ECO- Engineering

STUDENTS SOLUTIONS MANUAL

JUDITH A. PENNA

Indiana University Purdue University Indianapolis

COLLEGE ALGEBRA: GRAPHS AND MODELS

FIFTH EDITION

Marvin L. Bittinger Indiana University Purdue University Indianapolis

Judith A. Beecher Indiana University Purdue University Indianapolis

David J. Ellenbogen Community College of Vermont

Judith A. Penna

Indiana University Purdue University Indianapolis

Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal TorontoDelhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

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ISBN-13: 978-0-321-79125-2 ISBN-10: 0-321-79125-8

1 2 3 4 5 6 BRR 15 14 13 12 11

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Contents

Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Chapter 2. . . . . . . . . . . . . . . . . . . . . . . . . 63

Chapter 3. . . . . . . . . . . . . . . . . . . . . . . . . 93

Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 129

Chapter 5. . . . . . . . . . . . . . . . . . . . . . . . . 181

Chapter 6. . . . . . . . . . . . . . . . . . . . . . . . . 215

Chapter 7. . . . . . . . . . . . . . . . . . . . . . . . . 275

Chapter 8. . . . . . . . . . . . . . . . . . . . . . . . . 307

Copyright 2013 Pearson Education, Inc.

8

Chapter R

Basic Concepts of Algebra

Exercise Set R.1

1. Rational numbers: 3, 6, 2.45, 18.4, 11, 27, 56, 7, 0,163218

3. Irrational numbers: 3, 6 26, 7.151551555..., 35, 5 3

(Although there is a pattern in 7.151551555..., there is no repeating block of digits.)5. Whole numbers: 6, 3 27, 0, 16

7. Integers but not natural numbers: 11, 0

9. Rational numbers but not integers: 3, 2.45, 18.4, 56, 7218

11. This is a closed interval, so we use brackets. Interval no-tation is [5,5].

_505

13. This is a half-open interval. We use a parenthesis on the left and a bracket on the right. Interval notation is (3,1].

23. The endpoint 9 is included in the interval, so we use a bracket before the 9. The endpoint 4 is not included, so we use a parenthesis after the 4. Interval notation is [9,4).

25. Both endpoints are included in the interval, so we use brackets. Interval notation is [x,x+h].

27. The endpoint p is not included in the interval, so we use a parenthesis before the p. The interval is of unlimited ex-tent in the positive direction, so we use the innity symbol . Interval notation is (p,).

29. Since 6 is an element of the set of natural numbers, the statement is true.

31. Since 3.2 is not an element of the set of integers, the state-ment is false.

33. Since 11 is an element of the set of rational numbers, the statement is true.5

35. Since 11 is an element of the set of real numbers, the statement is false.

37. Since 24 is an element of the set of whole numbers, the statement is false.

39. Since 1.089 is not an element of the set of irrational num-bers, the statement is true.

_3 _1 0

15. This interval is of unlimited extent in the negative direc-tion, and the endpoint 2 is included. Interval notation is (,2].

_20

17. This interval is of unlimited extent in the positive direc-tion, and the endpoint 3.8 is not included. Interval nota-tion is (3.8,).

41. Since every whole number is an integer, the statement is true.

43. Since every rational number is a real number, the state-ment is true.

45. Since there are real numbers that are not integers, the statement is false.

47. The sentence 3 + y = y + 3 illustrates the commutative property of addition.

49. The sentence 3 1 = 3 illustrates the multiplicative 03.8identity property.

19. {x|7 < x}, or {x|x > 7}.

This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7,).

51. The sentence 5x = x5 illustrates the commutative prop-erty of multiplication.

53. The sentence 2(a+b) = (a+b)2 illustrates the commutative property of multiplication.

07

55. The sentence 6(m+n) = 6(n+m) illustrates the com-mutative property of addition.

21. The endpoints 0 and 5 are not included in the interval, so we use parentheses. Interval notation is (0,5).

57. The sentence 8 1 = 1 illustrates the multiplicative inverse

property.


255_5_5_31515__621___21__________yb1=m81zx24ab cb

59. The distance of 8.15 from 0 is 8.15, so |8.15| = 8.15.

61. The distance 295 from 0 is 295, so |295| = 295.

63. The distance of 97 from 0 is 97, so |97| = 97.

65. The distance of 0 from 0 is 0, so |0| = 0.

_ _67. The distance of 4 from 0 is 4, so _4_ = 4.

69. |14(8)| = |14+8| = |22| = 22, or |814| = |22| = 22

71. |3(9)| = |3+9| = |6| = 6, or |9(3)| = |9+3| = |6| = 6

73. |12.16.7| = |5.4| = 5.4, or

|6.712.1| = |5.4| = 5.4

Chapter R: Basic Concepts of Algebra

9. z0 z7 = z0+7 = z7, or z0 z7 = 1z7 = z7

11. 58 56 = 58+(6) = 52, or 25

13. m5 m5 = m5+5 = m0 = 1

15. y3 y7 = y3+(7) = y4, or y41

17. (x+3)4(x+3)2 = (x+3)4+(2) = (x+3)2

19. 33 38 3 = 33+8+1 = 36, or 729

21. 2x3 3x2 = 23x3+2 = 6x5

23. (3a5)(5a7) = 35a5+(7) = 15a12, or 15a12

______

75. _ 4 8 _ = _ 8 8 _ = _ 8 _ = 8 , or______

_15 3 _ _15 3_ _15 6_ _21_ 21 _ 8 4 _ _ 8 4_ _ 8 8_ _ 8 _ 8__=_++_____===

77. |70| = |7| = 7, or

|0(7)| = |0+7| = |7| = 7

25. (6x3y5)(7x2y9) = 6(7)x3+2y5+(9) =

42x1y4, or xy442

27. (2x)4(3x)3 = 24 x4 33 x3 = 1627x4+3 = 432x7

29. (2n)3(5n)2 = (2)3n3 52n2 = 825n3+2 =

200n5

79. Answers may vary. One such number is 0.124124412444....

81. Answers may vary. Since 101 = 0.0099 and 100 = 0.01, one such number is 0.00999.11

83. Since 12 +32 = 10, the hypotenuse of a right triangle with legs of lengths 1 unit and 3 units has a length of 10 units.

c2 = 12 +32 1c2 = 10c

3c =10

Exercise Set R.2

__ 1. 37 = 37 am = am , a = 011

3. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent.

x5 y4 y4 x5=

5. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent.

1 1266 t6 = m1n12 , or mn12mntt

7. 230 = 1 (For any nonzero real number, a0 = 1.)

3531. y31 = y3531 = y4

733. b12 = b712 = b19, or b19

2 23 35. x1y = x2(1)y21 = x3y3, or y3xyx

37. 4x5y8 = 4 x4(5)y38 = 8xy5, or yx32xy4 33285

39. (2x2y)4 = 24(x2)4y4 = 16x24y4 = 16x8y4

41. (2x3)5 = (2)5(x3)5 = (2)5x35 = 32x15

43. (5c1d2)2 = (5)2c1(2)d2(2) = c2d4c2d4(5)225

45. (3m4)3(2m5)4 = 33m12 24m20 = 2716m12+(20) = 432m8, or 432

_2x3y7 _3(2x3y7)323x9y21 z1 (z1)3 z347.===

8x9y21 , or 8y29z33

_10 8 7 _549.12a6b3c5= (2a4b5c2)5 = 25a20b25c10,

25or 32a20c10


Exercise Set R.23=

51. Convert 16,500,000 to scientic notation.

We want the decimal point to be positioned between the 1 and the 6, so we move it 7 places to the left. Since 16,500,000 is greater than 10, the exponent must be posi-tive.

16,500,000 = 1.65107

53. Convert 0.000000437 to scientic notation.

We want the decimal point to be positioned between the 4 and the 3, so we move it 7 places to the right. Since 0.000000437 is a number between 0 and 1, the exponent must be negative.

0.000000437 = 4.37107

55. Convert 234,600,000,000 to scientic notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 11 places to the left. Since 234,600,000,000 is greater than 10, the exponent must be positive.

234,600,000,000 = 2.3461011

57. Convert 0.00104 to scientic notation. We want the deci-mal point to be positioned between the 1 and the last 0, so we move it 3 places to the right. Since 0.00104 is a number between 0 and 1, the exponent must be negative.

0.00104 = 1.04103

59. Convert 0.00000000000000000000000000167 to scientic notation.

71.(4.2107)(3.2102)

= (4.23.2)(107 102)

= 13.44105This is not scientic notation. = (1.34410)105= 1.344106Writing scientic notation

73.(2.61018)(8.5107)

= (2.68.5)(1018 107)

= 22.11011This is not scientic notation. = (2.2110)1011= 2.211010

75.6.41076.4107 8.01068.0 106=

= 0.81013 This is not scientic notation.= (8101)1013

= 81014Writing scientic notation

77. 1.8103 7.2109

1.8103

7.2109

= 0.25106 This is not scientic notation.

= (2.5101)106

We want the decimal point to be positioned between the 1 and the 6, so we move it 27 places to the right. Since 0.00000000000000000000000000167 is a number between 0 and 1, the exponent must be negative.

0.00000000000000000000000000167 = 1.671027

61. Convert 7.6105 to decimal notation.

The exponent is positive, so the number is greater than 10. We move the decimal point 5 places to the right.

7.6105 = 760,000


The exponent is negative, so the number is between 0 and 1. We move the decimal point 7 places to the left.

1.09107 = 0.000000109



3.4961010 = 34,960,000,000



5.41108 = 0.0000000541



2.319108 = 231,900,000

= 2.5105

79. The average number of pieces of trash per mile is the total number of pieces of trash divided by the number of miles.

51.2 billion 5.121010 76 million 7.6107=

0.6737103

(6.737101)103 6.737102On average, there are about 6.737102 pieces of trash on each mile of roadway.

81. The number of people per square mile is the total number of people divided by the number of square miles.

38,0003.8104 0.75 7.5101=

0.50667105

(5.0667101)105 5.0667104There are about 5.0667104 people per square mile.

83. We multiply the number of light years by the number of miles in a light year.

4.225.881012 = 24.81361012

= (2.4813610)1012 = 2.481361013The distance from Earth to Alpha Centauri C is 2.481361013 mi.


444==n2n41+11+11+1

216

85. First nd the number of seconds in 1 hour: 1 hour = 1 hr 61 hrn 10mec = 3600 sec0 mi6 sin

The number of disintegrations produced in 1 hour is the number of disintegrations per second times the number of seconds in 1 hour.37 billion3600

= 37,000,000,0003600

= 3.71010 3.6103Writing scientic notation= (3.73.6)(1010 103)

= 13.321013Multiplying = (1.33210)1013= 1.3321014

One gram of radium produces 1.3321014 disintegrations in 1 hour.

87.= 53+832 +4(62)

= 53+832 +44Working inside parentheses = 53+89+44Evaluating 32= 15+72+16Multiplying

= 87+16Adding in order = 103from left to right

89.16442256

= 442256 Multiplying and dividing in order from left to right= 162256

= 8256


93. Since interest is compounded semiannually, n = 2. Substi-tute $3225 for P, 3.1% or 0.031 for i, 2 for n, and 4 for tin the compound interest formula. A = P_1+ i _nt

= $3225_1+ 0.031_24 Substituting

= $3225(1+0.0155)24 Dividing = $3225(1.0155)24 Adding= $3225(1.0155)8Multiplying 2 and 4

$3225(1.130939628) Evaluating the exponential expression $3647.2803Multiplying

$3647.28 Rounding to the nearest cent

95. Since interest is compounded quarterly, n = 4. Substitute $4100 for P, 2.3% or 0.023 for i, 4 for n, and 6 for t in thecompound interest formula. A = P_1+ i _nt

= $4100_1+ 0.023_46 Substituting

= $4100(1+0.00575)46 Dividing = $4100(1.00575)46 Adding= $4100(1.00575)24 Multiplying 4 and 6

$4100(1.147521919) Evaluating the exponential expression $4704.839868Multiplying

$4704.84 Rounding to the nearest cent

= 2048

91.4(86)2 43+28 31 +190

= 422 43+28Calculating in the numerator and in3+1

the denominator = 4443+28

= 1612+16

4+16

4 20

4 = 5

97. Substitute $250 for P, 0.05 for r and 27 for t and perform the resulting computation. _r _12t

S = P 12r

12 _1+ 0.05_1227 1 = $2500.0512

12 $170,797.30

99. Substitute $120,000 for S, 0.03 for r, and 18 for t and solve

for P. _r _12t

S = P 12r

12 _0.03_1218

$120,000 = P 10.03 122

$120,000 = P (1.0025)25 1 $120,000 P(285.94035)0.00

$419.67 P


Exercise Set R.35__

101. (xt x3t)2 = (x4t)2 = x4t2 = x8t

103. (ta+x txa)4 = (t2x)4 = t2x4 = t8x

_a b 3 (3xy )=27xy

2_3a 3b 2 105. (3xayb)2 9x2ay2b

= 3xayb 2

= 9x2ay2b

19.(y 3)(y +5)

= y2 +5y 3y 15 = y2 +2y 15

21.(x+6)(x+3)

= x2 +3x+6x+18

= x2 +9x+18

Using FOIL Collecting like terms

Using FOIL

Collecting like terms

Exercise Set R.323.

(2a+3)(a+5)

= 2a2 +10a+3a+15 Using FOIL

1. 7x3 4x2 +8x+5 = 7x3 +(4x2)+8x+5= 2a2 +13a+15Collecting like terms

Terms: 7x3, 4x2, 8x, 5

The degree of the term of highest degree, 7x3, is 3. Thus, the degree of the polynomial is 3.

3. 3a4b7a3b3 +5ab2 = 3a4b+(7a3b3)+5ab+(2) Terms: 3a4b, 7a3b3, 5ab, 2The degrees of the terms are 5, 6, 2, and, 0, respectively, so the degree of the polynomial is 6.

25.(2x+3y)(2x+y)

= 4x2 +2xy +6xy +3y2Using FOIL = 4x2 +8xy +3y2

27.(x+3)2

= x2 +2x3+32

[(A+B)2 = A2 +2AB +B2]

5.(3ab2 4a2b2ab+6)+

(ab2 5a2b+8ab+4)

= (31)ab2 +(45)a2b+(2+8)ab+(6+4) = 2ab2 9a2b+6ab+10

7.(2x+3y +z 7)+(4x2y z +8)+ (3x+y 2z 4)= (2+43)x+(32+1)y +(112)z+ (7+84)= 3x+2y 2z 3

9. (3x2 2xx3 +2)(5x2 8xx3 +4) = (3x2 2xx3 +2)+(5x2 +8x+x3 4)= (35)x2 +(2+8)x+(1+1)x3 +(24) = 2x2 +6x2

11. (x4 3x2 +4x)(3x3 +x2 5x+3) = (x4 3x2 +4x)+(3x3 x2 +5x3) = x4 3x3 +(31)x2 +(4+5)x3= x4 3x3 4x2 +9x3

13. (3a2)(7a4) = [3(7)](a2 a4) = 21a6

15. (6xy3)(9x4y2) = (69)(xx4)(y3 y2)

= 54x5y5

= x2 +6x+9

29.(y 5)2

= y2 2y 5+52

[(AB)2 = A2 2AB +B2] = y2 10y +25

31.(5x3)2

= (5x)2 25x3+32

[(AB)2 = A2 2AB +B2] = 25x2 30x+9

33.(2x+3y)2

= (2x)2 +2(2x)(3y)+(3y)2

[(A+B)2 = A2+2AB+B2] = 4x2 +12xy +9y2

35.(2x2 3y)2

= (2x2)2 2(2x2)(3y)+(3y)2

[(AB)2 = A2 2AB +B2] = 4x4 12x2y +9y2

37.(n+6)(n6)

= n2 62[(A+B)(AB) = A2 B2] = n2 36

39.(3y +4)(3y 4)

17.(ab)(2a3 ab+3b2)

= (ab)(2a3)+(ab)(ab)+(ab)(3b2)

= (3y)2 42[(A+B)(AB) = A2 B2]

= 9y2 16

Using the distributive property = 2a4 2a3ba2b+ab2 +3ab2 3b3Using the distributive property three more times

= 2a4 2a3ba2b+4ab2 3b3 Collecting like

terms

41.(3x2y)(3x+2y)

= (3x)2 (2y)2[(AB)(A+B) = A2B2]

= 9x2 4y2


6Chapter R: Basic Concepts of Algebra

43.(2x+3y +4)(2x+3y 4)

= [(2x+3y)+4][(2x+3y)4] = (2x+3y)2 42= 4x2 +12xy +9y2 16

45.(x+1)(x1)(x2 +1)

= (x2 1)(x2 +1)

15.x3 x2 5x+5

= x2(x1)5(x1) = (x1)(x2 5)

17.a3 3a2 2a+6

= a2(a3)2(a3)

= (a3)(a2 2)

= x4 1

47. (an +bn)(an bn) = (an)2 (bn)2 = a2n b2n

49. (an +bn)2 = (an)2 +2an bn +(bn)2 = a2n +2anbn +b2n

51.(x1)(x2 +x+1)(x3 +1)

= [(x1)x2 +(x1)x+(x1)1](x3 +1) = (x3 x2 +x2 x+x1)(x3 +1)= (x3 1)(x3 +1) = (x3)2 12= x6 1

53. (xab)a+b = x(ab)(a+b) = xa2b2

55.(a+b+c)2

19. w2 7w +10

We look for two numbers with a product of 10 and a sum of 7. By trial, we determine that they are 5 and 2.

w2 7w +10 = (w 5)(w 2)

21. x2 +6x+5


x2 +6x+5 = (x+1)(x+5)

23. t2 +8t+15


t2 +8t+15 = (t+3)(t+5)

25. x2 6xy 27y2


x2 6xy 27y2 = (x+3y)(x9y)

= (a+b+c)(a+b+c)

= (a+b+c)(a)+(a+b+c)(b)+(a+b+c)(c) = a2 +ab+ac+ab+b2 +bc+ac+bc+c2= a2 +b2 +c2 +2ab+2ac+2bc

Exercise Set R.4

1. 3x+18 = 3x+36 = 3(x+6)

3. 2z3 8z2 = 2z2 z 2z2 4 = 2z2(z 4)

5. 4a2 12a+16 = 4a2 43a+44 = 4(a2 3a+4)

7. a(b2)+c(b2) = (b2)(a+c)

27. 2n2 20n48 = 2(n2 10n24)

Now factor n2 10n24. We look for two numbers with a product of 24 and a sum of 10. By trial, we determine that they are 2 and 12. Then n2 10n24 =(n + 2)(n 12). We must include the common factor, 2, to have a factorization of the original trinomial.

2n2 20n48 = 2(n+2)(n12)

29. y2 4y 21


y2 4y 21 = (y +3)(y 7)

31. y4 9y3 +14y2 = y2(y2 9y +14)

9.3x3 x2 +18x6

= x2(3x1)+6(3x1) = (3x1)(x2 +6)

11.y3 y2 +2y 2

= y2(y 1)+2(y 1) = (y 1)(y2 +2)

13. 24x3 36x2 +72x108 = 12(2x3 3x2 +6x9)= 12[x2(2x3)+3(2x3)]

= 12(2x3)(x2 +3)

Now factor y2 9y + 14. Look for two numbers with a product of 14 and a sum of 9. The numbers are 2 and 7. Then y2 9y +14 = (y 2)(y 7). We must include the common factor, y2, in order to have a factorization of the original trinomial.

y4 9y3 +14y2 = y2(y 2)(y 7)

33. 2x3 2x2y 24xy2 = 2x(x2 xy 12y2)

Now factor x2 xy 12y2. Look for two numbers with a product of 12 and a sum of 1. The numbers are 4 and 3. Then x2xy12y2 = (x4y)(x+3y). We must include the common factor, 2x, in order to have a factorization of the original trinomial.

2x3 2x2y 24xy2 = 2x(x4y)(x+3y)


Exercise Set R.4

35. 2n2 +9n56

We use the FOIL method.

1. There is no common factor other than 1 or 1.

2. The factorization must be of the form (2n+)(n+).

3. Factor the constant term, 56. The possibilities are 156, 1(56), 228, 2(28), 416, 4(16), 7 8, and 7(8). The factors can be written in the opposite order as well: 56(1), 561, 28(2), 282, 16(4), 164, 8(7), and 87.

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term, 9n. By trial, we determine that the factorization is (2n7)(n+8).

37. 12x2 +11x+2

We use the grouping method.


2. Multiply the leading coecient and the constant: 122 = 24.

3. Try to factor 24 so that the sum of the factors is the coecient of the middle term, 11. The factors we want are 3 and 8.

4. Split the middle term using the numbers found in step (3):

11x = 3x+8x 5. Factor by grouping.12x2 +11x+2 = 12x2 +3x+8x+2

= 3x(4x+1)+2(4x+1) = (4x+1)(3x+2)

39. 4x2 +15x+9



2. The factorization must be of the form (4x+)(x+) or (2x+)(2x+).

3. Factor the constant term, 9. The possibilities are 19, 1(9), 33, and 3(3). The rst two pairs of factors can be written in the opposite order as well: 91, 9(1).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term, 15x. By trial, we determine that the factorization is (4x+3)(x+3).

41. 2y2 +y 6



2. Multiply the leading coecient and the constant: 2(6) = 12.


7


y = 4y 3y 5. Factor by grouping.2y2 +y 6 = 2y2 +4y 3y 6

= 2y(y +2)3(y +2) = (y +2)(2y 3)

43. 6a2 29ab+28b2



2. The factorization must be of the form (6x+)(x+) or (3x+)(2x+).

3. Factor the coecient of the last term, 28. The pos-sibilities are 1 28, 1(28), 2 14, 2(14), 4 7, and 4(7). The factors can be written in the op-posite order as well: 281, 28(1), 142, 14(2), 74, and 7(4).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term, 29. Observe that the second term of each bino-mial factor will contain a factor of b. By trial, we determine that the factorization is (3a4b)(2a7b).

45. 12a2 4a16

We will use the grouping method. 1. Factor out the common factor, 4.12a2 4a16 = 4(3a2 a4)

2. Now consider 3a2 a 4. Multiply the leading coecient and the constant: 3(4) = 12.



a = 4a+3a 5. Factor by grouping.3a2 a4 = 3a2 4a+3a4

= a(3a4)+(3a4) = (3a4)(a+1)We must include the common factor to get a factor-ization of the original trinomial.

12a2 4a16 = 4(3a4)(a+1)

47. z2 81 = z2 92 = (z +9)(z 9)

49. 16x2 9 = (4x)2 32 = (4x+3)(4x3)

51. 6x2 6y2 = 6(x2 y2) = 6(x+y)(xy)

53. 4xy4 4xz2 = 4x(y4 z2) = 4x[(y2)2 z2]= 4x(y2 +z)(y2 z)


8Chapter R: Basic Concepts of Algebra

55. 7pq4 7py4 = 7p(q4 y4)

= 7p[(q2)2 (y2)2]

= 7p(q2 +y2)(q2 y2)

= 7p(q2 +y2)(q +y)(q y)

57. x2 +12x+36 = x2 +2x6+62

= (x+6)2

87. x2 +9x+20

We look for two numbers with a product of 20 and a sum of 9. They are 4 and 5.

x2 +9x+20 = (x+4)(x+5)

89. y2 6y +5


59. 9z2 12z +4 = (3z)2 23z 2+22 = (3z 2)2y2 6y +5 = (y 5)(y 1)

61. 18x+16x2 = 12 214x+(4x)2 = (14x)2

63. a3 +24a2 +144a = a(a2 +24a+144)= a(a2 +2a12+122) = a(a+12)2

65. 4p2 8pq +4q2 = 4(p2 2pq +q2) = 4(pq)2

67. x3 +64 = x3 +43

= (x+4)(x2 4x+16)

69. m3 216 = m3 63

= (m6)(m2 +6m+36)

71. 8t3 +8 = 8(t3 +1) = 8(t3 +13)= 8(t+1)(t2 t+1)

73. 3a5 24a2 = 3a2(a3 8) = 3a2(a3 23)= 3a2(a2)(a2 +2a+4)

75. t6 +1 = (t2)3 +13

= (t2 +1)(t4 t2 +1)

77. 18a2b15ab2 = 3ab6a3ab5b

= 3ab(6a5b)

91. 2a2 +9a+4



2. The factorization must be of the form (2a+)(a+).

3. Factor the constant term, 4. The possibilities are 14, 1(4), and 22. The rst two pairs of factors can be written in the opposite order as well: 4 1, 4(1).

4. Find a pair of factors for which the sum of the out-side and the inside products is the middle term, 9a. By trial, we determine that the factorization is (2a+1)(a+4).

93. 6x2 +7x3



2. Multiply the leading coecient and the constant: 6(3) = 18.



7x = 9x2x 5. Factor by grouping.6x2 +7x3 = 6x2 +9x2x3

= 3x(2x+3)(2x+3)

= (2x+3)(3x1)

79. x3 4x2 +5x20 = x2(x4)+5(x4) = (x4)(x2 +5)

81. 8x2 32 = 8(x2 4)

= 8(x+2)(x2)

95. y2 18y +81 = y2 2y 9+92 = (y 9)2

97. 9z2 24z +16 = (3z)2 23z 4+42

= (3z 4)2

83. 4y2 5

There are no common factors. We might try to factor this polynomial as a dierence of squares, but there is no integer which yields 5 when squared. Thus, the polynomial is prime.

99. x2y2 14xy +49 = (xy)2 2xy 7+72 = (xy 7)2

101. 4ax2 +20ax56a = 4a(x2 +5x14)

= 4a(x+7)(x2)

85. m2 9n2 = m2 (3n)2

= (m+3n)(m3n)

103. 3z3 24 = 3(z3 8) = 3(z3 23)= 3(z 2)(z2 +2z +4)


Exercise Set R.5942___1_

105. 16a7b+54ab7 = 2ab(8a6 +27b6)= 2ab[(2a2)3 +(3b2)3]

= 2ab(2a2 +3b2)(4a4 6a2b2 +9b4)

107.y3 3y2 4y +12

129.(x+h)3 x3

= [(x+h)x][(x+h)2 +x(x+h)+x2]

= (x+hx)(x2 +2xh+h2 +x2 +xh+x2) = h(3x2 +3xh+h2)

131.(y 4)2 +5(y 4)24

= y2(y 3)4(y 3)

= (y 3)(y2 4)

= u2 +5u24Substituting u for y 4

= (u+8)(u3)

= (y 3)(y +2)(y 2)

109.x3 x2 +x1

= (y 4+8)(y 43)

= (y +4)(y 7)

Substituting y 4 for u

= x2(x1)+(x1) = (x1)(x2 +1)

111. 5m4 20 = 5(m4 4)

= 5(m2 +2)(m2 2)

113. 2x3 +6x2 8x24 = 2(x3 +3x2 4x12)= 2[x2(x+3)4(x+3)] = 2(x+3)(x2 4)= 2(x+3)(x+2)(x2)

115. 4c2 4cdd2 = (2c)2 22cdd2

= (2cd)2

133. x2n +5xn 24 = (xn)2 +5xn 24 = (xn +8)(xn 3)

135. x2 +ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)

137. 25y2m (x2n 2xn +1) = (5ym)2 (xn 1)2= [5ym +(xn 1)][5ym (xn 1)] = (5ym +xn 1)(5ym xn +1)

139.(y 1)4 (y 1)2

= (y 1)2[(y 1)2 1]

= (y 1)2[y2 2y +11]

117. m6 +8m3 20 = (m3)2 +8m3 20


m6 +8m3 20 = (m3 +10)(m3 2)

= (y 1)2(y2 2y) = y(y 1)2(y 2)

Exercise Set R.5

119. p64p4 = p(164p3)

= p[13 (4p)3]

1. x5 = 7

x = 12 Adding 5

= p(14p)(1+4p+16p2)The solution is 12.

121. y4 84+5y2 = y4 +5y2 84= u2 +5u84Substituting u for y2

3. 3x+4 = 8 3x = 12x = 4

Subtracting 4

Dividing by 3

= (u+12)(u7)The solution is 4. = (y2 +12)(y2 7) Substituting y2 for u 5. 5y 12 = 38228

123. y2 49 + 7y = y2 + 7y 49____= y + 7y 7_ _2 125. x2 +3x+ 4 = x2 +2x 2 + 2933

2 = x+ 23

_ _2 127. x2 x+ 4 = x2 2x 2 + 2111

2

= x 2

5y = 15 Adding 12

y = 3Dividing by 5 The solution is 3.

7. 6x15 = 45

6x = 60 Adding 15

x = 10 Dividing by 6 The solution is 10.

9. 5x10 = 45

5x = 55 Adding 10

x = 11 Dividing by 5

The solution is 11.


10Chapter R: Basic Concepts of Algebra5511277

11. 9t+4 = 5 9t = 9t = 1

Subtracting 4

Dividing by 9

27.24 = 5(2t+5) 24 = 10t+251 = 10tSubtracting 25

The solution is 1.

13. 8x+48 = 3x12

5x+48 = 12Subtracting 3x

10 = tDividing by 10

The solution is 10.

5x = 60Subtracting 48

x = 12Dividing by 5

29. 5y 4(2y 10) = 25

5y 8y +40 = 25

The solution is 12.3y +40 = 25Collecting like terms

15. 7y 1 = 235y 12y 1 = 2312y = 24

Adding 5y

Adding 1

3y = 15 y = 5The solution is 5.

Subtracting 40

Dividing by 3

y = 2Dividing by 12

The solution is 2.

31. 7(3x+6) = 11(x+2)

21x+42 = 11x2

17.3x4 = 5+12x

9x4 = 5Subtracting 12x 9x = 9Adding 4x = 1Dividing by 9 The solution is 1.

19. 54a = a13

55a = 13Subtracting a

5a = 18Subtracting 5 a = 18Dividing by 5

The solution is 18.

21. 3m7 = 13+m

2m7 = 13Subtracting m 2m = 6Adding 7m = 3Dividing by 2

The solution is 3.

21x+42 = 9xCollecting like terms 22x+42 = 9Adding x22x = 33Subtracting 42

x = 2Dividing by 22 The solution is 3.3

33. 4(3y 1)6 = 5(y +2) 12y 46 = 5y +1012y 10 = 5y +10Collecting like terms 7y 10 = 10Subtracting 5y7y = 20Adding 10

y = 20Dividing by 7

The solution is 20.

35.x2 +3x28 = 0

(x+7)(x4) = 0 Factoring

23. 113x = 5x+3 118x = 38x = 8

x = 1

Subtracting 5x

Subtracting 11

x+7 = 0 or x4 = 0 Principle of zero products x = 7 orx = 4The solutions are 7 and 4.

37.x2 +5x = 0

The solution is 1.x(x+5) = 0 Factoring

25. 2(x+7) = 5x+14 2x+14 = 5x+143x+14 = 14Subtracting 5x

x = 0 or x+5 = 0Principle of zero products x = 0 orx = 5The solutions are 0 and 5.

3x = 0Subtracting 14

x = 0

39.y2 +6y +9 = 0

(y +3)(y +3) = 0

The solution is 0.y +3 = 0 or y +3 = 0

y = 3 ory = 3 The solution is 3.


Exercise Set R.511 22323247 hhh

41.x2 +100 = 20x

x2 20x+100 = 0Subtracting 20x

53.x2 36 = 0

(x+6)(x6) = 0

(x10)(x10) = 0 x+6 = 0 or x6 = 0 x10 = 0 or x10 = 0 x = 6 or x = 6x = 10 orx = 10The solutions are 6 and 6. The solution is 10. 55. z2 = 144

43.x2 4x32 = 0

(x8)(x+4) = 0

z2 144 = 0

(z +12)(z 12) = 0

x8 = 0 or x+4 = 0z +12 = 0or z 12 = 0 x = 8 orx = 4 z = 12 orz = 12The solutions are 8 and 4.The solutions are 12 and 12.

45.3y2 +8y +4 = 0

(3y +2)(y +2) = 0

57. 2x2 20 = 0

2x2 = 20

3y +2 = 0or y +2 = 0 3y = 2 ory = 2

y = 3 ory = 2

The solutions are 3 and 2.

x2 = 10x = 10 or x = 10 Principle of square roots

The solutions are10 and 10, or 10.

59. 6z2 18 = 0

6z2 = 18

47. 12z2 +z = 6 12z2 +z 6 = 0(4z +3)(3z 2) = 0

z2 = 3z = 3 or z = 3


4z +3 = 0 or 3z 2 = 0 4z = 3 or 3z = 2

z = 4 orz = 3


49.12a2 28 = 5a

12a2 5a28 = 0

61.A = 2bh1

2A = bhMultiplying by 2 on both sides 2A = bDividing by h on both sides

63.P = 2l +2w

P 2l = 2wSubtracting 2l on both sides

(3a+4)(4a7) = 0P 2l = wDividing by 2 on both sides 3a+4 = 0or 4a7 = 02

3a = 4 or4a = 7

a = 3 ora = 447


51. 14 = x(x5) 14 = x2 5x0 = x2 5x14 0 = (x7)(x+2)

x7 = 0 or x+2 = 0 x = 7 orx = 2The solutions are 7 and 2.

65.A = 2h(b1 +b2)1

h = b1 +b2 Multiplying by h on both sides 2A b1 = b2, or2A2

2Ab1h = b2

67.V = 3r34

3V = 4r3 Multiplying by 3 on both sides

4r3 = Dividing by 4r3 on both sides3V


12Chapter R: Basic Concepts of Algebrax3az23236125.==9

69.F = 5C +32

F 32 = 5CSubtracting 32 on both sides9

9(F 32) = CMultiplying by 9 on both sides55

83. x{3x[2x(5x(7x1))]} = x+7 x{3x[2x(5x7x+1)]} = x+7 x{3x[2x(2x+1)]} = x+7 x{3x[2x+2x1]} = x+7x{3x[4x1]} = x+7

71. Ax+By = C

Ax = C By Subtracting By on both sides A = C By Dividing by x on both sides

73. 2w +2h+l = p

2h = p2w lSubtracting 2w and l

x{3x4x+1} = x+7 x{x+1} = x+7 x+x1 = x+72x1 = x+7 x1 = 7x = 8

The solution is 8.

h = p2w lDividing by 285.2

(5x2 +6x)(12x2 5x2) = 0

x(5x+6)(4x+1)(3x2) = 0

75. 2x3y = 6x = 0 or 5x+6 = 0or 4x+1 = 0or 3x2 = 0

3y = 62xSubtracting 2xx = 0 or5x = 6 or4x = 1 or3x = 2

y = 62x, orDividing by 3

2x6

3

x = 0 orx = 5 orx = 4 orx = 3612

The solutions are 0, 5, 4, and 3.

77.a = b+bcd

a = b(1+cd)

1+cd = b

Factoring

Dividing by 1+cd

87. 3x3 +6x2 27x54 = 0 3(x3 +2x2 9x18) = 03[x2(x+2)9(x+2)] = 0 Factoring by grouping

3(x+2)(x2 9) = 0

79.z = xy xy2

z = x(y y2) Factoring

y y2 = xDividing by y y2

81. 3[53(4t)]2 = 5[3(5t4)+8]26

3[512+3t]2 = 5[15t12+8]26

3(x+2)(x+3)(x3) = 0

x+2 = 0 or x+3 = 0 or x3 = 0 x = 2 or x = 3 or x = 3The solutions are 2, 3, and 3.

Exercise Set R.6

3[7+3t]2 = 5[15t4]26 21+9t2 = 75t2026 9t23 = 75t4666t23 = 46 66t = 23

t = 66

The solution is 66.

1. Since 5 is dened for all real numbers, the domain is {x|x is a real number}.3

3x3 x(x1)3.

The denominator is 0 when the factor x = 0 and also when x 1 = 0, or x = 1.The domain is {x|x is a real number and x = 0 and x = 1}.

x+5x+5

x2 +4x5(x+5)(x1)

We see that x+5 = 0 when x = 5 and x1 = 0 when x = 1. Thus, the domain is{x|x is a real number and x = 5 and x = 1}.

7. We rst factor the denominator completely. 7x2 28x+287x2 28x+28(x24)(x2+3x10)(x+2)(x2)(x+5)(x2)

We see that x + 2 = 0 when x = 2, x 2 = 0 when x = 2, and x + 5 = 0 when x = 5. Thus, the domain is {x|x is a real number and x=2 and x=2 and x=5}.


Exercise Set R.613(x===((==((=1=x+y +z+==2===+=+=

x2 4 (x+2)(x2) x+2 x2 4x+4 (x2)(x2) x29.==

x3 6x2 +9x x(x2 6x+9) x3 3x2 x2(x3)11.=

x/(x3) x3) x/x(x3)=

= x3

6y2 +12y 48 6(y2 +2y 8) 3y2 9y +6 3(y2 3y +2)13.=

2/3(y +4)(y2)

3/(y 1)(y2)

27. x2 y2x2 +xy +y2 x3 y3 x2 +2xy +y2

(x+y)(xy)(x2 +xy +y2) (xy)(x2 +xy +y2)(x+y)(x+y)=

1 (x+y)(xy)(x2 +xy +y2) x+y (x+y)(xy)(x2 +xy +y2)=

= x+y 1Removing a factor of 1 1x+y

29. (xy)2 z2 xy +z (x+y)2 z2 x+y z

15.

17.

19.

21.

23.

25.

2(y +4) y 1

4x1(x4) x2 +4x32 (x4)(x+8)=

= x+8, or x+811

r s r2 s2 (r s)(r2 s2) r +s (r s)2 (r +s)(r s)2=

(rs) rs)(r+s)1

(r+s)(rs)(rs) = 1

x2 +2x35 9x3 4x 3x3 2x27x+49

(x+7 (x5)(x)(3x+2)(3x2) x/x(3x2)(7)(x+7)=)

(x5)(3x+2) 7x=

4x2 +9x+2x2 1 x2 +x2 3x2 +x2

(4x+1)(x+2) x+1)(x1) (x+2)(x1)(3x2)(x+1) 4x+1=

3x2

m2 n2mn r +s r +s

m2 n2r +s r +s mn=

(m+n)(mn) r+s)

(r+s)(mn) = m+n

3x+12 (x+4)2 2x8 (x4)2

3x+12 (x4)2 2x8 (x+4)2=

3(x+4) x4)(x4) 2(x4)(x+4)(x+4) 3(x4)=

2(x+4)

31.

33.

35.

37.

(xy)2 z2 x+y z (x+y)2 z2 xy +z=

(xy +z)(xy z)(x+y z) (x+y +z)(x+y z)(xy +z) (xy +z)(x+y z) xy z (xy +z)(x+y z) x+y +z==

= 1 xy zRemoving a factor of 1

xy z x+y +z=

737+3

5x5x 5x 10

5x

/5 2 /5 x=

= x

4 3a 4+3a 3a+4 3a+4 3a+4+=

= 1(4+3a = 3a+4)

4z 8z, LCD is 8z 52353

4z 28z 10 3

8z8z

= 8z7

x+2 + x2 432

= x+2 + (x+2)(x2), LCD is (x+2)(x2) 3x2232

x+2 x2(x+2)(x2) 3x6 2(x+2)(x2)(x+2)(x2) 3x4(x+2)(x2)


14Chapter R: Basic Concepts of Algebra====+3x5y=+=+==+=+===+=+,=+=+=====++=++===+===, or

39.

41.

43.

45.

y2 y 20 y +447.y2

= (y +4)(y 5) y +4, LCD is (y +4)(y 5) y2y 5y2

(y +4)(y 5)y +4 y 5 y 2y 10(y +4)(y 5)(y +4)(y 5) y (2y 10)(y +4)(y 5) y 2y +10 (y +4)(y 5) y +10=

(y +4)(y 5)

3x5y

x+yx2 y249.

= x+y + (x+y)(xy), LCD is (x+y)(xy) 3xyx5yx+y xy (x+y)(xy) 3x3y x5y(x+y)(xy)(x+y)(xy) 4x8y(x+y)(xy)

y 1 + 1yy2

y12

y 11 1y51. y 2y 1y 1 y 2y 1

2x3y 3y 2x x1yxy

2x3y1 3y 2x x y2x3y2x3y

= 2x3y[x(y) = x+y]x+y

9x+27

3x2 2x83x2 +x4 9x+2 7(3x+4)(x2)(3x+4)(x1)

LCD is (3x+4)(x2)(x1) 9x+2x17x2(3x+4)(x2) x1(3x+4)(x1) x2 9x2 7x2 7x14(3x+4)(x2)(x1)(3x+4)(x1)(x2) 9x2 16(3x+4)(x2)(x1) (3x+4)(3x4) (3x+4)(x2)(x1)

3x4 (x2)(x1)

5a ab 4b ab a2 b2 a+b++

5a ab 4b ab (a+b)(ab) a+b++,

LCD is (a+b)(ab) 5aa+bab4babab a+b(a+b)(ab)a+b ab 5a2 +5ab ab 4ab4b2(a+b)(ab)(a+b)(ab)(a+b)(ab) 5a2 +10ab4b2(a+b)(ab)

7 x+8 3x2 x+2 4x2 44x+x2+

7x+83x2 x+2 (2+x)(2x) (2x)2+,

LCD is (2+x)(2x)2 7(2x)2x+82x2+x (2x)2 (2+x)(2x) 2x 3x2 2+x(2x)2 2+x

2828x+7x2(166xx2)+3x2+4x4 (2+x)(2x)22828x+7x2 16+6x+x2 +3x2 +4x4 (2+x)(2x)211x2 18x+811x2 18x+8

(2+x)(2x)2(x+2)(x2)2


Exercise Set R.6151xx+2=++=++====_xy=(8c8xyxy==

53.

55.

57.

2x+1 + 2x + x2 x2

1xx2 +2 x+1 2x (x+1)(x2)=++

11xx2 +2 x+11 2x (x+1)(x2)=++

1xx2 +2 x+1 x2 (x+1)(x2)=++,

LCD is (x+1)(x2) 1x2xx+1x2 +2x+1 x2x2 x+1(x+1)(x2) x2 x2 x x2 +2(x+1)(x2)(x+1)(x2)(x+1)(x2) x2x2 x+x2 +2(x+1)(x2)

= (x+1)(x2) = 00

abb abab a2 b2 b a2 b2=

ababab

b(a+b)(ab)

ab (ab)

/b (a+b)(ab) aa+b

xyxyyx = yx xy, LCM is xy y + xy + x _1111xy

= _y x_(xy) y + x (xy)11

x2 y2 x+y=

(x+y) xy)

(x+y)1 = xy

59.

61.

63.

2c+ c2 = c c2 + c2 1+ c 1 c + c2c2

c3 +8

c2 c+2=

c

c3 +8c c2 c+2=

(c+2 (c2 2c+4)/c /cc(c+2)=)

c2 2c+4 c=

x2 +xy +y2x2 +xy +y2 x2y2 x2 xy2 y=

yxyxxy

x2 +xy +y2 x3 y3 xy=

= (x2 +xy +y2) x3 y3

(x2 +xy +y2)(xy) (xy)(x2 +xy +y2) x2 +xy +y2xy x2 +xy +y2 xy==

= 1 xy xy xy=

aa1 a a a a a a+a1 a+ a a a + a=1=a11a1

a2 1

= a2 +1 aa

a2 1a

aa2 +1

a2 1

a2 +1


16++++++=++++====+11===+ ,5x+x=

65.

67.

69.

121x+32x3

x3 x+3 x3 x+3 x+3 x3 3 4 3 x+2 4 x1=

x1x+2x1 x+2x+2 x1

x+3+2(x3) (x3)(x+3) 3(x+2)4(x1) (x1)(x+2)=

x+3+2x6

(x3)(x+3) 3x+64x+4=

(x1)(x+2)

3x3

(x3)(x+3) x+10=

(x1)(x+2)

3x3(x1)(x+2) (x3)(x+3) x+10=

(3x3)(x1)(x+2) (x3)(x+3)(x+10) 3(x1)2(x+2) (x3)(x+3)(x+10)=, or

a1+aaa1+a 1a

1a a 1a a a 1a 1a a 1a 1+a a a=

a1+a a1+a1+a a a2 +(1a2)a(1a) (1a2)+a2 a(1+a)=

1a/(1+a) a/(1a) 1 1+a=

1a

121121a2 ab b2 a2 ab b2 a2b2 a2 b2 a2 b2 a2b21111=,

LCM is a2b2 b2 +2ab+a2

b2 a2 (b+a (b+a) (b+a)(ba) b+a=)

ba

73.

75.

77.

79.


(x+h)3 x3x3 +3x2h+3xh2 +h3 x3 h h=

3x2h+3xh2 +h3 h=

h/(3x2 +3xh+h2)

h/1

= 3x2 +3xh+h2

x+15 (x+1)+(x1) 5

x1 x1 (x+1)(x1)x+1=

x1x1 2x x1 5

x12 /2x(x1)5

12/(x1) = x5

n(n+1)(n+2) (n+1)(n+2) 23 2+

n(n+1)(n+2)(n+1)(n+2) 3

2323

LCD is 23

n(n+1)(n+2)+3(n+1)(n+2) 23=

= (n+1)(n+2)(n+3)Factoring the num-erator by grouping23

x2 9 5x2 15x+45 x2 +x x3 +27 x2 2x3 4+2x+

(x+3)(x3)(5)(x2 3x+9)x2 +x (x+3)(x2 3x+9)(x3)(x+1)4+2x (x+3)(x3)(x2 3x+9)5 x2 +x (x+3)(x3)(x2 3x+9) x+1 4+2x=+=+

2= 1 x+1 + 4+2x

5x2 +x x+1 2(2+x)=+

52(2+x)+(x2 +x)(x+1) 2(x+1)(2+x)=

20+10x+x3 +2x2 +x 2(x+1)(2+x)=

x3 +2x2 +11x+20

2(x+1)(2+x)

71.(x+h)2 x2x2 +2xh+h2 x2 h h

2xh+h2 h=

h/(2x+h) h/1=

= 2x+h

Exercise Set R.7

1. _(21)2 = |21| = 21

3. _9y2 = _(3y)2 = |3y| = 3|y|

5. _(a 2)2 = |a2|


Exercise Set R.717 ____39.===41.==43.==

7. 3 27x3 = 3 (3x)3 = 3x_

9. 4 81x8 = _(3x2)4 = |3x2| = 3x24

11. 5 32 = 5 25 = 2

13. 180 = 365 = 365 = 65 15. 72 = 362 = 362 = 62 17. 3 54 = 272 = 27 2 = 32 3333

19. 128c2d4 = 64c2d4 2 = |8cd2|2 = 82|c|d2 21.4 48x6y4 = 4 16x4y4 3x2 = |2xy| 4 3x2 = __

2|x||y| 4 3x2

23. x2 4x+4 = _(x2)2 = |x2|

25. 1535 = 1535 = 3557 = 52 37 =

52 37 = 5 21

27. 810 = 810 = 24 2 5 = 22 4 5 =

22 5 = 4 5

29. _2x3y12xy = _24x4y2 = _4x4y2 6 = 2x2y6 31. 3 3x2y36x = _108x3y = _27x3 4y = 3x 3 4y 33.3 2(x+ 4) 3 4(x+4)4 = 3 8(x+ 4)5 _333 _ __

= 3 8(x+4)3 (x+4)2 = 2(x+4) 3 (x+4)2_

_16 242 3 82 3 35.28 = 2 = 28m n8mnmn

37. 40xy = _40xy = _5y8x8x

45. 52+332 = 52+3162 = 5 2+34 2 = 5 2+12 2= (5+12) 2

= 17 2

47. 620445+80 = 645495+165

= 62 543 5+4 5 = 12 512 5+4 5

= (1212+4) 5

= 4 5 49.82x2 620x58x2

= 8x 26 45x5 4x2 2 = 8x 262 5x52x 2 = 8x 212 5x10x 2

= 2x 212 5x

51. _8+25__825_ = _8_2 _25_2

= 845 = 820= 12

53.(23+5)(335)

= 2 3 32 33 5+ 5 3 53 5 = 236 15+ 1535

= 66 15+ 1515

= 95 15

55. (25)2 = (2)2 225+52

= 210 2+25

= 2710 2

3 3x23 3x23 1 1__

3 24x524x58x32x_ _3 64a43 64a3 a

27b327b3

3 64a3 3 a 3 27b3=

4a 3 a

__3b7x37x2 x

36y636y6

x2 7x 36y6=_

x 7x

6y3

57. (56)2 = (5)2 256+(6)2 = 52 30+6= 112 30

59. We use the Pythagorean theorem. We have a = 47 and b = 25.c2 = a2 +b2

c2 = 472 +252 c2 = 2209+625 c2 = 2834c 53.2

The distance across the pond is about 53.2 yd.


18 a448267.===69.===== 223+1= 75.= 77.===79.34_ _===3=81.===83.=9(5)=85.== = a5a4aaa

61. a) h2 +_2_2 = a2Pythagorean theorema

2h2 + 4 = a2

h2 = 3a2

_h =3a2

h = 23a

b) Using the result of part (a) we have

A = 2 base height1

A = 2a 23 _2 + 2 = a_1aaa

A = 4 3 a2


73.236 = 236 23+6 112223+ 6

2 3+ 62 6 12 436=

2 3+ 62 62 3 126=

= 6 6, or 6666m+n

m nm nm+ n

6( m+ n) ( m)2 ( n)2=

6 m+6 n mn=

5050 210010

3323 23 2

63.x

xx xx2 +x2 = (82)2 Pythagorean theorem

2x2 = 128

_23 2 3 882 5 5 4 20 3 20 3 201111 1112111

33113333 95 95 9+5

3 33 3 9+ 5

22= 27+359315

x2 = 64 x = 8_3_3 _212121 7 7 7 49 49 7 65.7====

3 7753535 3333

3 253 253 53 1255 3 16 _16 _48 3 48 _333

993273 27

3 862 3 6

33

71.31 = 31 3+1

2( 3+1) 31=

= 2( 3+1) 2/(3+1)2

/21

=3+1

815

27+3 59 3 15

= 27+359315 a+ba+b ab 76

3a3aa b

( a)2 ( b)2

3a( a b) ab

3a a3a b

87. y5/6 = 6 y5_

89. 163/4 = (161/4)3 = _ 4 16_3 = 23 = 8

91. 1251/3 = 1251/3 = 3 125 = 5111

5/44 _

93. a5/4b3/4 = b3/4 = 4 b3 = 4 b3 , or a 4 b3

95. m5/3n7/3 = 3 m5 3 n7 = 3 m5n7 = mn2 3 m2n 97. 17 = 173/5 53

99. _ 5 12_4 = 124/5


Chapter R Review Exercisesxy___1= a77_7_7yy

101. 3 11 = _11_1/3 = (111/2)1/3 = 111/6 103. 5 3 5 = 51/2 51/3 = 51/2+1/3 = 55/6 _

105. 5 322 = 322/5 = (321/5)2 = 22 = 4

107. (2a3/2)(4a1/2) = 8a3/2+1/2 = 8a2

_ x6 _1/2 _ x6 _1/2 x3 x3b2 9b4 32b4 3b2 3109.==, or

2/3 5/6111. x1/3y1/2 = x2/3(1/3)y5/61/2 = xy1/3 = x 3 y

113. (m1/2n5/2)2/3 = m2 3 n2 3 = m1/3n5/3 = m 3 n5 = 3 mn5 = n 3 mn212 5 23

115. a3/4(a2/3 +a4/3) = a3/4+2/3 +a3/4+4/3 = a17/12 +a25/12 = 12 a17 + 12 a25 =

a 12 a5 +a2 12 a

117.3 62 = 61/321/2 = 62/623/6 = (6223)1/6

= 6 368

= 6 288

19

___a__

127.a a= a a/2= aa/2

Chapter R Review Exercises

1. True

3. True

5. Rational numbers: 7, 43, 9, 0, 64, 44, 7 , 102 7. Integers: 7, 43, 0, 64, 102433123

9. Natural numbers: 43, 64, 1023

11. (4,7]

__

13. The distance of 8 from 0 is 8, so _ 8_ = 8.__

15.32422 +6(31)

= 32422 +62 Working inside parentheses= 3244+62Evaluating 22

119.

121.

123.

125.

4 xy 3 x2y = (xy)1/4(x2y)1/3 = (xy)3/12(x2y)4/12 = _(xy)3(x2y)4_1/12 = _x3y3x8y4_1/12

= 12 x11y7 _a4a3 = _a4a3_1/3 = (a4a3/2)1/3 3

= (a11/2)1/3 = a11/6

= 6 a11 = a 6 a5

_(a+x)3 3 (a +x)2(a+x)3/2(a+x)2/3 4 a+x (a+x)1/4_ =

(a+x)26/12 (a+x)3/12=

= (a+x)23/12 = 12 (a+x)23_

= (a+x) 12 (a+x)11

_1+x2 + 1+x2

_ 2 1+x2 1 1+x2 1+x2 1+x2 1+x2=1+x+

(1+x2) 1+x21+x2 1+x2 1+x2=+

(2+x2) 1+x2

1+x2

= 616+12Multiplying

= 10+12Adding in order = 2from left to right



8.3105 = 0.000083

19. Convert 405,000 to scientic notation.

We want the decimal point to be positioned between the 4 and the rst 0, so we move it 5 places to the left. Since 405,000 is greater than 10, the exponent must be positive.

405,000 = 4.05105

21.(3.1105)(4.5103)

= (3.14.5)(105 103)

= 13.95102This is not scientic notation. = (1.39510)102= 1.395103Writing scientic notation

23. (3x4y5)(4x2y) = 3(4)x4+(2)y5+1 = 12x2y4, or 12x2 , or 12x244

25. 81 = 34 = 3 44


20 ba==b 1 3==

27.

ba1 b a ab1 a 1=1

= b a a a b b ab1a1b1

a a ab 1=

bb


49. 6(2x1) = 3(x+10) 12x6 = 3x10 12x6 = x7 13x6 = 713x = 1

x = 131

The solution is 13.

ab151. = ab1b

x2 x = 20 x2 x20 = 0(x5)(x+4) = 0

ab1b

aab1 (ab1 b)

a(ab1)

x5 = 0 or x+4 = 0 x = 5 orx = 4The solutions are 5 and 4.

= a53.

x(x2) = 3

x2 2x = 3

29. (37)(3+7) = (3)2 (7)2 = 37= 4

31. 85+ 25 = 85+ 25 5 = 85+ 255 5 555

= 8 5+5 5

= 13 5

x2 2x3 = 0 (x+1)(x3) = 0x+1 = 0 or x3 = 0 x = 1 or x = 3The solutions are 1 and 3.

55. n2 7 = 0

n2 = 7n = 7 or n = 7


33.(5a+4b)(2a3b)

= 10a2 15ab+8ab12b2

= 10a2 7ab12b2

57.xy = x+3 xy x = 3x(y 1) = 3

35. 32x4 40xy3 = 8x4x3 8x5y3 = 8x(4x3 5y3)

37. 24x+144+x2 = x2 +24x+144 = (x+12)2

39. 9x2 30x+25 = (3x5)2

41. 18x2 3x+6 = 3(6x2 x+2)

43. 6x3 +48 = 6(x3 +8)= 6(x+2)(x2 2x+4)

45. 2x2 +5x3 = (2x1)(x+3)

47. 5x7 = 3x9 2x7 = 92x = 2 x = 1The solution is 1.

x = y 1

59.x2 +9x+20 x2 +7x+12x4

= (x+5)(x+4) (x+4)(x+3)x4

LCD is (x+5)(x+4)(x+3)

x x+3 4 x+5 (x+5)(x+4) x+3 (x+4)(x+3) x+5=

x(x+3)4(x+5) (x+5)(x+4)(x+3) x2 +3x4x20 (x+5)(x+4)(x+3) x2 x20 (x+5)(x+4)(x+3) (x5)(x+4)===

(x+5)(x+4)(x+3) x5

(x+5)(x+3)


Chapter R Test_rr121+13.610x3

61._(a+b)3 3 a+ b 6 (a+b)7 _

(a+b)3/2(a+b)1/3 (a+b)7/6=

= (a+b)3/2+1/37/6

= (a+b)9/6+2/67/6 = (a+b)2/3= 3 (a+b)2

_ 32 16_ 32 16 _1/84 2 63.38 = 38 = 3 8m nmmnn

65. a = 8 and b = 17. Find c. c2 = a2 +b2

c2 = 82 +172 c2 = 64+289 c2 = 353c 18.8

The guy wire is about 18.8 ft long.

21

Chapter R Test

1. a) Whole numbers: 0, 3 8, 29 b) Irrational numbers:12

c) Integers but not natural numbers: 0, 5

d) Rational numbers but not integers: 67, 4 , 1.2613

2. |17.6| = 17.6

_15_ 15 _11_ 113.____=

4. |0| = 0

5. (3,6]

6. |96| = |15| = 15, or

|6(9)| = |6+9| = |15| = 15

67. 9x2 36y2 = 9(x2 4y2) = 9[x2 (2y)2]= 9(x+2y)(x2y)

Answer C is correct.

7. 3223 1243 = 3281243 = 41243= 433

69. Substitute $124,000 $20,000, or $104,000 for P, 0.0575 for r, and 12 30, or 360, for n and perform the resulting computation. __n

M = P 12 1+ 12 1+ 1_r_n

0.0575_0.0575_360 = $104,000120.0575 1201+__36

12 $606.92

71. (anbn)3 = (anbn)(anbn)2

= (an bn)(a2n 2anbn +b2n)

= a3n2a2nbn+anb2na2nbn+2anb2nb3n = a3n 3a2nbn +3anb2n b3n

73. m6n m3n = m3n(m3n 1) = m3n[(mn)3 13]= m3n(mn 1)(m2n +mn +1)

= 49 = 5

8. Position the decimal point 6 places to the left, between the 4 and the 5. Since 4,509,000 is a number greater than 10, the exponent must be positive.

4,509,000 = 4.509106

9. The exponent is negative, so the number is between 0 and 1. We move the decimal point 5 places to the left.

8.6105 = 0.000086

10.2.710 3 = 0.751074

= (7.5101)107 = 7.5106

11. x8 x5 = x8+5 = x3, or 1

12. (2y2)3(3y4)2 = 23y6 32y8 = 89y6+8 = 72y14

13.(3a5b4)(5a1b3)

= 35a5+(1) b4+3

= 15a4b1, or 15a4

14. (5xy47xy2+4x23)(3xy4+2xy22y+4) = (5xy47xy2+4x23)+(3xy42xy2+2y 4) = (5+3)xy4+(72)xy2+4x2+2y +(34) = 8xy4 9xy2 +4x2 +2y 7b


22Chapter R: Basic Concepts of Algebra 4433==x3==

15. (y 2)(3y +4) = 3y2 +4y 6y 8 = 3y2 2y 829. 3(y 5)+6 = 8(y +2)

16. (4x3)2 = (4x)2 24x3+32 = 16x2 24x+9

xyx xy y yxy xx y x+y x+y17.=

x2 y2 xy xy x+y=

x2 y2 xy x+y=

x2 y21 xy x+y=

(x+y (xy) xy(x+y)=)

xy xy=

18. 45 = 95 = 95 = 35 19. 3 56 = 87 = 87 = 27 33 33

20. 375+227 = 3253+293

= 35 3+23 3 = 15 3+6 3

= 21 3

21. 1810 = 1810 = 23325 = 23 5 = 6 5

22.(2+3)(523)

= 254 3+5 323 = 104 3+5 36

= 4+ 3

23. 8x2 18 = 2(4x2 9) = 2(2x+3)(2x3)

3y 15+6 = 8y 2 3y 9 = y +6 4y 9 = 64y = 15 y = 15

The solution is 15.

30. 2x2 +5x+3 = 0 (2x+3)(x+1) = 02x+3 = 0or x+1 = 0 2x = 3 orx = 1

x = 2 orx = 1


31. z2 11 = 0

z2 = 11z = 11 or z = 11


32. 8x+3y = 24

3y = 8x+24

y = 8x+24, or 3x+838

33. x2 +x6x2 25 x2 +8x+15 x2 4x+4

(x2 +x6)(x2 25) (x2 +8x+15)(x2 4x+4) (x+3) x2)(x+5)(x5)=(

(x+3)(x+5)(x2)(x2) x5

x2

24. y2 3y 18 = (y +3)(y 6)

25. 2n2 +5n12 = (2n3)(n+4)

26. x3 +10x2 +25x = x(x2 +10x+25) = x(x+5)2

27. m3 8 = (m2)(m2 +2m+4)

28. 7x4 = 24 7x = 28x = 4

The solution is 4.

34.x2 1 x2 +4x5

= (x+1)(x1) (x1)(x+5)x3

LCD is (x+1)(x1)(x+5)

x x+5 3 x+1 (x+1)(x1) x+5 (x1)(x+5) x+1=

x(x+5)3(x+1) (x+1)(x1)(x+5) x2 +5x3x3 (x+1)(x1)(x+5) x2 +2x3 (x+1)(x1)(x+5) (x+3)(x1)===

(x+1)(x1)(x+5) x+3(x+1)(x+5)


Chapter R Test23

557+335+53 7 3 7 3 7+ 3 493 35. ===

35+5 3 46

36. m3/8 = 8 m3

37. 6 35 = 35/6

38. a = 5 and b = 12. Find c. c2 = a2 +b2

c2 = 52 +122 c2 = 25+144 c2 = 169c = 13

The guy wire is 13 ft long.

39. (xy 1)2 = [(xy)1]2= (xy)2 2(xy)(1)+12 = x2 2xy +y2 2x+2y +1



_____2323_23

Chapter 1

Graphs, Functions, and Models

Exercise Set 1.1

1. Point A is located 5 units to the left of the y-axis and 4 units up from the x-axis, so its coordinates are (5,4).

Point B is located 2 units to the right of the y-axis and 2 units down from the x-axis, so its coordinates are (2,2).

Point Cis located 0 units to the right or left of the y-axis and 5 units down from the x-axis, so its coordinates are (0,5).

Point D is located 3 units to the right of the y-axis and 5 units up from the x-axis, so its coordinates are (3,5).

Point E is located 5 units to the left of the y-axis and 4 units down from the x-axis, so its coordinates are (5,4).

Point F is located 3 units to the right of the y-axis and 0 units up or down from the x-axis, so its coordinates are (3,0).

3. To graph (4,0) we move from the origin 4 units to the right of the y-axis. Since the second coordinate is 0, we do not move up or down from the x-axis.

To graph (3,5) we move from the origin 3 units to the left of the y-axis. Then we move 5 units down from the x-axis.

To graph (1,4) we move from the origin 1 unit to the left of the y-axis. Then we move 4 units up from the x-axis.

To graph (0,2) we do not move to the right or the left of the y-axis since the rst coordinate is 0. From the origin we move 2 units up.

To graph (2,2) we move from the origin 2 units to the right of the y-axis. Then we move 2 units down from the x-axis.

y

4

2 (0, 2)(4, 0)

_4 _224x _2

(_3, _5)_4

5. To graph (5,1) we move from the origin 5 units to the left of the y-axis. Then we move 1 unit up from the x-axis.

To graph (5,1) we move from the origin 5 units to the right of the y-axis. Then we move 1 unit up from the x-axis.

To graph (2,3) we move from the origin 2 units to the right of the y-axis. Then we move 3 units up from the x-axis.

To graph (2,1) we move from the origin 2 units to the right of the y-axis. Then we move 1 unit down from the x-axis.

To graph (0,1) we do not move to the right or the left of the y-axis since the rst coordinate is 0. From the origin we move 1 unit up.

y

4

2 (0, 1)

_4 _24 _2 (2, _1)x

_4

7. The rst coordinate represents the year and the cor-responding second coordinate represents the percentage of the U.S. population that is foreign-born.The or-dered pairs are (1970, 4.7%), (1980, 6.2%), (1990, 8.0%), (2000, 10.4%), (2010, 12.8%).

9. To determine whether (1,9) is a solution, substitute 1 for x and 9 for y.

y = 7x2

9 ? 7(1)2 _ 72_

9 _ 9TRUE

The equation 9 = 9 is true, so (1,9) is a solution.

To determine whether (0,2) is a solution, substitute 0 for x and 2 for y.

y = 7x2

2 ? 702

_ 02

2 _ 2FALSE

The equation 2 = 2 is false, so (0,2) is not a solution.

__11. To determine whether 3, 4 is a solution, substitute 3

for x and 4 for y. 6x4y = 1

6 3 4 4 ? 1 43 _23____

1 _ 1 TRUE__The equation 1 = 1 is true, so 3, 4 is a solution.


26Chapter 1: Graphs, Functions, and Models33_2141414_____2533_3______

__To determine whether 1, 2 is a solution, substitute 1 for

x and 2 for y. 6x4y = 1

614 2 ? 1 66 _3____

0 _ 1 FALSEThe equation 0 = 1 is false, so _1, 3_ is not a solution.

__13. To determine whether 2,5 is a solution, substitute

2 for a and 5 for b.

2a+5b = 3

____2 2 +5 5? 3

14 _5 _ 3 FALSEThe equation 5 = 3 is false, so _ 1,4_ is not a solu-

tion.__To determine whether 0, 5 is a solution, substitute 0 for

a and 5 for b.

2a+5b = 3

20+5 5 _ 3 0+3 _3?___

3 _ 3 TRUE__The equation 3 = 3 is true, so 0, 5 is a solution.

15. To determine whether (0.75,2.75) is a solution, substi-tute 0.75 for x and 2.75 for y.

x2 y2 = 3

(0.75)2 (2.75)2 ? 3

0.56257.5625 _

7 _ 3 FALSE

The equation 7 = 3 is false, so (0.75,2.75) is not a solution.

5x30 = 15 5x = 15x = 3

The x-intercept is (3,0).

To nd the y-intercept we replace x with 0 and solve for y.

503y = 15 3y = 15y = 5

The y-intercept is (0,5).

We plot the intercepts and draw the line that contains them. We could nd a third point as a check that the intercepts were found correctly.

y (0, 5)4

5x _ 3y _ _15 (_3, 0)2

_4 _22 4x _2

_4

19. Graph 2x+y = 4.

To nd the x-intercept we replace y with 0 and solve for x.2x+0 = 4

2x = 4 x = 2The x-intercept is (2,0).

To nd the y-intercept we replace x with 0 and solve for y.

20+y = 4 y = 4The y-intercept is (0,4).


y

2x _ y _ 4 4 (0, 4)

2(2, 0)


x2 y2 = 3

_4 _22 4x _2

_4

22 (1)2 ? 3

41 _

3 _ 3 TRUE


17. Graph 5x3y = 15.

To nd the x-intercept we replace y with 0 and solve for x.

21. Graph 4y 3x = 12.

To nd the x-intercept we replace y with 0 and solve for x.403x = 12

3x = 12 x = 4The x-intercept is (4,0).


Exercise Set 1.1273

To nd the y-intercept we replace x with 0 and solve fory

y.4x _ y _ 3 4y 30 = 122

4y = 12 y = 3The y-intercept is (0,3).

_4 _224x _2

_4


y

4y _ 3x _ 12 4 (0, 3) (_4, 0)2

_4 _22 4x _2

_4

23. Graph y = 3x+5.

We choose some values for x and nd the corresponding y-values.

When x = 3, y = 3x+5 = 3(3)+5 = 9+5 = 4.

27. Graph y = 4x+3.3

By choosing multiples of 4 for x, we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph.

x

y

(x,y)

4

6

(4,6)

0

3

(0,3)

4

0

(4,0)

y

4

2

When x = 1, y = 3x+5 = 3(1)+5 = 3+5 = 2. When x = 0, y = 3x+5 = 30+5 = 0+5 = 5We list these points in a table, plot them, and draw the graph.

_4 _22 4x _2_4 y _ __x _ 34

x

y

(x,y)

3

4

(3,4)

1

2

(1,2)

0

5

(0,5)

y

29. Graph 5x2y = 8.

We could solve for y rst. 5x2y = 82y = 5x+8 Subtracting 5x on both sides

y = 2x4 Multiplying by 2 on both sides51

6

y _ 3x _ 5

By choosing multiples of 2 for x we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph.

2

_424x _2

25. Graph xy = 3.

Make a table of values, plot the points in the table, and draw the graph.

x

y

(x,y)

2

5

(2,5)

0

3

(0,3)

3

0

(3,0)

x

y

(x,y)

0

4

(0,4)

2

1

(2,1)

4

6

(4,6)

y

4

2

_4 _22 4x _2

_4 5x _ 2y _ 8


28

31. Graph x4y = 5.

Chapter 1: Graphs, Functions, and Models

y


_4 _224x _2

x

y

(x,y)

3

2

(3,2)

1

1

(1,1)

5

0

(5,0)

y

4x _ 4y _ 5 2

_246x _2

_4

33. Graph 2x+5y = 10.

In this case, it is convenient to nd the intercepts along with a third point on the graph. Make a table of values, plot the points in the table, and draw the graph.

_4y _ _x2 _6

_8

37. Graph y = x2 3.


x

y

(x,y)

3

6

(3,6)

1

2

(1,2)

0

3

(0,3)

1

2

(1,2)

3

6

(3,6)

y

x

y

(x,y)

5

0

(5,0)

0

2

(0,2)

5

4

(5,4)

6

4

2y _ x2 _ 3

_4 _224x _2

y

42x _ 5y _ _10

2

39. Graph y = x2 +2x+3.


_6_22x

_4

35. Graph y = x2.


x

y

(x,y)

2

4

(2,4)

1

1

(1,1)

0

0

(0,0)

1

1

(1,1)

2

4

(2,4)

x

y

(x,y)

2

5

(2,5)

1

0

(1,0)

0

3

(0,3)

1

4

(1,4)

2

3

(2,3)

3

0

(3,0)

4

5

(4,5)

y y _ _x2 _ 2x _ 3

4

_8 _44 8x _4

_8

_12


Exercise Set 1.1

41. Graph (b) is the graph of y = 3x.

43. Graph (a) is the graph of y = x2 +2x+1.

45. Enter the equation, select the standard window, and graph the equation as described in the Graphing Calculator Man-ual that accompanies the text.

10

_10

47. First solve the equation for y: y = 4x + 7. Enter the equation in this form, select the standard window, and graph the equation as described in the Graphing Calcula-

29


y _ x2 _ 6 10

10

_10


y _ 2 _ x2

10

10

10

_10

49. Enter the equation, select the standard window, and graph the equation as described in the Graphing Calculator Man-

_10


y _ x2 _ 4x _ 2

10

10

_10

_10

1059. Standard window:

_10

51. First solve the equation for y. 2x+3y = 53y = 2x5

y = 2x5, or 3(2x5)31

Enter the equation in y = form, select the standard win-dow, and graph the equation as described in the Graphing Calculator Manual that accompanies the text.

2x _ 3y _ _5

10

[4,4,4,4]

_10

10

We see that the standard window is a better choice for this graph.

_10


(30Chapter 1: Graphs, Functions, and Models_ __77__1414____________________________,

61. Standard window:

[1,1,0.3,0.3], Xscl = 0.1, Yscl = 0.1

We see that [1,1,0.3,0.3] is a better choice for this graph.

63. Either point can be considered as (x1,y1). d =(45)2 +(69)2_

=(1)2 +(3)2 = 10 3.162

65. Either point can be considered as (x1,y1). d =(13(8))2 +(1(11))2_

=(5)2 +122 = 169 = 13

67. Either point can be considered as (x1,y1). d =(69)2 +(15)2_

=(3)2 +(6)2 = 45 6.708

69. Either point can be considered as (x1,y1). __2_

d =(88)2 + 11 11

=(16)2 +02 = 16

____ 2__2 71. d = 5 5 + 4 3332

__2

=02 + 3= 3

73. Either point can be considered as (x1,y1). d = (4.22.1)2 +[3(6.4)]2_

=(6.3)2 +(9.4)2 = 128.05 11.316

75. Either point can be considered as (x1,y1). d =(0a)2 +(0b)2 = a2 +b2_

77. First we nd the length of the diameter: d =(39)2 +(14)2_

=(12)2 +(5)2 = 169 = 13

The length of the radius is one-half the length of the di-ameter, or 1(13), or 6.5.

79. First we nd the distance between each pair of points. For (4,5) and (6,1):d = (46)2 +(51)2 = (10)2 +42 = 116_

For (4,5) and (8,5):

d =(4(8))2 +(5(5))2 = 42 +102 = 116

For (6,1) and (8,5):

d =(6(8))2 +(1(5))2

=142 +62 = 232Since (116)2 + (116)2 = (232)2, the points could be

the vertices of a right triangle.

81. First we nd the distance between each pair of points. For (4,3) and (0,5):d =(40)2 +(35)2 =(4)2 +(2)2 = 20_

For (4,3) and (3,4):

d =(43)2 +[3(4)]2 =(7)2 +72 = 98_

For (0,5) and (3,4):

d =(03)2 +[5(4)]2

=(3)2 +92 = 90The greatest distance is 98, so if the points are the ver-

tices of a right triangle, then it is the hypotenuse. But ( 20)2 + ( 90)2 = ( 98)2, so the points are not the ver-tices of a right triangle.

83. We use the midpoint formula.

4+(12), 9+(3)= 2,12= (4,6)8222


0+ 21 2 1 __ 2 , 2 2 = 2 , 2 = 5, 4505 211


6.1+3.8 3.8+(6.1)9.99.9 22 2 2,,==

(4.95,4.95)


6+(6), 5 2 8= 2 , 2=6, 22+131213


6 + 35 + 4 6 20 2 2 2213255,,13==

_5 13_

12 40

Copyright 2013 Pearson Education, Inc.2

Exercise Set 1.131_=,,___=,,=,,=,,157339111__ ___________ ___5__22_

93.y 8

6

4

2

_4 _246 x _2

101. The length of a radius is the distance between (1,4) and (3,7):

r =(13)2 +(47)2

=(4)2 +(3)2 = 25 = 5

(xh)2 +(y k)2 = r2 [x(1)]2 +(y 4)2 = 52(x+1)2 +(y 4)2 = 25

For the side with vertices (3,4) and (2,1): 3+2 4+(1)1 3222 2

For the side with vertices (2,1) and (5,2): 2+5 1+27 1____

222 2

For the side with vertices (5,2) and (0,7): 5+0 2+75 9____

222 2

For the side with vertices (0,7) and (3,4): 0+(3) 7+43 11____

222 2

For the quadrilateral whose vertices are the points found above, the diagonals have endpoints___ ____ _

2, 2 ,2, 2and2, 2 , 2, 2 . We nd the length of each of these diagonals.

For 2, 2 ,2, 2 :1539

_15_2_39_2 22 22d =+

=(3)2 +(3)2 = 18

For2, 2 , 2, 2 :713 11

_7_3__2_111_2 2 2 2 2d =+

=52 +(5)2 = 50

Since the diagonals do not have the same lengths, the mid-points are not vertices of a rectangle.


7+ 2 4+3 7+ 2 1 2 2 2 2=,,

97. Square the viewing window. For the graph shown, one possibility is [12,9,4,10].

99. (xh)2 +(y k)2 = r2_ _2

(x2)2 +(y 3)2 =3Substituting

(x2)2 +(y 3)2 = 259

103. The center is the midpoint of the diameter: 7+(3), 13+(11)= (2,1)__

Use the center and either endpoint of the diameter to nd the length of a radius. We use the point (7,13):

r =(72)2 +(131)2 = 52 +122 = 169 = 13

(xh)2 +(y k)2 = r2 (x2)2 +(y 1)2 = 132 (x2)2 +(y 1)2 = 169

105. Since the center is 2 units to the left of the y-axis and the circle is tangent to the y-axis, the length of a radius is 2.

(xh)2 +(y k)2 = r2 [x(2)]2 +(y 3)2 = 22(x+2)2 +(y 3)2 = 4

107.x2 +y2 = 4 (x0)2 +(y 0)2 = 22Center: (0,0); radius: 2

y

4

2

x2 _ y2 _ 4

_4 _2_2

_4

24x

109. x2 +(y 3)2 = 16 (x0)2 +(y 3)2 = 42Center: (0,3); radius: 4

y 86

4

2

_4 _2_2

24x

x2 _ (y _ 3)2 _ 16


32Chapter 1: Graphs, Functions, and Models__ _________31_____22+____

111. (x1)2 +(y 5)2 = 36

(x1)2 +(y 5)2 = 62

Then we have:(xh)2 +(y k)2 = r2

Center: (1,5); radius: 6

y

12

8

4

_8 _4_4

48x

(x _ 1)2 _ (y _ 5)2 _ 36

113.(x+4)2 +(y +5)2 = 9 [x(4)]2 +[y (5)]2 = 32Center: (4,5); radius: 3

y 2

_6 _4 _2_2

_4

_6

_8

2x

(x _ 4)2 _ (y _ 5)2 _ 9

115. From the graph we see that the center of the circle is (2,1) and the radius is 3. The equation of the circle is [x(2)]2 +(y 1)2 = 32, or (x+2)2 +(y 1)2 = 32.

117. From the graph we see that the center of the circle is (5,5) and the radius is 15. The equation of the circle is (x5)2 +[y(5)]2 = 152, or (x5)2 +(y+5)2 = 152.

119. If the point (p,q) is in the fourth quadrant, then p > 0 and q < 0. If p > 0, then p < 0 so both coordinates of the point (q,p) are negative and (q,p) is in the third quadrant.

121. Use the distance formula. Either point can be considered as (x1,y1).d = _(a+ha)2 +(a+ha)2=h2 +a+h2 a2 +ah+a =h2 + 2a+ h2a2 + ah _

Next we use the midpoint formula.

a+a+ha+ a+h2a+ha+ a+h 2 2 2 2=,,

123. First use the formula for the area of a circle to nd r2: A = r2

36 = r2

36 = r2

(x2)2 +[y (7)]2 = 36 (x2)2 +(y +7)2 = 36

125. Let (0,y) be the required point. We set the distance from (2,0) to (0,y) equal to the distance from (4,6) to (0,y) and solve for y.

[0(2)]2 +(y 0)2 =(04)2 +(y 6)2

4+y2 =16+y2 12y +36

4+y2 = 16+y2 12y +36 Squaring both sides48 = 12y

4 = y The point is (0,4).

127. a) When the circle is positioned on a coordinate system as shown in the text, the center lies on the y-axis and is equidistant from (4,0) and (0,2).

Let (0,y) be the coordinates of the center.

(40)2+(0y)2 = (00)2+(2y)2 42 +y2 = (2y)216+y2 = 44y +y2 12 = 4y3 = y

The center of the circle is (0,3).

b) Use the point (4,0) and the center (0,3) to nd the radius.(40)2 +[0(3)]2 = r2

25 = r2 5 = rThe radius is 5 ft.

129.x2 +y2 = 1 _ _2 __22+ 2? 1

4 + 4 _31__

1 _ 1 TRUE _ 23,2_ lies on the unit circle.1

131.x2 +y2 = 1 _ _2 _ _2 2+ 2? 1 2 2 ___

44 _

1 _ 1 TRUE _ 22, 22_ lies on the unit circle.

133. See the answer section in the text.


Exercise Set 1.2x+h+3111

Exercise Set 1.2

1. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range.


5. This correspondence is not a function, because there is a member of the domain (m) that corresponds to more than one member of the range (A and B).


9. This correspondence is a function, because each car has exactly one license number.

11. This correspondence is a function, because each integer less than 9 corresponds to exactly one multiple of 5.

13. This correspondence is not a function, because at least one student will have more than one neighboring seat occupied by another student.

15. The relation is a function, because no two ordered pairs have the same rst coordinate and dierent second coor-dinates.

The domain is the set of all rst coordinates: {2,3,4}.

The range is the set of all second coordinates: {10,15,20}.

17. The relation is not a function, because the ordered pairs (2,1) and (2,4) have the same rst coordinate and dif-ferent second coordinates.

The domain is the set of all rst coordinates: {7,2,0}.

The range is the set of all second coordinates: {3,1,4,7}.

19. The relation is a function, because no two ordered pairs have the same rst coordinate and dierent second coor-dinates.

The domain is the set of all rst coordinates: {2,0,2,4,3}.

The range is the set of all second coordinates: {1}.

21. g(x) = 3x2 2x+1

a) g(0) = 302 20+1 = 1

b) g(1) = 3(1)2 2(1)+1 = 6 c) g(3) = 332 23+1 = 22d) g(x) = 3(x)2 2(x)+1 = 3x2 +2x+1 e) g(1t) = 3(1t)2 2(1t)+1 =3(12t+t2)2(1t)+1 = 36t+3t22+2t+1 =

3t2 4t+2

33

23. g(x) = x3

a) g(2) = 23 = 8

b) g(2) = (2)3 = 8 c) g(x) = (x)3 = x3 d) g(3y) = (3y)3 = 27y3e) g(2+h) = (2+h)3 = 8+12h+6h2 +h3

25. g(x) = x+3x4

a) g(5) = 5+3 = 8541

b) g(4) = 4+7 = 044

c) g(3) = 3+3 = 0347

Since division by 0 is not dened, g(3) does not exist.

d) g(16.25) = 16.25+3 = 13.25 = 53 1.5283 e) g(x+h) = x+h416.25420.2581

27.g(x) = 1x2x

g(0) = 102 = 1 = 1 = 0000

g(1) = _1(1)2 = 11 = 1 = 11100

Since division by 0 is not dened, g(1) does not exist.

g(5) = 152 = 125 = 24555

Since24 is not dened as a real number, g(5) does not exist as a real number.

g_1_ = _2= 2_1_1

3

2 = 2 = 1 2 2 1 4 413 = 2 3 = 2 3 = 13, or 332 121 2

2 29.

Rounding to the nearest tenth, we see that g(2.1) 21.8, g(5.08) 130.4, and g(10.003) 468.3.

31. Graph f(x) = 2x+3.1

We select values for x and nd the corresponding values of f(x). Then we plot the points and connect them with a smooth curve.


(34Chapter 1: Graphs, Functions, and Models____

x

f(x)

(x,f(x))

4

1

(4,1)

0

3

(0,3)

2

4

(2,4)

y

4

2

_4 _224x _2

_4

f(x) _ 2 x _ 31

33. Graph f(x) = x2 +4.


y

4

2

_4 _224x _2

_4

f(x) _ _x2 _ 4

35. Graph f(x) = x 1.


y

4

2

_4 _224x _2

_4

f(x) _ _

x _ 1

37. From the graph we see that, when the input is 1, the output is 2, so h(1) = 2. When the input is 3, the output is 2, so h(3) = 2. When the input is 4, the output is 1, so h(4) = 1.

39. From the graph we see that, when the input is 4, the output is 3, so s(4) = 3. When the input is 2, the

output is 0, so s(2) = 0. When the input is 0, the output is 3, so s(0) = 3.

41. From the graph we see that, when the input is 1, the output is 2, so f(1) = 2. When the input is 0, the output is 0, so f(0) = 0. When the input is 1, the output is 2, so f(1) = 2.

43. This is not the graph of a function, because we can nd a vertical line that crosses the graph more than once.

y

x

45. This is the graph of a function, because there is no vertical line that crosses the graph more than once.

47. This is the graph of a function, because there is no vertical line that crosses the graph more than once.

49. This is not the graph of a function, because we can nd a vertical line that crosses the graph more than once.

51. We can substitute any real number for x. Thus, the do-main is the set of all real numbers, or (,).


55. The input 0 results in a denominator of 0. Thus, the do-main is {x|x = 0}, or (,0)(0,).

57. We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. We nd these inputs.

2x = 0 2 = xThe domain is {x|x = 2}, or (,2)(2,).

59. We nd the inputs that make the denominator 0: x2 4x5 = 0(x5)(x+1) = 0

x5 = 0 or x+1 = 0 x = 5 orx = 1The domain is {x|x = 5 and x = 1}, or (,1)(1,5)(5,).


63. We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. We nd these inputs.


(Exercise Set 1.235__

x2 7x = 079. x(x7) = 0

x = 0 or x7 = 0 x = 0 or x = 7

The domain is {x|x = 0 and x = 7}, or (,0) (0,7) (7,).


67. The inputs on the x-axis that correspond to points on the graph extend from 0 to 5, inclusive. Thus, the domain is {x|0 x 5}, or [0,5].

The outputs on the y-axis extend from 0 to 3, inclusive. Thus, the range is {y|0 y 3}, or [0,3].

69. The inputs on the x-axis that correspond to points on the graph extend from 2 to 2 inclusive. Thus, the domain is {x|2 x 2}, or [2,2].

The outputs on the y-axis extend from 1 to 1, inclusive. Thus, the range is {y|1 y 1}, or [1,1].

71. The graph extends to the left and to the right without bound. Thus, the domain is the set of all real numbers, or (,).

The only output is 3, so the range is {3}.

73. The inputs on the x-axis extend from 5 to 3, inclusive. Thus, the domain is [5,3].

The outputs on the y-axis extend from 2 to 2, inclusive. Thus, the range is [2,2].

75.

To nd the domain we look for the inputs on the x-axis that correspond to a point on the graph. We see that each point on the x-axis corresponds to a point on the graph so the domain is the set of all real numbers, or (,).

To nd the range we look for outputs on the y-axis. The number 0 is the smallest output, and every number greater than 0 is also an output. Thus, the range is [0,).

We see that each point on the x-axis except 3 corresponds to a point on the graph, so the domain is (,3)(3,). We also see that each point on the y-axis except 0 corre-sponds to an output, so the range is (,0)(0,).

81.

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (,).

Each point on the y-axis also corresponds to a point on the graph, so the range is the set of all real numbers, (,).

83.

The largest input on the x-axis is 7 and every number less than 7 is also an input. Thus, the domain is (,7].

The number 0 is the smallest output, and every number greater than 0 is also an output. Thus, the range is [0,).

85.

77.

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (,).

The largest output is 3 and every number less than 3 is also an output. Thus, the range is (,3].

87. a)In 2018, x = 20181985 = 33.

We see that each point on the x-axis corresponds to a point on the graph so the domain is the set of all real numbers, or (,). We also see that each point on the y-axis corresponds to an output so the range is the set of all real numbers, or (,).

V(33) = 0.4652(33)+10.87 $26.22 In 2025, x = 20251985 = 40.V(40) = 0.4652(40)+10.87 $29.48


(36Chapter 1: Graphs, Functions, and Models_____2__

b)Substitute 40 for V(x) and solve for x. 40 = 0.4652x+10.8729.13 = 0.4652x 63 xIt will take approximately $40 to equal the value of $1 in 1913 about 63 yr after 1985, or in 2048.

89. E(t) = 1000(100t)+580(100t)2

a) E(99.5) = 1000(10099.5)+580(10099.5)2 = 1000(0.5)+580(0.5)2= 500+580(0.25) = 500+145 = 645 m above sea levelb) E(100) = 1000(100100)+580(100100)2 = 10000+580(0)2 = 0+0= 0 m above sea level, or at sea level

91. To determine whether (0,7) is a solution, substitute 0 for x and 7 for y.

y = 0.5x+7

7 ? 0.5(0)+7 _ 0+7_

7 _ 7FALSE

The equation 7 = 7 is false, so (0,7) is not a solution.


95. Graph 2x5y = 10.


x

y

(x,y)

5

0

(5,0)

0

2

(0,2)

5

4

(5,4)

97. We nd the inputs for which 2x+5 is nonnegative. 2x+5 02x 5

x 5_ ____ Thus, the domain is x_x 2 , or 2, .__55

99. x+6 is not dened for values of x for which x + 6 is negative. We nd the inputs for which x+6 is nonnegative.

x+6 0 x 6We must also avoid inputs that make the denominator 0. (x+2)(x3) = 0x+2 = 0or x3 = 0

y = 0.5x+7x = 2 orx = 3

11 ? 0.5(8)+7

_ 4+7

11 _ 11TRUE


93. Graph y = (x1)2.


x

y

(x,y)

1

4

(1,4)

0

1

(0,1)

1

0

(1,0)

2

1

(2,1)

3

4

(3,4)

Then the domain is {x|x 6 and x = 2 and x = 3}, or [6,2)(2,3)(3,).

101. Answers may vary. Two possibilities are f(x) = x, g(x) = x+1 and f(x) = x2, g(x) = x2 4.

103. First nd the value of x for which x+3 = 1. x+3 = 1x = 4 Then we have:

g(x+3) = 2x+1

g(1) = g(4+3) = 2(4)+1 = 8+1 = 7

Exercise Set 1.3

1. a) Yes. Each input is 1 more than the one that pre-cedes it.

b) Yes. Each output is 3 more than the one that pre-cedes it.

c) Yes. Constant changes in inputs result in constant changes in outputs.

3. a) Yes. Each input is 15 more than the one that pre-cedes it.

b) No. The change in the outputs varies.

c) No. Constant changes in inputs do not result in constant changes in outputs.


Exercise Set 1.3336m =21=== 2111155__yy215505522

5. Two points on the line are (4,2) and (1,4).

y2 y1 4(2) 6 x2 x1 1(4) 5m ===

7. Two points on the line are (0,3) and (5,0).

m = x2 x1 = 50 = 5 , or 5yy210333

9. m = x2 x1 = 30 = 3 = 0yy12330

11. m = x2 x1 = 19 = 10 = 5y y122421

y2 y1 6(9) 15 x2 x1 44 013. m ===

Since division by 0 is not dened, the slope is not dened.

y2 y10.4(0.1)0.3 x2 x1 0.30.7 115. m ==== 0.3

y2 y1 2(2) 0 x2 x1 42 217. m ==== 0

__19. m = x2 x1 = 5 1 5 = 1 = 5 22yy211 5 6

y2 y1 5(13) 8 1 x2 x1 816 24 321. m ====

7(7)14 10(10) 023. m ==

Since division by 0 is not dened, the slope is not dened.

25. We have the points (4,3) and (2,15). y2 y1 153 12x x246____

27. We have the points5, 2and1, 2 .

111m = y2 y1 =22 = 6 = 6 5= 5 21 1 16x x6

____ 29. We have the points 6, 5 and 0, 5 .44

4 4m = x2 x1 = 60 = 6 = 0

31. y = 1.3x 5 is in the form y = mx + b with m = 1.3, so the slope is 1.3.

33. The graph of x = 2 is a vertical line, so the slope is not dened.

35. f(x) = 2x +3 is in the form y = mx +b with m = 2, so the slope is 2.111

37. y = 9x can be written as y = x+9, or y = 1x+9. Now we have an equation in the form y = mx + b with m = 1, so the slope is 1.

37

39. The graph of y = 0.7 is a horizontal line, so the slope is 0. (We also see this if we write the equation in the form y = 0x +0.7).

41. We have the points (2009, 945) and (1998, 425). We nd the average rate of change, or slope.945425520 20091998 11m == 47.3

The average rate of change in the revenue of the reworks industry in the U.S. from 1998 to 2009 was about $47.3 mil-lion per year.

43. We have the points (2000, 348,194) and (2010, 319,294). We nd the average rate of change, or slope.348,194319,29428,900 20002010 10m === 2890

The average rate of change in the population in St. Louis, Missouri, from 2000 to 2010 was 2890 people per year.

45. We have the points (2009, 21.0) and (2012, 25.0). We nd the average rate of change, or slope.21.025.04.0 20092012 3m == 1.3

The average rate of change in the sales of electric bikes in China from 2009 to 2012 was about 1.3 million bikes per year.

47. We have the points (1990, 3.1) and (2008, 5.5). We nd the average rate of change, or slope.

m = 20081990 = 18 0.135.53.12.4

The average rate of change in the consumption of broccoli per capita from 1990 to 2008 was about 0.13 lb per year.

49. y = 3x 7

The equation is in the form y = mx + b where m = 3 and b = 7. Thus, the slope is 3, and the y-intercept is (0,7).55

51. x = 52

This is the equation of a vertical line 2 unit to the left

of the y-axis. The slope is not dened, and there is no y-intercept.

53. f(x) = 5 2x, or f(x) = 2x +511

The second equation is in the form y = mx + b where m = 1 and b = 5. Thus, the slope is 1 and the y-

intercept is (0,5).

55. Solve the equation for y. 3x +2y = 102y = 3x +10

y = 2x +5 Slope: 3; y-intercept: (0,5)3

Copyright 2013 Pearson Education, Inc.2

38Chapter 1: Graphs, Functions, and Models2331121431110

57. y = 6 = 0x 6

Slope: 0; y-intercept: (0,6)

59. Solve the equation for y. 5y 4x = 85y = 4x +8

y = 5x + 5__ Slope: 5; y-intercept: 0, 54848

61. Solve the equation for y. 4y x +2 = 04y = x 2

67. First solve the equation for y. 3x 4y = 204y = 3x +20

y = 4x 53

Plot the y-intercept, (0,5). Then using the slope, 3,

start at (0,5) and nd another point by moving up 3 units and right 4 units. We have the point (4,2). We can move from the point (4,2) in a similar manner to get a third point, (8,1). Connect the three points to draw the graph.

y

y = 4x 2_ Slope: 4; y-intercept:0,21_1

63. Graph y = 2x 3.1

2 3x _ 4y _ 20

_4 _224x _2

_4

_6

Plot the y-intercept, (0,3). We can think of the slope as 1. Start at (0,3) and nd another point by moving

down 1 unit and right 2 units. We have the point (2,4).

We could also think of the slope as 2. Then we can start at (0,3) and get another point by moving up 1 unit and1

left 2 units. We have the point (2,2). Connect the three points to draw the graph.

y

4 y _ __x _ 31

2

69. First solve the equation for y. x +3y = 183y = x +18

y = 3x +61

Plot the y-intercept, (0,6). We can think of the slope as 1. Start at (0,6) and nd another point by moving down

1 unit and right 3 units. We have the point (3,5). We can move from the point (3,5) in a similar manner to get a third point, (6,4). Connect the three points and draw the graph.

_4 _224xy _2

_46

4

65. Graph f(x) = 3x 1.2 x _ 3y _ 18

Plot the y-intercept, (0,1). We can think of the slope as 3. Start at (0,1) and nd another point by moving

up 3 units and right 1 unit. We have the point (1,2). We can move from the point (1,2) in a similar manner to get a third point, (2,5). Connect the three points to draw the graph.

_4 _224x _2

71. a) y _ _x _ 11

300

y

4

08000 2 01

_4 _224x _2_4 f(x) _ 3x _ 1

b) P(0) = 33 0+1 = 1 atm

P(33) = 33 33+1 = 2 atm

P(1000) = 33 1000+1 = 3133 atm


Chapter 1 Mid-Chapter Mixed Review10214111011111122111135m ====______

P(5000) = 33 5000+1 = 15233 atm117

P(7000) = 33 7000+1 = 21333 atm

73. a) D(r) = 10r + 2111

The slope is 10.For each mph faster the car travels, it takes 11 ft

longer to stop.

b) y _ _x _ _111

100

39

87. False. For example, let f(x) = x + 1. Then f(c d) = cd+1, but f(c)f(d) = c+1(d+1) = cd.

89.f(x) = mx+b f(x+2) = f(x)+2m(x+2)+b = mx+b+2 mx+2m+b = mx+b+2 2m = 2m = 1

Thus, f(x) = 1x+b, or f(x) = x+b.

Chapter 1 Mid-Chapter Mixed Review

1. The statement is false. Thex-intercept of a line that passes through the originis (0 ,0).

01003. The statement is false. The line parallel to the y-axis that 0 passes through (5,25) is x = 5.11111112

c)D(

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