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8
÷
1
4
5
8\div 1 \frac{4}{5}
8
÷
1
5
4
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Math Problems
Grade 6
Add, subtract, multiply, or divide two fractions
Full solution
Q.
8
÷
1
4
5
8\div 1 \frac{4}{5}
8
÷
1
5
4
Convert to Improper Fraction:
Convert the mixed number to an improper
fraction
.
\newline
1
4
5
=
(
5
×
1
+
4
)
5
=
9
5
1 \frac{4}{5} = \frac{(5\times1 + 4)}{5} = \frac{9}{5}
1
5
4
=
5
(
5
×
1
+
4
)
=
5
9
Multiply by Reciprocal:
Perform the division by multiplying by the reciprocal of the fraction.
8
÷
9
5
=
8
×
5
9
8 \div \frac{9}{5} = 8 \times \frac{5}{9}
8
÷
5
9
=
8
×
9
5
Simplify Multiplication:
Simplify the multiplication.
8
×
5
9
=
40
9
8 \times \frac{5}{9} = \frac{40}{9}
8
×
9
5
=
9
40
Convert to Mixed Number:
Convert the improper fraction to a mixed number.
\newline
40
9
=
4
4
9
\frac{40}{9} = 4 \frac{4}{9}
9
40
=
4
9
4
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Question
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Divide:
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8^2 \div [(11 - 3) \cdot 2]
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6
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7
7
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Jeanne's bank account earns interest annually. The equation shows her starting balance of
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546.36
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546.36
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Question
Jeanne's bank account earns interest annually. The equation shows her starting balance of
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and her balance at the end of three years,
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. At what rate r did Jeanne earn interest?
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422.78
=
350
(
1
+
r
)
3
422.78=350(1+r)^{3}
422.78
=
350
(
1
+
r
)
3
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Posted 1 month ago
Question
A leaky
10
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10\text{-kg}
10
-kg
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14
m
14\, \text{m}
14
m
at a constant speed with a rope that weighs
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0.6\, \text{kg/m}
0.6
kg/m
. Initially the bucket contains
42
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42\, \text{kg}
42
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of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the
14
m
14\, \text{m}
14
m
level. How much work is done? (Use
9.8
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2
9.8\, \text{m/s}^2
9.8
m/s
2
for
g
g
g
.) Show how to approximate the required work (in
J
\text{J}
J
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x
x
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x
i
∗
x_i^*
x
i
∗
as
x
i
x_i
x
i
.)
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Question
A leaky
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−
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g
10-\mathrm{kg}
10
−
kg
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14
14
14
m at a constant speed with a rope that weighs
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g
/
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0.6 \mathrm{~kg} / \mathrm{m}
0.6
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/
m
. Initially the bucket contains
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14
−
m
14-\mathrm{m}
14
−
m
level. How much work is done? (Use
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m
/
s
2
9.8 \mathrm{~m} / \mathrm{s}^{2}
9.8
m
/
s
2
for g .)
\newline
Show how to approximate the required work (in J) by a Riemann sum. (Let
x
x
x
be the height in meters above the ground. Enter
x
i
∗
∗
x_{i}^{* *}
x
i
∗∗
as
x
i
x_{i}
x
i
)
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∞
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i
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1
n
(
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+
35.28
x
i
)
Δ
x
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lim
n
→
∞
∑
i
=
1
n
(
98
+
35.28
x
i
)
Δ
x
\newline
Express the work (in J) as an integral in terms of
x
x
x
(in m).
\newline
∫
0
14
(
98
+
35.28
x
\int_{0}^{14}(98+35.28 x
∫
0
14
(
98
+
35.28
x
\newline
Evaluate the integral (in J). (Round your answer to the nearest integer.)
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Question
Solve the equation.
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6 w^{7}=98,304
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Question
A leaky
10
-kg
10\text{-kg}
10
-kg
bucket is lifted from the ground to a height of
14
m
14\text{ m}
14
m
at a constant speed with a rope that weighs
0.6
kg/m
0.6\text{ kg/m}
0.6
kg/m
. Initially the bucket contains
42
kg
42\text{ kg}
42
kg
of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the
14
-m
14\text{-m}
14
-m
level. How much work is done? (Use
9.8
m/s
2
9.8\text{ m/s}^2
9.8
m/s
2
for
g
g
g
.) Show how to approximate the required work (in
J
\text{J}
J
) by a Riemann sum. (Let
x
x
x
be the height in meters above the ground. Enter
x
i
∗
x_i^*
x
i
∗
as
x
i
x_i
x
i
.)
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Posted 1 month ago
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