Identify the inequality with x = 3 as a solution. Select all that apply.(3 points)

Question

Responses

x + 3 < 4(x + 1) + 2
x + 3 < 4(x + 1) + 2

2x - 8 ≥ 7 - 3x
2x - 8 ≥ 7 - 3x

3(x + 2) < 14 - x
3(x + 2) < 14 - x

4(x - 1) < x + 5
4(x - 1) < x + 5

5 - 2( x + 1) > x
5 - 2( x + 1) > x

6(x - 4) ≤ 2x
6(x - 4) ≤ 2x
Question 2
The following inequality is solved for x.

Line A 4(x + 1) + 8 ≥ 4 + 2x

Line B 4x + 4 + 8 ≥ 4 + 2x

Line C 4x + 12 ≥ 4 + 2x

Line D 2x + 12 ≥ 4

Line E 2x ≥ - 8

Line F x ≥ -4

Which Line shows the inequality after you combine like terms?

(1 point)
Responses

Line B
Line B

Line C
Line C

Line D
Line D

Line E
Line E

Line F
Line F
Question 3
Mari is solving the inequality 4(x - 3) > 16. Her first step is 4x - 12 > 16.

Which step could be the next step? Select all that apply.

(3 points)
Responses

4x - 12 + 12 > 16 + 12
4x - 12 + 12 > 16 + 12

4x - 12 - (-12) > 16 - (-12)
4x - 12 - (-12) > 16 - (-12)

(14)(4x − 12) > 16 (1(4))
(14)(4x − 12) > 16 (1(4))

4x4 − 12 > 164
4x4 − 12 > 164

4(4x - 12) > 16(4)
4(4x - 12) > 16(4)
Question 4
Javon and Ivy are both given the equation 5 − 2x−13 ≤ 4
. Javon thinks the first step is − 2x−13 ≤ 9
. Ivy thinks the first step is −2x− 13 ≤ −1
. Who is incorrect and why?(1 point)
Responses

Ivy is incorrect because, on her first step, she did not add 5 (or equivalently subtract -5) from both sides of the equation.
Ivy is incorrect because, on her first step, she did not add 5 (or equivalently subtract -5) from both sides of the equation.

Both students are incorrect because, on their first step, they did not multiply by 3 on both sides of the equation.
Both students are incorrect because, on their first step, they did not multiply by 3 on both sides of the equation.

Both students are incorrect because, on their first step, they did not divide by 3 (or equivalently multiply by 13
) from both sides of the equation.
Both students are incorrect because, on their first step, they did not divide by 3 (or equivalently multiply by 1 third) from both sides of the equation.

Javon is incorrect because, on his first step, he did not subtract 5 ( or equivalently add -5) from both sides of the equation.
Javon is incorrect because, on his first step, he did not subtract 5 ( or equivalently add -5) from both sides of the equation.
Question 5
Consider the inequality 3( - 1 - x) ≤ -2x - 3x.

The solution set of the inequality is:

(1 point)
Responses

{x|x≥32
}
{x|xis greater than or equal to 3 halves}

{x|x≤32
}
{x|xis less than or equal to 3 halves}

{x|x≥23
}
{x|xis greater than or equal to 2 thirds}

{x|x≤23
}
{x|xis less than or equal to 2 thirds}
Question 6
Graph the previous solution set on a number line.(1 point)
Responses

Question 7
Solve the inequality: -2(2x - 4) ≤ 4(2 - x).(1 point)
Responses

x ≤ 0
x ≤ 0

x ≤ 4
x ≤ 4

x ≤ 8
x ≤ 8

All Real Numbers
All Real Numbers

No Solution
No Solution
Question 8
Which values are in the solution set of the inequality −23x + 13 ≥ −1 ?

Select all that apply.

(3 points)
Responses

19
19

20
20

21
21

22
22

23
23
Question 9
The compound inequality for this graph is



(3 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
x Response area5 Response area x Response area 18
Question 10
The above compound inequality is true only if _________ of the statement(s) is/are true.(1 point)
Responses

none
none

all
all

at least 1
at least 1
Question 11
Solve for x.

18 > 2x + 4 ≥ 4

(1 point)
Responses

14 < x ≤ 4
14 < x ≤ 4

7 > x > 8
7 > x > 8

7 > x ≥ 12
7 > x ≥ 12

7 > x ≥ 4
7 > x ≥ 4

7 > x ≥ 0
7 > x ≥ 0
Question 12
For which compound inequalities is 6 a solution? Select the statements that are true.(3 points)
Responses

-5 < x < 6
-5 < x < 6

-2 ≤ x ≤ 6
-2 ≤ x ≤ 6

x ≤ 2 or x ≥ 6
x ≤ 2 or x ≥ 6

x < 2 or x > 6
x < 2 or x > 6

x < -5 or x > 0
x < -5 or x > 0
Question 13
A student scored 85 in her Algebra class before she took the End of Course Exam (the EOC). The student wants her average to be between 80 and 90 inclusive after her EOC is entered into her grades. The EOC counts 1/5 of her overall grade and her class average counts 4/5 of her grade. Write and solve a compound inequality to find the possible score she will need to make on the EOC to get the average she wants for her final grade in the course.(1 point)
Responses

80 ≤85 + 85 +85 +85 + x5 ≤ 90
; 85 ≤ x ≤ 110
80 ≤85 + 85 +85 +85 + x5 ≤ 90
; 85 ≤ x ≤ 110

80 <85 + 85 +85 +85 + x5 < 90
; 85 < x < 110
80 <85 + 85 +85 +85 + x5 < 90
; 85 < x < 110

90 ≤85 + 85 +85 +85 + x5 ≤ 80
; 110 ≤ x ≤ 80
90 ≤85 + 85 +85 +85 + x5 ≤ 80
; 110 ≤ x ≤ 80

90 <85 + 85 +85 +85 + x5 < 80
; 110 < x < 85

Answer 1

80 ≤85 + 85 +85 +85 + x/5 ≤ 90
; 85 ≤ x ≤ 110