Advanced Algebra Topics COMPASS Review

Advanced Algebra Topics COMPASS Review You will be allowed to use a calculator on the COMPASS test. Acceptable calculators are:

basic calculators, scientific calculators, and graphing calculators up through the level of the TI-86.

1. If 423)( 2 xxxf , find )2(f .

a. 20 b. 2 c. -8 d. 12

2. If 23)( xxf and 14)( 2 xxxg , find: ))(( xgf

a. 3 23 14 11 2 x x x b. 34 6 1 x x c. 2 1 x x d. 2 7 3 x x


a. 2 7 3 x x b. 2 7 3 x x c. 2 1 x x d. 2 1 x x

4. If 23)( xxf and 14)( 2 xxxg , find: ))(( xfg

a. 23 8 1 x x b. 3 23 14 11 2 x x x c. 29 24 13 x x d. 2 1 x x

5. If 23)( xxf and 14)( 2 xxxg , find: )(xg

f

a. 2 4 1

3 2

x x

x b.

2

1

3x c.

1

2 d.

2

3 2

4 1

x

x x


a. 3 2x b. 3 23 14 11 2 x x x c. 23 12 1 x x d. 29 24 13 x x


a. 2 7 3 x x b. 2 4 1 x x c. 29 24 13 x x d. 23 12 1 x x

8. If 23)( xxf and 14)( 2 xxxg , find: ))2((gf 1

a. -12 b. -1 c. 1 d. -11

9. If 32)( xxf , find )(1 xf .

a. 1 3( )

2

xf x

b. 1( ) 2 3f x x c. 1 1( )

2 3f x

x

d. 1 1 1( )

2 3f x x

10. If 3 14)( xxf , find )(1 xf .

a.3

1 1( )

4

xf x

b. 31( ) 4 1f x x c. 1 3

1( ) 1

4f x x d. 1

3

1( )

4 1f x

x

11. If )(xf contains the point (4,1), then )(1 xf must contain what point?

a. (-4,1) b. (1,4) c. (4,-1) d. (1,1

4)

12. Find the domain of 2( ) 6 5f x x x

a. ( , ) b. [1,5] c. [3, ) d. [-4, )

13. Find the domain of 7

( )4 8

f xx

a. | 2x x b. | 0x x c. ( ,2) (2, ) d. ( ,0) (0, )


1)(

2

xx

xxf

a. ( ,0) (0, ) b. ( ,1) (1, ) c. ( , 2) ( 2,1) (1,4) (4, ) d. ( , 2) ( 2,4) (4, )

15. Find the domain of ( ) 3 9 1f x x

a. [ 3, ) b. ( 3, ) c. (0, ) d. [0,3]

16. If 12)3(

)32( 2

x

x

kxx, find the value of k .

a. k = –7 b. k = – 5 c. k = 5 d. k = 7

17. If 14 yx and 53 xz , find an expression for z in terms of y .

a. z = 4y + 1 b. z = 12y – 2 c. z = 12y – 19 d. z = 3y – 5

18. Find the vertex of the function, 563)( 2 xxxf .

a. (-3,-5) b. (1,-8) c. (-1,-2) d. (-1,-8)

19. What are the zeros of the function, 382)( 2 xxxf ?

a. 2 2 10 b. 10

82

c. 8 2 10 d. 10

22

20. What are the zeros of the function, 1892)( 23 xxxxf ?

a. x = 3, x = –3, x = 2 c. x = 3, x = –3, x = – 2

b. x = 0, x = –3, x = 3, x = –2, x = 2 d. x = 0, x = 9, x = 2, x = – 2

21. Find the sum of the solutions of 452 2 xx .

a. 5

2 b. 5 c.

5

4 d. 4

22. Write a cubic function that has ,2,0 xx and 5x as zeros.

a. 3 2( ) 7 10f x x x x b. 3 2( ) 7 10f x x x x c. 3 2( ) 3 10f x x x x d. 3 2( ) 3 10f x x x x

23. Write a quadratic function that has a vertex at (2,-3) and contains the point (-2,5).

a. 2( ) ( 2) 3f x x b. 21( ) 2 1

2f x x x c. 2( ) ( 2) 3f x x d. 2( ) 4 1f x x x

24. Simplify 3

2

2 )(a

a. 8

3a b. 4

3a c. 2

3a d. 1

3a

25. Simplify 8

62

6

3

ab

ba

a. 22ab b. 2

2

ab c.

22

a

b d.

2

2

ab

26. Simplify

2

1

5

3

a

a

a. 1

10a b. 3

10a c. 1

5a d. 2

5a

27. Simplify 7

32 )4(x

x

a. 212x b. 1312x c.

64

x

d. 1364x

28. Simplify 2

5

2

3

2

3

2

1

23 baba

a. 3 15

4 46a b b. 2 45a b c. 3 86a b d. 2 46a b

29. Simplify

3

63

52

3

2

yx

yx

a. 3

3

2

3

y

x b.

3

3

8

27

y

x c.

33

15

27

8

y

x d.

15

33

3

2

x

y

30. Evaluate 2

1

6

1

88

a. 4 b. 1

1264 c. 1

128 d. 16

31. Evaluate 3

2

3

1

33

a. 1 b. 3 c. 2

99 d. Not possible

32. Simplify, assuming that the variables represent positive numbers:3 2aa

a. a a b. a c. 6a a d. Not possible

33. Simplify, assuming that the variables represent positive numbers: 53 ba

a. 15 ab b. 15 5 3a b c. 8 ab d. Not possible

34. Simplify, assuming that the variables represent positive numbers: baba 42 82

a. 34a b b. 34a b c. 44a b d. 6 216a b

35. Simplify, assuming that the variables represent positive numbers:4 354 52 44 baba

a. 7 816a b b. 42 3 2 32a b a b c. 3 44a b a d. 42 32ab a

36. Rewrite using logarithmic notation: 1642

a. 16log 4 2 b.

4log 2 16 c. 4log 16 2 d.

2log 16 4

37. Rewrite using logarithmic notation: yM x

a. logx y M b. log y M x c. logM x y d. logM y x

38. Rewrite using exponential notation: 2100log10

a. 210 100 b. 102 100 c. 2100 10 d. 1002 10

39. Rewrite using exponential notation: 13

1log3

a. 3 1( 1)

3 b. 1 1

33

c. 1

33 1 d. 3

11

3

40. Evaluate without a calculator:

4

1log2

a. 1

2

b. 2 c. -2 d.

1

16

41. Expand:3

2

logz

yxa

a. 2 3loga x y z b. 2log 3loga ax y z c. 2log log 3loga a ax y z d. 1

2log log log3

a a ax y z

42. Express as a single logarithm: tzy 10101010 log8log6log43log2

a. 106log (3 )yzt b.

10log (6 4 6 8 )y z t c. 4

10 6 8

9log

y

z t

d. 108log (3 )y z t

43. Simplify i2

to i, -i, 1, or -1.

a. i b. -i c. 1 d. -1

44. Simplify 8i to i, -i, 1, or -1.

a. i b. -i c. 1 d. -1


a. i b. -i c. 1 d. -1

46. Add and write your answer in a+bi form: (1 + i) + (3 + 5i)

a. 4 + 6i b. -2 c. 4 – 6i d. 4 + 5i2

47. Subtract and write your answer in a+bi form:(3 – 4i) – (7 – i)

a. -4 – 5i b. 0 – 9i c. -4 – 3i d. 0 – 16i

48. Multiply and write your answer in a+bi form: 43 i

a. 0 + 12i b. 0 + 7i c. -12 d. 7

49. Multiply and write your answer in a+bi form: ii 52

a. 10i2 b. -10i c. -10 + 0i c. -10 + 0i

50. Multiply and write your answer in a+bi form: )52(7 ii

a. 35 + 14i b. 14 – 2i c. 2 – 35i d. -2 + 35i

51. Multiply and write your answer in a+bi form: )43)(21( ii

a. 3 + 8i b. 3 + 6i c. 5 – 10i d. -5 + 10i

52. Multiply and write your answer in a+bi form: 3)2( i

a. 8 – i3 b. 6 – i c. 2 – 11i d. 8 – 3i

53. Divide and write your answer in a+bi form: i3

2

a. 2

3 b. 6 + i c.

20

3i d. 6i


6

a. -1i b.

12 18

13 13i

c. 3 – 2i d. 12 – 18i

55. Divide and write your answer in a+bi form: 10 10

1 3

i

i

a. 2 + 4i b. -10 +

10

3i

c. 2 – 4i d. -40 – 2i

56. Solve for x over the complex number system: 0252 x

a. 5x i b. 5x c. 5x d. 5x

57. Solve for x over the complex number system: 422 xx

a. 1 3x i b. 2x c. 0,2x d. 2x i


a. 3, 4x b. 2 3x i c. 3,4x d. 2 3x

59. Write the first five terms of the sequence having the general term 2

1 1)1(

n

na n

n

a. 3 4 5 6

2, , , ,4 9 16 25

b. 3 4 5 6

1, , , ,9 16 25 36

c. 2 3 4

0,1, , ,9 16 25

d. 1 3 4 5 6

, , , ,2 4 5 16 25

60. A certain arithmetic sequence has 561 a and 2611 a . Find 17a .

a. 21 b. 18 c. 3 d. 8

61. A certain geometric sequence has 641 a and 3245 a . Find 7a .

a. 2059 b. 454 c. 1358 d. 1358

62. Find the sum of the first ten terms of an arithmetic sequence having 271 a and 9d .

a. 117 b. 1.8x1010 c. 675 d. 108

63. Find the sum of the first ten terms of a geometric sequence having 30721 a and 2

3r .

a. 41,472 b. 6157.5 c. 348,150 d. 118,098

64. Evaluate

23

1

)36(n

n .

a. 1587 b. 60 c. 1518 d. 135

65. Evaluate

14

1 2

18

n

n

a. 15 b. 4 c. 1 d. 64

66. Evaluate 9!

3! 6!

a. 1 b. 84 c. 1

2 d.

45

126

67. Give the entry in the first row and the first column for

62

30

43

5

05

73

28

a. 11 b. 5 c. -7 d. 47

68. Give the entry in the second row and the first column for

30

52

43

65

32

a. 21 b. 33 c. 24 d. 26

69. Evaluate 53

912

a. 87 b. -33 c. -87 d. 33

70. Evaluate

132

065

213

a. 1 b. 0 c. 66 d. 7

71. Give the z-value of the solution to the following system:

12

72

1154

zy

zx

yx

a. 2 b. -1 c. 1 d. 4

72. If r

aS

1, then r = ?

a. S a

S

b. S a c. 1

S

a

d. 1S a

73. If 2 3 1 6 8

5 6 0 2 15 17

k

, then k =?

a. 3 b. 4 c. 8 d. 3

2

74. If is defined to be: 1 yxyx and 315x , then x = ?

a. 26 b. 6.2 c. 2 d. 36

75. If 1

4 2

k=10, then k = ?

a. 5

4 b. 3 c. 10 d. 7

76. In the list 12,5

3,2.16,

8

0,,25,7,

2

1 , the sum of all the rational numbers is:

a. 28.2 b. 33.3 c. 28.3 d. 5.5

77. How many integers are in the list: 12,5

3,2.16,

8

0,,25,7,

2

1 ?

a. 4 b. 3 c. 1 d. 6

78. Find the fourth term in the expansion of 7)3( zy .

a. 4 335y z b. 4 32835y z c. 4 31701y z d. 4 32835y z

79. How many terms are in the expansion of 7)3( zy ?

a. 2 b. 6 c. 7 d. 8

80. A set containing five elements has how many subsets?

a. 15 b. 31 c. 32 d. 10

Answers to Advanced Algebra Topics COMPASS Review

1. a

2. c

3. b

4. b

5. d

6. c

7. c

8. d

9. a

10. a

11. b

12. a

13. c

14. d

15. a

16. a

17. b

18. d

19. d

20. c

21. a

22. d

23. b

24. b

25. c

26. a

27. d

28. d

29. c

30. a

31. b

32. c

33. b

34. a

35. d

36. c

37. d

38. a

39. b

40. c

41. d

42. c

43. d

44. c

45. b

46. a

47. c

48. a

49. c

50. a

51. d

52. c

53. c

54. b

55. a

56. a

57. a

58. b

59. a

60. d

61. d

62. c

63. c

64. a

65. a

66. b

67. c

68. a

69. d

70. d

71. b

72. a

73. a

74. c

75. d

76. b

77. b

78. d

79. d

80. c

2/2012

Solutions to Advanced Algebra Topics COMPASS Review

1. If 423)( 2 xxxf , find )2(f .

f(-2) = 3(-2)2 – 2(-2) + 4 = 12 + 4 + 4 = 20


(f + g) (x) = f(x) + g(x) = (3x – 2) + (x2 – 4x + 1) = x

2 – x – 1


(f – g)(x) = f(x) – g(x) = (3x – 2) – (x2 – 4x + 1) = – x

2 + 7x – 3


(f g)(x) = f(x) · g(x) = (3x – 2) (x2 – 4x + 1) = 3x

3 – 14x

2 + 11x – 2

5. If 23)( xxf and 14)( 2 xxxg , find: )(xg

f

)(xg

f

=

( )

( )

f x

g x =

2 4 1

3 2

x x

x


))(( xgf = ( ( ))f g x = 2( 4 1) f x x = 23( 4 1) 2 x x = 23 12 1 x x


))(( xfg = ( ( ))g f x = (3 2)g x = 2(3 2) 4(3 2) 1x x =

2(9 12 4) 12 8 1x x x =

29 24 13x x

8. If 23)( xxf and 14)( 2 xxxg , find: ))2((gf

))2((gf = ( 3)f = 3(–3) – 2 = – 9 – 2 = – 11

9. If 32)( xxf , find )(1 xf

32)( xxf Change the f(x) to y.

2 3y x Now reverse the roles of the x and y to form the inverse.

2 3x y This is the inverse, but now solve for y.

3

2

xy

Now name the function )(1 xf

)(1 xf = 3

2

x

10. If 3 14)( xxf , find )(1 xf .

3 14)( xxf

3 4 1y x Now reverse the roles of the x and y to form the inverse.

3 4 1x y This is the inverse, but now solve for y.

3

3 3 4 1x y

3 4 1x y

3 1

4

xy

)(1 xf =

3 1

4

x

11. If )(xf contains the point (4,1), then )(1 xf must contain what point?

The x and y values swap, so an ordered pair in )(1 xf is (1,4).

12. Find the domain of 2( ) 6 5f x x x

There will be restrictions on the domain if we have a square root (or any even root), if we have a variable in the

denominator of a fraction, or if we have a log. None of these occur in this function, so there are no restrictions on the

domain. The domain is ( , ) .


( )4 8

f xx

The denominator cannot equal zero.

4 8 0x

2x

The domain is ( ,2) (2, ) .


1)(

2

xx

xxf

The denominator cannot equal zero. 2 2 8 0x x

( 4)( 2) 0x x

4, 2x x

The domain is ( , 2) ( 2,4) (4, ) .

15. Find the domain of ( ) 3 9 1f x x

3 9x must be bigger than or equal to zero.

3 9 0x

3x

The domain is [ 3, ) .

16. If 12)3(

)32( 2

x

x

kxx, find the value of k .

Multiply by (x – 3) on both sides of the equation to get: 22 3 (2 1)( 3)x kx x x . 2 22 3 2 7 3x kx x x

So, k must be equal to -7.

17. If 14 yx and 53 xz , find an expression for z in terms of y .

Use substitution to get 3(4 1) 5z y .

12 2z y

18. Find the vertex of the function, 563)( 2 xxxf .

Method 1: Complete the square to obtain the standard form of 2( ) ( )f x a x h k , where the vertex is ( , )h k .

563)( 2 xxxf 2( ) 3( 2 ) 5f x x x 2( ) 3( 2 1) 5 3(1)f x x x

2( ) 3( 1) 8f x x

The vertex is ( 1, 8)

Method 2: x = 2

b

x

in 2( )f x ax bx c

563)( 2 xxxf

61

2(3)x

2( 1) 3( 1) 6( 1) 5f = – 8

The vertex is ( 1, 8) .

19. What are the zeros of the function, 382)( 2 xxxf ?

The zeros are the x-intercept values, where y = 0. 22 8 3y x x 20 2 8 3x x Since factoring is not possible, use the quadratic formula.

2( 8) ( 8) 4(2)(3)

2(2)x

4 1 10

2x

or

12 10

2x

20. What are the zeros of the function, 1892)( 23 xxxxf ?

This function is factorable by grouping: 2 2( ) ( 2) 9( 2) ( 9)( 2) ( 3)( 3)( 2)f x x x x x x x x x .

Set y = 0 to get zeros of 3, – 3 , and – 2.

21. Find the sum of the solutions of 452 2 xx .

The solutions of the equation are 5 57

4

.

5 57

4

+

5 57

4

=

5 5 5

4 4 2

22. Write a cubic function that has ,2,0 xx and 5x as zeros.

The zeros will create factors of ( )( 2)( 5)x x x .

Multiply these out to get 3 2( ) 3 10f x x x x .

23. Write a quadratic function that has a vertex at (2,-3) and contains the point (-2,5).

Use the standard form of a parabola: 2( ) ( )f x a x h k .

Insert the vertex: 2( ) ( 2) ( 3)f x a x .

Now use the other ordered pair: 25 ( 2 2) ( 3)a

1

2a

Now create the function using the a value: 21( ) ( 2) ( 3)

2f x x or f(x) = 21

2 12

x x

24. Simplify 3

2

2 )(a

Multiply the powers together.

25. Simplify 8

62

6

3

ab

ba

Take a factor of 3 out of the numerator and the denominator.

Subtract the powers of a to get 1a in the numerator.

Subtract the powers of b, starting in the denominator to get 2b in the denominator.

22

a

b

26. Simplify

2

1

5

3

a

a

Subtract the exponents: 3 1 6 5 1

5 2 10 10 10 . The result is

1

10a .

27. Simplify 7

32 )4(x

x

Take the power to a power first: 3 6

7

4 x

x.

Now deal with the negative exponent: 3 6 74 x x

Simplify: 1364x

28. Simplify 2

5

2

3

2

3

2

1

23 baba

Combine the constants, and combine the variables as appropriate. 1 3 3 5

2 2 2 26a b

= 2 46a b

29. Simplify

3

63

52

3

2

yx

yx

Deal with the outside power first, then the negative exponents, and then simplify. 3 6 15

3 9 18

2

3

x y

x y

=

3 15 18

3 6 9

3

2

y y

x x =

33

15

27

8

y

x

30. Evaluate 2

1

6

1

88

Add the exponents first: 1 1

6 28

= 2

38

Remember what the fractional exponent means and simplify: 2

3 8 = 22 = 4

31. Evaluate 3

2

3

1

33 1 2

3 33

= 3

32. Simplify, assuming that the variables represent positive numbers:3 2aa

21

32a a = 7

6a = 6 7a = 6a a

33. Simplify, assuming that the variables represent positive numbers: 53 ba 1 1

3 5a b = 5 3

15 15a b = 1

5 3 15( )a b = 15 5 3a b

34. Simplify, assuming that the variables represent positive numbers: baba 42 82

baba 42 82 = 6 216a b = 34a b

35. Simplify, assuming that the variables represent positive numbers:4 354 52 44 baba

2 5 5 34 4 4a a b b = 7 84 16a b = 2 342ab a

36. Rewrite using logarithmic notation: 1642

Remember the relationship between the exponential form of an equation and the logarithmic form:

logx

aa y y x to get 4log 16 2

37. Rewrite using logarithmic notation: yM x

Remember logx

aa y y x to get logM y x

38. Rewrite using exponential notation: 2100log10

Remember logx

aa y y x to get 210 100

39. Rewrite using exponential notation: 13

1log3

Remember logx

aa y y x to get 1 13

3

40. Evaluate without a calculator:

4

1log2

2

1log

4x

12

4

x 2

12

2

x 22 2x 2x

41. Expand:3

2

logz

yxa

Use the rules to expand a logarithm:

( ) log( ) log( )log ab a b

log( ) log( )a

log a bb

log logax a x

3

2

logz

yxa = 2 3log log loga a ax y z =

1

32log log loga a ax y z = 1

2log log log3

a a ax y z

42. Express as a single logarithm: tzy 10101010 log8log6log43log2

Use the rules above.

tzy 10101010 log8log6log43log2 = 2 4 6 8

10 10 10 10log 3 log log logy z t =

4 6 8

10 10 10 10log 9 log log logy z t = 4 6 8

10 10log (9 ) log ( )y z t = 4

10 6 8

9log

y

z t

43. Simplify i2

to i, -i, 1, or -1.

Remember that 2i is defined to be -1.


Remember 4 2 2 1 1 1i i i

So, 8i = 4 4i i = 1 1 1

45. Simplify 11i to i, -i, 1, or -1. 11i = 4 4 2i i i i = (1)(1)(-1)i = -i

46. Add and write your answer in a+bi form: (1 + i) + (3 + 5i)

(1 + i) + (3 + 5i) = (1+3) + (i+ 5i) = 8 + 6i

47. Subtract and write your answer in a+bi form:(3 – 4i) – (7 – i)

(3 – 4i) – (7 – i) = 3 – 4i – 7 + i = -4 – 3i

48. Multiply and write your answer in a+bi form: 43 i

43 i = (3)(4)i = 12i

49. Multiply and write your answer in a+bi form: ii 52

ii 52 = 210i = (10)(-1) = - 10 and in a+bi form: -10 + 0i

50. Multiply and write your answer in a+bi form: )52(7 ii

)52(7 ii = 214 35i i = 14 35( 1)i = 35 14i

51. Multiply and write your answer in a+bi form: )43)(21( ii

)43)(21( ii = 3 10 8( 1)i = 5 10i

52. Multiply and write your answer in a+bi form: 3)2( i

(2 )(2 )(2 )i i i = 2(2 )(4 4 )i i i = (2 )(3 4 )i i = 2 11i


2

2

3

i

i i =

2

2

3

i

i =

2

3( 1)

i

=

2

3

i = 0

2

3i


6

6 2 3

2 3 2 3

i

i i

=

2

6(2 3 )

4 9

i

i

=

12 18

13

i =

12 18

13 13i

55. Divide and write your answer in a+bi form: 10 10

1 3

i

i

10 10 1 3

1 3 1 3

i i

i i

=

2

2

10 40 30

1 9

i i

i

=

20 40

10

i =

20 40

10 10i = 2 4i


0252 x 2 25x

2 25x

1 25x

5x i

57. Solve for x over the complex number system: 422 xx 2 2 4 0x x

Use the quadratic formula: 2( 2) ( 2) 4(1)(4)

2(1)

=

2 12

2

=

2 2 3

2 2i = 1 3 i or 1 3i


122 x

2 12x

2 3x i

59. Write the first five terms of the sequence having the general term 2

1 1)1(

n

na n

n

1 1

1 2

1 1( 1)

1a =

21 2

1

2 1

2 2

2 1 3 3( 1) ( 1)

2 4 4a

3 1

3 2

3 1 4 4( 1) (1)

3 9 9a

4

5

16a and 5

6

25a

60. A certain arithmetic sequence has 561 a and 2611 a . Find 17a .

Use the formula for a particular term in an arithmetic sequence: 1 ( 1)( )na a n d

26 56 (11 1)( ) 26 56 10 3d d d

Since we now have 1a and d, use the formula to find

na : 56 ( 1)( 3) 56 3 3 59 3na n n n . Use this

formula to find 17a .

17 59 3(17) 8a

61. A certain geometric sequence has 641 a and 3245 a . Find 7a .

Use the formula for a particular term in a geometric sequence: 1

1

n

na a r

5 1 4 4324 64 5.0625 5.0625 1.5r r r r

Since we now have 1a and r, use the formula to find na :

13

642

n

na

6

7

364 729

2a

62. Find the sum of the first ten terms of an arithmetic sequence having 271 a and 9d .

Use the formula for the sum of an arithmetic sequence: 12 ( 1)2

n

nS a n d

10

102(27) (10 1)(9) 5 54 81 675

2S

63. Find the sum of the first ten terms of a geometric sequence having 30721 a and 2

3r .

Use the formula for the sum of a geometric sequence:

1

1

1

n

n

rS a

r

10

10

31

23072 348150

31

2

S

64. Evaluate

23

1

)36(n

n .

23

1

(6 3) (6 1 3) (6 2 3) (6 3 3) ... (6 23 3)n

n

= 3 + 9 + 15 + …+ 138.

We can see that the common difference is 6, so we will use the formula for the sum of an arithmetic series:

12 ( 1)2

n

nS a n d

23

232(3) (23 1)(6) 11.5 138 1587

2S

65. Evaluate

14

1 2

18

n

n

1 1 1 2 1 3 1 4 14

1

1 1 1 1 18 8 8 8 8 8 4 2 1 15

2 2 2 2 2

n

n

66. Evaluate 9!

3! 6!

9! 9 8 7 6! 9 8 784

3! 6! 3 2 1 6! 3 2 1

67. Give the entry in the first row and the first column for

62

30

43

5

05

73

28

Multiply each entry in the second matrix by 5. Add the first row first column entries together.

8 + -15 = -7

68. Give the entry in the second row and the first column for

30

52

43

65

32

Multiply the second row of the first matrix by the first column of the second to get this entry:

(– 2)(2) + (5)(5) = – 4 + 25 = 21

69. Evaluate 53

912

12 9(12)(5) ( 3)( 9) 60 27 33

3 5

70. Evaluate

132

065

213

Choose a row or column to work with, such as the third column.

3 1 25 6 3 1

5 6 0 (2) ( 1) (2)[ 15 ( 12)] ( 1)[ 18 ( 5)] (2)( 3) ( 1)( 13) 6 13 72 3 5 6

2 3 1

71. Give the z-value of the solution to the following system:

4 5 11

2 7

2 1

x y

x z

y z

Rewrite the system as:

4 5 11

2 7

2 1

x y

x z

y z

Multiply the 2nd

equation by -2 and add this new equation to the 1st equation to eliminate x .

Now we have two equations with two variables: 5 2 3

2 1

y z

y z

We want eliminate y now, so multiply the top equation by 2 and the bottom equation by 5. When we add them

together, we get z = -1.

72. If r

aS

1, then r = ?

(1 )1

a S aS S r a S rS a S a rS

r S

73. If 2 3 1 6 8

5 6 0 2 15 17

k

, then k =?

(2)(k) + (3)(0) = 6 and k = 3

74. If is defined to be: 1 yxyx and 315x , then x = ?

55 1x x

5 1 31x 5 32x

5 5 5 32x

2x

75. If 1

4 2

k=10, then k = ?

12 4 1 2 4

4 2

kk k

2k – 4 = 10

k = 7

76. In the list 12,5

3,2.16,

8

0,,25,7,

2

1 , the sum of all the rational numbers is:

1 0 3 1 325 16.2 12 5 0 16.2 12 33.3

2 8 5 2 5

77. How many integers are in the list: 12,5

3,2.16,

8

0,,25,7,

2

1 ?

025, , 12

8 are integers, since they are 5, 0, and 12. There are 3 integers.

78. Find the fourth term in the expansion of 7)3( zy .

Answer:

4 3 4 3 4 3

73 (35)(81) ( 1) 2835

3y z y z y z

79. How many terms are in the expansion of 7)3( zy ?

The number of terms is one more than the power, which makes it eight.

80. A set containing five elements has how many subsets?

There are 52 subsets.

Documents

Advanced Algebra Topics COMPASS Review