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Math Problems
Algebra 1
Compare linear and exponential growth
4
4
4
markers cost
$
7.04
\$ 7.04
$7.04
.
\newline
Which equation would help determine the cost of
7
7
7
markers?
\newline
Choose
1
1
1
answer:
\newline
(A)
4
7
=
$
7.04
x
\frac{4}{7}=\frac{\$ 7.04}{x}
7
4
=
x
$7.04
\newline
(B)
x
7
=
4
$
7.04
\frac{x}{7}=\frac{4}{\$ 7.04}
7
x
=
$7.04
4
\newline
(C)
7
x
=
$
7.04
4
\frac{7}{x}=\frac{\$ 7.04}{4}
x
7
=
4
$7.04
\newline
(D)
4
7
=
x
$
7.04
\frac{4}{7}=\frac{x}{\$ 7.04}
7
4
=
$7.04
x
\newline
(E) None of the above
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10
10
10
rubber stamps cost
$
10.30
\$ 10.30
$10.30
.
\newline
Which equation would help determine the cost of
2
2
2
rubber stamps?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
x
=
$
10.30
10
\frac{2}{x}=\frac{\$ 10.30}{10}
x
2
=
10
$10.30
\newline
(B)
x
2
=
10
$
10.30
\frac{x}{2}=\frac{10}{\$ 10.30}
2
x
=
$10.30
10
\newline
(C)
2
10
=
$
10.30
x
\frac{2}{10}=\frac{\$ 10.30}{x}
10
2
=
x
$10.30
\newline
(D)
2
$
10.30
=
x
10
\frac{2}{\$ 10.30}=\frac{x}{10}
$10.30
2
=
10
x
\newline
(E) None of the above
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What is the value of
A
A
A
when we rewrite
(
1
8
)
x
+
(
1
8
)
x
−
2
\left(\frac{1}{8}\right)^{x}+\left(\frac{1}{8}\right)^{x-2}
(
8
1
)
x
+
(
8
1
)
x
−
2
as
A
⋅
(
1
8
)
x
A \cdot\left(\frac{1}{8}\right)^{x}
A
⋅
(
8
1
)
x
?
\newline
A
=
A=
A
=
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What is the value of
A
A
A
when we rewrite
1.4
4
−
1.2
x
1.44^{-1.2 x}
1.4
4
−
1.2
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
144
12
A=\frac{144}{12}
A
=
12
144
\newline
(B)
A
=
1.4
4
−
1.2
A=1.44^{-1.2}
A
=
1.4
4
−
1.2
\newline
(C)
A
=
1.4
4
1.2
A=1.44^{1.2}
A
=
1.4
4
1.2
\newline
(D)
A
=
1
2
24
x
A=12^{24 x}
A
=
1
2
24
x
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What is the value of
A
A
A
when we rewrite
3
x
3^{x}
3
x
as
A
5
x
A^{5 x}
A
5
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
3
A=3
A
=
3
\newline
(B)
A
=
3
−
5
A=3^{-5}
A
=
3
−
5
\newline
(C)
A
=
3
1
5
A=3^{\frac{1}{5}}
A
=
3
5
1
\newline
(D)
A
=
3
−
1
5
A=3^{-\frac{1}{5}}
A
=
3
−
5
1
Get tutor help
What is the value of
A
A
A
when we rewrite
6
x
6^{x}
6
x
as
A
x
4
A^{\frac{x}{4}}
A
4
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
4
1
6
A=4^{\frac{1}{6}}
A
=
4
6
1
\newline
(B)
A
=
6
1
4
A=6^{\frac{1}{4}}
A
=
6
4
1
\newline
(C)
A
=
6
4
A=\frac{6}{4}
A
=
4
6
\newline
(D)
A
=
6
4
A=6^{4}
A
=
6
4
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What is the value of
A
A
A
when we rewrite
(
5
2
)
x
+
(
5
2
)
x
+
3
\left(\frac{5}{2}\right)^{x}+\left(\frac{5}{2}\right)^{x+3}
(
2
5
)
x
+
(
2
5
)
x
+
3
as A.
(
5
2
)
x
\left(\frac{5}{2}\right)^{x}
(
2
5
)
x
?
\newline
A
=
A=
A
=
Get tutor help
What is the value of
A
A
A
when we rewrite
(
2
3
)
x
+
4
−
(
2
3
)
x
\left(\frac{2}{3}\right)^{x+4}-\left(\frac{2}{3}\right)^{x}
(
3
2
)
x
+
4
−
(
3
2
)
x
as
A
⋅
(
2
3
)
x
A \cdot\left(\frac{2}{3}\right)^{x}
A
⋅
(
3
2
)
x
?
\newline
A
=
A=
A
=
Get tutor help
What is the value of
A
A
A
when we rewrite
4
31
x
4^{31 x}
4
31
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
4
31
A=\frac{4}{31}
A
=
31
4
\newline
(B)
A
=
4
31
A=4^{31}
A
=
4
31
\newline
(C)
A
=
4
⋅
31
A=4 \cdot 31
A
=
4
⋅
31
\newline
(D)
A
=
3
1
4
A=31^{4}
A
=
3
1
4
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What is the value of
A
A
A
when we rewrite
(
6
17
)
9
x
\left(\frac{6}{17}\right)^{9 x}
(
17
6
)
9
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
(
6
17
)
9
A=\left(\frac{6}{17}\right)^{9}
A
=
(
17
6
)
9
\newline
(B)
A
=
−
1
9
A=-\frac{1}{9}
A
=
−
9
1
\newline
(C)
A
=
54
17
A=\frac{54}{17}
A
=
17
54
\newline
(D)
A
=
(
17
6
)
9
A=\left(\frac{17}{6}\right)^{9}
A
=
(
6
17
)
9
Get tutor help
What is the value of
A
A
A
when we rewrite
8
x
8^{x}
8
x
as
A
−
x
9
A^{-\frac{x}{9}}
A
−
9
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
8
−
9
A=8^{-9}
A
=
8
−
9
\newline
(B)
A
=
−
9
8
A=-\frac{9}{8}
A
=
−
8
9
\newline
(C)
A
=
8
1
9
A=8^{\frac{1}{9}}
A
=
8
9
1
\newline
(D)
A
=
(
−
8
)
1
9
A=(-8)^{\frac{1}{9}}
A
=
(
−
8
)
9
1
Get tutor help
What is the value of
A
A
A
when we rewrite
(
1
7
)
x
−
2
−
(
1
7
)
x
\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^{x}
(
7
1
)
x
−
2
−
(
7
1
)
x
as
A
⋅
(
1
7
)
x
A \cdot\left(\frac{1}{7}\right)^{x}
A
⋅
(
7
1
)
x
?
\newline
A
=
A=
A
=
Get tutor help
Which exponential expression is equivalent to
(
c
7
)
5
(\sqrt[7]{c})^{5}
(
7
c
)
5
?
\newline
Choose
1
1
1
answer:
\newline
(A)
c
5
7
c^{\frac{5}{7}}
c
7
5
\newline
(B)
c
7
5
c^{\frac{7}{5}}
c
5
7
\newline
(C)
c
5
c
7
\frac{c^{5}}{c^{7}}
c
7
c
5
\newline
(D)
c
7
c
5
\frac{c^{7}}{c^{5}}
c
5
c
7
Get tutor help
Which expressions are equivalent to
(
d
1
8
)
5
\left(d^{\frac{1}{8}}\right)^{5}
(
d
8
1
)
5
?
\newline
Choose all answers that apply:
\newline
A.
(
d
5
)
1
8
\left(d^{5}\right)^{\frac{1}{8}}
(
d
5
)
8
1
\newline
B
(
d
5
)
8
\left(d^{5}\right)^{8}
(
d
5
)
8
\newline
C
(
d
8
)
5
(\sqrt[8]{d})^{5}
(
8
d
)
5
\newline
D None of the above
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Which expressions are equivalent to
(
v
−
1
)
1
9
\left(v^{-1}\right)^{\frac{1}{9}}
(
v
−
1
)
9
1
?
\newline
Choose all answers that apply:
\newline
A
(
v
9
)
−
1
(\sqrt[9]{v})^{-1}
(
9
v
)
−
1
\newline
B
(
v
1
9
)
−
1
\left(v^{\frac{1}{9}}\right)^{-1}
(
v
9
1
)
−
1
\newline
C
v
−
1
9
v^{-\frac{1}{9}}
v
−
9
1
\newline
D None of the above
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Which expressions are equivalent to
(
b
2
)
1
9
\left(b^{2}\right)^{\frac{1}{9}}
(
b
2
)
9
1
?
\newline
Choose all answers that apply:
\newline
A
(
b
−
1
9
)
2
\left(b^{-\frac{1}{9}}\right)^{2}
(
b
−
9
1
)
2
\newline
B
(
b
1
9
)
2
\left(b^{\frac{1}{9}}\right)^{2}
(
b
9
1
)
2
\newline
C
(
b
1
2
)
9
\left(b^{\frac{1}{2}}\right)^{9}
(
b
2
1
)
9
\newline
D None of the above
Get tutor help
Which exponential expression is equivalent to
(
y
5
)
4
(\sqrt[5]{y})^{4}
(
5
y
)
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
y
4
5
y^{\frac{4}{5}}
y
5
4
\newline
(B)
y
5
4
y^{\frac{5}{4}}
y
4
5
\newline
(C)
y
5
y
4
\frac{y^{5}}{y^{4}}
y
4
y
5
\newline
(D)
y
4
y
5
\frac{y^{4}}{y^{5}}
y
5
y
4
Get tutor help
Which exponential expression is equivalent to
(
a
7
)
2
(\sqrt[7]{a})^{2}
(
7
a
)
2
?
\newline
Choose
1
1
1
answer:
\newline
(A)
a
7
2
a^{\frac{7}{2}}
a
2
7
\newline
(B)
a
2
7
a^{\frac{2}{7}}
a
7
2
\newline
(C)
a
7
a
2
\frac{a^{7}}{a^{2}}
a
2
a
7
\newline
(D)
a
2
a
7
\frac{a^{2}}{a^{7}}
a
7
a
2
Get tutor help
Which expressions are equivalent to
k
−
1
6
k^{-\frac{1}{6}}
k
−
6
1
?
\newline
Choose all answers that apply:
\newline
A
1
k
6
\frac{1}{k^{6}}
k
6
1
\newline
B
(
k
−
1
)
1
6
\left(k^{-1}\right)^{\frac{1}{6}}
(
k
−
1
)
6
1
\newline
c
k
−
1
6
\sqrt[6]{k^{-1}}
6
k
−
1
\newline
D None of the above
Get tutor help
Find the value of
A
A
A
that makes the following equation true for all values of
x
x
x
.
\newline
0.
9
60
x
=
A
x
0.9^{60 x}=A^{x}
0.
9
60
x
=
A
x
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
0.9
⋅
60
A=0.9 \cdot 60
A
=
0.9
⋅
60
\newline
(B)
A
=
0.
9
60
A=0.9^{60}
A
=
0.
9
60
\newline
(C)
A
=
60
0.9
A=\frac{60}{0.9}
A
=
0.9
60
\newline
(D)
A
=
6
0
0.9
A=60^{0.9}
A
=
6
0
0.9
Get tutor help
What is the value of
A
A
A
when we rewrite
2
9
x
2
29^{\frac{x}{2}}
2
9
2
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
29
2
A=\frac{29}{2}
A
=
2
29
\newline
(B)
A
=
2
9
2
A=29^{2}
A
=
2
9
2
\newline
(C)
A
=
2
9
1
2
A=29^{\frac{1}{2}}
A
=
2
9
2
1
\newline
(D)
A
=
(
1
29
)
2
A=\left(\frac{1}{29}\right)^{2}
A
=
(
29
1
)
2
Get tutor help
What is the value of
A
A
A
when we rewrite
(
32
5
)
−
2
x
\left(\frac{32}{5}\right)^{-2 x}
(
5
32
)
−
2
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
−
64
5
A=-\frac{64}{5}
A
=
−
5
64
\newline
B
A
=
1
2
A=\frac{1}{2}
A
=
2
1
\newline
(C)
A
=
(
5
32
)
−
2
A=\left(\frac{5}{32}\right)^{-2}
A
=
(
32
5
)
−
2
\newline
(D)
A
=
(
32
5
)
−
2
A=\left(\frac{32}{5}\right)^{-2}
A
=
(
5
32
)
−
2
Get tutor help
What is the value of
A
A
A
when we rewrite
(
7
3
)
x
\left(\frac{7}{3}\right)^{x}
(
3
7
)
x
as
A
2
x
A^{2 x}
A
2
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
1
2
A=\frac{1}{2}
A
=
2
1
\newline
(B)
A
=
(
7
3
)
1
2
A=\left(\frac{7}{3}\right)^{\frac{1}{2}}
A
=
(
3
7
)
2
1
\newline
(C)
A
=
(
7
3
)
−
2
A=\left(\frac{7}{3}\right)^{-2}
A
=
(
3
7
)
−
2
\newline
(D)
A
=
3
7
A=\frac{3}{7}
A
=
7
3
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What is the expression for
f
(
x
)
f(x)
f
(
x
)
when we rewrite
(
1
49
)
x
⋅
(
1
7
)
6
x
+
11
\left(\frac{1}{49}\right)^{x} \cdot\left(\frac{1}{7}\right)^{6 x+11}
(
49
1
)
x
⋅
(
7
1
)
6
x
+
11
as
(
1
7
)
f
(
x
)
\left(\frac{1}{7}\right)^{f(x)}
(
7
1
)
f
(
x
)
?
\newline
f
(
x
)
=
f(x)=
f
(
x
)
=
Get tutor help
What is the value of
A
A
A
when we rewrite
3.1
4
159
x
3.14^{159 x}
3.1
4
159
x
as
A
x
A^{x}
A
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
3.14159
A=3.14159
A
=
3.14159
\newline
(B)
A
=
1.59
314
A=\frac{1.59}{314}
A
=
314
1.59
\newline
(C)
A
=
3.1
4
159
A=3.14^{159}
A
=
3.1
4
159
\newline
(D)
A
=
3.14
159
A=\frac{3.14}{159}
A
=
159
3.14
Get tutor help
Find the value of
A
A
A
that makes the following equation true for all values of
x
x
x
.
\newline
2
x
=
A
x
12
2^{x}=A^{\frac{x}{12}}
2
x
=
A
12
x
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
(
1
12
)
2
A=\left(\frac{1}{12}\right)^{2}
A
=
(
12
1
)
2
\newline
(B)
A
=
2
1
12
A=2^{\frac{1}{12}}
A
=
2
12
1
\newline
(C)
A
=
2
12
A=2^{12}
A
=
2
12
\newline
(D)
A
=
2
⋅
12
A=2 \cdot 12
A
=
2
⋅
12
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What is the value of
A
A
A
when we rewrite
1
5
x
15^{x}
1
5
x
as
A
x
6
A^{\frac{x}{6}}
A
6
x
?
\newline
Choose
1
1
1
answer:
\newline
(A)
A
=
1
5
−
6
A=15^{-6}
A
=
1
5
−
6
\newline
(B)
A
=
1
5
6
A=15^{6}
A
=
1
5
6
\newline
(C)
A
=
1
5
−
1
6
A=15^{-\frac{1}{6}}
A
=
1
5
−
6
1
\newline
(D)
A
=
15
6
A=\frac{15}{6}
A
=
6
15
Get tutor help
What is the expression for
f
(
x
)
f(x)
f
(
x
)
when we rewrite
3
6
x
⋅
6
x
−
2
36^{x} \cdot 6^{x-2}
3
6
x
⋅
6
x
−
2
as
6
f
(
x
)
6^{f(x)}
6
f
(
x
)
?
\newline
f
(
x
)
=
f(x)=
f
(
x
)
=
Get tutor help
What is the expression for
f
(
x
)
f(x)
f
(
x
)
when we rewrite
\newline
(
1
32
)
x
⋅
(
1
2
)
9
x
−
5
as
(
1
2
)
f
(
x
)
f
(
x
)
=
□
\begin{array}{l} \left(\frac{1}{32}\right)^{x} \cdot\left(\frac{1}{2}\right)^{9 x-5} \text { as }\left(\frac{1}{2}\right)^{f(x)} \\ f(x)=\square \end{array}
(
32
1
)
x
⋅
(
2
1
)
9
x
−
5
as
(
2
1
)
f
(
x
)
f
(
x
)
=
□
Get tutor help
What is the expression for
f
(
x
)
f(x)
f
(
x
)
when we rewrite
6
4
x
4
5
x
−
1
\frac{64^{x}}{4^{5 x-1}}
4
5
x
−
1
6
4
x
as
4
f
(
x
)
4^{f(x)}
4
f
(
x
)
?
\newline
f
(
x
)
=
f(x)=
f
(
x
)
=
Get tutor help
Let
f
(
x
)
=
3
x
f(x)=\frac{3}{x}
f
(
x
)
=
x
3
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
0
x=0
x
=
0
.
\newline
Choose
1
1
1
answer:
\newline
(A)
lim
x
→
0
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} f(x)=+\infty
lim
x
→
0
+
f
(
x
)
=
+
∞
and
lim
x
→
0
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} f(x)=+\infty
lim
x
→
0
−
f
(
x
)
=
+
∞
\newline
(B)
lim
x
→
0
+
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{+}} f(x)=+\infty
lim
x
→
0
+
f
(
x
)
=
+
∞
and
lim
x
→
0
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} f(x)=-\infty
lim
x
→
0
−
f
(
x
)
=
−
∞
\newline
(C)
lim
x
→
0
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} f(x)=-\infty
lim
x
→
0
+
f
(
x
)
=
−
∞
and
lim
x
→
0
−
f
(
x
)
=
+
∞
\lim _{x \rightarrow 0^{-}} f(x)=+\infty
lim
x
→
0
−
f
(
x
)
=
+
∞
\newline
(D)
lim
x
→
0
+
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{+}} f(x)=-\infty
lim
x
→
0
+
f
(
x
)
=
−
∞
and
lim
x
→
0
−
f
(
x
)
=
−
∞
\lim _{x \rightarrow 0^{-}} f(x)=-\infty
lim
x
→
0
−
f
(
x
)
=
−
∞
Get tutor help
Let
h
(
x
)
=
x
10
sin
2
(
x
+
1
)
h(x)=\frac{x}{10 \sin ^{2}(x+1)}
h
(
x
)
=
10
s
i
n
2
(
x
+
1
)
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)
lim
x
→
−
1
+
h
(
x
)
=
+
∞
\lim _{x \rightarrow-1^{+}} h(x)=+\infty
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\lim _{x \rightarrow-1^{-}} h(x)=+\infty
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(B)
lim
x
→
−
1
+
h
(
x
)
=
+
∞
\lim _{x \rightarrow-1^{+}} h(x)=+\infty
lim
x
→
−
1
+
h
(
x
)
=
+
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\lim _{x \rightarrow-1^{-}} h(x)=-\infty
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\newline
(C)
lim
x
→
−
1
+
h
(
x
)
=
−
∞
\lim _{x \rightarrow-1^{+}} h(x)=-\infty
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\lim _{x \rightarrow-1^{-}} h(x)=+\infty
lim
x
→
−
1
−
h
(
x
)
=
+
∞
\newline
(D)
lim
x
→
−
1
+
h
(
x
)
=
−
∞
\lim _{x \rightarrow-1^{+}} h(x)=-\infty
lim
x
→
−
1
+
h
(
x
)
=
−
∞
and
lim
x
→
−
1
−
h
(
x
)
=
−
∞
\lim _{x \rightarrow-1^{-}} h(x)=-\infty
lim
x
→
−
1
−
h
(
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
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
Solve the exponential equation for
x
x
x
.
\newline
5
2
x
+
7
=
2
5
1
3
x
=
\begin{array}{l} 5^{2 x+7}=25^{\frac{1}{3}} \\ x= \end{array}
5
2
x
+
7
=
2
5
3
1
x
=
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
3
2
x
3
=
8
x
−
12
x
=
□
\begin{array}{l} 32^{\frac{x}{3}}=8^{x-12} \\ x=\square \end{array}
3
2
3
x
=
8
x
−
12
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
(
1
25
)
5
x
+
1
=
(
1
125
)
3
x
+
6
x
=
□
\begin{array}{l} \left(\frac{1}{25}\right)^{5 x+1}=\left(\frac{1}{125}\right)^{3 x+6} \\ x=\square \end{array}
(
25
1
)
5
x
+
1
=
(
125
1
)
3
x
+
6
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
(
1
64
)
−
2
x
+
6
=
(
1
16
)
8
x
−
5
x
=
□
\begin{array}{l} \left(\frac{1}{64}\right)^{-2 x+6}=\left(\frac{1}{16}\right)^{8 x-5} \\ x=\square \end{array}
(
64
1
)
−
2
x
+
6
=
(
16
1
)
8
x
−
5
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
5
x
−
12
=
(
1
125
)
12
5
x
=
□
\begin{array}{l} 5^{x-12}=\left(\frac{1}{125}\right)^{\frac{12}{5}} \\ x=\square \end{array}
5
x
−
12
=
(
125
1
)
5
12
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
6
4
x
−
9
=
(
1
36
)
x
−
4
x
=
□
\begin{array}{l} 6^{4 x-9}=\left(\frac{1}{36}\right)^{x-4} \\ x=\square \end{array}
6
4
x
−
9
=
(
36
1
)
x
−
4
x
=
□
Get tutor help
Solve the exponential equation for
x
x
x
.
\newline
2
4
−
3
x
=
1
6
2
5
x
=
□
\begin{array}{l} 2^{4-3 x}=16^{\frac{2}{5}} \\ x=\square \end{array}
2
4
−
3
x
=
1
6
5
2
x
=
□
Get tutor help
Find
lim
x
→
−
5
h
(
x
)
\lim _{x \rightarrow-5} h(x)
lim
x
→
−
5
h
(
x
)
for
\newline
h
(
x
)
=
5
x
+
4
x
+
8
.
h(x)=\frac{5 x+4}{x+8} \text {. }
h
(
x
)
=
x
+
8
5
x
+
4
.
Get tutor help
Find
lim
x
→
1
h
(
x
)
\lim _{x \rightarrow 1} h(x)
lim
x
→
1
h
(
x
)
for
\newline
h
(
x
)
=
5
x
3
−
6
x
2
+
2
x
−
1
.
h(x)=5 x^{3}-6 x^{2}+2 x-1 \text {. }
h
(
x
)
=
5
x
3
−
6
x
2
+
2
x
−
1
.
Get tutor help
Which of the following functions are continuous at
x
=
−
3
x=-3
x
=
−
3
?
\newline
h
(
x
)
=
1
(
x
+
3
)
2
g
(
x
)
=
(
x
−
3
)
3
\begin{array}{l} h(x)=\frac{1}{(x+3)^{2}} \\ g(x)=(x-3)^{3} \end{array}
h
(
x
)
=
(
x
+
3
)
2
1
g
(
x
)
=
(
x
−
3
)
3
\newline
Choose
1
1
1
answer:
\newline
(A)
h
h
h
only
\newline
(B)
g
g
g
only
\newline
(C) Both
h
h
h
and
g
g
g
\newline
(D) Neither
h
h
h
nor
g
g
g
Get tutor help
Find
lim
x
→
−
4
x
2
−
15
x
2
−
16
\lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16}
lim
x
→
−
4
x
2
−
16
x
2
−
15
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
8
-\frac{1}{8}
−
8
1
\newline
(B)
1
8
\frac{1}{8}
8
1
\newline
(c)
19
20
\frac{19}{20}
20
19
\newline
(D) The limit doesn't exist
Get tutor help
f
(
x
)
=
{
1
x
+
1
for
−
5
<
x
2
x
for
x
>
−
4
f(x)=\left\{\begin{array}{ll}\frac{1}{x+1} & \text { for }-5<x \\ 2^{x} & \text { for } x>-4\end{array}\right.
f
(
x
)
=
{
x
+
1
1
2
x
for
−
5
<
x
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)
−
1
3
-\frac{1}{3}
−
3
1
\newline
(C)
1
16
\frac{1}{16}
16
1
\newline
(D) The limit doesn't exist.
Get tutor help
Find
lim
x
→
3
(
x
−
5
)
2
(
x
−
3
)
2
\lim _{x \rightarrow 3} \frac{(x-5)^{2}}{(x-3)^{2}}
lim
x
→
3
(
x
−
3
)
2
(
x
−
5
)
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
−
2
3
-\frac{2}{3}
−
3
2
\newline
(B)
0
0
0
\newline
(C)
11
3
\frac{11}{3}
3
11
\newline
D D The limit doesn't exist
Get tutor help
∑
k
=
0
99
2
(
3
)
k
≈
\sum_{k=0}^{99} 2(3)^{k} \approx
k
=
0
∑
99
2
(
3
)
k
≈
\newline
Choose
1
1
1
answer:
\newline
(A)
5.15
⋅
1
0
47
5.15 \cdot 10^{47}
5.15
⋅
1
0
47
\newline
(B)
3.80
⋅
1
0
30
3.80 \cdot 10^{30}
3.80
⋅
1
0
30
\newline
(C)
2.37
⋅
1
0
−
30
2.37 \cdot 10^{-30}
2.37
⋅
1
0
−
30
\newline
(D)
7.76
⋅
1
0
−
48
7.76 \cdot 10^{-48}
7.76
⋅
1
0
−
48
Get tutor help
∑
k
=
0
29
−
3
(
0.9
)
k
≈
\sum_{k=0}^{29}-3(0.9)^{k} \approx
k
=
0
∑
29
−
3
(
0.9
)
k
≈
\newline
Choose
1
1
1
answer:
\newline
(A)
−
4.63
⋅
1
0
13
-4.63 \cdot 10^{13}
−
4.63
⋅
1
0
13
\newline
(B)
−
28
-28
−
28
.
7
7
7
\newline
(C)
−
1
-1
−
1
.
65
65
65
\newline
(D)
−
0
-0
−
0
.
14
14
14
Get tutor help
∑
k
=
0
19
6
(
1.5
)
k
≈
\sum_{k=0}^{19} 6(1.5)^{k} \approx
k
=
0
∑
19
6
(
1.5
)
k
≈
\newline
Choose
1
1
1
answer:
\newline
(A)
7983
7983
7983
.
02
02
02
\newline
(B)
13
,
301.03
13,301.03
13
,
301.03
\newline
(C)
26
,
590.05
26,590.05
26
,
590.05
\newline
(D)
39
,
891.08
39,891.08
39
,
891.08
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