Water is leaking out of an inverted conical tank at a rate of 9,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)

Step-by-step explanation:

rate of leaking water = 9500 cm³/min

height of tank, H = 6 m = 600 cm

diameter of tank, = 4 m

radius of tank, R = 2 m = 200 cm

dh/dt = 20 cm/min

at any time, h = 2 m Let at that time radius of the tank is r.

According to the diagram

R / H = r / h

200 / 600 = r / h

r = h/3

The volume of tank is given by

Differentiate with respect to t

dV/dt = 279111.11 cm³/min

This is the volume of water increasing per minute in the tank.

Let C be the rate of volume of water pumped into the tank.

So, C = 279111.11 + 9500

C = 288611.11 cm³/min

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b