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
21 answers
Line B
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)
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)
The possible next steps could be:
4x - 12 + 12 > 16 + 12
4x - 12 - (-12) > 16 - (-12)
4(4x - 12) > 16(4)
4x - 12 + 12 > 16 + 12
4x - 12 - (-12) > 16 - (-12)
4(4x - 12) > 16(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 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.
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
}
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 3 halves}
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
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
x ≤ 4
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
Select all that apply.
(3 points)
Responses
19
19
20
20
21
21
22
22
23
Values 20, 21, 22, and 23 are in the solution set of the inequality −23x + 13 ≥ −1.
(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 area
5
Response area
x
Response area
18
<>
≤
≤
≥
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 area
5
Response area
x
Response area
18
<>
≤
≤
≥
Values 5 and 18 are in the solution set of the inequality −23x + 13 ≥ −1.
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
Responses
none
none
all
all
at least 1
"At least 1" of the statement(s) is/are true for the compound inequality to be true.
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
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 ≥ 4
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
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
The statements that are true and have 6 as a solution are:
-2 ≤ x ≤ 6
x ≤ 2 or x ≥ 6
-2 ≤ x ≤ 6
x ≤ 2 or x ≥ 6
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
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
80 ≤ 85 + 85 + 85 + 85 + x/5 ≤ 90
To solve for x, we can subtract 85 from all parts of the inequality:
-5 ≤ x/5 ≤ 5
Then, we can multiply all parts of the inequality by 5:
-25 ≤ x ≤ 25
Therefore, the possible score she will need to make on the EOC to get the average she wants for her final grade in the course is 85 ≤ x ≤ 110.
To solve for x, we can subtract 85 from all parts of the inequality:
-5 ≤ x/5 ≤ 5
Then, we can multiply all parts of the inequality by 5:
-25 ≤ x ≤ 25
Therefore, the possible score she will need to make on the EOC to get the average she wants for her final grade in the course is 85 ≤ x ≤ 110.