Gravel is being dumped from a conveyor belt at a rate of 35 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 12 ft high? (Round your answer to two decimal places.)

0.31 ft/s

Step-by-step explanation:

The volume of a cone is given by the formula:

V = πr²h/3

From the question, we are given the diameter and the height to be equal, thus;

r = h/2

Putting h/2 for r into the volume equation, we have;

V = (π(h/2)²h)/3

V = πh³/12

Using implicit derivatives,we have;

dV/dt = (πh²/4)(dh/dt)

From the question, we want to find out how fast is the height of the pile increasing. This is dh/dt.

We have;

dV/dt = 35 ft³/min and h = 12ft

Plugging in the relevant values, we have;

35 = (π×12²/4)(dh/dt)

dh/dt = (35 × 4)/(144 × π)

dh/dt = 0.3095 ft/s ≈ 0.31 ft/s

the answer is 0.000652422

but if you ment to ask what is 6131 divided by 4, it is 1532.75.

step-by-step explanation:

d. all of the above

step-by-step explanation: