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Let’s check out your problem:
Simplify the expression. Write your answers using
integers
or improper
fractions
.
\newline
2
k
−
1
2
(
1
2
k
+
2
)
2 k-\frac{1}{2}\left(\frac{1}{2} k+2\right)
2
k
−
2
1
(
2
1
k
+
2
)
\newline
Answer:
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Home
Math Problems
Grade 7
Powers with decimal and fractional bases
Full solution
Q.
Simplify the expression. Write your answers using integers or improper fractions.
\newline
2
k
−
1
2
(
1
2
k
+
2
)
2 k-\frac{1}{2}\left(\frac{1}{2} k+2\right)
2
k
−
2
1
(
2
1
k
+
2
)
\newline
Answer:
Distribute
(
1
/
2
)
(1/2)
(
1/2
)
:
Distribute the
(
1
/
2
)
(1/2)
(
1/2
)
across the terms inside the parentheses.
\newline
(
1
/
2
)
(
(
1
/
2
)
k
+
2
)
=
(
1
/
2
)
⋅
(
1
/
2
)
k
+
(
1
/
2
)
⋅
2
(1/2)((1/2)k + 2) = (1/2)\cdot(1/2)k + (1/2)\cdot2
(
1/2
)
((
1/2
)
k
+
2
)
=
(
1/2
)
⋅
(
1/2
)
k
+
(
1/2
)
⋅
2
Simplify multiplication:
Simplify the multiplication inside the parentheses.
\newline
(
1
2
)
∗
(
1
2
)
k
=
(
1
4
)
k
(\frac{1}{2})*(\frac{1}{2})k = (\frac{1}{4})k
(
2
1
)
∗
(
2
1
)
k
=
(
4
1
)
k
\newline
(
1
2
)
∗
2
=
1
(\frac{1}{2})*2 = 1
(
2
1
)
∗
2
=
1
\newline
So,
(
1
2
)
(
(
1
2
)
k
+
2
)
(\frac{1}{2})((\frac{1}{2})k + 2)
(
2
1
)
((
2
1
)
k
+
2
)
becomes
(
1
4
)
k
+
1
(\frac{1}{4})k + 1
(
4
1
)
k
+
1
Rewrite with simplification:
Rewrite the original expression with the simplified version of the parentheses.
\newline
2
k
−
(
1
2
)
(
1
2
k
+
2
)
2k - \left(\frac{1}{2}\right)\left(\frac{1}{2}k + 2\right)
2
k
−
(
2
1
)
(
2
1
k
+
2
)
becomes
2
k
−
(
1
4
k
+
1
)
2k - \left(\frac{1}{4}k + 1\right)
2
k
−
(
4
1
k
+
1
)
Distribute negative sign:
Distribute the negative sign across the terms in the parentheses.
\newline
2
k
−
(
(
1
/
4
)
k
+
1
)
=
2
k
−
(
1
/
4
)
k
−
1
2k - ((1/4)k + 1) = 2k - (1/4)k - 1
2
k
−
((
1/4
)
k
+
1
)
=
2
k
−
(
1/4
)
k
−
1
Combine like terms:
Combine like terms.
\newline
2
k
−
1
4
k
=
8
4
k
−
1
4
k
=
7
4
k
2k - \frac{1}{4}k = \frac{8}{4}k - \frac{1}{4}k = \frac{7}{4}k
2
k
−
4
1
k
=
4
8
k
−
4
1
k
=
4
7
k
\newline
So,
2
k
−
1
4
k
−
1
2k - \frac{1}{4}k - 1
2
k
−
4
1
k
−
1
becomes
7
4
k
−
1
\frac{7}{4}k - 1
4
7
k
−
1
Final simplified expression:
The expression is now simplified. The final simplified expression is
(
7
4
)
k
−
1
(\frac{7}{4})k - 1
(
4
7
)
k
−
1
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\newline
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\newline
x
2
=
0
x^2 = 0
x
2
=
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\newline
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\newline
x
=
x =
x
=
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x
x
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\newline
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\newline
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=
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Question
Solve for
x
x
x
.
\newline
6
=
7
x
6 = 7^x
6
=
7
x
\newline
x
=
x =
x
=
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Question
Select the equivalent expression.
\newline
(
(
8
−
5
)
/
(
2
−
2
)
)
−
4
=
?
((8^{-5})/(2^{-2}))^{-4}=?
((
8
−
5
)
/
(
2
−
2
)
)
−
4
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
(
8
20
)
/
(
2
8
)
(8^{20})/(2^{8})
(
8
20
)
/
(
2
8
)
\newline
(B)
\newline
(
2
6
)
/
(
8
9
)
(2^{6})/(8^{9})
(
2
6
)
/
(
8
9
)
\newline
(C)
\newline
(
1
)
/
(
8
⋅
2
2
)
(1)/(8\cdot 2^{2})
(
1
)
/
(
8
⋅
2
2
)
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Question
Select the equivalent expression.
\newline
(
b
7
4
5
)
−
3
=
?
\left(\frac{b^{7}}{4^{5}}\right)^{-3}=\,?
(
4
5
b
7
)
−
3
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
b
21
4
15
\frac{b^{21}}{4^{15}}
4
15
b
21
\newline
(B)
4
15
b
21
\frac{4^{15}}{b^{21}}
b
21
4
15
\newline
(C)
b
−
21
⋅
4
−
15
b^{-21}\cdot 4^{-15}
b
−
21
⋅
4
−
15
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Question
Select the equivalent expression.
\newline
(
3
3
⋅
6
6
)
−
3
=
(3^{3}\cdot6^{6})^{-3}=
(
3
3
⋅
6
6
)
−
3
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
3
9
⋅
6
18
\frac{1}{3^{9}\cdot6^{18}}
3
9
⋅
6
18
1
\newline
(B)
6
18
3
9
\frac{6^{18}}{3^{9}}
3
9
6
18
\newline
(C)
3
9
6
18
\frac{3^{9}}{6^{18}}
6
18
3
9
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Posted 1 year ago
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