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Examen

Solutions for Elementary and Intermediate Algebra, 4th edition by Carson, Jordan

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Solutions for Elementary and Intermediate Algebra, 4th edition by Carson, Jordan

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SOLUTIONS for
Elementary and Intermediate
TU

Algebra, 4th edition
Author (s): Tom Carson, Bill E. Jordan
VI
A
AP
PR
O
VE
D
?

, Chapter 1 32. The number 7.4 is located 0.4 =
4
of the way
10
Foundations of Algebra
ST
between 7 and 8, so we divide the space between 7
and 8 into 10 equal divisions and place a dot on
the 4th mark to the right of 7.
Exercise Set 1.1
2. {q, r, s, t, u, v, w, x, y, z}
4. {Alaska, Hawaii}
34. First divide the number line between −7 and −8
U
6. {2, 4, 6, 8, …} into tenths. The number −7.62 falls between
8. {16, 18, 20, 22, …} −7.6 and −7.7 on the number line. Subdivide this
section into hundredths and place a dot on the 2nd
10. {–2, –1, 0} mark to the left of −7.6 .
VI
12. Rational because 1 and 4 are integers.
14. Rational because −12 is an integer and all
integers are rational numbers.
π
A
16. Irrational because cannot be written as a ratio
4
of integers. 36. 6 = 6 because 6 is 6 units from 0 on a number
8 line.
18. Rational because −0.8 can be expressed as − ,
10 38. −8 = 8 because −8 is 8 units from 0 on a
AP
the ratio of two integers. number line.
20. Rational because 0.13 can be expressed as the 40. −4.5 = 4.5 because −4.5 is 4.5 units from 0 on a
13
fraction , the ratio of two integers. number line.
99
3 3 3 3
22. False. There are real numbers that are not rational 42. 2 = 2 because 2 is 2 units from 0 on a
5 5 5 5
PR
(irrational numbers).
number line.
24. False. There are real numbers that are not natural
3 44. −67.8 = 67.8 because −67.8 is 67.8 units from 0
numbers, such as 0, –2, , 0.6 , and π.
4 on a number line.
26. True 46. 2 < 7 because 2 is farther to the left on a number
line than 7.
1 1
O
28. The number 5 is located of the way between 48. −6 < 5 because −6 is farther to the left on a
2 2
5 and 6, so we divide the space between 5 and 6 number line than 5.
into 2 equal divisions and place a dot on the 1st 50. −19 < −7 because −19 is farther to the left on a
mark to the right of 5. number line than −7 .
VE
52. 0 > −5 because 0 is farther to the right on a
number line than −5 .
2 2 54. 2.63 < 3.75 because 2.63 is farther to the left on a
30. The number − is located of the way between number line than 3.75.
5 5
0 and −1 , so we divide the space between 0 and 56. −3.5 < −3.1 because −3.5 is farther to the left
−1 into 5 equal divisions and place a dot on the on a number line than −3.1 .
D
2nd mark to the left of 0.
?
Copyright © 2015 Pearson Education, Inc.

, 2 Chapter 1 Foundations of Algebra


5 1 5 1 5 9
58. 3 > 3 because 3 is farther to the right on 6. 8. 10.
6 4 6 4 8 16
ST
1
a number line than 3 . 5 ? 5 ⋅ 2 10
4 12. = ⇒ =
8 16 8 ⋅ 2 16
60. −4.1 = 4.1 because the absolute value of −4.1 The missing number is 10.
is equal to 4.1. 2 6 2⋅3 6
14. = ⇒ =
62. −10.4 > 3.2 because the absolute value of 5 ? 5 ⋅ 3 15
The missing number is 15.
U
−10.4 is equal to 10.4, which is farther to the
right on a number line than 3.2. 6 ? 6÷2 3
16. = ⇒ =
8 4 8÷ 2 4
64. −0.59 = 0.59 because the absolute value of
The missing number is 3.
VI
−0.59 and the absolute value of 0.59 are both
equal to 0.59. 27 9 27 ÷ 3 9
18. = ⇒ =
30 ? 30 ÷ 3 10
2 5 2 The missing number is 10.
66. 4 < 4 because 4 is farther to the left on
9 9 9
20. The LCD of 7 and 11 is 77.
A
5 5 ⋅11 55 3 ⋅ 7 21
a number line than the absolute value of 4 , = and =
9 7 ⋅11 77 11 ⋅ 7 77
5
which is equal to 4 . 22. The LCD of 8 and 12 is 24.
9
5 ⋅ 3 15 7 ⋅ 2 14
= =
AP
and
68. −10 > −8 because the absolute value of −10 8 ⋅ 3 24 12 ⋅ 2 24
is 10, the absolute value of −8 is 8, and 10 is 24. The LCD of 20 and 15 is 60.
farther to the right on a number line than 8.
9⋅3 27 7⋅4 28
− =− and − =−
70. −5.36 < 5.76 because the absolute value of 20 ⋅ 3 60 15 ⋅ 4 60
−5.36 is 5.36, the absolute value of 5.76 is 5.76, 26. The LCD of 21 and 14 is 42.
and 5.36 is farther to the left on a number line than 13 ⋅ 2 26 9⋅3 27
PR
5.76. − =− and − =−
21 ⋅ 2 42 14 ⋅ 3 42
9 7 28. 33 = 3 ⋅11
72. − > − because the absolute value of
11 11
30. 42 = 2 ⋅ 21 = 2 ⋅ 3 ⋅ 7
9 9 7 7
− is , the absolute value of − is , and 32. 48 = 2 ⋅ 24
11 11 11 11
9 = 2⋅8⋅3
O
is farther to the right on a number line than
11 = 2⋅ 2⋅ 4⋅3
7
. = 2⋅ 2⋅ 2⋅ 2⋅3
11
34. 810 = 2 ⋅ 405
VE
3 = 2 ⋅ 81 ⋅ 5
74. −12.6, −9.6,1, −1.3 , −2 , 2.9
4 = 2⋅9⋅9⋅5
1 1 = 2 ⋅ 3⋅ 3⋅ 3⋅ 3⋅ 5
76. −4 , −2 , −2, −0.13, 0.1 ,1.02, −1.06
8 4 48 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 4
36. = =
84 2 ⋅ 2 ⋅ 3 ⋅7 7
D
Exercise Set 1.2 42 2 ⋅ 3 ⋅ 7 6
38. = =
91 7 ⋅13 13
5 7
2. 4.
8 20
?
Copyright © 2015 Pearson Education, Inc.

, Instructor’s Solutions Manual 3


30 2 ⋅ 3 ⋅5 5 8 2 ⋅ 2 ⋅2 2
40. − =− =− 64. = =
54 2 ⋅ 3 ⋅3⋅3 9 60 2 ⋅ 2 ⋅ 3 ⋅ 5 15
ST
24 2 ⋅2⋅2⋅ 3 4 4 2⋅2 1
42. − =− =− 66. = =
162 2 ⋅ 3⋅3⋅3⋅ 3 27 12 2 ⋅ 2 ⋅ 3 3

44. Incorrect. 2 is not a factor of the numerator. 68. 47 Republicans + 2 Independents = 49 Not
49
46. Incorrect. The prime factorization of 108 should Democrats; of the Senate was not Democrat.
be 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 . 100
U
48. If 130 of the 250 calories come from fat, the 70. 6 + 12 + 6 = 24 atoms total
fraction of calories in a serving that comes from 12 + 6 = 18 not-carbon atoms
130 18 2 ⋅ 3 ⋅3 3
fat is . = =
24 2 ⋅ 2 ⋅ 2 ⋅ 3 4
VI
250
130 2 ⋅ 5 ⋅13 13
= =
250 2 ⋅ 5 ⋅ 5 ⋅ 5 25
Exercise Set 1.3
50. If 120 square feet of the 1830 square feet are used
A
as a home office, the fraction of her home that is 2. Commutative Property of Addition because the
120 order of the addends is changed.
used as an office is .
1830 4. Additive identity because the sum of a number
120 2 ⋅2⋅2⋅ 3 ⋅ 5 4 and 0 is that number.
= =
1830 2 ⋅ 3 ⋅ 5 ⋅ 61 61 6. Additive inverse because the sum of these
AP
52. There are 7 ⋅ 24 = 168 hours in one week. opposites is 0.
50 2 ⋅5⋅5 25 8. Associative Property of Addition because the
= = grouping is changed.
168 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7 84
25 10. Commutative Property of Addition because the
Carla spends of her week sleeping. order of the addends is changed.
84
PR
54. 50 + 40 + 18 + 4 = 112 hours for the listed 12. Additive inverse because the sum of the opposites
activities. The non-listed activities take −4.6 and 4.6 is 0.
168 − 112 = 56 hours. 14. 15 + 7 = 22
56 2⋅2⋅2⋅7 1
= = 16. −5 + (−7 ) = −12
168 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7 3
1 18. −5 + 16 = 11
Carla spends of her week away from all of the
3 20. −17 + 8 = −9
O
listed activities.
22. 29 + ( −7 ) = 22
310 2 ⋅ 5 ⋅ 31 31
56. = =
1000 2 ⋅ 2 ⋅ 2 ⋅ 5 ⋅ 5 ⋅ 5 100 24. −16 + 13 = −3
VE
58. 1000 − 310 = 690 non-victims; 9 5 9+5
26. + =
690 69 16 16 16
=
1000 100 14
=
16
60. a) 2008
2 ⋅7
26 2 ⋅13 13 =
b) = = 2 ⋅2⋅2⋅2
1000 2 ⋅ 2 ⋅ 2 ⋅ 5 ⋅ 5 ⋅ 5 500
D
7
=
9 3 ⋅3 3 8
62. = =
159 3 ⋅ 53 53
?
Copyright © 2015 Pearson Education, Inc.

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Subido en
18 de marzo de 2025
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