Sand falls onto a conical pile at the rate of 10 cubic feet per minute. The radius of the pile is always equal to one half it altitude. How fast is the altitude of the pile increasing when the pile is 5 feet high?

Either one of these.

Step-by-step explanation:

volumepile=1/3 (PI r^2)h

but r=h/2, so

volume=1/12 PI h^3

dv/dt=10 ft^3/min

but dv/dt=1/12 PI 3h^2 dh/dt

solve for dh/dt

This assumes you mean by "altitude" the height. If you mean altitude as slant height, you have to adjust the fromula

given: r = h/2

V = (1/3)π r^2 h

= (1/3)π (h/2)^2 (h)

= (1/12) π h^3

dV/dt = (1/4)π h^2 dh/dt

for the given data ...

10 = (1/4)π(25)dh/dt

dh/dt = 10(4)/((25π) = 1.6/π feet/min

answer

106 emergencies

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°