Download Unit 2

Practice Test
Unit 2 (Part 2)
2 – 6x + 8 = 0 and y = x + 3, then what
If
x
1
are the possible values of y ?
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x – 2 = 0, x – 4 = 0
x=2
x=4
Let x = 2: y = x + 3
Let x = 4: y = x + 3
=2+3
= 5
=4+3
= 7
2
(2x + 3y)2 – (2x – 3y)2
(2x + 3y)(2x + 3y) – (2x – 3y) (2x – 3y)
(4x2 + 6xy + 6xy + 9y2) – (4x2 – 6xy – 6xy + 9y2)
(4x2 + 12xy + 9y2) – (4x2 – 12xy + 9y2)
4x2 + 12xy + 9y2 – 4x2 + 12xy – 9y2
24xy
3
If x2 + y2 = 37 and xy = 24, what is
the value of (x – y)2?
(x – y)2 =
=
x2 + y2 = 37
=
=
=
xy = 24
=
=
(x – y)(x – y)
x2 – xy – xy + y2
x2 – 2xy + y2
x2 + y2 – 2xy
37 – 2xy
37 – 2(24)
37 – 48 = –11
4
If (–8x + 3)(–4x2 + 4x + 6) = ax3 + bx2 + cx + d
for all real values of x, what is the value of c ?
(–8x + 3)(–4x2 + 4x + 6)
–4x2
3
32x
–8x
3 –12x2
6
4x
–32x2 –48x
12x
18
32x3 – 44x2 – 36x + 18
c = –36
(Add matching colors)
5
If (y – 5)2 = 0, then find the
value of y2 – 2y ?
(y – 5)2 = 0
(y – 5)(y – 5) = 0
y–5=0, y–5=0
y=5
y=5
Find y2 – 2y
when y = 5
(5)2 – 2(5)
25 – 10
15
6
If x and y are positive integers, then which of
4x  6 y
the following must be equal to
?
2
2
4x  9 y
2
2(2 x  3 y )
4x  6 y


2
2
4 x  9 y (2 x  3 y )(2 x  3 y ) 2 x  3 y
7
Step 1 x + y
x
x y
y

  1
x x
x
Step 2
y 1
x
If  r , then reciprocal is

y
x r
Step 3
y
1
r 1
r +1
1
 1   
x
r
r r
r
8
3
1
a
If x =
and ab  0, then 3 
a
b
3
a
x=
b
3
x a
=
1 b
1  a3 = b  x
a3
= bx
Reciprocal
1
1

3
a
bx
Solve for m.
9
1 3
2
 2 
4 m
m
LCD = 4m2
4m  1 3  4m  2 
  2 
 
1 4 m 
1  m
2
2
4m2 12m2 8m2
 2 
4
m
m
m2 + 12 = 8m
m2 – 8m + 12 = 0
m2 – 8m + 12 = 0
(m – 2)(m – 6) = 0
m–2=0, m–6=0
m=2
m=6
If 4 x  3  5 , then find
the value of x + 4 ?
10

4x  3  5
4x  3

2
 5
2
4x – 3 = 25
+3 +3
4x
= 28
x
= 7
Find x + 4
when x = 7
7+4
11
11 If x is an integer and 7  x  2  8 , how
many different values of x are possible?
First, solve
inequality.
7  x2 8
7
2


x2

2
 8
49 < x – 2 < 64
+2
51 <
+2
x
+2
< 66
2
11 If x is an integer and 7  x  2  8 , how
many different values of x are possible?
51 < x < 66
Integers between
51 and 66
52 53 54 55 56
57 58 59 60 61
62 63 64 65
Answer: 14 Values
OR
66 – 51 = 15
15 – 1 = 14
11
Try this strategy
Use an inequality with two numbers
that are close together.
2 < m < 7
Integers between 2 and 7
3
4 5 6
Answer: 4 Values
OR
7–2=5
5–1=4
How many values
of m are possible?
10 < m
< 45
45 – 10 = 35
35 – 1 = 34
12
16  w  5

16  w
   5
2
2
16  w  25
–16
–16
w 9
 w
2
 9
w = 81
2
13 If a  8 and
b  64
a  8
 
a
2
  8
a = 64
b  3 a , then what is the
value of b ?
3
2
b 4
 b
2
 4
b = 16
2
14
–3x + 13 < –14
–13 –13
–3x
< –27
3x
3
x
27
>
3
>
9
15
10
If
 x , then which of the following
x
values could be x ?
Strategy: Test each answer by substituting for x.
A. –10
B. –5
10 < x
x
10 < –10
–10
–1 < –10
10 < x
x
10 < –5
–5
–2 < –5
NO
NO
15
10
If
 x , then which of the following
x
values could be x ?
C. 1
D. 2
E. 5
10 < x
x
10 < 1
1
10 < 1
10 < x
x
10 < 2
2
5 <2
10 < x
x
10 < 5
5
2 <5
NO
NO
YES
16
1
If 0  x  , then which of the
2
following statements must be true?
Strategy: Substitute ¼ for x in each answer.
1
2x = 1 2    1
4
21
  1
14
2
1
4
1
NO  1
2
xx
2
1 1
  
4 4
1
1

4 16
.25  .0625
.2500  .0625
2
x2
x
2
.25
.25 


2
2
.0625
.25 
2
.25  .03125
NO
16
1
If 0  x  , then which of the
2
following statements must be true?
Strategy: Substitute ¼ for x in each answer.
1
2
2
x  x 1 1
2x > 1 2  4   1
  
4 4
21
1


1
1
1
4



4 16
2
1
.25  .0625
4
1
NO .2500  .0625
NO 2  1
17
If x > y and y > 0 and xz < 0, then which
of the following must be true about all the
values of z?
y > 0  y = ( + ) positive
x > y  x = ( + ) positive
xz < 0 
xz = ( – ) negative
(+)z = ( – )
(+)(–) = ( – )
z is negative
z<0
18 If the sum of two integers x and k is less than x,
which of the following must be true?
x+k < x
–x
–x
k < 0
19 Twice the difference between a certain number and
its square root is 15 more than twice the number.
Which of the following equations represents the
statement above?
A.
B.
C.
D.
E.
2 N  N = 15  N
2 N  N = 15  2 N
2 N  N = 15  2 N


2 N  N + 15 = N
2 N  N +15  N


20 If a number is doubled and then increased by 10,
the result is 5 less than the square of the number.
Which of the following equations represents the
statement above?
A.
B.
C.
D.
E.
2 N + 10 = 5  N
2 N + 10 =
N 5
2  N + 10  =
N 5
2N + 10 = – 5
2
2N + 10 = N + 5
2
N
21
If 2 is subtracted from a number and this
difference is tripled, the result is 6 more than
the number. Find the number.
Let number = x
‘2 is subtracted from a number’ = x – 2
‘The difference is tripled’ = 3(x – 2)
3(x – 2) = x + 6
3x – 6 = x + 6
–x
–x
2x – 6 =
6
2x – 6 = 6
+6 +6
2x
= 12
x
= 6
22 If the sum of two consecutive odd integers
is 28, what is the product?
Let 1st integer = x
Let 2nd integer = x + 2
x + x + 2 = 28
2x + 2 =
–2
2x
=
x
=
28
–2
26
13
13
13+2 = 15
Sum
= 13 + 15
= 28
Product
= 13  15
= 195
23 If a positive integer is doubled and then
increased by 10, the result is 5 less than the
square of the integer. What is the integer?
Let integer = x
2x + 10 = x2 – 5
–2x – 10 –2x – 10
0 = x2 – 2x – 15
0 = (x + 3)(x – 5)
x+3=0,x–5=0
x = –3
x=5
–15
3 –5
–2
Two numbers with
Product of –15 and
Sum of –2
24 Jon buys one pencil and two pens for $3.50.
Lauren buys four pencils and three pens for $5.50.
How much would one pencil and one pen cost?
Cost of pencil = A
Cost of pen = B
Jon  1A + 2B = 3.50 (Multiply Jon by –4)
Lauren  4A + 3B = 5.50
Add
Jon  –4A – 8B = –14.00
Lauren  4A + 3B = 5.50
Equations
– 5B = –8.50
B = 1.70 (Pen cost)
24 Jon buys one pencil and two pens for $3.50.
Lauren buys four pencils and three pens for $5.50.
How much would one pencil and one pen cost?
Cost of pencil = A
B = 1.70 (Pen cost)
1A + 2B = 3.50
4A + 3B = 5.50
Cost of pencil and pen
= A + B
= 0.10 + 1.70 = 1.80
Cost of pen = B
Find A. Use one of
the original equations.
1A + 2B = 3.50
A + 2(1.70) = 3.50
A + 3.40 = 3.50
A
= 0.10
If y varies directly as x2, and y = 3 when
x = 3, what is the value of y when x is 6?
25
y varies directly as x
y varies directly as x2
y1 y2

x1 x2
y1
y2

2
2
x1  x2 
y1  3
x1  3
x1 
2
9
3 y

9 36
9  y  3  36
9 y  108
y  12
y2  ?
x2  6
 x2 
2
 36
26 Students receive 5 bonus points for every 2 community
service projects they perform. If Mark received 100
bonus points, how many projects did he perform?
Note: As the bonus points increase, the
community service projects should increase.
Direct
Variation
y1 y2

x1 x2
2
p
=
5 100
5  p = 2  100
5p = 200
p = 40
27
If it takes 4 men 3 hours each to pave a
playground, how many hours will it take
12 men to complete the same task?
Note: Increasing the number of men will decrease
the amount of time to complete the task.
Inverse
Variation
x1y1 = x2y2
M1H1 = M2H2
4 · 3 = 12 · H2
12 = 12H2
1 = H2
1 Hour
28
What is the value of
f(x) = 3x + 3x + 30 if x = 3 ?
f(3) =
3
3
+ 3(3) +
0
3
f(3) = 27 + 9 + 1
f(3) = 37
29
If f(x) = x + 2x, what is the
value of f(–2)?
f(–2) = –2 + 2–2
1
1
2 1
 2  2  2  

2
4
1 4
8 1
7
 2 4  1
     

 1 4 4
4 4
4
30
Find the domain for f ( x )  x  5
Note: We can only evaluate the square root
of numbers greater than or equal to zero.
Let expression inside radical be > 0.
x–5 > 0
+5 +5
x
> 5
31
The amount a restaurant owner pays for coffee beans is
directly proportional to the number of pounds of coffee
she buys. If she buys n pounds of coffee at d dollars
per pound, what is the total amount she pays, in
dollars, in terms of n and d.
n lb.
d dollars/lb.
total
1 lb.
$3 dollars/lb. 1  3 = $3
2 lb.
$3 dollars/lb. 2  3 = $6
3 lb.
$3 dollars/lb. 3  3 = $9
n lb.
$d dollars/lb.
nd
32
The cost of preparing for a book sale is $30.
If each book is sold for $3.00, express the
profit as a function of n, where n represents
the number of books sold.
Book # of books Preparation
Cost
sold
Cost
Profit
3
1
30
3(1) – 30 = 3–30 = –27
3
2
30
3(2) – 30 = 6–30 = –24
3
3
30
3(3) – 30 = 9–30 = –21
3
4
30
3(4) – 30 = 12–30 = –18
32
The cost of preparing for a book sale is $30.
If each book is sold for $3.00, express the
profit as a function of n, where n represents
the number of books sold.
Book # of books Preparation
Cost
sold
Cost
3
6
30
3
3
3
10
12
n
30
30
30
Profit
3(6) – 30 = 18–30 = –12
3(10) – 30 = 30–30 = 0
3(12) – 30 = 36–30 = 6
3n – 30
f(n) = 3n – 30
33
Morgan’s plant grew from 42 centimeters to 57
centimeters in a year. Linda’s plant, which was 59
centimeters at the beginning of the year, grew twice
as many centimeters as Morgan’s plant did during
the same year. How tall, in centimeters, was
Linda’s plant at the end of the year?
Step 1
Centimeters Morgan’s plant grew = 57 – 42 = 15 cm.
Step 2
Twice centimeters Morgan’s plant grew = 2(15) = 30 cm.
Step 3
Height of Linda’s plant = 59 cm + Step 2
= 59 cm + 30 cm = 89 cm.
34
If h( x)  3 x  1 , then h(x) is
I. Always Positive
II. Never Negative
III. Always an Integer
False
False
Note 1 We can only evaluate the square root
of numbers greater than or equal to zero.
Note 2
0 0
Note 3
8  2.83
35
For all numbers x and y, let xy be defined as
xy = xy + y2. What is the value of (31)1 ?
(31) = 31 + 12
= 3 + 1
=
4
(31)1 = 41
= 41 + 12
= 4 + 1
=
5
36
x  10
Which values are not in the domain of f ( x)  2
x  25
Let denominator = 0
Solve equation for x.
x2 – 25 = 0
(x – 5)(x + 5) = 0
x–5=0 , x+5=0
+5 +5
–5 –5
x = 5
x = –5
{–5,5}
37 The sign-up fee at a gym is $50. Members then must
pay $25 each month. Express the cost of using the
gym as a function of m, where m represents the
number of months the member participates.
Sign-up
Fee
# of
Months
Monthly
Cost
50
50
50
50
1
2
3
m
25(1)
25(2)
25(3)
25(m)
Gym
Cost
50 + 25(1) = 75
50 + 25(2) = 100
50 + 25(3) = 125
50 + 25(m)
f(m) = 50 + 25m
38 Each time Shannon pushes the button on a machine, a
bell rings 7 times. Each time she turns the switch on
the machine, the bell rings 3 times. During one hour,
Shannon caused the bell on the machine to ring 23
times. How many times did she push the button?
# of Rings
7(1) = 7
Total rings left
23 – 7 = 16 NO
# of Rings
7(2) = 14
3(3) = 9
Total rings left
23 – 14 = 9 YES
Pushed Button 1
There are 3 rings each time a switch is turned.
Thus, the number of rings must be a multiple of 3.
Pushed Button 2
Turned Switch
3
39 Which system of inequalities best represents the graph?
A. Wrong
2
y  x2
3
y  x  3
C.
B.
3
y  x2
2
y  x  3
D.
2
y  x2
3
y  x  3
2
y  x2
3
y  x  3
The dotted line has a negative slope.
The 2nd inequality for each answer has a negative slope.
The inequality sign should be < or > .
39 Which system of inequalities best represents the graph?
B.
3
y  x2
2
y  x  3
C.
2
y  x2
3
y  x  3
D. Wrong
2
y  x2
3
y  x  3
The solid line has a positive slope.
The 1st inequality for each answer has a positive slope.
Inequality sign should be < . The shading is below the line.
39 Which system of inequalities best represents the graph?
B.
C.
2
3
y  x2
y  x2
3
2
y  x  3
y  x  3
3
2

rise 2
m

run 3
- Answers B and C have

different slopes for
the solid line.
- Find two points on the line.
- Use rise run to get to each point on the line.
This will determine the slope and equation of the line.
40 Determine the solution of the system of inequalities.
y < –x – 1 Negative Slope (Shaded Below)
–2x + y > –2
y > 2x – 2 Positive Slope (Shaded Above)
A.
B.
C.
D.