A cylindrical tank of radius 2.5 feet is being drained of water at a rate of `0.25\ ft^{3}/\sec`. How fast is the height of the water decreasing?

Answers

Answer 1

Answer:

DONT PRESS THAT LINK

Step-by-step explanation:

Answer 2

The height of the water decreases at a rate of 0.012  ft/sec if a cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec.

What is a cylinder?

In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.

We know the volume of the cylinder is given by:

[tex]\rm V = \pi r^2 h[/tex]

It is given that:

A cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec

As we know,

The volume of the cylinder is given by:

V = πr²h

Let V is the water draining rate

V = 0.25 ft³/sec

The radius r = 2.5 feet

0.25 ft³/sec = π(2.5 feet)²(h)

h = [0.25/(2.5²π)]  [ft³/(sec×feet²]

h = 0.012  ft/sec

Thus, the height of the water decreases at a rate of 0.012  ft/sec if a cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec.

Learn more about the cylinder here:

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Answer and Step-by-step explanation:

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Step-by-step explanation:

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https://brainly.com/question/1859113

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Full Question

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Step-by-step explanation:

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Answers

Answer:

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Answer:

Step-by-step explanation:

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