Unit Test — Unit test Answers

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5

What is a correct first step in solving the inequality –4(3 – 5x)≥ –6x + 9?

A
–12 – 20x ≤ –6x + 9
B
–12 – 20x ≥ –6x + 9
C
–12 + 20x ≤ –6x + 9
D
–12 + 20x ≥ –6x + 9
6

On a coordinate plane, 2 lines are shown. Line H J has points (negative 4, negative 2) and (0, 4). Line F G has points (negative 4, 1) and (0, negative 2).

Question illustration
A
They are perpendicular because their slopes are equal.
B
They are perpendicular because their slopes are negative reciprocals.
C
They are not perpendicular because their slopes are equal.
D
They are not perpendicular because their slopes are not negative reciprocals.
9

Which equation represents the line that passes through (–6, 7) and (–3, 6)?

A
y = –x + 9
Option A
B
y = –x + 5
Option B
C
y = –3x – 11y
D
y = –3x + 25
11

On a coordinate plane, 2 lines are shown. Line C D has points (negative 2, 4) and (0, negative 4). Line F G has points (negative 4, 0) and (4, 2).

Question illustration
A
They are perpendicular because their slopes are equal.
B
They are perpendicular because their slopes are negative reciprocals.
C
They are not perpendicular because their slopes are equal.
D
They are not perpendicular because their slopes are negative reciprocals.
13

Which graph represents a function with direct variation?

A
Option
B
Option
C
Option
D
Option
14

Which equation represents a line that passes through (2, –) and has a slope of 3?

Question illustration
A
y – 2 = 3(x + )
Option A
B
y – 3 = 2(x + )
Option B
C
y + = 3(x – 2)
Option C
D
y + = 2(x – 3)
Option D
16

What is the value of x in the equation ?

Question illustration
A
StartFraction 2 Over 3 EndFraction left-parenthesis StartFraction one-half EndFraction. x plus 12 right-parenthesis equals left-parenthesis StartFraction one-half EndFraction left-parenthesis StartFraction one-third EndFraction x plus 14 right-parenthesis minus 3.
Option A
B
StartFraction one-half EndFraction left-parenthesis n minus 4 right-parenthesis minus 3 equals 3 minus left-parenthesis 2 n plus 3 right-parenthesis.
Option B
C
2
D
13
17

The given line segment has a midpoint at (3, 1).

Question illustration
A
y = x
Option A
B
y = x – 2
Option B
C
y = 3x
D
y = 3x − 8
18

Lily begins solving the equation 4(x – 1) – x = 3(x + 5) – 11. Her work is shown below.4(x – 1) – x = 3(x + 5) – 11 4x – 4 – x = 3x + 15 – 113x – 4 = 3x + 4

A
The equation has one solution: x = 0.
B
The equation has one solution: x = 8.
C
The equation has no solution.
D
The equation has infinite solutions.

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