Algebra and Trigonometry

692 Pages·6.154 MB·English

bitt_464818_ttl.qxd 2/8/07 11:53 AM Page 1
I ’
NSTRUCTOR S
S M
OLUTIONS ANUAL
J A. P
UDITH ENNA
Indiana University Purdue University Indianapolis
A T
LGEBRA AND RIGONOMETRY
T E
HIRD DITION
AND
P
RECALCULUS
T E
HIRD DITION
Judith A. Beecher
Indiana University Purdue University Indianapolis
Judith A. Penna
Indiana University Purdue University Indianapolis
Marvin L. Bittinger
Indiana University Purdue University Indianapolis
bitt_464818_ttl.qxd 2/8/07 11:53 AM Page 2
This work is protected by United States copyright laws and is provided solely
for the use of instructors in teaching their courses and assessing student
learning.Dissemination or sale of any part of this work (including on the
World Wide Web) will destroy the integrity of the work and is not permit-
ted.The work and materials from it should never be made available to
students except by instructors using the accompanying text in their
classes.All recipients of this work are expected to abide by these
restrictions and to honor the intended pedagogical purposes and the needs of
other instructors who rely on these materials.
Reproduced by Pearson Addison-Wesley from QuarkXPress®files.
Copyright ©2008 PearsonEducation, Inc.
Publishing as Pearson Addison-Wesley, 75 ArlingtonStreet, Boston,MA 02116.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or trans-
mitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, with-
out the prior written permission of the publisher. Printed in the United States of America.
ISBN-13: 978-0-321-46481-1
ISBN-10: 0-321-46481-8
1 2 3 4 5 6 OPM 10 09 08 07
Contents
Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 117
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 185
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 329
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 379
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 419
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 471
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 569
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 653
Chapter R
Basic Concepts of Algebra
16. ( 5, )
− ∞
Exercise Set R.1
1. Wholenumbers: √38,0,9,√25 (√38=2, √25=5) !5 0
2. Integers: 12, √38,0,9,√25 (√38=2, √25=5) 17. This interval is of unlimited extent in the positive direc-
− tion, and the endpoint 3.8 is not included. Interval nota-
3. Irrationalnumbers: √7,5.242242224..., √14, √55, √34 tionis(3.8, ).
− ∞
(Although there is a pattern in 5.242242224..., there is
norepeatingblockofdigits.)
0 3.8
4. Naturalnumbers: √38,9,√25
18. [√3, )
∞
7 2
5. Rational numbers: 12, 5.3, , √38, 0, 1.96, 9, 4 , [
5 − −3 − 3 -3 -2 -1 0 1 3 2 3
√25,
7
19. x7<x ,or xx>7 .
{ | } { | }
6. Realnumbers: Allofthem Thisintervalisofunlimitedextentinthepositivedirection
7 2 5 and the endpoint 7 is not included. Interval notation is
7. Rationalnumbersbutnotintegers: 5.3, , 1.96,4 , (7, ).
−3 − 3 7 ∞
8. Integersbutnotwholenumbers: 12
−
0 7
9. Integersbutnotrationalnumbers: 12,0
−
20. ( , 3)
7 −∞ −
10. Realnumbersbutnotintegers: √7,5.3, ,
−3
2 5
5.242242224..., √14, √55, 1.96,4 , √34, !3 0
− − 3 7
21. Theendpoints0and5arenotincludedintheinterval,so
11. This is a closed interval, so we use brackets. Interval no-
weuseparentheses. Intervalnotationis(0,5).
tationis[ 3,3].
−
22. [ 1,2]
−
!5!4!3!2!1 0 1 2 3 4 5
23. The endpoint 9 is included in the interval, so we use a
−
12. ( 4,4) bracket before the 9. The endpoint 4 is not included,
− so we use a parenth−esis after the 4. −Interval notation is
−
[ 9, 4).
− −
!4 0 4
24. ( 9, 5]
13. This is a half-open interval. We use a bracket on the − −
left and a parenthesis on the right. Interval notation is 25. Both endpoints are included in the interval, so we use
[ 4, 1). brackets. Intervalnotationis[x,x+h].
− −
26. (x,x+h]
!4 !1 0
27. Theendpointpisnotincludedintheinterval,soweusea
14. (1,6]
parenthesis before the p. The interval is of unlimited ex-
tentinthepositivedirection,soweusetheinfinitysymbol
. Intervalnotationis(p, ).
0 1 6 ∞ ∞
15. This interval is of unlimited extent in the negative direc- 28. ( ,q]
−∞
tion,andtheendpoint 2isincluded. Intervalnotationis
− 29. Since 6 is an element of the set of natural numbers, the
( , 2].
−∞ − statementistrue.
30. True
!2 0
2 Chapter R: Basic Concepts of Algebra
1
31. Since3.2isnotanelementofthesetofintegers,thestate- 57. Thesentence8 =1illustratesthemultiplicativeinverse
mentisfalse. · 8
property.
32. True
58. Distributiveproperty
11
33. Since is an element of the set of rational numbers, 59. Thedistanceof 7.1from0is7.1,so 7.1 =7.1.
− 5 − |− |
thestatementistrue.
60. 86.2
34. False
61. Thedistanceof347from0is347,so 347 =347.
| |
35. Since √11 is an element of the set of real numbers, the
62. 54
statementisfalse.
63. Thedistanceof √97from0is√97,so √97 =√97.
36. False − |− |
12
37. Since 24 is an element of the set of whole numbers, the 64.
19
statementisfalse.
65. Thedistanceof0from0is0,so 0 =0.
38. True | |
66. 15
39. Since1.089isnotanelementofthesetofirrationalnum-
5 5 5 5
bers,thestatementistrue. 67. Thedistanceof from0is ,so = .
4 4 4 4
! !
40. True ! !
68. √3 ! !
41. Since every whole number is an integer, the statement is
69. 5 6 = 11 =11,or
true. |− − | |− |
6 ( 5) = 6+5 = 11 =11
| − − | | | | |
42. False
70. 0 ( 2.5) = 2.5 =2.5,or
| − − | | |
43. Since every rational number is a real number, the state-
2.5 0 = 2.5 =2.5
mentistrue. |− − | |− |
71. 2 ( 8) = 2+8 = 6 =6,or
44. True |− − − | |− | | |
8 ( 2) = 8+2 = 6 =6
|− − − | |− | |− |
45. Since there are real numbers that are not integers, the
15 23 45 46 1 1
statementisfalse. 72. = = = ,or
8 − 12 24 − 24 − 24 24
!23 15! !46 45! ! 1 ! 1
46. False !! !!=!! !!=!! =!!
12 − 8 24 − 24 24 24
47. Thesentence6 x=x 6illustratesthecommutativeprop- !! !! !! !! !! !!
ertyofmultipli·cation.· 73. |!12.1−6.!7|=!|5.4|=5!.4,o!r !
6.7 12.1 = 5.4 =5.4
| − | |− |
48. Associativepropertyofaddition
74. 3 ( 14) = 3+14 = 11 =11,or
|− − − | |− | | |
49. The sentence 3 1 = 3 illustrates the multiplicative
14 ( 3) = 14+3 = 11 =11
− · −
identityproperty. |− − − | |− | |− |
3 15 6 15 21 21
50. Commutativepropertyofaddition 75. = = = , or
− 4 − 8 − 8 − 8 − 8 8
! ! ! ! ! !
! ! ! ! ! !
51. Thesentence5(ab)=(5a)billustratestheassociativeprop- !15 !3 ! 15 3! ! 15 !6 21 21
! ! != + ! =! +! = =
ertyofmultiplication. 8 − − 4 8 4 8 8 8 8
! " #! ! ! ! ! ! !
! ! ! ! ! ! ! !
52. Distributiveproperty 76. |!!− 3.4−10.2|!!=|−!! 13.6|=!! 13!!.6, or !! !! !!
10.2 ( 3.4) = 10.2+3.4 = 13.6 =13.6
53. Thesentence2(a+b)=(a+b)2illustratesthecommutative | − − | | | | |
propertyofmultiplication. 77. 7 0 = 7 =7, or
|− − | |− |
0 ( 7) = 0+7 = 7 =7
54. Additiveinverseproperty | − − | | | | |
78. 3 19 = 16 =16, or
55. Thesentence 6(m+n)= 6(n+m)illustratesthecom- | − | |− |
mutativeprop−ertyofadditi−on. 19 3 = 16 =16
| − | | |
56. Additiveidentityproperty 79. Provideanexample. Forinstance,16 (8 2)=16 4=4,
÷ ÷ ÷
but(16 8) 2=2 2=1.
÷ ÷ ÷
Exercise Set R.2 3
80. √aisarationalnumberwhenaisthesquareofarational 9. x9 x0=x9+0=x9
·
number. That is, √a is a rational number if there is a
rationalnumbercsuchthata=c2. 10. a0 a4=a0+4=a4
·
81. Answersmayvary. Onesuchnumberis 11. 58 5−6=58+(−6)=52,or25
·
0.124124412444....
1
12. 62 6 7=62+( 7)=6 5,or
82. Answers may vary. Since √2.01 1.418 and √2 · − − − 65
− ≈ − − ≈
1.414,onesuchnumberis 1.415. 13. m 5 m5=m 5+5=m0=1
− − − −
·
1
83. Answersmayvary. Since =0.0099and 14. n9 n−9=n9+(−9)=n0=1
−101 ·
1 1
= 0.01,onesuchnumberis 0.00999. 15. y3 y 7=y3+( 7)=y 4,or
−100 − − · − − − y4
84. Answersmayvary. Onesuchnumberis√5.995. 16. b 4 b12=b 4+12=b8
− −
·
85. Since12+32=10,thehypotenuseofarighttrianglewith 1
17. 73 7 5 7=73+( 5)+1=7 1,or
legsoflengths1unitand3unitshasalengthof√10units. · − · − − 7
!!!!c!!!!1 cc22 ==1102+32 1198.. 23x63·3−3x52·3=42=336+x(−3+5)2+=4=6x355
· · ·
3 c=√10
20. 3y4 4y3=3 4 y4+3=12y7
· · ·
21. ( 3a 5)(5a 7)= 3 5 a 5+( 7)= 15a 12,or
Exercise Set R.2 − − − − · · − − − −
15
−a12
1 1
1. 3−7= 37 a−m= am, a%=0 22. ( 6b−4)(2b−7)= 6 2 b−4+(−7)= 12b−11,or
" # − − · · −
12
1
2. =(5.9)4 −b11
(5.9) 4
−
23. (5a2b)(3a 3b4)=5 3 a2+( 3) b1+4=15a 1b5,or
− − −
3. Observe that each exponent is negative. We move each · · ·
15b5
factortotheothersideofthefractionbarandchangethe
a
signofeachexponent.
x−5 = y4 24. (4xy2)(3x−4y5)=4·3·x1+(−4)·y2+5=12x−3y7,or
y−4 x5 12y7
x3
4. Observe that each exponent is negative. We move each
factortotheothersideofthefractionbarandchangethe 25. (6x 3y5)( 7x2y 9)=6( 7)x 3+2y5+( 9)=
− − − −
signofeachexponent. − −
42
a−2 = b8 −42x−1y−4,or−xy4
b 8 a2
−
26. (8ab7)( 7a 5b2)=8( 7)a1+( 5)b7+2=
5. Observe that each exponent is negative. We move each − − − −
factortotheothersideofthefractionbarandchangethe 56b9
56a 4b9,or
signofeachexponent. − − − a4
m−1n−12 = t6 ,or t6 27. (2x)3(3x)2=23x3 32x2=8 9 x3+2=72x5
t 6 m1n12 mn12 · · ·
−
28. (4y)2(3y)3=16y2 27y3=432y5
6. Observe that each exponent is negative. We move each ·
factortotheothersideofthefractionbarandchangethe 29. ( 2n)3(5n)2=( 2)3n3 52n2= 8 25 n3+2=
signofeachexponent. − − · − · ·
200n5
x 9y 17 z11 −
− − =
z−11 x9y17 30. (2x)5(3x)2=25x5 32x2=32 9 x5+2=288x7
· · ·
7. 18◦=1 (Foranynonzerorealnumber,a0=1.) 31. bb4307 =b40−37=b3
4 0
8. "− 3# =1 32. aa3392 =a39−32=a7
4 Chapter R: Basic Concepts of Algebra
33. xx−165 =x−5−16=x−21,or x121 51. Convert405,000toscientificnotation.
We want the decimal point to be positioned between the
y 24 1 4 and the first 0, so we move it 5 places to the left. Since
34. y−21 =y−24−(−21)=y−24+21=y−3,or y3 405,000isgreaterthan10,theexponentmustbepositive.
−
405,000=4.05 105
x2y 2 x3 ×
35. x 1−y =x2−(−1)y−2−1=x3y−3,or y3 52. Position the decimal point 6 places to the left, between
−
the 1 and the 6. Since 1,670,000 is greater than 10, the
36. xx3−y1−y23 =x3−(−1)y−3−2=x4y−5,or xy54 exponent1,m67u0s,t0b0e0p=os1i.t6iv7e. 106
×
32x 4y3 32 8x
37. 4x−5y8 = 4 x−4−(−5)y3−8=8xy−5,or y5 53. Convert0.00000039toscientificnotation.
− We want the decimal point to be positioned between the
20a5b 2 20 4b 3 and the 9, so we move it 7 places to the right. Since
38. 5a7b−3 = 5 a5−7b−2−(−3)=4a−2b,or a2 0.00000039 is a number between 0 and 1, the exponent
−
mustbenegative.
39. (2ab2)3=23a3(b2)3=23a3b23=8a3b6
· 0.00000039=3.9 10 7
−
×
40. (4xy3)2=42x2(y3)2=16x2y6
54. Position the decimal point 4 places to the right, between
41. ( 2x3)5=( 2)5(x3)5=( 2)5x35= 32x15 the9andthe2. Since0.00092isanumberbetween0and
·
− − − − 1,theexponentmustbenegative.
42. ( 3x2)4=( 3)4(x2)4=81x8 0.00092=9.2 10 4
− − × −
43. ( 5c−1d−2)−2=( 5)−2c−1(−2)d−2(−2)= 55. Convert 234,600,000,000 to scientific notation. We want
− −
c2d4 c2d4 the decimal point to be positioned between the 2 and the
( 5)2 = 25 3,sowemoveit11placestotheleft. Since234,600,000,000
− isgreaterthan10,theexponentmustbepositive.
44. ( 4x−5z−2)−3=( 4)−3(x−5)−3(z−2)−3= 234,600,000,000=2.346 1011
− − ×
x15z6 x15z6
= 56. Positionthedecimalpoint9placestotheleft,betweenthe
( 4)3 64
− − 8 and the 9. Since 8,904,000,000 is greater than 10, the
45. (3m4)3(2m 5)4=33m12 24m 20= exponentmustbepositive.
− −
·
432 8,904,000,000=8.904 109
27 16m12+( 20)=432m 8,or ×
· − − m8
57. Convert 0.00104 to scientific notation. We want the deci-
46. (4n−1)2(2n3)3=42n−2 23n9=16 8 n−2+9= malpointtobepositionedbetweenthe1andthelast0,so
· · ·
128n7 wemoveit3placestotheright. Since0.00104isanumber
between0and1,theexponentmustbenegative.
47. 2x−3y7 3= (2x−3y7)3 = 23x−9y21 = 0.00104=1.04 10−3
z 1 (z 1)3 z 3 ×
" − # − −
8x 9y21 8y21z3 58. Position the decimal point 9 places to the right, between
− , or the5andthe1. Since0.00000000514isanumberbetween
z 3 x9
− 0and1,theexponentmustbenegative.
48. 3x5y−8 4= 81x20y−32, or 81x20z8 0.00000000514=5.14×10−9
z 2 z 8 y32
" − # −
59. Convert0.000016toscientificnotation.
49. "2142aa160bb−−38cc57#−5=(2a4b−5c2)−5=2−5a−20b25c−10, W1 aenwdanthteth6e, dsoecwimeaml povoeinittt5o pbleacpeossittoiontheedrbigethwt.eenSinthcee
b25 0.000016isanumberbetween0and1,theexponentmust
or 32a20c10 benegative.
0.000016=1.6 10 5
50. 12255pp182qq6−r1−41r522 −4=(5p4q−20r37)−4= 60. Position the decimal po×int 1−2 places to the left, between
$ %
q80 the ones. Since 1,137,000,000,000 is greater than 10, the
5 4p 16q80r 148,or
− − − 625p16r148 exponentmustbepositive.
1,137,000,000,000=1.137 1012
×
Exercise Set R.2 5
61. Convert8.3×10−5 todecimalnotation. 73. (2.6×10−18)(8.5×107)
Theexponentisnegative,sothenumberisbetween0and =(2.6 8.5) (10 18 107)
−
× × ×
1. Wemovethedecimalpoint5placestotheleft. =22.1 10 11 Thisisnotscientificnotation.
−
×
8.3×10−5=0.000083 =(2.21×10)×10−11
62. The exponent is positive, so the number is greater than =2.21 10−10
×
10. Wemovethedecimalpoint6placestotheright.
74. (6.4 1012)(3.7 10 5)=23.68 107
−
4.1 106=4,100,000 × × ×
× =(2.368 10) 107
× ×
63. Convert2.07 107 todecimalnotation. =2.368 108
× ×
The exponent is positive, so the number is greater than
6.4 10 7 6.4 10 7
10. Wemovethedecimalpoint7placestotheright. 75. × − = −
8.0 106 8.0 × 106
2.07×107=20,700,000 × =0.8 10−13 Thisisnotscientific
×
notation.
64. Theexponentisnegative,sothenumberisbetween0and
=(8 10 1) 10 13
1. Wemovethedecimalpoint6placestotheleft. × − × −
=8 10 14 Writingscientific
3.15×10−6=0.00000315 × − notation
65. Convert3.496 1010 todecimalnotation.
The exponent×is positive, so the number is greater than 76. 1.1×10−40 =0.55 1031
2.0 10 71 ×
10. Wemovethedecimalpoint10placestotheright. × −
=(5.5 10 1) 1031
−
3.496 1010=34,960,000,000 × ×
× =5.5 1030
×
66. The exponent is positive, so the number is greater than
1.8 10 3
10. Wemovethedecimalpoint11placestotheright. 77. × −
7.2 10 9
8.409 1011=840,900,000,000 × −
× = 1.8 10−3
67. Convert5.41 10−8 todecimalnotation. 7.2 × 10−9
× =0.25 106 Thisisnotscientificnotation.
Theexponentisnegative,sothenumberisbetween0and ×
1. Wemovethedecimalpoint8placestotheleft. =(2.5 10−1) 106
× ×
5.41 10 8=0.0000000541 =2.5 105
× − ×
68. T1.hWeeexmpoonveentthiesdneecgiamtiavle,psoointth1e0npulmacbeesrtiosbtheetwleefetn. 0and 78. 51..23×1100140 =0.25×10−6
×
6.27 10−10=0.000000000627 =(2.5×10−1)×10−6
× =2.5 10 7
−
69. Convert2.319 108 todecimalnotation. ×
×
79. We multiply the number of AUs from Earth to Pluto by
The exponent is positive, so the number is greater than
thenumberofmilesin1AU.
10. Wemovethedecimalpoint8placestotheright.
39 93million
2.319 108=231,900,000 ×
× =39 93,000,000
×
70. Theexponentisnegative,sothenumberisbetween0and =(3.9 10) (9.3 107) Writingscientificnotation
1. Wemovethedecimalpoint24placestotheleft. × × ×
=(3.9 9.3) (10 107)
1.67 10 24 g=0.00000000000000000000000167g × × ×
× − =36.27 108
×
71. (3.1 105)(4.5 10 3) =(3.627 10) 108
× × − × ×
=(3.1 4.5) (105 10 3) =3.627 109
× × × − ×
=13.95 102 Thisisnotscientificnotation. ThedistancefromEarthtoPlutoisabout3.627 109 mi.
× ×
=(1.395 10) 102
× × 80. 3.26 5.88 1012
=1.395 103 Writingscientificnotation × ×
× =19.1688 1012
×
72. (9.1 10−17)(8.2 103)=74.62 10−14 =(1.91688 10) 1012
× × × × ×
=(7.462 10) 10−14 =1.91688 1013 mi
× × ×
=7.462 10 13
−
×
6 Chapter R: Basic Concepts of Algebra
81. We multiply the diameter, in nanometers, by the number 87. 3 2 4 22+6(3 1)
· − · −
ofmetersin1nanometer. =3 2 4 22+6 2 Workinginside
360 0.000000001 · − · · parentheses
×
=(3.6 102) 10 9 =3 2 4 4+6 2 Evaluating22
× × − · − · ·
=3.6 (102 10 9) =6 16+12 Multiplying
× × − −
=3.6 10 7 = 10+12 Addinginorder
× − −
Thediameterofthewireis3.6 10 7 m. =2 fromlefttoright
−
×
88. 3[(2+4 22) 6(3 1)]
82. 30m36i5llion = 33.605×110062 =3[(2+4··4)−−6·2]−
×
8.2192 104 =3[(2+16) 6 2]
≈ × − ·
Onaverage,about8.2192 104 piecesofluggagewerelost =3[18 6 2]
× − ·
eachdayoftheyear. =3[18 12]
−
=3[6]
83. The average cost per mile is the total cost divided by the
numberofmiles. =18
$210 106 89. 16 4 4 2 256
× ÷ · ÷ ·
17.6 =4 4 2 256 Multiplyinganddividing
· ÷ ·
$210 106 inorderfromlefttoright
= ×
1.76 10 =16 2 256
× ÷ ·
$119 105 =8 256
≈ × ·
($1.19 102) 105 =2048
≈ × ×
$1.19 107
≈ × 90. 26 2−3 210 2−8
Theaveragecostpermilewasabout$1.19 107. · ÷ ÷
× =23 210 2−8
÷ ÷
412 4.12 102 =2−7 2−8
84. = × ÷
9,600,000 9.6 106 =2
×
≈0.43×10−4 91. 4(8−6)2−4·3+2· 8
(4.3 10−1) 10−4 31+190
≈ × ×
4.3 10 5 squaremiles 4 22 4 3+2 8
≈ × − = · − · · Calculatinginthe
3+1
numeratorandin
85. Firstfindthenumberofsecondsin1hour:
# thedenominator
60min 60sec
1hour=1"hr # =3600sec 4 4 4 3+2 8
× 1"hr × 1min = · − · ·
4
The number of disintegrations produced in 1 hour is the
16 12+16
numberofdisintegrationspersecondtimesthenumberof = −
4
secondsin1hour.
4+16
37billion 3600 =
× 4
=37,000,000,000 3600
× 20
=
=3.7 1010 3.6 103 Writingscientific 4
× × ×
notation =5
=(3.7 3.6) (1010 103)
× × × [4(8 6)2+4](3 2 8)
=13.32 1013 Multiplying 92. − − ·
× 22(23+5)
=(1.332 10) 1013 [4 22+4](3 16)
× × = · −
=1.332 1014 22(8+5)
×
Onegramofradiumproduces1.332 1014 disintegrations = [4·4+4](−13)
in1hour. × 22 13
·
[16+4]( 13)
= −
86. 2π 93,000,000 4 13
× ·
=2π 9.3 107 20( 13)
× × = −
58 107 52
≈ × 260
5.8 10) 107 = −
≈ × × 52
5.8 108 mi
≈ × = 5
−
Exercise Set R.2 7
93. Sinceinterestiscompoundedsemiannually,n=2. Substi- 100. t=65 25=40
−
tute $2125 for P, 6.2% or 0.062 for i, 2 for n, and 5 for t 0.04 1240
1+ · 1
inthecompoundinterestformula. S =$100 12 − $118,196.13
i nt $ 0.0%4 ≈
A=P 1+
n 12
 
=$2$125 1+% 0.062 2·5 Substituting 101. Substitute$120,000forS,0.06forr,and18fortandsolve
2 forP.
=$2125($1+0.031)2%·5 Dividing r 12t
1+ · 1
=$2125(1.031)2·5 Adding S =P 12r −
$ % 
=$2125(1.031)10 Multiplying2and5
12
 
$2125(1.357021264) Evaluatingthe 0.06 1218
≈ exponentialexpression 1+ · 1
$120,000=P 12 −
$2883.670185 Multiplying $ 0.0%6 
≈
$2883.67 Roundingtothenearestcent 12
≈  
(1.005)216 1
0.054 27 $120,000=P −
94. A=$9550 1+ · $13,867.23 0.05
2 ≈ * +
$120,000 P(387.3532)
$ % ≈
95. Sinceinterestiscompoundedquarterly,n=4. Substitute $309.79 P
$6700forP,4.5%or0.045fori,4forn,and6fortinthe ≈
compoundinterestformula. 102. t=70 30=40
−
i nt 0.045 1240
A=P 1+ 1+ · 1
n $200,000=P 12 −
$ % 0.045 46 $ 0.04%5 
=$6700 1+ · Substituting 12
4  
=$6700$(1+0.0112%5)4·6 Dividing P ≈$149.13
=$6700(1.01125)4·6 Adding 103. (xt x3t)2=(x4t)2=x4t·2=x8t
·
=$6700(1.01125)24 Multiplying4and6
104. (xy x y)3=(x0)3=13=1
−
$6700(1.307991226) Evaluatingthe ·
≈
exponentialexpression 105. (ta+x tx a)4=(t2x)4=t2x4=t8x
− ·
$8763.541217 Multiplying ·
≈
$8763.54 Roundingtothenearestcent 106. (mx−b nx+b)x(mbn−b)x
≈ ·
=(mx2 bxnx2+bx)(mbxn bx)
− −
0.058 49
96. A=$4875 1+ · $8185.56 =mx2nx2
4 ≈
$ %
97. Yes;findtheresultswithparenthesesandwithoutthem. (3xayb)3 2 27x3ay3b 2
107. =
( 3xayb)2 9x2ay2b
4·25÷(10−5)=4·25÷5=100÷5=20, , − - =,3xayb 2 -
but4 25 10 5=100 10 5=10 5=5.
· ÷ 1− ÷ − − 1 =.9x2ay2b/
98. No; x 2, or is positive for all x < 0 and x 1, or
− x2 − x xr 2 x2r 2 3
is negative for all x < 0. Partial confirmation can be ob- 108. − −
yt y4t
tainedbyinspectingthegraphsofy1=x−2 andy2=x−1 ,$ % $ % -
forx<0. = x2r x−4r −3
y2t y 8t
−
99. Substitute$250forP,0.05forr and27fortandperform ,$x 2r%$3 %-
theresultingcomputation. = − −
y 6t
−
S =P 1+ 1r2r12·t−1 =$yx168rt, o%rx6ry−18t
$ % 
12
 
0.05 1227
1+ · 1
=$250 12 −
$ 0.0%5 
12
 
$170,797.30
≈