Applied Calculus, 7th Edition by
Deborah Hughes-Hallett
Complete Chapter Solutions Manual
are included (Ch 1 to 10)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1 Functions And Change
2 Rate of Change: The Derivative
3 Shortcuts to Differentiation
4 Using the Derivative
5 Accumulated Change: The Definite Integral
6 Antiderivatives and Applications
7 Probability
8 Functions of Several Variables
9 Mathematical Modeling Using Differential Equations
10 Geometric Series
, PSA A / SOLUTIONS 1
Prerequisite Support Appendix A
Solutions for PSA A.1
EXERCISES
1. No. Reordering the expression on the left gives us 3� ⋅ �, or 3�2 , but not 4�.
2. Yes. The expression �(5�) can be reordered to 5� ⋅ �, which is equivalent to 5�2 .
3. No. Reordering the two terms gives 2� + � = � + 2�, but � + 2� is not equivalent to these.
4. Yes. The expression 5 − � is equivalent to 5 + (−�), which can be reordered to −� + 5.
5. No. The expression (2�)(2�) is equivalent to 4��, which is not equivalent to 2��.
6. No. The expression (2�)(5�) can be reordered as 2 ⋅ 5 ⋅ � ⋅ �, which is equivalent to 10��, not 7��.
7. Yes. The expression (3�)(4�)(2�) can be reordered as 3 ⋅ 4 ⋅ 2 ⋅ � ⋅ � ⋅ �, which is equivalent to 24���.
8. Yes. The expression ( + 3) + (� + 2) can be reordered as + � + 2 + 3, which is equivalent to ( + �) + 5.
9. Yes. Since 4 + � is equivalent to � + 4, the two expressions are equivalent.
10. Incorrect. For � = 2, the left side of the equation equals 32, but the right side of the equation equals 160.
11. Incorrect. For � = 2 and � = 1, the left side of the equation equals 16, but the right side equals 40.
12. Correct. Think of ℎ2 as an object. If you have 3 objects and 2 of the same object, you have 5 objects altogether.
13. Incorrect. For � = 2, the left side of the equation equals 14, but the right side equals 40.
14. We have (2� + 1) + (5� + 8) = (2� + 5�) + (1 + 8) = 7� + 9.
15. We have (4 − 2�) + (5� − 9) = (−2� + 5�) + (4 − 9) = 3� − 5.
16. We can combine the �2 terms in 3�2 − 2� 2 + 6�� − �2 to get 2�2 − 2� 2 + 6��.
17. We have (� + 1) + (� + 2) + (� + 3) = (� + � + �) + (1 + 2 + 3) = 3� + 6.
18. We can combine the �3 terms and the �� terms in �3 + 2�� − 4�3 + � − 2�� to get −3�3 + �.
19. We can combine the �4 terms in 5�4 + 5�3 − 3�4 to get 2�4 + 5�3 .
20. We have (7� + 1) + (5 − 3�) + (2� − 4) = (7� − 3� + 2�) + (1 + 5 − 4) = 6� + 2.
21. We have 5�2 + 5� + 3�2 = (5�2 + 3�2 ) + 5� = 8�2 + 5�.
22. There are no like terms to combine.
23. We have (2�)(3�) + 4� + 5� + (6�)(3�) = 6�� + 4� + 5� + 18�� = (6�� + 18��) + 4� + 5� = 24�� + 4� + 5�.
24. We have (2�)(5�) + (3�)(2�) + 5(3�) + �(3�) = 10�2 + 6�2 + 15� + 3�2 = (10�2 + 6�2 + 3�2 ) + 15� = 19�2 + 15�.
PROBLEMS
25. We regroup in the expression ( + 5) + (� − 3) + (� + 8) so that we can use the information that + � + � = 12. We have
( + 5) + (� − 3) + (� + 8) = + 5 + � + (−3) + � + 8
= + � + � + 5 + (−3) + 8
= ( + � + �) + (5 − 3 + 8)
= ( + � + �) + 10
= 12 + 10
= 22.
, 2 PSA A / SOLUTIONS
26. We regroup and reorder in the expression (� − 10) + (� + 8) + (� − 5) so that we can use the information that � + � + � = 25.
We have
(� − 10) + (� + 8) + (� − 5) = � + (−10) + � + 8 + � + (−5)
= � + � + � + (−10) + 8 + (−5)
= (� + � + �) + (−10 + 8 − 5)
= (� + � + �) − 7
= 25 − 7
= 18.
27. We regroup in the expression (3�)(2�)(5�) so that we can use the information that ��� = 100. We have
(3�)(2�)(5�) = (3 ⋅ 2 ⋅ 5)(���)
= 30(���)
= 30(100)
= 3000.
�
28. We regroup in the expression (2�)( )(6�) so that we can use the information that ��� = 20. We have
4
� 1
(2�)( )(6�) = 2 ⋅ � ⋅ ⋅ � ⋅ 6 ⋅ �
4 4
1
= (2 ⋅ ⋅ 6) ⋅ (���)
4
2⋅6
= ⋅ (���)
4
= 3(���)
= 3(20)
= 60.
29. We have
+ 2(� − ) − 3(� + �) = + (� − ) + (� − ) − (� + �) − (� + �) − (� + �)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
2 groups 3 groups
= − − +�+�−�−�−�−�−�−� regrouping terms
= − − � − 3� collecting like terms.
30. When the speed is � and the time is �, the distance traveled is � ⋅ �. We know the car travels 200 miles, so � ⋅ � = 200.
If the car travels half as fast, its new rate is (1∕2)�, and if it travels three times as long, its new time is 3�. We have
Distance = Speed × Time
( )
�
= ⋅ (3�)
2
( )
3
= ⋅ (��)
2
( )
3
= ⋅ (200)
2
= 300.
The car travels 300 miles.
Deborah Hughes-Hallett
Complete Chapter Solutions Manual
are included (Ch 1 to 10)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1 Functions And Change
2 Rate of Change: The Derivative
3 Shortcuts to Differentiation
4 Using the Derivative
5 Accumulated Change: The Definite Integral
6 Antiderivatives and Applications
7 Probability
8 Functions of Several Variables
9 Mathematical Modeling Using Differential Equations
10 Geometric Series
, PSA A / SOLUTIONS 1
Prerequisite Support Appendix A
Solutions for PSA A.1
EXERCISES
1. No. Reordering the expression on the left gives us 3� ⋅ �, or 3�2 , but not 4�.
2. Yes. The expression �(5�) can be reordered to 5� ⋅ �, which is equivalent to 5�2 .
3. No. Reordering the two terms gives 2� + � = � + 2�, but � + 2� is not equivalent to these.
4. Yes. The expression 5 − � is equivalent to 5 + (−�), which can be reordered to −� + 5.
5. No. The expression (2�)(2�) is equivalent to 4��, which is not equivalent to 2��.
6. No. The expression (2�)(5�) can be reordered as 2 ⋅ 5 ⋅ � ⋅ �, which is equivalent to 10��, not 7��.
7. Yes. The expression (3�)(4�)(2�) can be reordered as 3 ⋅ 4 ⋅ 2 ⋅ � ⋅ � ⋅ �, which is equivalent to 24���.
8. Yes. The expression ( + 3) + (� + 2) can be reordered as + � + 2 + 3, which is equivalent to ( + �) + 5.
9. Yes. Since 4 + � is equivalent to � + 4, the two expressions are equivalent.
10. Incorrect. For � = 2, the left side of the equation equals 32, but the right side of the equation equals 160.
11. Incorrect. For � = 2 and � = 1, the left side of the equation equals 16, but the right side equals 40.
12. Correct. Think of ℎ2 as an object. If you have 3 objects and 2 of the same object, you have 5 objects altogether.
13. Incorrect. For � = 2, the left side of the equation equals 14, but the right side equals 40.
14. We have (2� + 1) + (5� + 8) = (2� + 5�) + (1 + 8) = 7� + 9.
15. We have (4 − 2�) + (5� − 9) = (−2� + 5�) + (4 − 9) = 3� − 5.
16. We can combine the �2 terms in 3�2 − 2� 2 + 6�� − �2 to get 2�2 − 2� 2 + 6��.
17. We have (� + 1) + (� + 2) + (� + 3) = (� + � + �) + (1 + 2 + 3) = 3� + 6.
18. We can combine the �3 terms and the �� terms in �3 + 2�� − 4�3 + � − 2�� to get −3�3 + �.
19. We can combine the �4 terms in 5�4 + 5�3 − 3�4 to get 2�4 + 5�3 .
20. We have (7� + 1) + (5 − 3�) + (2� − 4) = (7� − 3� + 2�) + (1 + 5 − 4) = 6� + 2.
21. We have 5�2 + 5� + 3�2 = (5�2 + 3�2 ) + 5� = 8�2 + 5�.
22. There are no like terms to combine.
23. We have (2�)(3�) + 4� + 5� + (6�)(3�) = 6�� + 4� + 5� + 18�� = (6�� + 18��) + 4� + 5� = 24�� + 4� + 5�.
24. We have (2�)(5�) + (3�)(2�) + 5(3�) + �(3�) = 10�2 + 6�2 + 15� + 3�2 = (10�2 + 6�2 + 3�2 ) + 15� = 19�2 + 15�.
PROBLEMS
25. We regroup in the expression ( + 5) + (� − 3) + (� + 8) so that we can use the information that + � + � = 12. We have
( + 5) + (� − 3) + (� + 8) = + 5 + � + (−3) + � + 8
= + � + � + 5 + (−3) + 8
= ( + � + �) + (5 − 3 + 8)
= ( + � + �) + 10
= 12 + 10
= 22.
, 2 PSA A / SOLUTIONS
26. We regroup and reorder in the expression (� − 10) + (� + 8) + (� − 5) so that we can use the information that � + � + � = 25.
We have
(� − 10) + (� + 8) + (� − 5) = � + (−10) + � + 8 + � + (−5)
= � + � + � + (−10) + 8 + (−5)
= (� + � + �) + (−10 + 8 − 5)
= (� + � + �) − 7
= 25 − 7
= 18.
27. We regroup in the expression (3�)(2�)(5�) so that we can use the information that ��� = 100. We have
(3�)(2�)(5�) = (3 ⋅ 2 ⋅ 5)(���)
= 30(���)
= 30(100)
= 3000.
�
28. We regroup in the expression (2�)( )(6�) so that we can use the information that ��� = 20. We have
4
� 1
(2�)( )(6�) = 2 ⋅ � ⋅ ⋅ � ⋅ 6 ⋅ �
4 4
1
= (2 ⋅ ⋅ 6) ⋅ (���)
4
2⋅6
= ⋅ (���)
4
= 3(���)
= 3(20)
= 60.
29. We have
+ 2(� − ) − 3(� + �) = + (� − ) + (� − ) − (� + �) − (� + �) − (� + �)
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
2 groups 3 groups
= − − +�+�−�−�−�−�−�−� regrouping terms
= − − � − 3� collecting like terms.
30. When the speed is � and the time is �, the distance traveled is � ⋅ �. We know the car travels 200 miles, so � ⋅ � = 200.
If the car travels half as fast, its new rate is (1∕2)�, and if it travels three times as long, its new time is 3�. We have
Distance = Speed × Time
( )
�
= ⋅ (3�)
2
( )
3
= ⋅ (��)
2
( )
3
= ⋅ (200)
2
= 300.
The car travels 300 miles.
