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Klio2033 [76]
3 years ago
15

Can anyone help me with this math equation?

Mathematics
1 answer:
sergiy2304 [10]3 years ago
5 0

Answer:

yes I can help you. the answer to the problem is 4/35.

Step-by-step explanation:

1. convert any mixed number into fractions.

your equation becomes 4/7÷5/1. this equals 4×1 over 7×5. multiply normally and your result is 4/35

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Answer:

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Step-by-step explanation:


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Abbys school has a track it takes eight laps around the track to run 1 mile.
insens350 [35]
The correct answer to the following questions is this:

<span>Abbys school has a track it takes eight laps around the track to run 1 mile. 
This means that you have to run 8 laps in order to complete a 1 mile. 
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A. What fraction of a mile is two laps around the track? Show your work or explain how you know
Since 1 lap = 1/8 mile, then 2 laps would be 2/8 mile or 1/4 mile.

Abby ran 3 1/8 miles on Monday and 1 3/8 miles on Tuesday.
B. What was the total number of miles every right on Monday and Tuesday? show your work or explain how you know
On Monday, she ran 3 1/8 miles, 1 3/8 miles on Tuesday. The total number of miles is 3 1/8 + 1 3/8 = (3+1) (1/8 + 3/8) = 4 4/8 = 4 1/2 miles.

C. How many total lasted gym run on Monday and Tuesday? Show your work or explain how you know
This seems not clear regarding the gym.
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7 0
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Students at Westbrook middle school hold green team meetings
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What’s the whole question???!
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What is the correct definition for tan 0
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3 years ago
Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
lutik1710 [3]

Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

6 0
3 years ago
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