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Let’s check out your problem:
Divide.
\newline
1
3
7
1\frac{3}{7}
1
7
3
÷
\div
÷
2
2
2
= _____
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Math Problems
Grade 7
Divide fractions
Full solution
Q.
Divide.
\newline
1
3
7
1\frac{3}{7}
1
7
3
÷
\div
÷
2
2
2
= _____
Convert to Improper Fraction:
Convert the mixed number
1
3
7
1 \frac{3}{7}
1
7
3
to an improper
fraction
.
\newline
1
3
7
=
(
1
×
7
+
3
)
/
7
=
10
7
1 \frac{3}{7} = (1 \times 7 + 3)/7 = \frac{10}{7}
1
7
3
=
(
1
×
7
+
3
)
/7
=
7
10
Divide by
2
2
2
:
Divide the improper fraction by
2
2
2
.
\newline
10
7
÷
2
=
10
7
×
1
2
=
10
14
\frac{10}{7} \div 2 = \frac{10}{7} \times \frac{1}{2} = \frac{10}{14}
7
10
÷
2
=
7
10
×
2
1
=
14
10
Simplify Fraction:
Simplify the fraction
10
14
\frac{10}{14}
14
10
.
\newline
10
14
\frac{10}{14}
14
10
can be simplified by dividing both numerator and denominator by
2
2
2
, which gives
5
7
\frac{5}{7}
7
5
.
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\newline
2
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2
=
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Question
Multiply.
\newline
(
3
5
)
×
(
1
6
)
=
(\frac{3}{5})\times(\frac{1}{6})=
(
5
3
)
×
(
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)
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\newline
Submit
\newline
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Posted 8 months ago
Question
2
x
+
3
=
2
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+
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\newline
Find
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.
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Question
Simplify to a single power of
3
3
3
:
\newline
3
5
×
3
2
3^{5} \times 3^{2}
3
5
×
3
2
\newline
Answer:
3
□
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3
□
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Question
Simplify to a single power of
6
6
6
:
\newline
6
3
⋅
6
2
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6
3
⋅
6
2
\newline
Answer:
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□
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6
□
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Question
What value of
z
z
z
makes the equation below true?
\newline
9
z
−
2
=
79
9 z-2=79
9
z
−
2
=
79
\newline
3
3
3
\newline
9
9
9
\newline
13
13
13
\newline
79
79
79
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Question
What value of
y
y
y
makes the equation below true?
\newline
5
y
+
2
=
17
5 y+2=17
5
y
+
2
=
17
\newline
2
2
2
\newline
3
3
3
\newline
6
6
6
\newline
17
17
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