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Math Problems
Algebra 1
Multiplication with rational exponents
Write
7
4
⋅
49
\sqrt[4]{7} \cdot \sqrt{49}
4
7
⋅
49
using rational exponents.
\newline
A.
4
5
7
4^{\frac{5}{7}}
4
7
5
\newline
B.
7
5
4
7^{\frac{5}{4}}
7
4
5
\newline
C.
4
7
4^{7}
4
7
\newline
D.
7
4
7^{4}
7
4
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3
3
3
folders cost
$
2.91
\$ 2.91
$2.91
.
\newline
Which equation would help determine the cost of
2
2
2
folders?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
$
2.91
=
x
3
\frac{2}{\$ 2.91}=\frac{x}{3}
$2.91
2
=
3
x
\newline
(B)
2
x
=
3
$
2.91
\frac{2}{x}=\frac{3}{\$ 2.91}
x
2
=
$2.91
3
\newline
(C)
x
2
=
3
$
2.91
\frac{x}{2}=\frac{3}{\$ 2.91}
2
x
=
$2.91
3
\newline
(D)
2
x
=
$
2.91
3
\frac{2}{x}=\frac{\$ 2.91}{3}
x
2
=
3
$2.91
\newline
(E) None of the above
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7
15
−
2
15
\frac{7}{15}-\frac{2}{15}
15
7
−
15
2
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Which of the following is equivalent to
3
x
−
4
y
=
6
3 x-4 y=6
3
x
−
4
y
=
6
?
\newline
(A)
y
=
−
6
7
x
y=-\frac{6}{7} x
y
=
−
7
6
x
\newline
(B)
y
=
−
3
4
x
y=-\frac{3}{4} x
y
=
−
4
3
x
\newline
(C)
y
=
4
3
x
+
2
y=\frac{4}{3} x+2
y
=
3
4
x
+
2
\newline
(D)
y
=
3
4
x
−
3
2
y=\frac{3}{4} x-\frac{3}{2}
y
=
4
3
x
−
2
3
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5
4
5
+
(
−
3
2
7
)
5 \frac{4}{5}+\left(-3 \frac{2}{7}\right)
5
5
4
+
(
−
3
7
2
)
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Simplify:
\newline
x
8
x
3
\frac{x^{8}}{x^{3}}
x
3
x
8
\newline
(A)
x
5
x^{5}
x
5
\newline
(B)
x
−
5
x^{-5}
x
−
5
\newline
(C)
x
11
x^{11}
x
11
\newline
(D)
x
24
x^{24}
x
24
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Use the distributive property to remove the parentheses. Simplify your answer as much as possible.
\newline
15
(
1
3
w
−
2
5
)
15\left(\frac{1}{3} w-\frac{2}{5}\right)
15
(
3
1
w
−
5
2
)
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How can
7
×
11
×
13
+
13
7 \times 11 \times 13 + 13
7
×
11
×
13
+
13
be written as
13
×
(
7
×
11
+
1
)
13 \times (7 \times 11 + 1)
13
×
(
7
×
11
+
1
)
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Pick up the rational numbers from the following numbers.
\newline
6
7
,
−
1
2
,
0
,
1
0
,
100
0
\frac{6}{7}, \frac{-1}{2}, 0, \frac{1}{0}, \frac{100}{0}
7
6
,
2
−
1
,
0
,
0
1
,
0
100
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Simplify.
\newline
(
3
z
2
z
−
3
)
−
2
\left(\frac{3 z^{2}}{z^{-3}}\right)^{-2}
(
z
−
3
3
z
2
)
−
2
\newline
Write your answer using only positive exponents.
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5
5
5
.
5
6
5
5
\frac{5^{6}}{5^{5}}
5
5
5
6
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20
100
×
x
245
\frac{20}{100} \times \frac{x}{245}
100
20
×
245
x
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6
6
7
−
4
1
4
=
6 \frac{6}{7}-4 \frac{1}{4}=
6
7
6
−
4
4
1
=
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Which expression is equivalent to
1
1
−
26
11
?
\frac{11^{-26}}{11} ?
11
1
1
−
26
?
\newline
1
1
−
25
11^{-25}
1
1
−
25
\newline
1
1
1
−
26
\frac{1}{11^{-26}}
1
1
−
26
1
\newline
1
1
27
11^{27}
1
1
27
\newline
1
1
1
27
\frac{1}{11^{27}}
1
1
27
1
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405
x
3
y
3
5
x
−
1
y
4
=
\sqrt[4]{\frac{405 x^{3} y^{3}}{5 x^{-1} y}}=
4
5
x
−
1
y
405
x
3
y
3
=
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V
=
2
⋅
10
⋅
0.8
(
0.05
0.103
)
2
−
1
V = \frac{\sqrt{2 \cdot 10 \cdot 0.8}}{(\frac{0.05}{0.103})^{2}}-1
V
=
(
0.103
0.05
)
2
2
⋅
10
⋅
0.8
−
1
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3
x
2
+
4
x
3
=
\frac{3}{x^{2}}+\frac{4}{x^{3}}=
x
2
3
+
x
3
4
=
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1
1
1
)
[
−
8
3
+
−
13
12
)
×
5
6
=
(
−
8
3
×
5
6
)
+
(
−
13
12
×
5
6
)
\left[\frac{-8}{3}+\frac{-13}{12}\right) \times \frac{5}{6}=\left(-\frac{8}{3} \times \frac{5}{6}\right)+\left(-\frac{13}{12} \times \frac{5}{6}\right)
[
3
−
8
+
12
−
13
)
×
6
5
=
(
−
3
8
×
6
5
)
+
(
−
12
13
×
6
5
)
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Show that
(
−
2
5
+
4
9
)
+
(
−
3
4
)
=
−
2
5
+
{
4
9
+
(
−
3
4
)
}
\left(-\frac{2}{5}+\frac{4}{9}\right)+\left(-\frac{3}{4}\right)=-\frac{2}{5}+\left\{\frac{4}{9}+\left(-\frac{3}{4}\right)\right\}
(
−
5
2
+
9
4
)
+
(
−
4
3
)
=
−
5
2
+
{
9
4
+
(
−
4
3
)
}
.
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(ii)
(
−
3
29
÷
9
87
)
÷
−
1
7
\left(\frac{-3}{29} \div \frac{9}{87}\right) \div \frac{-1}{7}
(
29
−
3
÷
87
9
)
÷
7
−
1
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4
4
4
)
3
h
2
j
0
k
0
3
h
j
−
4
k
−
1
\frac{3 h^{2} j^{0} k^{0}}{3 h j^{-4} k^{-1}}
3
h
j
−
4
k
−
1
3
h
2
j
0
k
0
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x
4
3
x
2
3
\frac{x^{\frac{4}{3}}}{x^{\frac{2}{3}}}
x
3
2
x
3
4
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Simplify.
\newline
(
−
1057
10
+
1283
i
10
)
x
3
+
(
273
10
−
121
i
10
)
x
2
+
(
971
10
+
281
i
5
)
x
+
87
i
5
−
736
5
(-\frac{1057}{10}+\frac{1283 i}{10}) x^{3} + (\frac{273}{10}-\frac{121 i}{10}) x^{2} + (\frac{971}{10}+\frac{281 i}{5}) x + \frac{87 i}{5}-\frac{736}{5}
(
−
10
1057
+
10
1283
i
)
x
3
+
(
10
273
−
10
121
i
)
x
2
+
(
10
971
+
5
281
i
)
x
+
5
87
i
−
5
736
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Priya had
$
90
\$ 90
$90
. She spent
4
9
\frac{4}{9}
9
4
of her money and saved the remaining. How much money did Priya save?
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∫
(
(
4
⋅
x
−
4
)
⋅
(
x
2
−
2
⋅
x
−
15
)
8
)
d
x
\int\left((4 \cdot x-4) \cdot\left(x^{2}-2 \cdot x-15\right)^{8}\right) d x
∫
(
(
4
⋅
x
−
4
)
⋅
(
x
2
−
2
⋅
x
−
15
)
8
)
d
x
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(
−
23
5
)
⋅
[
11
⋅
(
−
5
23
)
]
\left(-\frac{23}{5}\right) \cdot\left[11 \cdot\left(-\frac{5}{23}\right)\right]
(
−
5
23
)
⋅
[
11
⋅
(
−
23
5
)
]
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(
x
−
4
y
2
)
−
3
⋅
x
4
y
2
\left(x^{-4} y^{2}\right)^{-3} \cdot x^{4} y^{2}
(
x
−
4
y
2
)
−
3
⋅
x
4
y
2
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The inverse of the function
f
(
x
)
=
x
+
1
x
−
2
f(x)=\frac{x+1}{x-2}
f
(
x
)
=
x
−
2
x
+
1
is
\newline
(
1
1
1
)
f
−
1
(
x
)
=
x
+
1
x
+
2
f^{-1}(x)=\frac{x+1}{x+2}
f
−
1
(
x
)
=
x
+
2
x
+
1
\newline
(
3
3
3
)
f
−
1
(
x
)
=
x
+
1
x
−
2
f^{-1}(x)=\frac{x+1}{x-2}
f
−
1
(
x
)
=
x
−
2
x
+
1
\newline
(
2
2
2
)
f
−
1
(
x
)
=
2
x
+
1
x
−
1
f^{-1}(x)=\frac{2 x+1}{x-1}
f
−
1
(
x
)
=
x
−
1
2
x
+
1
\newline
(
4
4
4
)
f
−
1
(
x
)
=
x
−
1
x
+
1
f^{-1}(x)=\frac{x-1}{x+1}
f
−
1
(
x
)
=
x
+
1
x
−
1
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The expression
(
m
2
m
(
1
3
)
)
−
(
1
2
)
\left(\frac{m^{2}}{m^{\left(\frac{1}{3}\right)}}\right)^{-\left(\frac{1}{2}\right)}
(
m
(
3
1
)
m
2
)
−
(
2
1
)
is equivalent to
\newline
(
1
1
1
)
−
m
5
1
-\sqrt[1]{m^{5}}
−
1
m
5
\newline
(
2
2
2
)
1
m
5
6
\frac{1}{\sqrt[6]{m^{5}}}
6
m
5
1
\newline
(
3
3
3
)
−
m
m
5
-m\sqrt[5]{m}
−
m
5
m
\newline
(
4
4
4
)
1
m
3
\frac{1}{\sqrt[3]{m}}
3
m
1
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The expression
(
m
2
m
1
3
)
−
1
2
\left(\frac{m^{2}}{m^{\frac{1}{3}}}\right)^{-\frac{1}{2}}
(
m
3
1
m
2
)
−
2
1
is equivalent to
\newline
(
1
1
1
)
m
5
16
\sqrt[16]{m^{5}}
16
m
5
\newline
(
3
3
3
)
−
m
m
5
-m \sqrt[5]{m}
−
m
5
m
\newline
(
2
2
2
)
1
m
5
6
\frac{1}{\sqrt[6]{m^{5}}}
6
m
5
1
\newline
(
4
4
4
)
1
m
m
3
\frac{1}{m \sqrt[3]{m}}
m
3
m
1
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Which expressions are equivalent to
\newline
4
−
2
⋅
7
−
2
4^{-2}\cdot7^{-2}
4
−
2
⋅
7
−
2
?
\newline
Choose
2
2
2
answers:
\newline
(A)
(
4
⋅
7
)
−
4
(4\cdot7)^{-4}
(
4
⋅
7
)
−
4
\newline
(B)
1
2
8
2
\frac{1}{28^{2}}
2
8
2
1
\newline
(C)
7
−
2
4
2
\frac{7^{-2}}{4^{2}}
4
2
7
−
2
\newline
(D)
(
4
⋅
7
)
4
(4\cdot7)^{4}
(
4
⋅
7
)
4
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Which expressions are equivalent to
4
−
2
⋅
7
−
2
4^{-2} \cdot 7^{-2}
4
−
2
⋅
7
−
2
?
\newline
Choose
2
2
2
answers:
\newline
A
(
4
⋅
7
)
−
4
(4 \cdot 7)^{-4}
(
4
⋅
7
)
−
4
\newline
B
1
2
8
2
\frac{1}{28^{2}}
2
8
2
1
\newline
c]
7
−
2
4
2
\frac{7^{-2}}{4^{2}}
4
2
7
−
2
\newline
D
(
4
⋅
7
)
4
(4 \cdot 7)^{4}
(
4
⋅
7
)
4
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Which expressions are equivalent to
4
−
3
4
−
1
\frac{4^{-3}}{4^{-1}}
4
−
1
4
−
3
?
\newline
Choose
2
2
2
answers:
\newline
(A)
4
1
4
3
\frac{4^{1}}{4^{3}}
4
3
4
1
\newline
(B)
1
4
2
\frac{1}{4^{2}}
4
2
1
\newline
c.
4
3
⋅
4
1
4^{3} \cdot 4^{1}
4
3
⋅
4
1
\newline
D
(
4
−
1
)
−
3
\left(4^{-1}\right)^{-3}
(
4
−
1
)
−
3
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Which expressions are equivalent to
4
−
3
4
−
1
\frac{4^{-3}}{4^{-1}}
4
−
1
4
−
3
?
\newline
Choose
2
2
2
answers:
\newline
A
4
1
4
3
\frac{4^{1}}{4^{3}}
4
3
4
1
\newline
B
1
4
2
\frac{1}{4^{2}}
4
2
1
\newline
c]
4
3
⋅
4
1
4^{3} \cdot 4^{1}
4
3
⋅
4
1
\newline
D
(
4
−
1
)
−
3
\left(4^{-1}\right)^{-3}
(
4
−
1
)
−
3
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Evaluate the limit and write your answer in simplest form:
\newline
lim
h
→
0
3
cos
(
4
π
3
+
h
)
−
3
cos
(
4
π
3
)
h
\lim _{h \rightarrow 0} \frac{3 \cos \left(\frac{4 \pi}{3}+h\right)-3 \cos \left(\frac{4 \pi}{3}\right)}{h}
h
→
0
lim
h
3
cos
(
3
4
π
+
h
)
−
3
cos
(
3
4
π
)
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2
1
2
+
4
1
5
2 \frac{1}{2}+4 \frac{1}{5}
2
2
1
+
4
5
1
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B.
7
6
−
□
6
=
2
6
\frac{7}{6}-\frac{\square}{6}=\frac{2}{6}
6
7
−
6
□
=
6
2
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Divide:
\newline
(
2.31
80
)
\left(\frac{2.31}{80}\right)
(
80
2.31
)
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1
4
+
1
2
\frac{1}{4}+\frac{1}{2}
4
1
+
2
1
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5
4
(
4
a
+
2
)
=
1
5
6
\frac{5}{4}(4 a+2)=1 \frac{5}{6}
4
5
(
4
a
+
2
)
=
1
6
5
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5
11
(
4
a
+
2
)
=
15
0.6
\frac{5}{11}(4 a+2)=\frac{15}{0.6}
11
5
(
4
a
+
2
)
=
0.6
15
\newline
What is the value of a?
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If
3
x
+
4
y
=
7
3 x+4 y=7
3
x
+
4
y
=
7
, which of the following expressions is equivalent to
y
y
y
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
3
x
+
7
-3 x+7
−
3
x
+
7
\newline
(B)
−
3
4
x
+
7
4
-\frac{3}{4} x+\frac{7}{4}
−
4
3
x
+
4
7
\newline
(C)
3
4
x
+
7
4
\frac{3}{4} x+\frac{7}{4}
4
3
x
+
4
7
\newline
(D)
3
x
+
7
3 x+7
3
x
+
7
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Which of the following is equivalent to
tan
π
5
\tan \frac{\pi}{5}
tan
5
π
?
\newline
tan
(
−
π
5
)
\tan \left(-\frac{\pi}{5}\right)
tan
(
−
5
π
)
\newline
tan
9
π
5
\tan \frac{9 \pi}{5}
tan
5
9
π
\newline
tan
4
π
5
\tan \frac{4 \pi}{5}
tan
5
4
π
\newline
tan
(
−
9
π
5
)
\tan \left(-\frac{9 \pi}{5}\right)
tan
(
−
5
9
π
)
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Which of the following is not equivalent to
sec
3
π
8
\sec \frac{3 \pi}{8}
sec
8
3
π
?
\newline
sec
(
−
3
π
8
)
\sec \left(-\frac{3 \pi}{8}\right)
sec
(
−
8
3
π
)
\newline
sec
19
π
8
\sec \frac{19 \pi}{8}
sec
8
19
π
\newline
sec
5
π
8
\sec \frac{5 \pi}{8}
sec
8
5
π
\newline
sec
13
π
8
\sec \frac{13 \pi}{8}
sec
8
13
π
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Which of the following is not equivalent to
cos
2
π
5
\cos \frac{2 \pi}{5}
cos
5
2
π
?
\newline
cos
7
π
5
\cos \frac{7 \pi}{5}
cos
5
7
π
\newline
cos
8
π
5
\cos \frac{8 \pi}{5}
cos
5
8
π
\newline
cos
12
π
5
\cos \frac{12 \pi}{5}
cos
5
12
π
\newline
cos
(
−
2
π
5
)
\cos \left(-\frac{2 \pi}{5}\right)
cos
(
−
5
2
π
)
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Which of the following is not equivalent to
cos
π
5
\cos \frac{\pi}{5}
cos
5
π
?
\newline
cos
11
π
5
\cos \frac{11 \pi}{5}
cos
5
11
π
\newline
cos
(
−
9
π
5
)
\cos \left(-\frac{9 \pi}{5}\right)
cos
(
−
5
9
π
)
\newline
cos
14
π
5
\cos \frac{14 \pi}{5}
cos
5
14
π
\newline
cos
9
π
5
\cos \frac{9 \pi}{5}
cos
5
9
π
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Which of the following is not equivalent to
csc
π
4
\csc \frac{\pi}{4}
csc
4
π
?
\newline
csc
9
π
4
\csc \frac{9 \pi}{4}
csc
4
9
π
\newline
csc
(
−
7
π
4
)
\csc \left(-\frac{7 \pi}{4}\right)
csc
(
−
4
7
π
)
\newline
csc
11
π
4
\csc \frac{11 \pi}{4}
csc
4
11
π
\newline
csc
(
−
π
4
)
\csc \left(-\frac{\pi}{4}\right)
csc
(
−
4
π
)
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Which of the following is not equivalent to
tan
π
4
\tan \frac{\pi}{4}
tan
4
π
?
\newline
tan
7
π
4
\tan \frac{7 \pi}{4}
tan
4
7
π
\newline
tan
9
π
4
\tan \frac{9 \pi}{4}
tan
4
9
π
\newline
tan
5
π
4
\tan \frac{5 \pi}{4}
tan
4
5
π
\newline
tan
(
−
7
π
4
)
\tan \left(-\frac{7 \pi}{4}\right)
tan
(
−
4
7
π
)
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Which of the following is not equivalent to
csc
2
π
5
?
\csc \frac{2 \pi}{5} ?
csc
5
2
π
?
\newline
csc
7
π
5
\csc \frac{7 \pi}{5}
csc
5
7
π
\newline
csc
13
π
5
\csc \frac{13 \pi}{5}
csc
5
13
π
\newline
csc
3
π
5
\csc \frac{3 \pi}{5}
csc
5
3
π
\newline
csc
(
−
8
π
5
)
\csc \left(-\frac{8 \pi}{5}\right)
csc
(
−
5
8
π
)
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Which of the following is not equivalent to
tan
2
π
7
\tan \frac{2 \pi}{7}
tan
7
2
π
?
\newline
tan
9
π
7
\tan \frac{9 \pi}{7}
tan
7
9
π
\newline
tan
16
π
7
\tan \frac{16 \pi}{7}
tan
7
16
π
\newline
tan
(
−
12
π
7
)
\tan \left(-\frac{12 \pi}{7}\right)
tan
(
−
7
12
π
)
\newline
tan
19
π
7
\tan \frac{19 \pi}{7}
tan
7
19
π
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