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SOLUTIONS MANUAL for College Algebra Essentials 1st Edition by Julie Miller. ISBN-13 978-0078035616.

Page 1

Contents Chapter R

Review of Prerequisites

R.1

Sets and the Real Number Line

1

R.2

Models, Algebraic Expressions, and Properties of Real Numbers

8

R.3

Integer Exponents and Scientific Notation

13

R.4

Rational Exponents and Radicals

20

R.5

Polynomials and Multiplication of Radicals

28

Problem Recognition Exercises: Simplifying Algebraic Expressions

35

R.6

Factoring

36

R.7

Rational Expressions and More Operations on Radicals

45

Chapter R Review Exercises

59

Chapter R Test

67

Chapter 1

Equations and Inequalities

1.1

Linear Equations and Rational Equations

71

1.2

Applications and Modeling with Linear Equations

85

1.3

Complex Numbers

96

1.4

Quadratic Equations

102

Problem Recognition Exercises: Simplifying Expressions Versus Solving Equations

118

1.5

Applications of Quadratic Equations

120

1.6

More Equations and Applications

127

1.7

Linear Inequalities and Compound Inequalities

143

1.8

Absolute Value Equations and Inequalities

152

Problem Recognition Exercises: Recognizing and Solving Equations and Inequalities

160

Chapter 1 Review Exercises

164

Chapter 1 Test

180

Chapter 1 Cumulative Review Exercises

186


Chapter 2

Functions and Graphs

2.1

The Rectangular Coordinate System and Graphing Utilities

189

2.2

Circles

202

2.3

Functions and Relations

212

2.4

Linear Equations in Two Variables and Linear Functions

220

2.5

Applications of Linear Equations and Modeling

234

Problem Recognition Exercises: Comparing Graphs of Equations

244

2.6

Transformations of Graphs

246

2.7

Analyzing Graphs of Functions and Piecewise-Defined Functions

258

2.8

Algebra of Functions and Function Composition

271

Chapter 2 Review Exercises

283

Chapter 2 Test

296

Chapter 2 Cumulative Review Exercises

300

Chapter 3

Polynomial and Rational Functions

3.1

Quadratic Functions and Applications

303

3.2

Introduction to Polynomial Functions

318

3.3

Division of Polynomials and the Remainder and Factor Theorems

329

3.4

Zeros of Polynomials

342

3.5

Rational Functions

359

Problem Recognition Exercises: Polynomial and Rational Functions

385

Polynomial and Rational Inequalities

387

Problem Recognition Exercises: Solving Equations and Inequalities

412

Variation

417

Chapter 3 Review Exercises

424

Chapter 3 Test

439

Chapter 3 Cumulative Review Exercises

448

3.6

3.7


Chapter 4

Exponential and Logarithmic Functions

4.1

Inverse Functions

451

4.2

Exponential Functions

460

4.3

Logarithmic Functions

471

Problem Recognition Exercises: Analyzing Functions

482

4.4

Properties of Logarithms

486

4.5

Exponential and Logarithmic Equations

493

4.6

Modeling with Exponential and Logarithmic Functions

511

Chapter 4 Review Exercises

521

Chapter 4 Test

531

Chapter 4 Cumulative Review Exercises

537

Chapter 5

Systems of Equations and Inequalities

5.1

Systems of Linear Equations in Two Variables and Applications

541

5.2

Systems of Linear Equations in Three Variables and Applications

559

5.3

Partial Fraction Decomposition

587

5.4

Systems of Nonlinear Equations in Two Variables

605

5.5

Inequalities and Systems of Inequalities in Two Variables

628

Problem Recognition Exercises: Equations and Inequalities in Two Variables

652

Linear Programming

656

Chapter 5 Review Exercises

668

Chapter 5 Test

685

Chapter 5 Cumulative Review Exercises

697

5.6


Section R.1

Chapter R Review of Prerequisites Section R.1 Sets and the Real Number Line 1. set

2. whole

3. natural

4. integers

5. set-builder

6. roster

7. rational

8. irrational

9. left

10. absolute value

5 , which is the 1 ratio of 5 and 1, so it is an element of  . True.

d. 5 can be written as

22. a.   1, 2,3, does not include

b.    0,1, 2,3, does not include

11. a  b or b  a

1 . 3

False. c.    , 2, 1,0,1, 2, does not include

12. base; exponent or power

1 . False. 3

13. square 14. 0; undefined

d.

15. 3 is an element of the set of natural numbers. 16.

1 . False. 3

1 is the ratio of 1 and 3, so it is an 3 element of  . True.

23. a.   1, 2,3, does not include 0.25 . False.

2 is an element of the set of rational 5 numbers.

b.    0,1, 2,3, does not include 0.25 . False.

17. 3.1 is not an element of the set of integers. 18.  is not an element of the set of rational numbers.

c.    , 2, 1,0,1, 2, does not include

19. The set of integers is a proper subset of the set of real numbers.

d. 0.25 is a repeating decimal, so it is an element of  . True.

20. The set of rational numbers is a proper subset of the set of real numbers.

24. a.    , 2, 1,0,1, 2, does not include  . False.

0.25 . False.

21. a.   1, 2,3, does not include 5 . False.

b.  cannot be written as the ratio of two integers, so it is not an element of  . False.

b.    0,1, 2,3, does not include 5 . False.

c.  cannot be represented by a terminating decimal or by a repeating decimal, so it is an element of  . True.

c.    , 2, 1,0,1, 2, .This includes 5 . True.

d.  is an element of  , so it is an element of  . True.

1


Chapter R Review of Prerequisites 25. The set given can be written in roster form as  , 10, 5,0,5,10, . The number 25 is an element of this set. True.

35. a. The first set is not a subset of the second, because the set of whole numbers does not include negative integers. False. b. The first set is a subset of the second, because all whole numbers are also integers. True.

26. The set given can be written in roster form as 1, 2,3,,9 . The number 7 is not an element of this set. False.

n , so all 1 integers are also rational numbers. True.

36. a. Any integer n can be written as

27. The number 0.8 is a repeating decimal, so it is rational. It can be written as 0.88 , and 0.88  0.8 . False.

b. The first set is not a subset of the second, because there all elements in the set of rational numbers that are not integers. False.

45 , so it 100 is rational. The number 0.45 can be written as 0.4545 , and 0.45  0.4545 . True.

28. The number 0.45 can be written as

37. a. The first set is not a subset of the second, because irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.

29. The set given can be written in roster form as  0, 2, 4,6, . The number 22 is an element of this set. True.

b. The first set is not a subset of the second, because irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.

30. The set given can be written in roster form as  , 8, 4, 0, 4,8, . The number 24 is an element of this set. True.

38. a. The first set is a subset of the second, because the real numbers are made up of rational and irrational numbers. True.

31. Irrational numbers are real numbers that cannot be represented as a ratio of integers, so a number cannot be both rational and irrational. False.

b. The first set is a subset of the second, because the real numbers are made up of rational and irrational numbers. True.

n , so all 1 integers are also rational numbers. True.

32. Any integer n can be written as

39. a. 6

b. 6

c. 12 , 6

33. a. The first set is a subset of the second, because all of the elements in the first set are elements of the second set. True.

d. 0.3,0.33, 0.9, 12,

b. The first set is not a subset of the second, because 0 is in the first set but not in the second set. False. 34. a. The first set is a subset of the second, because all of the elements in the first set are elements of the second set. True.

e.

5,

f.

5, 0.3, 0.33, 0.9, 12,

6

40. a. 1

b. The first set is not a subset of the second, because NM and TX are in the first set but not in the second set. False.

c. 0, 4 , 1

2

11 ,6 4

11  , 6, 4 6

b. 0, 1 d. 0, 4, 0.48,1,9.4


Section R.1 

41. a  5

42. b  6

5? 9   3 5 5 5? 9 3     3 5 5 3 25 ? 27 False   15 15

43. 3c  9

44. 8d  16

55.  7,   ;  x | x  7

45. m  4  70

46. n  7  4

56.  2,   ;  x | x  2

e. f.

2

 2

,  13

54.

, 0, 4, 0.48,1,  13,9.4

47. 3.14   ?

57.  , 4.1 ;  x | x  4.1

?

3.14  3.141592654 True

58.  ,  2.93 ;  x | x  2.93

?

48. 7   7 ?

59.  6, 0 ;  x | 6  x  0

?

60.  2, 8 ;  x | 2  x  8

7   2.645751311 True

49. 6.7  6.7

6.7 ⱨ 6.7 True

61.

 , 6

62.

 ,  4

63.

 7 1   ,  6 3

64.

 4 7  ,   3 4

65.

 4, 

66.

 3, 

?

50. 2.1   2.1

2.1ⱨ 2.1 True ?

6.15  6.15

51.

?

6.151515  6.155555 False ?

2.93  2.93

52.

?

2.933333  2.939393 False

53.

9 ? 11  7 8 9 8 ? 11 7     7 8 8 7 72 ? 77 False   56 56 

3

67.  x | 3  x  7

68.  x | 4  x  1

69.  x | x  6.7

70.  x | x  3.2

 3 71.  x x    5 

 7 72.  x x   8 


Chapter R Review of Prerequisites 73. 6    6  6

74. 4    4  4

88. 8  2  10  10 or 2   8  10  10

75. 0  0

76. 1  1

89. 8  3  8  3 or

3 8  8 3

77.

2 2  

 2  2  2  2

90. 11  5  11  5 or

78.

6 6  

 6  6  6  6

91. 6  2  2  6 or 2  6  2  6

79. a.   3    3

92. 3      3 or   3    3

b. 3      3       3

93. a. 42  4  4  16 b.  4   4   4  16

80. a. for m  11, m  11  m  11

2

b. for m  11, m  11    m  11  11  m

c. 42    4  4  16 d.

81. a. for x  2, x  2  x  2

f.

82. a. for t  6, t  6    t  6  t  6

4 is not a real number.

94. a. 92  9  9  81

b. for t  6, t  6  t  6

b.  9   9   9  81 2

z 5 z 5  1 z 5 z 5

c. 92    9  9  81 d.

b. for z  5, z  5   z  5 z 5   1  1 z 5 z 5 z 5 84. a. for x  7,

42

e.  4  2

b. for x  2, x  2    x  2   x  2

83. a. for z  5,

5  11  11  5

93

e.  9  3 f.

7x 7x  1 7x 7x

9 is not a real number.

95. a. 3 8  2 b. 3 8  2

b. for x  7, 7x 7x 7x   1  1 7  x   7  x 7x

 

c.  3 8   3 8  2

85. 1  6  5  5 or 6  1  5  5 86. 2  9  7  7 or 9  2  5  7

d.

100  10

e.

100 is not a real number.

f.  100  10

87. 3   4  7  7 or 4  3  7  7

96. a. 3 27  3 b. 3 27  3

4


Section R.1

2 106. 5  4  6  2  8   31  

c.  3 27   3 27  3

d.

49  7

 5  4  6  6   31  

e.

49 is not a real number.

 5   4  6  36  31 

2

 5   4  6  5  

f.  49  7

 5   4  30   5   26   21

3

2 2 2 8  2 97.        3 3 3 3 27

107.

52  32  25  9  16  4

4 4 4 64  4 98.        5 5 5 5 125

108.

62  82  36  64  100  10

99.  0.2   0.2   0.2   0.2   0.2

109.

 9  16    3  4  7  49

110.

 8  125    2  5  7  343

3

2

4

 0.0016 100.  0.1   0.1   0.1   0.1   0.1

3

3

2

3

2

3

3

4

2

2 7  2 2 7  111. 4      4       5 10   2 5 10 

 0.0001 101.

169 13  25 5

121 11  36 6

102.

 4 7  4      10 10 

 3  4      10  9  4  100

103. 20  12 36  32  2  20  12  36  9  2

 20  12  4  2  20  12  2  20  24  4

25

 1  2  1  2   1 1 112. 6         6      9 4  3   2  

 200  4  48  200  192  8

 6  12  3  5  18   2

2

1

 200  22  48

2

4 9 9    1 100 25

1   104. 200  22  6   4  200  22 12  4  2 

2 105. 6  12  3 1  6  18  

2

 4 1 9 1  6      4 9 9 4 9  4  6    36 36 

 5  6    36 

 6   12  3  25  18   6   12  3  7    6   12  21  6  9  3

1   6 5 5    1  36  6  6 

5


Chapter R Review of Prerequisites

113. 9  6  3  7  8  25

 4000  118. Wh  175   4000  16.1

 9  6  4  8  25

 4000   175   4016.1

 9   6  4  8   25  9   6  4   25

119. V 

 9  10  5 927

 8  2 4  3  5  9  8  2  4  2   9  8  2  4  2  9  8  12  3

124. A bracket is used if an endpoint to an interval is included in the set.

2  8

144  25  3  10 28

125.  represents the set of integers. All integers belong to  (the set of rational numbers). Therefore,  is a subset of  . However,  1 contains fractions and decimals, such as 2 and 0.2, that are not integers. Therefore, .

169  13 6  13  13 6 Undefined  0 116.

 121 cc

123. A parenthesis is used if an endpoint to an interval is not included in the set.

84  4

122  52  3  10

3

122. All integers can be written as the integer itself over the integer 1.

 8  2  6  3

115.

 14.44  8

121. All terminating decimals can be written as a decimal fraction—that is, a fraction with a denominator that is a power of 10.

 8  24  3  5   9

11  13  4  2

3

 1.0   250  120. V  6.0   10.2 L  0.5   293 

114. 8  2 4  2  5  5  9

  3.8  8 2

2

 173.6 lb

 9   6  4   25

2

 4  92  22  32   5 2  4  9 7  4   12  4 3   12  4 

25  4  9 3   3

20 Undefined 0

 4000  117. Wh   200   4000  5.5   4000    200   4005.5 

126. The statement 1   indicates that the set containing the element 1 is a subset of  . However, the statement 1   does not make sense, because the number 1 (not written in set brackets) does not represent a set. A quantity that is not a set cannot be a subset of another set. 127. If n  0 , then n  n  n  n  0

2

128. If n  0 , then n  n  n    n   n  n  2n

2

129. If n  0 , then n  n  n  n  2n

 119.5 lb

6


Section R.1 130. If n  0 , then n  n  n    n   n  n  0

141.

131. If n  0 , then  n   n 132. If n  0 , then  n     n   n 133. b  0

134. a  0

135. b  0

136. a  0

6744.25 142.

137. a is negative, b is positive, and c is negative. ab 2 So the sign of 3 can be represented by c

     2         , which is positive.       3

37,741.81 143.

138. a is negative, b is positive, and c is negative. a 2c So the sign of 4 can be represented by b

  2            , which is negative.       4 0.58

139. a is negative, b is positive, and c is negative. The sum of a and c must be negative. So the

144.

b  a  c can be represented by a2     3         , which is negative.     2 3

sign of

0.18 140. a is negative, b is positive, and c is negative. The even power of any nonzero number is

145. There are missing parentheses around the denominator.

 a  b  b  c  can positive. So the sign of b      be represented by     . Also,   2

4

either  a  b  or  b  c  can equal zero. So the sign of the expression can also be  0    0 or    0  0 . represented by   The sign is positive or zero.

7


Chapter R Review of Prerequisites Section R.2 Models, Algebraic Expressions, and Properties of Real Numbers 1.  5, 2 ;  x | 5  x  2

b. m  0.04 x 2  3.6 x  49

m  0.04  50  3.6  50  49 2

2.  1,   ;  x | x  1

m  31 mpg c. m  0.04 x 2  3.6 x  49

2  9  9  2 or 9  2  9  2

3.

m  0.04  75  3.6  75  49 2

4.   6  6   or 6    6   5. True

6. False

7. False

8. True

12  9  9 9.

2

m  4 mpg The model breaks down. The value of 4 mpg is not possible. 26. a. P  1718t 2  82,000t  10,000

P  1718  22  82,000  22  10,000 2

  2  3

P  982,000 bacteria

7  1

b. P  1718t 2  82,000t  10,000

12   9  3   2  3 2

12   6   2  3 2

 

10.

P  1718  36  82,000  36  10,000 2

6 6 12   18  3 6

102  82

 16  4 

2

P  735,000 bacteria

12  36   2  3

c. P  1718t 2  82,000t  10,000

6 12   54 66    11 6 6

100  64

 4  2

2

36

 2

2

P  1718  48  82,000  48  10,000 2

P  12,000 bacteria The model breaks down. It is impossible to have 12,000 bacteria.

6 3  4 2

27. a. E  27.45t  704.6

E  27.45  3  704.6

11. dependent

12. breakdown

13. term

14. coefficient

15. commutative

16. 0

17. 1

18. identity

19.  x

20. distributive

21. reciprocal

22. associative

23.  a  b   c

24. 0

28. a. W  150  1.2t

25. a. m  0.04 x  3.6 x  49

W  142.8 lb

E  $622.25 b. E  27.45t  704.6

E  27.45 15  704.6 E  $292.85 c. No. Eventually, the model would produce a negative expenditure. This is not possible.

W  150  1.2  6

2

m  0.04  35  3.6  35  49 2

m  28 mpg

8


Section R.2 43. a. C  159n  0.11159n 

b. W  150  1.2t

W  150  1.2 12

C  159n  17.49n C  176.49n

W  135.6 lb c. No. Eventually the model would produce a negative weight or a weight too small to sustain life. This is not possible. 29. Tn  Ts  16

30. K  C  273.15

31. P  W  M

32. S  P  D

33. a. J  C  1

b. C  J  1

34. a. R  F  9

b. F  R  9

35. D  0.25P

36. t  0.20c

37. v  2 gh

38. h  8 t

b. C  176.49n

C  176.49  4 C  $705.96 44. a. C  149n  0.16 149n   40

C  149n  23.84n  40 C  172.84n  40 b. C  172.84n  40

C  172.84  2  40 C  $385.68 45. 12 x 2 y 5  xy 4  9.2 xy 3

 12 x 2 y 5   xy 4  9.2 xy 3

39. a. C  0.12k  14.89

Terms: 12x y ,  xy 4 , 9.2xy 3 ; coefficients: 12, 1 , 9.2 2

b. C  0.12k  14.89

C  0.12 1200  14.89 C  $158.89

5

46. 6c 2 d  6.4cd 5  d 6

 6c 2 d  6.4cd 5  d 6

40. a. C  3.58n  13.50

Terms: 6c d , 6.4cd 5 , d 6 ; coefficients: 6, 6.4 , 1 2

b. 5000  2000  3000 The family used 3 thousand gal over the 2000-gal base. C  3.58n  13.50

47.

C  3.58  3  13.50 C  $24.24 41. a. C  640m  200n  500

5 1  5 m m 5 Term: ; coefficient: 5 m

2 1  2  n n 2 Term:  ; coefficient: 2 n

48. 

b. C  640m  200n  500

C  640 12  200  3  500 C  $8780

49. x y  z  1x y  z

42. a. C  105n  35 L  40

Term: x y  z ; coefficient: 1

b. C  105n  35 L  40

50. m 3  n  1m 3  n

C  105 12  35  2  40

Term: m 3  n ; coefficient: 1

C  $1370

51. 7  x  x  7

9


Chapter R Review of Prerequisites 52. 9  z  z  9

b.

53. 3  w  w   3

5  4   1 4  5

The multiplicative inverse is 54. 11  p  p   11

7 7 64. a.    0 9 9

1 1 55. y   y 3 3 56. z 

The additive inverse is

5 5  z 2 2

The multiplicative inverse is 

58.  c  4  5  c   4  5  c  9

b. 0   ?  1 The multiplicative inverse is undefined.

4  9   4 9 60.    p       p  1 p  p 9  4   9 4

66. a. 1   1  0 The additive inverse is 1 .

61. a. 8  8  0 The additive inverse is 8.

b. 1 1  1 The multiplicative inverse is 1.

 1 b. 8      1  8

67. a. 2.1   2.1  0 The additive inverse is 2.1 .

1 The multiplicative inverse is  . 8

 1  b. 2.1    1  2.1

62. a. 9   9  0 The additive inverse is 9 .

The multiplicative inverse is

 1 b. 9     1  9

1 10 or . 2.1 21

68. a. 4.5   4.5  0 The additive inverse is 4.5 .

1 The multiplicative inverse is . 9

 1  b. 4.5   1  4.5  The multiplicative inverse is 1 10 2 or  . 4.5 45 9

5  5    0 4  4 The additive inverse is 

9 . 7

65. a. 0  0  0 The additive inverse is 0.

1 1  5w    5 w  1w  w 5 5

63. a.

7 . 9

7  9 b.       1 9  7

57.  t  3  9  t   3  9  t  12

59.

4 . 5

5 . 4

69. 14 w3  3w3  w3   14  3  1 w3  16w3

10


Section R.2 70. 12t 5  t 5  6t 5  12  1  6 t 5  5t 5

77.   4 x     1  4 x       4 x  

71. 3.9 x3 y  2.2 xy 3  5.1x 3 y  4.7 xy 3

 3.9 x y  5.1x y  2.2 xy  4.7 xy 3

3

3

 9 x 3 y  6.9 xy 3

3 2

4

3 2

 24 x 2  16 x  8

 0.004m n  0.01m n  0.005m n 4

4

3 2

80. 6 5 y 2  3 y  4

 0.007 m3n 2

81.

82.

 

3 12 p 5 q  8 p 4 q 2  6 p 3q 4 3 3 3   12 p 5 q   8 p 4 q 2  6 p 3q 4 4 4 9  9 p 5 q  6 p 4 q 2  p 3q 2

 8w  16  14w  28  12  22w 84. 3  2 z  4  8  z  9  84

 6 z  12  8 z  72  84  14 z 85.   4u  8v   3  7u  2v   2v

 4u  8v  21u  6v  2v  25u  16v

2

86.  10 x  z   2  8 x  4 z   3 x

 12 p 3  18.3 p 2  24 p  6.6 76. 2 5a 4  a 3  0.4a  0.8

 

 10 x  z  16 x  8 z  3x  29 x  9 z

87. 12  4  8  2v   5  3w  4v    w

 2  5a  2   a  2  0.4a   2  0.8 4

 12  4  8  2v  15w  20v   w

3

 12  4  8  22v  15w  w

 10a  2a  0.8a  1.6 4

83. 2  4w  8  7  2w  4  12

  3   4 p   3   6.1 p   3  8 p   3  2.2

75. 3 4 p 3  6.1 p 2  8 p  2.2

2 6 x 2 y  18 yz 2  2 z 3 3 2 2 2   6 x 2 y   18 yz 2  2 z 3 3 3 3 4  4 x 2 y  12 yz 2  z 3 3

1 4 3 4 3 yz  y z  yz 4  y 4 z 10 4 2 1 4 3 3  yz  yz 4  y 4 z  y 4 z 10 4 2 1 4 10 3 2 3  yz   yz 4  y 4 z   y 4 z 10 10 4 2 2 1 4 10 3 6  yz   yz 4  y 4 z  y 4 z 10 10 4 4 11 4 3 4  yz  y z 10 4

3

 30 y 2  18 y  24

1 1 2 73. c 7 d  cd 7  c 7 d  2cd 7 3 2 5 1 7 2 7 1  c d  c d  cd 7  2cd 7 3 5 2 5 1 3 2 1 2   c 7 d   c 7 d  cd 7   2cd 7 5 3 3 5 2 2 5 7 6 7 1 7 4 7  c d  c d  cd  cd 15 15 2 2 1 3   c 7 d  cd 7 15 2

 6  5 y 2  6   3 y   6  4

 0.006m 4 n  0.002m3 n 2

74.

 

79. 8 3 x 2  2 x  1  8  3 x 2  8   2 x   8  1

72. 0.004m n  0.005m n  0.01m n  0.007 m n 4

1 1  1  78.   k  7   1  k   7    k  7 2  2 2 

3

3

 12  32  88v  60 w  w  88v  59w  20

11


Chapter R Review of Prerequisites 88. 6  2  9 z  6 y   8  y  z    11

97. W  2 L  3

 6  2  9 z  6 y  8 y  8 z   11

98. L  3S  5

 6  2  2 y  17 z   11

99. E2009  E2008  489,000

 6  4 y  34 z  11  4 y  34 z  5

100. C2011  C2010  0.25

2   89. 2 y 2  13  6 y 2  9  10  9 3  

1 1 c h 15 25 1 1 G   240   500 15 25 G  16  20  36 gal

101. G 

   2 y   4 y  9  9

 2 y 2  13  4 y 2  6  10  9 2

2

 2 y2  4 y2  9  9  6 y2 3   90. 6  5t 2  12  8t 2  5  11t 2 4  

1 1 c h 22 32 1 1 G   220   512 22 32 G  10  16  26 gal

102. G 

   6  11t  4  11t

 6  5t 2  9  6t 2  5  11t 2 2

2

 6  11t 2  4  11t 2  10

103. The commutative property of addition indicates that the order in which two quantities are added does not affect the sum. The associative property of addition indicates that the manner in which quantities are grouped under addition does not affect the sum.

b   6 6    3 91. 2a 2  1 2 92.

b 2  4ac 

 6 2  4  2 4

 36  32  4  2 93.

104. For any nonzero number a, the multiplicative 1 inverse is defined as . If a  0 , this would a 1 be which is undefined. 0

 x2  x1  2   y2  y1  2 

 1  2 2  1   4

 32   52  9  25  34

2

105.

y  y1 3.7  3.1 6.8    2 94. 2 x2  x1 2   1.4 3.4 95.

96.

1 2 1  22  2  r h     7   6 3 3 7  22 6468   49 6   308 21 21

106.

4 3 4  22  3 88  9  792  r     3  27  3 3 7  7 21   7

12


Section R.3 Section R.3 Integer Exponents and Scientific Notation 1. 

2.

0

1 1  5 x  2 y  15   2 y  20 5 10 2 1 3  x  y  3  y  2  x  y 1 5 5 5

 2  d.   p  1  3  0

 3 16. a.     1  7

1 5  3a  2b  6  12a  2b 3 6 2 5 7  a  b  2  10a  b  9a  b  2 3 3 3

b. 

30 1 1 1    30   1   7 7 7 7

3 3 3 3 c.  w0    w0   1   7 7 7 7

3. a  6    a  6  6  a

0

 3  d.   w  1  7 

4. 2  t    2  t   t  2 5. t  w  1.93

6. L  D  3

7. 1

8.  n

9. scientific

10. magnitude

11. m  n

12. quotient

17. a. 82 

1 1  2 8 64

b. 8 x 2  8 

1 8  2 2 x x

c.  8 x  

1

2

13. a. 8  1 0

1 64 x 2

1 1 1 d. 82  1 82  1  2  1   8 64 64

b. 80  1 80  11  1 c. 8 x 0  8  x 0  8 1  8

18. a. 7 2 

d.  8 x   1 0

14. a. 70  1 b. 7 0  1 70  11  1

1 1  2 7 49

b. 7 y 2  7 

1 7  2 2 y y

c.  7 y  

1

2

c. 7 y  7  y  7 1  7 0

8x

2

0

d.  7 y   1

7 y

2

1 49 y 2

0

d. 7 2  1  7 2  1

1 1 1  1   2 7 49 49

0

 2 15. a.     1  3 b. 

19. a.

20 1 1 1    20    1   3 3 3 3

1 1 q2     q2 1 1 1 q 2 q2

b. q 2 

2 2 2 2 c.  p 0    p 0   1   3 3 3 3

13

1 q2


Chapter R Review of Prerequisites 1 5 p3  2 q2 q

30.

1 2 5q 2 q  3 p3 p

31.

c. 5 p 3 q 2  5  p 3  q 2  5  p 3  d. 5 p 3q 2  5  p 3  q 2  5 

20. a.

1 t

 4

1 t4  1  t 4 1 1 t4

1 b. t 4  4 t

32.

1 11u 2 c. 11t 4u 2  11 t 4  u 2  11 4  u 2  4 t t d. 11t 4u 2  11 t 4  u 2  11 t 4 

1 11t 4  2 u2 u

21. 25  27  257  212 33. 22. 43  48  438  411 23. x 7  x 6  x 2  x 76 2  x11 24. y 3  y 7  y 4  y 37 4  y 8



25. 3c 2 d 7 4c 5 d  3  4  c 25  d 71

34.

 12  c 3  d 8

72 1 1  7 24  7 2  2  4 7 7 49 18k 2 p 9 18 k 2 p 9    27 k 5 p 2 27 k 5 p 2 2 2   k 25  p 92   k 3  p 7 3 3 2 1 2 p7   3  p7  3 3 k 3k

10a 3b11 10 a 3 b11    25a 7b3 25 a 7 b3 2 2   a 37  b113   a 4  b8 5 5 8 2 1 2b   4  b8  4 5 a 5a 2m 6 n 4 2 m 6 n 4    6m 2 n 1 6 m 2 n 1 1 1   m 6 2  n 4 1   m 4  n5 3 3 5 1 1 n   4  n5  3 m 3m 4

4 p 7 q 5 4 p 7 q 5   2  2  2  p 7 2  q 5 2 2 2 2p q 2 p q  2  p 5  q 7  2 

1 12d 8  12  3  d 8   3 c c



  4 4

26. 7 m 3 n 8 3m 5 n   7   3  m 3 5  n 81

35. 42

 21 m 8  n 7 1 1 21  21 8  7   8 7 m n mn

36. 23

27.

29.

23

6

  2 2

37.

y 3 y 6 y 36 y 3  2  2  y 32  y1  y y2 y y z 8 z12 z 812 z 4  3  3  z 4 3  z1  z z3 z z

5

35

p   p

15

2 7

27

 p 14 

  q

42

 q 8 

38. q 4 28.

3

1 7 2q 7 q  5 p5 p

2

1 p14

1 q8

39.  2cd    2  c   d   8c3 d 3 3

65 1 1  658  63  3  8 6 6 216

3

3

3

40.  8mn    8  m   n   64m 2 n 2 2

14

2

2

2


Section R.3

 7a    7  a   72 a 2  49a 2  7a  41.     b b2 b2 b2  b2 2

2

2

 36a 6b9  50.   9a 2b 4 

2

2

 4a 4b13

  

4 4 p4 q4 pq  p4q4   pq  42.       4 256 256  4 4

  4 

43. 4 x 2 y 3

  4  x   y   16 x y  16yx 2

2 2

2

3 2

6

4

4

6

   3  w   z 

3 5 2

44. 3w z

3 2

2

5 2

 4 x 3 z 5  51.   12 y 2 

9 z10  9w z  6 w 6 10

 7k  45.  2  n 

2

 7 kn 

 5w  46.   v  5

   7 k n

2 2

2

2

3

4

 3 z 2 w1  52.   15 p 4 

5 3

3

1 3

3

2

 1    4

3

 4 a b 2

3

 1    5

3

8 26

0

53.  2 y 

 64  64  1  1

3

1 16a8b 26

3

x   y  z  3 3

2 3

5 3

27 z15 x9 y 6 3

z  w   p  2 3

1 3

 53 z 6 w3 p 12 

 1     8 2  43  1  2

13 2

 a 8b 26

1     z 2 w1 p 4  5 

1 1 v3  53 w15v3  3  15  v3  5 w 125w15  1 47.    8

4 2

 33 x 9 y 6 z15 

   5  w   v 

 5w v

2

1

 1    3

1 1 4 n  2 n  49 k 49k 2 5 1 3

 4 2

2

 a  b 

2

1

1     x 3 y 2 z 5  3 

4

3

 4a 62b9 4

4 3

125w3 p12 z 6

 6 y z    2 y  6  y   z  3

2 8 2

3

2

2 2

8 2

  2  62  y 3  y 4  z16 3

 1 48.    9

2

 1    3

4

0

 1     92  34  1  27 

  2  62  y 7  z16 3

 81  81  1  1  16m n  49.  5 2   8m n  2

7

2

 2m 25 n7 2

 2m 3 n9   2  

2

1

 2

2

54.  15 z 

2

5z w  4

 2 y

2

3

 7

36 z16 9 z16  7 7 8 y 2y

6 3

  w 

  15 z 2  53 z 4

m  n  3 2

2

62 z16

3

6 3

  15  53  w18  z 2  z12 2

9 2

  15  53  w18  z10 2

 m 6 n 18 2

m6

m6   22 n18 4n18

15

53 z10

125 z10 5 z10    152 w18 225w18 9w18


Chapter R Review of Prerequisites  1 1 1 55.      2 3 6

3

 

 3   1   2   1   1                  3   2   2   3   6  

3

3

3

 3 2 1      6 6 6  1 1 1 56.       3 4 2

 2    6

3

 x 3 4 x 2   61.  x 6  

1

 x12 x 2 x 6

 

 x16

 1    3

2

 4 3 6       12 12 12 

2

 1     12 

2

 y 12 y 3 y 4

 y 11

2

 1    3

3 2 2

 1    6

3

 1 58.     3

 1    4

2

 23  33  63

 1    2

2

  3  4  2 2

2

2

2

2

1

 31 a 2b 4 6 12

3 a b 3

1

 3 ab 2

5

v w x  15

10

20

x 32 16 x 32   v 21w28 16  1 v 21w28 2

3 2 3

5

3 2 1

3 6

9

6

1 5

3 2

4

1 4 4

4

16

4 16

 143 v 6 w9 x 6  211 v 5 w3 x 2   7 

4

v w x  16

4 16

 143  21  7   v 15 w2 x8 4

1

   3a b  3

1

14v w x   7v w x 64.    21v w x  14 v w x    7   v w x  21 v w x

16  y 4 2 y 4   7 8 x7 x 3

5

 4281 x 7 y 4

 1   3  60.  2 4   5 4   3a b   a b 

42 v 2 w6 x 4 5  1 v 15 w10 x 20 4 8 12 8  2 vw x

 42  24   1  v 21w28 x 32

1

  4 x 3 y 5   8 x13 y 14   2 6 10 1 13 14 4 x y 8 x y

4 5

2

 42 v 2 w6 x 4  24 v 8 w12 x8   1

 9  16  4  3  1   8  59.  3 5   13 14   4x y   x y 

1

3 2 4

2

 8  27  216  197 2

1

4vw x   v w x 63.    2v w x  3

 1 57.    2

 y11

2

3

1

1

 

 y 2 6 y 3   62.  y 4  

  12  144 3

1

 x 16 1  16 x

 33  27

 4   1   3   1   6   1                     4   3   3   4   6   2  

5 4 1

143  21 x8

 74 v15 w

65.  3 x  5

14

5 4

 2

24 x8 v15 w2

 3x  5 2   3x  514 2   3x  512

3 a b

66.  2 y  7 z 

8

9a b8

16

4

 2 y  7 z 13   2 y  7 z  413 9  2 y  7z


Section R.3 67.  6v  7     6v  7    10 9

109

68.  4 x  9    4 x  9   5 11

511 

  6v  7 

c. Move the decimal point 0 places. The exponent on 10 is 0. 2.71  2.71  100

90

  4 x  9

55

73. a. Move the decimal point 1 place. The number is between 0 and 1, so use a negative power of 10. 0.86  8.6  101

1 1  1 2  4 22 21 1 1   1 2  4 4 2 1 2 4 8 16      4 4 4 4 4 31  4

69. 22  21  20  21  22 

b. Move the decimal point 0 places. The exponent on 10 is 0. 8.6  8.6  100 c. Move the decimal point 1 place. The number is greater than 10, so use a positive power of 10. 86  8.6  101

1 1  1 3  9 32 31 1 1   1 3  9 9 3 1 3 9 27 81      9 9 9 9 9 121  9

70. 32  31  30  31  32 

74. a. Move the decimal point 1 place. The number is between 0 and 1, so use a negative power of 10. 0.792  7.92  101 b. Move the decimal point 0 places. The exponent on 10 is 0. 7.92  7.92  100 c. Move the decimal point 1 place. The number is greater than 10, so use a positive power of 10. 79.2  7.92  101

71. a. Move the decimal point 5 places. The number is greater than 10, so use a positive power of 10. 350, 000  3.5  105 b. Move the decimal point 5 places. The number is between 0 and 1, so use a negative power of 10. 0.000 035  3.5  10 5

75. Move the decimal point 10 places. The number is greater than 10, so use a positive power of 10. 29,980,000,000 cm/sec  2.998  1010 cm/sec

c. Move the decimal point 0 places. The exponent on 10 is 0. 3.5  3.5  100

76. Move the decimal point 10 places. The number is greater than 10, so use a positive power of 10. 149,000, 000 km  1.49  108 km

72. a. Move the decimal point 3 places. The number is greater than 10, so use a positive power of 10. 2710  2.71  103

77. Move the decimal point 5 places. The number is between 0 and 1, so use a negative power of 10. 0.000 01 cm  1.0  105 cm

b. Move the decimal point 3 places. The number is between 0 and 1, so use a negative power of 10. 0.002 71  2.71  10 3

78. Move the decimal point 12 places. The number is between 0 and 1, so use a negative power of 10. 0.000 000 000 001 sec  1.0  10 12 sec

17


Chapter R Review of Prerequisites 79. Move the decimal point 0 places. The exponent on 10 is 0. 4.2 L  4.2  100 L

c. 101 is greater than 10. Move the decimal point 1 place to the right. 1.87  101  18.7

80. Move the decimal point 0 places. The exponent on 10 is 0. 8.25 hr  8.25  100 hr

85. 1.67  1021 molecules

81. a. 10 6 is between 0 and 1. Move the decimal point 6 places to the left. 2.61  10 6  0.000 002 61

86. 3.0  1012 bytes  3, 000, 000, 000, 000 bytes

 1,670,000,000,000,000,000,000 molecules

87. 7.0  106 m  0.000007 m

b. 106 is greater than 10. Move the decimal point 6 places to the right. 2.61  106  2, 610, 000

88. 4.7  10 7  0.000 000 47



 

89. 2  10 3 4  108   2 4  103 108

c. 10  1 The decimal point does not move. 2.61  100  2.61 0

 8  10



5

 

90. 3  104 2  101   3 2  104 101

2

82. a. 10 is between 0 and 1. Move the decimal point 2 places to the left. 3.52  102  0.035 2

 6  10

b. 102 is greater than 10. Move the decimal point 2 places to the right. 3.52  102  352

91.

8.4  106  8.4   106       2   4  104 2  2.1  10  2.1  10

c. 100  1 The decimal point does not move. 3.52  100  3.52

92.

6.8  1011  6.8   1011    2  108  3.4  103  3.4   103 



3

  

93. 6.2  1011 3  104   6.2 3  1011 104

1

83. a. 10 is between 0 and 1. Move the decimal point 1 place to the left. 6.718  10 1  0.6718

 18.6  10

15

 1.86  101  1015  1.86  1016

b. 100  1 The decimal point does not move. 6.718  100  6.718



  

94. 8.1  106 2  105   8.1 2  106 105

c. 101 is greater than 10. Move the decimal point 1 place to the right. 6.718  101  67.18

 16.2  1011

 1.62  101  1011  1.62  10

12

84. a. 101 is between 0 and 1. Move the decimal point 1 place to the left. 1.87  101  0.187

95.

b. 100  1 The decimal point does not move. 1.87  100  1.87

3.6  1014  3.6   10 14     5   105  5  105  0.72  1019

 7.2  101  10 19  7.2  10 20

18


Section 2.2 67. A circle is the set of all points in a plane that are equidistant from a fixed point called the center. 68. The center and radius can easily be identified from an equation of a circle written in standard form. 69. The calculator does not connect the pixels at the ends of the semicircles because of limited resolution.

73.  14, 39, 5 by  30, 5, 5

70. The screen is rectangular and is longer horizontally than vertically. If the x and y intervals are the same, then the graph will be elongated horizontally because there is greater distance horizontally over which to spread the graph. 71.  15.1,15.1,1 by  10,10, 1

74.  0.28, 0.18, 0.02 by  0.1, 0.2, 0.02

72.  15.1,15.1,1 by  10,10, 1

211


Chapter 2 Functions and Graphs

Section 2.3 Functions and Relations 1.

 x2  x1 2   y2  y1 2

r

2

2 1    4    1   2   2

49 85 85 9   4 4 2

1  Center:  ,  2 ; Radius: 2 

85 2

5. Substitute 0 for y: y  x 3  5 x 2  4 x  20

2.

0  x 3  5 x 2  4 x  20 0  x 3  4 x  5 x 2  20

 

0  x x2  4  5 x2  4

0  x  4  x  5 0   x  2 x  2 x  5 x  2 or x  2 or x  5 Substitute 0 for x: y  x 3  5 x 2  4 x  20

 x  x y  y2  3. C   1 2 , 1   2 2 

y   0  5  0  4  0  20  20

 2   1 7  3   1 10   1   ,   ,    , 5 2   2 2   2   2 r

2

3

2

x-intercepts:  2, 0 ,  2, 0 ,  5, 0 ; y-intercept:  0,  20

 x2  x1  2   y2  y1 2 2

1 2    2     7  5  2

6. Substitute 0 for y: y   x3  3x 2  x  3 0   x3  3x 2  x  3

9 25 5  4   4 4 2 5 1  Center:  , 5 ; Radius: 2  2

0  x3  3x 2  x  3 0  x3  x  3 x 2  3

 

0  x x2  1  3 x2  1

0  x  1  x  3

 x  x y  y2  4. C   1 2 , 1   2 2 

2

0   x  1 x  1 x  3 x  1 or x  1 or x  3 Substitute 0 for x: y   x3  3x 2  x  3

 3  4 5  1  1 4   1     ,    ,  2 ,   2  2  2 2  2

y    0  3  0   0  3  3 3

2

x-intercepts: 1, 0 ,  1, 0 ,  3, 0 ;

y-intercept:  0,  3

212


Section 2.3 7. relation; domain; y

25. No vertical line intersects the graph in more than one point. This relation is a function.

8. A relation defines y as a function of x if for each value of x in the domain, there is exactly one value of y in the range. 9. does not

10.  2, 4

11. y

12. f  x 

13. 5

26. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function. 27. No vertical line intersects the graph in more than one point. This relation is a function. 28. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function.

14. 5

15. a. {(Tom Hanks, 5), (Jack Nicholson, 12), (Sean Penn, 5), (Dustin Hoffman, 7)}

29. This mapping defines the set of ordered pairs:  8, 3 ,  8, 4 ,  8, 5 ,  8, 6 . All

b. {Tom Hanks, Jack Nicholson, Sean Penn, Dustin Hoffman} c. {5, 12, 7}

four ordered pairs have the same x value but different y values. This relation is not a function.

d. Yes

16. a. {(Albany, 285), (Denver, 5883), (Miami, 11), (San Francisco, 11)}

30. This mapping defines the set of ordered pairs:  3,11 ,  6,13 ,  6, 9 . Two

b. {Albany, Denver, Miami, San Francisco} c. {285, 5883, 11}

ordered pairs have the same x value but different y values. This relation is not a function.

d. Yes

17. a.  4, 3 ,  2,  3 , 1, 4 ,  3,  2 ,  3,1 b.  4,  2,1, 3

31. This mapping defines the set of ordered pairs: 1, 4 ,  2, 5 ,  3, 5 . No two ordered

c.  3,  3, 4,  2,1

pairs have the same x value but different y values. This relation is a function.

d. No 18. a.  2, 2 ,  1, 0 ,  0,  2 , 1, 0 , 1, 2 b.  2,  1, 0,1

32. This mapping defines the set of ordered pairs:  5,  4 ,  6,  4 ,  7,  4 ,  8,  4 .

c.  2, 0,  2

No two ordered pairs have the same x value but different y values. This relation is a function.

d. No 19. False

20. True 33.  x  1   y  5  25 This equation represents the graph of a circle with center  1,  5 and radius 5. This relation is not a function because it fails the vertical line test. 2

21. No vertical line intersects the graph in more than one point. This relation is a function. 22. No vertical line intersects the graph in more than one point. This relation is a function.

2

34.  x  3   y  4  1 This equation represents the graph of a circle with center  3,  4 and radius 1. This relation is not a function because it fails the vertical line test.

23. There is at least one vertical line that intersects the graph in more than one point. This relation is not a function.

2

24. No vertical line intersects the graph in more than one point. This relation is a function.

213

2


Chapter 2 Functions and Graphs 35. y  x  3

Two or more ordered pairs have the same x value but different y values. This relation is not a function.

38. a. y  x y  x

No vertical line intersects the graph in more than one point. This relation is a function.

36. y  x  4 No vertical line intersects the graph in more than one point. This relation is a function.

b.

x y y x y  x

No vertical line intersects the graph in more than one point. This relation is a function.

37. a. y  x

b.

x

y

y  x

Ordered pairs

0

0

y  0

1

1

y  1

2

2

y  2

3

3

y  3

 0, 0 1,  1  2,  2  3,  3

2

Two or more ordered pairs have the same x value but different y values. This relation is not a function.

No vertical line intersects the graph in more than one point. This relation is a function.

39.  4,1

x  y2

40.  7,  5

y2  x

41. f  x   x 2  3x

y x x

y

y x

0

0

y 00

1

1

y   1  1

4

2

y   4  2

9

3

y   9  3

a. f  2   2  3  2  4  6  2

Ordered pairs  0, 0

2

b. f  1   1  3  1  1  3  2 2

1,  1  4,  2  9,  3

c. f  0   0  3  0  0  0  0 2

d. f 1  1  3 1  1  3  4 2

e. f  2   2  3  2  4  6  10 2

214


Section 2.3 42. g  x  

49. k  5 

1 x

 5  1 

4

Undefined

a. g  2 

1 1   2 2

50. f  5   5  3  5  25  15  40

b. g  1 

1  1  1

51. k  8 

2

52. f  5   5  3  5  25  15  10 2

1  1 c. g      2  2  1    2

53. g  t  

1  1 d. g    2  2  1   2

e. g  2 

1 1  t  t

54. f  a    a   3  a   a 2  3a 2

55. k  a  b  

1 1   2 2

 a  b  1  a  b  1

56. h  a  b   5

43. h  x   5 a. h  2  5

57. f  a  4   a  4  3  a  4 2

b. h  1  5

c. h  0  5

 a 2  8a  16  3a  12

d. h 1  5

 a 2  11a  28

e. h  2  5

58. f  t  3   t  3  3  t  3 2

44. k  x   x  1 a. k  2 

 t 2  6t  9  3t  9  t 2  3t

 2  1  1

1 0 Undefined

59. g  0 

Undefined b. k  1 

 1  1  0  0

c. k  0 

 0  1  1  1

d. k 1 

1  1  2

e. k  3 

 3  1  4  2

45. g  3 

 8  1  9  3

1 1   3 3

1  1 47. g    3  3  1    3

60. k  10 

 10  1 

9

Undefined

46. h  7   5

61. f  9  7

62. f  1  6

63. f  3  4

64. f  2  3

65. f  x   6 when x  1 66. f  x   7 when x  9

48. h  7   5

67. f  x   3 when x  2

215


Chapter 2 Functions and Graphs 68. f  x   4 when x  3

d  2  18  2  36 Joe rides 36 mi in 2 hr.

 12

75. Solve h  x   0 :

Evaluate h  0 : h  x  x  8

 2  2 d    18    12  3  3 12 mi

h  x  x  8

r  x   250  x

x  8 x-intercepts:  8, 0 ,  8, 0 ;

x 8

Evaluate k  0 :

0 x 2

k  0    0  2

k  x   x  2

x 2

2

x  2 x-intercepts:  2, 0 ,  2, 0 ;

C  225   225  0.06  225  0.18  225

y-intercept:  0, 2

 225  13.5  40.5  279 If the cost of the food is $225, then the total bill including tax and tip is $279.

77. Solve p  x   0 :

Evaluate p  0 :

p  x    x 2  12

P  x   1.075  x  0.40 x 

p  x    x 2  12 p  0    0  12  12

0   x 2  12

P  60  1.075  60  0.40  60 

2

x 2  12

 1.075  60  24  1.075  84  90.30 If the cost of the book from the publisher is $60, the cost to the student after the bookstore markup and sales tax is $90.30.

0  2x  4

76. Solve k  x   0 : k  x   x  2

C  x   x  0.06 x  0.18 x

f  x  2 x  4

 8

y-intercept:  0, 8

r  x   250  x

73. Solve f  x   0 :

h  0   0  8

0  x 8

r 122  250  122  128 128 mi

72.

g  0  3  0  12

x4 x-intercept:  4, 0 ; y-intercept:  0, 12

r  50  250   50  200 After having driven 50 mi, Frank still has 200 mi remaining.

71.

0  3x  12

g  x   3x  12

3 x  12

40 2 b. 40 min is  hr. 60 3 d  t   18t

b.

Evaluate g  0 :

g  x   3x  12

69. a. d  t   18t

70. a.

74. Solve g  x   0 :

x   12  2 3



x-intercepts: 2 3, 0 , 2 3, 0 ; y-intercept:  0,12

Evaluate f  0 :

78. Solve q  x   0 :

f  x  2 x  4

q  x  x  8 2

f  0  2  0  4

0  x2  8

2x  4

 4 x2 x-intercept:  2, 0 ; y-intercept:  0,  4

x2  8 x   8  2 2

Evaluate q  0 : q  x  x2  8

q  0   0  8  8 2



x-intercepts: 2 2, 0 , 2 2, 0 ; y-intercept:  0, 8

216


Section 2.3 79. Solve r  x   0 : r  x  x  8 0  x 8 x 8  0

84. Evaluate B  0 :

Evaluate r  0 : r  x  x  8

B  x   0.12 x 2  0.86 x  1.6

r  0   0  8

B  0  0.12  0  0.86  0  1.6  1.6 2

 0,1.6 ; The y-intercept means that for

8

x8 x-intercept:  8, 0 ; y-intercept:  0, 8

80. Solve s  x   0 : s  x  x  3 0 x3 x3 0

x  0 (the year 2002), the number of Botox injection procedures in the United States was 1.6 million.

Evaluate s  0 :

85. Domain:  3,  2,  1, 2, 3 ;

s  x  x  3

Range:  4,  3, 3, 4, 5

s  0    0  3

86. Domain:  4,  2, 0, 2, 4 ;

3

x  3 x-intercept:  3, 0 ; y-intercept:  0, 3

81. Solve f  x   0 :

Evaluate f  0 :

f  x  x  2

f  x  x  2

0 x 2 x2

f  0 

Range:  2, 0, 2, 4, 6

87. Domain:  3,   ; Range: 1,   88. Domain:  , 4 ; Range:  , 4

 0  2

89. Domain:  ,   ; Range:  ,  

 2

x4 x-intercept:  4, 0 ; y-intercept:  0,  2

82. Solve g  x   0 :

Evaluate g  0 :

g  x   x  3

g  x   x  3

0  x 3

g  0    0  3

x 3

90. Domain:  ,   ; Range:  ,   91. Domain:  ,   ; Range:  3,   92. Domain:  ,   ; Range:  , 2 93. Domain:  5,1 ; Range:  1,1, 3

3

x9 x-intercept:  9, 0 ; y-intercept:  0, 3

94. Domain:  4, 5 ; Range:  2,1, 4 95. x  4  0 x4  , 4   4, 

83. Evaluate f  0 :

f  x   9.4 x  35.7 f  0  9.4  0  35.7

96. x  2  0

 35.7 0, 35.7   ; The y-intercept means that for x  0 (the year 2006), the average amount spent on video games per person in the United States was $35.70.

x2  , 2   2, 

217


Chapter 2 Functions and Graphs 106. 6  x  0  x  6 x6  , 6

97. 2t  7  0 2t  7

t

7 2

7  7    ,      ,  2 2

107. There is no restriction on x.  , 

98. 3a  4  0 3a  4

a

108. There is no restriction on x.  , 

4 3

109. a. f  2  4

4  4    ,      ,  3 3

b. f  3  2 c. f  x   1 for x  3 , x  1 , x  1

99. The denominator x  4 is always positive.

 , 

d. f  x   4 for x  2 , x  2

 10  e. x-intercepts:  0, 0 and   , 0  3 

100. The denominator x 2  3 is always positive.  , 

f. y-intercept:  0, 0

101. a  15  0 a  15  15, 

g. Domain:  ,   h. Range:  4,  

102. c  12  0 c  12  12,  

110. a. f  2  1 b. f  3  3 c. There are no x-values where f  x   1 .

103. 3  x  0  x  3

d. f  x   4 for all x on the interval  0, 2

x3  , 3

e. x-intercept:  1, 0 f. y-intercept:  0,  4

104. 6  x  0  x  6 x6  , 6

g. Domain:  ,   h. Range:  4   0,   111. a. f  2  0

105. 3  x  0  x  3 x3  , 3

b. f  3  5 c. f  x   1 for x  3 , x  1 , x  1

218


Section 2.3 d. f  x   4 for x  4 , x  0

119. f  x   3x 2  2

4  e. x-intercepts:  2, 0 and  , 0 3 

120. f  x   x  3

f. y-intercept:  0,  4

121. If two points in a set of ordered pairs are aligned vertically in a graph, then they have the same x-coordinate but different ycoordinates. This contradicts the definition of a function. Therefore, the points do not define y as a function of x.

g. Domain:  4,   h. Range:  4, 5 112. a. f  2  3

122. To find the x-intercept(s), find the real solutions to the equation f  x   0 . To find

b. f  3  2

the y-intercept, evaluate f  0 .

c. f  x   1 for x  0 , x  2

123. a. P  s   4s

d. f  x   4 for x  3 , x  5 e. x-intercept: 1, 0

b. A  s   s 2

f. y-intercept:  0,  1

P2  P c. A  P     or A  P    4 16

2

g. Domain:  , 5

d. P  A  4 A

h. Range:  4, 0   3

e.  d  s    s 2  s 2  2 s 2 2

113. r  x   x  400

d  s   2s 2

r  x   400  x

d  s  s 2

114. w  x   x  200

f.  s  d     s  d    d 2 2

w  x   200  x

2

2  s  d    d 2 2

115. P  x   x  x  x

2 d2  s  d    2

P  x  3x 116. 2 A  x   x  180

2 A  x   180  x A x 

180  x 2

sd  

d2 2

sd  

d

 d 2 g. P  d   4    2 

117. C  x   x  90

C  x   90  x

P  d   2 2d

118. S  x   x  180

S  x   180  x

219

2

or s  d  

d 2 2


Chapter 2 Functions and Graphs  d 2 h. A  d      2 

2

d f. A  d       2 A d  

2d 2 A d   4 d2 A d   2

d 2 4

C g. A  C       2 

124. a. C  r   2 r b. A  r    r 2 c. r  d  

2

d 2

AC  

C 2 4 2

AC  

C2 4

h. C  A  2

d. d  r   2r

2

A

2A  C  A  2  A C  A  2

e. C  d    d

Section 2.4 Linear Equations in Two Variables and Linear Functions 1. a. No vertical line intersects the graph in more than one point. This relation is a function. b.  ,  

c.  ,  

d.  2, 0

e.  0, 3

c.  ,  

d.  3, 0

e.  0,1

c.  3

d. None

e.  0, 3

 x  x y  y2  5. a. M   1 2 , 1  2   2   3 2  2 5  4 5    ,   2 2  

2. a. No vertical line intersects the graph in more than one point. This relation is a function. b.  ,  

b.  ,  

 4 2 3 5  ,    2 2    3 5   2 2,    3  

3. a. No, a vertical line is not a function. b.  3

c.  ,  

d.  3, 0

e. None

b. d 

 x2  x1  2   y2  y1  2

 2  3 2    4 5  5    2 2    5 5  2

2

4. a. No vertical line intersects the graph in more than one point. This relation is a function.

 8  125  133

220

2

2


Section 2.4 20. 2 x  y  4

 x  x y  y2  6. a. M   1 2 , 1   2 2 

y  2x  4

 6  2 6 3  5 3 ,   2 2  

x 3 2 1 0

 3 6 6 3  3 6  , , 3 3    2   2  2 

b. d 

 x2  x1 2   y2  y1  2

 2 6  6   5 3  3   6    4 3 2

2

y 2 0 2 4

2

x-intercept:  2, 0 ; y-intercept:  0, 4

2

21. 2 y  5 x  2

 6  48  54  3 6

7. scatter

8. linear

5 y   x 1 2

9. vertical

10. horizontal

x

11. True

12. m 

y2  y1 x2  x1

1

13. zero

14. undefined

15. slope; intercept

16. mx  b

17. m 

f  x2   f  x1  x2  x1

0 2 5 2

18. horizontal

2

0 2

4

22. 3 y  4 x  6

4 y   x2 3

4 y  3x  12

x 4

0

2  x-intercept:  , 0 ; y-intercept:  0,1 5 

19. 3 x  4 y  12 y

y 7 2 1

3 x3 4

x 1

y 0 3 2 3

0 3 2 3

9 2

y 10 3 2

0 2

3  x-intercept:  , 0 ; y-intercept:  0, 2 2 

x-intercept:  4, 0 ; y-intercept:  0, 3

221


Chapter 2 Functions and Graphs 23. x  6 x 6 6 6 6

26. 3x  2  4 3x  6 x2

y 2 0 2 4

x 2 2 2 2

y 2 0 2 4

x-intercept:  6, 0 ; y-intercept: None x-intercept:  2, 0 ; y-intercept: None

24. y  4 x 2 0 2 4

27. 0.02 x  0.05 y  0.1 2 x  5 y  10 5 y  2 x  10 2 y   x2 5

y 4 4 4 4

x 5 0

x-intercept: None; y-intercept:  0, 4 25. 5 y  1  11

1

5 y  10 y2

5

x 2 0 2 4

y 4 2 8 5 0

x-intercept:  5, 0 ; y-intercept:  0, 2

y 2 2 2 2

28. 0.03x  0.07 y  0.21 3x  7 y  21 7 y  3x  21 3 y   x3 7

x-intercept: None; y-intercept:  0, 2

x 1

0 1 7

y 24 7 3 18 7 0

x-intercept:  7, 0 ; y-intercept:  0, 3

222


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SOLUTIONS MANUAL for College Algebra Essentials 1st Edition by Julie Miller. ISBN-13 978-0078035616. by welldoneassistant - Issuu