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Math Problems
Grade 7
Integer inequalities with absolute values
An ant is on top of a tree
23
23
23
meters tall, directly above her home, which is
0
0
0
.
3
3
3
meters below the ground.
\newline
Which
2
2
2
of the following expressions represents the distance between the ant and her home?
\newline
Choose
2
2
2
answers:
\newline
(A)
∣
23
+
0.3
∣
|23+0.3|
∣23
+
0.3∣
\newline
(B)
∣
0.3
−
23
∣
|0.3-23|
∣0.3
−
23∣
\newline
(C)
∣
−
0.3
−
23
∣
|-0.3-23|
∣
−
0.3
−
23∣
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Fill in the blank.
\newline
∣
8.6
∣
−
−
−
8
|8.6|_{--}-8
∣8.6
∣
−−
−
8
\newline
Choose
1
1
1
answer:
\newline
(A)
<
<
<
\newline
(B)
>
>
>
\newline
(c)
=
=
=
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Find
lim
x
→
∞
4
x
2
−
3
x
x
3
\lim _{x \rightarrow \infty} \frac{4 x^{2}-3 x}{x^{3}}
lim
x
→
∞
x
3
4
x
2
−
3
x
.
\newline
Choose
1
1
1
answer:
\newline
(A)
4
4
4
\newline
(B)
0
0
0
\newline
(C)
−
3
-3
−
3
\newline
(D) The limit is unbounded
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Find the missing factor
A
A
A
that makes the equality true.
\newline
8
x
3
=
(
−
2
x
)
(
A
)
A
=
□
\begin{array}{l} 8 x^{3}=(-2 x)(A) \\ A=\square \end{array}
8
x
3
=
(
−
2
x
)
(
A
)
A
=
□
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Find the missing factor
G
G
G
that makes the equality true.
\newline
18
y
3
=
(
G
)
(
2
y
2
)
G
=
□
\begin{array}{l} 18 y^{3}=(G)\left(2 y^{2}\right) \\ G=\square \end{array}
18
y
3
=
(
G
)
(
2
y
2
)
G
=
□
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Find the missing factor
B
B
B
that makes the equality true.
\newline
21
y
4
=
(
B
)
(
7
y
3
)
B
=
□
\begin{array}{l} 21 y^{4}=(B)\left(7 y^{3}\right) \\ B=\square \end{array}
21
y
4
=
(
B
)
(
7
y
3
)
B
=
□
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Find the missing factor
B
B
B
that makes the equality true.
\newline
−
35
x
6
=
(
−
5
x
2
)
(
B
)
B
=
□
\begin{array}{l} -35 x^{6}=\left(-5 x^{2}\right)(B) \\ B=\square \end{array}
−
35
x
6
=
(
−
5
x
2
)
(
B
)
B
=
□
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Function
g
g
g
can be thought of as a scaled version of
f
(
x
)
=
∣
x
∣
f(x)=|x|
f
(
x
)
=
∣
x
∣
.
\newline
What is the equation for
g
(
x
)
g(x)
g
(
x
)
?
\newline
Choose
1
1
1
answer:
\newline
(A)
g
(
x
)
=
−
4
3
∣
x
∣
g(x)=-\frac{4}{3}|x|
g
(
x
)
=
−
3
4
∣
x
∣
\newline
(B)
g
(
x
)
=
4
3
∣
x
∣
g(x)=\frac{4}{3}|x|
g
(
x
)
=
3
4
∣
x
∣
\newline
(C)
g
(
x
)
=
−
3
4
∣
x
∣
g(x)=-\frac{3}{4}|x|
g
(
x
)
=
−
4
3
∣
x
∣
\newline
(D)
g
(
x
)
=
3
4
∣
x
∣
g(x)=\frac{3}{4}|x|
g
(
x
)
=
4
3
∣
x
∣
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is reflected across the
x
x
x
-axis and then scaled vertically by a factor of
5
3
\frac{5}{3}
3
5
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
5
3
∣
x
∣
y=-\frac{5}{3}|x|
y
=
−
3
5
∣
x
∣
\newline
(B)
y
=
3
5
∣
x
∣
y=\frac{3}{5}|x|
y
=
5
3
∣
x
∣
\newline
(C)
y
=
−
∣
x
−
5
∣
+
3
y=-|x-5|+3
y
=
−
∣
x
−
5∣
+
3
\newline
(D)
y
=
∣
x
−
3
∣
+
5
y=|x-3|+5
y
=
∣
x
−
3∣
+
5
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