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Full chapters Solution manual for Algebra and Trigonometry, 5th Edition James Stewart [ Instant Download Solution manual ]

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Full chapters Solution manual for Algebra and Trigonometry, 5th Edition James Stewart [ Instant Download Solution manual ]

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,Solution manual for Algebra and Trigonometry, 5th
Edition James Stewart
Notes
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,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

,6 CHAPTER P Prerequisites

86. (a) When L  60, x  8, and y  6, we have L  2 x  y  60  2 8  6  60  28  88. Because 88  108 the
post office will accept this package.
When L  48, x  24, and y  24, we have L  2 x  y  48  2 24  24  48  96  144, and since
144  108, the post office will not accept this package.
(b) If x  y  9, then L  2 9  9  108  L  36  108  L  72. So the length can be as long as 72 in.  6 ft.
m1 m2 m1 m m1n2  m2n1
87. Let x  and y  be rational numbers. Then x  y   2  ,
n1 n2 n1 n2 n1 n2
m m m n  m 2 n1 m m m m
xy 1  2  1 2 , and x  y  1  2  1 2 . This shows that the sum, difference, and product
n1 n2 n1 n2 n1 n2 n1n2
of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily
 
irrational; for example, 2  2  2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;
   
for example, 2   2  0 which is rational.
      
88. 12  2 is irrational. If it were rational, then by Exercise 6(a), the sum 12  2   12  2 would be rational, but
this is not the case.

Similarly, 12  2 is irrational.
(a) Following the hint, suppose that r  t  q, a rational number. Then by Exercise 6(a), the sum of the two rational
numbers r  t and r is rational. But r  t  r  t, which we know to be irrational. This is a contradiction, and
hence our original premise—that r  t is rational—was false.
a
(b) r is a nonzero rational number, so r  for some nonzero integers a and b. Let us assume that rt  q, a rational
b
c a c bc
number. Then by definition, q  for some integers c and d. But then r t  q  t  , whence t  , implying
d b d ad
that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number
and an irrational number is irrational.
89.
x 1 2 10 100 1000
1 1 1 1 1 1
x 2 10 100 1000
As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero.

x 1 05 01 001 0001
1 1 1 1 1 1
x 05  2 01  10 001  100 0001  1000
As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.

90. We can construct the number 2 on the number line by
Ï2
transferring the length of the hypotenuse of a right triangle 1
with legs of length 1 and 1.
 _1 0 1 Ï2 2 3
Similarly, to locate 5, we construct a right triangle with legs
of length 1 and 2. By the Pythagorean Theorem, the length Ï5
  1
of the hypotenuse is 12  22  5. Then transfer the
length of the hypotenuse to the number line. _1 0 1 2 Ï5 3

The square root of any rational number can be located on a
number line in this fashion.
The circle in the second figure in the text has circumference , so if we roll it along a number line one full rotation, we have
found  on the number line. Similarly, any rational multiple of  can be found this way.

, SECTION P.3 Integer Exponents and Scientific Notation 7

a  b  a  b abab
91. (a) Suppose that a  b, so max a b  a and a  b  a  b. Then   a.
2 2
On the other hand, if b  a, then max a b  b and a  b   a  b  b  a. In this case,
a  b  a  b abba
  b.
2 2
If a  b, then a  b  0 and the result is trivial.
a  b  a  b a  b  b  a
(b) If a  b, then min a b  a and a  b  b  a. In this case   a.
2 2
a  b  a  b
Similarly, if b  a, then  b; and if a  b, the result is trivial.
2
92. Answers will vary.
93. (a) Subtraction is not commutative. For example, 5  1  1  5.
(b) Division is not commutative. For example, 5  1  1  5.
(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes,
the result is not the same as if you proceed the other way around.
(d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result.
(e) Washing laundry and drying it are not commutative.
94. (a) If x  2 and y  3, then x  y  2  3  5  5 and x  y  2  3  5.
If x  2 and y  3, then x  y  5  5 and x  y  5.
If x  2 and y  3, then x  y  2  3  1 and x  y  5.
In each case, x  y  x  y and the Triangle Inequality is satisfied.
(b) Case 0: If either x or y is 0, the result is equality, trivially.
 
 xy if x and y are positive 
Case 1: If x and y have the same sign, then x  y   x  y.
  x  y if x and y are negative 
Case 2: If x and y have opposite signs, then suppose without loss of generality that x  0 and y  0. Then
x  y  x  y  x  y.



P.3 INTEGER EXPONENTS AND SCIENTIFIC NOTATION
1. Using exponential notation we can write the product 5  5  5  5  5  5 as 56 .
2. Yes, there is a difference: 54  5 5 5 5  625, while 54   5  5  5  5  625.
3. In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.
4. (a) When we multiply two powers with the same base, we add the exponents. So 34  35  39 .
35
(b) When we divide two powers with the same base, we subtract the exponents. So 2  33 .
3
 2
5. When we raise a power to a new power, we multiply the exponents. So 34  38 .
 1
1 1 1 1
6. (a) 21  (b) 23  (c) 2 (d) 3  23  8
2 8 2 2
7. To move a number raised to a power from numerator to denominator or from denominator to numerator change the sign of
1 1 a 3 1 6a 2
the exponent. So a 2  2 , 2  b2 , 2  3 2 , and 3  6a 2 b3 .
a b b a b b
8. Scientists express very large or very small numbers using scientific notation. In scientific notation, 8,300,000 is 83  106
and 00000327 is 327  105 .

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