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Algebra/Calculus Review Answers
ALGEBRA REVIEW
Add as indicated:
1. (2ac) +(–6ac) +(9ac) = 5ac
2. 3x +(–7x) = –4x
3. (–8a) –(–3a) –(2a) = –7a
4. (5x) –(6x) –(7x) = –8x
Add the two expressions in each problem
5. x +2y –8
3x –4y +9
4x –2y +1
6. 11m –7n +13
3m –8n –21
14m –15n –8
Subtract the two expressions in each problem
7. x +2y –8
– (3x –4y +9)
–2x +6y –17
9. 10a –17b +24c; 13a +14b –16c
8. 11m –7n +13
–(3m –8n –21)
8m +n +34
= –3a –31b +40c
Add the three expressions in each of the problems:
10. 7a –3b +11c; –14a +10b +10c; 8a +8b +13c = a +15b +34c
Combine like terms:
11. 3x +7y –3z +6xc –8y –7z +5 –1 = 3x –y –10z +6xc +4
12. 9xy +3x +4a –5ax +10a –7x +3yx +6xa = 12xy –4x +ax +14a
Remove the symbols of grouping and simplify by combining terms:
13. 3a –(b +c) +(a +b –c) = 4a –2c
14. –[x +(3 –x) –(4 +3x)] = 3x +1
15. –{5a –b –[3b –(c –2b +a) –4a] +c} = –5a +b +[3b –(c –2b +a) –4a] –c
= –5a +b +3b –c +2b -a –4a –c = –10a +6b –2c
Multiply as indicated:
16. (5)(-4)(-2) = 40
17. (3ab)(2a) = 6a2b
18. (6a2b)(3ab2) = 18a3b3
19. (3xy2)(5x2y)(xy) = 15x4y4
20. 3x2y(2xy2 +y) = 6x3y3 +3x2y2
21. –4mn(3 –5m +6mn +3n) = –12mn +20m2n –24m2n2 –12mn2
22. (a +3b)(3a2 +6ab +4b2) = 3a3 +15a2b +22ab2 +12b3
23. (3a +2)(a –2)(2a +1) = (3a2 –6a +2a –4)(2a +1) = 6a3 –8a2 –8a +3a2 –4a –4
= 6a3 –5a2 –12a –4
Divide as indicated:
24. x2/x2 = 1
25. –y3/y3 = –1
26. a8/a5 = a3
27. acz/ac = z
28. 12a4/6a = 2 a3
29. –9c3d4/3c5d3 = –3d/c2
30. 34a3b2/17a2b = 2ab
31. (9x2 –6x3 –3x4)/(-3x2) = –3 + 2x + x2
Perform the indicated operations, giving the results in the lowest term:
32.
3 10 7
  = 1/2
5 42 2
33. 5 
34.
5 x 28 y
= 4/3

7 y 15 x
35.
x
 4 = x/12
3
37.
xb
9ab 2 17 x 2
=

6
34 x 27ab
39.
a  b 2y  x
a b
2y  x
=
=1/2


x  2 y 2b  2a
 (2 y  x)  2(a  b)
36. 64 x 3 y 
8 xy3
24 x 2
=
3
y2
10 xy3 5 x 2 y 4 y 2
38.
=

6z
xz
3z 2
3 2
 = 10/13
13 3
Simplify the complex fractions and other expressions;
1 1
2

40. 3 5 = 15 = 1/21
4 14
2
5
5
5a  a 2
a(5  a)
42.
=
 a2
5
5a
1
a
a
1 2
3 2x


x
3
3
x
3x = 3  2 x  1 = 3  2 x
41.
=
3 x 4 x 12 x 2
4x
4x
1
43.
1
1
1
=
1
1
x
1
1
x 1
x
Solve for x, and check by substitution:
44. 8x –15 = 3x
5x = 15
x=3
46.
3
25 4
x
 x
4
2 3
25
25
x=
12
2
x=6
45. x –4 = 5 +2x
x = –9
47. (3x –1)(x +1) = 3x2
3x2 + 3x –x –1 = 3x2
x = 1/2
1

1
x
x 1

1
x 1

x  1  x 2x  1
x 1
Solve the following systems:
48.
50.
– 4x +5y +14 = 0
–2(– 2x +2y – 7 = 0)
y +28 = 0
y = –28
–4x +5(–28) +14 = 0
x = –31.5
3(3x + y – 6 = 0)
4x–3y –21 = 0
13x
–39 = 0
x=3
4(3) –3y –21 = 0
y=-3
49.
4xy – 2y –11 = 0
–2(2xy –3y – 4 = 0)
4y –3 = 0
y = 3/4
4x(3/4) –2(3/4) –11 = 0
x = 4.167
51.
3(4x –5y = 9)
–4(3x –4y = 8)
y = –5
4x –5(–5) = 9
x = –4
Simplify the complex fractions;
52.
0.34b 0.24 a 0.66
= 1.467b/a
0.24a 0.34b 0.76
53.
2.13a1.13b 0.34
= 6.265b/a
0.34a 2.13b 0.66
x10.5 x 20.5 x 2
54.
=
x10.5 x 20.5 x1
1
55.
2  C 3
 
3F 
1 F 
 
3C 
2
3
=2
C
F
CALCULUS REVIEW
Differentiate the following functions
56. Y = X2
dY
= 2X
dX
57 y = 6x3
dy
=18x2
dx
58. Z = 20Y0.5
dZ
= 10Y–0.5
dY
59. Z = 9 x = 3x1/2
dZ
= 3/2x–1/2
dx
60. y = 30
y’ = 0
61. t = 3x + 6x3
dt
= 3 + 18x2
dx
62. y = 7x0.5 + 12x1.5
dy
= 3.5x–0.5 + 18x0.5
dx
63. Y = 6X(6 + 2X2)
dY
= 6X(4X) + 6(6 + 2X2) = 24X2 + 36 + 12X2 = 36 + 36X2
dX
64. y = (2 + 6x)(2 + 5x2)
dy
= (2 + 6x)(10x) + (6)(2 + 5x2) = 12 + 20x + 90x2
dx
65. y = (x2 – 3x3)(x3 + 5)
dy
= (x2 – 3x3)(3x2) + (2x – 9x2)(x3 + 5)
dx
66. y = f(x) =
5x 2
3x  6
x 2  6x
67. y =
x3
f’(x) =
10 x(3x  6)  15 x 2 60 x  15 x 2
=
(3x  6) 2
(3x  6) 2
dy
(2 x  6)( x  3)  x 2  6 x
= f’(x) =
dx
( x  3) 2
Find the maximum and/or minimum values of the following functions. Be sure to state the
first and second order conditions.
68. y = 10 + 10x – 0.5x2
FOC
y’ = 10 – x = 0
x = 10
SOC
y” = –1
Maximum
69. T = – 2000S + 10S2
FOC
T’ = –2000 +20S = 0,
S = 100
SOC
T” = –20
Maximum
70. Z = 145 – 4Y – 0.3Y2
FOC
Z’ = –4 – 0.6S = 0,
S = – 6.67
SOC
Z” = –0.6
Maximum
71. X = – 145 + 4Y + 0.3Y2
FOC
X’ = +4 + 0.6Y = 0
S = 6.67
SOC
X” = +0.6
Minimum
Find the maximum and/or minimum values and the inflection point of the following functions.
Be sure to state the first and second order conditions. (Hint: the roots of the quadratic
equation: aX2 + bX + c = 0 are
 b  b 2  4ac
.)
X
2a
72. y = 5000 + 500x + 20x2 –0.1x3
FOC
y’ = 500 + 40x –0.3x2 = 0
x = -11.51,
SOC
x = 144.84
y” = 40 –0.6x
y” = 40 –0.6(-11.51) > 0 therefore minimum
y” = 40 –0.6(144.84) < 0 therefore maximum
Inflection point is where y” = 0
y” = 40 –0.6x = 0,
x = 6.67
73. z = -2000 +30x +10x2 – 0.3x3
FOC
y’ = 30 + 20x –0.9x2 = 0
x = -1.41,
SOC
x = 23.63
y” = 20 –1.8x
y” = 20 –1.8(-1.41) > 0 therefore minimum
y” = 20 –1.8(23.63) < 0 therefore maximum
Inflection point is where y” = 0
y” = 20 –1.8x = 0,
x = 11.11
74. y = 5000x –100x2 –0.5x3
FOC
y’ = 5000 –200x –1.5x2 = 0
x = –154.86,
SOC
x = 21.53
y” = –200 –3x
y” = –200 –3(-154.86) > 0 therefore minimum
y” = –200 –3(21.53) < 0 therefore maximum
Inflection point is where y” = 0
y” = –200 –3x = 0,
x = 66.67
75. y = 400x – 25x2 + 0.2x3
FOC
y’ = 400 –50x +0.6x2 = 0
x = 8.96,
SOC
x = 74.37
y” = –50 +1.2x
y” = –50 +1.2(8.96) < 0 therefore maximum
y” = –50 +1.2(74.37) > 0 therefore minimum
Inflection point is where y” = 0
y” = –50 +1.2x = 0,
x = 41.67