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

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3