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Masja [62]
3 years ago
5

A piston is seated at the top of a cylindrical chamber with radius 3 cm when it starts moving into the chamber at a constant spe

ed of 4 ​cm/s (see​ figure). What is the rate of change of the volume of the cylinder when the piston is 15 cm from the base of the​ chamber?
a) Let​ V, r, and h be the​volume, radius, and height of a​ cylinder, respectively. Write an equation relating​ V, r, and h.
b) Find the related rates equation.
c) When the piston is 15cm from the base of the​ chamber, the volume of the cylinder is changing at a rate of about
Mathematics
1 answer:
dlinn [17]3 years ago
4 0

Answer:

1. V = πr²h

2. dv/dt = πr²dh/dt

3. dv/dt = 113.04

Step-by-step explanation:

We have these information

Radius R = 3cm

Height = h

Volume = V

We have been given dh/dt = 4cm

1.

An equation relating V, r, h

Volume of cylinder = πr²h

2.

The related rates equation

dv/dt = πr²dh/dt

dv/dt = π3²dh/dt

3. Continuing from number 2 above, we have

dv/dt = π3²dh/dt

= π9x4

= 3.14x9x4

= 113.04

These are the answers to the questions

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Answer:

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Step-by-step explanation:

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Now, let's find the area of the circle and approximate pi to 3.14.

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7 0
3 years ago
You collect 140 total pieces of candy trick or treating and 35 of them are m&m's. What fraction of the candy is not m&m'
tekilochka [14]

\frac{3}{4} of the candy is not m&m's.

Step-by-step explanation:

Given,

Total pieces of candy = 140

Number of m&m's = 35

Fraction = \frac{Number\ of\ m&m}{Total\ pieces\ of\ candy}

Fraction = \frac{35}{140}

Both 35 and 140 are multiples of 7, therefore,

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As the number of m&m's fraction and not m&m's fraction will make a total of 1, therefore

fraction of m&m's + fraction of not m&m's = 1

fraction of not m&m's = 1 - fraction of m&m's

fraction of not m&m's = 1-\frac{1}{4}=\frac{4-1}{4}

fraction of not m&m's = \frac{3}{4}

\frac{3}{4} of the candy is not m&m's.

Keywords: fraction, subtraction

Learn more about fractions at:

  • brainly.com/question/8929610
  • brainly.com/question/8908016

#LearnwithBrainly

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