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Math Problems
Algebra 2
Domain and range of absolute value functions: equations
Solve for
x
x
x
.\begin{align*}x^{2}+12x+36&=0\newline x&=\square\end{align*}
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Given:
\newline
Universal Set
U
=
1
,
5
,
6
,
8
,
10
,
12
\mathrm{U}=1,5,6,8,10,12
U
=
1
,
5
,
6
,
8
,
10
,
12
\newline
Subset B
=
8
=8
=
8
\newline
What is the complement of Set B in Set U?
\newline
{
8
}
\{8\}
{
8
}
\newline
{
1
,
5
,
6
,
10
,
12
}
\{1,5,6,10,12\}
{
1
,
5
,
6
,
10
,
12
}
\newline
{
1
,
8
,
10
,
12
}
\{1,8,10,12\}
{
1
,
8
,
10
,
12
}
\newline
{
1
,
5
,
6
,
8
,
10
,
12
}
\{1,5,6,8,10,12\}
{
1
,
5
,
6
,
8
,
10
,
12
}
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Which of the following are integers?
\newline
Multi-select Choices:
\newline
(A)
3
3
3
\newline
(B)
8
8
8
\newline
(C)
π
\pi
π
\newline
(D)
2
7
\frac{2}{7}
7
2
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f
(
x
)
=
cos
[
e
x
–
sin
(
x
)
]
f(x)=\cos[e^x–\sin(x)]
f
(
x
)
=
cos
[
e
x
–
sin
(
x
)]
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{
g
(
1
)
=
−
5
g
(
2
)
=
3
g
(
n
)
=
g
(
n
−
2
)
+
g
(
n
−
1
)
\begin{cases} g(1) = -5 \\ g(2) = 3 \\ g(n) = g(n-2) + g(n-1) \end{cases}
⎩
⎨
⎧
g
(
1
)
=
−
5
g
(
2
)
=
3
g
(
n
)
=
g
(
n
−
2
)
+
g
(
n
−
1
)
\newline
g
(
3
)
=
□
g(3)=\square
g
(
3
)
=
□
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Solve for
v
v
v
:
−
9
8
=
v
−
1
2
,
v
=
□
\begin{align*} -\frac{9}{8} &= v - \frac{1}{2}, \ v &= \Box \end{align*}
−
8
9
=
v
−
2
1
,
v
=
□
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Solve for
r
r
r
.
−
10
−
5
r
=
−
6
r
−
1
r
=
…
\begin{array}{l} -10-5r=-6r-1 \ r=\dots \end{array}
−
10
−
5
r
=
−
6
r
−
1
r
=
…
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Find the expression for
f
(
x
)
f(x)
f
(
x
)
that makes the following equation true for all values of
x
x
x
.
\newline
8
1
x
9
5
x
−
8
=
9
f
(
x
)
f
(
x
)
=
□
\begin{array}{l} \frac{81^{x}}{9^{5 x-8}}=9^{f(x)} \\ f(x)=\square \end{array}
9
5
x
−
8
8
1
x
=
9
f
(
x
)
f
(
x
)
=
□
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Find the value of
A
A
A
that makes the following equation true for all values of
x
x
x
.
\newline
5
x
+
5
x
+
3
=
A
⋅
5
x
A
=
\begin{array}{l} 5^{x}+5^{x+3}=A \cdot 5^{x} \\ A= \end{array}
5
x
+
5
x
+
3
=
A
⋅
5
x
A
=
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Find the value of
A
A
A
that makes the following equation true for all values of
x
x
x
.
\newline
3
x
−
3
x
−
2
=
A
⋅
3
x
A
=
\begin{array}{l} 3^{x}-3^{x-2}=A \cdot 3^{x} \\ A= \end{array}
3
x
−
3
x
−
2
=
A
⋅
3
x
A
=
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Which of the following functions are continuous for all real numbers?
\newline
g
(
x
)
=
x
5
h
(
x
)
=
x
3
\begin{array}{l} g(x)=\sqrt[5]{x} \\ h(x)=\sqrt[3]{x} \end{array}
g
(
x
)
=
5
x
h
(
x
)
=
3
x
\newline
Choose
1
1
1
answer:
\newline
(A)
g
g
g
only
\newline
(B)
h
h
h
only
\newline
(C) Both
g
g
g
and
h
h
h
\newline
(D) Neither
g
g
g
nor
h
h
h
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Which of the following functions are continuous at
x
=
−
2
x=-2
x
=
−
2
?
\newline
h
(
x
)
=
x
+
1
3
f
(
x
)
=
x
+
1
4
\begin{array}{l} h(x)=\sqrt[3]{x+1} \\ f(x)=\sqrt[4]{x+1} \end{array}
h
(
x
)
=
3
x
+
1
f
(
x
)
=
4
x
+
1
\newline
Choose
1
1
1
answer:
\newline
(A)
h
h
h
only
\newline
(B)
f
f
f
only
\newline
(C) Both
h
h
h
and
f
f
f
\newline
(D) Neither
h
h
h
nor
f
f
f
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Which of the following functions are continuous at
x
=
0
x=0
x
=
0
?
\newline
f
(
x
)
=
tan
(
x
)
g
(
x
)
=
cot
(
x
)
\begin{array}{l} f(x)=\tan (x) \\ g(x)=\cot (x) \end{array}
f
(
x
)
=
tan
(
x
)
g
(
x
)
=
cot
(
x
)
\newline
Choose
1
1
1
answer:
\newline
(A)
f
f
f
only
\newline
(B)
g
g
g
only
\newline
(C) Both
f
f
f
and
g
g
g
\newline
(D) Neither
f
f
f
nor
g
g
g
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Which of the following functions are continuous at
x
=
1
x=1
x
=
1
?
\newline
f
(
x
)
=
e
x
−
1
g
(
x
)
=
ln
(
e
x
−
1
)
\begin{array}{l} f(x)=e^{x}-1 \\ g(x)=\ln \left(e^{x}-1\right) \end{array}
f
(
x
)
=
e
x
−
1
g
(
x
)
=
ln
(
e
x
−
1
)
\newline
Choose
1
1
1
answer:
\newline
(A)
f
f
f
only
\newline
(B)
g
g
g
only
\newline
(C) Both
f
f
f
and
g
g
g
\newline
(D) Neither
f
f
f
nor
g
g
g
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Which of the following functions are continuous at
x
=
3
x=3
x
=
3
?
\newline
g
(
x
)
=
ln
(
x
−
3
)
f
(
x
)
=
e
x
−
3
\begin{array}{l} g(x)=\ln (x-3) \\ f(x)=e^{x-3} \end{array}
g
(
x
)
=
ln
(
x
−
3
)
f
(
x
)
=
e
x
−
3
\newline
Choose
1
1
1
answer:
\newline
(A)
g
g
g
only
\newline
(B)
f
f
f
only
\newline
(C) Both
g
g
g
and
f
f
f
\newline
(D) Neither
g
g
g
nor
f
f
f
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h
(
x
)
=
{
cos
(
x
)
for
x
<
π
sin
(
x
)
for
x
≥
π
h(x)=\left\{\begin{array}{ll} \cos (x) & \text { for } x<\pi \\ \sin (x) & \text { for } x \geq \pi \end{array}\right.
h
(
x
)
=
{
cos
(
x
)
sin
(
x
)
for
x
<
π
for
x
≥
π
\newline
Find
lim
x
→
π
+
h
(
x
)
\lim _{x \rightarrow \pi^{+}} h(x)
lim
x
→
π
+
h
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
-1
−
1
\newline
(B)
0
0
0
\newline
(C)
1
1
1
\newline
(D) The limit doesn't exist.
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f
(
x
)
=
{
x
2
for
x
≤
0
ln
(
x
)
for
x
>
0
f(x)=\left\{\begin{array}{ll} x^{2} & \text { for } x \leq 0 \\ \ln (x) & \text { for } x>0 \end{array}\right.
f
(
x
)
=
{
x
2
ln
(
x
)
for
x
≤
0
for
x
>
0
\newline
Find
lim
x
→
1
f
(
x
)
\lim _{x \rightarrow 1} f(x)
lim
x
→
1
f
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
0
0
0
\newline
(B)
1
1
1
\newline
(C)
e
e
e
\newline
(D) The limit doesn't exist.
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f
(
x
)
=
{
2
x
for
0
<
x
≤
4
8
x
for
x
>
4
f(x)=\left\{\begin{array}{ll} 2^{x} & \text { for } 0<x \leq 4 \\ 8 \sqrt{x} & \text { for } x>4 \end{array}\right.
f
(
x
)
=
{
2
x
8
x
for
0
<
x
≤
4
for
x
>
4
\newline
Find
lim
x
→
4
f
(
x
)
\lim _{x \rightarrow 4} f(x)
lim
x
→
4
f
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
4
4
4
\newline
(B)
8
8
8
\newline
(C)
16
16
16
\newline
(D) The limit doesn't exist.
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g
(
x
)
=
{
2
x
−
1
for
−
8
≤
x
x
for
x
≥
1
g(x)=\left\{\begin{array}{ll} 2^{x}-1 & \text { for }-8 \leq x \\ \sqrt{x} & \text { for } x \geq 1 \end{array}\right.
g
(
x
)
=
{
2
x
−
1
x
for
−
8
≤
x
for
x
≥
1
\newline
Find
lim
x
→
4
g
(
x
)
\lim _{x \rightarrow 4} g(x)
lim
x
→
4
g
(
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
\newline
(B)
2
2
2
\newline
(C)
15
\mathbf{1 5}
15
\newline
(D) The limit doesn't exist.
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g
(
x
)
=
8
x
−
1
x
+
4
h
(
x
)
=
3
x
+
10
\begin{array}{l} g(x)=\frac{8 x-1}{x+4} \\ h(x)=3 x+10 \end{array}
g
(
x
)
=
x
+
4
8
x
−
1
h
(
x
)
=
3
x
+
10
\newline
Write
(
g
∘
h
)
(
x
)
(g \circ h)(x)
(
g
∘
h
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
g
∘
h
)
(
x
)
=
(g \circ h)(x)=
(
g
∘
h
)
(
x
)
=
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f
(
x
)
=
2
−
x
x
+
1
g
(
x
)
=
18
−
3
x
\begin{array}{l} f(x)=\frac{2-x}{x+1} \\ g(x)=18-3 x \end{array}
f
(
x
)
=
x
+
1
2
−
x
g
(
x
)
=
18
−
3
x
\newline
Write
(
f
∘
g
)
(
x
)
(f \circ g)(x)
(
f
∘
g
)
(
x
)
as an expression in terms of
x
x
x
.
\newline
(
f
∘
g
)
(
x
)
=
(f \circ g)(x)=
(
f
∘
g
)
(
x
)
=
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What is the midline equation of
\newline
y
=
−
8
cos
(
3
π
2
x
+
1
)
?
y
=
□
\begin{array}{l} y=-8 \cos \left(\frac{3 \pi}{2} x+1\right) ? \\ y=\square \end{array}
y
=
−
8
cos
(
2
3
π
x
+
1
)
?
y
=
□
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What is the midline equation of
\newline
y
=
2
sin
(
π
2
x
+
3
)
?
y
=
□
\begin{array}{l} y=2 \sin \left(\frac{\pi}{2} x+3\right) ? \\ y=\square \end{array}
y
=
2
sin
(
2
π
x
+
3
)
?
y
=
□
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What is the midline equation of the function
\newline
h
(
x
)
=
5
sin
(
4
x
−
2
)
−
3
?
y
=
□
\begin{array}{l} h(x)=5 \sin (4 x-2)-3 ? \\ y=\square \end{array}
h
(
x
)
=
5
sin
(
4
x
−
2
)
−
3
?
y
=
□
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What is the midline equation of
\newline
y
=
7
sin
(
3
π
4
x
−
π
4
)
+
6
?
y
=
□
\begin{array}{l} y=7 \sin \left(\frac{3 \pi}{4} x-\frac{\pi}{4}\right)+6 ? \\ y=\square \end{array}
y
=
7
sin
(
4
3
π
x
−
4
π
)
+
6
?
y
=
□
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What is the midline equation of the function
\newline
g
(
x
)
=
−
sin
(
8
x
−
3
)
+
5
?
y
=
□
\begin{array}{l} g(x)=-\sin (8 x-3)+5 ? \\ y=\square \end{array}
g
(
x
)
=
−
sin
(
8
x
−
3
)
+
5
?
y
=
□
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What is the midline equation of the function
\newline
g
(
x
)
=
3
sin
(
2
x
−
1
)
+
4
?
y
=
□
\begin{array}{l} g(x)=3 \sin (2 x-1)+4 ? \\ y=\square \end{array}
g
(
x
)
=
3
sin
(
2
x
−
1
)
+
4
?
y
=
□
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What is the midline equation of the function
\newline
h
(
x
)
=
−
4
cos
(
5
x
−
9
)
−
7
?
y
=
□
\begin{array}{l} h(x)=-4 \cos (5 x-9)-7 ? \\ y=\square \end{array}
h
(
x
)
=
−
4
cos
(
5
x
−
9
)
−
7
?
y
=
□
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What is the midline equation of the function
\newline
g
(
x
)
=
−
6
sin
(
3
π
x
+
4
)
−
2
?
y
=
□
\begin{array}{l} g(x)=-6 \sin (3 \pi x+4)-2 ? \\ y=\square \end{array}
g
(
x
)
=
−
6
sin
(
3
π
x
+
4
)
−
2
?
y
=
□
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z
=
−
60
−
12
i
Re
(
z
)
=
Im
(
z
)
=
\begin{array}{l}z=-60-12 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
z
=
−
60
−
12
i
Re
(
z
)
=
Im
(
z
)
=
Get tutor help
f
(
x
)
=
1
x
−
1
g
(
x
)
=
5
x
+
8
\begin{array}{l} f(x)=\frac{1}{x-1} \\ g(x)=5 x+8 \end{array}
f
(
x
)
=
x
−
1
1
g
(
x
)
=
5
x
+
8
\newline
The functions
f
f
f
and
g
g
g
are defined. What is the value of
f
(
g
(
−
1
)
)
f(g(-1))
f
(
g
(
−
1
))
?
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a
y
=
3
4
x
x
−
5
y
=
0
\begin{aligned} a y & =\frac{3}{4} x \\ x-5 y & =0 \end{aligned}
a
y
x
−
5
y
=
4
3
x
=
0
\newline
For what value of
a
a
a
does the system of linear equations in the variables
x
x
x
and
y
y
y
have infinitely many solutions?
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a
y
=
2
x
+
1
y
=
2
x
+
2
\begin{array}{r} a y=2 x+1 \\ y=2 x+2 \end{array}
a
y
=
2
x
+
1
y
=
2
x
+
2
\newline
Consider the system of equations, where
a
a
a
is a constant. For what value of
a
a
a
are there no
(
x
,
y
)
(x, y)
(
x
,
y
)
solutions?
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is shifted up by
7
7
7
units and to the left by
8
8
8
units.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
+
8
∣
−
7
y=|x+8|-7
y
=
∣
x
+
8∣
−
7
\newline
(B)
y
=
∣
x
+
8
∣
+
7
y=|x+8|+7
y
=
∣
x
+
8∣
+
7
\newline
(C)
y
=
∣
x
−
8
∣
+
7
y=|x-8|+7
y
=
∣
x
−
8∣
+
7
\newline
(D)
y
=
∣
x
−
8
∣
−
7
y=|x-8|-7
y
=
∣
x
−
8∣
−
7
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is shifted down by
6
6
6
units and to the left by
9
9
9
units.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
+
9
∣
−
6
y=|x+9|-6
y
=
∣
x
+
9∣
−
6
\newline
(B)
y
=
∣
x
−
9
∣
+
6
y=|x-9|+6
y
=
∣
x
−
9∣
+
6
\newline
(C)
y
=
∣
x
+
9
∣
+
6
y=|x+9|+6
y
=
∣
x
+
9∣
+
6
\newline
(D)
y
=
∣
x
−
9
∣
−
6
y=|x-9|-6
y
=
∣
x
−
9∣
−
6
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is shifted up by
4
4
4
units and to the right by
5
5
5
units.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
+
4
∣
−
5
y=|x+4|-5
y
=
∣
x
+
4∣
−
5
\newline
(B)
y
=
∣
x
+
5
∣
+
4
y=|x+5|+4
y
=
∣
x
+
5∣
+
4
\newline
(C)
y
=
∣
x
−
5
∣
+
4
y=|x-5|+4
y
=
∣
x
−
5∣
+
4
\newline
(D)
y
=
∣
x
+
4
∣
+
5
y=|x+4|+5
y
=
∣
x
+
4∣
+
5
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The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is shifted down by
9
9
9
units and to the right by
4
4
4
units.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
∣
x
−
4
∣
−
9
y=|x-4|-9
y
=
∣
x
−
4∣
−
9
\newline
(B)
y
=
∣
x
−
9
∣
+
4
y=|x-9|+4
y
=
∣
x
−
9∣
+
4
\newline
(C)
y
=
∣
x
−
9
∣
−
4
y=|x-9|-4
y
=
∣
x
−
9∣
−
4
\newline
(D)
y
=
∣
x
−
4
∣
+
9
y=|x-4|+9
y
=
∣
x
−
4∣
+
9
Get tutor help
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
3
8
\frac{3}{8}
8
3
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
8
3
∣
x
∣
y=\frac{8}{3}|x|
y
=
3
8
∣
x
∣
\newline
(B)
y
=
−
8
3
∣
x
∣
y=-\frac{8}{3}|x|
y
=
−
3
8
∣
x
∣
\newline
(C)
y
=
3
8
∣
x
∣
y=\frac{3}{8}|x|
y
=
8
3
∣
x
∣
\newline
(D)
y
=
−
3
8
∣
x
∣
y=-\frac{3}{8}|x|
y
=
−
8
3
∣
x
∣
Get tutor help
The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is scaled vertically by a factor of
7
2
\frac{7}{2}
2
7
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
−
2
7
∣
x
∣
y=-\frac{2}{7}|x|
y
=
−
7
2
∣
x
∣
\newline
(B)
y
=
−
7
2
∣
x
∣
y=-\frac{7}{2}|x|
y
=
−
2
7
∣
x
∣
\newline
(C)
y
=
2
7
∣
x
∣
y=\frac{2}{7}|x|
y
=
7
2
∣
x
∣
\newline
(D)
y
=
7
2
∣
x
∣
y=\frac{7}{2}|x|
y
=
2
7
∣
x
∣
Get tutor help
The graph of
y
=
∣
x
∣
y=|x|
y
=
∣
x
∣
is scaled vertically by a factor of
1
5
\frac{1}{5}
5
1
.
\newline
What is the equation of the new graph?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
=
1
5
∣
x
∣
y=\frac{1}{5}|x|
y
=
5
1
∣
x
∣
\newline
(B)
y
=
−
5
∣
x
∣
y=-5|x|
y
=
−
5∣
x
∣
\newline
(C)
y
=
∣
x
+
5
∣
y=|x+5|
y
=
∣
x
+
5∣
\newline
(D)
y
=
∣
x
−
5
∣
y=|x-5|
y
=
∣
x
−
5∣
Get tutor help
What is the range of this function?
\newline
y = |x|
\newline
Choices:
\newline
all real numbers
\text{all real numbers}
all real numbers
\newline
{
y
∣
y
≥
0
}
\{y \mid y \geq 0\}
{
y
∣
y
≥
0
}
\newline
{
y
∣
y
≤
0
}
\{y \mid y \leq 0\}
{
y
∣
y
≤
0
}
\newline
{
y
∣
y
>
0
}
\{y \mid y > 0\}
{
y
∣
y
>
0
}
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