Transcript Document

COMPASS Algebra
Practice Test A
• This practice test is 10 items long.
If x = -2 and y = 5, what is the value
of the expression 2x3 – 3xy ?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
What are the solutions to the
quadratic equation x2 – 2x = 48?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
Rationalize the denominator.
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
4
3
What is the equation of the line that
contains the points (-1, 1) and (2 , 7)?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
x  7 x  12
For all x  ±3,
?
2
x 9
2
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
Which is the complete
3
factorization of 5 y  125 y ?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
(3, -15)
What is the product of (4 2  5 5 )
and
( 2  3 5) ?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
1
If 3 x  8, then x  ?
2
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
1
x
3
If,
what is the value of
the expression 9 x 2  6 x  15 ?
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
For all b  0, 8b  16b  ...
4b
2
I know how to solve this problem.
a) Yes
b) No
c) I’m not sure
Answers
Algebra Practice Test A
1.
2.
3.
4.
5.
A
D
B
E
D
6. A
7. C
8. C
9. B
10.E
A1. If x = -2 and y = 5, what is the
value of the expression 2x3 – 3xy ?
 A.
 B.
 C.
 D.
 E.
14
46
54
-46
-54
2x3 – 3xy
2(-2)3 – 3(-2)(5)
2(-8) – 3(-10)
-16 – -30
-16+30 = 14
What are the solutions to the
quadratic equation x2 – 2x = 48?
A2.
 A.
 B.
 C.
 D.
 E.
-12, -4
-8, -6
6, 8
-6, 8
-8, 6
Set the equation equal to
zero.
x2 – 2x = 48
x2 – 2x – 48 = 0
Factor
(x + 6)(x – 8) = 0
Write a solution set.
x = {-6 , 8}
A3. Rationalize the denominator.
2
 A.
 D. 2
3
9
 B.
 C.
2 3
3
4
 E.
6
9
4
3
A3. Rationalize the denominator.
Rationalize the denominator
means to remove the radical from
the bottom of the fraction
2  3 2 3 2 3
4






3  3
3
3
9
First simplify
the expression.
Multiply the top and
bottom by a number
that will make the
denominator a
perfect square.
Then simplify the
denominator to get
rid of the radical.
4
3
This is a match
for B
A4. What is the equation of the line that
contains the points (-1, 1) and (2 , 7)?
 A.
y  5x  3
 D.
y  2x  5
 B.
y  3x  3
 E.
y  2x  3
 C.
3
y  x5
4
A4. What is the equation of the line that
contains the points (-1, 1) and (2 , 7)?
First calculate the slope m.
y2  y1
6
7 1
m
 2

x2  x1 2  (1) 3
Now find b
y  mx  b
7  2( 2)  b
7  4b
4 4
3b
Use Slope
Intercept
y  mx  b
Now put m and b together to
write the equation.
y  2x  3
This is a match
for E
x  7 x  12
A5. For all x  ±3,
?
2
x 9
2
 A.
 B.
 C.
x4
x 3
x4
x3
x4
x 3
 D.
 E.
x4
x3
 7 x  12
9
x  7 x  12
A5. For all x  ±3,
?
2
x 9
2
Rational Expression
Factor
x  7 x  12 ( x  3)( x  4) ( x  4)


2
( x  3)( x  3) ( x  3)
x 9
2
Cancel
This is a match
for D
A6. Which is the complete
factorization of 5 y 3  125 y ?
 A.
y(5 y  25)
2
5y3 – 125y
 B. 5 y 2 ( y  25)
Factor out the
common 5y
 C. 5 y ( y  5)( y  5)
5y(y2 – 25)
Difference of squares
2
 D. 5 y ( y  5)
5y(y + 5)(y – 5)
 E.
5 y( y  5)
2
A7. What is the product of (4 2  5 5 )
and ( 2  3 5 ) ?
A. 6 2  8 5
(4 2  5 5 )( 2  3 5 )
 B.
 C.
3 3
83  17 10
4 4  12 10  5 10  15 25
4( 2)  17 10  15(5)
8  17 10  75
 D.
74 10
 E.
40  13 10
83  17 10
1
A8. If 3 x  8, then x  ?
2
 A.
9
2
 B.
2
9
 C.
2
2
7
7
 D.
16
1
 E. 11
2
1
A8. If 3 x  8, then x  ?
2
Multiply by the reciprocal
First convert 3½
to an improper
7
2
2




fraction
  x 8  
2
7


7
1 6 1 7
16
3   
x
2 2 2 2
7
Convert 16/7 to a
mixed number
16 14 2
2
2
x
   2  2
7
7 7
7
7
This is a match
for C
1
x
3
A9. If,
what is the value of
the expression 9 x 2  6 x  15 ?
 A.
 B.
 C.
 D.
 E.
12
14
17
22
36
2
1
1
9   6   15
3
3
1
 9   2  15
9
 1  2 15  14
Rational Expression
A10. For all b  0, 8b  16b  ... Factor
2
 A.
 B.
 C.
 D.
 E.
4b
2
18b
8b  16b 4b(2b  4)

6b
4b
4b
Cancel
32b2 + 48b
2b2 + 4b
2b + 4
= 2b + 4