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Let’s check out your problem:
Solve for
k
k
k
.
\newline
k
4
+
3
=
14
k
=
□
\begin{array}{l} \frac{k}{4}+3=14 \\ k=\square \end{array}
4
k
+
3
=
14
k
=
□
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Math Problems
Grade 6
Multiply two decimals: where does the decimal point go?
Full solution
Q.
Solve for
k
k
k
.
\newline
k
4
+
3
=
14
k
=
□
\begin{array}{l} \frac{k}{4}+3=14 \\ k=\square \end{array}
4
k
+
3
=
14
k
=
□
Write Equation:
Write down the equation that needs to be solved.
\newline
(
k
4
)
+
3
=
14
(\frac{k}{4}) + 3 = 14
(
4
k
)
+
3
=
14
Subtract
3
3
3
:
Subtract
3
3
3
from both sides of the equation to isolate the term with
k
k
k
.
\newline
k
4
+
3
−
3
=
14
−
3
\frac{k}{4} + 3 - 3 = 14 - 3
4
k
+
3
−
3
=
14
−
3
\newline
k
4
=
11
\frac{k}{4} = 11
4
k
=
11
Multiply by
4
4
4
:
Multiply both sides of the equation by
4
4
4
to solve for
k
k
k
.
4
×
(
k
4
)
=
11
×
4
4 \times \left(\frac{k}{4}\right) = 11 \times 4
4
×
(
4
k
)
=
11
×
4
k
=
44
k = 44
k
=
44
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Question
Evaluate the following expression.
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f
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Solve for
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\newline
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=
\begin{array}{l} \frac{4}{z}=\frac{12}{5} \\ z= \end{array}
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Solve for
n
n
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Question
Solve for
r
r
r
.
\newline
4
r
=
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r
=
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\begin{array}{l} \frac{4}{r}=\frac{5}{7} \\ r=\square \end{array}
r
4
=
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5
r
=
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Question
Solve for
r
r
r
.
\newline
r
5
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=
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\begin{array}{l} \frac{r}{5}=\frac{4}{7} \\ r=\square \end{array}
5
r
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Question
Reduce to simplest form.
\newline
5
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=
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−
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Question
Reduce to simplest form.
\newline
−
9
12
−
(
−
7
4
)
=
-\frac{9}{12}-\left(-\frac{7}{4}\right)=
−
12
9
−
(
−
4
7
)
=
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Posted 9 months ago
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