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julsineya [31]
3 years ago
8

Ron's recycle shop had an old paper-shredding machine. Business was good, so he bought a new paper-shredding machine. The old ma

chine could shred a truckload of paper in 4 hours. The new machine could shred the same truckload in 2 hours. How long will it take to shred the same truckload of paper if Ron runs both shredders at the same time?
Mathematics
1 answer:
levacccp [35]3 years ago
8 0

Answer: 1 hour 20 minutes

Step-by-step explanation:

Given : The old machine could shred a truckload of paper in 4 hours.

Rate of work done by old machine = \dfrac{1}{4}

The new machine could shred the same truckload in 2 hours.

Rate of work done by new machine = \dfrac{1}{2}

Let 't' be the time taken by both of them working together , then we have the following equation:-

\dfrac{1}{t}=\dfrac{1}{4}+\dfrac{1}{2}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{1+2}{4}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{3}{4}\\\\\Rightarrow t=\dfrac{4}{3}=1\dfrac{1}{3}\ \text{ hours}

Since 1 hour = 60 minutes

Then, \dfrac{1}{3}\ \text{ hours}=\dfrac{1}{3}\times60=20\text{ minutes}

Hence, it will take 1 hour 20 minutes to shred the same truckload of paper if Ron runs both shredders at the same time.

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Answer:

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Step-by-step explanation:

Given (Missing Information):

y = x^\frac{3}{2}; y = 8; x=0

Required

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V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy

First solve for a and b.

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Substitute 8 for y

8 = x^\frac{3}{2}

Take 2/3 root of both sides

8^\frac{2}{3} = x^{\frac{3}{2}*\frac{2}{3}}

8^\frac{2}{3} = x

2^{3*\frac{2}{3}} = x

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4 =x

x = 4

This implies that:

a = 4

For x=0

This implies that:

b=0

So, we have:

V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy

V = 2\pi \int\limits^4_0 {p(y)h(y)} \, dy

The volume of the solid becomes:

V = 2\pi \int\limits^4_0 {x(8 - x^{\frac{3}{2}}}) \, dx

Open bracket

V = 2\pi \int\limits^4_0 {8x - x.x^{\frac{3}{2}}} \, dx

V = 2\pi \int\limits^4_0 {8x - x^{\frac{2+3}{2}}} \, dx

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Integrate

V = 2\pi  * [{\frac{8x^2}{2} - \frac{x^{1+\frac{5}{2}}}{1+\frac{5}{2}}]\vert^4_0

V = 2\pi  * [{4x^2 - \frac{x^{\frac{2+5}{2}}}{\frac{2+5}{2}}]\vert^4_0

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Substitute 4 and 0 for x

V = 2\pi  * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [{4*0^2 - \frac{2}{7}*0^{\frac{7}{2}}])

V = 2\pi  * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [0])

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V = 2\pi  * [{64 - \frac{2}{7}*2^2^{*\frac{7}{2}}]

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V = 2\pi  * [{64 - \frac{256}{7}]

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V = 2\pi  * [\frac{64*7-256}{7}]

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V = [\frac{2\pi  * 192}{7}]

V = \frac{\pi  * 384}{7}

V = \frac{384}{7}\pi

Hence, the required volume is:

Volume = \frac{384}{7}\pi

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