Water is drained out of an inverted cone, having the dimensions depicted in the figure below. if the water level drops at 6 ft divided by min, at what rate is water (in ft cubed divided by min) draining from the tank when the water depth is 3 ft?

the figure is a cone with a radius of 2 and a height of 6

-6π ft³/min ≈ -18.8 ft³/min

Step-by-step explanation:

Volume of a cone is:

V = ⅓ π r² h

Using similar triangles, we can relate the radius and height of the water to the radius and height of the tank.

r / h = R / H

r / h = 2 / 6

r = ⅓ h

Substituting:

V = ⅓ π (⅓ h)² h

V = π h³ / 27

Take the derivative with respect to time:

dV/dt = π h² / 9 dh/dt

Given that dh/dt = -6 ft/min, and h = 3 ft:

dV/dt = π (3)² / 9 (-6)

dV/dt = -6π

dV/dt ≈ -18.8 ft³/min

The water drains from the tank at a rate of 6π ft³/min or approximately 18.8 ft³/min.

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.