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Solutions Manual for Calculus Single and Multivariable 7th Edition By Hughes Hallett, McCallum / All Chapters / Latest Version 2024 A+$12.99
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Solutions Manual for Calculus Single and Multivariable 7th Edition By Hughes Hallett, McCallum / All Chapters / Latest Version 2024 A+
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Calculus Single and Multivariable 7th Edition
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Calculus Single And Multivariable 7th Edition
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Calculus: Single and Multivariable, 7e Student Solutions Manual
Solutions Manual for Calculus Single and Multivariable 7th Edition By Hughes Hallett, McCallum / All Chapters / Latest Version 2024 A+
Solutions Manual for Calculus Single and Multivariable 7th Edition By Hughes Hallett, McCallum / All Chapters / Latest Version 2024 A+
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Calculus: Single and Multivariable, 7th Edition by Deborah Hughes-Hallett, William G. McCallum, and Andrew M. Gleason, et al. All Chapters 1-21. (Complete Download). TEST BANK.
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Solutions Manual for
Calculus Single and
Multivariable. 7e
Hughes Hallett,
McCallum, Gleason,
Flath, Lock, Gordon,
Lomen, Lovelock,
Osgood (All Chapters
Download link at the
end of this file)
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, 1.1 SOLUTIONS 1
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since 𝑡 represents the number of years since 2010, we see that 𝑓 (5) represents the population of the city in 2015. In 2015,
the city’s population was 7 million.
2. Since 𝑇 = 𝑓 (𝑃 ), we see that 𝑓 (200) is the value of 𝑇 when 𝑃 = 200; that is, the thickness of pelican eggs when the
concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of workers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after which it goes downward; in other words, as the number of workers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to workers
getting in each other’s way as too many are crammed on the same line. Many other reasons are possible.
4. The slope is (1 − 0)∕(1 − 0) = 1. So the equation of the line is 𝑦 = 𝑥.
5. The slope is (3 − 2)∕(2 − 0) = 1∕2. So the equation of the line is 𝑦 = (1∕2)𝑥 + 2.
6. The slope is
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Now we know that 𝑦 = (1∕2)𝑥 + 𝑏. Using the point (−2, 1), we have 1 = −2∕2 + 𝑏, which yields 𝑏 = 2. Thus, the equation
of the line is 𝑦 = (1∕2)𝑥 + 2.
6−0
7. The slope is = 2 so the equation of the line is 𝑦 − 6 = 2(𝑥 − 2) or 𝑦 = 2𝑥 + 2.
2 − (−1)
5 5
8. Rewriting the equation as 𝑦 = − 𝑥 + 4 shows that the slope is − and the vertical intercept is 4.
2 2
9. Rewriting the equation as
12 2
𝑦 = − 𝑥+
7 7
shows that the line has slope −12∕7 and vertical intercept 2∕7.
10. Rewriting the equation of the line as
−2
−𝑦 = 𝑥−2
4
1
𝑦 = 𝑥 + 2,
2
we see the line has slope 1∕2 and vertical intercept 2.
11. Rewriting the equation of the line as
12 4
𝑦= 𝑥−
6 6
2
𝑦 = 2𝑥 − ,
3
we see that the line has slope 2 and vertical intercept −2∕3.
12. (a) is (V), because slope is positive, vertical intercept is negative
(b) is (IV), because slope is negative, vertical intercept is positive
(c) is (I), because slope is 0, vertical intercept is positive
(d) is (VI), because slope and vertical intercept are both negative
(e) is (II), because slope and vertical intercept are both positive
(f) is (III), because slope is positive, vertical intercept is 0
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,2 Chapter One /SOLUTIONS
13. (a) is (V), because slope is negative, vertical intercept is 0
(b) is (VI), because slope and vertical intercept are both positive
(c) is (I), because slope is negative, vertical intercept is positive
(d) is (IV), because slope is positive, vertical intercept is negative
(e) is (III), because slope and vertical intercept are both negative
(f) is (II), because slope is positive, vertical intercept is 0
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
−3 6 2
Slope = =− =− .
7.5 15 5
The 𝑦-intercept is at (0, 3), so a possible equation for the line is
2
𝑦 = − 𝑥 + 3.
5
(Answers may vary.)
15. 𝑦 − 𝑐 = 𝑚(𝑥 − 𝑎)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
29.2 − 27.8 1.4
Slope = = = 14.
5.3 − 5.2 0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
𝑦 − 27.8 = 14(𝑥 − 5.2)
which is equivalent to
𝑦 = 14𝑥 − 45.
17. 𝑦 = 5𝑥 − 3. Since the slope of this line is 5, we want a line with slope − 15 passing through the point (2, 1). The equation is
(𝑦 − 1) = − 51 (𝑥 − 2), or 𝑦 = − 51 𝑥 + 75 .
18. The line 𝑦 + 4𝑥 = 7 has slope −4. Therefore the parallel line has slope −4 and equation 𝑦 − 5 = −4(𝑥 − 1) or 𝑦 = −4𝑥 + 9.
−1
The perpendicular line has slope (−4) = 41 and equation 𝑦 − 5 = 14 (𝑥 − 1) or 𝑦 = 0.25𝑥 + 4.75.
19. The line parallel to 𝑦 = 𝑚𝑥 + 𝑐 also has slope 𝑚, so its equation is
𝑦 = 𝑚(𝑥 − 𝑎) + 𝑏.
The line perpendicular to 𝑦 = 𝑚𝑥 + 𝑐 has slope −1∕𝑚, so its equation will be
1
𝑦 = − (𝑥 − 𝑎) + 𝑏.
𝑚
20. Since the function goes from 𝑥 = 0 to 𝑥 = 4 and between 𝑦 = 0 and 𝑦 = 2, the domain is 0 ≤ 𝑥 ≤ 4 and the range is
0 ≤ 𝑦 ≤ 2.
21. Since 𝑥 goes from 1 to 5 and 𝑦 goes from 1 to 6, the domain is 1 ≤ 𝑥 ≤ 5 and the range is 1 ≤ 𝑦 ≤ 6.
22. Since the function goes from 𝑥 = −2 to 𝑥 = 2 and from 𝑦 = −2 to 𝑦 = 2, the domain is −2 ≤ 𝑥 ≤ 2 and the range is
−2 ≤ 𝑦 ≤ 2.
23. Since the function goes from 𝑥 = 0 to 𝑥 = 5 and between 𝑦 = 0 and 𝑦 = 4, the domain is 0 ≤ 𝑥 ≤ 5 and the range is
0 ≤ 𝑦 ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since 𝑥2 ≥ 0 for all 𝑥.
1
25. The domain is all 𝑥-values, as the denominator is never zero. The range is 0 < 𝑦 ≤ .
2
26. The value of 𝑓 (𝑡) is real provided 𝑡2 − 16 ≥ 0 or 𝑡2 ≥ 16. This occurs when either 𝑡 ≥ 4, or 𝑡 ≤ −4. Solving 𝑓 (𝑡) = 3, we
have
√
𝑡2 − 16 = 3
𝑡2 − 16 = 9
𝑡2 = 25
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, 1.1 SOLUTIONS 3
so
𝑡 = ±5.
4 3
27. We have 𝑉 = 𝑘𝑟3 . You may know that 𝑉 = 𝜋𝑟 .
3
𝑑
28. If distance is 𝑑, then 𝑣 = .
𝑡
29. For some constant 𝑘, we have 𝑆 = 𝑘ℎ2 .
30. We know that 𝐸 is proportional to 𝑣3 , so 𝐸 = 𝑘𝑣3 , for some constant 𝑘.
31. We know that 𝑁 is proportional to 1∕𝑙2 , so
𝑘
𝑁= , for some constant 𝑘.
𝑙2
Problems
32. (a) Each date, 𝑡, has a unique daily snowfall, 𝑆, associated with it. So snowfall is a function of date.
(b) On December 12, the snowfall was approximately 5 inches.
(c) On December 11, the snowfall was above 10 inches.
(d) Looking at the graph we see that the largest increase in the snowfall was between December 10 to December 11.
33. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in Figure 1.1:
𝑉 (thousand dollars)
(5, 6)
𝑎 (years)
Figure 1.1
(c) The vertical intercept is the value of 𝑉 when 𝑎 = 0, or the value of the car when it is new. The horizontal intercept is
the value of 𝑎 when 𝑉 = 0, or the age of the car when it is worth nothing.
34. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home.
(b) The story in (b) matches Graph (II), the flat tire story. Note the long period of time during which the distance from
home did not change (the horizontal part).
(c) The story in (c) matches Graph (III), in which the person started calmly but sped up later.
The first graph (I) does not match any of the given stories. In this picture, the person keeps going away from home,
but his speed decreases as time passes. So a story for this might be: I started walking to school at a good pace, but since I
stayed up all night studying calculus, I got more and more tired the farther I walked.
35. The year 2000 was 12 years before 2012 so 2000 corresponds to 𝑡 = 12. Thus, an expression that represents the statement
is:
𝑓 (12) = 7.049.
36. The year 2012 was 0 years before 2012 so 2012 corresponds to 𝑡 = 0. Thus, an expression that represents the statement is:
𝑓 (0) meters.
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