[ Full chapters Solution manual ] for College Algebra, 8th Edition Stewart -Instant Download Solution manual

,Solution manual for College Algebra, 8th Edition
James Stewart
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,PROLOGUE: Principles of Problem Solving

distance 1 1
1. Let r be the rate of the descent. We use the formula time  ; the ascent takes h, the descent takes h, and the
rate 15 r
2 1 1 1 1 1
total trip should take  h. Thus we have     0, which is impossible. So the car cannot go
30 15 15 r 15 r
fast enough to average 30 mi/h for the 2­mile trip.

2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price
decreases to 08P, and after another 20% discount, it becomes 08 08P  064P. Since 06P  064P, a 40% discount
is better.

3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since
the first cut produces 4 pieces, we get the formula f n  4  3 n  1, n  1. Since f 142  4  3 141  427, we
see that 142 parallel cuts produce 427 pieces.

4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two
amoebas into the vessel, it will take 60  3  57 minutes for the vessel to be full of amoebas.

5. The statement is false. Here is one particular counterexample:
Player A Player B
First half 1
1 hit in 99 at­bats: average  99 0 hit in 1 at­bat: average  01
Second half 1 hit in 1 at­bat: average  11 98 hits in 99 at­bats: average  98
99
Entire season 2
2 hits in 100 at­bats: average  100 99
99 hits in 100 at­bats: average  100

6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus,
any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup.
Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream cream
mixture being returned to the pitcher of cream. Suppose it is possible to separate
the cream and the coffee, as shown. Then you can see that the coffee going into the coffee

cream occupies the same volume as the cream that was left in the coffee.


Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream
cream 1 coffee y
is added to the coffee cup, the resulting mixture has the following ratios:  and  .
mixture y1 mixture y1
1
So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a
y1
y
spoonful of cream and spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is
y 1
1 y
1  of a spoonful. This is the same as the amount of coffee we added to the cream.
y 1 y1

7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius
by 1 foot, the new radius is r  1, so the new circumference is 2 r  1. Thus you need 2 r  1  2r  2 extra
feet of ribbon.
1

,2 Principles of Problem Solving

8. The north pole is such a point. And there are others: Consider a point a1 near the south pole such that the parallel passing
through a1 forms a circle C1 with circumference exactly one mile. Any point P1 exactly one mile north of the circle C1
along a meridian is a point satisfying the conditions in the problem: starting at P1 she walks one mile south to the point a1
on the circle C1 , then one mile east along C1 returning to the point a1 , then north for one mile to P1 . That’s not all. If a
point a2 (or a3 , a4 , a5 ,   ) is chosen near the south pole so that the parallel passing through it forms a circle C2 (C3 , C4 ,
C5 ,   ) with a circumference of exactly 12 mile ( 13 mi, 14 mi, 15 mi,   ), then the point P2 (P3 , P4 , P5 ,   ) one mile north
of a2 (a3 , a4 , a5 ,   ) along a meridian satisfies the conditions of the problem: she walks one mile south from P2 (P3 , P4 ,
P5 ,   ) arriving at a2 ( a3 , a4 , a5 ,   ) along the circle C2 (C3 , C4 , C5 ,   ), walks east along the circle for one mile thus
traversing the circle twice (three times, four times, five times,   ) returning to a2 (a3 , a4 , a5 ,   ), and then walks north one
mile to P2 ( P3 , P4 , P5 ,   ).

,CHAPTER P PREREQUISITES 1
P.1 Modeling the Real World with Algebra 1
P.2 Real Numbers 2
P.3 Integer Exponents and Scientific Notation 7
P.4 Rational Exponents and Radicals 12
P.5 Algebraic Expressions 16
P.6 Factoring 19
P.7 Rational Expressions 24
P.8 Solving Basic Equations 31
P.9 Modeling with Equations 36
Chapter P Review 42
Chapter P Test 48
¥ FOCUS ON MODELING: Making Optimal Decisions 51

,P PREREQUISITES

P.1 MODELING THE REAL WORLD WITH ALGEBRA
1. Using this model, we find that 15 cars have W  4 15  60 wheels. To find the number of cars that have a total of
W
W wheels, we write W  4X  X  .If the cars in a parking lot have a total of 124 wheels, we find that there are
4
X  124
4  31 cars in the lot.
2. If each gallon of gas costs $350, then x gallons of gas costs $35x. Thus, C  35x. We find that 12 gallons of gas would
cost C  35 12  $42.
3. If x  $120 and T  006x, then T  006 120  72. The sales tax is $720.
4. If x  62,000 and T  0005x, then T  0005 62,000  310. The wage tax is $310.
5. If   70, t  35, and d  t, then d  70  35  245. The car has traveled 245 miles.
 
6. V  r 2 h   32 5  45  1414 in3
N 240
7. (a) M    30 miles/gallon 8. (a) T  70  0003h  70  0003 1500  655 F
G 8
175 175 (b) 64  70  0003h  0003h  6  h  2000 ft
(b) 25  G  7 gallons
G  25   
9. (a) V  95S  95 4 km3  38 km3 10. (a) P  006s 3  006 123  1037 hp

(b) 19 km3  95S  S  2 km3 (b) 75  006s 3  s 3  125 so s  5 knots

11. (a) (b) We know that P  30 and we want to find d, so we solve the
Depth (ft) Pressure (lb/in2 ) equation 30  147  045d  153  045d 
0 045 0  147  147 153
d  340. Thus, if the pressure is 30 lb/in2 , the depth
10 045 10  147  192 045
20 045 20  147  237 is 34 ft.

30 045 30  147  282
40 045 40  147  327
50 045 50  147  372
60 045 60  147  417

12. (a) (b) We solve the equation 40x  120,000 
Population Water use (gal) 120,000
x  3000. Thus, the population is about 3000.
0 0 40
1000 40 1000  40,000
2000 40 2000  80,000
3000 40 3000  120,000
4000 40 4000  160,000
5000 40 5000  200,000
13. The number N of cents in q quarters is N  25q.
ab
14. The average A of two numbers, a and b, is A  .
2
1

,2 CHAPTER P Prerequisites

15. The cost C of purchasing x gallons of gas at $350 a gallon is C  35x.
16. The amount T of a 15% tip on a restaurant bill of x dollars is T  015x.
17. The distance d in miles that a car travels in t hours at 60 mi/h is d  60t.
d
18. The speed r of a boat that travels d miles in 3 hours is r  .
3
19. (a) $12  3 $1  $12  $3  $15
(b) The cost C, in dollars, of a pizza with n toppings is C  12  n.
(c) Using the model C  12  n with C  16, we get 16  12  n  n  4. So the pizza has four toppings.
20. (a) 3 30  280 010  90  28  $118
       
daily days cost miles
(b) The cost is    , so C  30n  01m.
rental rented per mile driven
(c) We have C  140 and n  3. Substituting, we get 140  30 3  01m  140  90  01m  50  01m 
m  500. So the rental was driven 500 miles.
21. (a) (i) For an all­electric car, the energy cost of driving x miles is Ce  004x.
(ii) For an average gasoline powered car, the energy cost of driving x miles is C g  012x.
(b) (i) The cost of driving 10,000 miles with an all­electric car is Ce  004 10,000  $400.
(ii) The cost of driving 10,000 miles with a gasoline powered car is C g  012 10,000  $1200.
22. (a) If the width is 20, then the length is 40, so the volume is 20  20  40  16,000 in3 .
(b) In terms of width, V  x  x  2x  2x 3 .
4a  3b  2c  1d  0 f 4a  3b  2c  d
23. (a) The GPA is  .
abcd  f abcd  f
(b) Using a  2  3  6, b  4, c  3  3  9, and d  f  0 in the formula from part (a), we find the GPA to be
463429 54
  284.
649 19


P.2 THE REAL NUMBERS
1. (a) The natural numbers are 1 2 3   .
(b) The numbers     3 2 1 0 are integers but not natural numbers.
p
(c) Any irreducible fraction with q  1 is rational but is not an integer. Examples: 32 ,  12
5 , 1729 .
23
q
p  
(d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e.
q
2. (a) ab  ba; Commutative Property of Multiplication
(b) a  b  c  a  b  c; Associative Property of Addition
(c) a b  c  ab  ac; Distributive Property
3. (a) In set­builder notation: x  3  x  5 (c) As a graph:
_3 5
(b) In interval notation: 3 5
4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.
5. The distance between a and b on the real line is d a b  b  a. So the distance between 5 and 2 is 2  5  7.
6. (a) If a  b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many
ba
numbers—including, for example a  n for every positive n. (If an interval extends to infinity in either or both
2
directions, then it obviously contains infinitely many numbers.)

, SECTION P.2 The Real Numbers 3

(b) No, because 5 6 does not include 5.
7. (a) No: a  b   b  a  b  a in general.
(b) No; by the Distributive Property, 2 a  5  2a  2 5  2a  10  2a  10.
8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive.
(b) Yes, b  a  a  b.

9. (a) Natural number: 100 10. (a) Natural numbers: 2, 9  3, 10

(b) Integers: 0, 100, 8 (b) Integers: 2,  100
2  50, 9  3, 10
   
(c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8 (c) Rational numbers: 45  92 , 13 , 16666     53 ,
 
(d) Irrational numbers: 7,  2,  100
2 , 9  3, 10
 
(d) Irrational numbers: 2, 314

11. Commutative Property of addition 12. Commutative Property of multiplication

13. Associative Property of addition 14. Distributive Property

15. Distributive Property 16. Distributive Property

17. Commutative Property of multiplication 18. Distributive Property

19. x  3  3  x 20. 7 3x  7  3 x

21. 4 A  B  4A  4B 22. 5x  5y  5 x  y

23. 2 x  y  2x  2y 24. a  b 5  5a  5b
 
25. 5 2x y  5  2 x y  10x y 26. 43 6y  43 6 y  8y

27.  52 2x  4y   52 2x  52 4y  5x  10y 28. 3a b  c  2d  3ab  3ac  6ad

29. (a) 23  57  14 15 29
21  21  21 30. (a) 25  38  16 15 1
40  40  40
5  3  10  9  1
(b) 12 (b) 32  58  16  36 15 4 25
8 24 24 24 24  24  24  24
  2
2
31. (a) 23 6  32  23  6  23  32  4  1  3 32. (a) 2  3  2  32  23  12  3  13  93  13  83
      2
3
(b) 3  14 1  45  12 4  4
1 5  4  13  1  13
5 5 4 5 20 2  1 2  1 2  1
(b) 15 23  51 21  51 21  10 45 9
10  12  3  3
10  15 10  5 10  5

33. (a) 2  3  6 and 2  72  7, so 3  72 34. (a) 3  23  2 and 3  067  201, so 23  067

(b) 6  7 (b) 23  067

(c) 35  72 (c) 06  06

35. (a) False 36. (a) False: 3  173205  17325.

(b) True (b) False

37. (a) True (b) False 38. (a) True (b) True

,4 CHAPTER P Prerequisites

39. (a) x  0 (b) t  4 40. (a) y  0 (b) z  3

(c) a   (d) 5  x  13 (c) b  8 (d) 0    17

(e) 3  p  5 (e) y    2

41. (a) A  B  1 2 3 4 5 6 7 8 42. (a) B  C  2 4 6 7 8 9 10

(b) A  B  2 4 6 (b) B  C  8

43. (a) A  C  1 2 3 4 5 6 7 8 9 10 44. (a) A  B  C  1 2 3 4 5 6 7 8 9 10

(b) A  C  7 (b) A  B  C  

45. (a) B  C  x  x  5 46. (a) A  C  x  1  x  5

(b) B  C  x  1  x  4 (b) A  B  x  2  x  4

47. 3 0  x  3  x  0 48. 2 8]  x  2  x  8


_3 0 2 8

   
49. [2 8  x  2  x  8 50. 6  12  x  6  x   12


2 8 1
_6 _ _2


51. [2   x  x  2 52.  1  x  x  1


2 1


53. x  1  x   1] 54. 1  x  2  x  [1 2]


1 1 2


55. 2  x  1  x  2 1] 56. x  5  x  [5 


_2 1 _5


57. x  1  x  1  58. 5  x  2  x  5 2


_1 _5 2


59. (a) [3 5] (b) 3 5] (c) 3  60. (a) [0 2 (b) 2 0] (c)  0]


61. 2 0  1 1  2 1 62. 2 0  1   1 0


_2 1 _1 0

, SECTION P.2 The Real Numbers 5

63. [4 6]  [0 8  [0 6] 64. [4 6]  [0 8  [4 8


0 6 _4 8


65.  4  4  66.  6]  2 10  2 6]


_4 4 2 6


67. (a) 50  50 68. (a) 2  8  6  6

(b) 13  13 (b) 8  2  8  2  6  6

69. (a) 6  4  6  4  2  2 70. (a) 2  12  2  12  10  10

(b) 1
1  1  1
1 (b) 1  1  1  1  1  1  1  0  1
   
   1 1
71. (a) 2  6  12  12 72. (a)  6
24    4   4
      
     5
(b)   13 15  5  5 (b)  712
127    5   1  1

73. 2  3  5  5 74. 25  15  4  4
        
7 1    49  5    54    18   18
75. (a) 17  2  15 76. (a)  15   21   105 105   105   35  35

(b) 21  3  21  3  24  24 (b) 38  57  38  57  19  19.
     
 3   12 55   67  67
(c)  10  11
8    40  40    40   40 (c) 26  18  26  18  08  08.

77. (a) Let x  0777   . So 10x  77777     x  07777     9x  7. Thus, x  79 .
(b) Let x  02888   . So 100x  288888     10x  28888     90x  26. Thus, x  26 13
90  45 .
(c) Let x  0575757   . So 100x  575757     x  05757     99x  57. Thus, x  57 19
99  33 .

78. (a) Let x  52323   . So 100x  5232323     1x  52323     99x  518. Thus, x  518
99 .
(b) Let x  13777   . So 100x  1377777     10x  137777     90x  124. Thus, x  124 62
90  45 .
(c) Let x  213535   . So 1000x  21353535     10x  213535     990x  2114. Thus, x  2114 1057
990  495 .
  
  

79.   3, so   3    3. 80. 2  1, so 1  2  2  1.

81. a  b, so a  b   a  b  b  a. 82. a  b  a  b  a  b  b  a  2b
83. (a) a is negative because a is positive.
(b) bc is positive because the product of two negative numbers is positive.
(c) a  ba  b is positive because it is the sum of two positive numbers.
(d) ab  ac is negative: each summand is the product of a positive number and a negative number, and the sum of two
negative numbers is negative.
84. (a) b is positive because b is negative.
(b) a  bc is positive because it is the sum of two positive numbers.
(c) c  a  c  a is negative because c and a are both negative.
(d) ab2 is positive because both a and b2 are positive.
85. Distributive Property