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Math Problems
Algebra 1
Compare linear, exponential, and quadratic growth
(
2
2
2
q^
5
5
5
+
7
7
7
q)+(
5
5
5
q^
5
5
5
+
3
3
3
q^
3
3
3
) Which of the following expressions is equivalent to the given expression? Choose
1
1
1
answer: Choose
1
1
1
answer:
\newline
(Choice A)
q
(
7
q
4
+
10
)
q(7q^4+10)
q
(
7
q
4
+
10
)
\newline
(Choice B)
q
(
7
q
4
+
10
q
2
)
q(7q^4+10q^2)
q
(
7
q
4
+
10
q
2
)
\newline
(Choice C)
q
(
2
q
4
+
12
q
3
+
3
q
2
)
q(2q^4+12q^3+3q^2)
q
(
2
q
4
+
12
q
3
+
3
q
2
)
\newline
(Choice D)
q
(
7
q
4
+
3
q
2
+
7
)
q(7q^4+3q^2+7)
q
(
7
q
4
+
3
q
2
+
7
)
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solve for
x
,
y
x,y
x
,
y
\newline
{
[
(
x
−
y
)
2
+
4
=
3
y
−
5
x
+
2
(
x
+
1
)
(
y
−
1
)
]
,
[
3
x
y
−
5
y
+
6
x
+
11
x
3
+
1
=
5
]
}
\left\{\left[(x-y)^{2}+4=3y-5x+2\sqrt{(x+1)(y-1)}\right],\left[\frac{3xy-5y+6x+11}{\sqrt{x^{3}+1}}=5\right]\right\}
{
[
(
x
−
y
)
2
+
4
=
3
y
−
5
x
+
2
(
x
+
1
)
(
y
−
1
)
]
,
[
x
3
+
1
3
x
y
−
5
y
+
6
x
+
11
=
5
]
}
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Find the derivatives of \begin{array} yy=\operatorname{arctan}^{2} x^{3} \\ \cot u=x^{3} d v=3 x\end{array}
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P
(
x
)
+
Q
(
x
)
−
S
(
x
)
=
(
2
x
3
−
9
x
+
2
)
+
(
3
x
2
+
x
−
3
)
−
(
x
3
−
2
x
2
+
x
+
7
)
\begin{array}{l} P(x)+Q(x)-S(x) \\ =\left(2 x^{3}-9 x+2\right)+\left(3 x^{2}+x-3\right)-\left(x^{3}-2 x^{2}+x+7\right)\end{array}
P
(
x
)
+
Q
(
x
)
−
S
(
x
)
=
(
2
x
3
−
9
x
+
2
)
+
(
3
x
2
+
x
−
3
)
−
(
x
3
−
2
x
2
+
x
+
7
)
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The highest recorded wind speed not associated with a tornado was recorded at Barrows Island, Australia, in the year
1996
1996
1996
. The wind gust of
220
kn
220 \, \text{kn}
220
kn
toppled the previous record held at Mount Washington. What was the wind speed in miles per hour
(
mi
hr
)
?
\left(\frac{\text{mi}}{\text{hr}}\right)?
(
hr
mi
)
?
(
1
kn
=
1.15
mi
hr
)
\left(1\,\text{kn} = 1.15\,\frac{\text{mi}}{\text{hr}}\right)
(
1
kn
=
1.15
hr
mi
)
\newline
□
\square
□
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−
2
−
x
=
y
2
−
4
y
+
10
11
-2-x=\frac{y^{2}-4y+10}{11}
−
2
−
x
=
11
y
2
−
4
y
+
10
\newline
−
11
x
+
3
y
=
62
-11 x+3y=62
−
11
x
+
3
y
=
62
\newline
If
(
x
1
,
y
1
)
(x_{1},y_{1})
(
x
1
,
y
1
)
and
x
2
,
y
2
)
x_{2},y_{2})
x
2
,
y
2
)
are two distinct solutions to the system of equations shown, what is the product of the
x
\ x
x
values of the two solutions
(
x
1
∗
x
2
)
(x_{1}*x_{2})
(
x
1
∗
x
2
)
?
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z
=
−
16
i
−
92.3
z=-16i-92.3
z
=
−
16
i
−
92.3
\newline
What are the real and imaginary parts of
z
z
z
?
\newline
Choose
1
1
1
answer:
\newline
(A)
Re
(
z
)
=
−
92.3
\text{Re}(z) = -92.3
Re
(
z
)
=
−
92.3
and
Im
(
z
)
=
−
16
i
\text{Im}(z) = -16i
Im
(
z
)
=
−
16
i
\newline
(B)
Re
(
z
)
=
−
92.3
\text{Re}(z) = -92.3
Re
(
z
)
=
−
92.3
and
Im
(
z
)
=
−
16
\text{Im}(z) = -16
Im
(
z
)
=
−
16
\newline
(C)
Re
(
z
)
=
−
16
i
\text{Re}(z) = -16i
Re
(
z
)
=
−
16
i
and
Im
(
z
)
=
−
92.3
\text{Im}(z) = -92.3
Im
(
z
)
=
−
92.3
\newline
(D)
Re
(
z
)
=
−
16
\text{Re}(z) = -16
Re
(
z
)
=
−
16
and
Im
(
z
)
=
−
92.3
\text{Im}(z) = -92.3
Im
(
z
)
=
−
92.3
Get tutor help
f
(
x
)
=
{
x
2
−
8
for
x
≠
2
4
for
x
=
2
f(x)=\left\{\begin{array}{lll} x^{2}-8 & \text { for } & x \neq 2 \\ 4 & \text { for } & x=2 \end{array}\right.
f
(
x
)
=
{
x
2
−
8
4
for
for
x
=
2
x
=
2
\newline
Find
f
(
4
)
f(4)
f
(
4
)
\newline
Answer:
\newline
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f
(
x
)
=
{
−
(
x
+
1
)
2
+
3
for
x
≤
−
1
−
2
x
+
2
for
−
1
<
x
<
3
−
4
for
x
≥
3
Find
f
(
3
)
\begin{array}{l} f(x)=\left\{\begin{array}{lll} -(x+1)^{2}+3 & \text { for } & x \leq-1 \\ -2 x+2 & \text { for } & -1<x<3 \\ -4 & \text { for } & x \geq 3 \end{array}\right. \\ \text { Find } f(3) \\ \end{array}
f
(
x
)
=
⎩
⎨
⎧
−
(
x
+
1
)
2
+
3
−
2
x
+
2
−
4
for
for
for
x
≤
−
1
−
1
<
x
<
3
x
≥
3
Find
f
(
3
)
\newline
Answer:
\newline
Get tutor help
B
=
[
2
3
1
3
]
and
A
=
[
−
1
3
−
1
0
]
B=\left[\begin{array}{ll} 2 & 3 \\ 1 & 3 \end{array}\right] \text { and } A=\left[\begin{array}{ll} -1 & 3 \\ -1 & 0 \end{array}\right]
B
=
[
2
1
3
3
]
and
A
=
[
−
1
−
1
3
0
]
\newline
Let
H
=
B
A
\mathrm{H}=\mathrm{BA}
H
=
BA
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
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D
=
[
2
0
4
1
]
and
C
=
[
2
4
4
4
2
4
]
\begin{array}{l} \mathrm{D}=\left[\begin{array}{ll} 2 & 0 \\ 4 & 1 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{lll} 2 & 4 & 4 \\ 4 & 2 & 4 \end{array}\right] \end{array}
D
=
[
2
4
0
1
]
and
C
=
[
2
4
4
2
4
4
]
\newline
Let
H
=
D
C
\mathrm{H}=\mathrm{DC}
H
=
DC
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
D
=
[
2
3
1
1
5
−
2
]
and
F
=
[
2
5
2
2
]
\mathrm{D}=\left[\begin{array}{rr} 2 & 3 \\ 1 & 1 \\ 5 & -2 \end{array}\right] \text { and } \mathrm{F}=\left[\begin{array}{ll} 2 & 5 \\ 2 & 2 \end{array}\right]
D
=
⎣
⎡
2
1
5
3
1
−
2
⎦
⎤
and
F
=
[
2
2
5
2
]
\newline
Let
H
=
D
F
\mathrm{H}=\mathrm{DF}
H
=
DF
. Find
H
\mathrm{H}
H
.
\newline
H
=
\mathbf{H}=
H
=
Get tutor help
f
′
(
x
)
=
4
e
x
and
f
(
2
)
=
16
+
4
e
2
.
f
(
0
)
=
□
\begin{array}{l}f^{\prime}(x)=4 e^{x} \text { and } f(2)=16+4 e^{2} . \\ f(0)=\square\end{array}
f
′
(
x
)
=
4
e
x
and
f
(
2
)
=
16
+
4
e
2
.
f
(
0
)
=
□
Get tutor help
g
(
x
)
=
sin
(
x
)
g
′
(
x
)
=
?
\begin{array}{l} g(x)=\sqrt{\sin (x)} \\ g^{\prime}(x)=? \end{array}
g
(
x
)
=
sin
(
x
)
g
′
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
cos
(
x
)
2
x
\frac{\cos (\sqrt{x})}{2 \sqrt{x}}
2
x
c
o
s
(
x
)
\newline
(B)
cos
(
x
)
\sqrt{\cos (x)}
cos
(
x
)
\newline
(C)
[
sin
(
x
)
]
−
1
2
2
\frac{[\sin (x)]^{-\frac{1}{2}}}{2}
2
[
s
i
n
(
x
)
]
−
2
1
\newline
(D)
cos
(
x
)
2
sin
(
x
)
\frac{\cos (x)}{2 \sqrt{\sin (x)}}
2
s
i
n
(
x
)
c
o
s
(
x
)
Get tutor help
Fred tried to evaluate an expression step by step.
\newline
(
−
10
+
6
)
−
5
=
−
4
−
5
Step
1
=
−
4
+
(
−
5
)
Step
2
=
9
Step
3
\begin{array}{l} (-10+6)-5 \\ =-4-5 \quad \text { Step } 1 \\ =-4+(-5) \quad \text { Step } 2 \\ =9 \quad \text { Step } 3 \\ \end{array}
(
−
10
+
6
)
−
5
=
−
4
−
5
Step
1
=
−
4
+
(
−
5
)
Step
2
=
9
Step
3
\newline
Find Fred's first mistake.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
10
-10
−
10
plus
6
6
6
equals
−
16
-16
−
16
, not
−
4
-4
−
4
.
\newline
(B) Fred was supposed to add
5
5
5
, not
−
5
-5
−
5
.
\newline
(C)
−
4
-4
−
4
plus
−
5
-5
−
5
equals
−
9
-9
−
9
, not
9
9
9
.
Get tutor help
Keisha tried to evaluate an expression step by step.
\newline
3
−
4
−
7
=
3
−
(
4
−
7
)
Step
1
=
3
−
(
−
3
)
Step
2
=
6
Step
3
\begin{array}{l} 3-4-7 \\ =3-(4-7) \quad \text { Step } 1 \\ =3-(-3) \quad \text { Step } 2 \\ =6 \quad \text { Step } 3 \\ \end{array}
3
−
4
−
7
=
3
−
(
4
−
7
)
Step
1
=
3
−
(
−
3
)
Step
2
=
6
Step
3
\newline
Find Keisha's first mistake.
\newline
Choose
1
1
1
answer:
\newline
(A) Changing the groups changes the difference's value.
\newline
(B)
4
4
4
minus
7
7
7
equals
3
3
3
, not
−
3
-3
−
3
\newline
(C)
3
3
3
minus
−
3
-3
−
3
equals
0
0
0
, not
6
6
6
.
Get tutor help
Solve for
x
x
x
.
\newline
3
x
−
2
3
x
+
1
=
1
3
x
=
\begin{array}{l} \frac{3 x-2}{3 x+1}=\frac{1}{3} \\ x= \end{array}
3
x
+
1
3
x
−
2
=
3
1
x
=
Get tutor help
Solve for
x
x
x
.
\newline
5
x
−
5
x
−
2
=
1
3
x
=
\begin{array}{l} \frac{5 x-5}{x-2}=\frac{1}{3} \\ x= \end{array}
x
−
2
5
x
−
5
=
3
1
x
=
Get tutor help
Let
g
(
x
)
=
2
x
+
3
g(x)=\frac{2}{x+3}
g
(
x
)
=
x
+
3
2
.
\newline
Select the correct description of the one-sided limits of
g
g
g
at
x
=
−
3
x=-3
x
=
−
3
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
+
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=+\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow-3^{-}} g(x)=-\infty \end{array}
lim
x
→
−
3
+
g
(
x
)
=
−
∞
and
lim
x
→
−
3
−
g
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
x
1
−
cos
(
x
−
2
)
f(x)=\frac{x}{1-\cos (x-2)}
f
(
x
)
=
1
−
c
o
s
(
x
−
2
)
x
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
2
x=2
x
=
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
2
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{+}} f(x)=+\infty
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{-}} f(x)=+\infty
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(B)
lim
x
→
2
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{+}} f(x)=+\infty
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{-}} f(x)=-\infty
lim
x
→
2
−
f
(
x
)
=
−
∞
\newline
(C)
lim
x
→
2
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{+}} f(x)=-\infty
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{-}} f(x)=+\infty
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(D)
lim
x
→
2
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{+}} f(x)=-\infty
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{-}} f(x)=-\infty
lim
x
→
2
−
f
(
x
)
=
−
∞
Get tutor help
Let
g
(
x
)
=
2
(
x
−
3
)
2
g(x)=\frac{2}{(x-3)^{2}}
g
(
x
)
=
(
x
−
3
)
2
2
.
\newline
Select the correct description of the one-sided limits of
g
g
g
at
x
=
3
x=3
x
=
3
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
3
+
g
(
x
)
=
+
∞
and
lim
x
→
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=+\infty \end{array}
lim
x
→
3
+
g
(
x
)
=
+
∞
and
lim
x
→
3
−
g
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
3
+
g
(
x
)
=
+
∞
and
lim
x
→
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=-\infty \end{array}
lim
x
→
3
+
g
(
x
)
=
+
∞
and
lim
x
→
3
−
g
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
3
+
g
(
x
)
=
−
∞
and
lim
x
→
3
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=+\infty \end{array}
lim
x
→
3
+
g
(
x
)
=
−
∞
and
lim
x
→
3
−
g
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
3
+
g
(
x
)
=
−
∞
and
lim
x
→
3
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 3^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 3^{-}} g(x)=-\infty \end{array}
lim
x
→
3
+
g
(
x
)
=
−
∞
and
lim
x
→
3
−
g
(
x
)
=
−
∞
Get tutor help
Let
h
(
x
)
=
1
−
x
tan
(
x
)
h(x)=\frac{1-x}{\tan (x)}
h
(
x
)
=
t
a
n
(
x
)
1
−
x
.
\newline
Select the correct description of the one-sided limits of
h
h
h
at
x
=
0
x=0
x
=
0
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
0
+
h
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} h(x)=+\infty
lim
x
→
0
+
h
(
x
)
=
+
∞
and
lim
x
→
0
−
h
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} h(x)=+\infty
lim
x
→
0
−
h
(
x
)
=
+
∞
\newline
(B)
lim
x
→
0
+
h
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} h(x)=+\infty
lim
x
→
0
+
h
(
x
)
=
+
∞
and
lim
x
→
0
−
h
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} h(x)=-\infty
lim
x
→
0
−
h
(
x
)
=
−
∞
\newline
(C)
lim
x
→
0
+
h
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} h(x)=-\infty
lim
x
→
0
+
h
(
x
)
=
−
∞
and
lim
x
→
0
−
h
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} h(x)=+\infty
lim
x
→
0
−
h
(
x
)
=
+
∞
\newline
(D)
lim
x
→
0
+
h
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} h(x)=-\infty
lim
x
→
0
+
h
(
x
)
=
−
∞
and
lim
x
→
0
−
h
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} h(x)=-\infty
lim
x
→
0
−
h
(
x
)
=
−
∞
Get tutor help
Let
h
(
x
)
=
−
2
x
(
x
+
1
)
2
h(x)=-\frac{2 x}{(x+1)^{2}}
h
(
x
)
=
−
(
x
+
1
)
2
2
x
.
\newline
Select the correct description of the one-sided limits of
h
h
h
at
x
=
−
1
x=-1
x
=
−
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=+\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} h(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} h(x)=-\infty \end{array}
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
−
1
(
x
−
1
)
2
f(x)=-\frac{1}{(x-1)^{2}}
f
(
x
)
=
−
(
x
−
1
)
2
1
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
1
x=1
x
=
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\newline
(c)
\newline
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
−
4
x
−
1
f(x)=-\frac{4}{x-1}
f
(
x
)
=
−
x
−
1
4
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
1
x=1
x
=
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
1
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 1^{+}} f(x)=+\infty
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 1^{-}} f(x)=+\infty
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(B)
lim
x
→
1
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 1^{+}} f(x)=+\infty
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 1^{-}} f(x)=-\infty
lim
x
→
1
−
f
(
x
)
=
−
∞
\newline
(C)
lim
x
→
1
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 1^{+}} f(x)=-\infty
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 1^{-}} f(x)=+\infty
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(D)
lim
x
→
1
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 1^{+}} f(x)=-\infty
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 1^{-}} f(x)=-\infty
lim
x
→
1
−
f
(
x
)
=
−
∞
Get tutor help
Let
g
(
x
)
=
1
tan
2
(
x
)
g(x)=\frac{1}{\tan ^{2}(x)}
g
(
x
)
=
t
a
n
2
(
x
)
1
.
\newline
Select the correct description of the one-sided limits of
g
g
g
at
x
=
0
x=0
x
=
0
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
0
+
g
(
x
)
=
+
∞
and
lim
x
→
0
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=+\infty \end{array}
lim
x
→
0
+
g
(
x
)
=
+
∞
and
lim
x
→
0
−
g
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
0
+
g
(
x
)
=
+
∞
and
lim
x
→
0
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=-\infty \end{array}
lim
x
→
0
+
g
(
x
)
=
+
∞
and
lim
x
→
0
−
g
(
x
)
=
−
∞
\newline
(c)
\newline
lim
x
→
0
+
g
(
x
)
=
−
∞
and
lim
x
→
0
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=+\infty \end{array}
lim
x
→
0
+
g
(
x
)
=
−
∞
and
lim
x
→
0
−
g
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
0
+
g
(
x
)
=
−
∞
and
lim
x
→
0
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 0^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 0^{-}} g(x)=-\infty \end{array}
lim
x
→
0
+
g
(
x
)
=
−
∞
and
lim
x
→
0
−
g
(
x
)
=
−
∞
Get tutor help
Let
g
(
x
)
=
−
x
3
sin
(
x
−
2
)
g(x)=-\frac{x}{3 \sin (x-2)}
g
(
x
)
=
−
3
s
i
n
(
x
−
2
)
x
.
\newline
Select the correct description of the one-sided limits of
g
g
g
at
x
=
2
x=2
x
=
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
2
+
g
(
x
)
=
+
∞
and
lim
x
→
2
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=+\infty \end{array}
lim
x
→
2
+
g
(
x
)
=
+
∞
and
lim
x
→
2
−
g
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
2
+
g
(
x
)
=
+
∞
and
lim
x
→
2
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=-\infty \end{array}
lim
x
→
2
+
g
(
x
)
=
+
∞
and
lim
x
→
2
−
g
(
x
)
=
−
∞
\newline
(c)
\newline
lim
x
→
2
+
g
(
x
)
=
−
∞
and
lim
x
→
2
−
g
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=+\infty \end{array}
lim
x
→
2
+
g
(
x
)
=
−
∞
and
lim
x
→
2
−
g
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
2
+
g
(
x
)
=
−
∞
and
lim
x
→
2
−
g
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} g(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} g(x)=-\infty \end{array}
lim
x
→
2
+
g
(
x
)
=
−
∞
and
lim
x
→
2
−
g
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
−
x
ln
2
(
x
−
1
)
f(x)=\frac{-x}{\ln ^{2}(x-1)}
f
(
x
)
=
l
n
2
(
x
−
1
)
−
x
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
2
x=2
x
=
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
2
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{+}} f(x)=+\infty
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{-}} f(x)=+\infty
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(B)
lim
x
→
2
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{+}} f(x)=+\infty
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{-}} f(x)=-\infty
lim
x
→
2
−
f
(
x
)
=
−
∞
\newline
(C)
lim
x
→
2
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{+}} f(x)=-\infty
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 2^{-}} f(x)=+\infty
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(D)
lim
x
→
2
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{+}} f(x)=-\infty
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 2^{-}} f(x)=-\infty
lim
x
→
2
−
f
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
x
5
sin
(
x
+
1
)
f(x)=\frac{x}{5 \sin (x+1)}
f
(
x
)
=
5
s
i
n
(
x
+
1
)
x
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
−
1
x=-1
x
=
−
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
−
1
+
f
(
x
)
=
+
∞
and
lim
x
→
−
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=+\infty \end{array}
lim
x
→
−
1
+
f
(
x
)
=
+
∞
and
lim
x
→
−
1
−
f
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
−
1
+
f
(
x
)
=
+
∞
and
lim
x
→
−
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=-\infty \end{array}
lim
x
→
−
1
+
f
(
x
)
=
+
∞
and
lim
x
→
−
1
−
f
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
−
1
+
f
(
x
)
=
−
∞
and
lim
x
→
−
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=+\infty \end{array}
lim
x
→
−
1
+
f
(
x
)
=
−
∞
and
lim
x
→
−
1
−
f
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
−
1
+
f
(
x
)
=
−
∞
and
lim
x
→
−
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow-1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow-1^{-}} f(x)=-\infty \end{array}
lim
x
→
−
1
+
f
(
x
)
=
−
∞
and
lim
x
→
−
1
−
f
(
x
)
=
−
∞
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
7
3
−
5
x
=
(
1
49
)
2
x
+
9
x
=
□
\begin{array}{l} 7^{3-5 x}=\left(\frac{1}{49}\right)^{2 x+9} \\ x=\square \end{array}
7
3
−
5
x
=
(
49
1
)
2
x
+
9
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
3
x
−
9
=
8
1
5
x
+
1
x
=
□
\begin{array}{l} 3^{x-9}=81^{5 x+1} \\ x=\square \end{array}
3
x
−
9
=
8
1
5
x
+
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
8
1
x
+
7
9
5
x
−
9
=
9
4
x
+
1
x
=
□
\begin{array}{l} \frac{81^{x+7}}{9^{5 x-9}}=9^{4 x+1} \\ x=\square \end{array}
9
5
x
−
9
8
1
x
+
7
=
9
4
x
+
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
9
3
x
−
10
=
(
1
81
)
1
6
x
=
□
\begin{array}{l} 9^{3 x-10}=\left(\frac{1}{81}\right)^{\frac{1}{6}} \\ x=\square \end{array}
9
3
x
−
10
=
(
81
1
)
6
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
3
4
x
+
5
=
8
1
−
3
4
x
=
□
\begin{array}{l} 3^{4 x+5}=81^{-\frac{3}{4}} \\ x=\square \end{array}
3
4
x
+
5
=
8
1
−
4
3
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
7
5
x
−
4
=
8
1
7
x
−
2
x
=
□
\begin{array}{l} 27^{5 x-4}=81^{7 x-2} \\ x=\square \end{array}
2
7
5
x
−
4
=
8
1
7
x
−
2
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
(
1
64
)
−
8
x
+
3
=
(
1
16
)
7
x
−
2
x
=
□
\begin{array}{l} \left(\frac{1}{64}\right)^{-8 x+3}=\left(\frac{1}{16}\right)^{7 x-2} \\ x=\square \end{array}
(
64
1
)
−
8
x
+
3
=
(
16
1
)
7
x
−
2
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
6
x
+
7
=
(
1
8
)
1
−
5
x
x
=
□
\begin{array}{l} 2^{6 x+7}=\left(\frac{1}{8}\right)^{1-5 x} \\ x=\square \end{array}
2
6
x
+
7
=
(
8
1
)
1
−
5
x
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
3
x
−
4
4
x
−
5
=
2
8
x
−
1
x
=
□
\begin{array}{l} \frac{2^{3 x-4}}{4^{x-5}}=2^{8 x-1} \\ x=\square \end{array}
4
x
−
5
2
3
x
−
4
=
2
8
x
−
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
(
1
4
)
9
x
−
5
=
3
2
x
+
8
x
=
□
\begin{array}{l} \left(\frac{1}{4}\right)^{9 x-5}=32^{x+8} \\ x=\square \end{array}
(
4
1
)
9
x
−
5
=
3
2
x
+
8
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
3
3
x
−
2
=
9
4
x
−
1
x
=
□
\begin{array}{l} 3^{3 x-2}=9^{4 x-1} \\ x=\square \end{array}
3
3
x
−
2
=
9
4
x
−
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
3
7
x
+
4
=
(
1
27
)
x
−
3
x
=
□
\begin{array}{l} 3^{7 x+4}=\left(\frac{1}{27}\right)^{x-3} \\ x=\square \end{array}
3
7
x
+
4
=
(
27
1
)
x
−
3
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
5
4
x
+
3
2
5
9
−
x
=
5
2
x
+
5
x
=
□
\begin{array}{l} \frac{5^{4 x+3}}{25^{9-x}}=5^{2 x+5} \\ x=\square \end{array}
2
5
9
−
x
5
4
x
+
3
=
5
2
x
+
5
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
3
x
+
5
=
6
4
x
−
7
x
=
□
\begin{array}{l} 2^{3 x+5}=64^{x-7} \\ x=\square \end{array}
2
3
x
+
5
=
6
4
x
−
7
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
8
−
x
3
2
x
+
12
=
2
3
x
−
7
x
=
□
\begin{array}{l} \frac{2^{8-x}}{32^{x+12}}=2^{3 x-7} \\ x=\square \end{array}
3
2
x
+
12
2
8
−
x
=
2
3
x
−
7
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
9
x
+
2
=
1
6
5
x
−
2
x
=
□
\begin{array}{l} 2^{9 x+2}=16^{5 x-2} \\ x=\square \end{array}
2
9
x
+
2
=
1
6
5
x
−
2
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
4
x
−
10
=
(
1
64
)
5
x
+
2
x
=
□
\begin{array}{l} 4^{x-10}=\left(\frac{1}{64}\right)^{5 x+2} \\ x=\square \end{array}
4
x
−
10
=
(
64
1
)
5
x
+
2
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
8
5
x
+
3
=
(
1
64
)
−
2
3
x
=
□
\begin{array}{l} 8^{5 x+3}=\left(\frac{1}{64}\right)^{-\frac{2}{3}} \\ x=\square \end{array}
8
5
x
+
3
=
(
64
1
)
−
3
2
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
7
3
x
−
7
=
9
6
x
−
1
x
=
□
\begin{array}{l} 27^{3 x-7}=9^{6 x-1} \\ x=\square \end{array}
2
7
3
x
−
7
=
9
6
x
−
1
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
4
9
4
x
−
1
7
x
−
6
=
7
8
x
+
3
x
=
□
\begin{array}{l} \frac{49^{4 x-1}}{7^{x-6}}=7^{8 x+3} \\ x=\square \end{array}
7
x
−
6
4
9
4
x
−
1
=
7
8
x
+
3
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
4
x
−
5
⋅
3
2
x
+
1
=
2
6
x
−
7
x
=
□
\begin{array}{l} 2^{4 x-5} \cdot 32^{x+1}=2^{6 x-7} \\ x=\square \end{array}
2
4
x
−
5
⋅
3
2
x
+
1
=
2
6
x
−
7
x
=
□
Get tutor help
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