ 44 │ 1936 │ 85184 │ 6.6332 │ 3.5303 │ 1.64345│ 22.7273 │138.23 │1520.53 │ └────┴────────┴────────┴────────┴────────┴────────┴──────────┴─────────┴──────────┘

210

Appendix I—POWERS, ROOTS, LOGARITHMS, ETC.

N Square Cube Square Cube Log. 1000 N = D.n. Root Root × Ramp. Circum. Area

45 2025 91125 6.7082 3.5569 1.65321 21.2232 141.37 1590.43 46 2116 97336 6.7823 3.5830 1.66276 21.7391 144.51 1661.90 47 2209 103823 6.8557 3.6088 1.67210 21.2766 147.65 1734.94 48 2304 110592 6.9282 3.6342 1.68124 20.8333 150.80 1809.56 49 2401 117649 7.0000 3.6593 1.69020 20.4082 153.94 1885.74

50 2500 125000 7.0711 3.6840 1.69897 20.0000 157.08 1963.50 51 2601 132651 7.1414 3.7084 1.70757 19.6078 160.22 2042.82 52 2704 140608 7.2111 3.7325 1.71600 19.2308 163.36 2123.72 53 2809 148877 7.2801 3.7563 1.72428 18.8679 166.50 2206.18 54 2916 157464 7.3485 3.7798 1.73239 18.5185 169.65 2290.22

55 3025 166375 7.4162 3.8030 1.74036 18.1818 172.79 2375.83 56 3136 175616 7.4833 3.8259 1.74819 17.8571 175.93 2463.01 57 3249 185193 7.5498 3.8485 1.75587 17.5439 179.07 2551.76 58 3364 195112 7.6158 3.8709 1.76343 17.2414 182.21 2642.08 59 3481 205379 7.6811 3.8930 1.77085 16.9492 185.35 2733.99

60 3600 216000 7.7460 3.9149 1.77815 16.6667 188.50 2827.43 61 3721 226981 7.8102 3.9365 1.78533 16.3934 191.64 2922.47 62 3844 238328 7.8740 3.9579 1.79239 16.1290 194.78 3019.07 63 3969 250047 7.9373 3.9791 1.79934 15.8730 197.92 3117.25 64 4096 262144 8.0000 4.0000 1.80618 15.6250 201.06 3216.99

65 4225 274625 8.0623 4.0207 1.81291 15.3846 204.20 3318.31 66 4356 287496 8.1240 4.0412 1.81954 15.1515 207.35 3421.19 67 4489 300763 8.1854 4.0615 1.82607 14.9254 210.49 3525.65 68 4624 314432 8.2462 4.0817 1.83251 14.7059 213.63 3631.68 69 4761 328509 8.3066 4.1016 1.83885 14.4928 216.77 3739.28

70 4900 343000 8.3666 4.1213 1.84510 14.2857 219.91 3848.45 71 5041 357911 8.4261 4.1408 1.85126 14.0845 223.05 3959.19 72 5184 373248 8.4853 4.1602 1.85733 13.8889 226.19 4071.50 73 5329 389017 8.5440 4.1793 1.86332 13.6986 229.34 4185.39 74 3476 405224 8.6023 4.1983 1.86923 13.5135 232.48 4300.84

75 5625 421875 8.6603 4.2172 1.87506 13.3333 235.62 4417.86 76 5776 438976 8.7178 4.2358 1.88081 13.1579 238.76 4536.46 77 5929 456533 8.7750 4.2543 1.88649 12.9870 241.90 4656.63 78 6084 474552 8.8318 4.2727 1.89209 12.8205 245.04 4778.36 79 6241 493039 8.8882 4.2908 1.89763 12.6582 248.19 4901.67

80 6400 512000 8.9443 4.3089 1.90309 12.5000 251.33 5026.55 81 6561 531441 9.0000 4.3267 1.90849 12.3457 254.47 5153.00 82 6724 551368 9.0554 4.3445 1.91381 12.1951 257.61 5281.02 83 6889 571787 9.1104 4.3621 1.91908 12.0482 260.75 5410.61 84 7056 592704 9.1652 4.3795 1.92428 11.9048 263.89 5541.77

85 7225 614125 9.2195 4.3968 1.92942 11.7647 267.04 5674.50 86 7396 636056 9.2736 4.4140 1.93450 11.6279 270.18 5808.80 87 7569 658503 9.3274 4.4310 1.93952 11.4943 273.32 5944.68 88 7744 681472 9.3808 4.4480 1.94448 11.3636 276.46 6082.12 89 7921 704969 9.4340 4.4647 1.94939 11.2360 279.60 6231.14

MATHEMATICS, VOLUME 1

N. Square Cube Square Cube Log. 1000 N = D.n. Root Root × Ramp. Circum. Area

90 8100 729000 9.4868 4.4814 1.95424 11.1111 282.74 6361.73 91 8281 753571 9.5394 4.4979 1.95904 10.9890 285.88 6503.88 92 8464 778688 9.5917 4.5144 1.96379 10.8696 289.03 6647.61 93 8649 804357 9.6437 4.5307 1.96848 10.7527 292.17 6792.91 94 8836 830584 9.6954 4.5468 1.97313 10.6383 295.31 6939.78

95 9025 857375 9.7468 4.5629 1.97772 10.5263 298.45 7088.22 96 9216 884736 9.7980 4.5789 1.98227 10.4167 301.59 7238.23 97 9409 912673 9.8489 4.5947 1.98677 10.3093 304.73 7389.81 98 9604 941192 9.8995 4.6104 1.99122 10.2041 307.88 7542.96 99 9801 970299 9.9499 4.6261 1.99564 10.1010 311.02 7697.69

100 10000 1000000 10.0000 4.6416 2.00000 10.00000 314.16 7833.98 101 10201 1030301 10.0499 4.6570 2.00432 9.90099 317.30 8011.85 102 10404 1061208 10.0995 4.6723 2.00860 9.80392 320.44 8171.28 103 10609 1092727 10.1489 4.6873 2.01284 9.70874 323.58 8332.59 104 10816 1124864 10.1980 4.7027 2.01703 9.61538 326.73 8494.87

105 11025 1157625 10.2470 4.7177 2.02119 9.52381 329.87 8659.01 106 11236 1191016 10.2956 4.7326 2.02531 9.43396 333.01 8824.73 107 11449 1225043 10.3441 4.7473 2.02938 9.34579 336.15 8992.12 108 11664 1259712 10.3923 4.7622 2.03342 9.25926 339.29 9160.88 109 11881 1295029 10.4403 4.7769 2.03743 9.17431 342.43 9331.32

110 12100 1331000 10.4881 4.7914 2.04139 9.09091 345.58 9503.32 111 12321 1367631 10.5357 4.8059 2.04532 9.00901 348.72 9676.89 112 12544 1404928 10.5830 4.8203 2.04922 8.93857 351.86 9852.03 113 12769 1442597 10.6301 4.8346 2.05308 8.84956 355.00 10028.7 114 12996 1481544 10.6771 4.8488 2.05690 8.77193 358.14 10207.0

115 13225 1520875 10.7238 4.8629 2.06070 8.69565 361.28 10386.9 116 13456 1560896 10.7703 4.8770 2.06446 8.62069 364.42 10568.3 117 13689 1601613 10.8167 4.8910 2.06819 8.54701 367.57 10751.3 118 13924 1643032 10.8628 4.9048 2.07188 8.47458 370.71 10935.8 119 14161 1685159 10.9087 4.9187 2.07555 8.40336 373.85 11122.0

120 14400 1728000 10.9545 4.9324 2.07918 8.33333 376.99 11309.7 121 14641 1771361 11.0000 4.9461 2.08279 8.26446 380.13 11499.0 122 14884 1815848 11.0454 4.9597 2.08636 8.19672 383.27 11689.9 123 15129 1860867 11.0905 4.9732 2.08991 8.13008 386.41 11882.3 124 15376 1906624 11.1355 4.9866 2.09342 8.06452 389.56 12076.3

125 15625 1953125 11.1803 5.0000 2.09691 8.00000 392.70 12271.8 126 15876 2000376 11.2250 5.0133 2.10037 7.93651 395.84 12469.0 127 16129 2048383 11.2694 5.0265 2.10380 7.87402 398.98 12667.7 128 16384 2097152 11.3137 5.0397 2.10721 7.81250 402.12 12868.0 129 16641 2146689 11.3578 5.0528 2.11059 7.75194 405.27 13069.8

130 16900 2197000 11.4018 5.0653 2.11394 7.69231 408.41 13273.2 131 17161 2248091 11.4455 5.0788 2.11727 7.63359 411.55 13478.2 132 17424 2299968 11.4891 5.0916 2.12057 7.57576 414.69 13684.8 133 17689 2352637 11.5326 5.1045 2.12385 7.51880 417.83 13892.9 134 17956 2406104 11.5758 5.1172 2.12710 7.46369 420.97 14103.6

APPENDIX II NATURAL SINES, COSINES, AND TANGENTS OF ANGLES FROM 0° to 90°

0°–14.9°

| Degs. | Function | 0.0° | 0.1° | 0.2° | 0.3° | 0.4° | 0.5° | 0.6° | 0.7° | 0.8° | 0.9° | |-------|----------|------|------|------|------|------|------|------|------|------|------| | 0 | sin cos tan | 0.0000 1.0000 0.0000 | 0.0017 1.0000 0.0017 | 0.0035 1.0000 0.0035 | 0.0052 1.0000 0.0052 | 0.0070 1.0000 0.0070 | 0.0087 1.0000 0.0087 | 0.0105 0.9999 0.0105 | 0.0122 0.9999 0.0122 | 0.0140 0.9999 0.0140 | 0.0157 0.9999 0.0157 | | 1 | sin cos tan | 0.0175 0.9998 0.0175 | 0.0192 0.9998 0.0192 | 0.0209 0.9998 0.0209 | 0.0227 0.9997 0.0227 | 0.0244 0.9997 0.0244 | 0.0262 0.9997 0.0262 | 0.0279 0.9996 0.0279 | 0.0297 0.9996 0.0297 | 0.0314 0.9995 0.0314 | 0.0332 0.9995 0.0332 | | 2 | sin cos tan | 0.0349 0.9994 0.0349 | 0.0366 0.9993 0.0366 | 0.0384 0.9993 0.0384 | 0.0401 0.9992 0.0401 | 0.0419 0.9991 0.0419 | 0.0436 0.9990 0.0436 | 0.0454 0.9989 0.0454 | 0.0471 0.9988 0.0471 | 0.0488 0.9988 0.0488 | 0.0506 0.9987 0.0506 | | 3 | sin cos tan | 0.0523 0.9986 0.0524 | 0.0541 0.9985 0.0541 | 0.0558 0.9984 0.0558 | 0.0576 0.9983 0.0576 | 0.0593 0.9982 0.0593 | 0.0610 0.9981 0.0610 | 0.0628 0.9980 0.0628 | 0.0645 0.9979 0.0645 | 0.0663 0.9978 0.0663 | 0.0680 0.9977 0.0680 | | 4 | sin cos tan | 0.0698 0.9976 0.0699 | 0.0715 0.9975 0.0715 | 0.0732 0.9974 0.0732 | 0.0750 0.9973 0.0750 | 0.0767 0.9972 0.0767 | 0.0785 0.9971 0.0785 | 0.0802 0.9969 0.0802 | 0.0819 0.9968 0.0819 | 0.0837 0.9967 0.0837 | 0.0854 0.9966 0.0854 | | 5 | sin cos tan | 0.0872 0.9962 0.0875 | 0.0889 0.9961 0.0892 | 0.0906 0.9960 0.0906 | 0.0924 0.9959 0.0924 | 0.0941 0.9957 0.0941 | 0.0958 0.9956 0.0958 | 0.0976 0.9954 0.0976 | 0.0993 0.9953 0.0993 | 0.1011 0.9951 0.1011 | 0.1028 0.9950 0.1028 | | 6 | sin cos tan | 0.1045 0.9945 0.1051 | 0.1063 0.9943 0.1063 | 0.1080 0.9942 0.1080 | 0.1097 0.9940 0.1097 | 0.1115 0.9938 0.1115 | 0.1132 0.9936 0.1132 | 0.1149 0.9934 0.1149 | 0.1167 0.9932 0.1167 | 0.1184 0.9930 0.1184 | 0.1201 0.9928 0.1201 | | 7 | sin cos tan | 0.1219 0.9925 0.1228 | 0.1236 0.9923 0.1236 | 0.1253 0.9921 0.1253 | 0.1271 0.9919 0.1271 | 0.1288 0.9917 0.1288 | 0.1305 0.9914 0.1305 | 0.1323 0.9912 0.1323 | 0.1340 0.9910 0.1340 | 0.1357 0.9907 0.1357 | 0.1374 0.9905 0.1374 | | 8 | sin cos tan | 0.1392 0.9903 0.1405 | 0.1409 0.9900 0.1409 | 0.1426 0.9898 0.1426 | 0.1444 0.9895 0.1444 | 0.1461 0.9893 0.1461 | 0.1478 0.9890 0.1478 | 0.1495 0.9888 0.1495 | 0.1513 0.9885 0.1513 | 0.1530 0.9882 0.1530 | 0.1547 0.9880 0.1547 | | 9 | sin cos tan | 0.1564 0.9877 0.1584 | 0.1582 0.9874 0.1582 | 0.1599 0.9871 0.1599 | 0.1616 0.9869 0.1616 | 0.1633 0.9866 0.1633 | 0.1650 0.9863 0.1650 | 0.1668 0.9860 0.1668 | 0.1685 0.9857 0.1685 | 0.1702 0.9854 0.1702 | 0.1719 0.9851 0.1719 | | 10 | sin cos tan | 0.1736 0.9848 0.1763 | 0.1754 0.9845 0.1754 | 0.1771 0.9842 0.1771 | 0.1788 0.9839 0.1788 | 0.1805 0.9836 0.1805 | 0.1822 0.9833 0.1822 | 0.1840 0.9829 0.1840 | 0.1857 0.9826 0.1857 | 0.1874 0.9823 0.1874 | 0.1891 0.9820 0.1891 | | 11 | sin cos tan | 0.1908 0.9816 0.1944 | 0.1925 0.9813 0.1925 | 0.1942 0.9810 0.1942 | 0.1959 0.9807 0.1959 | 0.1977 0.9803 0.1977 | 0.1994 0.9799 0.1994 | 0.2011 0.9796 0.2011 | 0.2028 0.9792 0.2028 | 0.2045 0.9789 0.2045 | 0.2062 0.9785 0.2062 | | 12 | sin cos tan | 0.2079 0.9781 0.2126 | 0.2096 0.9778 0.2096 | 0.2113 0.9774 0.2113 | 0.2130 0.9770 0.2130 | 0.2147 0.9766 0.2147 | 0.2164 0.9763 0.2164 | 0.2181 0.9759 0.2181 | 0.2198 0.9755 0.2198 | 0.2215 0.9751 0.2215 | 0.2233 0.9747 0.2233 | | 13 | sin cos tan | 0.2250 0.9744 0.2309 | 0.2267 0.9740 0.2267 | 0.2284 0.9736 0.2284 | 0.2300 0.9732 0.2300 | 0.2317 0.9728 0.2317 | 0.2334 0.9724 0.2334 | 0.2351 0.9720 0.2351 | 0.2368 0.9715 0.2368 | 0.2385 0.9711 0.2385 | 0.2402 0.9707 0.2402 | | 14 | sin cos tan | 0.2419 0.9703 0.2493 | 0.2436 0.9699 0.2436 | 0.2453 0.9694 0.2453 | 0.2470 0.9690 0.2470 | 0.2487 0.9686 0.2487 | 0.2504 0.9681 0.2504 | 0.2521 0.9677 0.2521 | 0.2538 0.9672 0.2538 | 0.2554 0.9668 0.2554 | 0.2571 0.9664 0.2571 |

| Degs. | Function | 0' | 6' | 12' | 18' | 24' | 30' | 36' | 42' | 48' | 54' |

213

MATHEMATICS, VOLUME 1

15°–29.9°

| Degs. | Function | 0.0° | 0.1° | 0.2° | 0.3° | 0.4° | 0.5° | 0.6° | 0.7° | 0.8° | 0.9° | |-------|----------|------|------|------|------|------|------|------|------|------|------| | 15 | sin cos tan | 0.2588 0.9563 0.2709 | 0.2605 0.9555 0.2605 | 0.2622 0.9550 0.2622 | 0.2639 0.9545 0.2639 | 0.2656 0.9541 0.2656 | 0.2672 0.9537 0.2672 | 0.2689 0.9532 0.2689 | 0.2706 0.9528 0.2706 | 0.2723 0.9524 0.2723 | 0.2740 0.9520 0.2740 | | 16 | sin cos tan | 0.2756 0.9613 0.2867 | 0.2773 0.9608 0.2773 | 0.2790 0.9603 0.2790 | 0.2807 0.9598 0.2807 | 0.2823 0.9594 0.2823 | 0.2840 0.9589 0.2840 | 0.2857 0.9584 0.2857 | 0.2874 0.9580 0.2874 | 0.2890 0.9575 0.2890 | 0.2907 0.9570 0.2907 | | 17 | sin cos tan | 0.2924 0.9563 0.3057 | 0.2940 0.9555 0.2940 | 0.2957 0.9553 0.2957 | 0.2974 0.9548 0.2974 | 0.2990 0.9543 0.2990 | 0.3007 0.9537 0.3007 | 0.3024 0.9532 0.3024 | 0.3040 0.9527 0.3040 | 0.3057 0.9521 0.3057 | 0.3074 0.9516 0.3074 | | 18 | sin cos tan | 0.3090 0.9511 0.3249 | 0.3107 0.9505 0.3107 | 0.3123 0.9500 0.3123 | 0.3140 0.9494 0.3140 | 0.3156 0.9489 0.3156 | 0.3173 0.9483 0.3173 | 0.3190 0.9478 0.3190 | 0.3206 0.9472 0.3206 | 0.3223 0.9466 0.3223 | 0.3240 0.9461 0.3240 | | 19 | sin cos tan | 0.3256 0.9455 0.3443 | 0.3272 0.9449 0.3272 | 0.3289 0.9444 0.3289 | 0.3305 0.9438 0.3305 | 0.3322 0.9432 0.3322 | 0.3338 0.9426 0.3338 | 0.3355 0.9421 0.3355 | 0.3371 0.9415 0.3371 | 0.3387 0.9409 0.3387 | 0.3404 0.9403 0.3404 | | 20 | sin cos tan | 0.3420 0.9397 0.3640 | 0.3437 0.9391 0.3437 | 0.3453 0.9385 0.3453 | 0.3469 0.9379 0.3469 | 0.3486 0.9373 0.3486 | 0.3502 0.9367 0.3502 | 0.3518 0.9361 0.3518 | 0.3535 0.9354 0.3535 | 0.3551 0.9348 0.3551 | 0.3567 0.9342 0.3567 | | 21 | sin cos tan | 0.3584 0.9336 0.3839 | 0.3600 0.9330 0.3600 | 0.3616 0.9323 0.3616 | 0.3633 0.9317 0.3633 | 0.3649 0.9311 0.3649 | 0.3665 0.9304 0.3665 | 0.3681 0.9298 0.3681 | 0.3697 0.9291 0.3697 | 0.3714 0.9285 0.3714 | 0.3730 0.9278 0.3730 | | 22 | sin cos tan | 0.3746 0.9272 0.4040 | 0.3762 0.9265 0.3762 | 0.3778 0.9259 0.3778 | 0.3795 0.9252 0.3795 | 0.3811 0.9245 0.3811 | 0.3827 0.9239 0.3827 | 0.3843 0.9232 0.3843 | 0.3859 0.9225 0.3859 | 0.3875 0.9219 0.3875 | 0.3891 0.9212 0.3891 | | 23 | sin cos tan | 0.3907 0.9205 0.4245 | 0.3923 0.9198 0.3923 | 0.3939 0.9191 0.3939 | 0.3955 0.9184 0.3955 | 0.3971 0.9178 0.3971 | 0.3987 0.9171 0.3987 | 0.4003 0.9164 0.4003 | 0.4019 0.9157 0.4019 | 0.4035 0.9150 0.4035 | 0.4050 0.9143 0.4050 | | 24 | sin cos tan | 0.4067 0.9135 0.4452 | 0.4083 0.9128 0.4083 | 0.4099 0.9121 0.4099 | 0.4115 0.9114 0.4115 | 0.4131 0.9107 0.4131 | 0.4147 0.9100 0.4147 | 0.4163 0.9092 0.4163 | 0.4179 0.9085 0.4179 | 0.4195 0.9078 0.4195 | 0.4210 0.9070 0.4210 | | 25 | sin cos tan | 0.4226 0.9063 0.4663 | 0.4242 0.9056 0.4242 | 0.4258 0.9048 0.4258 | 0.4274 0.9041 0.4274 | 0.4289 0.9033 0.4289 | 0.4305 0.9026 0.4305 | 0.4321 0.9018 0.4321 | 0.4337 0.9011 0.4337 | 0.4352 0.9003 0.4352 | 0.4368 0.8996 0.4368 | | 26 | sin cos tan | 0.4384 0.8988 0.4877 | 0.4399 0.8980 0.4399 | 0.4415 0.8973 0.4415 | 0.4431 0.8965 0.4431 | 0.4446 0.8957 0.4446 | 0.4462 0.8949 0.4462 | 0.4478 0.8942 0.4478 | 0.4493 0.8934 0.4493 | 0.4509 0.8926 0.4509 | 0.4524 0.8918 0.4524 | | 27 | sin cos tan | 0.4540 0.8910 0.5095 | 0.4555 0.8902 0.4555 | 0.4571 0.8894 0.4571 | 0.4586 0.8886 0.4586 | 0.4602 0.8878 0.4602 | 0.4617 0.8870 0.4617 | 0.4633 0.8862 0.4633 | 0.4648 0.8854 0.4648 | 0.4664 0.8846 0.4664 | 0.4679 0.8838 0.4679 | | 28 | sin cos tan | 0.4695 0.8829 0.5317 | 0.4710 0.8821 0.4710 | 0.4726 0.8813 0.4726 | 0.4741 0.8805 0.4741 | 0.4756 0.8796 0.4756 | 0.4772 0.8788 0.4772 | 0.4787 0.8780 0.4787 | 0.4802 0.8771 0.4802 | 0.4818 0.8763 0.4818 | 0.4833 0.8755 0.4833 | | 29 | sin cos tan | 0.4848 0.8746 0.5543 | 0.4863 0.8738 0.4863 | 0.4879 0.8729 0.4879 | 0.4894 0.8721 0.4894 | 0.4909 0.8712 0.4909 | 0.4924 0.8704 0.4924 | 0.4939 0.8695 0.4939 | 0.4955 0.8686 0.4955 | 0.4970 0.8678 0.4970 | 0.4985 0.8669 0.4985 |

| Degs. | Function | 0' | 6' | 12' | 18' | 24' | 30' | 36' | 42' | 48' | 54' |

214

Appendix II–NATURAL SINES, COSINES, AND TANGENTS

80°-44.0°

Deps. Function 0.0° 0.1° 0.2° 0.3° 0.4° 0.5° 0.6° 0.7° 0.8° 0.9°

30 sin 0.5000 0.5015 0.5030 0.5045 0.6060 0.5075 0.5090 0.5106 0.5120 0.5135 cos 0.8660 0.8652 0.8643 0.8635 0.8623 0.8616 0.8607 0.8599 0.8590 0.8581 tan 0.5774 0.5792 0.5810 0.5827 0.5845 0.5862 0.5880 0.5898 0.5915 0.5933

31 sin 0.5150 0.5165 0.5180 0.5195 0.5210 0.5225 0.5240 0.5255 0.5270 0.5284 cos 0.8572 0.8563 0.8554 0.8545 0.8536 0.8526 0.8517 0.8508 0.8499 0.8490 tan 0.6009 0.6032 0.6050 0.6068 0.6084 0.6102 0.6120 0.6138 0.6156 0.6174

32 sin 0.5299 0.5314 0.5329 0.5344 0.5358 0.5373 0.5388 0.5402 0.5417 0.5432 cos 0.8480 0.8471 0.8462 0.8453 0.8443 0.8434 0.8425 0.8415 0.8406 0.8396 tan 0.6249 0.6273 0.6289 0.6307 0.6325 0.6344 0.6362 0.6381 0.6400 0.6418

33 sin 0.5446 0.5461 0.5476 0.5490 0.5505 0.5519 0.5534 0.5548 0.5563 0.5577 cos 0.8387 0.8377 0.8368 0.8358 0.8348 0.8339 0.8329 0.8320 0.8310 0.8300 tan 0.6494 0.6512 0.6531 0.6549 0.6568 0.6587 0.6605 0.6624 0.6643 0.6661

34 sin 0.5592 0.5606 0.5621 0.5635 0.5650 0.5664 0.5678 0.5693 0.5707 0.5721 cos 0.8290 0.8281 0.8271 0.8261 0.8251 0.8241 0.8231 0.8221 0.8211 0.8201 tan 0.6745 0.6771 0.6796 0.6822 0.6847 0.6873 0.6899 0.6924 0.6950 0.6975

35 sin 0.5736 0.5750 0.5764 0.5779 0.5793 0.5807 0.5821 0.5835 0.5850 0.5864 cos 0.8192 0.8181 0.8171 0.8161 0.8151 0.8141 0.8131 0.8121 0.8111 0.8100 tan 0.7002 0.7028 0.7054 0.7080 0.7107 0.7133 0.7159 0.7186 0.7212 0.7239

36 sin 0.5878 0.5892 0.5906 0.5920 0.5934 0.5948 0.5962 0.5976 0.5990 0.6004 cos 0.8090 0.8080 0.8070 0.8059 0.8049 0.8039 0.8029 0.8018 0.8008 0.7997 tan 0.7265 0.7292 0.7319 0.7346 0.7373 0.7400 0.7427 0.7454 0.7481 0.7508

37 sin 0.6018 0.6032 0.6046 0.6060 0.6074 0.6088 0.6101 0.6115 0.6129 0.6143 cos 0.7986 0.7976 0.7965 0.7955 0.7944 0.7934 0.7923 0.7912 0.7902 0.7891 tan 0.7536 0.7563 0.7590 0.7618 0.7646 0.7673 0.7701 0.7729 0.7757 0.7785

38 sin 0.6157 0.6170 0.6184 0.6198 0.6211 0.6225 0.6239 0.6252 0.6266 0.6280 cos 0.7880 0.7869 0.7859 0.7848 0.7837 0.7826 0.7815 0.7804 0.7793 0.7782 tan 0.7813 0.7841 0.7869 0.7897 0.7925 0.7953 0.7981 0.8009 0.8037 0.8065

39 sin 0.6293 0.6307 0.6320 0.6334 0.6347 0.6361 0.6374 0.6388 0.6401 0.6414 cos 0.7771 0.7760 0.7749 0.7738 0.7727 0.7716 0.7705 0.7694 0.7683 0.7672 tan 0.8092 0.8121 0.8149 0.8177 0.8205 0.8233 0.8262 0.8290 0.8319 0.8347

40 sin 0.6428 0.6441 0.6455 0.6469 0.6483 0.6494 0.6508 0.6521 0.6534 0.6548 cos 0.7660 0.7649 0.7638 0.7627 0.7615 0.7604 0.7593 0.7581 0.7570 0.7559 tan 0.8391 0.8423 0.8451 0.8481 0.8511 0.8541 0.8571 0.8600 0.8632 0.8662

41 sin 0.6561 0.6574 0.6587 0.6600 0.6613 0.6626 0.6639 0.6652 0.6665 0.6678 cos 0.7547 0.7536 0.7524 0.7513 0.7501 0.7490 0.7478 0.7466 0.7455 0.7443 tan 0.8693 0.8724 0.8754 0.8785 0.8816 0.8847 0.8878 0.8910 0.8941 0.8972

42 sin 0.6691 0.6704 0.6717 0.6730 0.6743 0.6756 0.6769 0.6782 0.6794 0.6807 cos 0.7431 0.7420 0.7408 0.7396 0.7385 0.7373 0.7361 0.7349 0.7337 0.7325 tan 0.9004 0.9036 0.9067 0.9099 0.9131 0.9163 0.9195 0.9228 0.9260 0.9293

43 sin 0.6820 0.6833 0.6845 0.6858 0.6871 0.6884 0.6896 0.6909 0.6921 0.6934 cos 0.7313 0.7301 0.7290 0.7278 0.7266 0.7254 0.7242 0.7230 0.7218 0.7206 tan 0.9325 0.9358 0.9391 0.9424 0.9457 0.9490 0.9524 0.9557 0.9590 0.9624

44 sin 0.6947 0.6959 0.6972 0.6984 0.6997 0.7009 0.7022 0.7034 0.7046 0.7059 cos 0.7193 0.7181 0.7169 0.7157 0.7145 0.7133 0.7120 0.7108 0.7096 0.7083 tan 0.9657 0.9691 0.9725 0.9759 0.9793 0.9827 0.9861 0.9895 0.9930 0.9963

Deps. Function 0° 6° 12° 18° 24° 30° 36° 42° 48° 54°

MATHEMATICS, VOLUME 1

45°-60.0°

Deps. Function 0.0° 0.1° 0.2° 0.3° 0.4° 0.5° 0.6° 0.7° 0.8° 0.9°

45 sin 0.7071 0.7083 0.7096 0.7108 0.7120 0.7133 0.7145 0.7157 0.7169 0.7181 cos 0.7071 0.7059 0.7046 0.7034 0.7022 0.7009 0.6997 0.6984 0.6972 0.6959 tan 1.0000 1.0035 1.0070 1.0105 1.0141 1.0176 1.0212 1.0247 1.0283 1.0319

46 sin 0.7193 0.7206 0.7218 0.7230 0.7242 0.7254 0.7266 0.7278 0.7290 0.7302 cos 0.6947 0.6934 0.6921 0.6909 0.6896 0.6884 0.6871 0.6858 0.6845 0.6833 tan 1.0355 1.0392 1.0428 1.0464 1.0501 1.0538 1.0575 1.0612 1.0649 1.0686

47 sin 0.7314 0.7325 0.7337 0.7349 0.7361 0.7373 0.7385 0.7396 0.7408 0.7420 cos 0.6820 0.6807 0.6794 0.6782 0.6769 0.6756 0.6743 0.6730 0.6717 0.6704 tan 1.0724 1.0761 1.0799 1.0837 1.0875 1.0913 1.0951 1.0990 1.1028 1.1067

48 sin 0.7431 0.7443 0.7455 0.7466 0.7478 0.7490 0.7501 0.7513 0.7524 0.7536 cos 0.6691 0.6678 0.6665 0.6652 0.6639 0.6626 0.6613 0.6600 0.6587 0.6574 tan 1.1106 1.1144 1.1183 1.1222 1.1261 1.1300 1.1339 1.1378 1.1418 1.1457

49 sin 0.7547 0.7559 0.7570 0.7581 0.7593 0.7604 0.7615 0.7627 0.7638 0.7649 cos 0.6561 0.6548 0.6534 0.6521 0.6508 0.6494 0.6481 0.6468 0.6455 0.6441 tan 1.1504 1.1544 1.1584 1.1623 1.1663 1.1708 1.1748 1.1788 1.1828 1.1868

50 sin 0.7660 0.7672 0.7683 0.7694 0.7705 0.7716 0.7727 0.7738 0.7749 0.7760 cos 0.6428 0.6414 0.6401 0.6388 0.6374 0.6361 0.6347 0.6334 0.6320 0.6307 tan 1.1918 1.1960 1.2002 1.2045 1.2087 1.2130 1.2172 1.2215 1.2258 1.2301

51 sin 0.7771 0.7782 0.7793 0.7804 0.7815 0.7826 0.7837 0.7848 0.7859 0.7869 cos 0.6293 0.6280 0.6266 0.6252 0.6239 0.6225 0.6211 0.6198 0.6184 0.6170 tan 1.2349 1.2393 1.2437 1.2482 1.2527 1.2572 1.2617 1.2662 1.2708 1.2753

52 sin 0.7880 0.7891 0.7902 0.7912 0.7923 0.7934 0.7944 0.7955 0.7965 0.7976 cos 0.6157 0.6143 0.6129 0.6115 0.6101 0.6088 0.6074 0.6060 0.6046 0.6032 tan 1.2799 1.2846 1.2892 1.2938 1.2984 1.3032 1.3079 1.3127 1.3175 1.3222

53 sin 0.7986 0.7997 0.8007 0.8018 0.8029 0.8039 0.8049 0.8060 0.8070 0.8080 cos 0.6018 0.6004 0.5990 0.5976 0.5962 0.5948 0.5934 0.5920 0.5906 0.5892 tan 1.3270 1.3319 1.3367 1.3416 1.3465 1.3514 1.3564 1.3613 1.3663 1.3713

54 sin 0.8090 0.8100 0.8111 0.8121 0.8131 0.8141 0.8151 0.8161 0.8171 0.8181 cos 0.5878 0.5864 0.5850 0.5835 0.5821 0.5807 0.5793 0.5779 0.5764 0.5750 tan 1.3764 1.3814 1.3865 1.3916 1.3968 1.4019 1.4071 1.4122 1.4176 1.4229

55 sin 0.8192 0.8202 0.8211 0.8221 0.8231 0.8241 0.8251 0.8261 0.8271 0.8281 cos 0.5736 0.5721 0.5707 0.5693 0.5678 0.5664 0.5650 0.5635 0.5621 0.5607 tan 1.4281 1.4335 1.4388 1.4442 1.4496 1.4550 1.4605 1.4659 1.4715 1.4770

56 sin 0.8290 0.8300 0.8310 0.8320 0.8329 0.8339 0.8348 0.8358 0.8368 0.8377 cos 0.5592 0.5577 0.5563 0.5548 0.5534 0.5519 0.5505 0.5490 0.5476 0.5461 tan 1.4826 1.4882 1.4938 1.4994 1.5051 1.5108 1.5164 1.5222 1.5280 1.5339

57 sin 0.8387 0.8396 0.8406 0.8415 0.8425 0.8434 0.8443 0.8453 0.8462 0.8471 cos 0.5446 0.5432 0.5417 0.5402 0.5388 0.5373 0.5358 0.5344 0.5329 0.5314 tan 1.5399 1.5458 1.5517 1.5577 1.5637 1.5697 1.5757 1.5817 1.5878 1.5939

58 sin 0.8480 0.8490 0.8499 0.8508 0.8517 0.8526 0.8536 0.8545 0.8554 0.8563 cos 0.5299 0.5284 0.5270 0.5255 0.5240 0.5225 0.5210 0.5195 0.5181 0.5165 tan 1.6003 1.6066 1.6128 1.6191 1.6255 1.6319 1.6383 1.6447 1.6512 1.6577

59 sin 0.8572 0.8581 0.8590 0.8599 0.8607 0.8616 0.8625 0.8634 0.8643 0.8652 cos 0.5150 0.5135 0.5120 0.5106 0.5090 0.5075 0.5060 0.5045 0.5030 0.5015 tan 1.6643 1.6710 1.6777 1.6844 1.6912 1.6981 1.7050 1.7119 1.7189 1.7260

Deps. Function 0° 6° 12° 18° 24° 30° 36° 42° 48° 54°

Appendix II–NATURAL SINES, COSINES, AND TANGENTS

60°–74.9°

| Degs. | Function | 0.0' | 0.1' | 0.2' | 0.3' | 0.4' | 0.5' | 0.6' | 0.7' | 0.8' | 0.9' | |-------|----------|------|------|------|------|------|------|------|------|------|------| | 60 | sin cos tan | 0.8660 0.5000 1.7321 | 0.8665 0.4985 1.7341 | 0.8678 0.4970 1.7461 | 0.8689 0.4955 1.7532 | 0.8695 0.4939 1.7603 | 0.8704 0.4924 1.7675 | 0.8712 0.4909 1.7747 | 0.8721 0.4894 1.7820 | 0.8729 0.4879 1.7893 | 0.8734 0.4862 1.7966 | | 61 | sin cos tan | 0.8746 0.4848 1.8040 | 0.8755 0.4833 1.8115 | 0.8763 0.4818 1.8190 | 0.8771 0.4802 1.8265 | 0.8780 0.4787 1.8341 | 0.8788 0.4772 1.8418 | 0.8796 0.4756 1.8495 | 0.8805 0.4741 1.8572 | 0.8813 0.4726 1.8650 | 0.8821 0.4710 1.8728 | | 62 | sin cos tan | 0.8829 0.4695 1.8807 | 0.8838 0.4679 1.8887 | 0.8846 0.4664 1.8967 | 0.8854 0.4648 1.9047 | 0.8862 0.4633 1.9128 | 0.8870 0.4617 1.9210 | 0.8878 0.4602 1.9292 | 0.8886 0.4586 1.9375 | 0.8894 0.4571 1.9458 | 0.8902 0.4555 1.9542 | | 63 | sin cos tan | 0.8910 0.4540 1.9626 | 0.8918 0.4524 1.9711 | 0.8926 0.4509 1.9797 | 0.8934 0.4493 1.9883 | 0.8942 0.4478 1.9970 | 0.8949 0.4462 2.0057 | 0.8957 0.4446 2.0145 | 0.8965 0.4431 2.0233 | 0.8973 0.4415 2.0322 | 0.8980 0.4399 2.0411 | | 64 | sin cos tan | 0.8988 0.4384 2.0500 | 0.8996 0.4368 2.0590 | 0.9003 0.4352 2.0681 | 0.9011 0.4337 2.0772 | 0.9018 0.4321 2.0864 | 0.9026 0.4305 2.0957 | 0.9033 0.4289 2.1051 | 0.9041 0.4274 2.1145 | 0.9048 0.4258 2.1240 | 0.9056 0.4242 2.1335 | | 65 | sin cos tan | 0.9063 0.4226 2.1430 | 0.9070 0.4210 2.1527 | 0.9078 0.4195 2.1624 | 0.9085 0.4179 2.1721 | 0.9092 0.4163 2.1819 | 0.9100 0.4147 2.1918 | 0.9107 0.4131 2.2017 | 0.9114 0.4115 2.2118 | 0.9121 0.4099 2.2219 | 0.9128 0.4083 2.2321 | | 66 | sin cos tan | 0.9135 0.4067 2.2423 | 0.9143 0.4051 2.2527 | 0.9150 0.4035 2.2631 | 0.9157 0.4019 2.2736 | 0.9164 0.4003 2.2843 | 0.9171 0.3987 2.2950 | 0.9178 0.3971 2.3059 | 0.9184 0.3955 2.3169 | 0.9191 0.3939 2.3280 | 0.9198 0.3923 2.3393 | | 67 | sin cos tan | 0.9205 0.3907 2.3506 | 0.9212 0.3891 2.3621 | 0.9219 0.3875 2.3738 | 0.9225 0.3859 2.3855 | 0.9232 0.3843 2.3974 | 0.9239 0.3827 2.4095 | 0.9245 0.3811 2.4216 | 0.9252 0.3795 2.4341 | 0.9259 0.3778 2.4466 | 0.9265 0.3762 2.4593 | | 68 | sin cos tan | 0.9272 0.3746 2.4722 | 0.9278 0.3729 2.4851 | 0.9285 0.3714 2.4986 | 0.9291 0.3697 2.5118 | 0.9298 0.3681 2.5252 | 0.9304 0.3665 2.5386 | 0.9311 0.3649 2.5523 | 0.9317 0.3633 2.5660 | 0.9323 0.3616 2.5799 | 0.9330 0.3600 2.5939 | | 69 | sin cos tan | 0.9336 0.3584 2.6081 | 0.9342 0.3567 2.6228 | 0.9348 0.3551 2.6375 | 0.9354 0.3534 2.6524 | 0.9361 0.3518 2.6676 | 0.9367 0.3502 2.6828 | 0.9373 0.3486 2.6985 | 0.9379 0.3469 2.7143 | 0.9385 0.3453 2.7304 | 0.9391 0.3437 2.7465 | | 70 | sin cos tan | 0.9397 0.3420 2.7628 | 0.9403 0.3404 2.7796 | 0.9409 0.3387 2.7966 | 0.9415 0.3371 2.8133 | 0.9421 0.3355 2.8305 | 0.9426 0.3338 2.8478 | 0.9432 0.3322 2.8652 | 0.9438 0.3305 2.8829 | 0.9444 0.3289 2.9008 | 0.9449 0.3272 2.9187 | | 71 | sin cos tan | 0.9455 0.3256 2.9368 | 0.9461 0.3239 2.9552 | 0.9466 0.3223 2.9737 | 0.9472 0.3206 2.9925 | 0.9478 0.3190 3.0113 | 0.9483 0.3173 3.0303 | 0.9489 0.3157 3.0496 | 0.9494 0.3140 3.0689 | 0.9500 0.3123 3.0885 | 0.9505 0.3107 3.1084 | | 72 | sin cos tan | 0.9511 0.3090 3.1284 | 0.9516 0.3074 3.1487 | 0.9521 0.3057 3.1691 | 0.9527 0.3040 3.1897 | 0.9532 0.3024 3.2106 | 0.9537 0.3007 3.2317 | 0.9542 0.2990 3.2529 | 0.9548 0.2974 3.2745 | 0.9553 0.2957 3.2962 | 0.9558 0.2940 3.3181 | | 73 | sin cos tan | 0.9563 0.2924 3.3402 | 0.9568 0.2907 3.3627 | 0.9573 0.2890 3.3853 | 0.9578 0.2874 3.4082 | 0.9583 0.2857 3.4314 | 0.9588 0.2840 3.4548 | 0.9593 0.2823 3.4785 | 0.9598 0.2807 3.5024 | 0.9603 0.2790 3.5266 | 0.9608 0.2773 3.5511 | | 74 | sin cos tan | 0.9613 0.2756 3.5759 | 0.9617 0.2740 3.6009 | 0.9622 0.2723 3.6262 | 0.9627 0.2706 3.6518 | 0.9632 0.2689 3.6777 | 0.9636 0.2672 3.7040 | 0.9641 0.2656 3.7306 | 0.9646 0.2639 3.7575 | 0.9650 0.2622 3.7848 | 0.9655 0.2605 3.8124 |

| Degs. | Function | 0' | 6' | 12' | 18' | 24' | 30' | 36' | 42' | 48' | 54' | |-------|----------|-----|-----|------|------|------|------|------|------|------|------|

217

MATHEMATICS, VOLUME 1

75°–89.9°

| Degs. | Function | 0.0' | 0.1' | 0.2' | 0.3' | 0.4' | 0.5' | 0.6' | 0.7' | 0.8' | 0.9' | |-------|----------|------|------|------|------|------|------|------|------|------|------| | 75 | sin cos tan | 0.9659 0.2588 3.7321 | 0.9664 0.2571 3.7729 | 0.9668 0.2554 3.7844 | 0.9673 0.2538 3.8118 | 0.9677 0.2521 3.8391 | 0.9681 0.2504 3.8667 | 0.9686 0.2487 3.8947 | 0.9690 0.2470 3.9232 | 0.9694 0.2453 3.9520 | 0.9699 0.2436 3.9812 | | 76 | sin cos tan | 0.9703 0.2419 4.0108 | 0.9707 0.2402 4.0408 | 0.9711 0.2385 4.0713 | 0.9715 0.2368 4.1022 | 0.9720 0.2351 4.1335 | 0.9724 0.2334 4.1653 | 0.9728 0.2317 4.1976 | 0.9732 0.2300 4.2305 | 0.9736 0.2284 4.2637 | 0.9740 0.2267 4.2975 | | 77 | sin cos tan | 0.9744 0.2250 4.3315 | 0.9748 0.2233 4.3662 | 0.9751 0.2215 4.4015 | 0.9755 0.2198 4.4372 | 0.9759 0.2181 4.4737 | 0.9763 0.2164 4.5107 | 0.9767 0.2147 4.5483 | 0.9770 0.2130 4.5864 | 0.9774 0.2113 4.6252 | 0.9778 0.2096 4.6646 | | 78 | sin cos tan | 0.9781 0.2079 4.7046 | 0.9785 0.2062 4.7453 | 0.9789 0.2045 4.7867 | 0.9792 0.2028 4.8288 | 0.9796 0.2011 4.8716 | 0.9799 0.1994 4.9152 | 0.9803 0.1977 4.9594 | 0.9806 0.1959 5.0045 | 0.9810 0.1942 5.0504 | 0.9813 0.1925 5.0970 | | 79 | sin cos tan | 0.9816 0.1908 5.1446 | 0.9820 0.1891 5.1929 | 0.9823 0.1874 5.2422 | 0.9827 0.1857 5.2924 | 0.9830 0.1840 5.3435 | 0.9833 0.1823 5.3955 | 0.9836 0.1806 5.4486 | 0.9840 0.1789 5.5026 | 0.9843 0.1771 5.5578 | 0.9846 0.1754 5.6140 | | 80 | sin cos tan | 0.9848 0.1736 5.6713 | 0.9851 0.1719 5.7297 | 0.9854 0.1702 5.7894 | 0.9857 0.1685 5.8502 | 0.9860 0.1668 5.9124 | 0.9863 0.1650 5.9758 | 0.9866 0.1633 6.0086 | 0.9869 0.1616 6.1084 | 0.9871 0.1599 6.1698 | 0.9874 0.1582 6.2325 | | 81 | sin cos tan | 0.9877 0.1564 6.3138 | 0.9880 0.1547 6.3764 | 0.9882 0.1530 6.4596 | 0.9885 0.1512 6.5435 | 0.9888 0.1495 6.6288 | 0.9890 0.1478 6.7145 | 0.9893 0.1461 6.8007 | 0.9895 0.1444 6.8880 | 0.9898 0.1426 6.9762 | 0.9900 0.1409 7.0654 | | 82 | sin cos tan | 0.9903 0.1392 7.1154 | 0.9905 0.1374 7.2066 | 0.9907 0.1357 7.2990 | 0.9910 0.1340 7.3927 | 0.9912 0.1323 7.4876 | 0.9914 0.1305 7.5838 | 0.9917 0.1288 7.6812 | 0.9919 0.1271 7.7800 | 0.9921 0.1253 7.8802 | 0.9923 0.1236 7.9818 | | 83 | sin cos tan | 0.9925 0.1219 8.0847 | 0.9928 0.1201 8.1892 | 0.9930 0.1184 8.2953 | 0.9932 0.1167 8.4029 | 0.9934 0.1149 8.5126 | 0.9936 0.1132 8.6238 | 0.9939 0.1115 8.7367 | 0.9941 0.1097 8.8514 | 0.9943 0.1080 8.9679 | 0.9945 0.1063 9.0862 | | 84 | sin cos tan | 0.9945 0.1045 9.2052 | 0.9947 0.1028 9.3270 | 0.9949 0.1011 9.4504 | 0.9951 0.0993 9.5757 | 0.9952 0.0976 9.7030 | 0.9954 0.0958 9.8325 | 0.9956 0.0941 9.9641 | 0.9957 0.0924 10.08 | 0.9959 0.0906 10.20 | 0.9960 0.0889 10.30 | | 85 | sin cos tan | 0.9962 0.0872 11.43 | 0.9963 0.0854 11.47 | 0.9965 0.0837 11.91 | 0.9966 0.0819 12.16 | 0.9968 0.0802 12.43 | 0.9969 0.0785 12.71 | 0.9971 0.0767 13.00 | 0.9972 0.0750 13.30 | 0.9973 0.0732 13.63 | 0.9974 0.0715 13.90 | | 86 | sin cos tan | 0.9976 0.0698 14.30 | 0.9977 0.0680 14.57 | 0.9978 0.0663 15.05 | 0.9979 0.0645 15.46 | 0.9980 0.0628 15.89 | 0.9981 0.0610 16.35 | 0.9982 0.0593 16.83 | 0.9983 0.0576 17.30 | 0.9984 0.0558 17.89 | 0.9985 0.0541 18.48 | | 87 | sin cos tan | 0.9986 0.0523 19.08 | 0.9987 0.0506 19.74 | 0.9988 0.0488 20.21 | 0.9989 0.0471 20.81 | 0.9990 0.0454 21.47 | 0.9990 0.0436 22.02 | 0.9991 0.0419 22.77 | 0.9992 0.0401 23.46 | 0.9993 0.0384 24.27 | 0.9994 0.0366 25.27 | | 88 | sin cos tan | 0.9994 0.0349 26.03 | 0.9995 0.0332 26.77 | 0.9995 0.0314 27.31 | 0.9996 0.0297 28.64 | 0.9997 0.0279 29.30 | 0.9997 0.0262 30.42 | 0.9998 0.0244 31.82 | 0.9998 0.0227 33.69 | 0.9999 0.0209 35.00 | 0.9999 0.0192 36.59 | | 89 | sin cos tan | 0.9998 0.0175 57.29 | 0.9999 0.0157 63.66 | 0.9999 0.0140 71.62 | 0.9999 0.0123 81.85 | 1.000 0.0105 95.49 | 1.000 0.0087 114.6 | 1.000 0.0070 143.2 | 1.000 0.0052 191.0 | 1.000 0.0035 573.0 | 1.000 ... ... |

| Degs. | Function | 0' | 6' | 12' | 18' | 24' | 30' | 36' | 42' | 48' | 54' | |-------|----------|-----|-----|------|------|------|------|------|------|------|------|

218

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APPENDIX V FORMULAS

Areas Areas

A = s² The area of a square is equal to A = 4πr² The square area of a sphere is the square of a side. equal to 4 times pi times the radius squared. A = ½ h The area of a triangle is equal to b Volumes one half the base times the height. V = e³ The volume of a cube equals the cube of an edge. A = πr² The area of a circle is equal to the radius squared times pi. V = Bh The volume of a rectangular solid or cylinder equals the area of A = lw The area of a rectangle is equal the base times the height. to the length times the width. V = ⁴⁄₃ πr³ The volume of a sphere equals ⁴⁄₃ A = Ch The lateral area of a cylinder is equal to the circumference of pi times the radius cubed. the base times the height.

221

INDEX

Absolute value, 21 Centigrade thermometer, 19 Accuracy, 15, 59 Changing: Addend, 7 common fractions to decimals, 49 Adding: fractions to percent, 55 complex numbers, 164 integers to percent, 55 decimals, 51 percent to a decimal, 56 fractions, 118 Characteristic, logarithms, 83 signed numbers, 21 Checking accuracy, 14 unlike fractions, 35 Chord of a circle, 188 Addition: Circle, 187 and subtraction, 7 Circular cylinder, 194-195 method for solving simultaneous equations, Circumference of a circle, 187 135 Coefficients, literal, 125, 136 Adjacent angles, 182 Combined variation, 150 Algebraic: Combining: expressions, 99 radicals, 74 fractions, 117 terms, 100 sum, 99 Common: Alternation in a proportion, 144 denominator, 34 Altitude of a triangle, 183 factors, 111 Angles, 182 fractions, 28, 49 Apex of a triangle, 183 logarithms, 81 Approximate numbers, 61 Commutative laws, 26, 98 Arabic numerals, 1 Complement of an angle, 182 Arbitrary constant, 120 Completing the square, 169 Areas: Complex: circle, 189 decimal, 47 quadrilateral, 186 fraction, 43-44 triangle, 184 numbers, 158-163 Associative laws, 26, 98 plane, 161 Axioms of equality, 25 Components of logarithms, 83 Composite number, 17 Base of: Concentric circles, 189 exponent, 65 Conditional equation, 121-122 number system, 2 Conjugates of complex numbers, 165 solid, 193 Constant: triangle, 183 definition, 120 of proportionality, 147 Binary number system, 3 Construction, geometric, 190 Binomial factors, 113 Coordinates, 130-131 Bisecting an angle, 191 Counting, 1 Borrow process, 7, 8 Cube: Broken lines, 181 by slide rule, 94 geometric, 104 Calculating with approximate numbers, 61 root, 79, 95 Calipers, vernier, 64 Cancellation, 38 Dashed line, 181 Carry and borrow, 7 Decimal: Celsius thermometer, 19 adding, 51 Center of a circle, 192

222

INDEX

Decimal--Continued: complex, 47 divisors, 53 equivalent, 47 fractions, 45 mixed, 47 multiplying, 51-52 nonterminating, 50 number system, 2, 45 points, 13, 15 power of, 66 reducing, 47 system, 2 Degree: angular, 182 of an equation, 121 Denominate numbers, 9, 15 Denominator, definition, 28 Dependence, 151 Dependent variable, 151 Developing formulas, 154 Diameter: circle, 187 sphere, 198 Difference: answer in subtraction, 7 of two squares, 113 Digit positions: binary, 3 decimal, 2 Digits, significant, 60 Direction of measurement, 19 Directly proportional, 147 Direct variation, 146 Discriminant, 176 Distributive law, 27, 99 Dividend, 11 Dividing: a line into equal segments, 190 approximate numbers, 61 by powers of ten, 54 complex numbers, 165 decimals, 52 denominate numbers, 15 Divisibility, test for, 18 Division: fractions, 40 general, 10 in fraction form, 25 methods, 13 signed numbers, 24 synthetic, 110 Divisor, 11 Double roots, 177

Edge of a prism, 194 Element: cylinder, 194 set, 4 Ellipses, 192 Ellipsis, definition, 5 End zeros in multiplication, 13 Equality axioms, 25 Equal or double roots, 177 Equations, plotting, 131 Equilateral triangle, 185 Equivalent: decimal, 47 fraction, 29 Error: percent of, 59 relative, 60 Estimation, 14, 58 Evaluating: formulas, 153 radicals, 78 Exponential form, 80 Exponents: and radicals, 102 definition, 65 fractional, 70 laws of, 67 literal, 112 Extremes of a proportion, 142

Faces of a solid, 193 Factor, 11, 17 Factoring: definition, 111 method of solving quadratic equations, 168 radicals, 75 trinomials, 115 Fixed constant, 120 Formulas: developing, 154 evaluating, 153 graphing, 156 table of, 221 translating, 155 Fractional: exponents, 70 percents, 57 Fractions: algebraic: dividing, 117 multiplying, 117 changing to decimals, 49 complex, 43-44 equivalent, 29

223

MATHEMATICS, VOLUME 1

Fractions--Continued: fundamental rule, 30 improper, 28, 31 in equations, 125 measurement, 29 negative, 32 partitive, 29 power of, 66 proper, 28 reducing, 31, 116 Function: general, 151 trigonometric, 202, 213 Fundamental rule of fractions, 30

General form of a linear equation, 126 Geometric: classification of angles, 182 figures, 183-190 Graphical: interpretation of roots, 179 representation of complex numbers, 160 solution of quadratic equations, 172 Graphing: formulas, 156 general, 20 inequalitites, 129, 139 Great circle, 198 Greater than (symbol), 20, 128 Greatest common divisor, 34 Grouping: for multiplication, 11 symbols, 101

Hemisphere, 198 Highest common factor, 34 Horizontal lines, 181 Hypotenuse, 199

Identity, 121 Imaginary: number, 66, 159 root, 176 unit, 159 Improper fraction, 28, 31 Independent variable, 151 Index of a root, 65 Inequalities, 128 Inequalities in two variables, 139 Infinite sets, 6 Integers, 1 Intercepts, definition, 132 Interpolation, 203 Interpreting equations, 137 Interpreting roots by graphs, 179

Intersecting lines, 182 Inversely proportional, 148 Inverse ratio, 142 Inverse variation, 148 Inversion in a proportion, 144 Irrational: number, 77, 158 root, 178 Irregular pyramid, 196 Isosceles triangle, 185

Joint variation, 149

Lateral: area, pyramid, 197 edge, prism, 194 Laws: associative, 26 commutative, 26 distributive, 27 exponents, 67 sines, 208 Least common multiple, 34 Less than (symbol), 20, 128 Like: fractions, 33 signs, adding, 21 Line: general, 161 parallel, 137 segment, 5 Linear equation, 121, 126 Literal: coefficient, 124, 136 exponent, 112 Logarithm: definition, 80 natural, 81 Lowest common denominator, 34

Mantissa, 83, 85 Mathematical symbols, 219 Maximum and minimum points, 174 Means of a proportion, 142 Measurement: fraction, 29 principles of, 58 Mental: calculation, 10 multiplication, 107 Micrometer: scale, 61 settings, 62 vernier, 64 Minimum and maximum points, 174

224

INDEX

Minuend, 7 Mixed: decimal, 47 number, 28, 32 Monomial multiplication, 103 Multiples, 17 Multplicand, 11 Multiplication: fractions, 37 general, 10 grouping, 11 Multiplier, 11 Multiplying: approximate numbers, 61 complex numbers, 164 decimals, 51-52 denominate numbers, 15 signed numbers, 23

Natural logarithms, 81 Negative: exponents, 69 fractions, 32 logarithms, 83 numbers, 19 Nonterminating decimals, 50 Number: set, 4 systems, 2, 3 Number line: fractions, 28 general, 5, 20 Numerals, 1 Numerator, definition, 28 Numerical coefficient, definition, 100

Oblique: line, 181 triangle, 185, 207 Obtuse: angle, 182 triangle, 185 One as an exponent, 69 Operation: with decimals, 50 with inequalities, 128 Operator i, 160-161 Operators, 158 Order: of operations, 16 properties of numbers, 128 Orientation of lines, 181

Parabola, 174 Paralleliped, 193

Parallel lines, 181 Parallelogram, 186 Parentheses, removing, 101 Partial products, 12 Partitive fractions, 29 Percent: changing numbers to, 55 changing to decimal, 56 definition, 55 fractional, 57 of error, 59 Percentage cases, 56 Perimeter: quadrilateral, 186 triangle, 184 Perpendicular: at any point on a line, 191 bisector of a line, 191 lines, 181 Pi (π), 188 Place value, 1, 2, 46 Placing decimal points, 13, 15 Plotting: complex numbers, 162 coordinates, 131 equations, 131 inequalities, 139 Points and lines, 5 Polar form, 163 Polynomials, 104-106 Positional notation, 2 Positive: and negative numbers, 20 integers, 4 Powers and roots, 65 Powers of: fractions, 66 negative integers, 65 ten, 52, 54, 71-73 Precision, 58 Prime: factor, 17 number, 152 Principles of: measurement, 58 verniers, 63 Prims, 193 Product: general, 11 of sum and difference, 106 Proper fraction, 29 Proportion, 142 Proportionality constant, 147 Pure imaginaries, 161 Pythagorean Theorem, 199

225

MATHEMATICS, VOLUME 1

Quadrant, definition, 131 Quadratic: definition, 167 equations, 172, 179 formula, 170-172 Quadrilateral, 186 Quotient, 11

Radical, 73, 102 Radicand, definition, 74 Radius: circle, 187 sphere, 198 Ratio: definition, 141 trigonometric, 201 Rational: number, 28, 77, 158 roots, 178 Rationalizing denominators, 77, 106 Ray, geometric, 5 Reading: decimals, 47 micrometers, 62 slide rule scales, 87 Real numbers, 86, 158 Reciprocals, 73 Rectangle, 186 Rectangular: coordinates, 19, 130 prism, 193 Reducing: decimals, 47 fractions, 31, 116 Regrouping, 7 Regular pyramid, 196 Relative error, 60 Remainder, 14 Removing parentheses, 101 Rhombus, 186 Right: angle, 182 circular cone, 196 cylinder, 194 prism, 193 triangle, 185, 199 Roots: equal, 177 imaginary, 177 of an equation, 65, 168 rational, 178 Rounding off, 47

Scientific notation, 71 Sector of a circle, 188

Segment of a circle, 188 Sense reversal, inequalities, 129 Sets: comprising points and lines, 5 elements of, 4 infinite, 6 Sides of a triangle, 183 Signed numbers, 19, 23 Significant digits, 60, 73 Similar triangles, 200 Simplifying radicals, 75 Simultaneous: equations, 133 inequalities, 140 Sines, law of, 208 Slide rule: description, 86 operation, 88-97 Solid figures, 193 Solving: linear equations, 122-124 oblique triangles, 208 Special: exponents, 69 products, 106 triangles, 204-250 Spheres, 197-198 Square: geometric, 186 of a sum or difference, 108 root, 78, 92 Squaring: by slide rule, 91 complex numbers, 165 Straight and curved lines, 181 Subject of a formula, 152 Subscripts, 152 Subsets, 4 Substitution method for solving systems of equations, 136 Subtracting: by borrowing, 8 complex numbers, 164 decimals, 51 fractions, 118 general, 7 mentally, 10 signed numbers, 22 Subtrahend, 7 Sum: angles of a triangle, 185 general, 7 Supplement of an angle, 182 Surface area: prism, 194

226

INDEX

Surface area--Continued: sphere, 198 Symbols: grouping, 101 in formulas, 152 mathematical, 219 Synthetic division, 110 System of equations, 133

Tangent to a circle, 187 Terms: and coefficiente, 99 of a proportion, 142 Test for divisibility, 18 Thermometer, 19 Three percentage cases, 56 Translating formulas, 155 Trapezoid, 187 Trial quotients, 14 Triangle: general, 183-186 similar, 200 special, 204-205 Triangular prism, 193 Trigonometric: ratios, 201 tables, 202, 213 Trinomial: factoring, 115 squares, 114

Uneven division, 14 Unit, imaginary, 159

Unlike: fractions, 33 signs, adding, 21

Variable, 120, 151 Variation: combined, 150 general, 146 joint, 149 Vector representation of complex numbers, 163 Verbal problems, 138-139, 179 Vernier: caliper, 64 general, 61-64 measurements, 63 micrometer, 64 principle, 63 Vertex: angle, 182 triangle, 183 Vertical: angle, 182 line, 181 Volume: prism, 194 pyramid, 197 sphere, 198

Weights and measures, 220 Whole numbers, 1

Zero as an exponent, 69

* U.S. GOVERNMENT PRINTING OFFICE: 1997 — 532 - 154 / 60033

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Assignment 1

Number Systems and Sets; Positive Integers

Textbook Assignment: Chapters 1, 2 (7-18)

1-1. Which of the following Navy personnel have a need for this course? 1. Storekeepers 2. Damage Controlmen 3. Electronics Technicians 4. All of the above

1-2. An understanding of mathematics has an important effect upon an officer's potential success.

1-3. Which of the following groups of symbols could represent a number? 1. 7x2 2. 8/4 3. 68-22 4. All of the above

1-4. Which of the following could be classified as an integer? 1. 9 2. 76 3. 5,280 4. All of the above

1-5. In our numbering system the place value of each digit is 1. higher than that of the digits to the left of it 2. lower than that of the digits to the left of it 3. the same as that of the digits to the right of it 4. lower than that of the digits to the right of it

1-6. What does the zero mean in the number 6,509? 1. There are no tens. 2. It causes the 5 to stand for 50. 3. There are no hundreds. 4. It causes the 9 to stand for 9 tens and not 9 ones.

1-7. Which figure is in the "thousands" place in the number 850,179? 1. 0 2. 1 3. 5 4. 8

1-8. Which figure is in the "ten thousands" place in the number 516,789,240? 1. 6 2. 7 3. 8 4. 9

1-9. How is the number 15,026,745 read? 1. Fifteen million, twenty-six thousand, seven hundred forty-five 2. Fifteen billion, twenty-six thousand, seven hundred and forty-five 3. One billion, five million, twenty-six thousand, seven hundred forty-five 4. Fifteen million, two hundred sixty thousand, seven hundred and forty-five

● Numbers written in systems other than the decimal system should have the base noted as a subscript, that is, 204₅ is a base five number, 111₂ is a base two of binary number,etc.

1-10. The 2 in the number 214₅ means 1. two 125's 2. two 100's 3. two 25's 4. two 5's

1-11. How many digit symbols are used to perform all the calculations involved in the binary system? 1. 2 2. 5 3. 8 4. 10

1-12. The binary equivalent of decimal 6 is 1. 011 2. 101 3. 110 4. 111

1-13. The binary equivalent of decimal 15 is 1. 0110 2. 0111 3. 1001 4. 1111

6

1-14. The number system which is most commonly used in digital computers is the 1. octal 2. binary 3. ternary 4. decimal

1-15. A collection of symbols which have at least one common characteristic is called a 1. set 2. group 3. series 4. quantity

1-16. Which of the following is a set of the even positive integers less than 10? 1. {2, 4, 6, 8} 2. {0, 2, 4, 6, 8} 3. {2, 4, 6, 8, 10} 4. {0, 2, 4, 6, 8, 10}

1-17. Which of the following correctly designates a set of the names of days of the week which contain the letter "u" and are between Sunday and Saturday? 1. {Tuesday, Thursday} 2. {Sunday, Tuesday, Thursday} 3. {Tuesday, Thursday, Saturday} 4. {Sunday, Tuesday, Thursday, Saturday}

1-18. A group of symbols encompassing a part of a set is called a 1. group 2. subset 3. subgroup 4. quantity

1-19. What is used in mathematics to indicate that a pattern continues indefinitely? 1. A dot 2. A line 3. Three dots 4. Three hyphens

1-20. Although a dot is used to represent it, a point actually has no length, width, nor thickness.

1-21. In mathematics, which statement describes an ideal line? 1. It has length and thickness, but no width. 2. It has length and width, but no thickness. 3. It has length, but no width nor thickness. 4. It has length, width, and thickness.

1-22. A mathematical plane is determined by three points which do not lie on the same line.

1-23. A mathematical line may be considered as a subset of a plane surface.

1-24. When a series of points with no space between them begins at a point and is extended infinitely in one direction only, how may the series be identified? 1. As a ray 2. As a half-line 3. As a line segment 4. As either 1 or 2 above

1-25. A set of points comprising a plane has subsets called 1. rays 2. lines 3. line segments 4. all of the above

1-26. Uses of a scale include 1. tire gage 2. yard stick 3. thermometer 4. all of the above

1-27. If the inch between 1 and 2 on a one-foot rule could be stretched to 1 mile, how many numbers could be inserted between the 1 and the 2? 1. 1,760 2. 5,280 3. None 4. An infinite quantity

1-28. In the subtraction problem, 12 - 3 = 9, the numeral 3 is called the 1. addend 2. minuend 3. subtrahend 4. quotient

1-29. The most common method of arranging the addends 467, 1382, and 17 in a vertical column for addition is 1. 17 467 1382

2. 467 1382 17

3. 467 1382 17

4. 1000 + 300 + 80 + 2 400 + 60 + 7 10 + 7

7

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1-54. Determine the proper terminology associated with division in the following problem 28 3/85 6 25 24 1

1. 3 is divisor 3. 3 is divisor 28 is remainder 28 is quotient 85 is dividend 85 is dividend 1 is quotient 1 is remainder

2. 3 is dividend 4. 3 is dividend 28 is quotient 28 is divisor 85 is divisor 85 is quotient 1 is remainder 1 is remainder

1-55. If a single digit divisor is too large to be contained in the first digit of a four digit dividend, in a division problem, you should 1. divide the first two digits of the dividend by the divisor and place the quotient over the first digit of the dividend 2. divide the second digit of the dividend by the divisor and place the quotient over the first digit of the dividend 3. divide the first two digits of the dividend by the divisor and place the quotient over the second digit of the dividend 4. divide the second digit of the dividend by the divisor and place the quotient over the second digit of the dividend

1-56. How would you estimate the first digit of the quotient resulting from dividing 3094 by 68? 1. Divide 30 by 6 2. Divide 309 by 65 3. Divide 309 by 70 4. Divide 310 by 60

1-57. In dividing a four-digit number by a two- digit number, the trial quotient is found to be too small when 1. the trial quotient is greater than 10 2. the trial quotient is smaller than 10 3. the new dividend is smaller than the divisor 4. the new dividend is as large or larger than the divisor

1-58. Which of the following divisions is an exact division? 1. 457 ÷ 9 2. 221 ÷ 13 3. 396 ÷ 13 4. 745 ÷ 25

1-59. What is the remainder when 259 is divided by 8? 1. 1 2. 3 3. 5 4. 7

1-60. How much is 22,308 divided by 74? 1. 3,114 with a remainder of 44 2. 287 with a remainder of 70 3. 301 with a remainder of 34 4. 31 with a remainder of 34

1-61. If a group of 39 people is to be divided into 4-member teams, how should this division be expressed? 1. 8 teams, R 7 2. 9 teams, R 3 3. 9 teams, R 4 4. 10 teams

1-62. The purpose for maintaining proper vertical alignment in division is to assist in placing the decimal point in the quotient.

1-63. A division has been accurately performed when the dividend equals the product of the 1. remainder times the divisor plus the quotient 2. remainder times the quotient plus the divisor 3. quotient times the divisor plus the remainder 4. quotient times the remainder plus the divisor

10

1-64. The product, in simplified form, of the multiplication, problem; 4 (2 hours 22 minutes 32 seconds) is 1. 8 hours 88 minutes 128 seconds 2. 8 hours 90 minutes 8 seconds 3. 9 hours 28 minutes 8 seconds 4. 9 hours 30 minutes 8 seconds 1-65. The product of 12 miles and 13 miles is 1. 156 miles 2. miles 3. 156 milessquare 4. 156 1-66. The product of 2 feet 8 inches times 3 feet 4 inches may be found by 1. multiplying 2-feet times 3 feet then multiplying 8 inches times 4 inches 2. multiplying 3 feet times 2 feet 8 inches then multiplying 4 inches times 2 feet 8 inches 3. converting 2 feet 8 inches to 2 feet 2 3 and 3 feet 4 inches to 3 then multiplying 1 feet and3 4. changing 2 feet 8 inches to 3 feet and 3 feet 4 inches to 4 feet and then multiplying 1-67. If a pipe 22 feet 6 inches long is cut into 3 equal lengths, how long are the pieces? (Neglect the width of the saw cuts.) 1. 7 feet 1 inch 2. 7 feet 3 inches 3. 7 feet 4 inches 4. 7 feet 6 inches 1-68. The result of dividing 23 hours 31 minutes 20 seconds by 5 is 1. 43 hours 6 minutes 4 seconds1 5 2. 4 hours 0 minutes 4224 seconds 3. 4 hours 42 minutes 16 seconds 4. 4 hours 6 minutes 4 seconds 1-69. In which of the following series of operations is the order in which the operations are performed important? l . 2 + 3 + 5 2. (3) (9) (7) 3. 6 x 8 x 9 4. 48 ÷ 6 x 3 1-70. The answer to the problem 24 ÷ 4 ÷ 3 ÷ 2 is 1. not defined 2. 1 3. 4 4. 9 1-71. The order of operations is important if division or multiplication is involved with other operations, Use the rules pertaining to a series of mixed operations to calculate the value of 6 x 4 + 8 ÷ 2. 1. 16 2. 28 3. 36 4. 48 1-72. The number 36 is a multiple of 1, 2, 3, 4, 6, 9, 18, 36, and 1. 10 2. 11 3. 12 4. 13 1-73. An odd number when divided by 2 produces a remainder of 1. 1-74. The number 7 is a factor of l. 11 2. 17 3. 24 4. 35 1-75. What is the value of x if x = (3)(0)(4)(6)? l. 0 2. 18 3. 24 4. 72 11

Assignment 2

Positive Integers; Signed Numbers; Common Fractions

Textbook Assignment: Chapters 2 (17, 18), 3, 4 (28, 29)

2-1. The number 19 is a composite number.

2-2. A factor of a composite integer is never larger than the integer.

2-3. A prime number is any number that is divisible only by 1. 0 2. 2 3. itself and 1 4. another prime number

2-4. The prime factorization of 40 is 1. 2·5·8 2. 2·2·10 3. 5·8·40 4. 2·2·2·5

2-5. An odd number can have no even factors.

2-6. All even numbers are divisible by 2.

2-7. Any number ending in 3 is divisible by 3.

2-8. In which of the following groups is each of the four numbers divisible by 4? 1. 704; 820; 710; 414 2. 91,276; 902; 414; 612 3. 8,924; 672; 14,000; 818 4. 3,724; 716; 1,312; 81,728

2-9. The numbers 90 and 802 are both divisible by 5.

2-10. No odd number is divisible by 6.

2-11. In which of the following groups is each of the three numbers divisible by 8? 1. 240; 896; 830 2. 217,120; 112,112; 4,098 3. 637,168; 78,126; 111,736 4. 215,240; 817,896; 425,800

2-12. Which of the following numbers is divisible by both 3 and 8? 1. 16,807 2. 33,120 3. 49,928 4. 62,412

2-13. If any number is multiplied by 9, the sum of the digits of the product is divisible by 9.

2-14. A number without a sign is considered to be either negative or positive depending on the problem.

2-15. The minus sign (-) may indicate either the operation of subtraction or that a number is negative.

2-16. A minus sign placed in front of a reading taken from a mercurial thermometer indi- cates that the 1. actual temperature is less than the reading 2. present reading is less than the previous reading 3. mercury has fallen below the scale of the thermometer 4. temperature is a number of units below a zero reference point

2-17. If the reading on a thermometer is +10 centigrade and the temperature falls 20 centigrade, the new reading will be 1. +10 centigrade 2. 0° centigrade 3. -10 centigrade 4. -2° centigrade

-3

-2

-1 A 1 2 3 -3 -2 -1

-1

•C -2

-3 •B

Figure 2A.--Rectangular coordinate system.

12

● In answering items 2-18 and 2-19 refer to figure 2B.

2-18. To get from point A to point B, as indi- cated by the arrows on the rectangular coordinate system, a person must go in a 1. negative direction and then in a posit- ive direction 2. negative direction and then in a nega- tive direction 3. positive direction and then in a nega- tive direction 4. positive direction and then in a posi- tive direction

2-19. The signs associated with point C with re- gard to the horizontal and vertical direc- tions, respectively, are 1. +, - 2. +, + 3. -, - 4. -, +

2-20. The "greater than" symbol (>) and the "less than" symbol (<) always point toward the smaller number.

2-21. Which of the following groups of mathe- matical statements is true? 1. If A > 0 and B < 0, then A<B 2. If A > 0 and B > 0, then A<B 3. If A < 0 and B > 0, then B>A 4. If A > 0 and B > 0, then B>A

● In answering items 2-22 and 2-23, refer to figure 2B.

2-22. What is the absolute value of a number? 1. The number times itself 2. 1 divided by the number 3. The value of the number without regard to sign 4. The value of a number with the fractional or decimal part disregarded

2-23. What is the distance between point A and point B? 1. -1 2. 1 3. 3 4. 7

2-24. The absolute value of 5 is greater than the absolute value of -7.

2-25. What is the general rule for adding two or more negative numbers? 1. Find the sum of the numbers, disregard- ing the sign. 2. Find the product of the absolute values of the numbers, taking the sign of the largest. 3. Find the sum of the absolute values of the numbers and place a minus sign be- fore the result. 4. Find the difference between the absolute values of the numbers and place a minus sign before the result.

2-26. What is the general rule for adding a posi- tive and a negative number? 1. Find the difference between the recipro- cal values of the numbers and place a plus sign before the result. 2. Find the difference between the absolute values of the numbers and prefix the sign of the number having the larger absolute value. 3. Find the sum of the absolute values of the numbers and place a plus sign before the result. 4. Find the sum of the absolute values of the numbers and prefix the sign of the number having the larger absolute value.

2-27. What result do you obtain when you add -6 and -7? 1. +13 2. +1 3. -1 4. -13

2-28. What result do you obtain when you take the sum of -26 and +8? 1. +34 2. +18 3. -18 4. -34

2-29. What is the result of subtracting +3 from -10? 1. +13 2. +7 3. -7 4. -13

- + -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 A B

Figure 2B.--Signed numbers line.

13

2-30. In solving a subtraction problem involving signed numbers, what is the correct proce- dure to use in terms of the number line? 1. Find the subtrahend on the number line then move the number of units of the minuend in the direction opposite in sign of the minuend. 2. Find the subtrahend on the number line then move the number of units of the minuend in the direction of the sign of the minuend. 3. Find the minuend on the number line then move the number of units of the subtra- hend in the direction opposite in sign of the subtrahend. 4. Find the minuend on the number line then move the number of units of the subtra- hend in the direction of the sign of the subtrahend.

2-31. What is the result of subtracting +14 from +6? 1. +20 2. +8 3. -8 4. -20

2-32. What is the result of the operation indi- cated by the expression (-3) - (-4)? 1. +7 2. +1 3. -1 4. -7

2-33. What is the result of the operation indi- cated by the expression (+9) - (-5)? 1. +14 2. +4 3. -4 4. -14

2-34. What is the result of the operation indi- cated by the expression (-6) + (-4)? 1. -10 2. -2 3. +2 4. +10

2-35. What is the product of -5 and 4? 1. +20 2. +9 3. -1 4. -20

2-36. What is the product of 5 and -4? 1. +20 2. +9 3. -1 4. -20

2-37. What is the product of -5 and -4? 1. +20 2. +9 3. -1 4. -20

2-38. To multiply +6 by +3 means to 1. add +3 to 2. add +6 to +3 3. add (+3) + (+3) + (+3) 4. add (+6) + (+6) + (+6)

2-39. What is the general rule for determining the sign of the product of two numbers that are opposite in sign? 1. The sign is negative when the larger number is positive and positive when the larger number is negative. 2. The sign is positive when the larger number is positive and negative when the larger number is negative. 3. The sign is always negative. 4. The sign is always positive.

2-40. What is the rule for the sign of the product of two negative numbers? 1. The sign is negative unless both numbers have equal value. 2. The sign is negative in all cases except those in which one of the numbers is less than one. 3. The sign is always negative. 4. The sign is always positive.

2-41. Which of the following products is always negative? 1. The product of more than two numbers 2. The product of an even number of negative numbers 3. The product of an odd number of negative numbers 4. The product of more than two negative numbers

2-42. Why is the product of -3 and -4 a positive 12? 1. Adding -4 three times produces movement in the negative direction. 2. Adding -4 three times produces movement in the positive direction. 3. Taking away -4 three times produces movement in the positive direction. 4. Taking away -4 three times produces movement in the negative direction.

2-43. What is the product of -2, -4, +8, and -1.5? 1. +0.5 2. -15.5 3. +96 4. -96

2-44. Which of the following statements concern- ing the relation of the process of division to other processes in mathematics is correct? 1. Division is a short way of adding. 2. Division is the basis of subtraction. 3. Division is the opposite or inverse of subtraction. 4. Division is the opposite or inverse of multiplication.

14

2-45. What is the general rule for determining the sign of the result when dividing a number by another number opposite in sign? 1. The sign of the result is always positive. 2. The sign of the result is always negative. 3. The sign of the result is positive only when the sign of the larger number is negative. 4. The sign of the result is positive only when the sign of the smaller number is negative.

2-46. What is the quotient of 18 divided by -3? 1. +15 2. +6 3. -6 4. -21

2-47. What is the quotient of -18 divided by -3? 1. +15 2. +6 3. -6 4. -21

2-48. What is the result of performing the division indicated by -2/3 ? 1. +12 2. +5 3. -5 4. -45

2-49. What is the quotient of -12 divided by +3? 1. +36 2. -9 3. -4 4. +4

2-50. What is the result of (6)(-4)/8 ? 1. -4 2. -3 3. +3 4. +6

2-51. The expressions 8 - (6 + 2 - 4) and 8 - 6 + 2 - 4 have the same value.

2-52. Which statement is true about the fraction - 3/5 ? 1. The sign of the 3 is negative; the 5 has no sign. 2. The sign of the 5 is negative; the 3 has no sign. 3. The sign of the 3 is negative; the sign of the 5 is negative; the sign of the fraction is negative. 4. The sign of the 3 is positive; the sign of the 5 is positive; the sign of the fraction is negative.

2-53. The simplest form of -6/-7 is 1. - 6/7 3. -6/-7 2. -6/+7 4. +6/-7

2-54. The fraction -8/-12 is equivalent to 1. -3/4 3. +2/3 2. -2/3 4. +3/2

2-55. A statement which requires proof to verify its truth is considered an axiom.

2-56. Which of the following illustrates an axiom of equality? 1. 7 = 4 + 3. 2. If a > b, then b < a. 3. If a = b, then a + 4 = b + 4. 4. If a = 2b and b = 2, then a = 4.

2-57. Incorrect usage of the multiplication axiom is illustrated by 1. 2(4) = 2(3+1) 2. 3(5+6) = 3(1+2+8) 3. 7(6) = 7(2+5) 4. both 2 and 3 above

2-58. Is it true or false that an axiom is used in the following development? Assume 4y = 28. Divide both sides of this equation by 4 (that is, 4y/4 = 28/4). There- fore, y = 7.

2-59. The sum of the numbers 4, 3, and 2 may be found by adding 4 and 3 and then adding 2 or by adding 4 to the sum of 3 and 2. These two procedures produce equivalent results as stated by the 1. associative law of multiplication 2. commutative law of multiplication 3. distributive law 4. associative law of addition

2-60. Which of the following expressions is equivalent to 7-(8-4) ? 1. (7-8)-4 2. 7-8-4 3. 7+[(-8) + (-4)] 4. 7+[(-8) + 4]

15

2-61. A column of figures may be added from either the top to the bottom or from the bottom to the top. This fact results from the 1. distributive law 2. commutative law of addition 3. associative law of addition 4. associative law of multiplication

2-62. The commutative law of multiplication means that the product of two or more numbers is the same regardless of the order of multiplication.

2-63. All of the following expressions are equivalent to 2(3 + 4 + 5) except 1. 2·3 + 2·4 + 2·5 2. (4 + 3 + 5)2 3. 2(12) 4. 2(5 + 4) + 3

2-64. Common fractions and integers constitute a subset of the real numbers called the rational numbers.

2-65. The denominator of a fraction tells how many and the numerator tells what kind.

2-66. If 5 seconds is expressed as a fraction of an hour, what number will be in the denom- inator of the fraction? 1. 25 2. 3,600 3. 30,400 4. 126,420

2-67. If 120 feet is expressed as a fraction of a mile, what number will appear in the numerator of the fraction? 1. 120 2. 5,280 3. 5,280/120 4. (120 × 5,280)

2-68. The rational number 7 can be written as the ratio of two integers.

2-69. Which of the following is a proper fraction? 1. 52/3 3. 6/3 2. 9/4 4. 19/21

2-70. Which of the following is an improper fraction? 1. 1-3/7 3. 4/5 2. 3/5 4. 8/7

2-71. Which of the following fractions would not normally be written as a mixed number? 1. 4/3 3. 9/3 2. 5/2 4. 11/2

2-72. Which of the following mixed numbers has the largest numerator when it is written as an improper fraction? 1. 1-1/32 3. 11-1/4 2. 5-3/8 4. 14-2/3

2-73. If an angle of 360° is divided into 30° sectors, what fraction of the 360° angle is represented by 1 sector? 1. 1/24 3. 1/6 2. 1/12 4. 1/3

2-74. If a stick one yard long is divided into 72 equal parts, what is the length of each part? 1. 1/72 inch 3. 1/2 inch 2. 1/24 inch 4. 2 inches

2-75. The probability of rolling a 3 on a 6-faced die is 1/6. This relationship can also be expressed as 1. 1 part in 6 2. the ratio of 1 to 6 3. 1 time out of 6 4. each of the above

16

Assignment 3

Common Fractions; Decimals

Textbook Assignment: Chapters 4 (29-44), 5 (45-51)

3-1. Assume that a man measure a 3/4 -inch block 3-6. Which of the following fractions cannot be four times. Each time he uses different reduced to lower terms? graduation on his ruler. Which of his measurements is incorrect? 1. 39/52 3. 89/121 1. 6/8 in. 3. 25/32 in. 2. 48/72 4. 144/256 2. 12/16 in. 4. 48/64 in. 3-7. The fraction 78/234, when reduced to lowest 3-2. Which of the following fractions is not terms, becomes equivalent to 1/2 ? 1. 1/3 3. 39/117 1. 0.5/1 3. 327/652 2. 13/39 4. 78/234 2. 125/250 4. 4/11 (or 4/8-2/2) 3-8. It is incorrect mathematically to compute 3-3. The fundamental rule of fractions states with improper fractions. that adding the same number to both terms of a fraction does not change the value of 3-9. The improper fraction 22/8 when changed to a the fraction. mixed number and reduced to lowest terms becomes 3-4. How may the fraction 3/8 be changed to twenty-fourths? 1. 11/4 3. 2 5/8 1. Multiply the numerator by 3 2. 2 3/4 4. 2 6/8 2. Multiply the denominator by 3 3. Multiply both terms of the fraction 3-10. The mixed number 3 7/9 is equivalent to the by 3 improper fraction 4. Multiply both terms of the fraction by 24 1. 11/3 3. 30/9 3-5. What fraction with a numerator of 8 is 2. 16/9 4. 34/9 equivalent to the fraction 2/6 ? 1. 3/8 3. 8/24 2. 8/3 4. 8/48

3-11. Multiplying each term of a fraction by 3-17. Which of the following groups of fractions -1 has the same effect as has the smallest least common denominator? 1. multiplying the fraction by 1 2. multiplying the fraction by -1 1. 2/5, 2/9, 7/18 3. 2/6, 7/12, 2/3 3. changing the sign of the numerator only 2. 3/8, 5/16, 3/4 4. 20/25, 13/15, 9/10 4. changing the sign in front of the fraction 3-18. What method do you use in finding a common denominator for a series of fractions? 3-12. Which one of the following four fractions 1. Add all the denominators. differs in value from the other three? 2. Cancel the common factors in numerators and denominators. 1. -3/4 3. -2/4 3. Find the smallest number that all of the numerators will go into. 2. -3/4 4. -3/-4 4. Find the smallest number that is divisible by all the denominators. 3-13. Fractions must always be changed into like fractions before they can 3-19. Find the LCM of the following numbers: be added. 180 = 5·2²·3²; 210 = 5·2·3·7; 225 = 5²·3² [Note: 2³ means 2 is taken as a factor three times] 3-14. The person who states that 9/24 + 1/2 + 5/24 = 15/24 1. 5²·2·3·7 has failed to observe that 2. 5²·2²·3²·7 1. to find the sum of two or more fractions, 3. 5⁴·2³·3²·7 the numerators should be added to obtain the numerator and the denominators 4. 180·210·225 should be added to obtain the denominator 2. fractions must be reduced to lowest 3-20. Find the greatest common divisor (GCD) of terms before they are added 6, 15, and 21. 3. fractions cannot be added without raising them to higher terms 1. 3 4. quantities to be added must be expressed 2. 6 in common denominators 3. 210 4. 3³·2·5·7 3-15. The fractions 4/9 and 4/9 are unlike fractions. 3-21. Find the GCD of the numbers 3-16. The least common multiple (LCM) of 60, 36, 120 = 2³·3·5; 140 = 2²·7·5; 42 = 2·3·7 and 18 is 1. 2 1. 180 2. 6 2. 540 3. 2³·3·5·7 3. 1080 4. 120·140·42 4. 2160

3·22. What lowest common denominator (LCD) would be used to add 1/9, 2/7, 3/32, and -2/25?

1. 1575 2. 2205 3. 3²·7·9·5 4. 9·7·35·25

Figure 3A.--Illustration for addition of fractions.

3-23. What is the sum of 2¼ and 12⅚?

1. 4⅝ 3. 4⁷₁₂

2. 4⅜ 4. 4⁷₁₂

3-24. In figure 3A, find the distance along the fence ABCDEF.

1. 4¹₃/₁₆ mi 3. 5¹/₈ mi

2. 4⅜ mi 4. 5⅜ mi

3-25. How much is ⅖ subtracted from 1⅝? 12 8

1. 1⅛ 3. 1⁷/₂₄

2. 1⁷/₈ 4. ⅜

3-26. In figure 3B, what is the length of the dimension marked Y on the sketch of the machine bolt?

1. 1⁶⁷/₆₄ in. 3. 2¹¹/₆₄ in.

2. 2⁵/₆₄ in. 4. 2⅞/₆₄ in.

3-27. How much is ¹/₂₄ of 6?

1. 4 3. ¼

2. ¼ 4. 1/144

3-28. What is the product of 3½ × ⅖? 12

1. 1¹¹/₂₄ 3. ²⁷/₉

2. 1¼ 4. 8⅛

3-29. The answer to the problem ⅔ × ⁸/₉ × ⅖ × ⅖ = 6 3 2

is wrong because 1. the sum of 3 and 2 is 5 2. a mistake was made in division 3. the wrong numbers were divided out 4. the numerator was omitted in the answer

3-30. What is the answer to the problem ⅖ × ¼ × ⅔ 2 when reduced to lowest terms?

1. ⅖ 3. ¹⁰/₂₄

2. ⁵/₁₂ 4. ¹⁵/₃₆

Figure 3B.--Sketch of a machine bolt.

3-31. Use canceling to determine the product of ⁹⁶/₁₄₄ × 38.

1. 23 2. 25⅔ 3. 27⅖ 4. 37⅓

19

3-32. The common denominator method of dividing fractions is illustrated by the example

1. 4/3 ÷ 1/2 = 4 ÷ 4 = 8/3 3

2. 4/3 ÷ 1/2 = 4/3 × 2 = 8/3

3. 4/3 ÷ 1/2 = 4/3 × 2/1 = 8/3

4. 4/3 ÷ 1/2 = 8/6 ÷ 3/6 = 8 ÷ 3 = 8/3

3-33. What is the simplest way to find the reciprocal of a fraction? 1. Divide the fraction by 1. 2. Subtract the fraction from 1. 3. Divide the numerator by the denominator. 4. Interchange the numerator and denominator.

3-34. The reciprocal of 3¼ is

1. ²²/₇ 3. ¹³/₄

2. ²⁷/₂₂ 4. 7⅓

3-35. The reciprocal of 50 is greater than the reciprocal of 25.

3-36. The reciprocal method of dividing fractions is illustrated by the example

1. 3/8 ÷ 1/3 = 1/8 × 1/1 = 1/8

2. 3/8 ÷ 1/3 = 3/8 × 3/1 = 9/8

3. 3/8 ÷ 1/3 = 9/24 ÷ 8/24 = 9 ÷ 8 = 9/8

4. 3/8 ÷ 1/3 = 1/3 × 1/8 × 1/8 = 1

3-37. How much is 4 divided by 1⅜?

1. 3¼ 3. 3²/₉

2. 3⁶/₉ 4. 4⅛/₇

3-38. How much is c⅔ divided by 4⅔? 3

1. 1⁵/₇ 3. 1½

2. ¹⁶/₇ 4. ³¹/₉

3-39. How much is 8⅟₅ divided by 6?

1. 1¹¹/₃₀ 3. 2⁷/₁₀

2. 1⁵/₁₂ 4. 49⅕

3-40. Using the formula R_t = 1/(1/R₁ + 1/R₂), find R_t

when R₁ = 6 and R₂ = 3.

1. ⅑ 3. 2

2. ½ 4. 9

3-41. The fraction ¾ is in the form of a decimal fraction.

3-42. Decimal fractions are expressed in terms of 1. twentieths and powers of one-twentieth 2. twelfths and multiples of twelfths 3. tenths and powers of one-tenth 4. fifths and multiples of fifths

3-43. In the number 89.654 the nine is in which place? 1. Ten thousands 2. Thousands 3. Hundreds 4. Tens

3-44. In the decimal fraction 0.03672 the seven is in which place? 1. Tenths 2. Hundredths 3. Thousandths 4. Ten-thousandths

3-45. Which of the following measurements made in a machine shop is the largest? 1. 3.0070 2. 3.1340 3. 3.0988 4. 3.2100

20

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D C

0.312" 0.563" 1.219"

H O ↓4" ←0.469"→↑

←0.813"→ ←1.938"→

3C.--Dimensions of machine part.

● In answering items 3-72 through 3-75 3-74. What is the length of the shaft from point refer to figure 3C. A to point B? 1. 1.485 in. 3-72. What is the total length of the machine 2. 1.688 in. part? 3. 1.713 in. 1. 2.012 in. 4. 1.918 in. 2. 2.759 in. 3. 2.771 in. 3-75. What is the thickness of the shoulder from 4. 3.001 in. point C to point D? 1. 0.346 in. 3-73. What is the distance from point B to 2. 0.351 in. point J? 3. 0.435 in. 1. 0.212 in. 4. 0.456 in. 2. 0.692 in. 3. 0.719 in. 4. 1.019 in.

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Assignment 4

Decimals; Percentage and Measurement; Exponents and Radicals

Textbook Assignment: Chapters 5 (51-54), 6, 7 (65)

_______________________________________________________________________________

4-1. In multiplying a decimal by a whole number, 4-8. It is decided to terminate a quotient at the location of the decimal point in the the second decimal place. To determine the product is determined by counting the num- correct rounding, the division should be ber of places in the whole number. carried to 1. at least to the fourth decimal place 4-2. Assuming an allowance of 0.010 inch for 2. the third decimal place each saw cut, what length of stock would be 3. the second decimal place required to produce 16 machine parts if 4. the first decimal place each part is to be 2.002 inches long? [Hint: There are 17 saw cuts.] 4-9. How much is 19.37 divided by 5 carried to 1. 362.06 in. 3. 36.206 in. three decimal places? 2. 216.36 in. 4. 21.636 in. 1. 0.387 3. 3.870 2. 3.670 4. 3.874 4-3. When two decimals are multiplied, placing the decimal point is in effect multiplying 4-10. The quotient of 0.00243 divided by 18 is the numerators. 1. 0.0000135 2. 0.000135 4-4. A rule for placing the decimal point in 3. 0.00135 multiplying decimals is that the number of 4. 0.0135 decimal places in the answer is equal to 1. the number of decimal places in the 4-11. Moving the decimal point two places to the multiplier right in the dividend and the divisor in 2. the number of decimal places in both the example the multiplicand the multiplicand 5.10 3. twice the number of decimal places in 1.25/6.3810 both the multiplicand the multiplicand 4. the number of decimal places in the is equivalent to multiplicand less the number of decimal 1. dividing the quotient by 100 places in the multiplier 2. multiplying the quotient by 100 3. dividing both 1.25 and 6.381 by 100 4-5. The product of 40.6 and 0.18 is 4. multiplying both 1.25 and 6.381 by 100 1. 40.78 3. 7.308 2. 0.7308 4. 73.08 4-12. The quotient of 0.02146 ÷ 0.012 rounded to three decimal places is 4-6. A rod that is 39 inches long at 65° C 1. 1.788 expands 0.000133 inch for each inch of its 2. 1.789 length for every degree of temperature in- 3. 17.882 crease. How much would its length increase 4. 17.883 if its temperature rose to 600° C? 1. 0.0005187 in. 3. 0.027557 in. 4-13. To divide a number by 100, move the decimal 2. 0.007155 in. 4. 0.2775045 in. point two placesto the right.

4-7. What is the product of 17.250 and 10.000? 4-14. Division by 10 is the same as multiplica- 1. 17.2500000 3. 172,500 tion by 0.1. 2. 17,500 4. 1,725,000

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4-15. Division by 0.1 is the same as multiplication by 10.

4-16. 85% is equivalent to 1. 0.85

2. 85 100

3. 85 parts out of 100 4. all of the above

4-17. Percent is more often used to represent absolute values than relative values.

4-18. In computations, percents are normally changed to decimal forms.

4-19. The decimal fraction 0.00014 can be written as 1. 0.000014% 2. 0.014% 3. 0.14% 4. 0.14%

4-20. Which is the largest of the following expressions? 1. 0.00401% 2. 0.0401% 3. 0.401% 4. 4.01%

4-21. How is the number 25 expressed as a percent? 1. 0.25% 2. 25% 3. 250% 4. 2500%

4-22. A percentage greater than 100 has no meaning.

4-23. If the 1965 production was 200% of the 1964 production and the 1964 production was 50 tons, the 1965 production was 1. increased by 200% 2. 150 tons 3. 100 tons 4. 50 tons

4-24. There was a 200% increase in production from 1964 to 1965. If the 1964 production was 400 tons, the total 1965 production was 1. 400 tons 2. 600 tons 3. 800 tons 4. 1200 tons

4-25. If a baseball player's batting average is quoted as 265, how often has he gotten a hit? 1. 0.265% of the time 2. 2.65 times out of 100 3. 26.5% of the time 4. 265% of the time

4-26. To change a percent to a decimal, drop the percent sign and add two zeros.

4-27. To change a percent larger than 100 percent to a decimal, drop the percent sign and 1. move the decimal point two places to the left 2. move the decimal point two places to the right 3. subtract 100 percent from the original figure and change the result to a decimal 4. divide the original figure by 10 before changing it to a decimal

4-28. What is the rate, base, and percentage of the equation R = 75%? B 1. Rate 75%, base 4, percentage 3 2. Rate 75%, base 3, percentage 4 3. Rate 3, base 4, percentage 75% 4. Rate 4, base 3, percentage 75%

4-29. The result of finding 22% of 44 is 1. 200 2. 20.0 3. 9.68 4. 6.6

4-30. A 20-gallon tank has 5 gallons left in it. What percent of the tank is empty? 1. 15 percent 2. 25 percent 3. 50 percent 4. 75 percent

4-31. Which decimal expresses 1/4?

1. 0.0025 3. 0.25

2. 0.025 4. 2.5

25

4-32. Under certain conditions the speed of an aircraft is 15 percent faster than that registered by the speed indicator on the instrument panel. If the indicator shows a speed of 285 miles per hour, what is the actual speed of the aircraft? 1. 248 mph 2. 307 mph 3. 320 mph 4. 328 mph

4-33. What is the correct procedure for finding what percent one number is of another number? 1. Multiply the two numbers together and divide by 100. 2. Divide the smaller number by the larger number and divide again by 100. 3. Divide one number by the other, expressing the quotient in decimal form; then multiply this decimal by 100. 4. Divide one number by the other, expressing the quotient in decimal form; then divide this decimal by 100.

4-34. The number 2 is what percent of 400?

1. 0.05%

2. 1/2%

3. 50% 4. 200%

4-35. Three percent of what number is 9? 1. 0.0033 2. 0.27 3. 30 4. 300

4-36. Bill's income is 70% of John's income. If Bill's income is $4,900 a year, what is John's income? 1. $3,430 2. $6,370 3. $7,000 4. $7,500

4-37. Assume that your ship is leaving a harbor in heavy seas, and that 65 percent of the crew of 325 men have never been to sea before. It is expected that about 40 percent of the new men will become seasick. Approximately how many of the new men are likely to become seasick? 1. 50 2. 85 3. 128 4. 215

4-38. 1/4 is what percent of 3/4 ?

1. 50% 2. 200% 3. 0.5% 4. 0.005%

4-39. It is not necessary to round off the decimal equivalent of 3, 0.375, when adding it to the decimal equivalent of 7/4, 0.75.

4-40. The concepts of precision and accuracy are necessary considerations when dealing with measurements.

4-41. The maximum probable error in an instrument marked off in hundredths of an inch is 1. 0.5 in. 2. 0.05 in. 3. 0.005 in. 4. 0.0005 in.

4-42. Precision of a measurement refers to the size of the smallest division on the scale.

4-43. The precision of the sum of the numbers 4.2, 4.23, 4.236, and 4.2367, which were determined from measurements with four different instruments, is no greater than the precision of 1. 4.2 2. 4.23 3. 4.236 4. 4.2367

4-44. If the numbers 29.138, 19.21, 130.68, and 84.9823 are accurate only to the last place shown, what is the best way to add them? 1. Round off all numbers to one decimal place and then add. 2. Round off all numbers to two decimal places and then add. 3. Round off all numbers to three decimal places and then add. 4. Do not round off any of the numbers, but add them as given.

4-45. To subtract 8.173 grams from 13.62 grams, round 8.173 grams to two decimal places.

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4-74. Which base when raised to the fifth power gives an answer of thirty-two? 1. 1 2. 2 3. 4 4. 8

4-75. The exponent of 4 is 0.

29

Assignment 5

Exponents and Radicals

Textbook Assignment: Chapter 7 (65-77)

5-1. Another name for the square root sign, √, is the 1. extraction sign 2. radian sign 3. factor sign 4. radical sign

5-2. Which root of 27 is indicated by the expression∛27? 1. The first root 2. The cube root 3. The square root 4. The quadratic root

5-3. What is the meaning of ⁴√625? 1. The number which when multiplied by itself 4 times is equal to the square root of 625 2. One-half of the square root of 625 3. The fourth root of 625 4. 625 to the fourth power

5-4. Which number is the cube root of 8? 1. 2 2. 8/3 3. 24 4. 512

5-5. Which number is the third power of 1.5? 1. 0.015 2. 0.5 3. 2.25 4. 3.375

5-6. If a negative number is raised to the 54th power, what is the sign of the result? 1. It depends on the base. 2. It fluctuates. 3. It is positive. 4. It is negative.

5-7. What type number is the square root of 28? 1. Rational 2. Imaginary 3. Integral 4. Real

5-8. What type number is the square root of -9? 1. Irrational 2. Integral 3. Real 4. Imaginary

5-9. The cube of -1/2 can be written as 1. (-1/2)³ 3. (-1)³/(2)³ 2. -(1/2)³ 4. any of the above

5-10. In which other way may (3/7)⁴ be written? 1. 3/7⁴ 3. 4/7 2. 3⁴/7⁴ 4. 4/3

5-11. The fifth real root of -1/32 is equivalent to 1. -1/2 3. (1/2) 2. -1/2 4. Each of the above

5-12. If a three-decimal place number is raised to the fourth power, the result will have seven decimal places.

5-13. What is the value of ⁶√m⁹? 1. 1 2. m 3. m³ 4. m⁶

30

5-14. What is the product of 2² x 2³? 1. 2⁵ 2. 2⁶ 3. 4⁵ 4. 4⁶

5-15. If exponents are added in the multiplication process, the bases must be equal.

5-16. Using the law of exponents for multiplication, what is the product of 3⁶ x 4³, if any? 1. 3⁹ 2. 4⁹ 3. 12⁹ 4. It is impossible to indicate a product under this condition.

5-17. To divide one power of a base by another power of the same base, raise the base to the power found by subtracting the exponent in the divisor from the exponent in the dividend.

5-18. What is the quotient obtained by dividing 8⁵ by 8³? 1. 1⁵ 2. 8⁻² 3. 8¹·⁵ 4. 8²

5-19. What does the example (2³)⁴ illustrate? 1. The multiplication of a root by a power 2. The addition of two exponents 3. The law of double exponent 4. The power of a power

5-20. The example (4³)² is interpreted to mean 1. 4⁵ 2. 4³·4³·4³ 3. 4·4·4·4·4·4 4. 4·4·4·4·4·4·4·4

5-21. Which of the following is the result obtained when the term 5³ is cubed? 1. ³/₅³ 3. 5⁷ 2. 5⁹ 4. 5¹²

5-22. The expression (2·2·2·2)³ is equivalent to 1. 2⁸ 2. 2⁹ 3. 18 4. 216

5-23. What is the value of (2⁴·4²)²? 1. 4¹⁶ 2. 8¹⁶ 3. 256 4. (256)²

5-24. What is the value of (³⁶/₁₀₂)³? 1. 1 3. ⁷⁷ 2. ⅓ 4. ⁶²/₁₇₂₈

5-25. The expression (4.5)⁰ is greater than 4⁰.

5-26. What is the sum of 9⁰ plus 9¹? 1. 0 2. 1 3. 9 4. 10

5-27. The expression 7⁻¹ has the same value as 1/7

5-28. The zero power of 2 plus the zero power of 5 equals 1. zero 2. 1 3. 2 4. the zero power of 7

5-29. Which of the following expressions is equivalent to 7/4·2? 1. 1 + 7/4² 3. (⁷/₇)² 2. 4²/7 4. 7(4²)

5-30. A negative exponent has no meaning but is introduced to complete the set of exponents.

5-31. Of the expressions, 2⁻⁷, (1/2)⁷, -(-2)⁻⁷, and 1/(2)⁷, which are equal? 1. 2⁻⁷ and (1/2)⁷ 2. -(–2)⁻⁷ and 1/(2)⁷ 3. (1/2)⁷, 2⁻⁷, and -(–2)⁻⁷ 4. All are equal.

31

5-32. What is the value of 1/₅₋₄ ? 1. 5⁴ 3. 4⁵ 2. 1/625 4. (1/5)⁴

5-33. What is the relationship between √16 and (16)½? 1. √16 > (16)½ 2. √16 < (16)½ 3. (16)½ is 4 more than √16 4. They are equal.

5-34. The expression (4³)⅓ is equal to 1. ¹/4³ 3. (1/3)(4³) 2. 4 4. 4(³√3)

5-35. What is the value of 3²⁵⁹? 1. 4 3. 5/⁹√24 2. 8 4. 16⅓

5-36. The expression 3⁵ can be changed to another form by the steps 1. 3⁵ = ¹²/₂ = 3⁵ 2. 3⁵ = ⁵√3⁷ = ⁷√9 3. 3⁵ = 3² ⁺ ½ = 3² · 3½ = 3¹ = 3 4. 3⁵ = 3² ⁺ ½ = 3² · 3½ = 3² √3

5-37. The value of (8⁰)⁵ is the same as (8⁵)⁰.

5-38. What is 7³ expressed in radical form? 1. ⁷√72 3. ⁷√27 2. ⁷√73 4. ⁷√73

5-39. What is the decimal equivalent of 10⁻⁵? 1. 0.01 2. 0.001 3. 0.0001 4. 0.00001

5-40. An expression written in scientific notation must contain a number between 1 and 10.

5-41. The number 4,980 written in scientific notation is 1. 498 x 10¹ 2. 49.8 x 10² 3. 4.98 x 10³ 4. 0.498 x 10⁴

5-42. The number 0.0214 written in scientific notation is 1. 2.14 x 10⁻² 2. 2.14 x 10⁻³ 3. 0.214 x 10⁻¹ 4. 0.214 x 10⁻²

5-43. The number 97,200 is equivalent to 1. 972 x 10³ 2. 97.2 x 10⁴ 3. 9.72 x 10⁴ 4. 0.972 x 10⁵

5-44. Which of the following expressions represents an intermediate step in simplifying 718 x 0.0003 by using powers of 10? 0.0085 x 75,000 1. (7.18 x 3) x (10² + 10⁻⁴H)/(8.5 x 7.5) x (10⁻³ + 10⁴) 2. 7.18 x 3/(8.5 x 7.5) x 10² x 10⁻⁴ x 10⁻³ x 10⁴ 3. 7.18 x 10² x 3 x 10⁻⁴/(8.5 x 10⁻³ x 7.5 x 10⁴) 4. 7.18 x 10² x 3 x 10⁻⁴/(8.5 x 10⁻³ x 7.5 x 10⁴)

5-45. Using scientific notation, how would the term 0.00000123 be expressed? 1. 0.123 x 10⁻⁶ 2. 0.123 x 10⁵ 3. 1.23 x 10⁻⁶ 4. 1.23 x 10⁶

32

5-46. The reciprocal of a number is 1 divided by that number. For example, the reciprocal of 50 is 1/50, which can be expressed as

1/(5 x 10^1) = 10^-1/5 = 10 x 10^-2/5 = 2 x 10^-2.

Which of the following expressions represents a step in evaluating the reciprocal of 4,500 x 0.000028?

1. 1/(4.5 x 10^-3 x 2.8 x 10^5)

2. (4.5 x 2.8)/(10^-3 x 10^5)

3. (10^-3 x 10^5)/(4.5 x 2.8)

4. (10^3 x 10^-5)/(4.5 x 2.8)

5-47. Using scientific notation, how may the computation involving 100,000 x 0.00027 x 0.015 be expressed? 1. 2.7 x 0.015 2. 0.27 x 1.5 x 10^1 3. 2.7 x 1.5 x 10^-1 4. 2.7 x 1.5 x 10^-2

5-48. The factor 4^7 in the denominator of a fraction is equivalent to a factor of 4^-7 in the numerator of the fraction.

5-49. How may (12,000 x 0.018 x 3.6)/(90 x 400) be expressed in scientific notation? 1. 3.6 x 10^-2 2. 2.16 x 10^0 3. 2.16 x 10^-2 4. 0.0036 x 10^-3

5-50. What is the number of significant digits in 6,458 x 10^-3? 1. Two 2. Three 3. Four 4. Five

5-51. If all the terms in the example 0.00058 x 41.7 x 0.005169 0.0000029 x 2.16 x 0.8343 are expressed in scientific notation and rounded to one more digit than the number of significant digits in the least accurate term, which of the following, having been rounded to the number of significant digits in the least accurate term, is the correct solution? 1. 2.4 x 10^1 2. 2.04 x 10^-1 3. 3.01 x 10^-1 4. 3.01 x 10^0

5-52. The reciprocal of 2,500 x 0.002 x 0.04 can be expressed as

1. 5 x 1 2. 5 x 10^-1 3. 5 x 10^-2 4. 5 x 10^2

5-53. How is (10,000 x 0.003 x 20)^3 expressed in scientific notation? [Note: Round the answer to the number of significant digits in the least accurate term of the problem] 1. 2 x 10^6 2. 2 x 10^8 3. 2.16 x 10^6 4. 2.16 x 10^8

5-54. In the expression ∛18, the index is 3 and the radicand is 18.

5-55. When the index of a radical expression is not written, it is understood to be a 3.

5-56. What is the meaning of the expression +√9? 1. The plus is used before the minus. 2. The minus is used before the plus. 3. The value of the radical is ambiguous. 4. The square root of 9 can either be +3 or -3.

5-57. What is the value of the radicand in the expression ∛25? 1. 2 2. 3 3. 6 4. 25

5-58. In the expression 6x^2, six is a coefficient of x^2.

5-59. The radicals 8√7, (1/2)√(3/7), and √3 can be combined by addition.

5-60. What is the sum of 2∛10 plus (1/2)√10?

1. √10 3. 2(1/2)√10

2. 2√10 + 5 4. 2(5/20)

5-61. The answer to the problem √5 - (1/2)√5 is

1. (1/2)√5 3. (3/2)√5

2. -(1/2)√5 4. -(3/2)√5

5-62. In order for two radicals to be multiplied they must have the same index and radicand.

5-63. The product of ∛7 and √6 is ∛42.

5-64. Which of the following radical expressions are equal?

1. ∛(3/8) - ∛(3/2) 3. (∛2/√7) - (∛2 x 1/∛7)

2. (∛3/∛7) - (1/∛1) x (1/∛3) 4. (∛2/∛7) - √(3/7) - ∛(3/7)

5-65. The value of the expression (2∛3 · √3)/3 in the most simplified form is 1. 2 2. (2√3)/3 3. 2√3 4. (2√5)/3

5-66. The radicals ∛3 and ∛k cannot be combined into one radical by addition.

5-67. The computational steps ∛16 = ∛(8·∛2 = 2∛2 may be used in solving for the cube root of 16.

5-68. Which of the following simplifications is in error? 1. ∛18 = √9·√2 = 3√2 2. ∛(250) - √(125)·√2 = 5√2 3. ∛16 = √(8) · ∛(2) = 2 4. √51 = √49 + 2 = 7√2

5-69. While 64 is a perfect third power, it is not a perfect sixth power.

5-70. The expressions ∛7 and 7^(1/3) are equivalent.

5-71. Which expression is obtained by simplifying the radical√(1,375)? 1. 105 2. 15√7 3. 45√7 4. 3^2·5^2·2√7

5-72. Using fractional exponents, how may the radical ∜(4) be written? 1. 4^3.7 3. 4^7.3

2. 4^(3/7) 4. 4^(7/3)

5-73. What is the prime factorization of √(3^2 · 7^3 · 5^5)?

1. 3√(7^3·5^5) 3. 105√(5·7)

2. 3·7·5^2√35 4. 3·7·5^2·√(3·5·7)

5-74. What is the solution of the expression (∛27)/(√9)? 1. 1 3. √3 2. √3 4. (∜21)/√9

5-75. (3)/(2√3) is an example of a rational number.

Assignment 6

Exponents and Radicals; Logarithms

Textbook Assignment: Chapters 7 (77-79),8(80-86)

6-1. The square root of 7 is an example of an irrational number.

6-2. Rationalizing the denominator is the process whereby an irrational number in the denominator of a fraction is changed to a rational number.

6-3. To rationalize the denominator of 4/∛3, multiply both numerator and denominator by 1. 1. 2. 4 3. ∛3 4. 4/∛3

6-4. To rationalize the denominator of 5√3/3√3, multiply both numerator and denominator by 1. √2/2 2. 3√3 3. √3 4. 1

6-5. What is the form of the fraction 3/(2√3) after the denominator has been rationalized? 1. 3√5/10 2. 3√5/2 3. 15/2 4. 3/10

6-6. What is the proper way to group the digits of the number 418.796 when preparing to calculate its square root? 1. 4|18. 796 2. 4'|18.7'|96 3. 41'|8.7'|96 4. 0'6'|18. 79'|60

6-7. In calculating the square root of 4,096, the first digit in the answer is the greatest number whose square is contained in 40, that is, the square is either equal to 40 or is less than 40.

● Note that in the square root process each trial division is obtained by multiplying the quotient by 20. For example, 2 7. 1 6 √7'|38. 00 00

20.2 = 40|3 38 47|3 29 20.27 = 540 9 00 541 9 7 20.271=5420 3 59 00 5426 3 25 56 33 44

Therefore, √738 = 27.2 (rounded to tenths)

6-8. What is the square root of 324? 1. 17.62 2. 17.94 3. 18.00 4. 22.00

6-9. In the following problem the process of taking the square root is correct to the point to which it has been carried. √63'83.09 140 69 141 168 161 7

6-10. What is the error in the following square root calculation? 9 4.0 √88'20.00 81 180 720 720

1. The trial divisor was not adjusted to form a true divisor. 2. The digits were not properly grouped. 3. There is an error in multiplication. 4. The decimal point is not properly aligned.

6-11. The decimal point in a square root calculation is kept aligned as in long division with the exception that alignment is accomplished with pairs of digits rather than with single digits.

6-12. What is the square root of 15,129? 1. 102.3 2. 123 3. 390.1 4. 393

6-13. What is the square root of 816.7 correct to the nearest tenth? 1. 9.0 2. 28.5 3. 28.6 4. 29.9

6-14. If the square root of 54 is 7.35, the square root of 5,400 is 73.5.

6-15. If the square root of 3,812 is 61.741, the square root of 38,120 is 617.41.

6-16. If the cube root of 89 is 4.46, the cube root of 0.089 is 1. 0.00446 2. 0.0446 3. 0.446 4. 44.6

6-17. In the expression 3^4 = 81, which number may be interpreted as a logarithm? 1. 3 2. 4 3. 64 4. 81

6-18. What is the logarithmic form of the expression, 2^5 = 32? 1. log₂ 5 = 32 2. log₅ 32 = 5 3. log₅ 32 = 2 4. log₃₂ 5 = 2

● Refer to table 8-1 in your textbook in answering items 6-19 and 6-20.

6-19. What base is used in the system of logarithms in which the logarithm of the number 16 is 1? 1. 2 2. 2 3. 4 4. 16

6-20. What is the value of x if log₉ 9 = x? 1. 2 2. 3 3. 9 4. 27

6-21. Since a logarithm is an exponent, multiplication using logarithms is reduced to a problem of addition of logarithms.

6-22. Refer to table 8-2 in your textbook. Which of the following is correct in the multiplication of 16 × 128? 1. log₈ 16 = 2 log₈ 128 = 7 log₈of the product = 9

2. log₂ 16 = 4 log₂ 128 = 7 log₈of the product = 11

3. log₂ 16 = 4 log₂ 128 = 7 log₈of the product = 28

4. There is not enough information given in the table to work this problem.

6-23. What number is used as the base of the system of logarithms for most ordinary computations? 1. 2 2. 2.3026 3. 2.71828 4. 10

6-24. When the word log is used without a subscript, it is understood that the base 10 is to be used.

6-25. Assume that you have used a formula involving natural logarithms and the answer you have found is in x = 0.29366. You can find the value of x by first applying the correct conversion factor to obtain 1. log x = 0.123582 2. log x = 0.127537 3. log x = 0.158243 4. log x = 0.675416

● The characteristic of a number may be determined by writing the number in scientific notation. The resulting exponent is the characteristic. For example, in log .0078, write .0078 as 7.8x10⁻³. The characteristic is then -3. For log 256, write 256 as 2.56x10². The characteristic is then 2.

● Items 6-26 through 6-60 refer to common logarithms unless otherwise indicated.

6-26. What is the log of 0.00001? 1. -5 2. -4 3. -3 4. -1/-5

6-27. What is the common logarithm of 100,000? 1. 3 2. 5 3. 7 4. 10

6-28. The log of a number between 100 and 1,000 is between 1. 0 and 1 2. 1 and 2 3. 2 and 3 4. 3 and 4

6-29. Refer to table 8-3 in your textbook. Be- tween what logarithms may the logarithm of 0.0004 be located? 1. -2 and -3 2. -3 and -4 3. -4 and -5 4. 3 and 4

6-30. If 2 = 10^0.30103, and 2 x 5 = 10^1, to what power must 10 be raised to equal 5? 1. 0.47712 2. 0.60206 3. 0.69897 4. 0.90309

6-31. If the characteristic of a logarithm is 1, the associated number must be between 1. 0 and 1 2. 1 and 10 3. 10 and 100 4. 100 and 1,000

6-32. For any number greater than 1, the characteristic is one less than the number of digits in the whole number portion of the number.

6-33. What is the characteristic of 72,319? 1. 3 2. 4 3. 5 4. 6

6-34. Which of the following numbers has a characteristic of -4? 1. 0.00001 2. 0.00095 3. 0.10005 4. 0.40008

6-35. What is the value of the mantissa in the expression log 0.0054 = 7.73239 - 10? 1. 7.73239 2. 0.0054 3. 0.73239 4. 7.73239

6-36. The mantissa for the numerical sequence 165 is 0.21748. Which of the following logarithms can be used to express the decimal fraction 0.0001657 1. 4.21748 2. 0.21748 - 4 3. 6.21748 - 10 4. All of the above

6-37. If the mantissa for the number sequence 17900 is 0.25285, what is the log of 1799? 1. 1.25285 2. 2.25285 3. 4.25285 4. 8.25285 - 10

⦿ In answering items 6-38 through 6-41, refer to the table of logarithms in Appendix I.

6-38. What is the log of 70? 1. 0.1213 2. 1.1213 3. 1.84510 4. 2.84510

6-39. What is the log of 2,700? 1. 0.43136 2. 1.43136 3. 2.43136 4. 3.43136

6-40. What is the log of 0.0024? 1. 0.38021 2. 7.38021 - 10 3. 8.38021 - 10 4. 9.38021 - 10

6-41. What is the log of 1? 1. 0.00000 2. 0.10000 3. 0.10000 - 10 4. 9.00000 - 10

⦿ An antilogarithm is a number which corresponds to a logarithm; for example, in log 5.2 = .716, 5.2 is said to be the antilogarithm of .716. Mathematically, antilog .716 = 5.2. Generalizing, for log N = L; N is the antilogarithm and L is the logarithm. Finding the antilogarithm is the reverse process of finding the logarithm, that is, rather than determining the characteristic and mantissa of a number, the number must be determined given the characteristic and mantissa.

EXAMPLE: Find the antilogarithm of 2.9345

SOLUTION: 1. Find the mantissa .9345 in column six of Appendix I. This mantissa corresponds to the digit sequence 86.

37

2. Since the characteristic of the original logarithm is 2 then the antilogarithm written in scientific notation is 8.6 x 10^2 or antilog 2.9345 = 860

6-42. If log 12 = 1.07918 then the antilog equals 1. .3333 2. 1.07918 3. 10 4. 12

⦿ Refer to Appendix I in answering items 6-43 and 6-44.

6-43. If log A = 1.83251 then A equals 1. -3010 2. 6.8 3. 30.10 4. 68

6-44. If antilog 3.62325 = B then B equals 1. .5563 2. 420 3. 4200 4. 5563

⦿ The logarithm of a product is equal to the sum of the logarithms of the factors, that is log (a·b·c·d)= log a + log b + log c + log d

EXAMPLE: Find the product of 3·4 using logarithms.

SOLUTION: 1. Log (3·4)=log 3 + log 4 =.47712 + .60206 =1.07918

2. Antilog 1.07918 = product, or antilog 1.07918 = 1.2 x 10¹ = 12 Therefore 3 x 4 = 12.

6-45. Since log 40 = 1.60206 and log 5 = .69897 , the log of the product of 40 x 5 is equivalent to 1. log 40 x log 5 2. 1.60206 x .69897 3. 1.60206 + .69897 4. log 1.60206 + log .69897

⦿ Refer to Appendix I in answering items 6-46 through 6-48

6-46. Use logarithms to find the product of 28 x 20. What is the mantissa which must be used to find the digit sequence for the product? 1. .27481 2. .30103 3. .74819 4. 1.30103

6-47. The antilog of what value must be used to find the product of 59 x 38 ? 1. 3.35063 2. 3.77085 3. 4.35063 4. 4.77085

6-48. Use logarithms to find the product of 2900 and 3000. The product equals 1. 7.7 x 10⁶ 2. 8.7 x 10⁶ 3. 7.7 x 10⁷ 4. 8.7 x 10⁷

⦿ The logarithm of a power of a number is equal to the product of the power and the logarithm of the number. That is, log A^n = n log A

Note that A^n = A x A x A x A

and log A^n= log A + log A + log A+ log A

| taking logarithms | | of both sides of | or log A^n = 4 log A | the equations |

EXAMPLE: Find the value of 3^4, using logarithms.

SOLUTION: 1. Log 3^4 = 4 log 3 = 4(.47712) = 1.90848

2. Antilog 1.90848 = answer, or antilog 1.90848 = 8.1 x 10¹ = 81

6-49. Log 16^32 is equal to which of the following? 1. log 16 + log 32 2. log 16 x log 32 3. 16 log 32 4. 32 log 16

6-50. Using logarithms, find the approximate value of 5^8. 1. 3.9 x 10^5 2. 3.9 x 10^6 3. 4.0 x 10^5 4. 4.0 x 10^6

6-51. The log 26·38^n is equivalent to 1. 104 log 38 2. log 26 x 4 log 38 3. log 26 + log 4 + log 38 4. log 26 + 4 log 38

38

6-52. The antilog of what value is used to find the approximate product of 22·43° ? (Refer to Appendix I)

1. 10.14324 2. 11.14324 3. 12.15683 4. 13.15683

• Note that log a/b = log a·b⁻¹ = log a + log b⁻¹ = log a - 1 log b = log a - log b

Therefore, the logarithm of the quotient of two numbers equals the logarithm of the dividend minus the logarithm of the divisor. EXAMPLE: Find the value of 24/8 using logarithms.

SOLUTION:

1. Log 24/8 = log 24 - log 8 = 1.38021 - .90308 = .47713

2. Antilog .47713 = quotient, or antilog .47713 = 3 x 10⁰ = 3 x 1 = 3 Therefore 24/8 = 3

6-53. Since log 12 = 1.07918 and log 2 = .30103, the log of the quotient of 12/2 is equivalent to

1. .30103 - 1.07918 2. 1.07918 - .30103 3. log 1.07918 : log .30103 4. log 12 : log 2

• Refer to Appendix I in answering items 6-54 through 6-56.

6-54. Use logarithms to find the quotient of 84/45 · What is the mantissa used to find the digit sequence of the quotient ?

1. .25528 2. .30103 3. .36000 4. .56170

6-55. The antilog of what value must be used to find the quotient of 540/36 ?

1. 1.28869 2. 1.17609 3. 3.17609 4. 3.28869

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6-56. Use logarithms to find the quotient of 780000/30 · The quotient equals

1. 2.3 x 10³ 2. 2.6 x 10³ 3. 2.3 x 10⁴ 4. 2.6 x 10⁴

6-57. Log 22²/6 is equivalent to

1. 3 log 22 - 6 2. 3 log 22 - log 6 3. log 3 x log 22 - log 6 4. log 22 + log 3 - log 6

6-58. Which expression below is equi- valent to log _5(14⁵)_ ? (6(13²)

1. 5 log .5 + log14 - log 6 x 2 log 13 2. log.5 + 5 log 14 - log 6 x 2 log 13 3. log.5 x 5 log 14 - (log 6 + 2 log13) 4. log.5 + 5 log 14 - (log 6 + 2 log13)

6-59. The expression log __4__ is equi- 7-3 7 valent to

1. 4 log 7⁻³ 2. log 4 - 3 log 7 3. log 4 + 3 log 7 4. 3 log 7 - 4

6-60. The antilog of what value is used to find the approximate result of 13·4² ?

(Refer to Appendix I)

1. .63132 2. 1.47296 3. 1.63132 4. 2.47296

Assignment 7 Fundamentals of Algebra; Factoring Polynomials Textbook Assignment: Chapters 9, 10 (111-117) 7-1. The literal numbers a, x, and p are more general than the numbers 9, 8, and 7. 7-2. The commutative law for addition is il- lustrated by the equation 1. ab = ba 2 . a+a=b+b 3 . a+b=b+a 4. a(b + c) = ab + ac 7-3. The associative law of multiplication is illustrated by the equation 1. abc = acb = cba 2. a x (b + c) = c x (b + a) 3. a (b + c + d) = ab + ac + ad 4. a· (b·c) = a·b·c = (a·b)·c 7-4. If a = 2, b = -3, and c = 4, the algebraic sum of a - b - c equals 1. -5 2. 1 3. 3 4. 9 7-5. If r = 1, s = 3, t = 12, and x = 15, what is the value of the expression 2rx ?t- s 1. - 3 3. 2 2. -2 4. 3 7-6. The algebraic expression is considered to be three numbers. 7-7. What is the value of the algebraic ex- pression 5x2 - 2xy + (3x) 2 when x = 2 and y = -3? 1. 44 2. 68 3. 124 4. 148 7-8. What name is given to the algebraic ex- pression ax - 2bx + cx2 - 4? 1. Monomial 2. Binomial 3. Trinomial 4. Polynomial 7-9. Which statement is true regarding the monomial 17xyz? 1. 17 is the coefficient of xyz. 2. 17x is the coefficient of yz. 3. 17xy is the coefficient of z. 4. Each of the above is a true statement. 7-10. In the expression xy, what is the co- efficient of xy? 1. 1 2. x 3. y 4. x y 7-11. What is the coefficient of x 2 y in the expression x 2y - ab? 1. 1 2. 2 3. x 2 4. y 7-12. Two terms of an expression are said to be like if they contain 1. at least one factor in common 2. the same numerical coefficient 3. the same literal factors with only their exponents different 4. the same literal factors raised to the same powers 7-13. The like terms in the expression 2 ac 2 - 2bc 2 + ac 2 - 2c 2 are 1. 2ac 2 and 2bc 2 2. 2ac 2 and ac 2 3. 2ac 2 and 2c 2 4. 2bc 2 and 2c 2 40

7-14. What is the result when the expression

6x - 4x²

is simplified? 1. 2x² 2. 2(1 - x) 3. 2x(3 - 2x) 4. 2x(x²) - 3

7-15. What is the correct procedure for combin- ing the like terms ?cd² and -cd²? 1. Add -1 to 7 and use this sum as the coefficient of cd². 2. Subtract -1 from 7 and use this dif- ference as the coefficient of cd². 3. Add 1 to 7 and use this sum as the co- efficient of -cd². 4. Add -1 to 7 and use this sum as the coefficient of -cd².

7-16. Which of the following expressions are equivalent, if any? A. 6x - (x - y + 4) B. 6x - [x - y + 4] C. 6x - [x - y + 4] D. 6x - x - y + 4 1. A and B only are equivalent. 2. A, B, and C only are equivalent. 3. All are equivalent. 4. None are equivalent.

7-17. After the removal of parentheses, the expression 6 - (-x + y - z) becomes 1. 6 - x - y + z 2. 6 - x + y + z 3. 6 + x - y + z 4. 6 - x - y + z

7-18. What is the result of removing parentheses and brackets from the expression

(x - y) - [3x - (4 - 8x)]?

1. -4 + 4x - y 2. -4 + 14x - y 3. 4 - 12x - y 4. 4 - 14x + y

7-19. If an expression in parentheses is pre- ceded by a minus sign, what happens to the signs of its terms when the paren- theses are removed? 1. The signs remain unchanged. 2. The plus signs are changed to minus signs and the minus signs are left un- changed. 3. The minus signs are changed to plus signs and the plus signs are left un- changed. 4. The plus signs are changed to minus signs and the minus signs are changed to plus signs.

7-20. Which of the following expressions is equivalent to the expression

8x - y - 7 + 14x?

1. (8x - y) - (7 + 14x) 2. (8x - y) - (7 - 14x) 3. -(8x - y) + (7 - 14x) 4. -(8x - y) + (7 + 14x)

7-21. When the polynomial 6x - z - y + 4 is grouped by enclosing the first two terms in parentheses preceded by a minus sign, and the last two terms in parentheses preceded by a plus sign, what is its appearance? 1. -(6x - z) + (y + 4) 2. -(6x + z) + (-y + 4) 3. -(6x - z) + (-y - 4) 4. -(6x + z) + (y - 4)

7-22. The product of x^y and x^z is

1. x^y² 3. x(y + z)

2. 2x^yz 4. 2x(y + z)

7-23. What is the product of 7r³st² and 5re²t³?

1. 12r²s³t 3. 35r²st

2. 12r⁴st⁵ 4. 35r⁴s³t⁵

7-24. If two monomials contain a common literal factor and one monomial is divided by the other, the common literal factor in the quotient will have an exponent that is equal to the 1. sum of the exponents of the factor in the two monomials 2. product of the exponents of the factor in the two monomials 3. exponent of the factor in the numerator minus the exponent of the factor in the denominator 4. exponent of the factor in the numerator divided by the exponent of the factor in the denominator

7-25. The product of 6a²b and 14a³b²is

1. 20a⁵b³ 3. 84s⁵b³

2. 20a⁶b³ 4. 84a⁵b²

7-26. The quotient resulting from dividing

16x⁴y²z by (-8x³y²z) is

1. -2xy 3. -2z⁻¹y

2. 2x⁻¹y 4. -2xyz⁰

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7-27. What is the sum of 11p - 7q - r, 3p + q - 9r, and -p - q + 2r? 1. 13p - 9q + r 2. 13p - 7q - 8r 3. 14p + q - 6r 4. 14p - q + 10r

7-28. What is the result of subtracting -3r - s + 6t from 5r + 2s - t? 1. 2r - s - 7t 2. 2r + s + 5t 3. 8r + a - 5t 4. 8r + 3s - 7t

7-29. What is the result when the expression

(3x - 2y + 4) - (-6x + y - 5)

is simplified? 1. -3x - y - 1 2. -3x - y + 1 3. 9x - y - 1 4. 9x - 3y + 9

7-30. What is the product when (-4y + x - 7) is multiplied by x? 1. 4xy + x² - 7x 2. -4xy + x² + 7x 3. 4xy - x² + 7x 4. -4xy + x² - 7x

7-31. What is the product of

(p - q) and (r - s - t)?

1. pr - ps - pt - qr + qs + qt 2. pr - ps + pt - qr + qs - qt 3. pr + ps - pt + qr - qs - qt 4. pr + ps + pt + qr + qs - qt

7-32. What is the result of multiplying

(5v - 7) by (2v + 9)?

1. 10v² + 31v - 63 2. 10v² + 31v + 2 3. 10v - 31v + 63 4. 10v² + 59v + 63

7-33. What is the product when 9x² - 6x + 1 is multiplied by 3x + 2? 1. 27x³ - 36x² + 15x - 2 2. 27x³ - 36x² - 15x + 2 3. 27x³ - 9x - 2 4. 27x³ - 9x + 2

7-34. The product

(x + y)(x + y) = x² + 2xy + y²

is called the product of the sum and difference of two numbers.

7-35. The product of (x + 3)(x - 4) is x² - x - 12.

7-36. The product of x⁸ - y⁸ and x⁴ + y⁴ is

1. x⁸ - y⁸ 3. x⁸ - 2x⁴y⁴ + y⁸

2. x¹⁶ - y¹⁶ 4. x¹⁶ - 2x⁴y⁴ + y¹⁶

7-37. The denominator of the fraction

x ___________ y - √z

can be rationalized by 1. multiplying both the numerator and denominator by y + √z 2. multiplying both the numerator and denominator by y - √z 3. multiplying both the numerator and denominator by √z - y 4. squaring both the numerator and denominator

7-38. Rationalizing the binomial denominator of which of the following fractions results in the fraction equaling 8√5 - 16 ?

1. 8 3. 8 √5 - 2 √2 + 5

2. 8 4. 8 √2 - 5 √5 + 2

7-39. Which of the following statements is true of the square of the difference of two numbers but not of the square of the sum of the same numbers? 1. There is no middle term. 2. The sign of the middle term is negative. 3. The middle term is equal to the prod- uct of the numbers. 4. The middle term is four times the product of the numbers.

7-40. Why is the product of the sum and the difference of two numbers considered a special product? 1. The middle term is always irrational. 2. The sign of the middle term is always negative. 3. The value of the first term is always equal to the middle term. 4. The product can be written without going through the whole multiplica- tion process.

7-41. The square of (6 - √3) is

1. 9 3. 36 - 10√3

2. 24 - 12√3 4. 39 - 12√3

42

7-42. Both (6x + 8 - y)÷ 6 and 6 ÷ (6x + 8 -y) may be solved by the distributive method.

7-43. What is the result of dividing

27x⁴y³z⁴ + 9x²y⁴z⁶ - 12x³yz

by 3x²yz?

1. 9x²yz + 3yz - 4xyz 2. 9xy²z² + 3y²z⁴ - 4x 3. 9x²y³z³ + 3y³z⁵ - 4x 4. 9x³y³z³ + 3y³z⁴ - 4xy

7-44. Which of the following factors is a monomial factor of

p²qr³ - pr²s + pq²rs²?

1. pr 3. p²r

2. qr² 4. pqs³

7-45. What is the numerical coefficient of the xy term in the expression obtained by dividing

2x³ + 5x²y + xy² - 8y³ by x - y?

1. 2 3. 7

2. 5 4. 8

7-46. Division of

6x⁴ - 28x³ + 19x² - 14x - 1 by 3x - 2

leaves a remainder of

1. 0 3. -2 3x - 2

2. -1 4. -9 3x - 2 3x - 2

7-47. The polynomial 4 - 6x³ + 8x - x² arranged in the order of descending powers of x is 1. 4 + 8x - x² - 6x³ 2. -6x³ + 8x - x² - 4 3. -6x³ - x² + 8x + 4 4. 8x + 4 - (6x³ + x²)

7-48. What is the quotient when

x³ - 8x² + 16x - 5 is divided by x - 5? 1. x² - x + 5 2. x² - 3x + 2 3. x² - 3x + 1 4. x² + 13x - 49 + 240 (x - 5)

7-49. The division problem

x² - 3x + 4 x - 2 x³ - 5x² + 10x - 8 x³ - 2x² ________ -3x² + 10x -3x² + 6x _________ 4x - 8 4x - 8

can be condensed to which of the following?

1. x - 2 x² - 3x + 4 x³ - 5x² + 10x - 8 - 2x² - 6x + 8 - 3x² + 4x 0

2. x - 2 x² - 3x + 4 x³ - 5x² + 10x - 8 - 2x² - 6x - 8 - 3x² + 4x 0

3. x - 2 x² - 3x + 4 x³ - 5x² + 10x - 8 - 2x² + 6x - 8 - 3x² + 4x 0

4. x - 2 x² - 3x + 4 x³ - 3x² + 10x - 8 - 2x² + 6x - 8

7-50. The division problem in item 7-49 can be further condensed to

1. -2 1 - 3 4 1 - 5 10 - 8 - 2 -6 - 8 1 - 3 4 0

2. -2 1 - 3 4 1 - 5 10 - 8 - 2 -6 - 8 1 - 3 4 0

3. -2 1 - 3 4 1 - 5 10 - 8 - 2 -6 - 8 1 - 3 4 0

4. -2 1 - 3 4 1 - 5 10 - 8 1 - 3 4 - 8

7-51. If 2x³ - 6x⁴ + 8x is divided by x - 7, the results of each successive step of multiplication and subtraction are determined by the 1. x's 2. 2, -6, and 8 3. 2, -6, 8, and -7 4. 2, -6, 8, and the x's

43

7-52. Synthetic division is limited to divisors of the form x - a and x² - a.

7-53. The division of x³ - 5x² + 11x - 15 by x - 3 using synthetic division becomes

-3 | 1 - 5 11 - 15 | - 3 6 - 15 | 1 - 2 5 0

7-54. In the form

2 | 3 - 4 1 6 | 6 - 20 42 | 3 -10 21 - 36

the result of dividing 3x³ - 4x² + x + 6 by x + 2 is 1. 3x² - 4x + 1 + 6/(x + 2) 2. 3x² - 10x + 21 - 36/(x + 2) 3. 6x² - 20x + 42 + x + 2 4. 6x² - 20x + 42 - 36/(x + 2)

7-55. The prime factors of x² - 16 are x + 4 and x - 4.

7-56. The prime factors of x³ - 9x are 1. x³, -9x 2. x, x² - 9 3. x, -3, +3 4. x, x - 3, x + 3

7-57. Express the polynomial 5x²y - 10xy + 25xy² as a product of prime factors.

1. 5x²y²(1 - 2x²y-2 + 5) 2. y(5x² - 10x + 25xy) 3. x(3xy - 10y + 5y²) 4. 5xy(x - 2 + 5y)

7-58. Factor 2y^(t+s) - 4y^t into prime factors.

1. 2(y^(t+s) - 2y^t) 3. 2y^(t+s) - 4y^t 2. 2y(y^s - 2y^s) 4. 2y^t(y^s - 2)

7-59. The algebraic form ab + ac - xb - xc is equivalent to the form (a - x)(b + c).

7-60. What are the factors of 4x - 2y + xy - 8? 1. (4 - y)(x + 2) 2. (y - 4)(x + 2) 3. (y + 4)(z - x) 4. (y + 4)(x - 2)

7-61. What is the factored form of

98x³ - 32xy²?

1. 2(7x + 4y)(7x - 4y) 2. 4(7x + 2y)(7x - 2y) 3. 2x(7x + 4y)(7x - 4y) 4. 4x(7x + 2y)(7x - 2y)

7-62. What are the factors of the expression

2a³b - 8ab³?

1. 2ab(2b - a)(2b + a) 2. 2ab(a - 2b)(a + 2b) 3. 2a²b²(a - b)(a + b) 4. 2a²b(a - 2b)(a + 2b)

7-63. What are the prime factors of 81x⁴ - 1? 1. (9x² + 1)(9x² + 1) 2. (9x² - 1)(3x + 1)(3x + 1) 3. (3x - 1)(3x + 1)(9x² + 1) 4. (3x - 1)(3x + 1)(3x + 1)(3x + 1)

7-64. What are the factors of 8a³ - 1? 1. (2a + 1)(4a² + 2a + 1) 2. (2a + 1)(4a² - 2a + 1) 3. (2a - 1)(4a² - 2a + 1) 4. (2a - 1)(4a² + 2a + 1)

7-65. Which of the following trinomials is a perfect square? 1. 36t² + 9s² - 9st 2. 36t² + 9s² - 18st 3. 36t² + 9s² + 18st 4. 36t² + 9s² - 36st

7-66. In order for the incomplete trinomial

16x² + 25y² + ?

to be a trinomial square, the missing term must be 1. ±40xy 2. ±400x²y² 3. ±20xy 4. ±60xy

7-67. Assume that a trinomial can be factored into two binomials that have a common term. The unlike terms will be opposite in sign and the positive one will be numerically smaller than the negative one only if the 1. second and third terms of the trinomial are both positive 2. second and third terms of the trinomial are both negative 3. second term of the trinomial is positive and the third term is negative 4. second term of the trinomial is negative and the third term is positive

44

7-68. What are the factors of x² - x - 20? 1. (x - 4)(x - 5) 2. (x - 4)(x + 5) 3. (x - 5)(x + 4) 4. (x - 21)(x + 1)

7-69. What are the factors of x² - x + 30? 1. (x - 6)(x + 5) 2. (x - 6)(x - 5) 3. (x + 6)(x - 5) 4. It cannot be factored.

7-70. What are the factors of y² + 48y - 100? 1. (y - 50)(y - 2) 2. (y - 50)(y + 2) 3. (y + 50)(y - 2) 4. It cannot be factored.

7-71. What are the factors of the trinomial expression

6m² - 13m + 6?

1. (3m + 2)(2m + 3) 2. (3m - 2)(2m + 3) 3. (3m - 2)(2m - 3) 4. It cannot be factored.

7-72. A fraction can be reduced to lower terms only if the 1. numerator is greater than the denominator 2. denominator is greater than the numerator 3. numerator and denominator both contain numerical factors 4. numerator and denominator contain common factors

7-73. Which of the following fractions takes the form

2a/7x²z²

after it has been reduced to its lowest terms?

1. 12a²xz/42xy³z³ 3. 36a²x²z²/69xy³ 2. 12a²x/49x²z² 4. 36ax³z²/56a²x²z³

7-74. What is the result when the fraction

y² - 9/(y² + 6y + 9)

is reduced to its lowest terms?

1. -1/6y 3. (y - 3)/(y + 3) 2. (y + 3)/(y - 3) 4. It cannot be reduced.

7-75. What is the product when

(2m² - 5m - 12)/(2m + m³ + 3m²)

is multiplied by

(m² + m)/(7m - 3 + 6m²)?

1. m(m - 4)/((3m - 1)(m + 2)) 2. (m - 4)/((3m - 1)(m + 2)) 3. (m + 1)/((3m - 1)(m - 2)) 4. m(m + 1)/((3m - 1)(m + 2))

45

Assignment 8

Factoring Polynomials; Linear Equations in One Variable; Linear Equations in Two Variables

Textbook Assignment: Chapters 10 (117-119), 11, 12

8-1. What is the quotient when

p2 - p - 12 10p2 + 3pq - q2

is divided by

p2 - 2p - 8 5p2 + 6pq - q2 ?

1. (p + 3)(p + q) (2p + q)(p + 2)

2. (p + 3)(p - q) (2p + q)(p + 2)

3. (p + 3)(p + q) (p + q)(2p + 2)

4. (p + 3)(2p + q) (p + q)(2p + 2)

8-2. What is the sum of the fractions

m + 5 2m2 + 5m - 3

and

2m - 11 2m2 + 13m - 7 ?

1. (3m + 1)(m + 2) (2m + 1)(m + 3)(m + 7)

2. (3m + 1)(m + 2) (2m - 1)(m + 3)(m + 7)

3. (3m - 1)(m - 2) (2m - 1)(m - 3)(m - 7)

4. (3m - 1)(m + 2) (2m + 1)(m + 3)(m - 7)

8-3. What is the remainder when

2a - 3 a2 + 5a - 14

is subtracted from

2a - 5 a2 + 4a - 12 ?

1. -17 (a + 6)(a + 7)(a - 2)

2. -17 (a + 6)(a - 7)(a + 2)

3. -17 (a - 6)(a + 7)(a + 2)

4. -17 (a - 6)(a - 7)(a - 2)

8-4. What is the least common denominator of the fractions

7 a + 3 , a2 - 2a - 15 , and a2 - 16 ? a2 - 5a - 5

1. a2 + 4a + 3 2. a2 - 4a - 5 3. a3 - a2 - 17a - 15 4. a3 - 9a2 + 15a + 25

8-5. An equation is an expression of equality.

8-6. What is the fixed constant in the equation bx + 4y = c? 1. b 2. c 3. y 4. 4

46

8-7. Which of the following letters normally represents an arbitrary constant? 1. a 2. x 3. y 4. w

8-8. How many variable terms are in the equation 2x + 3y + 5 = -z ?

1. Four 2. Three 3. Two 4. One

8-9. For how many values of x does 2x - 7 = 18? 1. One 2. Two 3. Three 4. Indefinite number

5-10. What is the degree of the equation 3x - x(2 + y) = y?

[Hint: Simplify the equation first by performing the indicated multiplication.] 1. First 2. Second 3. Third 4. Fourth

8-11. The equation which is first degree in the variables x and y is called a linear equation because its graph is a straight line.

8-12. Which of the following expressions is an identity? 1. 2x - 7 = 0 2. 9a + 2a - 5 3. a(a + b) = a2 - ab 4. c(10 - b) = -bc + 10c

8-13. An identity is true for one value of the variable whereas a conditional equation is true for many values of the variable.

8-14. A conditional equation reduces to an identity when the variable is replaced by 1. the arbitrary constant in the equation 2. the fixed constant in the equation 3. an arbitrary numerical value 4. a root of the equation

8-15. In solving a linear equation by manipulating both sides of the equation, division by zero is not permitted.

8-16. What is the value of x in the equation x + 1 = 1? 1. 0 2. 1 3. 2 4. Not solvable

8-17. What is the value of y in the equation y 7 = 6?

1. 6 7

3. 7

2. 6

4. 42

8-18. What is the value of x in the equation 7x = 19?

1. 2.6

3. 7 19

2. 2.9

4. 19 7

8-19. What is the value of x in the equation x 7 + 12 = 15? 1. 7 2. 14 3. 21 4. 28

8-20. The equation 8x - 3 = 37 can be solved by 1. dividing both members by 8 and then adding 3 to both members 2. adding 3 to both members and then dividing both members by 8 3. subtracting 3 from both members and then dividing both members by 8 4. subtracting 37 from both members and then dividing both members by 8

8-21. What is the value of x in the equation x - 4 = 2x - 8?

1. -4 3

2. -4

3. 4 3

4. 4

47

8-22. To solve the equation

5z --- - 2z = 8z + 1, 7

you first multiply both members by 7. What steps must you take next?

1. Add 9z to both members and divide both members by 7. 2. Subtract 19z from both members and divide both members by 75. 3. Subtract 56z from both members and divide both members by -65. 4. Subtract 19z from both members and divide both members by 37.

8-23. What is the value of x in the equation a + b = 0? 1. a - b 2. b - a -b 3. --- a -a 4. --- b

8-24. Which of the following procedures can be used to solve the equation cy + d = 10 - dy for y? 1. Add dy to both members, factor out the coefficient of y, and then divide both members by c + d. 2. Add dy to both members, factor out the coefficient of y, and then divide both members by c - d. 3. Add dy -7 to both members, factor out the coefficient of y, and then divide both members by c + d. 4. Add dy -7 to both members, factor out the coefficient of y, and then divide both members by c - d.

8-25. What is the value of x in the equation

2x - [x - (3x + 7)] = 17 + 2(x + 4)?

1. -3 2. 2 3. 9 4. 15

8-26. What step must you take first when you solve the equation 8y 1 y --- = --- - ----? 7 2 12

1. Divide both members by y. 2. Multiply both members by 12. 3. Multiply both members by 36. 1 4. Subtract --- from both members. 2

8-27. What is the numerical value of y in the equation y y --- + 4 = --- ? 2 3

1. 12 2. 24 3. -12 4. -24

8-28. The general form, ax + b = 0, does not represent 7x - [4 - (6 + x)] = 4.

8-29. What are the values of the constants a and b when the equation

9 - (2 - 8x) = 14x + 12

is simplified and put into the general form ax + b = 0? 1. a = 10, b = -13 2. a = 12, b = 7 3. a = -8, b = 12 4. a = -6, b = -5

8-30. Assume that two resistors connected in parallel have a resistance of 240 ohms. The markings on R₁ indicate that it has a resistance of 400 ohms. Overheating has burned off the marking of R₂. The total resistance of two resistors con- nected in parallel is given by the formula 1 1 1 --- = --- + --- Rt R₁ R₂ What is the resistance of R₂? 1. 600 ohms 2. 800 ohms 3. 1,000 ohms 4. 1,200 ohms

8-31. Which equation expresses the following statement? Three numbers are such that the second is twice the first and the third is seven less than three times the second and the sum of the three is 38. 1. 9x = 63 2. x + 2x + 6x = 31 3. x + 2x + 3(2x - 7) = 38 4. x + 2x + (6x - 7) = 38

8-32. The inequality x + 7 > 15 has 1. a finite number of solutions 2. a solution of x = 8 3. a solution set 4. one solution

8-33. The two inequalities 5 > 3 and 3 < 5 have the same sense.

8-34. If the same negative number is added to both sides of an inequality, the sense of the inequality is reversed.

8-35. How may the value of x be identified in the inequality x + 4 < - 7? 1. x > - 11 2. x < - 3 3. x < - 11 4. x < - 15

8-36. If both members of an inequality are multiplied by a positive number less than one, the sense of the inequality is re- versed.

4d

8-37. How may the value of x be identified in the inequality 4 - x > 8? 1. x > - 4 2. x < - 4 3. x < - 12 4. x > - 12

8-38. If x² > 9, the solution set includes ±3.

8-39. In the Cartesian or rectangular coordi- nate system, the x and y axes intersect at an angle of 1. 0° 2. 45° 3. 60° 4. 90°

8-40. In writing the coordinates of a point in the Cartesian coordinate system, the y coordinate is always written second.

8-41. What are the coordinates of the origin point? 1. (1, 0) 2. (1, 1) 3. (0, 0) 4. (0, 1)

8-42. What are the coordinates of a point lo- cated 3 units to the right of the y-axis and 7 units to the right of the y-axis in a Cartesian coordinate system? 1. Ordinate is 3, abscissa is 7. 2. Ordinate is -3, abscissa is 7. 3. Ordinate is 7, abscissa is -3. 4. Ordinate is -7, abscissa is -3.

8-43. Refer to figure 12-1 in your textbook. In a rectangular coordinate system, the point with coordinates (2, -1) is lo- cated in which quadrant? 1. First 2. Second 3. Third 4. Fourth

8-44. When the rectangular coordinates of a point are squared, numbers result that represent the rectangular coordinates of another point. The two points will be in the same quadrant only if the original point was in which quadrant? 1. First 2. Second 3. Third 4. Fourth

8-45. The quadrant in which both x and y are negative is the 1. first 2. second 3. third 4. fourth

8-46. The graph of x - y = 4 reveals then an infinite number of points satisfy the equation.

8-47. The x-coordinate of a point lying on the graph of the equation 3x + y = 2 is 9. What is the y-coordinate of the point? 1. -21 2. -24 3. -25 4. -29

8-48. Which of the following pairs of coordi- nates are the coordinates of a point lying on the graph of the equation 3y - 2x = -1? 1. (-5, 2) 2. (1, -3) 3. (8, 6) 4. (11, 7)

8-49. How many points do the graphs of the equations 6x - 3y = 7 and 12x - 6y = 14 have in common? 1. None 2. 1 3. 2 4. An infinite number

8-50. Assume that a line is drawn through the points with coordinates (-2, 10) and (5, -11). Which of the following equa- tions is the equation of that line? 1. x - 7 = -9 2. 2x + 2y = 5 3. 3x + y = 4 4. 4x - 3y = -11

8-51. Which of the following pairs of numbers is a solution of the equation 17x + 8y = -15? 1 1. x = --, y = -3 2 1 2. x = -1, y = - 7 1 8 3. x = --, y = - 3 -3 9 4. x = -3, y = - 4

8-52. If both point A and point B satisfy both linear equation 1 and linear equation 2, then equation 1 and equation 2 represent the same straight line.

8-53. To find the y intercept of a linear equation, let y = 0.

8-54. Which of the following equations has a graph with an x intercept of 4 and a y intercept of 6? 1. 3x + 2y = 12 2. y - 2x = 4 3. 2y - 3x = 6 4. 6x + 4y = 2

49

8-55. An equation that contains y as the only variable always has a graph that 1. passes through the origin 2. has an x intercept 3. lies parallel to the x-axis 4. lies parallel to the y-axis

8-56. What is the graph of the equation x = 0? 1. The x-axis 2. The y-axis 3. A horizontal line above the x-axis 4. A vertical line to the right of the y-axis

8-57. If two linear equations are satisfied by the point (3, 4) and if this point is the only point satisfying both equations, then the two lines are parallel.

8-58. The graphical method of solving simul- taneous equations is an exact method.

8-59. In solving a pair of simultaneous linear equations by the addition method, you must first 1. add the two equations 2. subtract one equation from the other 3. make the coefficients of both vari- ables the same in both equations 4. make the coefficients of one of the variables the same, except for sign, in both equations

8-60. One way to eliminate x from the equations 3x + 2y = 6 and 5x + 7y = 43 when solv- ing by means of the addition method is to 1. multiply the first equation by 5 and the second equation by -3 2. multiply the first equation by 3 and the second equation by 5 3. subtract the first equation from the second 4. subtract the second equation from the first

8-61. In eliminating y from the simultaneous equations 2x + 7y = 3 3x - 5y = 31 by the method of addition, the resulting equation is 1. 5x = 34 2. 29x = 176 3. 31x = 232 4. 31x = 372

8-62. The substitution method of solving simul- taneous linear equations involves replac- ing one of the variables in one equation with 1. the variable's coefficient in the other equation 2. its value in terms of the other variable equation 3. the other equation 4. the constant from the other equation

8-63. In solving the system of equations, 2x - 3y = 4 x + 2y = 5 by the substitution method, which of the following is a correct substitution for a variable from one equation into the other? 1. x = 2y + 6 2. x = 3y + 4 3. y = 4 - 2x 3 4. y = 5 - x 2

8-64. After substituting for one variable in the method of substitution, the original simultaneous equations are re- duced to one equation in one variable.

8-65. Which of the following sets of expres- sions is a solution of the simultaneous equations 4x + y = a 3x + 2y = b?

1. x = 2a - b, y = 4b - 3a 5 5 2. x = a - b, y = b - a 5 5 3. x = 2a - 3b, y = 3a - 4b 5 5 4. x = 3a - 2b, y = 4b - a 5 5

8-66. A form of the first-degree equation in two variables is 1. x² + py² = q 2. x + py² = q 3. x² + py = q 4. x + py = q

8-67. Where do the graphs of a pair of simul- taneous equations intersect if the con- stant term of each equation is zero? 1. On the positive x-axis 2. On the negative y-axis 3. At the origin 4. At infinity

8-68. Refer to figure 12-8 in your textbook which shows the graphs of two parallel linear equations. Which of the follow- ing statements concerning the nature of the two equations is true? 1. The equations contain an x term only. 2. The equations are identical except for the value of the constant term. 3. The equations have the same constant term. 4. The equations both have zero con- stant terms.

8-69. How many solutions does a pair of simul- taneous equations in two variables have if the graphs of the equations are parallel? 1. None 2. 1 3. 2 4. An indefinite number

8-70. Which of the following sets of simul- taneous equations cannot be solved? 1. x - 2y = 0 and 2x + 7 = 0 2. 2x + 3y = 4 and 2x + 5y = 8 3. 2x + y = 9 and 2x + y = 18 4. 2x + 3y = 5 and 2x + 7y = 10

8-71. A hawser was originally 70 feet long. After it was cut into two pieces, one piece was 8 feet longer than the other. What are the two equations you must solve to find the length of the two pieces? 1. a + b = 8 and a - b = 70 2. a + b = 70 and a - b = 8 3. a + b = 78 and a - b = 62 4. a + b = 62 and a - b = 78

8-72. A man takes his boat up a river at 8 miles per hour and returns to his start- ing point at his top speed of 22 miles per hour. If he must complete his trip in 3 hours, about how far up the river can he go? 1. 16 mi 2. 16.7 mi 3. 17 mi 4. 17.6 mi

8-73. When two batteries are connected in series, their combined voltage is 120 volts. If the batteries are wired so that their resulting voltage is equal to the difference between their voltages rather than their sum, the combined voltage is 60 volts. What are the voltages of the two battery units? 1. 20 v and 100 v 2. 30 v and 90 v 3. 40 v and 80 v 4. 50 v and 70 v

8-74. The plot of the inequality x - y > 5 is made by first plotting 1. random points 2. many points 3. two points 4. x - y = 5

8-75. The plot of the simultaneous solution of x + y > 5 and x - y > 1 is the overlap of the plot of the individual areas x + y > 5 and x - y > 1.

Assignment 9

Ratio, Proportion and Variation; Dependence, Functions and Formulas; Complex Numbers

Textbook Assignment: Chapters 13, 14, 15 (158-164)

9-1. A radio operator has a top code speed of 45 words per minute. How fast can he take code compared with an average operator who has a speed of 36 words per minute?

1. 1¼ times as fast

2. 1¼ times as fast

3. 1¾ times as fast

4. 1⅘ times as fast

9-2. When ratios are used to compare two quan- tities, the quantities must be stated in the same units.

9-3. Which of the following ratios is equivalent to the inverse of the ratio 42:48?

1. ⅞ 3. ⅞

2. ⁴⁄₃ 4. ⁸⁄₇

9-4. Which of the following expresses the ratio of 8 ft 3 in. to 3 in., when reduced to its lowest terms?

1. 2 ft 9 in. 3. ³³⁄₁

2. ⁸ ft 3 in. 4. ⁹⁹⁄₁ 3 in.

9-5. The usual methods of expressing the same proportion are

1. ⁶⁄₃ = ¹⁄₂; 3:6 = 1:2; 3:6::1:2

2. ³⁄₆ = ¹⁄₂; 3:2 = 1:6; ³⁄₁ = ⁶⁄₂

3. 3:6::1:2; ³⁄₆ = ¹⁄₂; 3:6 = 1:2

4. 3 - 6 = 1 - 2; ³⁄₆ = ¹⁄₂; 3:6 = 1:2

9-6. What are the means in the proportion 2:3 = 10:15? 1. 2 and 3 3. 3 and 10

2. 2 and 10 4. 3 and 15

9-7. In the proportion ᵃ⁄ᵦ = ᶜ⁄ᵈ the extremes are

1. a and c 3. c and b

2. a and d 4. d and c

9-8. In a proportion, where does the factor of proportionality appear? 1. In the means 2. In the extremes 3. In the numerators of both ratios 4. In the numerator and denominator of one ratio

9-9. What is the factor of proportionality in

the proportion ⁹⁄₁₇ = ⁶³⁄₁₁₉?

1. 3 3. 7

2. 4 4. 8

9-10. The proportion ˣ⁄₂ = -²⁄₁₆ is equivalent to

the equation 1. 9x = 8 3. 16x = ²⁄₂

2. 9x = 32 4. 16x = 18

9-11. What is the value of x in the proportion ³⁄ₓ = -ˣ⁄₁₁?

1. 14 3. ± √14

2. 33 4. ± √33

52

9-12. What is the third proportional in a pro- portion whose first proportional is 15 and whose mean proportional is 105? [Hint: In a proportion with equal inner terms; xy = zw. z is called the third proportional.] 1. 585 2. 615 3. 735 4. 1,575

9-13. What is the value of x in the proportion ᵃ⁄ₓ = ᵇ⁄ᶜ ?

1. ᵃ⁄ᵦᶜ

2. ᵃᶜ⁄ᵦ

3. ᵇ⁄ᵃᶜ

4. ᵇᶜ⁄ᵃ

9-14. If the numbers a, b, c, and d are in pro- portion in the arrangement ᵃ⁄ᵦ = ᶜ⁄ᵈ, there are other arrangements of a, b, c, and d which will form a proportion.

9-15. Which of the following proportions is ob- tained from the proportion ᵃ⁄₂ = ᶻ⁄ᵇ by inversion?

1. ²⁄ₐ = ᵇ⁄ᶻ 3. ²⁄ₐ = ᵇ⁄ᶻ

2. ᵃ⁄ᵇ = ᵇ⁄₂ 4. ᶻ⁄₂ = ²⁄ᵦ

9-16. If ᵐ⁄ₙ = ᵖ⁄q, the proportion formed by alter- nation is 1. m:p = n:q 2. n:m = q:p 3. n:q = m:p 4. p:m = q:n

9-17. A post and a flag pole, both vertical to the ground, cast shadows of 3 ft and 10 ft, respectively. If the post is 6 ft 6 in. tall, how tall is the flag pole? 1. 19 ft 6 in. 2. 20 ft 3. 21 ft 8 in. 4. 65 ft

9-18. A 3-foot scale model is made of a ship that is actually 180 feet long. Let x = the scale length of a mast that is actually 40 feet long. Which of the following propor- tions will express the relationship between the mast length of the model and the mast length of the actual ship? 1. ¹⁸⁰⁄₃ = ˣ⁄₄₀

2. ³⁄₄₀ = ¹⁸⁰⁄ₓ

3. ³⁄ₓ = ⁴⁰⁄₁₈₀

4. All of the above proportions

9-19. How many seconds does a chronometer lose in a week if it loses 9.0 seconds in 30 days? 1. 2.1 sec 2. 2.4 sec 3. 3.3 sec 4. 3.8 sec

9-20. If a destroyer travels at the rate of 24 knots, how long does it take to go 84 nautical miles? 1. 3.1 hr 2. 3.2 hr 3. 3.3 hr 4. 3.5 hr

9-21. The depth of water at a beach increases smoothly from zero at the shore to 120 feet at a distance of 2,000 yards out. How close to the shore can a small craft with a draught of 3 feet come without running aground? [Note: 2,000 yards must be converted to feet before working the problem.] 1. 90 ft 2. 150 ft 3. 160 ft 4. 180 ft

9-22. A ship's radar screen has a target blip showing inside the 20,000-yard range circle. If the radius of the 20,000-yard range circle is 5 inches, and the blip is 3¼ inches from the center of the screen, how far is the target from the ship? 1. 11,000 yd 2. 13,000 yd 3. 15,000 yd 4. 17,000 yd

53

9-23. Which of the following is a situation in which x varies directly as y?

1. x = 4 y

3. xy = 6

2. x = xy

4. x = 7 y 16

9-24. If x is directly proportional to y as in x = ky, the expression k is called the 1. mean proportional 2. arbitrary constant 3. third proportional 4. constant of proportionality

9-25. How may the rule that the perimeter (P) of a square is directly proportional to the length of the side (S) be expressed?

1. P = 4S

3. P = kS²

2. P = k/S

4. P = k + S

9-26. As x increases, y decreases but the product xy is constant, is an expression of inverse variation involving variables x and y.

9-27. Which of the following expressions illus- trates inverse variation?

1. x = ky

3. x = k y

2. kx = y

4. x = k + y

9-28. If 8 men can do a job in 7 days, 24 men can do the same job in how many days?

1. 2⅓

3. 14

2. 3

4. 21

9-29. How does F vary in the equation F = kmm' d² 1. Jointly as m and m', and inversely as k 2. Jointly as m and m', and inversely as d² 3. Jointly as k and d, and inversely as m 4. Jointly as k and d², and inversely as m and m'

9-30. Which of the following is an example of joint variation?

1. A = LW

3. C = 2πD

2. A = πr²

4. P = 4S

9-31. The power dissipated by a resistor varies directly as the square of the applied volt- age and inversely as the magnitude of the resistance. How will the amount of dissi- pated power change if the resistance is doubled and the voltage is halved?

1. It will increase to twice its original value. 2. It will increase to four times its original value. 3. It will decrease to one-eighth its original value. 4. It will decrease to one-fourth its original value.

9-32. What is the dependent variable in the formula for the area of a circle, A = πr²?

1. r

3. π

2. r²

4. A

9-33. In standard practice for any formula of the form y = 2z, y is the independent variable.

9-34. If the length of a rectangle is tripled and the width is multiplied by five, the area is multiplied by

1. 3

3. 8

2. 5

4. 15

9-35. What is meant by the notation y = f(x)? 1. y is f times x. 2. y is bigger than x. 3. y is a function of x. 4. y is not related to x.

9-36. For the function y = 1, which of the fol- x lowing statements is true? 1. If x is halved, y is halved. 2. If x is doubled, y is halved. 3. If x is doubled, y is doubled. 4. If x is increased, y is increased.

9-37. In the formula

X₀ = 10⁶ 2πfC ,

which of the following changes will re- sult in a decrease in X₀? 1. A decrease in C 2. A decrease in f 3. An increase in C 4. Each of the above

54

9-38. What is the subject of the formula

V = 4 πr³? 3

1. r

3. --1-- 3πr³

2. r³

4. V

9-39. A formula differs from an equation in that the subject of a formula normally appears 1. on both sides of the equality sign 2. together with other variables 3. without subscripts 4. only once

9-40. The symbols R₁ and R₂ differ and are read R prime and R double prime.

9-41. Which of the following formulas can be derived from the formula

V = 1 πr²h? 3

1. r = √(3V) √h

3. π = 3V rh

2. h = 3 πr²V 3

4. r = 2 Vhπ 2

9-42. How should the formula

T = r² + L 4

be rewritten so that the new subject is r?

1. ± 2√T + L

3. 4(T - L)

2. ± 2√T - L

4. ± 4√T - L

9-43. What is the value of A given by the formula A = 2s² + 4sh if s = 19 and h = 25?

1. 2,622 2. 2,648 3. 2,662 4. 2,688

9-44. If e = e₁R₂ , and e = 75, e₁ = 120, o R₁ + R₂ o and R₂ = 5,000, what is the value of R₁?

1. 1,000 2. 2,000 3. 3,000 4. 4,000

9-45. When L equals length and H equals height, both in feet, R equals gallons required per one square foot of area, and Q equals quantity of paint in gallons, what formula should be used in estimating the total quantity of paint required to paint a large bulkhead?

1. Q = LH + R

3. Q = LHR

2. Q = LR + H

4. Q = __LH__ R

9-46. If V = the volume of a cylinder, h = its height, and r = its radius, which is the formula which indicates that the volume of a cylinder is equal to π times the square of the radius times the height?

1. V = πr² h

3. V = πrh²

2. V = r²h π

4. V = πr²h

Height in feet (H) Time in seconds (t)

2,600 5 4,400 10 5,400 15 5,600 20

Table 9A.--Height-time relationships.

9-47. Table 9A shows the relationship between the height reached by a shell fired with an upward velocity of 600 feet per second and the time of flight. Which formula expresses this relationship?

1. H = 600t + 16t² 2. H = 600t - 16t² 3. H = 600c² + 16t 4. H = 600c² - 16t

55

9-48. The formula R₁R₂ Rₜ = ------- R₁ + R₂

shows the relationship between the total resistance Rₜ of a parallel circuit and the two individual resistances, R₁ and R₂. When you solve this formula for R₁, you get a formula that says R₁ is equal to the 1. sum of Rₜ and the total resistance di- vided by the product of R₂ and the total resistance 2. sum of R₂ and the total resistance di- vided by R₂ minus the total resistance 3. product of R₂ and the total resistance divided by the sum of R₂ and the total resistance 4. product of R₂ and the total resistance divided by R₂ minus the total resistance

9-49. Which of the following completes the table?

│ V │ O │ π/3 │ 4π/3 │ 3π │ 16π/3 │ 25π/3 │ ? │ ├───┼───┼─────┼──────┼────┼───────┼───────┼───┤ │ r │ O │ 1 │ 2 │ 3 │ 4 │ 5 │ 6 │

1. 10π 3. 35π/3

2. 12π 4. 40π/3

9-50. Which of the following statements is re- flected by the formula A = 1/2(b₁ + b₂)h?

1. The area of a figure is twice the height and the sum of the bases. 2. The area of a figure equals one half of the height times the sum of base 1 and base 2. 3. The area of a figure is two less than the height multiplied by the sum of the bases. 4. The area of a figure equals the height multiplied by base 1 added to one half of base 2.

[GRAPH: Y-axis labeled "T (TIME IN HOURS)" with scale 0-4, X-axis labeled "C (WEIGHT OF CARGO IN TONS)" with scale 0-50. Linear relationship shown starting near origin]

Figure 9A.--Graph showing weight-time relationships.

9-51. The graph in figure 9A shows that the time required to load a certain type of cargo varies with the amount loaded. What formu- la for the time consumed can be derived from the graph?

1. T = 1/2(C + 1) 3. T = 1/2(C/10 + 1)

2. T = 1/2(C + 1/2) 4. T = 1/2(C/10 - 1)

9-52. Figure 14-5 in your textbook depicts two ships leaving port at the same time. If the last ship leaves port at 1000 hours on a constant heading and the 2nd ship leaves port at 1100 hours on a different constant heading, how many miles will each ship have traveled when they are the same distance from port?

1. 10 miles 2. 30 miles 3. 50 miles 4. 65 miles

9-53. The real number system consists of 1. real and imaginary numbers 2. complex, rational, and irrational 3. integers, irrational numbers, and complex numbers 4. rational, irrational, positive and negative numbers

9-54. An imaginary number occurs in the solution of the equation x² - 9 = 0.

9-55. In the imaginary concept, both i and j are used to represent the square root of minus 1.

9-56. Reduce √-27 to its simplest form. 1. i√27 2. 3√i 3. 3i√3 4. -3√3

9-57. When the expression √-98 is reduced to its simplest form, what is its appearance? 1. i√98 3. 7i√2

2. 7√2 4. 7√2i

9-58. Imaginary numbers are a product of the imagnation and have no physical meaning.

9-59. Which of the following numbers is real and greater than zero? 1. i 2. i² 3. i³ 4. i⁴

56

9-60. Which of the following equations yields an imaginary number when solved for x? 1. x(x + 2) = 2x + 5 2. x(x + 2) = 2x - 5 3. x(2x + 1) = x + 1 4. x(2x - 1) = 1 - x

9-61. What is the result when the expression i¹⁰ is reduced to its basic value? 1. -i 2. i 3. -1 4. 1

9-62. The simplest method of expressing i⁴³ is 1. -i 2. i 3. -1 4. 1

9-63. Operation with -1 is equivalent to a rotation of how many degrees? 1. 0 2. 90 3. 180 4. 360

9-64. The rotation resulting from multiplication by 2i is twice the rotation resulting from multiplication by i.

9-65. Multiplying a number successively by i four times results in a rotation of how many degrees? 1. 60 2. 90 3. 180 4. 360

9-66. In the complex plane, the vertical axis is called the axis of imaginaries.

● In answering items 9-67 through 9-69, refer to figure 9B.

AXIS OF IMAGINARIES A↑ │ ●B │ C──┼──D AXIS OF │ REALS │ ●E│ │●F

Figure 9B.--Numbers plotted in the complex plane.

9-67. The product of which two numbers is a real number less than zero? 1. A and D 2. A and F 3. C and D 4. C and F

9-68. Which number is a pure imaginary? 1. A 2. B 3. C 4. D

9-69. Which number is of the form a + bi, where both a and b are less than zero and the number a is a real part and bi is the imaginary part of the complex number? 1. A 2. B 3. D 4. E

9-70. In which quadrant is the complex number -3 + 4i plotted? 1. First 2. Second 3. Third 4. Fourth

9-71. The complex number 0 + 3i is a pure imaginary.

9-72. What is the proper procedure for finding the length of the vector representing a number in the complex plane? 1. Add the real and imaginary coefficients and square the sum. 2. Multiply the real and imaginary coef- ficients and take the square root of the product. 3. Square the real and imaginary coeffi- cients, add the squares, and take the square root of this sum. 4. Square the imaginary coefficient, multi- ply by the real coefficient, and take the square root of this product.

9-73. A vector represents both direction and magnitude.

9-74. The length of a vector represented by -3 -4i is 1. 3 units 2. 4 units 3. 5 units 4. 7 units

9-75. Refer to figures 15-8 and 15-9 in your textbook. If a vector is designated by the value -6 + 8i, where is it located? 1. Left of the Y axis and above the X axis 2. Left of the Y axis and below the X axis 3. Right of the Y axis and above the X axis 4. Right of the Y axis and below the X axis

57

Assignment 10

Complex Numbers; Quadratic Equations in One Variable; Plane Figures

Textbook Assignment: Chapters 15 (164-166), 16, 17 (181-186)

10-1. The coefficient of the imaginary part of the sum of two complex numbers is equal to 1. sum of the real and imaginary coeffi- cients of the two complex numbers 2. sum of the real coefficients of the imaginary parts of the two complex numbers 3. difference of the imaginary coeffi- cients of the two complex numbers 4. product of the imaginary coefficients of the two complex numbers

10-2. What is the sum of 1 + i and 7 - 6i? 1. -6 + 7i 2. 7 + i 3. 8 - 5i 4. 8 - 6i

10-3. What is the product of 1 + √-7 and -3 - √-11 ? 1. -3 + √77 - i(3√7 + √11) 2. -3 - √77 + i(3√7 + √11) 3. -3 + √77 + i(3√7 - √11) 4. -3 - √77 +i(-3√7 + √11)

10-4. What is the simplified product of 3 + i and 3 - i? 1. 8 2. 9 + i² 3. 9 - 6i 4. 10

10-5. How is the conjugate of a complex number formed? 1. By changing the sign of the real part 2. By changing the sign of the imaginary part 3. By multiplying the real part by i 4. By multiplying the imaginary part by i

10-6. Both the sum and product of two conju- gate complex numbers are real numbers.

10-7. Which of the following numbers can be represented as the product of two con- jugate complex numbers? 1. 3 + 12i 2. 8 - 12i 3. 8 4. 12i

10-8. When dividing one complex number by another you should first multiply the 1. dividend and divisor by the conjugate of the dividend 2. dividend and divisor by the conjugate of the divisor 3. dividend and divisor by i 4. dividend and divisor by 1 - i

10-9. Which of the following numbers is equal to 8 - i ? 2 + i

1. 1 + i 2. 1 - 3i 3. 2 + 2i 4. 3 - 2i

10-10. When the expression

2 + i i

is simplified, what is the result? 1. 2i 3. 2i + 1 2. 1 - 2i 4. 2i - 1

10-11. What determines the degree of an equa- tion that contains various powers of x, but no other variables? 1. The number of terms in the equation 2. The number of different powers of x that appear in the equation 3. The highest power of x that appears in the equation 4. The coefficient of the highest power of x that appears in the equation

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10-12. What is the coefficient of the constant term in the equation

2x⁰ - 5x² + 3x¹ = 0?

1. +2 2. -5 3. +3 4. 0

10-13. What are the coefficients of the equa- tion

½(x² - 12) + 2x = x² - 38 2

when it is simplified and put into gen- eral form? 1. a = 1, b = -1, c = 2 2. a = 1, b = -1, c = -2 3. a = 1, b = 2, c = 1 4. a = -1, b = -2, c = 1

10-14. The expression ax² + bx + c = 0 repre- sents a quadratic equation except when 1. a = c 2. b = 0 3. b = c 4. b = 0

10-15. If an equation contains only the single variable x, the number of solutions is determined by the 1. number of terms in the equation 2. number of different powers of x that appear in the equation 3. highest power of x that appears in the equation 4. coefficient of the highest power of x that appears in the equation

10-16. The equation x³ - 8 = 0 has how many roots? 1. One 2. Two 3. Three 4. Four

10-17. Which of the following values of x are roots of the equation

x² - 8x + 15 = 0?

1. x = 1 and x = -7 2. x = 2 and x = -5 3. x = -2 and x = 6 4. x = 3 and x = 5

10-18. The factoring method of solving quadratic equations is based upon the fact that 1. every equation can be factored 2. every quadratic equation can be factored 3. the product of two factors is zero only if at least one of the factors is zero 4. the product of two factors is always greater than zero

10-19. What are the two roots of x² - 100 = 0? 1. 2, -50 3. 10, 10

2. 5, 20 4. 10, -10

10-20. Assume that you wish to solve a second degree equation by the factoring method. What is your next step after you sepa- rate the equation into its factored form? 1. Set the product of the factors equal to zero and solve for x. 2. Set each factor equal to zero and solve both equations for x. 3. Set the factors equal to each other and solve for x. 4. Set the factors equal to the original equation and solve for x.

10-21. To solve the quadratic equation

3x² - 17x - 28 = 0

by the factoring method, the equation is first factored into 1. (x - 7)(3x + 4) = 0 2. (x + 7)(3x - 4) = 0 3. (3x - 7)(x + 4) = 0 4. (3x + 7)(x - 4) = 0

10-22. One solution of the equation x² - 0.07x + 0.0006 = 0 is x = 0.01.

10-23. What are the roots when the quadratic equation

2x - 48 = -x²

is factored?

1. 12, 4 3. 4, -12

2. 6, 8 4. -8, 6

10-24. What is the relationship between the first-degree term and the constant term in the perfect square trinomial

x² + bx + c = 0?

1. The constant term is the square of the coefficient of the first-degree term. 2. The constant term is the square of ½ the coefficient of the first degree term. 3. The coefficient of the first-degree term is the square root of the con- stant term. 4. There is no predictable relationship between the first-degree term and the constant term.

10-25. What happens to the original constant term of a quadratic equation when you solve the equation by completing the square? 1. It is squared. 2. It is divided by 2 and squared. 3. It is multiplied by half the coeffi- cient of the x term. 4. It is placed on the right side of the equation.

59

10-26. When you solve the equation x² + 8x - 9 = 0

by the method of completing the square, what number do you add to both sides of the equation after you put the constant term in its proper position? 1. 8 2. 16 3. 32 4. 64

10-27. When you solve a quadratic equation by the method of completing the square, you must first make the coefficient of the x² term equal to 1. zero 2. one 3. half the coefficient of the x term 4. the square of half the coefficient of the x term

10-28. When the quadratic equation

x² + 6x - 1 = 0

is solved by completing the square, what results after taking the square root of both sides of the equation? 1. x + 3 = √10 2. x + 3 = ±√10 3. x + 3 = ±√10 4. (x + 3)² = √10

10-29. When the quadratic equation x²+ 3x + 3/4 is completed by completing the square, what are the roots of the equation? 1. 3/2 ± √31 3. ± 3/2

2. 3/2 ± 31/12 4. -3/2 ± 1/2√(3/1)

10-30. Completing the square, like factoring, cannot be used to solve every quadratic equation.

10-31. What number appears under the radical sign when you use the quadratic formula to solve the equation

2x² - 11x + 1 = 0?

1. 37 3. 113

2. 81 4. 146

10-32. When the equation ax² + bx + c = 0 is solved by completing the square, the resulting quadratic formula represents the solution of all quadratics.

10-33. Which of the following is derived when the equation

3x² - 5x + 4 = 0

is solved? 1. 5 + i √23 3. 5 + i √23 6 6

2. 5 - i √23 4. 5 + √73 6 6

10-34. Which of the following may be used to solve the quadratic equation

2x² - x - 2 = 0? 1. 1 + √17 3. -1 - √17 4 4

2. 1 - √17 4. Both 1 and 2 above 4

10-35. A quadratic equation having real roots may be solved by which of the following methods? 1. Completion of the square and graphing 2. Completion of the square and factor- ing 3. The quadratic formula and graphing 4. All of the above

10-36. In order to graph the expression 2x² + 4x + 3, it is first necessary to 1. divide the expression by x² 2. subtract 3 from the expression 3. divide the expression by 2x 4. let 2x² + 4x + 3 equal a second variable

10-37. The roots of ax² + bx + c = 0 lie on the graph of the equation y = ax² + bx + c at the points where 1. x = y 2. the graph has a maximum or minimum 3. the graph crosses the x-axis 4. the graph crosses the y-axis

10-38. Which of the following statements de- scribes the curve of the equation y = -4(3x + 1) - 5x²?

1. The curve opens upward and crosses the x-axis at 2 points to the left of the y-axis. 2. The curve opens upward and crosses the x-axis at 2 points to the right of the y-axis. 3. The curve opens downward and crosses the x-axis at 2 points to the left of the y-axis. 4. The curve opens downward and crosses the x-axis at 2 points to the right of the y-axis.

60

10-39. When the value of a in the quadratic

y = ax² + bx + c

is negative, the parabola formed by graphing has a minimum value.

● In answering items 10-40 through 10-42, refer to pages 174 and 175 of the text and the following information.

The relationship between the time of flight (in seconds) and the altitude (in feet) of a projectile is given approximately by the quadratic formula

a = vt - 16t²

where v is the muzzle velocity of the projec- tile. Assuming that a gun having a muzzle velocity of 400 ft per second is fired, the formula for altitude becomes

a = 400t - 16t²

10-40. How long after the gun is fired will the projectile hit a surface target? 1. 20 sec 2. 25 sec 3. 30 sec 4. 35 sec

10-41. The negative coefficient of the t² term of the formula indicates that the pro- jectile will have a 1. Maximum altitude 2. Minimum altitude 3. Constantly decreasing altitude 4. Constantly decreasing speed

10-42. What will be the maximum altitude reached by the projectile? 1. 750 ft 2. 1,500 ft 3. 1,750 ft 4. 2,500 ft

10-43. What is the smallest value that y can have if y = x² + 10x + 32? 1. -3 2. 1 3. 5 4. 7

10-44. What is the x-coordinate of the point on the graph of the equation y = 3x² - 2x + 17 that is closest to the x-axis?

1. 1/4 3. 2/3

2. 1/3 4. 3/4

10-45. The general form of the quadratic equa- tion has imaginary roots whenever 1. b² is less than 4ac 2. b² is greater than 4ac 3. 4ac is less than zero 4. 4ac is greater than zero

10-46. What is the other root of a quadratic equation when one of its roots is 2 + i√3? 1. -2 + 3i 2. 2 - i√3 3. 2 - i√3 4. -2 - √3

10-47. Which of the following statements de- scribes the curve of a second degree equation with a discriminant that is equal to zero and an x² term coefficient that is positive? 1. The curve reaches a maximum below the x-axis. 2. The curve reaches a maximum on the x-axis. 3. The curve reaches a minimum above the x-axis. 4. The curve reaches a minimum on the x-axis.

10-48. A quadratic equation that can be separated into two identical factors always has a discriminant that is 1. less than zero 2. equal to zero 3. greater than zero 4. not a perfect square

10-49. Which of the following sets of co- efficients will give the equation

ax² + bx + c = 0

roots that are rational, unequal, and that do not contain an imaginary term? 1. a = 2, b = 9, c = 7 2. a = -2, b = 2, c = 8 3. a = 4, b = 6, c = 5 4. a = 6, b = -3, c = -1

10-50. The roots of x² + 4x + 4 = 0 are 1. equal 2. unequal 3. positive 4. imaginary

10-51. The roots of the quadratic equation

x² + x + 1 = 0

are 1. real 2. equal 3. rational 4. imaginary

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10-58. It two lines intersect and form four equal angles, the lines are said to be

1. oblique 3. concurrent 2. parallel 4. perpendicular

10-59. An acute angle may be defined as an angle of

1. 90 degrees 3. less than 90 degrees 2. 180 degrees 4. more than 90 degrees

10-60. A straight angle is an angle of

1. 90 degrees 3. less than 180 degrees 2. 180 degrees 4. more than 180 degrees

10-61. In figure 17-6 in your textbook, which of the following angles are called vertical angles?

1. 1 and 2 3. 2 and 3 2. 1 and 3 4. 3 and 4

10-62. Which of the following angles are comple- mentary?

1. 43 degrees and 47 degrees 2. 60 degrees and 60 degrees 3. 90 degrees and 30 degrees 4. 100 degrees and 80 degrees

10-63. Which angle has a value twice its own supplement?

1. 60 degrees 3. 100 degrees 2. 80 degrees 4. 120 degrees

10-64. A square cannot be classified as a polygon because it only has four equal angles.

10-65. Which of the following is not a part of a triangle?

1. Arc 3. Base 2. Apex 4. Vertex

10-66. The altitude of any triangle, when drawn, will always lie inside the triangle.

10-67. What is the area of a triangle whose base is 2 feet and whose height is 8 inches?

1. 8 sq ft 3. 96 sq in 2. 16 sq ft 4. 192 sq in

y \ / \ B / \ / \ \ / / \ A \ / D / x \ \ C / / \___________________/

Figure 10A.--Graph of four equations.

● In answering 10-52 and 10-53, refer to figure 10A.

10-52. Which curve is the graph of an equation that has a double root?

1. A 3. C 2. B 4. D

10-53. Which graph is the graph of an equation that has zero as one of its roots?

1. A 3. C 2. B 4. D

10-54. How many points do the x-axis and the graph of the equation

y = ax² + bx + c

have in common when

b² = 4ac?

1. None 3. Two 2. One 4. An infinite number

10-55. How many points, if any, do the x-axis and the graph of a quadratic equation have in common when the discriminant of the equation is less than zero?

1. None 3. Two 2. One 4. An infinite number

10-56. Which of the following is a line segment?

1. ●_______●

2. /\ / \

3. ●_____ ___●

4. ~~~~ ~~~

10-57. A part BC of the circumference of a circle is designated as

1. broken line BC 3. arc BC 2. dashed line BC 4. line BC

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10-70. If two sides of a triangle are 8 units each, the triangle is classified as

1. right 2. scalene 3. isosceles 4. equilateral

10-71. An equilateral triangle is also an isosceles triangle.

10-72. A right triangle with a 10-degree angle also includes

1. an obtuse angle 2. a supplementary angle 3. a 60 degree angle 4. an 80 degree angle

10-73. If a diagonal of any quadrilateral is drawn, it always divides the quadri- lateral into two

1. equal triangles 2. triangles having equal bases 3. triangles neither of which is isosceles 4. triangles neither of which is equilateral

10-74. A quadrilateral is a special example of a parallelogram.

10-75. Rhombus is the name given to a parallelogram whose four sides have equal length.

|----5 in.---|---4 in.---| |D /B /A | /90° / | / / 8 in| / / | / / | / / | / / |_____/___________/ C

Figure 10B

● In answering item 10-68 refer to figure 10B. [Hint: Area of triangle ABC + area of triangle BCD = area of triangle ACD]

10-68. What is the area of triangle ABC?

1. 16 sq in. 2. 20 sq in. 3. 36 sq in. 4. 72 sq in.

10-69. When a triangle has sides of 15, 20, and 25 units, it is classified as which type?

1. Right 2. Acute 3. Isosceles 4. Equilateral

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Assignment 11

Plane Figures; Geometric Construction and Solid Figures; Numerical Trigonometry

Textbook Assignment: Chapters 17 (186-189), 18, 19

11-1. Refer to Figure 17-14 in your textbook. In the parallelogram, which of the following is true? 1. Angle DAB equals angle BCD. 2. AD equals BC. 3. DC is parallel to AB. 4. Each of the above is true.

11-2. A rectangle is a parallelogram whose four angles are 90° each.

11-3. Since every square is a rectangle and every rectangle is a parallelogram, it logically follows that every parallelo- gram is a square.

11-4. The formula for calculating the area of a parallelogram is 1. A = bh 2. A = ²/₂bh 3. A = 2h + 2b 4. A = ½h(b₁ + b₂)

11-5. Refer to figure 17-15 in your textbook. If the nonparallel sides of a trapezoid are extended until they meet, a triangle is formed.

11-6. In square units, what is the area of a trapezoid whose height is 8 units and whose bases are 10 and 12 units respec- tively? 1. 88 2. 100 3. 108 4. 176

11-7. Which of the following terms does not apply to a circle? 1. Circumference 2. Diameter 3. Chord 4. Side

11-8. The diameter of a circle is a chord of the circle.

11-9. That part of a circle cut off by two radii is called 1. an arc 2. a sector 3. a portion 4. a segment

11-10. When π is used in calculations, its value is considered to be 1. exactly 3.14 2. exactly the diameter divided by the circumference 3. approximately the radius times the circumference 4. approximately 3.14

11-11. A circle, whose radius is 10 units, has a circumference of 1. 78.5 units 2. 31.4 units 3. 62.8 units 4. 314.0 units

11-12. A circle whose diameter is 10 units has an area of 1. 31.4 square units 2. 78.5 square units 3. 157.0 square units 4. 314.0 square units

11-13. Two concentric circles have radii of 5 in. and 10 in. respectively. The area of the ring between the two circles is 1. 78.5 sq in. 2. 235.5 sq in. 3. 225.0 sq in. 4. 942.0 sq in.

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11-20. Any radius may be chosen to draw the circle whose center is located at point O.

11-21. Line EC is drawn before line ED.

11-22. The angle formed by lines EC and CD is a right angle.

11-23. Refer to figure 18-5 in your textbook and assume that the intersection of the two construction arcs between points A and B is point X. In bisecting the angle AOB, point X is determined by which successive steps? 1. Carefully draw the angle AOX so that it is exactly equal to the angle BOX. 2. First construct the segment OC to equal the segment CX, then construct OD equal to DX. 3. First construct the segment OC to equal the segment OD, then construct CX equal to DX. 4. First construct the segment OC to equal the segment CX, then construct CX to be slightly greater than DX.

11-24. Any triangle whose sides are in the ratio of 3 to 4 to 5 is a right tri- angle.

11-25. Using a compass and ruler to construct a 60-degree angle, which of the follow- ing instructions should you follow? 1. Trisect a straight angle. 2. Draw a 3-4-5 right triangle. 3. Draw an equilateral triangle. 4. Bisect an angle of an equilateral triangle.

11-26. Using a compass and ruler, which of the following operations could you perform? 1. Bisect a right angle. 2. Construct a right triangle contain- ing two equal sides. 3. Bisect a 30-degree angle. 4. You could do each of the above.

11-27. To find the center of a circle, how many perpendicular bisectors of chords of the circle must be drawn? 1. One 2. Two 3. Three 4. Four

11-28. The perpendicular bisector of any chord of a circle passes through the center of the circle.

H | | C |D |\_ |/ B |\_ |/ | \_ |/ A \|/ |________|________|________|____ F A' B' C' D' G

Figure 11A.--Dividing line FG into equal segments.

● In answering items 11-14 through 11-16 refer to figure 11A.

11-14. When figure 11A was being constructed, the line segment FH was drawn. When line segment FG, the line to be divided, had been drawn, what was the next step? 1. Points C and H were connected. 2. Lines parallel to HC were drawn. 3. Line FH was marked off into 5 equal segments. 4. Line FG was marked off into 5 equal segments.

11-15. Relative to the figure, which of the following statements is correct? 1. FA equals AB equals BC. 2. FA' equals A'B' equals B'C'. 3. HC is parallel to BB'. 4. Each of the above is correct.

11-16. The line segment FH can be longer than the given line segment FG.

11-17. When a line segment is to be bisected by the geometric method, the arcs to be employed must have a radius of what length? 1. Half as long as the line segment 2. Twice as long as the line segment 3. More than half the length of the line segment 4. Less than half the length of the line segment

11-18. The instrument used in mathematical construction to draw circles is called a compass.

● In answering items 11-19 through 11-22, refer to figure 18-4 in your textbook.

11-19. In order to construct a perpendicular to line segment AB at point C, point O was chosen as a convenient point above the line at which to begin the construction.

65

11-29. Refer to figure 18-9 in your textbook. In constructing the ellipse, the dis- tances ab and ac are equal respectively to 1. AB, DC 2. DC, AB 3. ½AB, ½DC 4. ½DC, ⅔AB

11-30. A cylinder is one example of a prism.

11-31. Which of the following gives a particu- lar type prism its name? 1. Lateral faces 2. Edges 3. Base 4. Size

11-32. How many faces does a parallelepiped have? 1. 5 2. 6 3. 8 4. Either 6 or 8

11-33. What is the total surface area of a triangular prism eight inches long with each base a right triangle whose sides measure three inches, four inches, and five inches, respectively? 1. 96 sq in. 2. 96 sq in. 3. 102 sq in. 4. 108 sq in.

11-34. Refer to figure 18-11 in your text. A right prism whose dimensions are 3 in. by 4 in. by 1 ft has a volume of how many cu in.? 1. 12 cu in. 2. 19 cu in. 3. 24 cu in. 4. 144 cu in.

11-35. A straight line moving at right angles to its length and such that its lower end traces a closed curve always generates a 1. solid 2. prism 3. cylinder 4. circular cylinder

11-36. What is the volume of a right circular cylinder whose base has a radius of 10 inches and whose height is 21 inches? 1. 1318 cu in. 2. 2198 cu in. 3. 6594 cu in. 4. 8792 cu in.

11-37. What is the lateral area of a pyramid with a 6-sided base measuring 8 inches on a side if the slant height is 23 inches? 1. 600 sq in. 2. 800 sq in. 3. 1200 sq in. 4. 1600 sq in.

11-38. Find the lateral area of a right circu- lar cone whose slant height is 10 inches and whose base has a radius of 8 inches. 1. 160 sq in. 2. 251.2 sq in. 3. 502.4 sq in. 4. 2009.6 sq in.

11-39. How much material is needed to make both the base and lateral surface of a right circular cone whose base has a radius of 5 inches and whose slant height is 8 inches? 1. 135.60 sq in. 2. 175.20 sq in. 3. 178.50 sq in. 4. 204.10 sq in.

11-40. What is the volume of a right circular cone whose height is 9 inches and whose base has a diameter of 2 inches? 1. 9.42 cu in. 2. 28.26 cu in. 3. 37.68 cu in. 4. 56.52 cu in.

11-41. Which of the following solid figures has the greatest volume? 1. Sphere, with a radius of 2 inches 2. Right rectangular prism, with dimen- sions of 2, 3, and 4 inches 3. Right cylinder, with base radius of 2 inches and height of 2 inches 4. Pyramid, with base a square of 3 inches on a side and height of 7 inches

11-42. The volume of a sphere whose radius is m is given by the formula 1. 4πm² 2. ⅔πm² 3. ⁴⁄₃πm³ 4. ⁴⁄₃m³

11-43. The amount of material required to cover a spherical ball whose radius is 2 inches is 1. 12.56 sq in. 2. 25.12 sq in. 3. 33.49 sq in. 4. 50.24 sq in.

11-58. What is the value of angle A when line BC = 8.693 units and line AC = 10 units? 1. 41° 2. 42° 3. 51.9° 4. 60.4°

11-59. The sine of an angle of 49 degrees, 48 minutes is 1. 0.6455 2. 0.7536 3. 0.7638 4. 1.1833

11-60. The value of the tangent of an angle of 18 degrees 18 minutes is 1. 0.3288 2. 0.3249 3. 0.3307 4. 0.9494

11-61. Using the method of interpolation, what is the angle whose sine is 0.1573? 1. 8.5° 2. 8.0° 3. 9°2' 4. 9°3'

11-62. What is the angle whose cosine is 0.4186, rounded to the nearest minute? 1. 65°20' 2. 65°22' 3. 65°23' 4. 65°24'

11-63. Using interpolation, what is the tangent of 37 degrees, 21 minutes? 1. 0.7618 2. 0.7632 3. 0.7640 4. 0.7646

11-64. Using interpolation, what is the value of sin 16.58°? 1. 0.2845 2. 0.2849 3. 0.2854 4. 0.2856

11-65. In a 30° -60° -90° triangle, the hypotenuse is twice the side opposite the 60° angle.

● The legs of a right triangle are defined to be those two sides of a right triangle which lie opposite acute angles.

11-66. In a 30° -60° -90° triangle, if the shorter of the two legs is 8 inches what is the length of the longer leg? 1. 16 in. 2. 16√3 in. 3. 12√3 in. 4. 8√3 in.

11-67. What is the altitude of an equilateral triangle whose sides are 10 inches? 1. 5 in. 2. √70 in. 3. 5√3 in. 4. 5√2 in.

11-68. What is the length of the sides of a 45°-90° triangle whose hypotenuse is 10 units? 1. 5 2. 5√2 3. 5√2/2 4. 10√2

11-69. A triangle with sides of 6, 8, and 10 units respectively is a right triangle.

11-70. Which of the following triangles is not a right triangle? 1. A triangle with sides 2, 2√5, and 4 2. A triangle with sides 3, 3, and 3√7 3. A triangle with sides 4, 6, and 9 4. A triangle with sides 10, 24, and 26

11-71. If the distance from the top of a pole to a point on the ground is feet from its base is 20 feet, what is the height of the pole? 1. 12 ft 2. 15 ft 3. 18 ft 4. 24 ft

11-72. The six trigonometric ratios are based on what type triangle? 1. Acute 2. Obtuse 3. Oblique 4. Right

● In answering items 11-73 through 11-75 refer to the trigonometric functions in Appendix II.

11-73. Refer to figure 19-18 in your text. If angle A is 30 degrees and angle C is 40 degrees, what is the length of side AC if side BC is 18 units in length? 1. 32.18 2. 33.83 3. 34.57 4. 35.46

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11-74. Refer to figure 19-19 in your textbook. If angle BAC is 40 degrees and side AC is 80 feet long, what is the length of the side BD of triangle CBD if angle BCD has a value of 60 degrees? 1. 120 ft 2. 125.7 ft 3. 130.2 ft 4. 132.2 ft

11-75. Refer to figure 19-21 (A) in your text- book. When angles A, B, and C are 50, 70, and 60 degrees respectively and side b is 12 units, what is the value of side c? 1. 8.4 2. 10.9 3. 11.1 4. 11.7

*U.S. Government Printing Office: 1996 - 532-181/40072 69