basic algebraic operations - pearson...

1Basic Algebraic Operations

Interest in things such as the land on whichthey lived, the structures they built, andthe motion of the planets led people in

early civilizations to keep records and to cre-ate methods of counting and measuring. Inturn, some of the early ideas of arithmetic,geometry, and trigonometry were developed.From such beginnings, mathematics hasplayed a key role in the great advances in sci-ence and technology.

Often, mathematical methods were developedfrom studies made in sciences, such as astron-omy and physics, to better describe and un-derstand the subject being studied. Some ofthese methods resulted from the needs in a particular area of application.

Many people were interested in the mathitself, and added to what was then known.Although this additional mathematicalknowledge may not have been related toapplications at the time it was developed, itoften later became useful in applied areas.

In the chapter introductions that follow,examples of the interaction of technology andmathematics are given. From these examplesand the text material, it is hoped you will bet-ter understand the important role that mathhas had and still has in technology. In thistext, there are applications from technologiesincluding (but not limited to) aeronautical,business, communications, electricity, elec-tronics, engineering, environmental, heat andair conditioning, mechanical, medical, mete-orology, petroleum, product design, solar, andspace. To solve the applied problems in thistext will require a knowledge of the mathemat-ics presented but will not require prior knowl-edge of the field of application.

We begin by reviewing the concepts that dealwith numbers and symbols. This will enableus to develop topics in algebra, an under-standing of which is essential for progress inother areas such as geometry, trigonometry,and calculus.

1.1 Numbers

1.2 Fundamental Operationsof Algebra

1.3 Calculators and ApproximateNumbers

1.4 Exponents

1.5 Scientific Notation

1.6 Roots and Radicals

1.7 Addition and Subtraction ofAlgebraic Expressions

1.8 Multiplication of AlgebraicExpressions

1.9 Division of AlgebraicExpressions

1.10 Solving Equations

1.11 Formulas and LiteralEquations

1.12 Applied Word Problems

CHAPTER EQUATIONS

REVIEW EXERCISES

PRACTICE TEST

1

The Great Pyramid at Giza in Egypt was built about 4500years ago.

In the 1500s, 1600s, and 1700s, discoveriesin astronomy and the need for more accu-rate maps and instruments in navigationwere very important in leading scientistsand mathematicians to develop usefulnew ideas and methods in mathematics.

Late in the 1800s, scientists werestudying the nature of light. This ledto a mathematical prediction of theexistence of radio waves, now used inmany types of communication. Also,in the 1900s and 2000s, mathematicshas been vital to the development ofelectronics and space travel.

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2 CHAPTER 1 Basic Algebraic Operations

1.1 Numbers

Real Number System • Number Line •Absolute Value • Signs of Inequality •Reciprocal • Denominate Numbers •Literal Numbers

In technology and science, as well as in everyday life, we use the very familiar count-ing numbers 1, 2, 3, and so on. They are also called natural numbers or positiveintegers. The negative integers and so on are also very necessary anduseful in mathematics and its applications. The integers include the positive integersand the negative integers and zero, which is neither positive nor negative. This meansthe integers are the numbers and so on.

To specify parts of a quantity, rational numbers are used. A rational number is anynumber that can be represented by the division of one integer by another nonzero inte-ger. Another type of number, an irrational number, cannot be written as the divisionof one integer by another.

E X A M P L E 1 Identifying rational numbers and irrational numbers

The numbers 5 and are integers. They are also rational numbers since they can bewritten as and respectively. Normally, we do not write the 1’s in the denominators.

The numbers and are rational numbers because the numerator and the de-nominator of each are integers.

The numbers and are irrational numbers. It is not possible to find two inte-gers, one divided by the other, to represent either of these numbers. It can be shownthat square roots (and other roots) that cannot be expressed exactly in decimal formare irrational. Also, is sometimes used as an approximation for but it is not equal exactly to . We must remember that is rational and is irrational.

The decimal number 1.5 is rational since it can be written as . Any such terminatingdecimal is rational. The number where the 6’s continue on indefinitely, isrational since we may write it as . In fact, any repeating decimal (in decimal form, aspecific sequence of digits is repeated indefinitely) is rational. The decimal number

is a repeating decimal where the sequence of digits 732 is repeatedindefinitely ■

The integers, the rational numbers, and the irrational numbers, including all suchnumbers that are positive, negative, or zero, make up the real number system (seeFig. 1.1). There are times we will encounter an imaginary number, the name given tothe square root of a negative number. Imaginary numbers are not real numbers andwill be discussed in Chapter 12. However, unless specifically noted, we will use realnumbers. Until Chapter 12, it will be necessary to only recognize imaginary numberswhen they occur.

Also in Chapter 12, we will consider complex numbers, which include both the realnumbers and imaginary numbers. See Exercise 37 of this section.

E X A M P L E 2 Identifying real numbers and imaginary numbers

The number 7 is an integer. It is also rational since and it is a real number sincethe real numbers include all the rational numbers.

The number is irrational, and it is real since the real numbers include all theirrational numbers.

The numbers and are imaginary numbers.The number is rational and real. The number is irrational and real.The number is irrational and real. The number is imaginary. ■

1- 32

p6

- 17- 37

- 1-71-10

3p

7 =71,

10.673 273 273 2 Á =112116652.

0.673 273 273 2 Á

23

0.6666 Á ,

32

p227p

p,227

p12

- 113

58

- 191 ,5

1

-19

Á , -3, -2, -1, 0, 1, 2, 3,

-1, -2, -3,

■ A notation that is often used for repeatingdecimals is to place a bar over the digits thatrepeat. Using this notation we can write

and .23 = 0.61121

1665 = 0.6732

■ Real numbers and imaginary numbers areboth included in the complex number system.See Exercise 37.

Real Numbers

Rationalnumbers

Integers

Irrationalnumbers

Fig. 1.1

■ Irrational numbers were discussed by theGreek mathematician Pythagoras in about 540 B.C.E.

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1.1 Numbers 3

A fraction may contain any number or symbol representing a number in its numer-ator or in its denominator. The fraction indicates the division of the numerator by thedenominator, as we previously indicated in writing rational numbers. Therefore, a frac-tion may be a number that is rational, irrational, or imaginary.

E X A M P L E 3 Fractions

The numbers and are fractions, and they are rational.

The numbers and are fractions, but they are not rational numbers. It is notpossible to express either as one integer divided by another integer.

The number is a fraction, and it is an imaginary number. ■

Real numbers may be represented by points on a line. We draw a horizontal line anddesignate some point on it by , which we call the origin (see Fig. 1.2). The integerzero is located at this point. Equal intervals are marked to the right of the origin, andthe positive integers are placed at these positions. The other positive rational numbersare located between the integers. The points that cannot be defined as rational numbersrepresent irrational numbers. We cannot tell whether a given point represents a rationalnumber or an irrational number unless it is specifically marked to indicate its value.

O

1- 56

6p

129

- 32

27

■ Fractions were used by early Egyptians andBabylonians. They were used for calculationsthat involved parts of measurements, property,and possessions.

The Number Line

��2��11 1.7� 26

5194

49

�6 �5 �4 �3 �2 �1 10

OriginNegative direction Positive direction

2 3 4 5 6

Fig. 1.2

The negative numbers are located on the number line by starting at the origin andmarking off equal intervals to the left, which is the negative direction. As shown inFig. 1.2, the positive numbers are to the right of the origin and the negative numbersare to the left of the origin. Representing numbers in this way is especially useful forgraphical methods.

We next define another important concept of a number. The absolute value of a pos-itive number is the number itself, and the absolute value of a negative number is thecorresponding positive number. On the number line, we may interpret the absolutevalue of a number as the distance (which is always positive) between the origin and thenumber. Absolute value is denoted by writing the number between vertical lines, asshown in the following example.

E X A M P L E 4 Absolute value

The absolute value of 6 is 6, and the absolute value of is 7. We write these asand See Fig. 1.3.ƒ -7 ƒ = 7.ƒ6 ƒ = 6

-7Practice Exercises

1. 2. - ` - 3

4` = ?ƒ -4.2 ƒ = ?

0�4 4 8�8

6 units7 units��7� � 7 �6� � 6

Fig. 1.3

Other examples are since ■ƒ -9 ƒ = 9.- ƒ -9 ƒ = -9

ƒ -5.29 ƒ = 5.29,- ƒp ƒ = -p,ƒ0 ƒ = 0,ƒ - 12 ƒ = 12,ƒ75 ƒ =

75,

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On the number line, if a first number is to the right of a second number, then the firstnumber is said to be greater than the second. If the first number is to the left of the sec-ond, it is less than the second number. The symbol designates “is greater than,” andthe symbol designates “is less than.” These are called signs of inequality. See Fig. 1.4.

E X A M P L E 5 Signs of inequality

6

7

■ The symbols , , and were introducedby English mathematicians in the late 1500s.

76=

0�4 4 62�2

3 is to theleft of 6

2 is to the right of

2 � �4 3 � 6

�4 0 � �45 � 9 �3 � �7 �1 � 0

Pointed toward smaller number

Fig. 1.4

Practice Exercises

Place the correct sign of inequality or between the given numbers.

3. 4 4. 0 -3-5

)76(

Every number, except zero, has a reciprocal. The reciprocal of a number is 1 di-vided by the number.

E X A M P L E 6 Reciprocal

The reciprocal of 7 is The reciprocal of is

invert denominator and multiply (from arithmetic)

The reciprocal of 0.5 is The reciprocal of is Note that the negativesign is retained in the reciprocal of a negative number.

We showed the multiplication of 1 and as We could also show it as

or We will often find the form with parentheses is preferable. ■

In applications, numbers that represent a measurement and are written with units ofmeasurement are called denominate numbers. The next example illustrates the use ofunits and the symbols that represent them.

E X A M P L E 7 Denominate numbers

To show that a certain HDTV set weights 28 kilograms, we write the weight as 28 kg.To show that a giant redwood tree is 110 metres high, we write the height as 110 m.To show that the speed of a rocket is 1500 metres per second, we write the speed

as 1500 (Note the use of s for second. We use s rather than sec.)To show that the area of a computer chip is 0.75 square centimetre, we write the

area as 0.75 cm2.(We will not use sq cm.)To show that the volume of water in a glass tube is 25 cubic centimetres, we write

the volume as (We will not use cu cm nor cc.) ■

It is usually more convenient to state definitions and operations on numbers in ageneral form. To do this, we represent the numbers by letters, called literal numbers.For example, if we want to say “If a first number is to the right of a second number onthe number line, then the first number is greater than the second number,” we can write“If is to the right of on the number line, then ” Another example of using aliteral number is “The reciprocal of is ”

Certain literal numbers may take on any allowable value, whereas other literal num-bers represent the same value throughout the discussion. Those literal numbers thatmay vary in a given problem are called variables, and those literal numbers that areheld fixed are called constants.

1>n.na 7 b.ba

25 cm3.

m>s.

1 A 32 B .1 # 3

21 *32.3

2

-1p.-p

10.5 = 2.

123

= 1 *

3

2=

3

2

23

17.

■

■ For reference, see Appendix B for units ofmeasurement and the symbols used for them.

Literal Numbers

➝

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1.1 Numbers 5

E X A M P L E 8 Variables and constants

(a) The resistance of an electric resistor is . The current in the resistor equals thevoltage divided by , written as For this resistor, and may takeon various values, and is fixed. This means and are variables and is a con-stant. For a different resistor, the value of may differ.R

RVIRVII = V>R.RV

IR

(b) The fixed cost for a calculator manufacturer to operate a certain plant is dollarsper day, and it costs dollars to produce each calculator. The total daily cost toproduce calculators is

Here, and are variables, and and are constants, and the product of and is shown as an. For another plant, the values of and would probably differ.

If specific numerical values of and are known, say per calculator andthen Thus, constants may be numerical or literal. ■C = 7n + 3000.b = $3000,

a = $7baba

nabanC

C = an + b

nCa

b

EXERCISES 1.1

In Exercises 1–4, make the given changes in the indicated examplesof this section, and then answer the given questions.

1. In the first line of Example 1, change the 5 to and the to14. What other changes must then be made in the first paragraph?

2. In Example 4, change the 6 to What other changes must thenbe made in the first paragraph?

3. In the left figure of Example 5, change the 2 to What otherchanges must then be made?

4. In Example 6, change the to What other changes must thenbe made?

In Exercises 5 and 6, designate each of the given numbers as being aninteger, rational, irrational, real, or imaginary. (More than one desig-nation may be correct.)

5. 6.

In Exercises 7 and 8, find the absolute value of each number.

7. 8.

In Exercises 9–16, insert the correct sign of inequality ( or ) be-tween the given numbers.

9. 6 8 10. 7 5

11. 12. 0

13. 14.

15. 16. 0.2

In Exercises 17 and 18, find the reciprocal of each number.

17. 3, 18. -

1

3, 0.25, x

y

b-

4

13,

-0.6-

1

2-

1

3

-1.42- 12- ƒ -3 ƒ-4

-4-3.2p

67

-0.857, 12, -

19

43, -4, -

p

2

- 2-6, -2.33, 17

33, 1-4, -

p

6

32.2

3

-6.

-6.

-19-3

In Exercises 19 and 20, locate each number on a number line as inFig. 1.2.

19. 20.

In Exercises 21–44, solve the given problems. Refer to Appendix B forunits of measurement and their symbols.

21. Is an absolute value always positive? Explain.

22. Is 2.17 rational? Explain.

23. What is the reciprocal of the reciprocal of any positive or negativenumber?

24. Find a rational number between and that can be writ-ten with a denominator of 11 and an integer in the numerator.

25. Find a rational number between 0.13 and 0.14 that can be writtenwith a numerator of 3 and an integer in the denominator.

26. If and is ?

27. List the following numbers in numerical order, starting with thesmallest:

28. List the following numbers in numerical order, starting with thesmallest:

29. If and are positive integers and what type of numberis represented by the following?

(a) (b) (c)

30. If and represent positive integers, what kind of number is rep-resented by (a) (b) and (c) ?

31. For any positive or negative integer: (a) Is its absolute valuealways an integer? (b) Is its reciprocal always a rational number?

32. For any positive or negative rational number: (a) Is its absolutevalue always a rational number? (b) Is its reciprocal always arational number?

33. Describe the location of a number on the number line when(a) and (b) x 6 -4.x 7 0

x

a * ba>b,a + b,ba

b - a

b + aa - bb - a

b 7 a,ba

15, - 110, - ƒ -6 ƒ , -4, 0.25, ƒ -p ƒ .

-1, 9, p, 15, ƒ -8 ƒ , - ƒ -3 ƒ , -3.1.

ƒb - a ƒ 6 ƒb ƒ - ƒa ƒa 7 0,b 7 a

-1.0-0.9

-

12

2, 2p, 123

192.5, -

12

5, 13

W

W

W

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34. Describe the location of a number on the number line when(a) and (b)

35. For a number describe the location on the number line ofthe reciprocal of .

36. For a number describe the location on the number line ofthe number with a value of

37. A complex number is defined as where and are realnumbers and For what values of and is the com-plex number a real number? (All real numbers and allimaginary numbers are also complex numbers.)

38. A sensitive gauge measures the total weight of a container andthe water that forms in it as vapor condenses. It is found that

where is the weight of the container and isthe time of condensation. Identify the variables and constants.

39. In an electric circuit, the reciprocal of the total capacitance of twocapacitors in series is the sum of the reciprocals of the capaci-tances. Find the total capacitance of two capacitances of 0.0040 Fand 0.0010 F connected in series.

40. Alternating-current (ac) voltages change rapidly between positiveand negative values. If a voltage of 100 V changes to which is greater in absolute value?

-200 V,

tcw = c10.1t + 1,

w

a + bjbaj = 1-1.

baa + bj,

ƒx ƒ .x 6 0,

xx 7 1,

ƒx ƒ 7 2.ƒx ƒ 6 1x 41. The memory of a certain computer has bits in each byte. Ex-

press the number of bits in kilobytes in an equation. (A bit isa single digit, and bits are grouped in bytes in order to representspecial characters. Generally, there are 8 bits per byte. If neces-sary, see Appendix B for the meaning of kilo.)

42. The computer design of the base of a truss is m long. Later it isredesigned and shortened by cm. Give an equation for thelength , in centimetres, of the base in the second design.

43. In a laboratory report, a student wrote “ ” Is thisstatement correct? Explain.

44. After 5 s, the pressure on a valve is less than 600 kPa. Using torepresent time and to represent pressure, this statement can bewritten “for s, kPa.” In this way, write the state-ment “when the current in a circuit is less than 4 A, the voltage

is greater than 12 V.”

Answers to Practice Exercises

1. 4.2 2. 3. 4. 76-

3

4

VI

p 6 600 t 7 5p

t

-20°C 7 -30°C.

Ly

x

nNa

1.2 Fundamental Operations of Algebra

Fundamental Laws of Algebra • Operationson Positive and Negative Numbers • Orderof Operations • Operations with Zero

If two numbers are added, it does not matter in which order they are added. (For exam-ple, and or ) This statement, generalized andaccepted as being correct for all possible combinations of numbers being added, iscalled the commutative law for addition. It states that the sum of two numbers is thesame, regardless of the order in which they are added. We make no attempt to provethis law in general, but accept that it is true.

In the same way, we have the associative law for addition, which states that the sumof three or more numbers is the same, regardless of the way in which they are groupedfor addition. For example,

The laws just stated for addition are also true for multiplication. Therefore, theproduct of two numbers is the same, regardless of the order in which they are multi-plied, and the product of three or more numbers is the same, regardless of the wayin which they are grouped for multiplication. For example, and

Another very important law is the distributive law. It states that the product of onenumber and the sum of two or more other numbers is equal to the sum of the productsof the first number and each of the other numbers of the sum. For example,

In this case, it can be seen that the total is 30 on each side.In practice, these fundamental laws of algebra are used naturally without thinking

about them, except perhaps for the distributive law.

514 + 22 = 5 * 4 + 5 * 2

5 * 14 * 22 = 15 * 42 * 2.2 * 5 = 5 * 2,

3 + 15 + 62 = 13 + 52 + 6.

5 + 3 = 3 + 5.3 + 5 = 8,5 + 3 = 8

The Commutative and Associative Laws

The Distributive Law

■ Note carefully the difference: associative law: distributive law: 5 * 14 + 225 * 14 * 22

➝➝

W

W

W

W

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1.2 Fundamental Operations of Algebra 7

Not all operations are commutative and associative. For example, division is notcommutative, since the order of division of two numbers does matter. For instance,

( is read “does not equal)”. (Also, see Exercise 50.)Using literal numbers, the fundamental laws of algebra are as follows:

Commutative law of addition:Associative law of addition:Commutative law of multiplication:Associative law of multiplication:Distributive law:

Each of these laws is an example of an identity, in that the expression to the left of thesign equals the expression to the right for any value of each of , , and .

OPERATIONS ON POSITIVE AND NEGATIVE NUMBERSWhen using the basic operations (addition, subtraction, multiplication, division) onpositive and negative numbers, we determine the result to be either positive or negativeaccording to the following rules.

Addition of two numbers of the same sign Add their absolute values and assignthe sum their common sign.

E X A M P L E 1 Adding numbers of the same sign

(a) the sum of two positive numbers is positive

(b) the sum of two negative numbers is negative

The negative number is placed in parentheses since it is also preceded by a plussign showing addition. It is not necessary to place the in parentheses. ■

Addition of two numbers of different signs Subtract the number of smaller ab-solute value from the number of larger absolute value and assign to the result thesign of the number of larger absolute value.

E X A M P L E 2 Adding numbers of different signs

(a) the negative has the larger absolute value

(b)

(c) the positive has the larger absolute value

(d)

the subtraction of absolute values ■

Subtraction of one number from another Change the sign of the number beingsubtracted and change the subtraction to addition. Perform the addition.

E X A M P L E 3 Subtracting positive and negative numbers

(a)

Note that after changing the subtraction to addition, and changing the sign of 6 tomake it , we have precisely the same illustration as Example 2(a).

(b)

Note that after changing the subtraction to addition, and changing the sign of 6 tomake it , we have precisely the same illustration as Example 1(b).

(c)

This shows that subtracting a number from itself results in zero, even if the num-ber is negative. Therefore, subtracting a negative number is equivalent to adding apositive number of the same absolute value. ■

-a - 1-a2 = -a + a = 0

-6

-2 - 6 = -2 + 1-62 = -12 + 62 = -8

-6

2 - 6 = 2 + 1-62 = -16 - 22 = -4

-2 + 6 = 6 - 2 = 4

66 + 1-22 = 6 - 2 = 4

-6 + 2 = -16 - 22 = -4

62 + 1-62 = -16 - 22 = -4

-2-6

-2 + 1-62 = -12 + 62 = -8

2 + 6 = 8

cba=

a1b � c2 � ab � aca1bc2 � 1ab2c

ab � baa � 1b � c2 � 1a � b2 � c

a � b � b � a

Z65 Z

56

➝➝

➝➝

➝

■ From Section 1.1, we recall that a positivenumber is preceded by no sign. Therefore, inusing these rules, we show the “sign” of a pos-itive number by simply writing the number itself.

Subtraction of a Negative Number

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Multiplication and division of two numbers The product (or quotient) of twonumbers of the same sign is positive. The product (or quotient) of two numbers ofdifferent signs is negative.

E X A M P L E 4 Multiplying and dividing positive and negative numbers

(a)

(b)

(c)

(d)■

ORDER OF OPERATIONSOften, how we are to combine numbers is clear by grouping the numbers using sym-bols such as parentheses, the bar, ____ , between the numerator and denominatorof a fraction, and vertical lines for absolute value. Otherwise, for an expression inwhich there are several operations, we use the following order of operations.

1 2,

12

-3= -

12

3= -4-31122 = -13 * 122 = -36

-12

3= -

12

3= -431-122 = -13 * 122 = -36

-12

-3= 4-31-122 = 3 * 12 = 36

12

3= 431122 = 3 * 12 = 36

result is positive if bothnumbers are positive

result is positive if bothnumbers are negative

result is negative if onenumber is positive and the other is negative

E X A M P L E 5 Order of operations

(a) is evaluated by first adding and then dividing. The grouping of is clearly shown by the parentheses. Therefore,

(b) is evaluated by first dividing 20 by 2 and then adding. No specificgrouping is shown, and therefore the division is done before the addition. Thismeans

(c) is evaluated by first multiplying 2 by 3 and then subtracting. We donot first subtract 2 from 16. Therefore,

(d) is evaluated by first dividing 16 by 2 and then multiplying. From leftto right, the division occurs first. Therefore,

(e) is evaluated by first performing the subtractions within theabsolute value vertical bars, then evaluating the absolute values, and then subtract-ing. This means that ■|3 - 5| - |-3 - 6| = |-2| - |-9| = 2 - 9 = -7.

|3 - 5| - |-3 - 6|16 , 2 * 4 = 8 * 4 = 32.

16 , 2 * 4

16 - 2 * 3 = 16 - 6 = 10.16 - 2 * 3

20 , 2 + 3 = 10 + 3 = 13.

20 , 2 + 3

5 = 4.20 , 12 + 32 = 20 ,

2 + 32 + 320 , 12 + 32

Practice Exercises

Evaluate: 1.2. 16 , 12 * 42

12 - 6 , 2

When evaluating expressions, it is generally more convenient to change the opera-tions and numbers so that the result is found by the addition and subtraction of positivenumbers. When this is done, we must remember that

(1.1)

(1.2)a - 1-b2 = a + b

a + 1-b2 = a - b

■ Note that whereas20 , 2 + 3 =

202 + 3.

20 , 12 + 32 =20

2 + 3,

ORDER OF OPERATIONS

1. Operations within specific groupings are done first.

2. Perform multiplications and divisions (from left to right).

3. Then perform additions and subtractions (from left to right).

CAUTION �

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1.2 Fundamental Operations of Algebra 9

Practice Exercises

Evaluate: 3.

4.ƒ5 - 15 ƒ

2-

-9

3

21-32 -

4 - 8

2

E X A M P L E 6 Evaluating numerical expressions

(a) using Eq. (1.1)

(b) using Eq. (1.2)

(c)

(d)-12

2 - 8+

5 - 1

21-12 =

-12

-6+

4

-2= 2 + 1-22 = 2 - 2 = 0

|3 - 15|

-2-

8

4 - 6=

12

-2-

8

-2= -6 - (-4) = -6 + 4 = -2

18

-6+ 5 - 1-22(3) = -3 + 5 - (-6) = 2 + 6 = 8

7 + 1-32 - 6 = 7 - 3 - 6 = 4 - 6 = -2

In illustration (b), we see that the division and multiplication were done before theaddition and subtraction. In (c) and (d), we see that the groupings were evaluated first.Then we did the divisions, and finally the subtraction and addition. ■

E X A M P L E 7 Evaluating in an application

A 1500-kg van going at ran head-on into a 1000-kg car going at Aninsurance investigator determined the velocity of the vehicles immediately after thecollision from the following calculation. See Fig. 1.5.

The numerator and the denominator must be evaluated before the division is performed.The multiplications in the numerator are performed first, followed by the addition inthe denominator and the subtraction in the numerator. ■

OPERATIONS WITH ZEROSince operations with zero tend to cause some difficulty, we will show them here.

If is a real number, the operations of addition, subtraction, multiplication, anddivision with zero are as follows:

( means “is not equal to”)

E X A M P L E 8 Operations with zero

(a) (b) (c) 0 - 4 = -4-6 - 0 = -65 + 0 = 5

Z 0 � a �0a

� 0 1if a � 02 a : 0 � 0

a � 0 � a 0 � a � �a

a � 0 � a

a

=

40 000

2500= 16 km/h

15001402 + 1100021-202

1500 + 1000=

60 000 + 1-20 000)

1500 + 1000=

60 000 - 20 000

2500

20 km>h.40 km>h

(d) (e) (f) ■

Note that there is no result defined for division by zero. To understand the reason forthis, consider the results for and

If then This cannot be true because for any value of Thus,

division by zero is undefined

(The special case of is termed indeterminate. If then which is truefor any value of . Therefore, no specific value of can be determined.)bb

0 = 0 * b,00 = b,0

0

b.0 * b = 00 * b = 6.60 = b,

6

2= 3 since 2 * 3 = 6

60.6

2

5 * 0

7=

0

7= 0

0

-3= 0

0

6= 0

1000 kg

20 km/h

1500 kg

40 km/h

16 km/h

Fig. 1.5

NOTE �

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


E X A M P L E 9 Division by zero is undefined

see bottom of page 9

■

The operations with zero will not cause any difficulty if we remember to

never divide by zero

Division by zero is the only undefined basic operation. All the other operations withzero may be performed as for any other number.

2

5, 0 is undefined

8

0 is undefined

7 * 0

0 * 6 is indeterminate

CAUTION �

➝

➝

➝

EXERCISES 1.2

In Exercises 1–4, make the given changes in the indicated examplesof this section, and then solve the resulting problems.

1. In Example 5(c), change 3 to and then evaluate.

2. In Example 6(b), change 18 to and then evaluate.

3. In Example 6(d), interchange the 2 and 8 in the first denominatorand then evaluate.

4. In the rightmost illustration in Example 9, interchange the 6 andthe 0 above the 6. Is any other change needed?

In Exercises 5–36, evaluate each of the given expressions by perform-ing the indicated operations.

5. 6. 7.

8. 9. 10.

11. 12. 13.

14. 15. 16.

17. 18. 19.

20. 21. 22.

23. 24. 25.

26. 27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

In Exercises 37–44, determine which of the fundamental laws of alge-bra is demonstrated.

37. 38.

39. 40. 415 * p) = (4 * 521p2613 + 12 = 6132 + 61126 + 8 = 8 + 66172 = 7162

201-122 - 401-15298 - ƒ -98 ƒ

31-92 - 21-323 - 10

-71-32 +

6

-3- 1-92-7 -

ƒ -14 ƒ

212 - 32 - 3 ƒ6 - 8 ƒ

-18

3-

4 - 6

-1

24

3 + 1-52 - 41-92

7 - ƒ -5 ƒ

-11-22301-621-22 , 10 - 402

-10 - 1-621-82-21-62 + ` 8

-2`20 + 8 , 4

8 - 31-427 - 7

7 - 7

17 - 7

7 - 7

17 - 72 , 15 - 72-9 - ƒ 2 - 10 ƒ

-64

-2 ƒ 4 - 8 ƒ

212 - 72 , 1031-42162-21421-5228

-716 - 52-6120 - 102

-3

-9

3

-71-52-913281-328 - (-4)-19 - 1-16218 - 21

-3 + 9-4 + 1-728 + 1-42

-18

1-32

41. 42.

43.

44.

In Exercises 45–48, for numbers and , determine which of the fol-lowing expressions equals the given expression.(a) (b) (c) (d)

45. 46.

47. 48.

In Exercises 49–64, answer the given questions. Refer to Appendix Bfor units of measurement and their symbols.

49. (a) What is the sign of the product of an even number of negativenumbers? (b) What is the sign of the product of an odd number ofnegative numbers?

50. Is subtraction commutative? Explain.

51. Explain why the following definition of the absolute value of areal number is either correct or incorrect (the symbol means“is equal to or greater than”: If then if then ).

52. Explain what is the error if the expression is eval-uated as 27. What is the correct value?

53. Describe the values of and for which (a) and

(b)

54. Describe the values of and for which (a) and (b)

55. Some solar energy systems are used to supplement the utilitypower company power supplied to a home such that the meterruns backward if the solar energy being generated is greater thanthe energy being used. With such a system, if the solar power av-erages 1.5 kW for a 3.0-h period and only is used dur-ing this period, what will be the change in the meter reading forthis period?

56. A baseball player’s batting average (total number of hits dividedby total number of at-bats) is expressed in decimal form from0.000 (no hits for all at-bats) to 1.000 (one hit for each at-bat). Aplayer’s batting average is often shown as 0.000 before the firstat-bat of the season. Is this a correct batting average? Explain.

2.1 kW # h

ƒx - y ƒ = ƒx ƒ + ƒy ƒ .ƒx + y ƒ = ƒx ƒ + ƒy ƒyx

x - y

x - y= 1.

-xy = 1yx

24 - 6 , 2 # 3

ƒx ƒ = -xx 6 0,ƒx ƒ = x;x Ú 0,

Úx

-a - 1-b2-b - 1-a2b - 1-a2-a + 1-b2

-a - bb - aa - ba + b

ba

13 * 62 * 7 = 7 * 13 * 62115 * 32 * 9 = 15 * 13 * 92

813 - 22 = 8132 - 81223 + 15 + 92 = 13 + 52 + 9

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W

W

W

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.3 Calculators and Approximate Numbers 11

57. The daily high temperatures (in ) in the Falkland Islands in thesouthern Atlantic Ocean during the first week in July were re-corded as 7, 3, , , , 4, and 6. What was the average dailytemperature for the week? (Divide the algebraic sum of the read-ings by the number of readings.)

58. A flare is shot up from the top of a tower. Distances above theflare gun are positive and those below it are negative. After 5 s thevertical distance (in m) of the flare from the flare gun is found byevaluating Find this distance.

59. Find the sum of the voltages of the batteries shown in Fig. 1.6.Note the directions in which they are connected.

(20)(5) + (-5)(25).

-1-3-2

°C 61. One oil-well drilling rig drills 100 m deep the first day and 200 mdeeper the second day. A second rig drills 200 m deep the first dayand 100 m deeper the second day. In showing that the total depthdrilled by each rig was the same, state what fundamental law ofalgebra is illustrated.

62. A water tank leaks 12 L each hour for 7 h, and a second tank leaks7 L each hour for 12 h. In showing that the total amount leaked isthe same for the two tanks, what fundamental law of algebra is il-lustrated?

63. Each of four persons spends 8 min browsing one website and6 min browsing a second website. Set up the expression for thetotal time these persons spent browsing these websites. Whatfundamental law of algebra is illustrated?

64. A jet travels relative to the air. The wind is blowing atIf the jet travels with the wind for 3 h, set up the expres-

sion for the distance traveled. What fundamental law of algebra isillustrated?


1. 9 2. 2 3. 4. 8-4

50 km>h.600 km>h

You will be doing many of your calculations on a calculator, and a graphing calculatorcan be used for these calculations and many other operations. In this text, we will re-strict our coverage of calculator use to graphing calculators because a scientific calcu-lator cannot perform many of the required operations we will cover.

A brief discussion of the graphing calculator appears in Appendix C, and samplecalculator screens appear throughout the book. Since there are many models of graph-ing calculators, the notation and screen appearance for many operations will differfrom one model to another. You should practice using your calculator and review itsmanual to be sure how it is used. Following is an example of a basic calculation doneon a graphing calculator.

E X A M P L E 1 Calculating on a graphing calculator

In order to calculate the value of the numbers are entered as fol-lows. The calculator will perform the multiplication first, following the order of opera-tions shown in Section 1.2. The sign of is entered using the key, before3.58 is entered. The display on the calculator screen is shown in Fig. 1.7.

38.3 12.9 3.58 keystrokes

This means that Note in the display that the negative sign of is smaller and a little higher to dis-

tinguish it from the minus sign for subtraction. Also note the * shown for multiplication;the asterisk is the standard computer symbol for multiplication. ■

Looking back into Section 1.2, we see that the minus sign is used in two differentways: (1) to indicate subtraction and (2) to designate a negative number. This is clearlyshown on a graphing calculator because there is a key for each purpose. The - key is used for subtraction, and the key is used before a number to make it negative.(-)

-3.5838.3 - 12.91-3.582 = 84.482.

ENTER(-)�-

(-)-3.58

38.3 - 12.91-3.582,

1.3 Calculators and Approximate Numbers

Graphing Calculators • ApproximateNumbers • Significant Digits •Accuracy and Precision • RoundingOff • Operations with ApproximateNumbers • Estimating Results

■ The calculator screens shown with text ma-terial are for a TI-83 or TI-84. They are intendedonly as an illustration of a calculator screen forthe particular operation. Screens for othermodels may differ.

Fig. 1.6

� � � � � � � � � �

6 V �2 V 8 V �5 V 3 V

■ Some calculator keys on different modelsare labeled differently. For example, on somemodels, the EXE key is equivalent to the ENTER key.

Fig. 1.7

60. The electric current was measured in a given ac circuit at equalintervals as 0.7 mA, and Whatwas the change in the current between (a) the first two readings,(b) the middle two readings, and (c) the last two readings?

-0.6 mA.-0.9 mA,-0.2 mA,

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


We will first use a graphing calculator for the purpose of graphing in Section 3.5.Before then, we will show some calculational uses of a graphing calculator.

APPROXIMATE NUMBERS AND SIGNIFICANT DIGITSMost numbers in technical and scientific work are approximate numbers, having beendetermined by some measurement. Certain other numbers are exact numbers, havingbeen determined by a definition or counting process.

E X A M P L E 2 Approximate numbers and exact numbers

If a voltage on a voltmeter is read as 116 V, the 116 is approximate. Another voltmetermight show the voltage as 115.7 V. However, the voltage cannot be determined exactly.

If a computer prints out the number of names on a list of 97, this 97 is exact. Weknow it is not 96 or 98. Since 97 was found from precise counting, it is exact.

By definition, min, and the 60 and the 1 are exact. ■

An approximate number may have to include some zeros to properly locate thedecimal point. Except for these zeros, all other digits are called significant digits.

E X A M P L E 3 Significant digits

All numbers in this example are assumed to be approximate.34.7 has three significant digits.0.039 has two significant digits. The zeros properly locate the decimal point.706.1 has four significant digits. The zero is not used for the location of the deci-

mal point. It shows the number of tens in 706.1. 5.90 has three significant digits. The zero is not necessary as a placeholder and

should not be written unless it is significant.1400 has two significant digits, unless information is known about the number

that makes either or both zeros significant. (A temperature shown as has twosignificant digits. If a price list gives all costs in dollars, a price shown as $1400 hasfour significant digits.) Without such information, we assume that the zeros areplaceholders for proper location of the decimal point.

Other approximate numbers with the number of significant digits are 0.0005 (one),960 000 (two), 0.0709 (three), 1.070 (four), and 700.00 (five). ■

From Example 3, we see that all nonzero digits are significant. Also, zeros not usedas placeholders (for location of the decimal point) are significant.

In calculations with approximate numbers, the number of significant digits and theposition of the decimal point are important. The accuracy of a number refers to the num-ber of significant digits it has, whereas the precision of a number refers to the decimalposition of the last significant digit.

E X A M P L E 4 Accuracy and precision

An electric current is measured as 0.31 A on one ammeter and as 0.312 A on anotherammeter. Here, 0.312 is more precise since its last digit represents thousandths and0.31 is expressed only to hundredths. Also, 0.312 is more accurate since it has threesignificant digits and 0.31 has only two.

A concrete driveway is 130 m long and 0.1 m thick. Here, 130 is more accurate (twosignificant digits) and 0.1 is more precise (expressed to tenths). ■

The last significant digit of an approximate number is not exact. It has usually beendetermined by estimating or rounding off. However, it is not off by more than one-halfof a unit in its place value.

1400°C

60 s = 1

CAUTION �

■ To show that zeros at the end of a wholenumber are significant, a notation that can beused is to place a bar over the last significantzero. Using this notation, is shown tohave four significant digits.

78 000

■ Calculator keystrokes will generally not beshown, except as they appear in the displayscreens. They may vary from one model toanother.

M01_WASH42254_09_SE_C01_2p.qxd 5/19/10 5:51 PM


E X A M P L E 5 Meaning of the last digit of an approximate number

When we write the voltage in Example 2 as 115.7 V, we are saying that the voltage is atleast 115.65 V and no more than 115.75 V. Any value between these two, rounded offto tenths, would be expressed as 115.7 V.

In changing the fraction to the approximate decimal value 0.667, we are sayingthat the value is between 0.6665 and 0.6675. ■

To round off a number to a specified number of significant digits, discard all digitsto the right of the last significant digit (replace them with zeros if needed to properlyplace the decimal point). If the first digit discarded is 5 or more, increase the last sig-nificant digit by 1 (round up). If the first digit discarded is less than 5, do not changethe last significant digit (round down).

E X A M P L E 6 Rounding off

70 360 rounded off to three significant digits is 70 400. Here, 3 is the third significantdigit and the next digit is 6. Since we add 1 to 3 and the result, 4, becomes thethird significant digit of the approximation. The 6 is then replaced with a zero in orderto keep the decimal point in the proper position.

70 430 rounded off to three significant digits, or to the nearest hundred, is 70 400.Here the 3 is replaced with a zero.

187.35 rounded off to four significant digits, or to tenths, is 187.4.187.349 rounded off to four significant digits is 187.3. We do not round up the 4

and then round up the 3.35.003 rounded off to four significant digits is 35.00. We do not discard the zeros since

they are significant and are not used only to properly place the decimal point. ■

6 7 5,

23

■ On graphing calculators, it is possible to setthe number of decimal places (to the right ofthe decimal point) to which results will berounded off.

CAUTION �

NOTE �

Practice Exercises

Round off each number to three significantdigits.

1. 2015 2. 0.3004

OPERATIONS WITH APPROXIMATE NUMBERSWhen performing operations on approximate numbers, we must not express the resultto an accuracy or precision that is not valid. Consider the following examples.

E X A M P L E 7 Application of precision

A pipe is made in two sections. One is measured as 16.3 m long and the other as 0.927 mlong. What is the total length of the two sections together?

It may appear that we simply add the numbers as shown at the left. However, bothnumbers are approximate, and adding the smallest possible values and the largest pos-sible values, the result differs by 0.1 (17.2 and 17.3) when rounded off to tenths.Rounded off to hundredths (17.18 and 17.28), they do not agree at all since the tenthsdigit is different. Thus, we get a good approximation for the total length if it is roundedoff to tenths, the precision of the least precise length, and it is written as 17.2 m. ■

E X A M P L E 8 Application of accuracy

We find the area of the rectangular piece of land in Fig. 1.8 by multiplying the length, 207.54 m, by the width, 81.4 m. Using a calculator, we find that

This apparently means the area is However, the area should not be expressed with this accuracy. Since the length and

width are both approximate, we have

least possible area

greatest possible area

These values agree when rounded off to three significant digits ( ) but do notagree when rounded off to a greater accuracy. Thus, we conclude that the result is accu-rate only to three significant digits, the accuracy of the least accurate measurement, andthat the area is written as ■16 900 m2.

16 900 m2

1207.545 m2181.45 m2 = 16 904.540 25 m2

1207.535 m2181.35 m2 = 16 882.972 25 m2

16 893.756 m2.181.42 = 16 893.756.1207.542

16 900 m2

0.005 m

81.4 m

0.05 m207.54 m

Fig. 1.8

smallest values largest values16.3 m 16.25 m 16.35 m0.927 m 0.9265 m 0.9275 m

17.227 m 17.1765 m 17.2775 m

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


OPERATIONS WITH APPROXIMATE NUMBERS

1. When approximate numbers are added or subtracted, the result is expressedwith the precision of the least precise number.

2. When approximate numbers are multiplied or divided, the result is expressedwith the accuracy of the least accurate number.

3. When the root of an approximate number is found, the result is expressed withthe accuracy of the number.

4. When approximate numbers and exact numbers are involved, the accuracy ofthe result is limited only by the approximate numbers.

Following are the rules used in expressing the result when we perform basic opera-tions on approximate numbers. They are based on reasoning similar to that shown inExamples 7 and 8.

■ When rounding off a number, it may seemdifficult to discard the extra digits. However, ifyou keep those digits, you show a number withtoo great an accuracy, and it is incorrect to do so.

CAUTION �

NOTE �

Always express the result of a calculation with the proper accuracy or precision.When using a calculator, if additional digits are displayed, round off the final result (do not round off in any of the intermediate steps).

E X A M P L E 9 Adding approximate numbers

Find the sum of the approximate numbers 73.2, 8.0627, and 93.57.Showing the addition in the standard way and using a calculator, we have

least precise number (expressed to tenths) 8.0627

73.2

➝

final display must be rounded to tenths

Therefore, the sum of these approximate numbers is 174.8. ■

E X A M P L E 1 0 Combined operations

In finding the product of the approximate numbers 2.4832 and 30.5 on a calculator, thefinal display shows 75.7376. However, since 30.5 has only three significant digits, theproduct is 75.7.

In Example 1, we calculated that We know thatIf these numbers are approximate,

we must round off the result to tenths, which means the sum is 84.5. We see thatwhere there is a combination of operations, the final operation determines how thefinal result is to be rounded off. ■

E X A M P L E 1 1 Operations with exact numbers and approximate numbers

Using the exact number 600 and the approximate number 2.7, we express the result totenths if the numbers are added or subtracted. If they are multiplied or divided, we ex-press the result to two significant digits. Since 600 is exact, the accuracy of the resultdepends only on the approximate number 2.7.

■

A note regarding the equal sign is in order. We will use it for its defined meaningof “equals exactly” and when the result is an approximate number that has been properly rounded off. Although where means “equals approximately,” we write since 5.27 has been properly rounded off.127.8 = 5.27,

L127.8 L 5.27,

(=)

600 , 2.7 = 220 600 * 2.7 = 1600

600 - 2.7 = 597.3 600 + 2.7 = 602.7

38.3 - 12.91-3.582 = 38.3 + 46.182 = 84.482.38.3 - 12.91-3.582 = 84.482.

➝174.8327

93.57

Practice Exercise

Evaluate using a calculator.

3.(Numbers areapproximate.)40.5 +

3275

-60.041

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


You should make a rough estimate of the result when using a calculator. An estima-tion may prevent accepting an incorrect result after using an incorrect calculatorsequence, particularly if the calculator result is far from the estimated value.

E X A M P L E 1 2 Estimating results

In Example 1, we found that

using exact numbers

When using the calculator, if we forgot to make 3.58 negative, the display would be7.882, or if we incorrectly entered 38.3 as 83.3, the display would be 129.482.However, if we estimate the result as

we know that a result of or 129.482 cannot be correct.When estimating, we can often use one-significant-digit approximations. If the cal-

culator result is far from the estimate, we should do the calculation again. ■

-7.882

40 - 101-42 = 80

-

38.3 - 12.91-3.582 = 84.482

EXERCISES 1.3

In Exercises 1–4, make the given changes in the indicated examplesof this section, and then solve the given problems.

1. In Example 3, change 0.039 (the second number discussed) to0.390. Is there any change in the conclusion?

2. In the last paragraph of Example 6, change 35.003 to 35.303 andthen find the result.

3. In the first paragraph of Example 10, change 2.4832 to 2.483 andthen find the result.

4. In Example 12, change 12.9 to 21.9 and then find the estimatedvalue.

In Exercises 5–8, determine whether the given numbers are approximateor exact.

5. A car with 8 cylinders travels at

6. A computer chip 0.002 mm thick is priced at $7.50.

7. In 24 h there are 1440 min.

8. A calculator has 50 keys, and its battery lasted for 50 h of use.

In Exercises 9–14, determine the number of significant digits in eachof the given approximate numbers.

9. 107; 3004 10. 3600; 730 11. 6.80; 6.08

12. 0.8735; 0.0075 13. 3000; 3000.1 14. 1.00; 0.01

In Exercises 15–20, determine which of the pair of approximate num-bers is (a) more precise and (b) more accurate.

15. 30.8; 0.01 16. 0.041; 7.673 17. 0.1; 78.0

18. 7040; 0.004 19. 7000; 0.004 20. 50.060; 8.914

In Exercises 21–28, round off the given approximate numbers (a) tothree significant digits and (b) to two significant digits.

21. 4.936 22. 80.53 23. 50 893 24. 7.004

25. 9549 26. 30.96 27. 0.9449 28. 0.9999

55 km>h.

In Exercises 29–36, assume that all numbers are approximate. (a) Es-timate the result and (b) perform the indicated operations on a calcu-lator and compare with the estimate.

29. 30.

31. 32.

33. 34.

35. 36.

In Exercises 37–40, perform the indicated operations. The first num-ber is approximate, and the second number is exact.

37. 38.

39. 40.

In Exercises 41–44, answer the given questions. Refer to Appendix Bfor units of measurement and their symbols.

41. The manual for a heart monitor lists the frequency of the ultra-sound wave as 2.75 MHz. What are the least possible and thegreatest possible frequencies?

42. A car manufacturer states that the engine displacement for a cer-tain model is What should be the least possible andgreatest possible displacements?

43. A flash of lightning struck a tower 5.23 km from a person. Thethunder was heard 15 s later. The person calculated the speed ofsound and reported it as What is wrong with thisconclusion?

44. A technician records 4.4 s as the time for a robot arm to swingfrom the extreme left to the extreme right, 2.72 s as the time forthe return swing, and 1.68 s as the difference in these times. Whatis wrong with this conclusion?

348.7 m>s.

2400 cm3.

8.62 , 17283.142165217.311 - 22.980.9788 + 14.9

1

0.5926+

3.6957

2.935 - 1.054

3872

503.1-

2.056 * 309.6

395.2

0.693 78 + 0.049 97

257.4 * 3.216

23.962 * 0.015 37

10.965 - 8.249

0.3275

1.096 * 0.500 850.0350 -

0.0450

1.909

3.64117.06212.78 + 1.0495 - 1.633

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In Exercises 45–60, perform the calculations on a calculator.

45. Evaluate: (a) (b)

46. Evaluate: (a) (b)

47. Evaluate: (a) (b) (c) (d) (e) Compare with operations with zero on page 9.

48. Evaluate: (a) and (b) and(c) Explain why the displays differ.

49. Enter a positive integer (five or six digits is suggested) and thenrearrange the same digits to form another integer . Evaluate

. What type of number is the result?

50. Enter the digits in the order 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, using be-tween them any of the operations ( ) that will lead to aresult of 100.

51. Show that is not equal exactly to (a) 3.1416, or (b) 22 7.

52. At some point in the decimal equivalent of a rational number,some sequence of digits will start repeating endlessly. An ir-rational number never has an endlessly repeating sequence of digits. Find the decimal equivalents of (a) and (b) Notethe repetition for and that no such repetition occurs for

53. Following Exercise 52, show that the decimal equivalents of thefollowing fractions indicate they are rational: (a) 1 3 (b) 5 11(c) 2 5. What is the repeating part of the decimal in (c)?

54. Following Exercise 52, show that the decimal equivalent of thefraction 124 990 indicates that it is rational. Why is the last digitdifferent?

/

///

p.8>33p.8>33

/p

+ , - , * , ,

1x - y2 , 9y

x

0 , 00.0001 , 0.00012 , 02 , 0.0001

2 , 02 * 00 - 22 - 02 + 0

6.03 , 12.25 + 1.7726.03 , 2.25 + 1.77

12.2 + 3.82 * 4.52.2 + 3.8 * 4.5

55. In 3 successive days, a home solar system produced 32.4 MJ,26.704 MJ, and 36.23 MJ of energy. What was the total energyproduced in these 3 days?

56. Two jets flew at and respectively. Howmuch faster was the second jet?

57. If 1 K of computer memory has 1024 bytes, how many bytes arethere in 256 K of memory? (All numbers are exact.)

58. Find the voltage in a certain electric circuit by multiplying thesum of the resistances and by the cur-rent 3.55 A.

59. The percent of alcohol in a certain car engine coolant is found by

performing the calculation Find this percent

of alcohol. The number 100 is exact.

60. The tension (in N) in a pulley cable lifting a certain crate was

found by calculating the value of where the 1 is

exact. Calculate the tension.


1. 2020 2. 0.300 3. -14.0

50.4519.8021 + 100.9 , 23

,

100140.63 + 52.962105.30 + 52.96

.

101.23 Æ15.2 Æ, 5.64 Æ,

1450 km>h,938 km>h

1.4 Exponents

Positive Integer Exponents • Zero andNegative Exponents • Order ofOperations • Evaluating AlgebraicExpressions

In mathematics and its applications, we often have a number multiplied by itself sev-eral times. To show this type of product, we use the notation where is the numberand is the number of times it appears. In the expression the number is called thebase, and is called the exponent; in words, is read as “the nth power of a.”

E X A M P L E 1 Meaning of exponents

(a) the fifth power of 4

(b) the fourth power of �2

(c) the second power of , called “ squared”

(d) the third power of , called “ cubed” ■

We now state the basic operations with exponents using positive integers as expo-nents. Therefore, with and as positive integers, we have the following operations:

(1.3)

(1.4)

(1.5)

(1.6)1ab2n = anbn aa

bbn

=

an

bn (b Z 0)

1am2n = amn

am

an = am-n (m 7 n, a Z 0) am

an =

1

an-m (m 6 n, a Z 0)

am* an

= am+n

nm

15

15a 1

5b a 1

5b a 1

5b = a 1

5b3

aaa * a = a2

1-221-221-221-22 = 1-2244 * 4 * 4 * 4 * 4 = 45

annaan,n

aan,

➝

➝

■ Two forms are shown for Eqs. (1.4) in orderthat the resulting exponent is a positive integer.We consider negative and zero exponents afterthe next three examples.

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1.4 Exponents 17

E X A M P L E 2 Illustrating Eqs. (1.3) and (1.4)

Using Eq. (1.3): Using the meaning of exponents:

add exponents( factors of )( factors of )

factors of

Using first form Eq. (1.4): Using the meaning of exponents:

Using second form Eq. (1.4): Using the meaning of exponents:

■

E X A M P L E 3 Illustrating Eqs. (1.5) and (1.6)

Using Eq. (1.5): Using the meaning of exponents:

multiply exponents

Using first form Eq. (1.6): Using the meaning of exponents:

Using second form Eq. (1.6): Using the meaning of exponents:

■aa

bb3

= aa

bb aa

bb aa

bb =

a3

b3aa

bb3

=

a3

b3

1ab23 = 1ab21ab21ab2 = a3b31ab23 = a3b3

1a523 = 1a521a521a52 = a5+5+5= a151a523 = a5132

= a15

5 7 3

a3

a5 =

a1

* a1

* a1

a1

* a1

* a1

* a * a=

1

a2

a3

a5 =

1

a5-3 =

1

a2

a5

a3 =

a1

* a1

* a * a * a1

a1

* a1

* a1

= a2a5

a3 = a5-3= a2

5 7 3

a3* a5

= 1a * a * a21a * a * a * a * a2 = a8a3* a5

= a3+5= a8

a8a5a3■ In , which equals , each is

called a factor. A more general definition offactor is given in Section 1.7.

aa * a * aa3

➝ ➝

➝

➝

■ Here we are using the fact that (not zero)divided by itself equals 1, or a>a = 1.

a

➝ ➝

➝➝

➝

CAUTION � When an expression involves a product or a quotient of different bases, only expo-nents of the same base may be combined. Consider the following example.

E X A M P L E 4 Other illustrations of exponents

(a)

exponent add exponents of of

(b) add exponents of

(c)

(d)1ry322r1y224 =

r2y6

ry8 =

r

y2

13 * 22413 * 523 =

3424

3353 =

3 * 24

53

xax21ax23 = ax21a3x32 = a4x5

1a

1-x223 = 31-12x243 = 1-1231x223 = -x6

➝ ➝ ➝

➝

➝

➝

➝

Practice Exercises

Use Eqs. (1.3) – (1.6) to simplify the givenexpressions.

1. 2.12c2513cd22ax31-ax22

In illustration (b), note that means a times the square of and does not mean, whereas does mean ■a3x3.1ax23a

2x 2

xax 2CAUTION �

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


E X A M P L E 5 Exponents in an application

In the analysis of the deflection of a beam (the amount it bends), the expression thatfollows is simplified as shown.

is the length of the beam, and is the force applied to it. and are constants re-lated to the beam. In simplifying this expression, we combined exponents of and di-vided out the 2 that was in the numerator and in the denominator. ■

ZERO AND NEGATIVE EXPONENTSIf we let in Eqs. (1.4), we would have Also,since any nonzero quantity divided by itself equals 1. Therefore, for Eqs. (1.4) to hold,when we havem = n,

am>am= 1,am>am

= am-m= a0.n = m

LIEPL

=

21PL1L22

21132142142EI

=

PL3

48EI

1

2 a PL

4EIb a2

3b aL

2b2

=

1

2 a PL

4EIb a2

3b aL2

22 b

■ Although positive exponents are generallypreferred in a final result, there are some casesin which zero or negative exponents are to beused. Also, negative exponents are very usefulin some operations that we will use later.

(1.7)a0= 1 (a Z 0)

(1.8)

then all of the laws of exponents will hold for negative integers.

a-n=

1

an (a Z 0)

Equation (1.7) states that any nonzero expression raised to the zero power is 1. Zeroexponents can be used with any of the operations for exponents.

E X A M P L E 6 Zero as an exponent

(a) (b) (c) (d)

(e) (f) (g)

We note in illustration that only is raised to the zero power. If the quantity wereraised to the zero power, it would be written as ■

If we apply the first form of Eqs. (1.4) to the case where the resulting expo-nent is negative. This leads to the definition of a negative exponent.

E X A M P L E 7 Basis for negative exponents

Applying both forms of Eqs. (1.4) to we have

If these results are to be consistent, then ■

Following the reasoning of Example 7, if we define

a-5=

1a5.

a2

a7 = a2-7= a-5 and a2

a7 =

1

a7-2 =

1

a5

a2>a7,

n 7 m,

12t20.2tt(g)

b0= 1

2t0= 2112 = 21a2b0c22 = a4c21ax + b20 = 1

12x20 = 1-1-320 = -11-320 = 150= 1

➝

➝ ➝

➝ ➝ ➝

➝ ➝

➝

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

As stated in Section 1.3, in evaluating an expression on a calculator, we should alsoestimate its value as in Example 10. For an estimate, note that a negative number raisedto an even power gives a positive value and a negative number raised to an odd powergives a negative value.

1.4 Exponents 19

E X A M P L E 8 Negative exponents

(a) (b) (c)

(d) a a3t

b2xb-2

=

1a3t2-2

1b2x2-2 =

1b2x221a3t22 =

b4x2

a6t2

1

a-3 = a34-2=

1

42 =

1

163-1

=

1

3 change signs of exponents➝ ➝ ➝

➝

➝

Practice Exercises

Simplify: 3. 4.13x2-1

2a-2

-70

c-3

From illustration (a), where we see that the reciprocal of a number (not 0)is In illustration (e), only the has the exponent of ■

ORDER OF OPERATIONSIn Section 1.2, we saw that it is necessary to follow a particular order of operationswhen performing the basic operations on numbers. Since raising a number to a poweris a form of multiplication, this operation is performed before additions and subtrac-tions. In fact, it is performed before multiplications and divisions.

-1.xx-1.x3-1

= 1>3,CAUTION �

ORDER OF OPERATIONS

1. Operations within specific groupings

2. Powers

3. Multiplications and divisions (from left to right)

4. Additions and subtractions (from left to right)

E X A M P L E 9 Order of operations

Since there were no specific groupings, we first squared and Next, we foundthe product in the last term. Finally, the subtractions were performed. Note care-fully that we did not change the sign of before we squared it. ■

EVALUATING ALGEBRAIC EXPRESSIONS

An algebraic expression is evaluated by substituting given values of the literal num-

bers in the expression and calculating the result. On a calculator, the key is usedto square numbers, and the or key is used for other powers.

E X A M P L E 1 0 Evaluating using exponents

The distance (in m) that an object falls in 4.2 s is found by substituting 4.2 for in theexpression We show this as

substituting

estimation

The result is rounded off to two significant digits (the accuracy of ). The calculatorwill square 4.2 before multiplying. See Fig. 1.9. ■

t

51422 = 80➝4.9014.222 = 86 m

t = 4.2 s

4.90t2.t

xy�

x2

-12192 -3.-1

= 8 - 1 - 18 = -11

8 - 1-122 - 21-322 = 8 - 1 - 2192

■ The use of exponents is taken up in moredetail in Chapter 11.

CAUTION �

➝

➝

■ On many calculators, there is a specific keyor key sequence to evaluate x3.

Fig. 1.9

■ When less than half of a calculator screenis needed, the figure for that calculator screenwill show only a partial screen.

(e) 3x-1= 3a 1

xb =

3

x

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


E X A M P L E 1 1 Power of a number

Using the meaning of a power of a number, we have

■

E X A M P L E 1 2 Evaluating in an application

A wire made of a special alloy has an electric resistance (in ) given bywhere (in °C) is the temperature (between and 4°C). Find

for and Substituting these values, we have

estimation:

Note in the estimation that ■

Graphing calculators generally use computer symbols in the display for some of theoperations to be performed. These symbols are as follows:

�

Therefore, to calculate the value of we use the key sequence

20 6 200 5 3 4

with the result of 79 shown in the display of Fig. 1.10. Note carefully that 200 isdivided only by 5. If it were divided by then we would use parentheses andshow the expression to be evaluated as 20 * 6 + 200>15 - 342.5 - 34,

��

20 * 6 + 200>5 - 34,

Multiplication: * Division: > Powers:

1-323 = -27.

0.8 + 0.011-323 = 0.8 + 0.011-272 = 0.53 = 0.566 Æ

R = 0.838 + 0.01151-2.8723T = -2.87°C.a = 0.838 ÆR

-4°CT0.0115T3,R = a +ÆR

1-224 = 16 1-225 = -32 1-226 = 64 1-227 = -128

1-222 = 1-221-22 = 4 1-223 = 1-221-221-22 = -8

CAUTION �

EXERCISES 1.4

In Exercises 1–4, make the given changes in the indicated examplesof this section, and then simplify the resulting expression.

1. In Example 4(a), change to

2. In Example 6(d), change to

3. In Example 8(d), interchange the and

4. In Example 9, change to

In Exercises 5–48, simplify the given expressions. Express results withpositive exponents only.

5. 6. 7. 8.

9. 10. 11. 12.

13. 14. 15. 16.

17. 18. 19. 20.

21. 22. 23. 24.

25. 26. 27. 28.

29. 30. 31. 32. 1-y3251- t2271

- t-48

1

R-2

-w-56-1-1-220-3x0

6v018a20a 3

n3b3ax2

2b4

aF

tb20a 2

bb313r2231aT2230

1ax2512p231x8231P2243s

s4

n5

7n9

2x6

x

m5

m3

3k5k2b4b2y2y7x3x4

1-123.1-122b2.a3

2x0.12x201-x322.1-x223

33. 34. 35. 36.

37. 38. 39. 40.

41. 42. 43.

44. 45. 46.

47. 48.

In Exercises 49–56, evaluate the given expressions. In Exercises55–60, all numbers are approximate.

49. 50.

51. 52.

53. 54.

55.

56. 0.5131-2.7782 - 1-3.6723 + 0.8894>11.89 - 1.09222.381-60.722 - 2540>1.173

+ 0.8065126.13- 9.8842

15.662- 1-4.01724

1.0441-3.6823.071-1.862

1-1.8624 + 1.596

-0.7112- 1-0.80926-1-26.522 - 1-9.8523

6 + 1-225 - 1-2218271-42 - 1-522

1nRT-2232

R-2T32

15n2T5

3n-1T6

a2b2

y5b-2a4x-1

a-1b-3

ax-21-a2x231-8g-1s32213m-2n42-21p0x2a-12-1

13t2-1

3t-1

1n2241n422

x2x3

1x2232v4

12v24

2i40i-70-

L-3

L-5-1-c42-412v22-6

Fig. 1.10

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.5 Scientific Notation 21

1.5 Scientific Notation

Meaning of Scientific Notation •Changing Numbers to and from Scientific Notation • Scientific Notation on a Calculator

In technical and scientific work, we often encounter numbers that are either verylarge or very small. Examples of such numbers are: (a) Radio and televisionsignals travel at about 30 000 000 000 cm s; (b) The mass of the earth is about6 000 000 000 000 000 000 000 000 kg; (c) A typical individual fiber in a fiber-opticcommunications cable has a diameter of 0.000 005 m; (d) some X-rays have a wave-length of about 0.000 000 095 cm.

As we can see, writing numbers such as these is inconvenient in ordinary notation.Although calculators and computers can often handle such numbers, there is a moreefficient way of expressing numbers, in order to work more easily with them. A con-venient and useful notation, called scientific notation, is used to represent these or anyother numbers.

A number in scientific notation is expressed as the product of a number greater thanor equal to 1 and less than 10, and a power of 10, and is written as

where and is an integer. (The symbol means “is less than or equal to.”)

E X A M P L E 1 Scientific notation

(a) (b)

between and

(c) ■

From Example 1, we see how a number is changed from ordinary notation to scien-tific notation. The decimal point is moved so that only one nonzero digit is to its left.The number of places moved is the power of 10 (k), which is positive if the decimalpoint is moved to the left and negative if moved to the right.

E X A M P L E 2 Changing numbers to scientific notation

(a) (b) (c)

4 places to left 0 places 3 places to right ■

0.005 03 = 5.03 * 10-36.82 = 6.82 * 10034 000 = 3.4 * 104

0.005 03 =

5.03

1000=

5.03

103 = 5.03 * 10-3

101

6.82 = 6.82 * 1 = 6.82 * 10034 000 = 3.4 * 10 000 = 3.4 * 104

…k1 … P 6 10

P * 10k

/

In Exercises 57–68, perform the indicated operations.

57. Does represent the reciprocal of ?

58. Does equal 1? Explain.

59. If then what does equal?

60. Is for any negative value of a? Explain.

61. If a is a positive integer, simplify .

62. If a and b are positive integers, simplify

63. In developing the “big bang” theory of the origin of the uni-verse, the expression arises. Simplifythis expression.

64. In studying planetary motion, the expression arises. Simplify this expression.

65. In designing a cam for a pump, the expression is

used. Simplify this expression.

pa r

2b3a 4

3pr2b

1GmM21mr2-11r-221kT>1hc2231GkThc22c

1ya-b # ya+b22.

1xa # x-a25a-2

6 a-1

a12a3= 5,

a0.2 - 5 -1

10-2b0

xa 1

x-1b-1

66. For a certain integrated electric circuit, it is necessary to simplify

the expression Perform this simplification.

67. If $2500 is invested at 4.2 interest, compounded quarterly, theamount in the account after 6 years is Calculate this amount (the 1 is exact).

68. In designing a building, it was determined that the forces actingon an I beam would deflect the beam an amount (in cm), given by

where x is the distance (in m) from one

end of the beam. Find the deflection for (The 1000and 20 are exact.)


1. 2. 3. 4.a2

6x-c325c3

32d2=

32c3

9d2a3x5

x = 6.85 m.

x11000 - 20x2+ x32

1850,

250011 + 0.042>4224.%

gM

2pfC12pfM22.

W

W

➝➝ ➝

➝➝

■ Television was invented in the 1920s andfirst used commercially in the 1940s.

The use of fiber optics was developed in the1950s.

X-rays were discovered by Roentgen in 1895.

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

The key sequence is 3 6 4.74 12 . See Fig. 1.11. ■ENTEREEEE


To change a number from scientific notation to ordinary notation, we reverse theprocedure used in Example 2.

E X A M P L E 3 Changing numbers to ordinary notation

➝

Practice Exercises

1. Change to ordinary notation.2. Change 235 to scientific notation.

2.35 * 10-3

E X A M P L E 6 Scientific notation on a calculator

The wavelength (in m) of the light in a red laser beam can be found from the follow-ing calculation. Note the significant digits in the numerator.

= 6.33 * 10-7 m l =

3 000 000

4 740 000 000 000=

3.00 * 106

4.74 * 1012

l

■ Another important use of scientific notationis shown by the metric system use of prefixesas shown in Appendix B.

■ The number cannot be enteredon a scientific calculator in the form

. It could be entered in thisform on a graphing calculator.750 000 000 000

7.50 * 1011

Fig. 1.11,

(a) To change to ordinarynotation, we move the decimal pointsix places to the right, including addi-tional zeros to properly locate thedecimal point.

6 places to right

5.83 * 106= 5 830 000

5.83 * 106 (b) To change to ordinary no-tation, we must move the decimal pointthree places to the left, again includingadditional zeros to properly locate thedecimal point.

3 places to left ■

8.06 * 10-3= 0.008 06

8.06 * 10-3

➝

Scientific notation provides a practical way to handle calculations with very large orvery small numbers, particularly products, quotients, and powers. First, all numbers areexpressed in scientific notation. Then the calculation can be performed on numbers be-tween 1 and 10, using the laws of exponents to find the power of 10 in the result. In thisway, scientific notation gives an important application of the use of exponents.

E X A M P L E 4 Scientific notation in calculations

In designing a computer, it was determined that it would be able to process 803 000 bitsof data in 0.000 005 25 s. (See Exercise 41 of Exercises 1.1 for a brief note on com-puter data.) The rate of processing the data is

As shown, it is proper to leave the result (rounded off) in scientific notation. Thismethod is useful when using a calculator and then estimating the result. In this case,the estimate is ■

Another advantage of scientific notation is that the precise number of significantdigits of a number can be shown directly, even when the final significant digit is 0.

E X A M P L E 5 Scientific notation and significant digits

In evaluating if we know that 750 000 000 000 has three significantdigits, we can show the significant digits by writing

If the answer is to be written in scientific notation, we should write it as the productof a number between 1 and 10 and a power of 10. This means we should rewrite it as

■

We can enter numbers in scientific notation on a calculator, as well as have the cal-culator give results automatically in scientific notation. See the next example.

56.3 * 1022= 15.63 * 102110222 = 5.63 * 1023

750 000 000 0002= 17.50 * 101122 = 7.502

* 102*11= 56.3 * 1022

750 000 000 0002,

18 * 1052 , 15 * 10-62 = 1.6 * 1011.

803 000

0.000 005 25=

8.03 * 105

5.25 * 10-6= a 8.03

5.25b * 1011

= 1.53 * 1011 bits>s5 - 1-62 = 11 ➝

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.5 Scientific Notation 23

EXERCISES 1.5

In Exercises 1 and 2, make the given changes in the indicated exam-ples of this section, and then rewrite the number as directed.

1. In Example 3(b), change the exponent to 3 and then write thenumber in ordinary notation.

2. In Example 5, change the exponent 2 to and then write the re-sult in scientific notation.

In Exercises 3–10, change the numbers from scientific notation toordinary notation.

3. 4. 5. 6.

7. 8. 9. 10.

In Exercises 11–20, change the numbers from ordinary notation toscientific notation.

11. 4000 12. 56 000 13. 0.0087 14. 0.7

15. 6.09 16. 100 17. 0.063 18. 0.000 090 8

19. 1 20. 10

In Exercises 21–24, perform the indicated calculations using a calcu-lator and by first expressing all numbers in scientific notation.

21. 22.

23. 24.

In Exercises 25–28, perform the indicated calculations and then checkthe result using a calculator. Assume that all numbers are exact.

25. 26.

27. 28.

In Exercises 29–36, perform the indicated calculations using a calcu-lator. All numbers are approximate.

29. 30.

31. 32.

33. 34.

35.

36.

In Exercises 37–44, change numbers in ordinary notation to scientificnotation or change numbers in scientific notation to ordinary nota-tion. See Appendix B for an explanation of symbols used.

37. The power plant at Grand Coulee Dam produces 6 500 000 kW ofpower.

38. A certain computer has 85 000 000 000 bytes (85 gigabytes) ofmemory.

39. A fiber-optic system requires 0.000 003 W of power.

40. A red blood cell measures 0.0075 mm across.

41. The frequency of a certain TV signal is 2 000 000 000 Hz.

19.907 * 107211.08 * 10122213.603 * 10-52120542

13.69 * 10-7214.61 * 102120.0504

17.309 * 10-1225.984312.5036 * 10-20213.642 * 10-8212.736 * 1052

0.004 52

2430197 10020.0732167102

0.001 3410.02312

0.000 056 913 190 000212801865 0002143.82

12 * 10-162-511.2 * 1029235.3 * 1012

- 3.7 * 10102 * 10-35+ 3 * 10-34

0.000 03

6 000 000

88 000

0.0004

50 00010.006228 00012 000 000 0002

1 * 10-11.86 * 108 * 1003.23 * 100

9.61 * 10-52.01 * 10-36.8 * 1074.5 * 104

-1

-3

42. A parsec, a unit used in astronomy, is about

43. The power of the signal of a laser beam probe is

44. The electrical force between two electrons is about times the gravitational force between them.

In Exercises 45–48, solve the given problems.

45. Engineering notation expresses a number as where has one, two, or three digits to the left of the decimal point, and is an integral multiple of . Write the following numbers in engi-neering notation. (a) (b) (c)

46. See Exercise 45. Write the following numbers in engineeringnotation. (a) 8 090 000 (b) 809 000 (c) 0.0809

47. A googol is defined as 1 followed by 100 zeros. (a) Write thisnumber in scientific notaion. (b) A googolplex is defined as 10 tothe googol power. Write this number using powers of 10, and notthe word googol. (Note the name of the Internet company.)

48. The number of electrons in the universe has been estimated atHow many times greater is a googol than the estimated

number of electrons in the universe? (See Exercise 47.)

In Exercises 49–52, perform the indicated calculations.

49. A computer can do an addition in How long does ittake to perform additions?

50. Uranium is used in nuclear reactors to generate electricity. Aboutof the uranium disintegrates each day. How

much of 0.085 mg of uranium disintegrates in a day?

51. A TV signal travels at for a total of from the station transmitter to a satellite and then to a receiverdish. How long does it take the signal to go from the transmitterto the dish?

52. (a) Determine the number of seconds in a day in scientific nota-tion. (b) Using the result of part (a), determine the number of sec-onds in a century (assume 365.24 days year).

In Exercises 53–56, perform the indicated calculations by first ex-pressing all numbers in scientific notation.

53. One atomic mass unit (amu) is If one oxygenatom has 16 amu (an exact number), what is the mass of 125 000 000 oxygen atoms?

54. The rate of energy radiation (in W) from an object is found byevaluating the expression where is the thermodynamictemperature. Find this value for the human body, for which

and

55. In a microwave receiver circuit, the resistance of a wire 1 mlong is given by where is the diameter of the wire.Find if and

56. The average distance between the sun and earth is about 149 600 000 km, and this distance is called an astronomicalunit (AU). It takes light about 499.0 s to travel 1 AU. What is thespeed of light? Compare this with the speed of the TV signal inExercise 51.


1. 0.002 35 2. 2.35 * 102

d = 0.000 079 98 m.k = 0.000 000 021 96 Æ # m2RdR = k>d2,

R

T = 303 K.k = 0.000 000 057 W>K4

TkT4,

1.66 * 10-27 kg.

/

7.37 * 104 km3.00 * 105 km>s0.000 000 039%

5.6 * 1067.5 * 10-15 s.

1079.

230.2323003

kPP * 10k,

2.4 * 10-43

1.6 * 10-12 W.

3.086 * 1016 m.

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


E X A M P L E 1 Roots of numbers

(a) (the square root of 2) (b) (the cube root of 2)

(c) (the fourth root of 2) (d) (the seventh root of 6) ■

To have a single defined value for all roots (not just square roots) and to consideronly real-number roots, we define the principal nth root of to be positive if is posi-tive and to be negative if is negative and is odd. (If is negative and n is even, theroots are not real.)

E X A M P L E 2 Principal nth root

(a) (b)

(c) since (d) since odd

(e) (f) (g) ■

Another property of square roots is developed by noting illustrations such asIn general, this property states that the

square root of a product of positive numbers is the product of their square roots.

(1.9)

This property is used in simplifying radicals. It is most useful if either a or b is aperfect square, which is the square of a rational number.

E X A M P L E 3 Simplifying square roots

(a)

perfect squares simplest form

(b)

(c)

(Note that the square root of the square of a positive number is that number.) ■

24 * 102= 142102

= 21102 = 20

175 = 2(25)(3) = 12513 = 513

18 = 2(4)(2) = 1412 = 212

1ab = 1a1b (a and b positive real numbers)

136 = 14 * 9 = 14 * 19 = 2 * 3 = 6.

- 13 27 = -(+3) = -313 -27 = -3- 14 256 = -4

0.22= 0.0410.04 = 0.233

= 2723 27 = 3

- 164 = -81169 = 13 11169 Z -132

anaaa

17 614 2

13 212

At times, we have to find the square root of a number, or maybe some other root of anumber, such as a cube root. This means we must find a number that when squared, orcubed, and so on equals some given number. For example, to find the square root of 9,we must find a number that when squared equals 9. In this case, either 3 or is ananswer. Therefore, either 3 or is a square root of 9 since and

To have a general notation for the square root and have it represent one number, wedefine the principal square root of to be positive if is positive and represent it by

This means and not .-319 = 31a.aa

1-322 = 9.32= 9-3

-3

1.6 Roots and Radicals

Principal nth Root • SimplifyingRadicals • Using a Calculator •Imaginary Numbers

➝➝ ➝

➝

➝ ➝

➝

➝

The general notation for the principal th root of is (When do notwrite the 2 for .) The sign is called a radical sign.1n

n = 2,1n a.an■ Unless we state otherwise, when we referto the root of a number, it is the principal root.

➝

➝

➝ ➝➝

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.6 Roots and Radicals 25

Fig. 1.12

■ Try this one on your calculator:112345678987654321

In order to represent the square root of a number exactly, use Eq. (1.9) to write it insimplest form. However, a decimal approximation is often acceptable, and we use the

key on a calculator. (We will show another way of finding the root of a number inChapter 11.)

E X A M P L E 4 Using a calculator in an application

After reaching its greatest height, the time (in s) for a rocket to fall m is found byevaluating Find the time for the rocket to fall 360 m.

The calculator evaluation is shown in Fig. 1.12. From the display, we see that therocket takes 8.5 s to fall 360 m. The result is rounded off to two significant digits, theaccuracy of 0.45 (an approximate number). ■

0.451h.h

1x

In simplifying a radical, all operations under a radical sign must be done beforefinding the root. The horizontal bar groups the numbers under it.

E X A M P L E 5 More on simplifying square roots

(a) first perform the addition

However, is not .

(b) but

is not ■

In defining the principal square root, we did not define the square root of a negativenumber. However, in Section 1.1, we defined the square root of a negative number tobe an imaginary number. More generally, the even root of a negative number is animaginary number, and the odd root of a negative number is a negative real number.

E X A M P L E 6 Imaginary roots and real roots

even even odd

is imaginary is imaginary (a real number) ■

A much more detailed coverage of imaginary numbers, roots, and radicals is takenup in Chapters 11 and 12.

13 -64 = -414 -2431-64

222+ 262

= 2 + 6 = 8.222+ 62

222+ 62

= 14 + 36 = 140 = 14110 = 2110,

116 � 19 � 4 � 3 � 7116 � 9

= 5

16 + 9 116 + 9 = 125

➝ ➝ ➝

➝ ➝ ➝

Practice Exercises

Simplify:

1. 2. 136 + 144112

CAUTION �

EXERCISES 1.6

In Exercises 1–4, make the given changes in the indicated examplesof this section and then solve the given problems.

1. In Example 2(b), change the square root to a cube root and thenevaluate.

2. In Example 3(b), change to and explainwhether or not this would be a better expression to use.

3. In Example 5(a), change the to and then evaluate.

4. In the first illustration of Example 6, place a sign before theradical. Is there any other change in the statement?

In Exercises 5–36, simplify the given expressions. In each of 5–9 and12–21, the result is an integer.

5. 6. 7. 8.

9. 10. 11. 12.

13. 14. 15. 16. 15 -3213 -21624 1623 125

- 190010.0910.25- 149

- 136- 11211225181

-

*+

2115215221252132

17. 18. 19. 20.

21. 22. 23. 24. 15011200- 132A - 14 53 B4A15 -23 B5A - 13 -47 B3A23 31 B3A15 B2

25. 26. 27. 28.

29. 30. 31. 32.

33. 34.

35. 36.

In Exercises 37–44, find the value of each square root by use of a cal-culator. Each number is approximate.

37. 38. 39. 40.

41. (a) (b)

42. (a) (b) 110.6276 + 12.1609110.6276 + 2.1609

11296 + 1230411296 + 2304

10.062710.472913762185.4

282- 42232

+ 92

125 + 144136 + 64

2525 243

31144

72181

3214924 9223 82

281 * 102A80

7 - 3

1108

22184

W

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


43. (a) (b)

44. (a) (b)


45. The speed (in ) of a car that skids to a stop on dry pavementis often estimated by where is the length (in m) of theskid marks. Estimate the speed if m.

46. The resistance in an amplifier circuit is found by evaluat-

ing Find the resistance for and

47. The speed (in ) of sound in seawater is found by evaluatingfor Pa and Find

this speed, which is important in locating underwater objects us-ing sonar.

48. The terminal speed (in ) of a skydiver can be approximated bywhere is the mass (in kg) of the skydiver. Calculate the

terminal speed (after reaching this speed, the skydiver’s speed re-mains fairly constant before opening the parachute) of a 75-kgskydiver.

49. An HDTV screen is 93.0 cm wide and 52.1 cm high. The lengthof a diagonal (the dimension used to describe it—from one cor-ner to the opposite corner) is found by evaluating where is the width and is the height. Find the diagonal.hw

2w2+ h2,

m140m,m>s

d = 1.03 * 103 kg>m3.B = 2.18 * 1091B>dm>s

X = 2.875 Æ.Z = 5.362 Æ2Z2

- X2.

s = 46s1207s,

km>h

23.6252+ 20.614223.6252

+ 0.6142

20.04292- 20.0183220.04292

- 0.01832 50. A car costs $22 000 new and is worth $15 000 2 years later. The annual rate of depreciation is found by evaluating

where is the cost and is the value after2 years. At what rate did the car depreciate? (100 and 1 are exact.)

51. Is it always true that ? Explain.

52. For what values of is (a) (b) and(c) ?

53. A graphing calculator has a specific key sequence to find cuberoots. Using a calculator, find and .

54. A graphing calculator has a specific key sequence to find throots. Using a calculator, find and .

55. A graphing calculator can determine if a number is real or imagi-nary. Use the setting for numbers (see Exercise 37 onpage 6), where represents on a calculator. Use a calculatorto determine if the following numbers are real or imaginary:(a) (b)

56. Use a calculator to determine if the following numbers are real orimaginary (see Exercise 55): (a) (b)


1. 2. 615213

14 -6415 -32

13 -641-64

1-1ia + bi

27-38227 0.382n

23 -0.21423 2140

x 6 1xx = 1x,x 7 1x,x

2a2= a

VC100 A1 - 2V>C B ,

1.7 Addition and Subtraction of Algebraic Expressions

Algebraic Expressions • Terms •Factors • Polynomials • SimilarTerms • Simplifying • Symbols of Grouping

Since we use letters to represent numbers, we can see that all operations that can beused on numbers can also be used on literal numbers. In this section, we discuss themethods for adding and subtracting literal numbers.

Addition, subtraction, multiplication, division, and taking of roots are known asalgebraic operations. Any combination of numbers and literal symbols that resultsfrom algebraic operations is known as an algebraic expression.

When an algebraic expression consists of several parts connected by plus signs andminus signs, each part (along with its sign) is known as a term of the expression. If agiven expression is made up of the product of a number of quantities, each of thesequantities, or any product of them, is called a factor of the expression.

It is very important to distinguish clearly between terms and factors, because someoperations that are valid for terms are not valid for factors, and conversely. Some of thecommon errors in handling algebraic expressions occur because these operations arenot handled properly.

E X A M P L E 1 Terms and factors

In the study of the motion of a rocket, the following algebraic expression may be used.

terms

factors

This expression has three terms: , and . The first term, , has a factorof and two factors of Any product of these factors is also a factor of Thismeans other factors are and itself. ■gt2t2,gt,

gt2.t.ggt22s-2vtgt2,

gt2- 2vt + 2s

CAUTION �

➝ ➝ ➝ ➝ ➝ ➝ ➝

➝ ➝ ➝

W

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.7 Addition and Subtraction of Algebraic Expressions 27

E X A M P L E 2 Terms and factors

is an expression with terms and

The term has individual factors of 7, , and as well asproducts of these factors. The factor has two terms, and .

The term has factors 2, 3, , and . The negativesign in front can be treated as a factor of . The factor has three terms, 5,

, and . ■

A polynomial is an algebraic expression with only nonnegative integer exponentson one or more variables, and has no variable in a denominator. The degree of a termis the sum of the exponents of the variables of that term, and the degree of the polyno-mial is the degree of the term of highest degree.

A multinomial is any algebraic expression of more than one term. Terms like and can be included in a multinomial, but not in a polynomial. (Since ,the exponent is negative.

E X A M P L E 3 Polynomials

Some examples of polynomials are as follows:

(a) (degree 2) (b) (degree 6) (c) (degree 1)

(d) (degree 4) (add exponents of and )

(e) (degree 0)

From (c), note that a single term can be a polynomial, and from (e), note that a constantcan be a polynomial. The expressions in (a), (b), and (d) are also multinomials.

The expression is a multinomial, but not a polynomial becauseof the square root term. ■

A polynomial with one term is called a monomial. A polynomial with two terms iscalled a binomial, and one with three terms is called a trinomial. The numerical fac-tor is called the numerical coefficient (or simply coefficient) of the term. All termsthat differ at most in their numerical coefficients are known as similar or like terms.That is, similar terms have the same variables with the same exponents.

E X A M P L E 4 Monomial, binomial, trinomial

(a) is a monomial. The numerical coefficient is 7.

(b) is a binomial. The numerical coefficient of the first term is 3, and thenumerical coefficient of the second term is . Note that the sign is attached tothe coefficient.

(c) is a trinomial. The coefficients of the first two terms are 8 and

(d) is a polynomial with four terms (no special name). ■

E X A M P L E 5 Similar terms

(a) is a trinomial. The first and third terms are similar since theydiffer only in their numerical coefficients. The middle term is not similar to theothers since it has a factor of a.

(b) is a binomial. The terms are not similar since the first term has twofactors of x, and the second term has only one factor of x.

(c) is a polynomial. The commutative law tells us that, which means the first two terms are similar. ■

In adding and subtracting algebraic expressions, we combine similar terms into asingle term. The simplified expression will contain only terms that are not similar.

x2y3= y3x2

3x2y3- 5y3x2

+ x2- 2y3

4x2- 3x

8b - 6ab + 81b

x2y2- 2x + 3y - 9

-1.8cx3- x + 2

-63ab - 6a

7x4

x2+ 1y + 2 - 8

1-6 = -6x02-6

yxxy3+ 7x - 3

3x2x6- x4x2

- 5x + 3

1>x = x-11x1>x

-3yx5 + x - 3y-115 + x - 3y2x-6x15 + x - 3y2 2yx2x2

+ 2y1x2

+ 2y2,a7a1x2+ 2y2

-6x15 + x - 3y2. 7a1x2+ 2y27a1x2

+ 2y2 - 6x15 + x - 3y2

■ In practice, the numerical coefficient isusually called the coefficient. However, a moregeneral definition is: any factor of a product isthe coefficient of the other factors.

■ In Chapter 11, we will see that roots areequivalent to noninteger exponents.

M01_WASH42254_09_SE_C01_2p.qxd 5/19/10 5:53 PM


add like terms ■ = 8a + 4c

CAUTION �

➝ ➝➝

Practice Exercises

Use the distributive law.

1. 2. -31-2r + s2312a + y)

Practice Exercise

3. Simplify 2x - 3(4y - x)

➝

➝➝➝

➝

➝➝➝

➝➝

■ Note in each case that the parentheses areremoved and the sign before the parentheses isalso removed.

■ = 64 - 16x + x2

1618 - x2 - 218x - x22 - 164 - 16x + x22 = 128 - 16x - 16x + 2x2- 64 + 16x - x2

➝➝➝

E X A M P L E 6 Simplifying expressions

(a) Add similar terms—result has unlike terms

(b) cannot be simplified since there are no like terms.

(c) commutative law6a + 5c + 2a - c = 6a + 2a + 5c - c

6a2- 7a + 8ax

3x + 2x - 5y = 5x - 5y

➝

➝■ Some calculators can display algebraicexpressions and perform algebraic operations.

To group terms in an algebraic expression, we use symbols of grouping. In this text,we use parentheses, ( ), brackets, [ ], and braces, The bar, which is used with radi-cals and fractions, also groups terms. In earlier sections, we used parentheses and the bar.

When adding and subtracting algebraic expressions, it may be necessary to removesymbols of grouping. To do so, we must change the sign of every term within the sym-bols if the grouping is preceded by a minus sign. If the symbols of grouping are precededby a plus sign, each term within the symbols retains its original sign. This is a result ofthe distributive law,

E X A M P L E 7 Symbols of grouping

(a) use distributive law

(b) treat sign as

note change of signs

Normally, would be written simply as ■

In Example 7(a), we see that This is true for any values ofand , and is therefore an identity (see p. 7). In fact, an expression and any proper

form to which it may be changed form an identity when they are shown to be equal.

E X A M P L E 8 Simplifying: signs before parentheses

sign before parentheses

(a) use distributive law

signs retained

sign before parentheses

(b) use distributive law

signs changed

(c) use distributive law

signs changed ■

E X A M P L E 9 Simplifying in an application

In designing a certain machine part, it is necessary to perform the following simplification.

3c - 1-2b + c2 = 3c + 2b - c = 2b + 2c

2b = +2b

3c - 12b - c2 = 3c - 2b + c = -2b + 4c

-

2b = +2b

3c + 12b - c2 = 3c + 2b - c = 2b + 2c

+

xa21a + 2x2 = 2a + 4x.

a.+a

= -a + 3c

= 1-121+a2 + 1-121-3c2-1- -1+a - 3c2 = 1-121+a - 3c2

= 2a + 4x

21a + 2x2 = 2a + 212x2

a1b + c2 = ab + ac.

5 6.

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.7 Addition and Subtraction of Algebraic Expressions 29

At times, we have expressions in which more than one symbol of grouping is to beremoved in the simplification. Normally, when several symbols of grouping are to beremoved, it is more convenient to remove the innermost symbols first.

One of the most common errors made is changing the sign of only the first termwhen removing symbols of grouping preceded by a minus sign. Remember, if the sym-bols are preceded by a minus sign, we must change the sign of every term.

NOTE �

CAUTION �

E X A M P L E 1 0 Several symbols of grouping

(a)

remove parentheses

remove brackets

(b)

remove parentheses

remove brackets

remove braces ■

Calculators use only parentheses for grouping symbols, and we often need to useone set of parentheses within another set. These are called nested parentheses. In thenext example, note that the innermost parentheses are removed first.

E X A M P L E 1 1 Nested parentheses

■= -x - 2= 2 - 3x + 10 - 14 + 2x

= 2 - 13x - 10 + 14 - 2x2 2 - 13x - 215 - 17 - x222 = 2 - 13x - 215 - 7 + x22

= 5a2b - 2a - 2b

= 3a2b - a + 2a2b - a - 2b

= 3a2b - 5a - 2a2b + a + 2b63a2b - 53a - 12a2b - a24 + 2b6 = 3a2b - 53a - 2a2b + a4 + 2b6

= 5s

= 3ax - ax + 5s - 2ax

3ax - 3ax - 15s - 2ax24 = 3ax - 3ax - 5s + 2ax4 ➝

➝

➝

➝

➝

EXERCISES 1.7

In Exercises 1–4, make the given changes in the indicated examples ofthis section, and then solve the resulting problems.


2. In Example 8(a), change the sign before from to .


4. In Example 10(b), change to

In Exercises 5–54, simplify the given algebraic expressions.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34. 9v - 36 - 1v - 42 + 4v4aZ - 33 - 1aZ + 424-23-31x - 2y2 + 4y4-23-x - 2a - 1a - x2433-3 - 1a - 424234 - 1t2

- 524-31A - B2 - 1B - A24-316 - n2 - 12n - 324-15t + a22 - 213a2

- 2st2-716 - 3j2 - 21j + 4231a - b2 - 21a - 2b2312r + s2 - 1-5s - r221x - 2y2 + 15x - y2-1t - 2u2 + 13u - t214x - y2 - 1-2x - 4y21a - 32 + 15 - 6a21A + 1h - 21A2 - 31A2 - 3 - 14 - 5a22a - 1b - a2v - 14 - 5x + 2v25 + 13 - 4n + p2s + 13s - 4 - s2xy2

- 3x2y2+ 2xy2a2b - a2b2

- 2a2b

x - 2y + 3x - y + z2F - 2T - 2 + 3F - T

4C + L - 6C2y - y + 4x

6t - 3t - 4t5x + 7x - 4x

32a2b - 1a + 2b246.5a -

53a - 12a2b - a24 + 2b62ax4.31ax - 5s2 -

3ax - 15s - 2ax24-+12b - c2

2y.2x

35.

36.

37.

38.

39.

40.

41. 42.

43.

44.

45.

46.

47. In determining the size of a V belt to be used with an engine, theexpression is used. Simplify this expression.

48. When finding the current in a transistor circuit, the expressionis used. Simplify this expression. (The num-

bers below the ’s are subscripts. Different subscripts denote dif-ferent variables.)

49. Research on a plastic building material leads to

Simplify this expression.

50. One car goes 30 for hours, and a second car goes40 for hours. Find the expression for the sum of thedistances traveled by the two cars.

51. A shipment contains film cartridges for 15 exposures each andcartridges for 25 exposures each. What is the total number

of photographs that can be taken with the film from this shipment?x + 10

x

t + 2km>ht - 1km>h

2 AB -23 a B D - C AB +

43 a B - AB -

23 a B D .C AB +

43 a B +

ii1 - 12 - 3i22 + i2

3D - 1D - d2352.1e - 1.33f - 21e - 5f246-434R - 2.51Z - 2R2 - 1.512R - Z24a2

- 21x - 5 - 17 - 21a2- 2x2 - 3x22

-13t - 17 + 2t - 15t - 6222-2F + 2112F - 12 - 525V2

- 16 - 12V2+ 322

-5-3-1x - 2a2 - b4 - 1a - x26-25-14 - x22 - 33 + 14 - x2246-14 - 1LC2 - 3151LC - 72 - 161LC + 2245p - 1q - 2p2 - 33q - 1p - q247y - 5y - 32y - 1x - y2468c - 55 - 32 - 13 + 4c246

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


52. Each of two stores has mouse pads costing $3 each andmouse pads costing $2 each. How much more is the total

value of the $3 mouse pads than the $2 mouse pads in the twostores?

53. For the expressions and find (a) thesum, and (b) the difference if the second is subtracted from thefirst.

54. For the following expressions, subtract the third from the sum ofthe first two: 3a2

+ b - c3, 2c3- 2b - a2, 4c3

- 4b + 3.

3y - x2- b2x2

- y + 2a

n - 22n + 1 In Exercises 55 and 56, answer the given questions.

55. For any real numbers and is it true that Explain.

56. Is subtraction associative? That is, in general, does equal Explain.


1. 2. 3. 5x - 12y6r - 3s6a + 3y

a - 1b - c2?1a - b2 - c

ƒa - b ƒ = ƒb - a ƒ?b,a

1.8 Multiplication of Algebraic Expressions

Multiplying Monomials • Products ofMonomials and Polynomials • Powers of Polynomials

To find the product of two or more monomials, we use the laws of exponents as givenin Section 1.4 and the laws for multiplying signed numbers as stated in Section 1.2. Wefirst multiply the numerical coefficients to find the numerical coefficient of the prod-uct. Then we multiply the literal numbers, remembering that the exponents may becombined only if the base is the same.

E X A M P L E 1 Multiplying monomials

(a)

(b)

(c) ■

If a product contains a monomial that is raised to a power, we must first raise it tothe indicated power before proceeding with the multiplication.

E X A M P L E 2 Product containing power of a monomial

(a)

(b) ■

We find the product of a monomial and a polynomial by using the distributive law,which states that we multiply each term of the polynomial by the monomial. In doingso, we must be careful to give the correct sign to each term of the product.

E X A M P L E 3 Product of monomial and polynomial

(a)

(b) ■

It is generally not necessary to write out the middle step as it appears in the preced-ing example. We write the answer directly. For instance, Example 3(a) would appear as2ax13ax2

- 4yz2 = 6a2x3- 8axyz.

5cy21-7cx - ac2 = 15cy221-7cx2 + 15cy221-ac2 = -35c2xy2- 5ac2y2

2ax13ax2- 4yz2 = 2ax13ax22 + 12ax21-4yz2 = 6a2x3

- 8axyz

2s31-st42214s2t2 = 2s31s2t8214s2t2 = 8s7t9

312a2x231-ax2 = 318a6x321-ax2 = -24a7x4

2xy1-6cx2213xcy22 = -36c2x4y3

1-2b2y321-9aby52 = 18ab3y8

3c51-4c22 = -12c7 multiply numerical coefficientsand add exponents of c

➝➝

add exponents of same base

➝

➝

➝➝

W

W

➝➝

➝➝ ➝

Practice Exercises

Perform the indicated multiplications.

1. 2. -5x2y312xy - y422a3b1-6ab22

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1.8 Multiplication of Algebraic Expressions 31

We find the product of two polynomials by using the distributive law. The result isthat we multiply each term of one polynomial by each term of the other and add the re-sults. Again we must be careful to give each term of the product its correct sign.

E X A M P L E 4 Product of polynomials

■

Finding the power of a polynomial is equivalent to using the polynomial as a factorthe number of times indicated by the exponent. It is sometimes convenient to write thepower of a polynomial in this form before multiplying.

E X A M P L E 5 Power of a polynomial

two factors

(a)

(b) the exponent 3 indicates three factors

We should note in illustration (a) that

is not equal to

since the term is not included. We must follow the proper procedure and not sim-ply square each of the terms within the parentheses. ■

E X A M P L E 6 Products in an application

An expression used with a lens of a certain telescope is simplified as shown.

■ = ab2+ as2

+ bs2

= a3+ a2b + a2b + ab2

- a3+ as2

- 2a2b + bs2

= a1a2+ ab + ab + b22 + a3

- 2a3+ as2

- 2a2b + bs2

= a1a + b21a + b2 + a3- 12a3

- as2+ 2a2b - bs22

a1a + b22 + a3- 1a + b212a2

- s22

10x

x2 � 251x � 522

= 8a3- 12a2b + 6ab2

- b3

= 8a3- 8a2b + 2ab2

- 4a2b + 4ab2- b3

= 12a - b214a2- 4ab + b22

= 12a - b214a2- 2ab - 2ab + b22

12a - b23 = 12a - b212a - b212a - b2= x2

+ 10x + 25

1x + 522 = 1x + 521x + 52 = x2+ 5x + 5x + 25

= x2+ 3x - 2x - 6 = x2

+ x - 6

1x - 221x + 32 = x1x2 + x132 + 1-221x2 + 1-22132■ Note that, using the distributive law,

leads to the same result.1x - 221321x - 221x + 32 = 1x - 221x2 +

➝➝

➝➝

CAUTION �

Practice Exercises

Perform the indicated multiplications.

3. 4. 13u + 2v2212s - 5t21s + 4t2➝

EXERCISES 1.8

In Exercises 1–4, make the given changes in the indicated examplesof this section, and then solve the resulting problems.

1. In Example 2(b), change the factor to .

2. In Example 3(a), change the factor to .

3. In Example 4, change the factor to .

4. In Example 5(b), change the exponent 3 to 2.

In Exercises 5–64, perform the indicated multiplications.

5. 6. 7.

8. 9. 10. 6pq313pq22212ax2221-2ax2-2cs21-4cs22-ac21acx3212xy21x2y321a221ax2

1x - 321x + 32-2ax2ax

1-st4231-st42211. 12. 13.

14. 15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30. 13y2+ 2212y - 921x2

- 1212x + 5215p - 2q21p + 8q212s + 7t213s - 5t214w2

- 3213w2- 1212a - b213a - 2b2

14t1 + t2212t1 - 3t221x + 5212x - 121a + 721a + 121x - 321x + 52-21-3st3213s - 4t2ax1cx221x + y32-4c21-9gc - 2c + g223M1-M - N + 22

a2bc12ac - 3b2c25m1m2n + 3mn2-3b12b2- b2

-3s1s2- 5t22x1p - q2i21R + 2r2

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


31.

32.

33. 34.

35. 36.

37. 38.

39. 40. 41.

42. 43. 44.

45. 46.

47. 48.

49. 50.

51. Let and to show that (a) and (b) . ( means “does not equal”)

52. Evaluate the product by expressing it as .

53. Square an integer between 1 and 9 and subtract 1 from the result.Explain why the result is the product of the integer before and theinteger after the one you chose.

54. Explain how, by appropriate grouping, the product is easier to find. Find the

product.

55. By multiplication, show that is not equal to .

56. By multiplication, show that

57. In finding the value of a certain savings account, the expressionis used. Multiply out this expression.P11 + 0.01r22

1x + y21x2- xy + y22 = x3

+ y3.

x3+ y31x + y23

1x - 221x + 321x + 221x - 32

1100 + 221100 - 22198211022

Z1x - y22 Z x2- y2

1x + y22 Z x2+ y2y = 4x = 3

31x - 2221x + 22423T1T + 2212T - 1213x - c22312 + x213 - x21x - 12313R + 42221x + 8221b - 6x2221xyz - 22217m + 1221x1 + 3x2221x - 3y2212x - 522

ax1x + 4217 - x222L1L + 1214 - L22n15 - n216n + 52-313 - 2T213T + 22-51y - 321y + 6221a + 121a - 92

12a + 3b + 1212a + 3b - 121x - 2y - 421x - 2y + 42 58. The force due to the liquid in a parabolic container leads to the

expression . Multiply out this expression.

59. In using aircraft radar, the expression arises. Simplify this expression.

60. In calculating the temperature variation of an industrial area, theexpression arises. Perform the indi-cated multiplication.

61. In a particular computer design containing circuit elements,switches are needed. Find the expression for the number of

switches needed for circuit elements.

62. Simplify the expression , whicharises when analyzing the energy radiation from an object.

63. In finding the maximum power in part of a microwave transmit-ter circuit, the expression is used.Multiply and simplify.

64. In determining the deflection of a certain steel beam, the ex-pression is used. Multiply andsimplify.


1. 2.

3. 4. 9u2+ 12uv + 4v22s2

+ 3st - 20t2

-10x3y4+ 5x2y7

-12a4b3

27x2- 241x - 622 - 1x - 1223

1R1 + R222 - 2R21R1 + R22

1T2- 10021T - 1021T + 102

n + 100n2

n

12T3+ 321T2

- T - 32

12R - X22 - 1R2+ X22

w11 - x214 - x22

1.9 Division of Algebraic Expressions

Dividing Monomials • Dividing by aMonomial • Dividing One Polynomialby Another

To find the quotient of one monomial divided by another, we use the laws of exponentsand the laws for dividing signed numbers. Again, the exponents may be combined onlyif the base is the same.

E X A M P L E 1 Dividing monomials

(a) (b)

(c)

As shown in illustration (c), we use only positive exponents in the final result unlessthere are specific instructions otherwise. ■

From arithmetic, we may show how a multinomial is to be divided by a monomial.

When adding fractions , we have .

Looking at this from right to left, we see that the quotient of a multinomial divided by amonomial is found by dividing each term of the multinomial by the monomial andadding the results. This can be shown as

Be careful: Although , we must note that is notca

�cb

.c

a � ba + b

c=

a

c+

b

c

a + b

c=

a

c+

b

c

2

7+

3

7=

2 + 3

7Asay

27 and

37 B

-6a2xy2

2axy4 = - a 6

2ba2-1x1-1

y4-2 = -

3a

y2

16x3y5

4xy2 =

16

41x3-121y5-22 = 4x2y33c7

c2 = 3c7-2= 3c5

divide subtract coefficients exponents

➝➝

➝

CAUTION �

■ This is an identity and is valid for all valuesof and , and all values of except zero(which would make it undefined).

cba

W

W

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1.9 Division of Algebraic Expressions 33

We set up the division as we would for long division in arithmetic. Then, follow-ing the procedure outlined above, we have the following:

subtract

subtract

0

The remainder is zero and the quotient is . Note that when we subtracted from , we obtained ■4x.x

-3x3x + 2

4x - 2 x - (-3x) = 4x

4x - 26x2- 6x2

= 0

3x(2x - 1) 6x2- 3x

2x - 1 � 6x2+ x - 2

3x + 2

■ Until you are familiar with the method, it isrecommended that you do write out the middlesteps.

■ This is similar to long division of numbers.

E X A M P L E 2 Dividing by a monomial

(a)

(b)

We usually do not write out the middle step as shown in these illustrations. The divi-sions of the terms of the numerator by the denominator are usually done by inspection(mentally), and the result is shown as it appears in the next example. ■

= 2x - 4xy + 1

4x3y - 8x3y2+ 2x2y

2x2y=

4x3y

2x2y-

8x3y2

2x2y+

2x2y

2x2y

4a2+ 8a

2a=

4a2

2a+

8a

2a= 2a + 4

➝➝

➝

➝ each term ofnumerator dividedby denominator

➝➝➝

Practice Exercise

Divide: 1.4ax2

- 6a2x

2ax

E X A M P L E 3 Application of dividing by a monomial

The expression is used when analyzing the operation of an irriga-

tion pump. Performing the indicated division, we have

■

DIVISION OF ONE POLYNOMIAL BY ANOTHERTo divide one polynomial by another, use the following steps. (1) Arrange the dividend(the polynomial to be divided) and the divisor in descending powers of the variable.(2) Divide the first term of the dividend by the first term of the divisor. The result is thefirst term of the quotient. (3) Multiply the entire divisor by the first term of the quotientand subtract the product from the dividend. (4) Divide the first term of this difference bythe first term of the divisor. This gives the second term of the quotient. (5) Multiply thisterm by the entire divisor and subtract the product from the first difference. (6) Repeatthis process until the remainder is zero or a term of lower degree than the divisor.

E X A M P L E 4 Dividing one polynomial by another

Perform the division .

aThis division can also be indicated in the fractional form 6x2

+ x - 2

2x - 1.b

16x2+ x - 22 , 12x - 12

2p + v2d + 2ydg

2dg=

p

dg+

v2

2g+ y

2p + v2d + 2ydg

2dg

CAUTION �

divide first term of dividendby first term of divisor

6x2

2x

➝➝

➝

➝

Practice Exercise

Divide: 2. 16x2+ 7x - 32 , 13x - 12

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


E X A M P L E 5 Quotient with a remainder

Perform the division . Since there is no -term in the dividend, we

should leave space for any -terms that might arise (which we will show as 0 ).

divisor dividend

subtract

subtract

remainder

Since the degree of the remainder is less than that of the divisor, we nowshow the quotient in this case as . ■2x + 1 +

2x + 4

4x2

- 1

2x + 4

2x + 4

4x2- 1

4x2+ 2x + 3

8x3- 2x

4x2- 1 � 8x3

+ 4x2+ 0x + 3

2x + 1

xx

x8x3

+ 4x2+ 3

4x2- 1

8x3

4x2= 2x

4x2

4x2= 1

➝

0x - (-2x2 = 2x ➝

EXERCISES 1.9

In Exercises 1–4, make the given changes in the indicated examplesof this section and then perform the indicated divisions.

1. In Example 1(c), change the denominator to .

2. In Example 2(b), change the denominator to .

3. In Example 4, change the dividend to .

4. In Example 5, change the sign of the middle term of the numera-tor from to .

In Exercises 5–24, perform the indicated divisions.

5. 6. 7. 8.

9. 10. 11.

12. 13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

In Exercises 25–42, perform the indicated divisions. Express theanswer as shown in Example 5 when applicable.

25. 26.

27. 28.

29.

30. 16 + 7y + 6y22 , 12y + 12(x - 14x2

+ 8x3) , (2x - 3)

(2x2- 5x - 7) , (x + 1)(x2

- 3x + 2) , (x - 2)

13t2- 7t + 42 , 1t - 1212x2

+ 7x + 32 , 1x + 32

3a1F + T2b2- 1F + T2

a1F + T26y2n

- 4ayn+1

2yn

2xn+2+ 4axn

2xn3ab2

- 6ab3+ 9a2b2

9a2b2

91aB24 - 6aB4

3aB3

2pfL - pfR2

pfR

a2x1x22 + ax3

1 - ax1

ax1

4pq3+ 8p2q2

- 16pq5

4pq2

-5a2n - 10an2

5an

3rst - 6r2st2

3rs

2m2n - 6mn

2m

3a2x + 6xy

3x

12a2b

13ab222

61ax22-ax2

15sT218s2T3210s3T2

115x2214bx212y230bxy

51mn5

17m2n2

-16r3t5

-4r5t

-18b7c3

bc2

8x3y2

-2xy

-+

6x2- 7x + 2

2xy2

-2a2xy5

31.

32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

In Exercises 43–52, perform the indicated divisions.

43. When is divided by , the quotient is .Find .

44. When is divided by , the remainder is zero.Find .

45. By division show that is not equal to .

46. By division show that is not equal to .

47. In the optical theory dealing with lasers, the following expression

arises: . ( is the Greek letter mu.)

48. In finding the total resistance ofthe resistors shown in Fig. 1.13,the following expression is used.

6R1 + 6R2 + R1R2

6R1R2

m8A5

+ 4A3m2E2- Am4E4

8A4

x2+ y2x3

+ y3

x + y

x3x4+ 1

x + 1

k3x + 46x2

- x + k

c2x + 1x + c2x2

- 9x - 5

a4+ b4

a2- 2ab + 2b2

x2- y2

+ 2yz - z2

x + y - z

3r2- 5rR + 2R2

r - 3R

x2- 2xy + y2

x - y

D3- 1

D - 1

x3+ 8

x + 2

6T3+ T2

+ 2

3T2- T + 2

2a4+ 4a2

- 16

a2- 2

3x3+ 19x2

+ 13x - 20

3x - 2

x3+ 3x2

- 4x - 12

x + 2

16x2- 5x - 92 , 13x - 42

14Z2- 5Z - 72 , 14Z + 32

Fig. 1.13

R2

R1

6 �

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1.10 Solving Equations 35

49. When analyzing the potential energy associated with gravitational

forces, the expression arises. Perform

the indicated division.

50. A computer model shows that the temperature change ina certain freezing unit is found by using the expression

. Perform the indicated division.

51. In analyzing the displacement of a certain valve, the expression

is used. Find the reciprocal of this expression and

then perform the indicated division.

s2- 2s - 2

s4+ 4

3T3- 8T2

+ 8

T - 2

T

GMm31R + r2 - 1R - r242rR

52. The voltage and resistance in a certain electric circuit varywith time such that the current is given by the expression

. By performing the indicated division,

find the expression for the current.


1. 2. 2x + 32x - 3a

2t3+ 94t2

- 290t + 500

2t + 100

1.10 Solving Equations

Types of Equations • Solving Basic Types of Equations • Checking the Solution •First Steps • Ratio and Proportion

In this section, we show how algebraic operations are used in solving equations. In thefollowing sections, we show some of the important applications of equations.

An equation is an algebraic statement that two algebraic expressions are equal.Any value of the literal number representing the unknown that produces equality whensubstituted in the equation is said to satisfy the equation.

E X A M P L E 1 Valid values for equations

The equation is true only if . Substituting 3 for in the equa-tion, we have , or ; substituting , we have , whichis not correct.

This equation is valid for only one value of the unknown. An equation valid onlyfor certain values of the unknown is a conditional equation. In this section, nearly allequations we solve will be conditional equations that are satisfied by only one valueof the unknown. ■

E X A M P L E 2 Identity and contradiction

(a) The equation is true for all values of x. For example,substituting in the equation, we have , or .Substituting , we have , or .An equation valid for all values of the unknown is an identity.

(b) The equation is not true for any value of For any value of wetry, we find that the left side is 4 greater than the right side. Such an equation iscalled a contradiction. ■

To solve an equation, we find the values of the unknown that satisfy it. There is onebasic rule to follow when solving an equation:

Perform the same operation on both sides of the equation.

We do this to isolate the unknown and thus to find its value.By performing the same operation on both sides of an equation, the two sides

remain equal. Thus,

we may add the same number to both sides, subtract the same number from bothsides, multiply both sides by the same number, or divide both sides by the samenumber (not zero).

xx.x + 5 = x + 1

-3 = -31-122 - 4 = 1-1 - 221-1 + 22x = -15 = 532

- 4 = 13 - 2213 + 22x = 3x2

- 4 = 1x - 221x + 22

1 = 3x = 24 = 43132 - 5 = 3 + 1xx = 33x - 5 = x + 1

■ We have previously noted identities onpages 7 and 28.

■ Equations can be solved on most graphingcalculators. An estimate (or guess) of theanswer may be required to find the solution.See Exercises 43 and 44.

■ Although we may multiply both sides of anequation by zero, this produces , which isnot useful in finding the solution.

0 = 0

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E X A M P L E 3 Basic operations used in solving

In each of the following equations, we may isolate x, and thereby solve the equation,by performing the indicated operation.

add to both sides subtract from multiply both divide both both sides sides by sides by

Each solution should be checked by substitution in the original equation. ■

E X A M P L E 4 Operations used for solution; checking

Solve the equation .We are to perform basic operations to both sides of the equation to finally isolate

t on one side. The steps to be followed are suggested by the form of the equation.

original equation

add to both sides

combine like terms

divide both sides by

simplify

Therefore, we conclude that . Checking in the original equation, we have

The solution checks. ■

2182 - 7 � 9, 16 - 7 � 9, 9 = 9

t = 8

t = 8

2 2t

2=

16

2

2t = 16

7 2t - 7 + 7 = 9 + 7

2t - 7 = 9

2t - 7 = 9

x = 4x = 36x = 9x = 15

3x

3=

12

33ax

3b = 31122x + 3 - 3 = 12 - 3x - 3 + 3 = 12 + 3

3333

3x = 12x

3= 12x + 3 = 12x - 3 = 12

NOTE �

NOTE �

■ The word algebra comes from Arabic andmeans “a restoration.” It refers to the fact thatwhen a number has been added to one side ofan equation, the same number must be addedto the other side to maintain equality.

E X A M P L E 5 First remove parentheses

Solve the equation .

parentheses removed

x-terms combined on right

3x added to both sides —by inspection

7 added to both sides —by inspection

both sides divided by 4 —by inspection

Checking in the original equation, we obtain (after simplifying) . ■

Note that we always check in the original equation. This is done since errors mayhave been made in finding the later equations.

From these examples, we see that the following steps are used in solving the basicequations of this section.

-134 = -

134

x =154

4x = 15

4x - 7 = 8

x - 7 = -3x + 8

x - 7 = 3x - 6x + 8

x - 7 = 3x - 16x - 82

Practice Exercises

Solve for

1.

2. 215 - x2 = x - 8

3x + 4 = x - 6

x.

■ With simpler numbers, many basic steps aredone by inspection and not actually writtendown.

PROCEDURE FOR SOLVING EQUATIONS

1. Remove grouping symbols (distributive law).

2. Combine any like terms on each side (also after step 3).

3. Perform the same operations on both sides until result is obtained.

4. Check the solution in the original equation.

x =

■ Many other types of equations require moreadvanced methods for solving. These areconsidered in later chapters.

■ Note that the solution generally requires acombination of basic operations.

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1.10 Solving Equations 37

Practice Exercise

3. If the ratio of 2 to 5 equals the ratio of to 30, find .xx

If an equation contains numbers not easily combined by inspection, the best proce-dure is to first solve for the unknown and then perform the calculation.

E X A M P L E 6 First solve for unknown – then calculate

When finding the current i (in A) in a certain radio circuit, the following equation andsolution are used.

evaluate

The calculator solution of this equation, using the Solver feature, is shown in Fig. 1.14.When doing the calculation indicated above in the solution for i, be careful to groupthe numbers in the denominator for the division. Also, be sure to round off the resultas shown above, but do not round off values before the final calculation. ■

The quotient is also called the ratio of to An equation stating that two ra-tios are equal is called a proportion. Since a proportion is an equation, if one of thenumbers is unknown, we can solve for its value as with any equation. Usually, this isdone by noting the denominators and multiplying each side by a number that will clearthe fractions.

E X A M P L E 7 Ratio

If the ratio of to 8 equals the ratio of 3 to 4, we have the proportion

We can solve this equation by multiplying both sides by 8. This gives

Substituting into the original proportion gives the proportion . Since theseratios are equal, the solution checks. ■

E X A M P L E 8 Applied proportion

The ratio of electric current (in A) to the voltage across a resistor is constant. Iffor find for .

Since the ratio of and is constant, we have the following solution.

Checking in the original equation, we have . Since thevalue of is rounded up, the solution checks. ■

We will use the terms ratio and proportion (particularly ratio) when studying trigonom-etry in Chapter 4. A detailed discussion of ratio and proportion is found in Chapter 18.A general method of solving equations involving fractions is given in Chapter 6.

I0.0253 A>V = 0.0254 A>V

I = 2.08 A

2.08 A = I

82.0a 1.52

60.0b = 82.0a I

82.0b

1.52

60.0=

I

82.0

VIV = 82.0 VIV = 60.0 V,I = 1.52 A

VI

68 =

34x = 6

8ax

8b = 8a 3

4b , or x = 6

x

8=

3

4

x

b.aa>b

= 0.003 36 A

i =

8.8510.003 162 - 0.0595

-0.525 - 8.85

1-0.525 - 8.852i = 8.8510.003 162 - 0.0595

0.0595 - 0.525i - 8.85i - 8.8510.003 162 = 0

0.0595 - 0.525i - 8.851i + 0.003 162 = 0

NOTE �

note how the aboveprocedure is followed

Fig. 1.14

Ratio and Proportion

■ Generally, units of measurement will not beshown in intermediate steps. The proper unitswill be shown with the data and final result.

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


EXERCISES 1.10

In Exercises 1–4, make the given changes in the indicated examplesof this section and then solve the resulting problems.

1. In Example 3, change 12 to in each of the four illustrationsand then solve.

2. In Example 4, change to and then solve.

3. In Example 5, change to and then solve.

4. In Example 8, for the same given values of and , find for.

In Exercises 5–40, solve the given equations.

5. 6. 7.

8. 9. 10.

11. 12. 13.

14. 15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

In Exercises 33–40, all numbers are approximate.

33. 34.

35. 36.

37. 38.

39. 40.


41. Identify each of the following equations as a conditional equa-tion, an identity, or a contradiction.(a) (b)

42. For what values of is the equation a conditionalequation? Explain.

2x + a = 2xa

2x - 3 = 3 - 2x2x + 3 = 3 + 2x

276x

17.0=

1360

46.4

165

223=

13V

15

3.0

7.0=

R

42

x

2.0=

17

6.0

27.515.17 - 1.44x2 = 73.4-0.241C - 0.502 = 0.63

1.9t = 0.514.0 - t2 - 0.85.8 - 0.31x - 6.02 = 0.5x

2 - ƒx ƒ = 4ƒx ƒ - 1 = 8

2x =

3 - 517 - 3x24

4x - 21x - 423

= 8

3 - 612 - 3t2 = t - 57 - 311 - 2p2 = 4 + 2p

1.5x - 0.31x - 42 = 60.1x - 0.51x - 22 = 2

417 - F2 = -721x - 32 = -x

5 - 1x + 22 = 5x6 - 1r - 42 = 2r

314 - n2 = -n21s - 42 = s

6 + 4L = 5 - 3L3x + 7 = x

8 - 5t = 185 - 2y = -35D - 2 = 13

3t + 5 = -42x = 124E = -20

x

-4= 2

t

2= -5s + 6 = -3

x + 5 = 4x - 4 = -1x - 2 = 7

V = 48.0 VIVI

18 - 6x216x - 827 - 2t2t - 7

-12

43. Solve the equation of Example 5 by using the Equation Solver ofa graphing calculator.

44. Solve the equation of Example 6 by using the Equation Solver ofa graphing calculator.

45. In finding the speed (in ) at which a certain person wastraveling, the equation is to besolved. Find the speed.

46. In finding the rate (in ) at which a polluted stream is flow-ing, the equation is used. Find

47. In finding the maximum operating temperature (in °C) for acomputer integrated circuit, the equation isused. Find the temperature.

48. To find the voltage in a circuit in a TV remote-control unit, theequation is used. Find .

49. In blending two gasolines of different octanes, in order to find the number of litres of one octane needed, the equation

is used. Find , giventhat 0.06 and 0.09 are exact and the first zero of 2000 is significant.

50. In order to find the distance such that the weights are balancedon the lever shown in Fig. 1.15, the equation

must be solved. Find .(3 is exact.)

x38.518.25 - 3x221013x2 = 55.3x +

x

n0.0612000 - n2 = 0.091200020.14n +n

V1.12V - 0.67110.5 - V2 = 0V

1.1 = 1T - 762>40T

v.1515.5 + v2 = 2415.5 - v2km>hv

2.0v + 40 = 2.51v + 5.02km>hv

3x x8.25 m

210 N 55.3 N 38.5 N

Fig. 1.15

51. The manufacturer of a certain car powered partly by gas andpartly by batteries (a hybrid car) claims it can go 2050 km on onefull tank of 55 L. How far can the car go on 22 L?

52. A person 1.8 m tall is photographed with a 35-mm camera, andthe film image is 20 mm. Under the same conditions, how tall is aperson whose film image is 16 mm?


1. 2. 6 3. 12-5

1.11 Formulas and Literal Equations

Formulas • Literal Equations •Subscripts • Solve for Symbol beforeSubstituting Numerical Values

■ Einstein published his first paper on relativityin 1905.

An important application of equations is in the use of formulas that are found in geom-etry and nearly all fields of science and technology. A formula is an equation that ex-presses the relationship between two or more related quantities. For example,Einstein’s famous formula shows the equivalence of energy to the mass of an object and the speed of light c.

mEE = mc2

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1.11 Formulas and Literal Equations 39

NOTE �

We can solve a formula for a particular symbol just as we solve any equation. That is,we isolate the required symbol by using algebraic operations on literal numbers.

E X A M P L E 1 Solving for symbol in formula

In Einstein’s formula , solve for .

divide both sides by

switch sides to place at left

The required symbol is usually placed on the left, as shown. ■

E X A M P L E 2 Symbol with subscript in formula

A formula relating acceleration , velocity initial velocity , and time is. Solve for

subtracted from both sides

both sides divided by and then sides switched ■

E X A M P L E 3 Symbol in capital and in lower case

In the study of the forces on a certain beam, the equation is used.

Solve for

multiply both sides by 8

simplify right side

remove parentheses

subtract from both sides

divide both sides by and switch sides ■

E X A M P L E 4 Formula with groupings

The effect of temperature upon measurements is important when measurements must bemade with great accuracy. The volume of a special precision container at temperature

in terms of the volume at temperature is given by ,where depends on the material of which the container is made. Solve for .

Since we are to solve for , we must isolate the term containing . This can bedone by first removing the grouping symbols.

original equation

remove parentheses

remove brackets

subtract and add to both sides

divide both sides by and switch sides

■

To determine the values of any literal number in an expression for which we knowvalues of the other literal numbers, we should first solve for the required symbol andthen evaluate.

bV0 T =

V - V0 + bT0V0

bV0

bT0V0V0 V - V0 + bT0V0 = bTV0

V = V0 + bTV0 - bT0V0

V = V031 + bT - bT04 V = V031 + b1T - T024

TTTb

V = V031 + b1T - T024T0V0TV

2L P =

8W - wL2

2L

wL2 8W - wL2= 2LP

8W = wL2+ 2LP

8W = L1wL + 2P2 8W =

8L1wL + 2P28

P.

W =

L1wL + 2P28

a t =

v - v0

a

v0 v - v0 = at

t.v = v0 + atv0v,a

m m =

E

c2

c2 E

c2 = m

mE = mc2

■ The subscript makes a different literalsymbol from . (We have used subscripts in afew of the earlier exercises.)

vv00

■ Be careful! Just as subscripts can indicatedifferent literal numbers, a capital letter and thesame letter in lowercase are different literalnumbers. In this example, and are different.This is shown in several of the exercises for thissection.

wW

Practice Exercises

Solve for the indicated letter. Each comesfrom the indicated area of study.

1. , for (robotics)

2. , for p (economics)P = n1p - c2lu = kA + l

NOTE �

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E X A M P L E 5 Solve for symbol before substituting

The electric resistance (in ) of a resistor changes with the temperature (in °C)according to , where is the resistance at 0°C. For a given resistor,

and . Find the value of for .We first solve for and then substitute the given values.

Now substituting, we have

estimation:

rounded off ■

800 - 700

0.00517002 =

1

0.035= 30

= 34.9°C

T =

825 - 712

10.004 55217122

T =

R - R0

aR0

R - R0 = R0aT

R = R0 + R0aT

TR = 825 ÆTa = 0.004 55>°CR0 = 712 Æ

R0R = R0 + R0aTTÆR

EXERCISES 1.11

In Exercises 1–4, solve for the given letter from the indicated exampleof this section.

1. For the formula in Example 2, solve for .



4. For the formula in Example 5, solve for . (Do not evaluate.)

In Exercises 5–40, each of the given formulas arises in the technicalor scientific area of study shown. Solve for the indicated letter.

5. , for (electricity)

6. , for (chemistry)

7. , for (surveying)

8. , for (air conditioning)

9. , for (machine design)

10. , for (mechanics)

11. , for (hydrodynamics)

12. , for (nuclear physics)

13. , for (jet engine design)

14. , for (spectroscopy)

15. , for (medical technology)

16. , for (fluid flow)

17. , for (traffic flow)

18. , for (electricity)£L =

N£

i

dT =

c + d

v

C2p + dv2= 2d(C - W)

act2= 0.3t - ac

Lu = -

eL

2m

tA =

Rt

PV

I2Q = 2I + A + S

hp = pa + dgh

TP = 2pTf

LQ = SLd2

QW = SdT - Q

g1rL = g2 - g1

TPV = nRT

RE = IR

a

T0

w

a

19. , for (kinetic energy)

20. , for (photography)

21. , for (pulleys)

22. , for (ballistics)

23. , for (electronics)

24. , for (photography)

25. , for (engineering)

26. , for (oil drilling)

27. , for (air temperature)

28. , for (population growth)

29. , for (refrigeration)

30. , for (pressure gauges)

31. , for (machine design)

32. , for (computer access time)

33. , for (pulleys)

34. , for (electronics)

35. , for (jet engine power)

36. , for (refrigeration)

37. , for (electronics)eC =

2eAk1k2

d1k1 + k22S2W = T1S1 - S22 - Q

V2P =

V11V2 - V12gJ

mI =

VR2 + VR111 + m2R1R2

r1L = p1r1 + r22 + 2x1 + x2

hta = tc + 11 - h2tm

N1N = N1T - N211 - T2y2p - pa = dg1y2 - y12

Q2Q1 = P1Q2 - Q12rp2 = p1 + rp111 - p12

hT2 = T1 -

h

100

T1T = 31T2 - T12sN = r1A - s2

MA1 = A1M + 12VC2

0 = C2111 + 2V2

Mv =

V1m + M2m

Ma =

2mg

M + 2m

df =

F

d - F

m2K1

K2=

m1 + m2

m1

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.12 Applied Word Problems 41

L

b BFig. 1.16

1.12 Applied Word Problems

Procedure for Solving WordProblems • Identifying the UnknownQuantities • Setting Up the ProperEquation • Examples of SolvingWord Problems

Many applied problems are at first word problems, and we must put them into mathe-matical terms for solution. Usually, the most difficult part in solving a word problem isidentifying the information needed for setting up the equation that leads to the solution.To do this, you must read the problem carefully to be sure that you understand all ofthe terms and expressions used. Following is an approach you should use.

PROCEDURE FOR SOLVING WORD PROBLEMS

1. Read the statement of the problem. First, read it quickly for a generaloverview. Then reread slowly and carefully, listing the information given.

2. Clearly identify the unknown quantities and then assign an appropriate letterto represent one of them, stating this choice clearly.

3. Specify the other unknown quantities in terms of the one in step 2.

4. If possible, make a sketch using the known and unknown quantities.

5. Analyze the statement of the problem and write the necessary equation. This is often the most difficult step because some of the information may beimplied and not explicitly stated. Again, a very careful reading of the state-ment is necessary.

6. Solve the equation, clearly stating the solution.

7. Check the solution with the original statement of the problem.

■ See Appendix A, page A-1, for a variation tothe method outlined in these steps. You mightfind it helpful.

CAUTION �

38. , for (beam deflection)

39. , for (property deprecation)

40. , for (atomic theory)

In Exercises 41–46, find the indicated values.

41. The efficiency (in ) of a certain engine is given by the formula. Find if and K.

42. A formula used in determining the total transmitted power in an AM radio signal is . Find if

and .

43. A formula relating the Fahrenheit temperature and the Celsiustemperature is . Find the Celsius temperature thatcorresponds to 90.2°F.

44. In forestry, a formula used to determine the volume of a log is, where is the length of the log and and are

the areas of the ends. Find (in ) if , ,and . See Fig. 1.16.B = 0.244 m2

L = 4.91 mV = 1.09 m3m2bbBLV =

12L1B + b2

V

F =95C + 32C

F

m = 0.925Pt = 680 WPcPt = Pc11 + 0.500 m22

Pt

T2 = 850e = 45%T1e = T2/1T1 + T22%e

Bp

P=

AI

B + AI

nV = Ca1 -

n

Nb

Ld =

3LPx2- Px3

6EI

45. The voltage across resistance

is , where is the

voltage across resistances and. See Fig. 1.17. Find (in )

if , ,and .

46. The efficiency of a computer multiprocessor compilation is

given by , where is the number of processors

and is the fraction of the compilation that can be performedby the available parallel processors. Find for and

.

In Exercises 47 and 48, set up the required formula and solve for theindicated letter.

47. One missile travels at a speed of for 4 h, and anothermissile travels at a speed of for hours. If they travel a to-tal of km, solve the resulting formula for

48. A microwave transmitter can handle telephone communica-tions, and 15 separate cables can handle connections each. If thecombined system can handle connections, solve for .


1. 2.P + nc

nu - kA

yCy

x

t.dt + 2v1

v2 km>h

q = 0.83E = 0.66p

q

pE =

1

q + p11 - q2E

V = 12.0 VV1 = 6.30 VR1 = 3.56 Æ

ÆR2R2

R1

VV1 =

VR1

R1 + R2

R1V1

R1V1

V

R2

I

Fig. 1.17

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


Read the following examples very carefully and note just how the outlined proce-dure is followed.

E X A M P L E 1 Sum of forces on a beam

A 78-N beam is supported at each end. The supporting force at one end is 10 N morethan at the other end. Find the forces.

Since the force at each end is required, we write

step 2

as a way of establishing the unknown for the equation. Any appropriate letter could beused, and we could have let it represent the larger force.

Also, since the other force is 10 N more, we write

step 3

We now draw the sketch in Fig. 1.18. step 4

F + 10 = the larger force (in N)

let F = the smaller force (in N)■ Be sure to carefully identify your choice forthe unknown. In most problems, there is reallya choice. Using the word let clearly shows thata specific choice has been made.

F 10 N F � 10

Fig. 1.18

■ The statement after “let (or some otherappropriate letter) ” should be clear. It shouldcompletely define the chosen unknown.

=

x

Since the forces at each end of the beam support the weight of the beam, we havethe equation

step 5

This equation can now be solved:

step 6

Thus, the smaller force is 34 N, and the larger force is 44 N. This checks with the orig-inal statement of the problem. step 7 ■

E X A M P L E 2 Total resistance of electric resistors

In designing an electric circuit, it is found that 34 resistors with a total resistance of 56 are required. Two different resistances, 1.5 and 2.0 , are used. How many ofeach are in the circuit?

Since we want to find the number of each resistance, we

Also, since there are 34 resistors in all,

We also know that the total resistance of all resistors is 56 . This means that

1.5 number 2.0 each each number

total resistance of all resistors

total resistance total resistanceof resistors of resistors

Therefore, there are 24 1.5- resistors and 10 2.0- resistors. The total resistanceof these is . We see that this checks with thestatement of the problem. ■

2411.52 + 1012.02 = 36 + 20 = 56 Æ

ÆÆ

x = 24

-0.5x = -12

1.5x + 68 - 2.0x = 56

2.0-Æ1.5-Æ

+ 2.0134 - x2 = 561.5x

ÆÆ

Æ

34 - x = number of 2.0-Æ resistors

let x = number of 1.5-Æ resistors

ÆÆÆ

F = 34 N

2F = 68

F + 1F + 102 = 78

➝ ➝ ➝ ➝➝

■ “Let resistors” is incomplete. Wewant to find out how many of them there are.

x = 1.5-Æ

■ See Appendix A, page A-1, for a “sketch”that might be used with this example.

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

1.12 Applied Word Problems 43

Therefore, the sample can be divided into 24 slides with 5 mg each, or 20 slides with6 mg each. Since the total mass, 120 mg, is the same for each set of slides, the solutionchecks with the statement of the problem. ■

E X A M P L E 4 Distance traveled – space travel

A space shuttle maneuvers so that it may “capture” an already orbiting satellite that is6000 km ahead. If the satellite is moving at and the shuttle is moving at

, how long will it take the shuttle to reach the satellite? (All digits shownare significant.)

First, we let the time for the shuttle to reach the satellite. Then, using thefact that the shuttle must go 6000 km farther in the same time, we draw the sketchin Fig. 1.19. Next, we use the formula . This leadsto the following equation and solution.

speed of speed ofshuttle time satellite time

distance traveled distance between distance traveledby shuttle at beginning by satellite

This means that it will take the shuttle 2.400 h to reach the satellite. In 2.400 h, theshuttle will travel 70 800 km, and the satellite will travel 64 800 km. We see that thesolution checks with the statement of the problem. ■

t = 2.400 h

2500t = 6000

27 000t+6000=29 500t

(d = rt)distance = rate * time

t =

29 500 km>h 27 000 km>h

Practice Exercise

1. Solve the problem in Example 3 byletting number of slides with 6 mg.y =

➝ ➝ ➝ ➝

Shuttle

6000 km

Satellite

Fig. 1.19

■ A maneuver similar to the one in thisexample was used to repair the Hubble spacetelescope in 1994 and 2001.

E X A M P L E 3 Number of medical slides

A medical researcher finds that a given sample of an experimental drug can be dividedinto 4 more slides with 5 mg each than with 6 mg each. How many slides with 5 mgeach can be made up?

We are asked to find the number of slides with 5 mg, and therefore we

Since the sample may be divided into 4 more slides with 5 mg each than of 6 mgeach, we know that

Since it is the same sample that is to be divided, the total mass of the drug on eachtype is the same. This means

x - 4 = number of slides with 6 mg

let x = number of slides with 5 mg

mg number mg numbereach each

total mass total mass-mg slides -mg slides

-x = -24 or x = 24

5x = 6x - 24

65

61x - 42=5x

65

➝ ➝➝ ➝

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


➝ ➝ ➝■ “Let methanol” is incomplete. We wantto find out the volume (in L) that is to be added.

x =

➝

72506.00%

7250 � x10.0%

x100%

Litres of methanol

� �

Fig. 1.20

EXERCISES 1.12

In Exercises 1–4, make the given changes in the indicated examplesof this section and then solve the resulting problems.

1. In Example 2, in the second line, change 2.0 to 2.5 .

2. In Example 3, in the second line, change “4 more slides” to “3 more slides.”

3. In Example 4, in the second line, change to.

4. In Example 5, change “pure methanol” to “of a blend with 50.0methanol.”

In Exercises 5–32, set up an appropriate equation and solve. Data areaccurate to two sig. digits unless greater accuracy is given.

5. A certain new car costs $5000 more than the same model new carcost 6 years ago. Together a new model today and six years agocost $49 000. What was the cost of each?

6. The flow of one stream into a lake is more than the flowof a second stream. In 1 h, flow into the lake fromthe two streams. What is the flow rate of each?

7. Approximately 4.5 million wrecked cars are recycled in two con-secutive years. There were 700 000 more recycled the second yearthan the first year. How many are recycled each year?

8. A business website was accessed as often on the first day of a pro-motion as on the next two days combined. The first day hits werefour times those on the third. The second day hits were 4000 morethan the third. How many were there each day?

9. Petroleum rights to 140 hectares of land are leased for $37 000.Part of the land leases for $200 per hectare, and the reminder for$300 per hectare. How much is leased at each price?

10. A vial contains 2000 mg, which is to be used for two dosages.One patient is to be administered 660 mg more than another. Howmuch should be administered to each?

4.14 * 105 m345 m3>s

%

27 500 km>h27 000 km>h

ÆÆ

11. After installing a pollution control device, a car’s exhaust con-tained the same amount of pollutant after 5.0 h as it had in 3.0 h.Before the installation the exhaust contained (parts permillion per hour) of the pollutant. By how much did the device re-duce the emission?

12. Three meshed spur gears have a total of 107 teeth. If the secondgear has 13 more teeth than the first and the third has 15 moreteeth than the second, how many teeth does each have?

13. In the design of a bridge, an engineer determines that four fewer18-m girders are needed for the span length than 15-m girders.How many 18-m girders are needed?

14. A fuel oil storage depot had an 8-week supply on hand. How-ever, cold weather caused the supply to be used in 6 weeks when20 kL extra were used each week. How many litres were in theoriginal supply?

15. The sum of three electric currents that come together at a point ina circuit is zero. If the second current is twice the first and thethird current is A more than the first, what are the currents?(The sign indicates the direction of flow.)

16. A delivery firm uses one fleet of trucks on daily routes of 8 h. Asecond fleet, with five more trucks than the first, is used on dailyroutes of 6 h. Budget allotments allow for 198 h of daily deliverytime. How many trucks are in each fleet?

17. A natural gas pipeline feeds into three smaller pipelines, each ofwhich is 2.6 km longer than the main pipeline. The total length ofthe four pipelines is 35.4 km. How long is each section?

18. At 100 efficiency two generators would produce 750 MW ofpower. At efficiencies of 65 and 75 , they produce 530 MW.At 100 efficiency, what power would each produce?

19. A person has a total of 54 CDs and DVDs, which cost a total of$876. Each CD cost $15 and each DVD cost $18. How many ofeach does the person have?

%%%

%

9.2 m

150 ppm>h

E X A M P L E 5 Mixture – gasoline and methanol

A refinery has 7250 L of a gasoline-methanol blend that is 6.00% methanol. How muchpure methanol must be added so that the resulting blend is 10.0% methanol?

First, the number of litres of methanol to be added. The total volume ofmethanol in the final blend is the volume in the original blend plus that which isadded. This total volume is to be 10.0 of the final blend. See Fig. 1.20.

original volume final volume

methanol in methanol methanol inoriginal mixture added final mixture

, to be added

Checking (to three sig. digits), there would be 757 L of methanol of a total 7570 L.■

x = 322 L 0.900x = 290

435 + x = 725 + 0.100x

0.10017250 + x2=x+0.060017250210%5.00%

%

let x =

M01_WASH42254_09_SE_C01_2p.qxd 5/19/10 5:54 PM

Chapter Equations 45

20. A person won a lottery prize of $20 000, from which 25 was de-ducted for taxes. The remainder was invested, partly for a 40gain, and the rest for a 10 loss. How much was each investmentif there was a $2000 net investment gain?

21. A ski lift takes a skier up a slope at . The skier then skisdown the slope at . If a round trip takes 24 min, howlong is the slope?

22. Before being put out of service, the supersonic jet Concorde madea trip averaging less than the speed of sound for 1.00 h,and more than the speed of sound for 3.00 h. If the tripcovered 5740 km, what is the speed of sound?

23. Trains at each end of the 50.0-km-long Eurotunnel under the En-glish Channel start at the same time into the tunnel. Find theirspeeds if the train from France travels 8.0 faster than thetrain from England and they pass in 17.0 min. See Fig. 1.21.

km>h

400 km>h100 km>h

150 m>min50 m>min

%%

% car travels at , and the second car travels at .How long does it take the first car to overtake the second, andwhich car will be ahead after eight laps?

26. A computer chip manufacturer produces two types of chips. Intesting a total of 6100 chips of both types, 0.50 of one type and0.80 of the other type were defective. If a total of 38 defectivechips were found, how many of each type were tested?

27. Two gasoline distributors, A and B, are 367 km apart on a high-way. A charges and B charges . Each charges

per kilometre for delivery. Where on Interstate 80 is thecost to the customer the same?

28. An outboard engine uses a gasoline-oil fuel mixture in the ratioof 15 to 1. How much gasoline must be mixed with a gasoline-oilmixture, which is 75 gasoline, to make 8.0 L of the mixture forthe outboard engine?

29. A car’s radiator contains 12 L of antifreeze at a 25 concentra-tion. How many litres must be drained and then replaced by pureantifreeze to bring the concentration to 50 (the manufacturer’s“safe” level)?

30. How much sand must be added to 250 kg of a cement mixture thatis 22 sand to have a mixture that is 25 sand?

31. To pass a 20-m long semitrailer traveling at in 10 howfast must a 5.0-m long car go?

32. An earthquake emits primary waves moving at and sec-ondary waves moving at . How far from the epicenter ofthe earthquake is the seismic station if the two waves arrive at thestation 2.0 min apart?

5.0 km>s8.0 km>s

s,70 km>h%%

%

%

%

0.16¢>L$1.68>L$1.80>L

%%

73.0 m>s79.0 m>s

17.0 min 50.0 km

English ChannelEngland France

Fig. 1.21

24. An executive would arrive 10.0 min early for an appointment iftraveling at 60 , or 5.0 min early if traveling at .How much time is there until the appointment?

25. One lap at the Canadian Grand Prix in Montreal is 4.36 km. In arace, a car stalls and then starts 30.0 s after a second car. The first

45.0 km>hkm>h

CHAPTER 1 EQUATIONS

Commutative law of addition:

Associative law of addition: a + 1b + c2 = 1a + b2 + c

a + b = b + a Commutative law of multiplication:

Associative law of multiplication: a1bc2 = 1ab2cab = ba

Distributive law:

(1.1) (1.2)

(1.3)

(1.5)

(1.7)

and positive real numbers (1.9))b(a1ab = 1a1b

a0= 1 (a Z 0)

1am2n = amn

am

an = am-n (m 7 n, a Z 0),

am* an

= am+n a - 1-b2 = a + b a + 1-b2 = a - b

a1b + c2 = ab + ac

(1.8)a-n=

1

an (a Z 0) (1.6)1ab2n = anbn, aa

bbn

=

an

bn (b Z 0)

(1.4)am

an =

1

an-m (m 6 n, a Z 0)

M01_WASH42254_09_SE_C01_2p.qxd 5/19/10 5:56 PM


CHAPTER 1 REVIEW EXERCISES

In Exercises 1–12, evaluate the given expressions.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

In Exercises 13–20, simplify the given expressions. Where appropri-ate, express results with positive exponents only.

13. 14. 15.

16. 17. 18.

19. 20.

In Exercises 21–24, for each number, (a) determine the number ofsignificant digits and (b) round off each to two significant digits.

21. 8840 22. 21 450 23. 9.040 24. 0.700

In Exercises 25–28, evaluate the given expressions. All numbers areapproximate.

25. 26.

27. 28.

In Exercises 29–60, perform the indicated operations.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39.

40.

41.

42.

43. 44.

45. 46.

47.

48.

49. 50.27s3t2

- 18s4t + 9s2t

9s2t

12p3q2- 4p4q + 6pq5

2p4q

3x32y - r - 41s - 2r243p31q - p2 - 2p11 - 3q24

-s14s - 3t22-3y1x - 4y221x - 3212x2

- 3x + 1212x + 121x2- x - 32

x2+ 3b + 31b - y2 - 312b - y + z24

2xy - 53z - 35xy - 17z - 6xy2463b - 33a - 1a - 3b24 + 4a

4R - 32r - 13R - 4r24

4a2x3- 8ax4

-2ax2

2h3k2- 6h4k5

2h2k

12r - 9s221x + 8221C - 4D212C - D212x - 121x + 52-12x - b2 - 31-x - 5b26LC - 13 - LC2xy - y - 5y - 4xya - 3ab - 2a + ab

1

0.035 68+

37,466

29.632

10.1958 + 2.844

3.14216522

8.896 * 10-12

3.5954 + 6.044937.3 - 16.9211.06722

19 + 36145

-35x-1y1x2y25xy-1

-16N-21NT22-2N0T-1

15p4q2r

5pq5r

-3mn-5t18m-3n4213a0b-2231-2rt222

- 24 16 + (16)2(17)2- 23 8

- 181 + 144116 - 164

-1-322 -

-8

1-22 - ƒ -4 ƒ

18

3 - 5- 1-422

3 - 5 ƒ -3 - 2 ƒ -

12

-4-5 - ƒ21-62 ƒ +

-15

3

1-921-1221-4224

1-521621-421-22132

6 - 8 - 1-421-22 + 1-52 - 3

51.

52.

53. 54.

55. 56.

57.

58.

59. 60.

In Exercises 61–72, solve the given equations.

61. 62.

63. 64.

65. 66.

67. 68.

69.

70.

71.

72.

In Exercises 73–82, change numbers in ordinary notation to scientificnotation or change numbers in scientific notation to ordinary notation.(See Appendix B for an explanation of the symbols that are used.)

73. A certain computer has 60 000 000 000 bytes of memory.

74. The escape velocity (the velocity required to leave the earth’sgravitational field) is about .

75. When pictures of the surface of Mars were transmitted to theearth from the Pathfinder mission in 1997, Mars was about 192 000 000 km from earth.

76. Police radar has a frequency of .

77. Among the stars nearest the earth, Centaurus A is aboutkm away.

78. Before its destruction in 2001, the World Trade Center hadnearly of office space. (See Exercise 32, page 128.)

79. The faintest sound that can be heard has an intensity of about.

80. An optical coating on glass to reduce reflections is about 0.000 000 15 m thick.

81. The maximum safe level of radiation in the air of a home due toradon gas is . (Bq is the symbol for bequerel,the metric unit of radioactivity, where .)

82. A certain virus was measured to have a diameter of about 0.000 000 18 m.

1 Bq = 1 decay>s1.5 * 10-1 Bq>L

10-12 W>m2

106 m2

4.05 * 1013

1.02 * 109 Hz

40 000 km>h

0.25016.721 - 2.44x2 = 2.08

2.7 + 2.012.1x - 3.42 = 0.1

-18 - x2 = x - 212 - x23t - 217 - t2 = 512t + 12

2 ƒx ƒ - 1 = 32s + 413 - s2 = 6

-21-4 - y2 = 3y6x - 5 = 31x - 42

21N - 423

=

5

4

5x

7=

3

2

4y - 3 = 5y + 73x + 1 = x - 8

6x2+ 5xy - 4y2

2x - y

2y3+ 9y2

- 7y + 5

2y - 1

11 - 2x21x - 32 - 1x + 4214 - 3x2-351r + s - t2 - 2313r - 2s2 - 1t - 2s246

8x3- 14x + 3

2x + 3

4x4+ 10x3

+ 18x - 1

x + 3

w3- 4w2

+ 7w - 12

w - 3

3x3- 7x2

+ 11x - 3

3x - 1

14x2- 412 , 12x + 72

12x2+ 7x - 302 , 1x + 62

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

Review Exercises 47

In Exercises 83–96, solve for the indicated letter. Where noted, thegiven formula arises in the technical or scientific area of study.


84. , for (astronomy: black holes)

85. , for (mechanics)

86. , for (thermodynamics)

87. , for (moments of forces)


89. , for (optics)

90. , for (physics: momentum)

91. , for (mechanics: gears)

92. , for (solar heating)

93. , for (thermal resistance)

94. , for (radar design)

95. , for (mechanics: beams)

96. , for (thermal expansion)

In Exercises 97–102, perform the indicated calculations.

97. A computer’s memory is bytes, and that of a model20 years older is bytes. What is the ratio of the newercomputer’s memory to the older computer’s memory?

98. The time (in s) for an object to fall metres is given by the ex-pression . How long does it take a person to fall 22 mfrom a sixth-floor window into a net while escaping a fire?

99. The CN Tower in Toronto is 0.553 km high. The Sears Tower inChicago is 443 m high. How much higher is the CN Tower thanthe Sears Tower?

100. The time (in s) it takes a computer to check memory cells isfound by evaluating . Find the time to check 48 cells.

101. The combined electric resistance of two parallel resistors is

found by evaluating the expression . Evaluate this for

and .

102. The distance (in m) from the earth for which the gravitationalforce of the earth on a spacecraft equals the gravitational forceof the sun on it is found by evaluating , where

and are the masses of the earth and sun, respectively. Findthis distance for and .

In Exercises 103–106, simplify the given expressions.

103. One transmitter antenna is cm long, and another ism long. What is the sum, in centimetres, of their lengths?

104. In finding the value of an annuity, the expression is used. Multiply out this expression.11 + i221Ai - R2

1x + 2a21x - 2a2

M = 1.99 * 1030 kgm = 5.98 * 1024 kgMm

1.5 * 10111m>M

R2 = 0.0590 ÆR1 = 0.0275 Æ

R1R2

R1 + R2

1n>265022n

0.451hh

6.4 * 1045.25 * 1010

T2V = V031 + 3a1T2 - T124ad = kx2331a + b2 - x4

lZ2a1 -

l

2ab = k

T2R =

A1T2 - T12H

Bq =

KA1B - C2L

N2N1 = T1N2 - N32 + N3

Mmu = 1m + M2vnd = 1n - 12A

RV = IR + Ir

qPp + Qq = Rr

pf = p1c - 12 - c1p - 12EP =

p2EI

L2

GR =

2GM

c2

ZR = n2Z

105. A computer analysis of the velocity of a link in an industrialrobot leads to the expression . Simplifythis expression.

106. When analyzing the motion of a communications satellite,

the expression is used. Perform the indi-

cated division.

In Exercises 107–114 solve the given problems.

107. Does the value of change if the parenthesesare removed?

108. Does the value of change if the parenthesesare removed?

109. In solving the equation , what conclu-sion can be made?

110. In solving the equation , what conclusioncan be made?

111. Show that .

112. Is division associative? That is, is it true that?

113. What is the ratio of to ?

114. What is the ratio of to ?

In Exercises 115–128, solve the given problems. All data are accurateto two significant digits unless greater accuracy is given.

115. Two computer software programs cost $190 together. If onecosts $72 more than the other, what is the cost of each?

116. A sponsor pays a total of $9500 to run a commercial on twodifferent TV stations. One station charges $1100 more than theother. What does each charge to run the commercial?

117. Three chemical reactions each produce oxygen. If the firstproduces twice that of the second, the third produces twice thatof the first, and the combined total is what volume isproduced by each?

118. In testing the rate at which a polluted stream flows, a boat thattravels at 5.5 in still water took 5.0 h to go downstreambetween two points, and it took 8.0 h to go upstream between thesame two points. What is the rate of flow of the stream?

119. The voltage across a resistor equals the current times the resis-tance. In a microprocessor circuit, one resistor is 1200 greaterthan another. The sum of the voltages across them is 12.0 mV.Find the resistances if the current is 2.4 A in each.

120. An air sample contains 4.0 ppm (parts per million) of two pollu-tants. The concentration of one is four times the other. What arethe concentrations?

121. One road crew constructs 450 m of road bed in 12 h. If anothercrew works at the same rate, how long will it take them to con-struct another 250 m of road bed?

122. The fuel for a two-cycle motorboat engine is a mixture of gaso-line and oil in the ratio of 15 to 1. How many liters of each arein 6.6 L of mixture?

m

Æ

km>h

560 cm3,

1414 + 36

2 * 1048 * 10-3

1a , b2 , c = a , 1b , c21if b Z 0, c Z 02

1x - y23 = -1y - x237 - 12 - x2 = x + 2

x - 13 - x2 = 2x - 3

(3 * 18) , 9 - 6

3 * 18 , 19 - 62

k2r - 2h2k + h2rv2

k2r

41t + h2 - 21t + h22

W

W

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM


115 km/h

30 min

175 km/h

Fig. 1.22

123. A ship enters the Red Sea from the Suez Canal, moving south at28.0 . Two hours later, a second ship enters the Red Sea atthe southern end, moving north at 35.0 . If the Red Sea is2230 km long, when will the ships pass?

124. A helicopter used in fighting a forest fire travels at 175 from the fire to a pond and 115 with water from the pondto the fire. If a round-trip takes 30 min, how long does it takefrom the pond to the fire? See Fig. 1.22.

km>hkm>h

km>hkm>h 126. How much water must be added to 20.0 mL of a 60.0% saline

solution in order to make a 45.0% saline solution?

127. An architect plans to have 25% of the floor area of a house inceramic tile. In all but the kitchen and entry, there are 205 offloor area, 15% of which is tile. What area can be planned forthe kitchen and entry if each has an all-tile floor?

128. A karat equals part of gold in an alloy (for example,9-karat gold is gold). How many grams of 9-karat goldmust be mixed with 18-karat gold to get 200 g of 14-karat gold?

Writing Exercise129. In calculating the simple interest earned by an investment, the

equation is used, where is the value after aninitial principal is invested for years at interest rate Solvefor , and then evaluate for , , and

years. Write a paragraph or two explaining (a) yourmethod for solving for , and (b) the calculator steps used toevaluate , noting the use of parentheses.r

rt = 4.000

P0 = $6250P = $7625rrr.tP0

PP = P0 + P0rt

9>241>24

m2

CHAPTER 1 PRACTICE TEST

In Problems 1–5, evaluate the given expressions. In Problems 3 and 5,the numbers are approximate.

1. 2. 3.

4. 5.

In Problems 6–12, perform the indicated operations and simplify.When exponents are used, use only positive exponents in the result.

6. 7. 8.

9. 10.

11. 12.

13. Solve for :

14. Solve for :

15. Express 0.000 003 6 in scientific notation.

31x - 32 = x - 12 - 3d2x

5y - 21y - 42 = 7y

3x - 34x - 13 - 2x2412x - 321x + 72

6x2- 13x + 7

2x - 1

8a3x2- 4a2x4

-2ax2

3m21am - 2m3212x + 32212a0b-2c32-3

346.4 - 23.5

287.7-

0.9443

13.46210.10921+621-22 - 31-12

ƒ2 - 5 ƒ

3.372 * 10-3

7.526 * 1012

1721-321-221-6210219 + 16

16. List the numbers , , , , and 0.3 in numerical order.

17. What fundamental law is illustrated by?

18. (a) How many significant digits are in the number 3.0450?

(b) Round it off to two significant digits.

19. If dollars is deposited in a bank that compounds interest times a year, the value of the account after years is found byevaluating , where is the annual interest rate. Findthe value of an account for which , , ,and years (values are exact).

20. In finding the illuminance from a light source, the expressionis used. Simplify this expression.

21. The equation is used when studyingthermal expansion. Solve for .

22. An alloy weighing 20 N is 30% copper. How many newtons ofanother alloy, which is 80% copper, must be added for the finalalloy to be 60% copper?

t2

L = L031 + a1t2 - t12481100 - x22 + x2

t = 3n = 2i = 5%P = $1000

iP11 + i>n2ntt

nP

3182315 + 82 = 3152 +12-pƒ -4 ƒ-3

125. One grade of oil has 0.50% of an additive, and a higher gradehas 0.75% of the additive. How many liters of each must be usedto have 1000 L of a mixture with 0.65% of the additive?

M01_WASH42254_09_SE_C01.qxd 1/17/09 6:23 PM

2When building the pyramids nearly

5000 years ago, and today whenusing MRI (magnetic resonance

imaging) to detect a tumor in a human being,the size and shape of an object are measured.Since geometry deals with size and shape,the topics and methods of geometry areimportant in many of the applications oftechnology.

Many of the methods of measuring geometricobjects were known in ancient times, andmost of the geometry used in technology hasbeen known for hundreds of years. In about300 B.C.E., the Greek mathematician Euclid(who lived and taught in Alexandria, Egypt)organized what was known in geometry. Headded many new ideas in a 13-volume set ofwritings known as the Elements. Centurieslater it was translated into various languages,

and today is second only to the Bible asthe most published book in history.

The study of geometry includes the propertiesand measurements of angles, lines, and sur-faces and the basic figures they form. In thischapter, we review the more important meth-ods and formulas for calculating the importantgeometric measures, such as area and volume.Technical applications are included from areassuch as architecture, construction, instrumen-tation, surveying and civil engineering, me-chanical design, and product design of varioustypes, as well as other areas of engineering.

Geometric figures and concepts are also basicto the development of many areas of mathe-matics, such as graphing and trigonometry.We will start our study of graphs in Chapter 3and trigonometry in Chapter 4.

2.1 Lines and Angles

2.2 Triangles

2.3 Quadrilaterals

2.4 Circles

2.5 Measurement of IrregularAreas

2.6 Solid Geometric Figures

CHAPTER EQUATIONS

REVIEW EXERCISES

PRACTICE TEST

49

In Section 2.5, we see how to find an excel-lent approximation of the area of an irregu-lar geometric figure, such as a lake. At theleft is a satellite photograph of Lake On-tario, one of the Great Lakes between theUnited States and Canada.

Geometry

M02_WASH42254_09_SE_C02.qxd 1/17/09 6:25 PM

50 CHAPTER 2 Geometry

E X A M P L E 1 Basic angles

Figure 2.1(a) shows a right angle (marked as ). The vertex of the angle is point , and theray is the half-line . Figure 2.1(b) shows a straight angle. Figure 2.1(c) shows an acuteangle, denoted as (or or ). In Fig. 2.1(d), is an obtuse angle.∠G∠FED∠DEF∠E

BABJ

If two lines intersect such that the angle between them is a right angle, the lines areperpendicular. Lines in the same plane that do not intersect are parallel. These areillustrated in the following example.

E X A M P L E 2 Parallel lines and perpendicular lines

In Fig. 2.2(a), lines and are perpendicular (which is shown as ) sincethey meet in a right angle (again, shown as ) at .

In Fig. 2.2(b), lines and are drawn so they do not meet, even if extended.Therefore, these lines are parallel (which can be shown as ).

In Fig. 2.2(c), and AB 7 DC.AB � BC, DC � BC,AB 7 CD

CDABBJ

AC � DEDEAC

C

A

(a) (b) (c) (d)

E D G

F

180°

B

Fig. 2.1

Name of angle Measure of angle

Right angle 90

Straight angle 180

Acute angle Between 0 and 90

Obtuse angle Between 90 and 180°°

°°

°

°

■ Recognizing complementary angles isimportant in trigonometry.

D

E

C

(a)

BA

Fig. 2.2

It is not possible to define every term we use. In geometry, the meanings of point, line,and plane are accepted without being defined. These terms give us a starting point forthe definitions of other useful geometric terms.

The amount of rotation of a ray (or half-line) about its endpoint is called an angle.A ray is that part of a line (the word line means “straight line”) to one side of a fixedpoint on the line. The fixed point is the vertex of the angle. One complete rotation of aray is an angle with a measure of 360 degrees, written as . Some special types ofangles are as follows:

360°

2.1 Lines and Angles

Basic Angles • Parallel Lines andPerpendicular Lines • SupplementaryAngles and Complementary Angles •Angles Between Intersecting Lines •Segments of Transversals

■ The use of 360 comes from the ancientBabylonians and their number system based on60 rather than 10, as we use today. However,the specific reason for the choice of 360 is notknown. (One theory is that 360 is divisibleby many smaller numbers and is close to thenumber of days in a year.)

(b)

BA

DC

A B(c)

CD

It will be important to recognize perpendicular sides and parallel sides in manyof the geometric figures in later sections. ■

If the sum of the measures of two angles is , then the angles are calledsupplementary angles. Each angle is the supplement of the other. If the sum of themeasures of two angles is , the angles are called complementary angles. Each isthe complement of the other.

90°

180°

■

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2.1 Lines and Angles 51

Practice Exercise

1. What is the measure of the complementof in Fig. 2.3(a)?∠A

E X A M P L E 3 Supplementary angles and complementary angles

(a) In Fig. 2.3(a), and in Fig. 2.3(b), Sinceand are supplementary angles.

(b) In Fig. 2.4, we see that is a right angle, or Sinceis the complement of

(or is the complement of ).∠POR∠ROQ∠ROQ∠POR + ∠ROQ = ∠POQ = 90°, ∠POR

∠POQ = 90°.∠POQ

∠DEF55° + 125° = 180°, ∠BAC∠DEF = 125°.∠BAC = 55°,

A B D

C F

E(b)(a)

Supplementaryangles55° 125°

Fig. 2.3

Q

PO

RComplementaryangles

Fig. 2.4

It is often necessary to refer to certain special pairs of angles. Two angles that havea common vertex and a side common between them are known as adjacent angles. Iftwo lines cross to form equal angles on opposite sides of the point of intersection,which is the common vertex, these angles are called vertical angles.

E X A M P L E 4 Adjacent angles and vertical angles

(a) In Fig. 2.5, and have a common vertex at and the commonside between them. This means that and are adjacent angles.

(b) In Fig. 2.6, lines and intersect at point . Here, andare vertical angles, and they are equal. Also, and

are vertical angles and are equal. ■

∠AOD∠BOC∠BOD∠AOCOCDAB

∠CAD∠BACA∠CAD∠BAC

■

D

BA

C

Adjacentangles

Fig. 2.5

A

C

OB

D

Verticalangles

Fig. 2.6

Practice Exercise

2. In Fig. 2.7, if then ∠5 = ?∠2 = 42°,

A B

F

E

Transversal

2

4

13

5 687

C D

Fig. 2.7

d

b

c

a

Fig. 2.8

We should also be able to identify the sides of an angle that are adjacent to theangle. In Fig. 2.5, sides and are adjacent to and in Fig. 2.6, sides and are adjacent to Identifying sides adjacent and opposite an angle in atriangle is important in trigonometry.

In a plane, if a line crosses two or more parallel or nonparallel lines, it is called atransversal. In Fig. 2.7, and the transversal of these two parallel lines is theline .EF

AB 7 CD,

∠BOD.ODOB∠BAC,ACAB

When a transversal crosses a pair of parallel lines, certain pairs of equal angles re-sult. In Fig. 2.7, the corresponding angles are equal (that is,

and ). Also, the alternate-interior angles are equal (and ), and the alternate-exterior angles are equal ( and

).When more than two parallel lines are crossed by two transversals, such as is shown

in Fig. 2.8, the segments of the transversals between the same two parallel lines arecalled corresponding segments. A useful theorem is that the ratios of correspondingsegments of the transversals are equal. In Fig. 2.8, this means that

∠2 = ∠7∠1 = ∠8∠4 = ∠5

∠3 = ∠6∠4 = ∠8∠3 = ∠7,∠1 = ∠5, ∠2 = ∠6,

(2.1)a

b=

c

d

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E X A M P L E 5 Segments of transversals

In Fig. 2.9, part of the beam structure within a building is shown. The vertical beams areparallel. From the distances between beams that are shown, determine the distance between the middle and right vertical beams.

Using Eq. (2.1), we have

rounded off■

x =

65517752565

= 898 mm

655

565=

x

775

xx

655 mm

565 mm 775 mm

Fig. 2.9

EXERCISES 2.1

In Exercises 1–4, answer the given questions about the indicatedexamples of this section.

1. In Example 2, what is the measure of in Fig. 2.2(a)?

2. In Example 3(b), if in Fig. 2.4, what is the meas-ure of ?

3. In Example 4, how many different pairs of adjacent angles arethere in Fig. 2.6?

4. In Example 5, if the segments of 655 mm and 775 mm are inter-changed, (a) what is the answer, and (b) is the beam along which

is measured more nearly vertical or more nearly horizontal?

In Exercises 5–12, identify the indicated angles and sides in Fig. 2.10.In Exercises 9 and 10, also find the measures of the indicated angles.

5. Two acute angles 6. Two right angles

7. The straight angle 8. The obtuse angle

9. If find its complement.

10. If find its supplement.

11. The sides adjacent to

12. The acute angle adjacent to

In Exercises 13–15, use Fig. 2.11. In Exercises 16–18, use Fig. 2.12.Find the measures of the indicated angles.

13. 14. 15.

16. 17. 18. ∠5∠4∠3

∠BOD∠AOC∠AOB

∠DBC.

∠DBC

∠CBD = 65°,

∠CBD = 65°,

x

∠QOR∠POR = 32°

∠ABE

In Exercises 19–22, find the measures of the angles in Fig. 2.13.

19. 20. 21. 22. ∠4∠3∠2∠1

A C

E

B

D

Fig. 2.10

C

A

B

DO

50°

Fig. 2.11

145°

1 � 2

4

2

3

1

5

Fig. 2.12

A

C

B

D

62°

3 4

AB ‖ CD

2

1

Fig. 2.13

A44°

C

E

B

D

F

Fig. 2.14

In Exercises 23–28, find the measures of the angles in the truss shownin Fig. 2.14. A truss is a rigid support structure that is used in the con-struction of buildings and bridges.

23. 24. 25.

26. 27. 28.

In Exercises 29–32, find the indicated distances between the straightirrigation ditches shown in Fig. 2.15. The vertical ditches are parallel.

29. 30. 31. 32. dcba

∠ADE∠DFE∠DBE

∠DEB∠ABE∠BDF

3.20 m

3.05 m

4.75 m6.25 m

5.05 mc

a b

d

Fig. 2.15


33. A steam pipe is connected in sec-tions and as shown in Fig. 2.16. What is the angle be-tween and if ?AB 7 CDCDBC

CDAB, BC,C D

AB

47°Fig. 2.16

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2.2 Triangles 53

34. Part of a laser beam striking asurface is reflected and the remain-der passes through (see Fig. 2.17).Find the angle between the surfaceand the part that passes through.

36. An electric circuit board has equally spaced parallel wires withconnections at points , , and , as shown in Fig. 2.19. How faris from , if ?BC = 2.15 cmCA

CBAC

A BO

Laserbeam 28°

?

AB ⊥ CO

Fig. 2.17

35. Find the distance on Dundas St. W between Dufferin St. andOssington Ave. in Toronto, as shown in Fig. 2.18. The north–southstreets are parallel.

Bloor St. W

St.

Cla

rens

Ave

.

Duf

feri

n St

.

Oss

ingt

on A

ve.

550 m

590 m

Dundas St. W

860 m

A

B

CFig. 2.19

In Exercises 37–40, solve the given problems related to Fig 2.20.

37.

38.

39. Based on Exercise 38, whatconclusion can be drawnabout a closed geometric figure like the one with ver-tices at , , and

40. Based on Exercises 38 and 39, what conclusion can be drawn abouta closed geometric figure like the one with vertices , , , and


1. 2. 138°35°

D?CBA

D?BA

∠4 + ∠2 + ∠5 = ?

∠1 + ∠2 + ∠3 = ?

Fig. 2.20

2.2 Triangles

Types and Properties of Triangles •Perimeter and Area • Hero’s Formula •Pythagorean Theorem • Similar Triangles

When part of a plane is bounded and closed by straight-line segments, it is called apolygon, and it is named according to the number of sides it has. A triangle has threesides, a quadrilateral has four sides, a pentagon has five sides, a hexagon has sixsides, and so on. The most important polygons are the triangle, which we consider inthis section, and the quadrilateral, which we study in the next section.

TYPES AND PROPERTIES OF TRIANGLESIn a scalene triangle, no two sides are equal in length. In an isosceles triangle, two ofthe sides are equal in length, and the two base angles (the angles opposite the equalsides) are equal. In an equilateral triangle, the three sides are equal in length, and eachof the three angles is .

The most important triangle in technical applications is the right triangle. In a righttriangle, one of the angles is a right angle. The side opposite the right angle is thehypotenuse, and the other two sides are called legs.

E X A M P L E 1 Types of triangles

Figure 2.21(a) shows a scalene triangle; each side is of a different length. Figure 2.21(b)shows an isosceles triangle with two equal sides of 2 m and equal base angles of .Figure 2.21(c) shows an equilateral triangle with equal sides of 5 cm and equal angles of

. Figure 2.21(d) shows a right triangle. The hypotenuse is side AB.60°

40°

60°

■ The properties of the triangle are importantin the study of trigonometry, which we start inChapter 4.

B

C A

(a)

5 cm

(b)

2 m

(c) (d)

Hypotenuse5 cm 5 cm

5 cm

2 m4 cm6 cm

60°60°40° 40°

60°

Fig. 2.21

W

WFig. 2.18

■

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One very important property of a triangle is that

the sum of the measures of the three angles of a triangle is .

In the next example, we show this property by using material from Section 2.1.

E X A M P L E 2 Sum of angles of triangle

In Fig. 2.22, since and constitute a straight angle,

Also, by noting alternate interior angles, we see that and There-fore, by substitution, we have

see Exercises 37–40 on page 53

Therefore, if two of the angles of a triangle are known, the third may be found by sub-tracting the sum of the first two from ■

E X A M P L E 3 Application of sum of angles

An airplane is flying north and then makes a turn to the west. Later, it makes an-other left turn of . What is the angle of a third left turn that will cause the plane toagain fly north? See Fig. 2.23.

From Fig. 2.23, we see that the interior angle of the triangle at is the supple-ment of , or . Since the sum of the measures of the interior angles of the tri-angle is , the interior angle at is

The required angle is the supplement of which is ■

A line segment drawn from a vertex of a triangle to the midpoint of the opposite sideis called a median of the triangle. A basic property of a triangle is that the three mediansmeet at a single point, called the centroid of the triangle. See Fig. 2.24. Also, the threeangle bisectors (lines from the vertices that divide the angles in half) meet at a commonpoint. See Fig. 2.25.

120°.60°,

∠B = 180° - 190° + 30°2 = 60°

B180°30°150°

A

150°90°

180°.

∠4 + ∠2 + ∠5 = 180°

∠3 = ∠5.∠1 = ∠4

∠1 + ∠2 + ∠3 = 180°

∠3∠1, ∠2,

180°NOTE �

DE C

A B

2

4 5

31

AB ‖ EC

Fig. 2.22

A

B

30° 90° N

W

?

150°

Fig. 2.23

MediansCentroid

Fig. 2.24

Angle bisectors

Fig. 2.25

Altitudes

Fig. 2.26

Practice Exercise

1. If the triangle in Fig. 2.24 is isosceles andthe vertex angle (at the left) is , whatare the base angles?

30°

An altitude (or height) of a triangle is the line segment drawn from a vertex perpen-dicular to the opposite side (or its extension), which is called the base of the triangle.The three altitudes of a triangle meet at a common point. See Fig. 2.26. The three com-mon points of the medians, angle bisectors, and altitudes are generally not the samepoint for a given triangle.

PERIMETER AND AREA OF A TRIANGLEWe now consider two of the most basic measures of a plane geometric figure. The firstof these is its perimeter, which is the total distance around it. In the following exam-ple, we find the perimeter of a triangle.

M02_WASH42254_09_SE_C02.qxd 1/17/09 6:25 PM

2.2 Triangles 55

4.89 m

3.22 m2.56 m

Fig. 2.27

E X A M P L E 4 Perimeter of triangle

Find the perimeter of a triangle with sides 2.56 m, 3.22 m, and 4.89 m. See Fig. 2.27.Using the definition of perimeter, for this triangle we have

Therefore, the distance around the triangle is 10.67 m. We express the results tohundredths since each side is given to hundredths. ■

The second important measure of a geometric figure is its area. Although the con-cept of area is primarily intuitive, it is easily defined and calculated for the basic geo-metric figures. Area gives a measure of the surface of the figure, just as perimeter givesthe measure of the distance around it.

The area of a triangle of base and altitude ishbA

p = 2.56 + 3.22 + 4.89 = 10.67 m

p

(2.2)A =12 bh

(2.3)where s =

12 1a + b + c2

A = 1s1s - a21s - b21s - c2,

In Eq. (2.3), , , and are the lengths of the sides, and is one-half of theperimeter.

scba

5.75 cm

16.2 cm 16.2 cm(a) (b)

Fig. 2.28

■ Named for Hero (or Heron), a first-centuryGreek mathematician.

E X A M P L E 6 Application of Hero’s formula

A surveyor measures the sides of a triangular parcel of land betweentwo intersecting straight roads to be 206 m, 293 m, and 187 m. Findthe area of this parcel (see Fig. 2.29).

To use Eq. (2.3), we first find :

Now, substituting in Eq. (2.3), we have

= 19 100 m2

A = 13431343 - 20621343 - 29321343 - 1872

=12 16862 = 343 m

s =12 1206 + 293 + 1872

s

206 m

187 m293 m

Fig. 2.29

The calculator solution is shown in Fig. 2.30. Note that the value of was stored inmemory (as ) and then used to find . Using the calculator, it is not necessary towrite down anything but the final result, properly rounded off. ■

AxsPractice Exercise

2. What is the perimeter of the triangle inFig. 2.29?

Fig. 2.30

The following example illustrates the use of Eq. (2.2).

E X A M P L E 5 Area of triangle

Find the areas of the triangles in Fig. 2.28(a) and Fig. 2.28(b).Even though the triangles are of different shapes, the base of each is 16.2 cm, and

the altitude of each is 5.75 cm. Therefore, the area of each triangle is

■

Another formula for the area of a triangle that is particularly useful when we havea triangle with three known sides and no right angle is Hero’s formula, which is

A =12 bh =

12 116.2215.752 = 46.6 cm2

hb

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NOTE � For two triangles that are similar, if one property is true, then the other is also true. Intwo similar triangles, the corresponding sides are the sides, one in each triangle, thatare between the same pair of equal corresponding angles.

E X A M P L E 8 Similar triangles

In Fig. 2.34, a pair of similar triangles are shown. They are similar even though thecorresponding parts are not in the same position relative to the page. Using standardsymbols, we can write where means “triangle” and means“is similar to.”

The pairs of corresponding angles are and and and and Thismeans and

The pairs of corresponding sides are and and and andIn order to show that these corresponding sides are proportional, we write

■AB

A¿B¿

=

BC

B¿C¿

=

AC

A¿C¿

A¿C¿.ACB¿C¿,BCA¿B¿,AB

C = C¿.A = A¿, B = B¿,C¿.CB¿,BA¿,A

'^ÂBC '

Â¿B¿C¿,

Pole

Guy wire

3.20

m

2.65 m

AB

C

Fig. 2.32

ac

b

Fig. 2.31

THE PYTHAGOREAN THEOREMAs we have noted, one of the most important geometric figures in technical applica-tions is the right triangle. A very important property of a right triangle is given by thePythagorean theorem, which states that

in a right triangle, the square of the length of the hypotenuse equals the sum of thesquares of the lengths of the other two sides.

If is the length of the hypotenuse and and are the lengths of the other two sides(see Fig. 2.31), the Pythagorean theorem is

bac

E X A M P L E 7 Application of Pythagorean theorem

A pole is perpendicular to the level ground around it. A guy wire is attached3.20 m up the pole and at a point on the ground, 2.65 m from the pole. Howlong is the guy wire?

We sketch the pole and guy wire as shown in Fig. 2.32. Using thePythagorean theorem, and then substituting, we have

The guy wire is 4.15 m long. The calculator solution is shown in Fig 2.33.Note the use of parentheses to group ■

SIMILAR TRIANGLESThe perimeter and area of a triangle are measures of its size. We now consider the shapeof triangles.

Two triangles are similar if they have the same shape (but not necessarily the samesize). There are two very important properties of similar triangles.

2.652+ 3.202.

AC = 22.652+ 3.202

= 4.15 m

= 2.652+ 3.202

AC2= AB2

+ BC2

(2.4)c2= a2

+ b2

■ Named for the Greek mathematicianPythagoras (sixth century B.C.E.)

■ Calculators can be programmed to performspecific calculations. See Appendix C for agraphing calculator program PYTHAGTH. It canbe used to find a side of a right triangle, giventhe other two sides.

PROPERTIES OF SIMILAR TRIANGLES

1. The corresponding angles of similar triangles are equal.

2. The corresponding sides of similar triangles are proportional.

BC A' C '

B'

A

Fig. 2.34

sides of

sides of Â'B'C'

➝

ÂBC

➝

Fig. 2.33

M02_WASH42254_09_SE_C02.qxd 1/17/09 6:25 PM

2.2 Triangles 57

24 m 3.0 m

4.0

mh

Fig. 2.35

Practice Exercise

3. In Fig. 2.35, knowing the value of ,what is the distance between the top ofthe silo and the end of its shadow?

h

One of the most practical uses of similar geometric figures is that of scale drawings.Maps, charts, architectural blueprints, engineering sketches, and most drawings thatappear in books (including many that have already appeared, and will appear, in thistext) are familiar examples of scale drawings.

In any scale drawing, all distances are drawn at a certain ratio of the distances thatthey represent, and all angles are drawn equal to the angles they represent. Note the dis-tances and angles shown in Fig. 2.36 in the following example.

E X A M P L E 1 0 Scale drawing

In drawing a map of the area shown in Fig. 2.36, a scale of is used. Inmeasuring the distance between Chicago and Toronto on the map, we find it to be 3.5 cm.The actual distance between Chicago and Toronto is found from the proportion

scale

If we did not have the scale but knew that the distance between Chicago andToronto is 700 km, then by measuring distances on the map between Chicagoand Toronto (3.5 cm) and between Toronto and Philadelphia (2.7 cm), we couldfind the distance between Toronto and Philadelphia. It is found from the follow-ing proportion, determined by use of similar triangles:

■

Similarity requires equal angles and proportional sides. If the corresponding anglesand the corresponding sides of two triangles are equal, the two triangles arecongruent. As a result of this definition, the areas and perimeters of congruent trian-gles are also equal. Informally, we can say that similar triangles have the same shape,whereas congruent triangles have the same shape and same size.

y =

2.7170023.5

= 540 km

700 km

3.5 cm=

y

2.7 cm

x

3.5 cm=

200 km

1 cm or x = 700 km

actual distance ➝distance on map ➝

x

1 cm = 200 km

➝

Chicago Philadelphia

2.7 cm3.5 cm

Toronto

Fig. 2.36

If we know that two triangles are similar, we can use the two basic properties ofsimilar triangles to find the unknown parts of one triangle from the known parts ofthe other triangle. The next example illustrates this in a practical application.

E X A M P L E 9 Application of similar triangles

On level ground, a silo casts a shadow 24 m long. At the same time, a nearby verticalpole 4.0 m high casts a shadow 3.0 m long. How tall is the silo? See Fig. 2.35.

The rays of the sun are essentially parallel. The two triangles in Fig. 2.35 are sim-ilar since each has a right angle and the angles at the tops are equal. The other anglesmust be equal since the sum of the angles is . The lengths of the hypotenusesare of no importance in this problem, so we use only the other sides in stating theratios of corresponding sides. Denoting the height of the silo as , we have

We conclude that the silo is 32 m high. ■

h

4.0=

24

3.0 , h = 32 m

h

180°

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E X A M P L E 1 1 Similar and congruent triangles

A right triangle with legs of 2 m and 4 m is congruentto any other right triangle with legs of 2 m and 4 m Itis also similar to any right triangle with legs of 5 mand 10 m, or any other right triangle that has legs inthe ratio of 1 to 2, since the corresponding sides areproportional. See Fig. 2.37. ■

2 m

4 m

2 m

4 m

5 m

10 m

SimilarCongruent

Fig. 2.37

EXERCISES 2.2

In Exercises 1–4, answer the given questions about the indicated ex-amples of this section.

1. In Example 2, if and in Fig. 2.22, what isthe measure of ?

2. In Example 5, change 16.2 cm to 61.2 cm. What is the answer?

3. In Example 7, change 2.65 m to 6.25 m. What is the answer?

4. In Example 9, interchange 4.0 m and 3.0 m. What is the answer?

In Exercises 5–8, determine in the indicated figures.

5. Fig. 2.38(a)

6. Fig. 2.38(b)

7. Fig. 2.38(c)

8. Fig. 2.38(d)

∠A

∠2∠5 = 45°∠1 = 70°

In Exercises 9–16, find the area of each triangle.

9. Fig. 2.39(a) 10. Fig. 2.39(b)

11. Fig. 2.39(c) 12. Fig. 2.39(d)

(a)

84°

40°A B

(b)

48°

AC

B

C

(c)

66°

44

CB(d)

110° 33

A B

C

A

Fig. 2.38

2.2 m

7.6 m

7.62 mm

16.0 mm

205 cm 322 cm

415 cm23.5 m 68.4 m

86.2 m

(a) (b)

(d)(c)

Fig. 2.39

13. Right triangle with legs 3.46 cm and 2.55 cm.

14. Right triangle with legs 234 mm and 342 mm.

15. Isosceles triangle, equal sides of 0.986 m, third side of 0.884 m

16. Equilateral triangle of sides 320 dm

In Exercises 17–20, find the perimeter of each triangle.

17. Fig. 2.39(c) 18. Fig. 2.39(d)

19. An equilateral triangle of sides 21.5 cm

20. Isosceles triangle, equal sides of 2.45 mm, third side of 3.22 mm.

In Exercises 21–24, find the third side of the right triangle shown inFig. 2.40 for the given values.

21.

22.

23.

24.

In Exercises 25–28, use the right triangle in Fig. 2.41.

25. Find

26. Find side .

27. Find the perimeter.

28. Find the area.


29. What is the angle between the bisectors of the acute angles of aright triangle?

30. If the midpoints of the sides of an isosceles triangle are joined,another triangle is formed. What do you conclude about this innertriangle?

31. For what type of triangle is the centroid the same as the intersec-tion of altitudes and the intersection of angle bisectors?

32. Is it possible that the altitudes of a triangle meet, when extended,outside the triangle? Explain.

c

∠B.

b = 0.474 km, c = 0.836 km

a = 175 cm, c = 551 cm

a = 2.48 m, b = 1.45 m

a = 13.8 mm, b = 22.7 mmac

b

Fig. 2.40

90.5 cm

23°

38.4 cm

B

c

Fig. 2.41

W

W

M02_WASH42254_09_SE_C02.qxd 1/17/09 6:25 PM

2.2 Triangles 59

W

W

33. The altitude to the hypotenuse of a right triangle divides the trian-gle into two smaller triangles. What do you conclude about thesetwo triangles?

34. The altitude to the hypotenuse of a right triangle divides the trian-gle into two smaller triangles. What do you conclude about theoriginal triangle and the two new triangles? Explain.

35. In Fig. 2.42, show that

36. In Fig. 2.43, show that

37. In Fig. 2.42, if and find .

38. In Fig. 2.43, if and find .

39. The angle between the roof sections of an A-frame house is .What is the angle between either roof section and a horizontalrafter?

40. A transmitting tower is supported by a wire that makes an angleof with the level ground. What is the angle between the towerand the wire?

41. A wall pennant is in the shape of an isosceles triangle. If eachequal side is 76.6 cm long and the third side is 30.6 cm, what isthe area of the pennant?

42. The Bermuda Triangle is sometimes defined as an equilateral trian-gle 1600 km on a side, with vertices in Bermuda, Puerto Rico, andthe Florida coast. Assuming it is flat, what is its approximate area?

43. The sail of a sailboat is in the shape of a right triangle with sidesof 3.2 m, 6.0 m, and 6.8 m. What is the area of the sail?

44. An observer is 550 m horizontally from the launch pad of a rocket.After the rocket has ascended 750 m, how far is it from the observer?

45. The base of a 6.0-m ladder is 1.8 m from a wall. How far up on thewall does the ladder reach?

46. The beach shade shown in Fig. 2.44 is made of

triangularsections. Find . (In a

triangle, the sideopposite the angle isone-half the hypotenuse.)

47. A rectangular room is 18 m long, 12 m wide, and 8.0 m high.What is the length of the longest diagonal from one corner to an-other corner of the room?

30°30°-60°-90°

x30°-60°-90°

52°

50°

ABAC = 12,AD = 9

LMMO = 12,KN = 15, MN = 9,

D

B

C A

Fig. 2.43

M

L

K

O

N

Fig. 2.42

^ACB '^ADC.

^MKL '^MNO.

48. On a blueprint, a hallway is 45.6 cm long. The scale isHow long is the hallway?

49. Two parallel guy wires are attached to a vertical pole 4.5 m and5.4 m above the ground. They are secured on the level ground atpoints 1.2 m apart. How long are the guy wires?

50. The two sections of a folding door, hinged in the middle, are atright angles. If each section is 0.85 m wide, how far are the hingesfrom the far edge of the other section?

51. A 1.5-m wall stands 0.75 m from a building. The ends of astraight pole touch the building and the ground 2.0 m from thewall. A point on the pole touches the top of the wall. How long isthe pole? See Fig. 2.45.

52. To find the width of a river, a surveyor places markers at ,, and , as shown in Fig. 2.46. The markers are placed such

that and How wide is the river?

53. A water pumping station is to be built on a river at point in orderto deliver water to points and . See Fig. 2.47. The design re-quires that so that the total length of piping thatwill be needed is a minimum. Find this minimum length of pipe.

54. The cross section of a drainage trough has the shape of an isosce-les triangle whose depth is 12 cm less than its width. If the depthis increased by 16 cm and the width remains the same, the area ofthe cross section is increased by Find the original depthand width. See Fig. 2.48.


1. 2. 686 m 3. 40 m75°

160 cm2.

16 cm

x � 12 cm

x

Fig. 2.48

A

B

P

C D

10.0 km

6.00 km

12.0 km

Fig. 2.47

∠APD = ∠BPCBA

P

AB = 80.0 m.DC = 312 m,BC = 50.0 m,AB 7 ED,DC

B,AED

DE

B

C

A

Fig. 2.46

0.75 m

1.5 mWall B

uild

ing

Pole

2.0 m

Fig. 2.45

1.2 cm = 1.0 m.

2.00 m

x

Fig. 2.44

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PERIMETER AND AREA OF A QUADRILATERALThe perimeter of a quadrilateral is the sum of the lengths of the four sides.

E X A M P L E 2 Application of perimeter

An architect designs a room with a rectangular window 920 mm high and 540 mmwide, with another window above in the shape of an equilateral triangle, 540 mm on aside. See Fig. 2.52. How much molding is needed for these windows?

The length of molding is the sum of the perimeters of the windows. For the rec-tangular window, the opposite sides are equal, which means the perimeter is twicethe length plus twice the width . For the equilateral triangle, the perimeter isthree times the side . Therefore, the length of molding is

mm ■

We could write down formulas for the perimeters of the different kinds of trian-gles and quadrilaterals. However, if we remember the meaning of perimeter as beingthe total distance around a geometric figure, such formulas are not necessary.

= 4540

= 219202 + 215402 + 315402 L = 2l + 2w + 3s

Lswl

A quadrilateral is a closed plane figure with four sides, and these four sides form fourinterior angles. A general quadrilateral is shown in Fig. 2.49.

A diagonal of a polygon is a straight line segment joining any two nonadjacent ver-tices. The dashed line is one of two diagonals of the quadrilateral in Fig. 2.50.

TYPES OF QUADRILATERALSA parallelogram is a quadrilateral in which opposite sides are parallel. In a parallelo-gram, opposite sides are equal and opposite angles are equal. A rhombus is a parallel-ogram with four equal sides.

A rectangle is a parallelogram in which intersecting sides are perpendicular, whichmeans that all four interior angles are right angles. In a rectangle, the longer side is usu-ally called the length, and the shorter side is called the width. A square is a rectanglewith four equal sides.

A trapezoid is a quadrilateral in which two sides are parallel. The parallel sides arecalled the bases of the trapezoid.

E X A M P L E 1 Types of quadrilaterals

A parallelogram is shown in Fig. 2.51(a). Opposite sides are equal in length, as areopposite sides . A rhombus with equal sides is shown in Fig. 2.51(b). A rectangle isshown in Fig. 2.51(c). The length is labeled , and the width is labeled . A squarewith equal sides is shown in Fig. 2.51(d). A trapezoid with bases and is shownin Fig. 2.51(e).

b2b1swl

sba

2.3 Quadrilaterals

Types and Properties ofQuadrilaterals • Perimeter and Area

Fig. 2.49 Fig. 2.50

b1

b2s

s

ss

(e)(d)

l

l

w w

(c)

s

s

ss

(b)

a

a

bb

(a)Fig. 2.51 ■

Practice Exercise

1. Develop a formula for the length of adiagonal for the rectangle in Fig. 2.51(c).

d

920 mm

540 mm

540 mm

Fig. 2.52

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2.3 Quadrilaterals 61

Practice Exercise

2. If we did develop perimeter formulas,what is the formula for the perimeter ofa rhombus of side ?s

p

Square of side (Fig. 2.53) (2.5)

Rectangle of length and width (Fig. 2.54) (2.6)

Parallelogram of base and height (Fig. 2.55) (2.7)

Trapezoid of bases and and height (Fig. 2.56) (2.8)hb2b1A =12 h(b1 + b2)

hbA = bh

wlA = lw

sA = s2

Since a rectangle, a square, and a rhombus are special types of parallelograms, the areaof these figures can be found from Eq (2.7). The area of a trapezoid is of importancewhen we find areas of irregular geometric figures in Section 2.5.

E X A M P L E 3 Application of area

A city park is designed with lawn areas in the shape of a right triangle, a parallelogram,and a trapezoid, as shown in Fig. 2.57, with walkways between them. Find the area ofeach section of lawn and the total lawn area.

The total lawn area is about ■

E X A M P L E 4 Perimeter in a word problem

The length of a rectangular computer chip is 2.0 mm longer than its width. Find thedimensions of the chip if its perimeter is 26.4 mm.

Since the dimensions, the length and the width, are required, let the width of thechip. Since the length is 2.0 mm more than the width, we know that thelength of the chip. See Fig. 2.58.

Since the perimeter of a rectangle is twice the length plus twice the width, wehave the equation

since the perimeter is given as 26.4 mm. This is the equation we need.Solving this equation, we have

Therefore, the length is 7.6 mm and the width is 5.6 mm. These values check with thestatements of the original problem. ■

w = 5.6 mm and w + 2.0 = 7.6 mm

4w = 22.4

2w + 4.0 + 2w = 26.4

21w + 2.02 + 2w = 26.4

w + 2.0 =

w =

7200 m2.

A3 =12 h1b1 + b22 =

12 1452172 + 352 = 2400 m2

A2 = bh = 17221452 = 3200 m2

A1 =12 bh =

12 17221452 = 1600 m2

s

Fig. 2.53

Fig. 2.57

l

w

Fig. 2.54

b

h

Fig. 2.55

b2

b1

h

Fig. 2.56

35 m

45 m

72 m72 m

45 m72 m

45 m

For the areas of the square, rectangle, parallelogram, and trapezoid, we have the fol-lowing formulas.

w � 2.0

w

Fig. 2.58

■ The computer microprocessor chip was firstcommercially available in 1971.

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EXERCISES 2.3

In Exercises 1–4, make the given changes in the indicated examplesof this section and then solve the given problems.

1. In Example 1, interchange the lengths of and in Fig. 2.51(e).What type of quadrilateral is the resulting figure?

2. In Example 2, change the equilateral triangle of side 540 mm to asquare of side 540 mm and then find the length of molding.

3. In Example 3, change the dimension of 45 m to 55 m in each fig-ure and then find the area.

4. In Example 4, change 2.0 mm to 3.0 mm, and then solve.

In Exercises 5–12, find the perimeter of each figure.

5. Square: side of 65 m 6. Rhombus: side of 2.46 km

7. Rectangle:

8. Rectangle:

9. Parallelogram in Fig. 2.59 10. Parallelogram in Fig. 2.60

11. Trapezoid in Fig. 2.61 12. Trapezoid in Fig. 2.62

l = 142 cm, w = 126 cm

l = 0.920 mm, w = 0.742 mm

b2b1

In Exercises 13–20, find the area of each figure.

13. Square: 14. Square:

15. Rectangle:

16. Rectangle:

17. Parallelogram in Fig. 2.59 18. Parallelogram in Fig. 2.60

19. Trapezoid in Fig. 2.61 20. Trapezoid in Fig. 2.62

In Exercises 21–24, set up a formula for the indicated perimeter orarea. (Do not include dashed lines.)

21. The perimeter of the figure in Fig. 2.63 (a parallelogram and asquare attached)

22. The perimeter of the figure in Fig. 2.64 (two trapezoids attached)

23. Area of figure in Fig. 2.63

24. Area of figure in Fig. 2.64

l = 142 cm, w = 126 cm

l = 0.920 km, w = 0.742 km

s = 15.6 ms = 2.7 mm

a

b

h

Fig. 2.63

a

a

b

b

Fig. 2.64


25. If the angle between adjacent sides of a parallelogram is ,what conclusion can you make about the parallelogram?

26. What conclusion can you make about the two triangles formed bythe sides and diagonal of a parallelogram? Explain.

27. Find the area of a square whose diagonal is 24.0 cm.

28. In a trapezoid, find the angle between the bisectors of the twoangles formed by the bases and one nonparallel side.

29. Noting how a diagonal of a rhombus divides an interior angle,explain why the automobile jack in Fig. 2.65 is in the shape of arhombus.

90°

Fig. 2.65

12 mm

16 mm

Fig. 2.66

W

W

W

30. Part of an electric circuit is wired in the configuration of a rhom-bus and one of its altitudes as shown in Fig. 2.66. What is thelength of wire in this part of the circuit?

31. A walkway 3.0 m wide is constructed along the outside edge of asquare courtyard. If the perimeter of the courtyard is 320 m, whatis the perimeter of the square formed by the outer edge of thewalkway?

32. An architect designs a rectangular window such that the width ofthe window is 450 mm less than the height. If the perimeter of thewindow is 4500 mm, what are its dimensions?

33. A designer plans the top of a rectangular workbench to be fourtimes as long as it is wide and then determines that if the width is1500 mm greater and the length is 4500 mm less, it would be asquare. What are its dimensions?

34. A beam support in a building is in the shape of a parallelogram, asshown in Fig. 2.67. Find the area of the side of the beam shown.

3.50 m

1.80 m

Fig. 2.67

9300 mm

4200 mm

5300 mm

3300 mm 3300 mm1200 mm

Fig. 2.68

35. Each of two walls (with rectangular windows) of an A-framehouse has the shape of a trapezoid as shown in Fig. 2.68. If a litreof paint covers 12 m2, how much paint is required to paint thesewalls? (All data are accurate to two significant digits.)

2.5 m 2.7 m

3.7 m

Fig. 2.59

12.6 mm 14.2 mm

27.3 mm

Fig. 2.60

29.8 dm36.2 dm 44.0 dm

73.0 dm

61.2 dm

Fig. 2.61

201 cm272 cm 223 cm

672 cm

392 cm

Fig. 2.62

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2.4 Circles 63

W

38. A rectangular security area is enclosed on one side by a wall, andthe other sides are fenced. The length of the wall is twice thewidth of the area. The total cost of building the wall and fence is$13 200. If the wall costs and the fence costs find the dimensions of the area.

39. What is the sum of the measures of the interior angles of a quadri-lateral? Explain.

40. Find a formula for the area of a rhombus in terms of its diagonalsand (See Exercise 29.)


1. 2. p = 4sd = 2l2+ w2

d2.d1

$5.00>m,$50.00>m

2.4 Circles

Properties of Circles • Tangent and Secant Lines • Circumference and Area •Circular Arcs and Angles • Radian Measure of an Angle

36. Six equal trapezoidal sec-tions form a conferencetable in the shape of ahexagon, with a hexagonalopening in the middle. See Fig. 2.69. From thedimensions shown, findthe area of the table top.

150 cm

75.0 cm

75.0 cm75.0 cm

37. A fenced section of a ranch is in the shape of a quadrilateral whosesides are 1.74 km, 1.46 km, 2.27 km, and 1.86 km, the last two sidesbeing perpendicular to each other. Find the area of the section.

Diameter

Center

Radius Radius

Fig. 2.70

B

OA

Fig. 2.72

(2.9)

(2.10) A = pr2 Area of a circle of radius r

c = 2pr Circumference of a circle of radius r

Here, equals approximately 3.1416. In using a calculator, can be entered to a muchgreater accuracy by using the key.p

pp

■ The symbol (the Greek letter pi), which weuse as a number, was first used in this way as anumber in the 1700s.

p

■ On the TI-83 and TI-84, values are stored inmemory using up to 14 digits with a two-digitexponent.

Fig. 2.69

The next geometric figure we consider is the circle. All points on a circle are at thesame distance from a fixed point, the center of the circle. The distance from the centerto a point on the circle is the radius of the circle. The distance between two points onthe circle on a line through the center is the diameter. Therefore, the diameter istwice the radius , or See Fig. 2.70.

There are also certain types of lines associated with a circle. A chord is a line seg-ment having its endpoints on the circle. A tangent is a line that touches (does not passthrough) the circle at one point. A secant is a line that passes through two points ofthe circle. See Fig. 2.71.

An important property of a tangent is that a tangent is perpendicular to the radiusdrawn to the point of contact. This is illustrated in the following example.

E X A M P L E 1 Tangent line perpendicular to radius

In Fig. 2.72, is the center of the circle, and is tangent at . If find

Since the center is is a radius of the circle. A tangent is perpendicular to aradius at the point of tangency, which means so that

Since the sum of the angles of a triangle is , we have

■

CIRCUMFERENCE AND AREA OF A CIRCLEThe perimeter of a circle is called the circumference. The formulas for the circumfer-ence and area of a circle are as follows:

∠AOB = 180° - 115° = 65°

180°

∠OAB + ∠OBA = 25° + 90° = 115°

∠ABO = 90°O, OB

∠AOB.∠OAB = 25°,BABO

d = 2r.rd

Secant

Tangent

Chord

Fig. 2.71

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E X A M P L E 2 Application of area

A circular oil spill has a diameter of 2.4 km. It is to be enclosed within special flexibletubing. What is the area of the spill, and how long must the tubing be? See Fig. 2.73.

Since , Using Eq. (2.10), the area is

The length of tubing needed is the circumference of the circle. Therefore,

note that

rounded off ■

Many applied problems involve a combination of geometric figures. The followingexample illustrates one such combination.

E X A M P L E 3 Application of perimeter and area

A machine part is a square of side 3.25 cm with a quarter-circle removed (see Fig.2.74). Find the perimeter and the area of one side of the part.

Setting up a formula for the perimeter, we add the two sides of length to one-fourth of the circumference of a circle with radius . For the area, we subtract the areaof one-fourth of a circle from the area of the square. This gives

bottom circular quarterand left section square circle

where is the side of the square and the radius of the circle. Evaluating, we have

The calculator solution for and is shown in Fig. 2.75. Note that is stored (as )in order to shorten the solution. ■

xsAp

A = 3.252-

p13.25224

= 2.27 cm2

p = 213.252 +

p13.2522

= 11.6 cm

s

p = 2s +

2ps

4= 2s +

ps

2 A = s2

-

ps2

4

ss

= 7.5 km

c = pd c = 2pr = 2p11.22

= 4.5 km2

A = pr2= p11.222

r = d>2 = 1.2 km.d = 2rOil

Spill

Tubing

2.4 km

Fig. 2.73

s � 3.25 cm

s

Fig. 2.74

Fig. 2.75

➝

➝ ➝

➝

Practice Exercises

1. Find the circumference of a circle with aradius of 20.0 cm.

2. Find the area of the circle in Exercise 1.

AB

C

O

Sector

Segment

Arc

Centralangle

Fig. 2.76

CIRCULAR ARCS AND ANGLESAn arc is part of a circle, and an angle formed at the center by two radii is a central an-gle. The measure of an arc is the same as the central angle between the ends of the radiithat define the arc. A sector of a circle is the region bounded by two radii and the arcthey intercept. A segment of a circle is the region bounded by a chord and its arc. (Thereare two possible segments for a given chord. The smaller region is a minor segment, andthe larger region is a major segment.) These are illustrated in the following example.

E X A M P L E 4 Sector and segment

In Fig. 2.76, a sector of the circle is between radii and and arc (denotedas ). If the measure of the central angle at between the radii is , the measure of

is .A segment of the circle is the region between chord and arc ■

An inscribed angle of an arc is one for which the endpoints of the arc are points onthe sides of the angle and for which the vertex is a point (not an endpoint) of the arc.An important property of a circle is that the measure of an inscribed angle is one-halfof its intercepted arc.

1BCµ 2.BCBC

70°ABµ 70°OAB

µ ABOBOA

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2.4 Circles 65

E X A M P L E 5 Inscribed angle

(a) In the circle shown in Fig. 2.77, is inscribed in and itintercepts . If then

(b) In the circle shown in Fig. 2.78, is a diameter, and is inscribed in the semicircular Since From this we conclude that an angle inscribed in a semicircle is aright angle. ■

RADIAN MEASURE OF AN ANGLETo this point, we have measured all angles in degrees. There is another measure of anangle, the radian, that is defined in terms of an arc of a circle. We will find it of impor-tance when we study trigonometry.

If a central angle of a circle intercepts an arc equal in length to the radius of the cir-cle, the measure of the central angle is defined as 1 radian. See Fig. 2.79. The radiuscan be marked off along the circumference times (about 6.283 times). Thus,

(where rad is the symbol for radian), and the basic relationship betweenradians and degrees is2p rad = 360°

2p

PSQ¬

= 180°, ∠PRQ = 90°.PRQ¬

.∠PRQPQ

∠ABC = 30°.ACµ

= 60°,ACµ ABC

¬,∠ABC

A

C

B

Inscribedangle

Inte

rcep

ted

arc

Fig. 2.77

R

QPO

S

Fig. 2.78

(2.11)p rad = 180°

r

r

Arc lengthequals radius

1 rad

Fig. 2.79

Practice Exercise

3. Express the angle in radian measure.85.0°

E X A M P L E 6 Radian measure of an angle

(a) If we divide each side of Eq. (2.11) by we get

where the result has been rounded off.

(b) To change an angle of to radian measure, we have

118.2° = 118.2°ap rad

180°b = 2.06 rad

118.2°

1 rad = 57.3°

p,

Multiplying by the unit that remains is rad, since degrees “cancel.”We will review radian measure again when we study trigonometry. ■

p rad>180°,118.2°

EXERCISES 2.4


1. In Example 1, if in Fig. 2.72, then what is themeasure of ?

2. In the first line of Example 2, if “diameter” is changed to “radius,”what are the results?

3. In Example 3, if the machine part is the unshaded part (rather thanthe shaded part) of Fig. 2.74, what are the results?

4. In Example 5(a), if in Fig. 2.77, then what is themeasure of ?AC

µ ∠ABC = 25°

∠OAB∠AOB = 72°

In Exercises 5–8, refer to the circle with center at in Fig. 2.80. Iden-tify the following.

5. (a) A secant line(b) A tangent line

6. (a) Two chords(b) An inscribed

angle

7. (a) Two perpen-dicular lines

(b) An isoscelestriangle

8. (a) A segment(b) A sector with an acute central angle

O

OA B C

F

E

D

Fig. 2.80

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44. The cross section of a large circular conduit has seven smallerequal circular conduits within it. The conduits are tangent to eachother as shown in Fig. 2.87. What fraction of the large conduit isoccupied by the seven smaller conduits?

45. Find the area of the room in the plan shown in Fig. 2.88.

O

A

C

TBFig. 2.81

12 000 mm

8100 mm

320 mm

Fig. 2.88

60°

80°

A B

C

Fig. 2.82

r

Fig. 2.83

ra

b

Fig. 2.84

90 cm180 cm

Fig. 2.86 Fig. 2.87

W

W

In Exercises 9–12, find the circumference of the circle with the givenradius or diameter.

9. 10.

11. 12. dm

In Exercises 13–16, find the area of the circle with the given radius ordiameter.

13. km 14.

15. 16. mm

In Exercises 17–20, refer to Fig. 2.81, where is a diameter, is atangent line at , and . Determine the indicated angles.

17.

18.

19.

20.

In Exercises 21–24, refer to Fig. 2.82. Determine the indicated arcsand angles.

21.

22.

23.

24.

In Exercises 25–28, change the given angles to radian measure.

25. 26. 27. 28.

In Exercises 29–32, find a formula for the indicated perimeter or area.

323.0°125.2°60.0°22.5°

∠ACB

∠ABC

ABµ

BCµ

∠BTC

∠CAB

∠BCT

∠CBT

∠ABC = 65°BTBAB

d = 1256d = 2.33 m

r = 45.8 cmr = 0.0952

d = 8.2d = 23.1 mm

r = 0.563 mr = 275 cm

35. In a circle, a chord connects the ends of two perpendicular radiiof 6.00 cm. What is the area of the minor segment?

36. In Fig. 2.85, chords andare parallel. What is the

relation between and? Explain.

37. The radius of the earth’s equator is 6370 km. What is the circum-ference?

38. As a ball bearing rolls along a straight track, it makes 11.0 revo-lutions while traveling a distance of 109 mm. Find its radius.

39. The rim on a basketball hoop has an inside diameter of 45.7 cm.The largest cross section of a basketball has a diameter of 30.5 cm.What is the ratio of the cross sectional area of the basketball to thearea of the hoop?

40. With no change in the speed of flow, by what factor should the di-ameter of a pipe be increased in order to double the amount ofwater that flows through the pipe?

41. Using a tape measure, the circumference of a tree is found to be112 cm. What is the diameter of the tree (assuming a circularcross section)?

42. What is the area of the largest circle that can be cut from a rectan-gular plate 21.2 cm by 15.8 cm?

43. A window designed between semicircular regions is shown inFig. 2.86. Find the area of the window.

^CDE^ABC

DEAB

29. The perimeter of the quarter-circle in Fig. 2.83

30. The perimeter of the figure in Fig. 2.84. A quarter-circle isattached to a triangle.

31. The area of the segment of the quarter-circle in Fig. 2.83

32. The area of the figure in Fig. 2.84


33. Describe the location of the midpoints of a set of parallel chordsof a circle.

34. The measure of on a circle of radius is . What is thelength of the arc in terms of and p?r

45°rABµ

A

D

B

E

C

Fig. 2.85

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2.5 Measurement of Irregular Areas 67

46. Find the length of the pulley belt shown in Fig. 2.89 if the beltcrosses at right angles. The radius of each pulley wheel is 5.50 cm.

48. Part of a circular gear with 24 teeth is shown in Fig. 2.90. Find theindicated angle.

Fig. 2.89


1. 126 cm 2. 3. 1.48 rad1260 cm2

20�

x

Fig. 2.90

47. The velocity of an object moving in a circular path is directed tan-gent to the circle in which it is moving. A stone on a string movesin a vertical circle, and the string breaks after 5.5 revolutions. Ifthe string was initially in a vertical position, in what directiondoes the stone move after the string breaks? Explain.

W

Therefore, the approximate area is

To this point, the figures for which we have found areas are well defined, and the areascan be found by direct use of a specific formula. In practice, however, it may be neces-sary to find the area of a figure with an irregular perimeter or one for which there isno specific formula. In this section, we show two methods of finding a very goodapproximation of such an area. These methods are particularly useful in technical areassuch as surveying, architecture, and mechanical design.

THE TRAPEZOIDAL RULEThe first method is based on dividing the required area into trapezoids withequal heights. Considering the area shown in Fig. 2.91, we draw parallel lines at

equal intervals between the edges of the area. We then join the ends of theseparallel line segments to form adjacent trapezoids. The sum of the areas of thetrapezoids gives a good approximation to the required area.

Calling the lengths of the parallel lines and the height ofeach trapezoid (the distance between the parallel lines), the total area is thesum of areas of all the trapezoids. This gives us

Ahy0, y1, y2, . . . , yn

n

2.5 Measurement of Irregular Areas

Trapezoidal Rule • Simpson’s Rule

first second third next-to-last last trapezoid trapezoid trapezoid trapezoid trapezoid

=

h

21y0 + y1 + y1 + y2 + y2 + y3 +

Á+ yn-2 + yn-1 + yn-1 + yn2

A =

h

21y0 + y12 +

h

21y1 + y22 +

h

21y2 + y32 +

Á+

h

21yn-2 + yn-12 +

h

21yn-1 + yn2

(2.12)A =

h

2 1y0 + 2y1 + 2y2 +

Á+ 2yn-1 + yn2■ Note carefully that the values of and

are not multiplied by 2.yny0

h hy2y1

y0 yn � 2 yn � 1yn

Fig. 2.91

Equation (2.12) is known as the trapezoidal rule. The following examples illustrateits use.

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When approximating the area with trapezoids, we omit small parts of the area forsome trapezoids and include small extra areas for other trapezoids. The omitted areasoften approximate the extra areas, which makes the approximation better. Also, the useof smaller intervals improves the approximation since the total omitted area or total ex-tra area is smaller.

E X A M P L E 2 Application of trapezoidal rule

From a satellite photograph of Lake Ontario (as shown on page 49), one of the GreatLakes between the United States and Canada, measurements of the width of the lakewere made along its length, starting at the west end, at 26.0-km intervals. The widthsare shown in Fig. 2.93 and are given in the following table.

E X A M P L E 1 Application of trapezoidal rule

A plate cam for opening and closing a valve is shown in Fig. 2.92. Widths of the faceof the cam are shown at 2.00-cm intervals from one end of the cam. Find the area of theface of the cam.

From the figure, we see that

(In making such measurements, often a -value at one end—or both ends—is zero.In such a case, the end “trapezoid” is actually a triangle.) From the given informa-tion in this example, Therefore, using the trapezoidal rule, Eq. (2.12),we have

The area of the face of the cam is approximately ■26.3 cm2.

= 26.3 cm2

A =

2.00

2 32.56 + 213.822 + 213.252 + 212.952 + 211.852 + 0.004

h = 2.00 cm.

y

y3 = 2.95 cm y4 = 1.85 cm y5 = 0.00 cm

y0 = 2.56 cm y1 = 3.82 cm y2 = 3.25 cm

2.56

cm

3.82

cm

2.00 cm

0.00

cm

3.25

cm

2.95

cm

1.85

cm

Fig. 2.92

■ See the chapter introduction.

RochesterNiagara Falls

Lake OntarioToronto

Fig. 2.93

Distance from West End (km) 0.0 26.0 52.0 78.0 104 130 156Width (km) 0.0 46.7 52.1 59.2 60.4 65.7 73.9

Distance from West End (km) 182 208 234 260 286 312Width (km) 87.0 75.5 66.4 86.1 77.0 0.0

Here, we see that and Therefore, using the trapezoidal rule, the approximate area of Lake Ontario is found as follows:

The area of Lake Ontario is actually ■19 477 km2.

= 19 500 km2

+ 2187.02 + 2175.52 + 2166.42 + 2186.12 + 2177.02 + 0.04 A =

26.0

2 30.0 + 2146.72 + 2152.12 + 2159.22 + 2160.42 + 2165.72 + 2173.92

yn = 0.0 km.y0 = 0.0 km, y1 = 46.7 km, y2 = 52.1 km, . . . ,

Practice Exercise

1. In Example 2, use only the distances fromwest end of lake (in km) 0.0, 52.0, 104,156, 208, 260, and 312. Calculate thearea and compare with the answer in Example 2.

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2.5 Measurement of Irregular Areas 69

Parabola

Fig. 2.94

yn � 2y2y1y0

hh

yn � 1

yn

Fig. 2.95

■ Named for the English mathematicianThomas Simpson (1710–1761).

SIMPSON’S RULEFor the second method of measuring an irregular area, we also draw parallel lines atequal intervals between the edges of the area. We then join the ends of these parallellines with curved arcs. This takes into account the fact that the perimeters of most fig-ures are curved. The arcs used in this method are not arcs of a circle, but arcs of aparabola. A parabola is shown in Fig. 2.94 and is discussed in detail in Chapter 21.(Examples of parabolas are (1) the path of a ball that has been thrown and (2) the crosssection of a microwave “dish.”)

The development of this method requires advanced mathematics. Therefore, we willsimply state the formula to be used. It might be noted that the form of the equation issimilar to that of the trapezoidal rule.

The approximate area of the geometric figure shown in Fig. 2.95 is given by

Equation (2.13) is known as Simpson’s rule. In using Eq. (2.13), the number n of in-tervals of width must be even.

E X A M P L E 3 Application of Simpson’s rule

A parking lot is proposed for a riverfront area in a town. The town engineer measuredthe widths of the area at 30.0-m (three sig. digits) intervals, as shown in Fig. 2.96. Findthe area available for parking.

First, we see that there are six intervals, which means Eq. (2.13) may be used.With and we have

■

For most areas, Simpson’s rule gives a somewhat better approximation than thetrapezoidal rule. The accuracy of Simpson’s rule is also usually improved by usingsmaller intervals.

E X A M P L E 4 Application of Simpson’s rule

From an aerial photograph, a cartographer determines the widths of Easter Island (in thePacific Ocean) at 1.50-km intervals as shown in Fig. 2.97. The widths found are as follows:

= 21 900 m2

A =

30.0

3 3124 + 411472 + 211162 + 411152 + 21872 + 411172 + 1444

h = 30.0 m,y0 = 124 m, y1 = 147 m, . . . , y6 = 144 m,

h

River

Street

124

m

147

m

116

m 115

m

117

m 144

m

87 m

Fig. 2.96

Distance from South End (km) 0 1.50 3.00 4.50 6.00 7.50 9.00 10.5 12.0 13.5 15.0Width (km) 0 4.8 5.7 10.5 15.2 18.5 18.8 17.9 11.3 8.8 3.1

Since there are ten intervals, Simpson’s rule may be used. From thetable, we have the following values:

and Using Simpson’s rule, the cartogra-pher would approximate the area of Easter Island as follows:

■ + 2118.82 + 4117.92 + 2111.32 + 418.82 + 3.12 = 174 km2

A =

1.50

310 + 414.82 + 215.72 + 4110.52 + 2115.22 + 4118.52

h = 1.5.y9 = 8.8, y10 = 3.1,y0 = 0, y1 = 4.8, y2 = 5.7, Á ,

y0 � 0

Easter Island

1.50 km

y10

y9

y1

y2

Fig. 2.97

(2.13)A =

h

31y0 + 4y1 + 2y2 + 4y3 +

Á+ 2yn-2 + 4yn-1 + yn2

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9. Using aerial photography, the widths of an area burned by a for-est fire were measured at 0.5-km intervals, as shown in the fol-lowing table:

Determine the area burned by the fire by using the trapezoidal rule.

10. Find the area burned by the forest fire of Exercise 9, usingSimpson’s rule.

11. A cartographer measured the width of Kruger National Park (and gamereserve) in South Africa at 6.0-mmintervals on a map, as shown in Fig. 2.101. The widths are shown in the list that follows. Find the area of the park if the scale of the map is

12. The widths of an oval-shaped floor were measured at 1.5-m inter-vals, as shown in the following table:

Find the area of the floor by using Simpson’s rule.

13. The widths of the baseball playing area in Boston’s Fenway Parkat 14-m intervals are shown in Fig. 2.102. Find the playing areausing the trapezoidal rule.

14. Find the playing area ofFenway Park (see Exer-cise 13) by Simpson’srule.

y8 = 8 mm y9 = 3 mmy6 = 9 mm y7 = 12 mmy4 = 13 mm y5 = 10 mmy2 = 7 mm y3 = 11 mmy0 = 7 mm y1 = 15 mm

1.0 mm = 6.0 km.

Distance (km) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Width (km) 0.6 2.2 4.7 3.1 3.6 1.6 2.2 1.5 0.8

EXERCISES 2.5

In Exercises 1 and 2, answer the given questions related to the indi-cated examples of this section.

1. In Example 1, if widths of the face of the same cam were given at1.00-cm intervals (five more widths would be included), from themethods of this section, what is probably the most accurate wayof finding the area? Explain.

2. In Example 4, if you use only the data from the south end of (in km)0, 3.00, 6.00, 9.00, 12.0, and 15.0, would you choose the trapezoidalrule or Simpson’s rule to calculate the area? Explain. Do not calcu-late the area for these data.

In Exercises 3 and 4, answer the given questions related to Fig. 2.98.

3. Which should be more accuratefor finding the area, the trape-zoidal rule or Simpson’s rule?Explain.

4. If the trapezoidal rule is used tofind the area, will the result prob-ably be too high, about right, ortoo little? Explain.

In Exercises 5–16, calculate the indicated areas. All data are accurateto at least two significant digits.

5. The widths of a kidney-shaped swimming pool were measured at2.0-m intervals, as shown in Fig. 2.99. Calculate the surface areaof the pool, using the trapezoidal rule.

6. Calculate the surface area of the swimming pool in Fig. 2.99, us-ing Simpson’s rule.

7. The widths of a cross section of an airplane wing are measured at0.30-m intervals, as shown in Fig. 2.100. Calculate the area of thecross section, using Simpson’s rule.

8. Calculate the area of the cross section of the airplane wing inFig. 2.100, using the trapezoidal rule.

2.0 m0.0

m

6.4

m

7.4

m

7.0

m

6.1

m

5.2

m

5.0

m

5.1

m

0.0

m

Fig. 2.99

Fig. 2.98

y9

y8

y1

y0

6.0 mm

Fig. 2.101

W

W

W

W

Distance (m) 0.0 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0

Width (m) 0.0 5.0 7.2 8.3 8.6 8.3 7.2 5.0 0.0

The Green Monster72 m

89 m

102 m

107 m

119 m

125 m

128 m

110 m52 mFig. 2.102

0.00 m

0.16 m

0.23 m0.30 m0.

35 m

0.20

m

0.30

m

0.32

m

Fig. 2.100

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2.6 Solid Geometric Figures 71

15. Soundings taken across a river channel give the following depthswith the corresponding distances from one shore.

Find the area of the cross section of the channel using Simpson’srule.

16. The widths of a bell crank are measured at 2.0-cm intervals, asshown in Fig. 2.103. Find the area of the bell crank if the two con-nector holes are each 2.50 cm in diameter.

3.5 cm6.0 cm7.6 cm10.8 cm

16.2 cm18.6 cm19.0 cm

17.8 cm12.5 cm

2.0 cm

8.2 cmFig. 2.103

Distance (m) 0 50 100 150 200 250 300 350 400 450 500

Depth (m) 5 12 17 21 22 25 26 16 10 8 0

Distance (cm) 0.000 0.250 0.500 0.750 1.000 1.250 1.500 1.750 2.000

Length (cm) 0.000 1.323 1.732 1.936 2.000 1.936 1.732 1.323 0.000

In Exercises 17–20, calculate the area of the circle by the indicatedmethod.

The lengths of parallel chords of a circle that are 0.250 cm apart aregiven in the following table. The diameter of the circle is 2.000 cm.The distance shown is the distance from one end of a diameter.

Using the formula the area of the circle is

17. Find the area of the circle using the trapezoidal rule and only thevalues of distance of 0.000 cm, 0.500 cm, 1.000 cm, 1.500 cm,and 2.000 cm with the corresponding values of the chord lengths.Explain why the value found is less than

18. Find the area of the circle using the trapezoidal rule and all valuesin the table. Explain why the value found is closer to thanthe value found in Exercise 17.

19. Find the area of the circle using Simpson’s rule and the same tablevalues as in Exercise 17. Explain why the value found is closer to

than the value found in Exercise 17.

20. Find the area of the circle using Simpson’s rule and all values inthe table. Explain why the value found is closer to thanthe value found in Exercise 19.

Answer to Practice Exercise

1. 18 100 km2

3.14 cm2

3.14 cm2

3.14 cm2

3.14 cm2.

3.14 cm2.A = pr2,

W

W

W

W

We now review the formulas for the volume and surface area of some basic solid geo-metric figures. Just as area is a measure of the surface of a plane geometric figure,volume is a measure of the space occupied by a solid geometric figure.

One of the most common solid figures is the rectangular solid. This figure has sixsides (faces), and opposite sides are rectangles. All intersecting sides are perpendicularto each other. The bases of the rectangular solid are the top and bottom faces. A cubeis a rectangular solid with all six faces being equal squares.

A right circular cylinder is generated by rotating a rectangle about one of its sides.Each base is a circle, and the cylindrical surface is perpendicular to each of the bases.The height is one side of the rectangle, and the radius of the base is the other side.

A right circular cone is generated by rotating a right triangle about one of its legs.The base is a circle, and the slant height is the hypotenuse of the right triangle. Theheight is one leg of the right triangle, and the radius of the base is the other leg.

The bases of a right prism are equal and parallel polygons, and the sides are rec-tangles. The height of a prism is the perpendicular distance between bases. The baseof a pyramid is a polygon, and the other faces, the lateral faces, are triangles thatmeet at a common point, the vertex. A regular pyramid has congruent triangles forits lateral faces.

A sphere is generated by rotating a circle about a diameter. The radius is a line seg-ment joining the center and a point on the sphere. The diameter is a line segmentthrough the center and having its endpoints on the sphere.

2.6 Solid Geometric Figures

Rectangular Solid • Cylinder • Prism •Cone • Pyramid • Sphere • Frustum

M02_WASH42254_09_SE_C02.qxd 1/17/09 6:25 PM


h

l

w

Fig. 2.104e

Fig. 2.105

h

r

Fig. 2.106 h

Fig. 2.107

hs

r

Fig. 2.108

h

s

Fig. 2.109

r

Fig. 2.110

Equation (2.21) is valid for any prism, and Eq. (2.26) is valid for any pyramid. Thereare other types of cylinders and cones, but we restrict our attention to right circularcylinders and right circular cones, and we will often use “cylinder” or “cone” when re-ferring to them.

The frustum of a cone or pyramid is the solid figure that remains after the top is cutoff by a plane parallel to the base. Figure 2.111 shows the frustum of a cone.

E X A M P L E 1 Volume of rectangular solid

What volume of concrete is needed for a driveway 25.0 m long, 2.75 m wide, and 0.100 mthick?

The driveway is a rectangular solid for which andUsing Eq. (2.14), we have

■ = 6.88 m3

V = 125.0212.75210.1002h = 0.100 m.

l = 25.0 m, w = 2.75 m,

Fig. 2.111

In the following formulas, represents the volume, represents the total surfacearea, represents the lateral surface area (bases not included), represents the areaof the base, and represents the perimeter of the base.p

BSAV

(2.29)A = 4pr2

(2.28)Sphere (Fig. 2.110)V =

4

3 pr3

(2.27)S =

1

2 ps

(2.26)Regular pyramid (Fig. 2.109)V =

1

3 Bh

(2.25)S = prs

(2.24)A = pr2+ prs

(2.23)Right circular cone (Fig. 2.108)V =

1

3 pr2h

(2.22)S = ph

(2.21)Right prism (Fig. 2.107)V = Bh

(2.20)S = 2prh

(2.19)A = 2pr2+ 2prh

(2.18)Right circular cylinder (Fig. 2.106)V = pr2h

(2.17)A = 6e2

(2.16)Cube (Fig. 2.105)V = e3

(2.15)A = 2lw + 2lh + 2wh

(2.14)Rectangular solid (Fig. 2.104)V = lwh

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2.6 Solid Geometric Figures 73

E X A M P L E 2 Total surface area of right circular cone

Calculate the total surface area of a right circular cone for which the radius and the height is See Fig. 2.112.

To find the total surface area using Eq. (2.24), we need the radius and the slantheight of the cone. Therefore, we must first find . The radius and height are legsof a right triangle, and the slant height is the hypotenuse. To find , we use thePythagorean theorem:

Pythagorean theorem

solve for

Now, calculating the total surface area, we have

Eq. (2.24)

substituting

= 1040 cm2

= p111.922 + p111.92115.82 A = pr2

+ prs

= 211.92+ 10.42

= 15.8 cm

ss = 2r2+ h2

s2= r2

+ h2

sss

h = 10.4 cm.r = 11.9 cm

r � 11.9 cm

h � 10.4 cm

Fig. 2.112

Fig. 2.113

Practice Exercise

1. Find the volume of the cone in Example 2. Note in the calculator display in Fig. 2.113 that we stored as , and did not have toactually record its value. ■

E X A M P L E 3 Volume of combination of solids

A grain storage building is in the shape of a cylinder surmounted by a hemisphere (halfa sphere). See Fig. 2.114. Find the volume of grain that can be stored if the height ofthe cylinder is 40.0 m and its radius is 12.0 m.

The total volume of the structure is the volume of the cylinder plus the volume ofthe hemisphere. By the construction we see that the radius of the hemisphere is thesame as the radius of the cylinder. Therefore,

cylinder hemisphere

■ = 21 700 m3

= p112.022140.02 +

2

3 p112.023

V = pr2h +

1

2 a4

3pr3b = pr2h +

2

3 pr3

xs

h � 40.0 m

r � 12.0 m

Fig. 2.114

EXERCISES 2.6


1. In Example 1, if the length is doubled and the thickness is dou-bled, by what factor is the volume changed?

2. In Example 2, if the value of the slant height cm is giveninstead of the height, what is the height?

3. In Example 2, if the radius is halved and the height is doubled,what is the volume?

4. In Example 3, if is halved, what is the volume?

In Exercises 5–20, find the volume or area of each solid figure for thegiven values. See Figs. 2.104 to 2.110.

5. Volume of cube: e = 7.15 cm

h

s = 17.5

6. Volume of right circular cylinder:

7. Total surface area of right circular cylinder: m,m

8. Area of sphere:

9. Volume of sphere:

10. Volume of right circular cone:

11. Lateral area of right circular cone:

12. Lateral area of regular pyramid:

13. Volume of regular pyramid: square base of side 76 cm,

14. Volume of right prism: square base of side 29.0 cm, h = 11.2 cm

h = 130 cm

s = 272 mp = 345 m,

r = 78.0 cm, s = 83.8 cm

r = 25.1 mm, h = 5.66 mm

r = 0.877 m

r = 0.067 mm

h = 233r = 689

r = 23.5 cm, h = 48.4 cm

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15. Lateral area of regular prism: equilateral triangle base of side1.092 m,

16. Lateral area of right circular cylinder: ,

17. Volume of hemisphere:

18. Volume of regular pyramid: square base of side 22.4 m,

19. Total surface area of right circular cone:

20. Total surface area of pyramid: All faces and base are equilateraltriangles of side 3.67 dm.


21. Eq. (2.28) expresses the volume of a sphere in terms of theradius . Express in terms of the diameter .

22. Derive a formula for the total surface area of a hemisphericalvolume of radius (curved surface and flat surface).

23. The radius of a cylinder is twice as long as the radius of a cone, andthe height of the cylinder is half as long as the height of the cone.What is the ratio of the volume of the cylinder to that of the cone?

24. The base area of a cone is one-fourth of the total area. Find theratio of the radius to the slant height.

25. In designing a weather balloon, it is decided to double the dia-meter of the balloon so that it can carry a heavier instrument load.What is the ratio of the final surface area to the original surfacearea?

26. During a rainfall of 3.00 cm, what weight of water falls on an areaof ? Each cubic metre of water weighs 9800 N.

27. A rectangular box is to be used to store radioactive materials. Theinside of the box is 12.0 cm long, 9.50 cm wide, and 8.75 cmdeep. What is the area of sheet lead that must be used to line theinside of the box?

28. A swimming pool is 15.0 m wide, 24.0 m long, 1.00 m deep at oneend, and 2.60 m deep at the other end. How many cubic metres of water can it hold? (The slope on the bottom is constant.) See Fig.2.115.

1.00 km2

rA

dVrV

h = 0.274 cmr = 3.39 cm,

s = 14.2 m

diameter = 0.83 cm

h = 347 mmdiameter = 250 mm

h = 1.025 m30. A glass prism used in the study of

optics has a right triangular base.The legs of the triangle are 3.00 cmand 4.00 cm. The prism is 8.50 cmhigh. What is the total surface areaof the prism? See Fig. 2.116.

1.00 m2.60 m

15.0 m

24.0 m

Fig. 2.115

8.50 cm4.00 cm

3.00 cm

Fig. 2.116

31. The Great Pyramid of Egypt has a square base approximately 230 m on a side. The height of the pyramid is about 150 m. Whatis its volume? See Fig. 2.117.

32. A paper cup is in the shape of a cone as shown in Fig. 2.118.What is the surface area of the cup?

33. Spaceship Earth (shown in Fig. 2.119) at Epcot Center inFlorida is a sphere of 50.3 m in diameter. What is the volume ofSpaceship Earth?

34. A propane tank is constructed in the shape of a cylinder with ahemisphere at each end, as shown in Fig. 2.120. Find the volumeof the tank.

150 m

Fig. 2.117

9.20 cm

8.90 cm

Fig. 2.118

50.3 m

Fig. 2.119

1.98 m

1.22 m

Fig. 2.120

35. A special wedge in the shape of a regular pyramid has a squarebase 16.0 mm on a side. The height of the wedge is 40.0 mm.What is the total surface area of the wedge?

29. The Alaskan oil pipeline is 1200 km long and has a diameter of 1.2 m. What is the maximum volume of the pipeline?

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Chapter Equations 75

36. A lawn roller is a cylinder 0.96.0 m long and 0.60 m in diameter.How many revolutions of the roller are needed to roll 76 m2 oflawn?

37. The circumference of a basketball is about 75.7 cm. What is itsvolume?

38. What is the area of a paper label that is to cover the lateral surfaceof a cylindrical can 8.50 cm in diameter and 11.5 cm high? Theends of the label will overlap 0.50 cm. when the label is placed onthe can.

39. The side view of a rivet is shown in Fig. 2.121. It is a conical parton a cylindrical part. Find the volume of the rivet.

40. A dipstick is made to measure the volume remaining in the coni-cal container shown in Fig. 2.122. How far below the full mark(at the top of the container) on the stick should the mark for half-full be placed?

2.75 cm0.625 cm

1.25 cm 0.625 cm

Fig. 2.121

12.0 cm

18.0 cm

Fig. 2.122

Answer to Practice Exercise

1. 1540 cm3

Line segments Fig. 2.8 (2.1)

Triangle (2.2)

Hero’s formula (2.3)where

Pythagorean theorem Fig. 2.31 (2.4)

Square Fig. 2.53 (2.5)

Rectangle Fig. 2.54 (2.6)

Parallelogram Fig. 2.55 (2.7)

Trapezoid Fig. 2.56 (2.8)

Circle (2.9)

(2.10)

Radians Fig. 2.79 (2.11)

Trapezoidal rule Fig. 2.91 (2.12)

Simpson’s rule Fig. 2.95 (2.13)A =

h

3 1y0 + 4y1 + 2y2 + 4y3 +

Á+ 2yn-2 + 4yn-1 + yn2

A =

h

2 1y0 + 2y1 + 2y2 +

Á+ 2yn-1 + yn2

p rad = 180°

A = pr2

c = 2pr

A =12 h1b1 + b22

A = bh

A = lw

A = s2

c2= a2

+ b2

s =12 1a + b + c2

A = 1s1s - a21s - b21s - c2,A =

12 bh

a

b=

c

d

CHAPTER 2 EQUATIONS

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Rectangular solid Fig. 2.104 (2.14)

(2.15)

Cube Fig. 2.105 (2.16)

(2.17)

Right circular cylinder Fig. 2.106 (2.18)

(2.19)

(2.20)

Right prism Fig. 2.107 (2.21)

(2.22)

Right circular cone Fig. 2.108 (2.23)

(2.24)

(2.25)

Regular pyramid Fig. 2.109 (2.26)

(2.27)

Sphere Fig. 2.110 (2.28)

(2.29)A = 4pr2

V =43 pr3

S =12 ps

V =13 Bh

S = prs

A = pr2+ prs

V =13 pr2h

S = ph

V = Bh

S = 2prh

A = 2pr2+ 2prh

V = pr2h

A = 6e2

V = e3

A = 2lw + 2lh + 2wh

V = lwh

CHAPTER 2 REVIEW EXERCISES

In Exercises 1–4, use Fig. 2.123. Determine the indicated angles.

1.

2.

3.

4.

In Exercises 5–12, find the indicated sides of the right triangle shownin Fig. 2.124.

5. 6.

7. 8.

9.

10.

11.

12.

In Exercises 13–20, find the perimeter or area of the indicated figure.

13. Perimeter: equilateral triangle of side 8.5 mm

14. Perimeter: rhombus of side 15.2 cm

15. Area: triangle,

16. Area: triangle of sides 175 cm, 138 cm, 119 cm

17. Circumference of circle: d = 98.4 mm

b = 3.25 m, h = 1.88 m

a = 0.782, c = 0.885, b = ?

b = 29.3, c = 36.1, a = ?

a = 126, b = 25.1, c = ?

a = 6.30, b = 3.80, c = ?

b = 56, c = 65, a = ?a = 400, b = 580, c = ?

a = 14, b = 48, c = ?a = 9, b = 40, c = ?

∠EGI

∠DGH

∠EGF

I H

D

B

FE

A

CG

148°AB ‖ CD

Fig. 2.123

∠CGE

18. Perimeter: rectangle,

19. Area: trapezoid,

20. Area: circle,

In Exercises 21–24, find the volume of the indicated solid geometricfigure.

21. Prism: base is right triangle with legs 26.0 cm and 34.0 cm,height is 14.0 cm

22. Cylinder: base radius 36.0 cm, height 2.40 cm

23. Pyramid: base area height 125 m

24. Sphere: diameter 22.1 mm

In Exercises 25–28, find the surface area of the indicated solid geo-metric figure.

25. Total area of cube of edge 0.520 m

26. Total area of cylinder: base diameter 1.20 cm, height 5.80 cm

27. Lateral area of cone: base radius 1.82 mm, height 11.5 mm

28. Total area of sphere:

In Exercises 29–32, use Fig. 2.125. Lineis tangent to the circle with center at

Find the indicated angles.

29. 30.

31. 32. ∠ABT∠BTC

∠TAB∠BTA

O.CT

d = 12 760 km

3850 m2,

d = 32.8 m

b1 = 67.2 cm, b2 = 126.7 cm, h = 34.2 cm

w = 1.86 dml = 2.98 dm,

ca

bFig. 2.124

C

TO

A

B

50°

Fig. 2.125

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Review Exercises 77

In Exercises 33–36, use Fig. 2.126. Given thatand find the indicated angle and lengths.

33.

34.

35.

36.

In Exercises 37–40, find the formulas for the indicated perimetersand areas.

37. Perimeter of Fig. 2.127 (a right triangle and semicircle attached)

38. Perimeter of Fig. 2.128 (a square with a quarter circle at each end)

39. Area of Fig. 2.127 40. Area of Fig. 2.128

AE

BE

AD

∠ABE

∠ADC = 53°,CD = 6,AB = 4, BC = 4, 50. An airplane is 640 m directly above one end of a 3200-m run-

way. How far is the plane from the glide-slope indicator on theground at the other end of the runway?

51. A machine part is in the shape of a square with equilateral tri-angles attached to two sides (see Fig. 2.130). Find the perimeterof the machine part.

2ab

Fig. 2.127

s

Fig. 2.128

In Exercises 41–46, answer the given questions.

41. Is a square also a rectangle, a parallelogram, and a rhombus?

42. If the measures of two angles of one triangle equal the measuresof two angles of a second triangle, are the two triangles similar?

43. If the dimensions of a plane geometric figure are each multi-plied by by how much is the area multiplied? Explain, usinga circle to illustrate.

44. If the dimensions of a solid geometric figure are each multipliedby by how much is the volume multiplied? Explain, using acube to illustrate.

45. What is an equation relating chordsegments and shown in Fig. 2.129? The dashed chords are anaid in the solution.

46. From a common point, two line segments are tangent to thesame circle. If the angle between the lines segments is , whatis the angle between the two radii of the circle drawn from thepoints of tangency?


47. A tooth on a saw is in the shape of an isosceles triangle. If theangle at the point is , find the two base angles.

48. A lead sphere 1.50 cm in diameter is flattened into a circularsheet 14.0 cm in diameter. How thick is the sheet?

49. A ramp for the disabled is designed so that it rises 1.2 m over ahorizontal distance of 7.8 m. How long is the ramp?

38°

36°

da, b, c,

n,

n,

a b

dc

Fig. 2.129

2 cm

Fig. 2.130

4.50 m

Fig. 2.131

52. A patio is designed with semicircular areas attached to a square,as shown in Fig. 2.131. Find the area of the patio.

53. A radio transmitting tower is supported by guy wires. The towerand three parallel guy wires are shown in Fig. 2.132. Find thedistance along the tower.AB

B

A

42 m

38 m 54 m

Fig. 2.132

B

A

Main Street

First S

treet

Fig. 2.133

54. Find the areas of lots and in Fig. 2.133. has a frontage onMain St. of 140 m, and has a frontage on Main St. of 84 m.The boundary between lots is 120 m.

55. To find the height of a flagpole, a person places a mirror at ,as shown in Fig. 2.134. The person’s eyes at are 160 cm abovethe ground at . From physics, it is known that

If and find the height of the flagpole.

BFMB = 4.5 m,AM = 120 cm∠BMF.∠AME =A

EM

BABA

Mirror

2A

E

M B

F

1

Fig. 2.134

B C

D E

A

Fig. 2.135

56. The metal support in the form of shown in Fig. 2.135 isstrengthened by brace , which is parallel to . How long isthe brace if and ?BC = 33 cmAB = 24 cm, AD = 16 cm,

BCDE^ABC

W

W

AE

B

C

DFig. 2.126

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57. A typical scale for an aerial photograph is In a 20.0-by-25.0-cm photograph with this scale, what is the longestdistance (in km) between two locations in the photograph?

58. For a hydraulic press, the mechanical advantage is the ratio ofthe large piston area to the small piston area. Find the mechan-ical advantage if the pistons have diameters of 3.10 cm and2.25 cm.

59. The diameter of the earth is 12 700 km, and a satellite is in orbitat an altitude of 340 km. How far does the satellite travel in onerotation about the earth?

60. The roof of the Louisiana Superdome in New Orleans is sup-ported by a circular steel tension ring 651 m in circumference.Find the area covered by the roof.

61. A rectangular piece of wallboard with two holes cut out forheating ducts is shown in Fig. 2.136. What is the area of theremaining piece?

1>18 450. 66. A horizontal cross section of a concrete bridge pier is a regularhexagon (six sides, all equal in length, and all internal angles areequal), each side of which is 2.50 m long. If the height of thepier is 6.75 m, what is the volume of concrete in the pier?

67. A railroad track 1000.00 m long expands 0.20 m (20 cm) duringthe afternoon (due to an increase in temperature of about ).Assuming that the track cannot move at either end and that theincrease in length causes a bend straight up in the middle of thetrack, how high is the top of the bend?

68. Two persons are talking to each other on cellular phones. Ifthe angle between their signals at the tower is as shown inFig. 2.138, how far apart are they?

90°,

17°C

2400 mm

1200 mm

350 mm

350 mm

Fig. 2.136

62. The diameter of the sun is the diameter of theearth is and the distance from the earth to the sun(center to center) is What is the distance from thecenter of the earth to the end of the shadow due to the rays fromthe sun?

63. Using aerial photography, the width of an oil spill is measured at250-m intervals, as shown in Fig. 2.137. Using Simpson’s rule,find the area of the oil spill.

1.50 * 108 km.1.27 * 104 km,

1.38 * 106 km,

190 m260 m 250 m

220

m

530

m 480

m 320

m

350

m

730

m

560

m

240

m

510

m

Fig. 2.137

64. To build a highway, it is necessary to cut through a hill. A sur-veyor measured the cross-sectional areas at 250-m intervalsthrough the cut as shown in the following table. Using the trape-zoidal rule, determine the volume of soil to be removed. 70. A tent is in the shape of a regular pyramid surmounted on a

cube. If the edge of the cube is 2.50 m and the total height of thetent is 3.25 m, find the area of the material used in making thetent (not including any floor area).

71. The diagonal of a HDTV screen is 107 cm. If the ratio of thewidth of the screen to the height of the screen is 16 to 9, what arethe width and height?

69. A hot-water tank is in the shape of a right circular cylinder sur-mounted by a hemisphere as shown in Fig. 2.139. How manylitres does the tank hold? ( contains 1000 L.)1.00 m3

65. The Hubble space telescope is within a cylinder 4.3 m in diame-ter and 13 m long. What is the volume within this cylinder?

Dist. (m) 0 250 500 750 1000 1250 1500 1750

Area (m2) 560 1780 4650 6730 5600 6280 2260 230

2.4 km

3.7 km

Fig. 2.138

0.760 m

2.05 m

Fig. 2.139

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Practice Test 79

72. A satellite is 1590 km directly above the center of the eye ofa circular (approximately) hurricane that has formed in theAtlantic Ocean. The distance from the satellite to the edge ofthe hurricane is 1620 km. What area does the hurricane cover?Neglect the curvature of the Earth and any possible depth of thehurricane.

Writing Exercise73. The Pentagon, headquarters of the U.S. Department of Defense,

is the world’s largest office building. It is a regular pentagon(five sides, all equal in length, and all interior angles are equal)281 m on a side, with a diagonal of length 454 m. Using thesedata, draw a sketch and write one or two paragraphs to explainhow to find the area covered within the outside perimeter of thePentagon. (What is the area?)

CHAPTER 2 PRACTICE TEST

1. In Fig. 2.140, determine

2. In Fig. 2.140, determine ∠2.

∠1. 8. Find the surface area of a tennis ball whose circumference is21.0 cm.

9. Find the volume of a right circular cone of radius 2.08 m andheight 1.78 m.

10. In Fig. 2.142, find

11. In Fig. 2.142, find ∠2.

∠1.2

1

52°

C D

A B

AB ‖ CDFig. 2.140

12. In Fig. 2.143, find the perimeter of the figure shown. It is asquare with a semicircle removed.

13. In Fig. 2.143, find the area of the figure shown.

14. The width of a marshy area is measured at 50-ft intervals, withthe results shown in the following table. Using the trapezoidalrule, find the area of the marsh. (All data accurate to two or moresignificant digits.)

3. A tree is 2.4 m high and casts a shadow 3.0 m long. At the sametime, a telephone pole casts a shadow 7.6 m long. How tall is thepole?

4. Find the area of a triangle with sides of 2.46 cm, 3.65 cm, and4.07 cm.

5. What is the diagonal distance between corners of a rectangularroom 3810 mm wide and 5180 mm long?

6. What is the area of the trapezoid shown in Fig. 2.141?

23.5 mm

49.8 mm

23.5 mm15.6 mm 13.8 mm

Fig. 2.141

64°1

2C

BOA

Fig. 2.142

2.25 cm

Fig. 2.143

Distance (m) 0 50 100 150 200 250 300

Width (m) 0 90 145 260 205 110 20

7. The edge of a cube is 4.50 cm. What are (a) the surface area and(b) the volume of the cube?

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basic algebraic operations - pearson...

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