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Sholpan [36]
3 years ago
15

Find the exact volume of the cylinder. 1471 m3 2871 m3 5670 m3 11271 m3

Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer:

hmmmm........

Step-by-step explanation:

radius is 4/2=2m

volume of the cylinder is equal

to πr^h

i.e, π(2m)^2×7m

≈88 m^3

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What is the conjecture sentence for the pattern 1, ,4, 9 ,16?
nexus9112 [7]

Answer:

Square numbers

Step-by-step explanation:

if 1 is squared, you get 1.

if 2 is squared, you get 4.

if 3 is squared, you get 9.

if 4 is squared, you get 16.

And thus square numbers.

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Write 9 decimals with three decimal places that when rounded to the nearest tenth round to 1.3
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1. The mechanics at Lincoln Automotive are reborning a 6 in deep cylinder to fit a new piston. The machine they are using increa
Firdavs [7]

Answer:

0.0239\frac{in^{3}}{min}

Step-by-step explanation:

In order to solve this problem, we must start by drawing a diagram of the cylinder. (See attached picture)

This diagram will help us visualize the problem better.

So we start by determining what data we already know:

Height=6in

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Radius = 1.9 in (because the radius is half the length of the diameter)

The problem also states that the radius will increase on thousandth of an inch every 3 minutes. We can find the velocity at which the radius is increasing with this data:

r'=\frac{1/1000in}{3min}

which yields:

r'=\frac{1}{3000}\frac{in}{min}

with this information we can start solving the problem.

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V=\pi r^{2}h

where V is the volumen, r is the radius, h is the height and π is a mathematical constant equal approximately to 3.1416.

Now, the height of the cylinder will not change at any time during the reborning, so we can directly substitute the provided height, so we get:

V=\pi r^{2}(6)

or

V=6 \pi r^{2}

We can now take the derivative to this formula so we get:

\frac{dV}{dt}=2(6)\pi r \frac{dr}{dt}

Which simplifies to:

\frac{dV}{dt}=12\pi r \frac{dr}{dt}

We can now substitute the data provided by the problem to get:

\frac{dV}{dt}=12\pi (1.9) (\frac{1}{3000})

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\frac{dV}{dt}=0.0239\frac{in^{3}}{min}

3 0
3 years ago
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bixtya [17]
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