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Welcome to Bytelearn!
Let’s check out your problem:
Given x>0 and y>0 , select the expression that is equivalent to
\newline
64
x
8
y
5
\sqrt{64 x^{8} y^{5}}
64
x
8
y
5
\newline
32
x
4
y
5
2
32 x^{4} y^{\frac{5}{2}}
32
x
4
y
2
5
\newline
32
x
1
4
y
2
5
32 x^{\frac{1}{4}} y^{\frac{2}{5}}
32
x
4
1
y
5
2
\newline
8
x
1
4
y
2
5
8 x^{\frac{1}{4}} \boldsymbol{y}^{\frac{2}{5}}
8
x
4
1
y
5
2
\newline
8
x
4
y
5
2
8 x^{4} y^{\frac{5}{2}}
8
x
4
y
2
5
View step-by-step help
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Math Problems
Precalculus
Operations with rational exponents
Full solution
Q.
Given
x
>
0
x>0
x
>
0
and
y
>
0
y>0
y
>
0
, select the expression that is equivalent to
\newline
64
x
8
y
5
\sqrt{64 x^{8} y^{5}}
64
x
8
y
5
\newline
32
x
4
y
5
2
32 x^{4} y^{\frac{5}{2}}
32
x
4
y
2
5
\newline
32
x
1
4
y
2
5
32 x^{\frac{1}{4}} y^{\frac{2}{5}}
32
x
4
1
y
5
2
\newline
8
x
1
4
y
2
5
8 x^{\frac{1}{4}} \boldsymbol{y}^{\frac{2}{5}}
8
x
4
1
y
5
2
\newline
8
x
4
y
5
2
8 x^{4} y^{\frac{5}{2}}
8
x
4
y
2
5
Identify expression:
Identify the expression to simplify.
\newline
We need to simplify the
square root
of the expression
64
x
8
y
5
64x^{8}y^{5}
64
x
8
y
5
.
Simplify
64
64
64
:
Simplify the square root of
64
64
64
. The square root of
64
64
64
is
8
8
8
because
8
×
8
=
64
8 \times 8 = 64
8
×
8
=
64
.
Simplify
x
8
x^{8}
x
8
:
Simplify the square root of
x
8
x^{8}
x
8
.
\newline
The square root of
x
8
x^{8}
x
8
is
x
4
x^{4}
x
4
because
(
x
4
)
2
=
x
8
(x^{4})^2 = x^{8}
(
x
4
)
2
=
x
8
.
Simplify
y
5
y^{5}
y
5
:
Simplify the square root of
y
5
y^{5}
y
5
. The square root of
y
5
y^{5}
y
5
is
y
(
5
2
)
y^{\left(\frac{5}{2}\right)}
y
(
2
5
)
because
(
y
(
5
2
)
)
2
=
y
5
\left(y^{\left(\frac{5}{2}\right)}\right)^2 = y^{5}
(
y
(
2
5
)
)
2
=
y
5
.
Combine results:
Combine the simplified parts.
\newline
Combine the results from steps
2
2
2
,
3
3
3
, and
4
4
4
to get the simplified expression:
8
x
4
y
(
5
/
2
)
8x^{4}y^{(5/2)}
8
x
4
y
(
5/2
)
.
More problems from Operations with rational exponents
Question
f
(
x
)
=
6
x
+
5
2
−
14
+
5
x
f(x)=\frac{6 x+5}{2-\sqrt{14+5 x}}
f
(
x
)
=
2
−
14
+
5
x
6
x
+
5
\newline
We want to find
lim
x
→
−
2
f
(
x
)
\lim _{x \rightarrow-2} f(x)
lim
x
→
−
2
f
(
x
)
.
\newline
What happens when we use direct substitution?
\newline
Choose
1
1
1
answer:
\newline
(A) The limit exists, and we found it!
\newline
(B) The limit doesn't exist (probably an asymptote).
\newline
(C) The result is indeterminate.
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Posted 1 year ago
Question
Let
x
4
+
y
2
=
17
x^{4}+y^{2}=17
x
4
+
y
2
=
17
.
\newline
What is the value of
d
2
y
d
x
2
\frac{d^{2} y}{d x^{2}}
d
x
2
d
2
y
at the point
(
−
2
,
1
)
(-2,1)
(
−
2
,
1
)
?
\newline
Give an exact number.
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Posted 1 year ago
Question
What is the area of the region bound by the graphs of
f
(
x
)
=
x
−
2
,
g
(
x
)
=
14
−
x
f(x)=\sqrt{x-2}, g(x)=14-x
f
(
x
)
=
x
−
2
,
g
(
x
)
=
14
−
x
, and
x
=
2
x=2
x
=
2
?
\newline
Choose
1
1
1
answer:
\newline
(A)
19
6
\frac{19}{6}
6
19
\newline
(B)
99
2
\frac{99}{2}
2
99
\newline
(C)
151
2
\frac{151}{2}
2
151
\newline
(D)
45
2
\frac{45}{2}
2
45
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Question
Let
g
(
x
)
=
x
4
3
g(x)=x^{\frac{4}{3}}
g
(
x
)
=
x
3
4
.
\newline
g
′
(
27
)
=
g^{\prime}(27)=
g
′
(
27
)
=
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Posted 1 year ago
Question
Simplify
ln
(
1
e
)
\ln \left(\frac{1}{\sqrt{e}}\right)
ln
(
e
1
)
\newline
Answer:
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Question
Simplify
ln
(
1
e
)
\ln \left(\frac{1}{e}\right)
ln
(
e
1
)
\newline
Answer:
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Question
y
=
(
sin
−
1
(
5
x
2
)
)
3
y=(\sin^{-1}(5x^{2}))^{3}
y
=
(
sin
−
1
(
5
x
2
)
)
3
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(
10
x
3
−
5
x
2
−
6
x
)
÷
(
2
x
2
)
\left(10x^{3}-5x^{2}-6x\right)\div\left(2x^{2}\right)
(
10
x
3
−
5
x
2
−
6
x
)
÷
(
2
x
2
)
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Posted 1 year ago
Question
y
=
cos
4
x
sin
4
x
y=\cos^{4}x\sin^{4}x
y
=
cos
4
x
sin
4
x
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Question
3
5
−
3
10
\frac{3}{5}-\frac{3}{10}
5
3
−
10
3
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