Water is leaking out of an inverted conical tank at a rate of 10600.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 m and the the diameter at the top is 6.0 m. If the water level is rising at a rate of 27.0 cm/min when the height of the water is 1.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Water is leaking out of an inverted conical tank at a rate of 11900.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 meters and the diameter at the top is 5.5meters. If the water level is rising at a rate of 21.0 centimeters per minute when the height of the water is 3.5meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. 

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Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dVdt=R-11900.0 Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 13pr2h

To find the rate at which water is being pumped into the tank, we need to determine the rate at which the volume of water in the tank is changing.

Let's start by finding an expression for the volume of water in the tank as a function of its height.

We can consider a small change in height Δh and calculate the corresponding change in volume ΔV.

The tank is in the shape of an inverted cone, so we can use the formula for the volume of a cone:

V = (1/3) * π * r^2 * h,

where V is the volume, π is a mathematical constant (approximately equal to 3.14159), r is the radius of the base, and h is the height.

In this case, the radius r is half of the diameter, so r = 6.0 m / 2 = 3.0 m.

Let's rewrite the volume formula in terms of h only:

V = (1/3) * π * (3.0)^2 * h.

Now, let's differentiate both sides of the equation to find the rate at which the volume is changing with respect to time:

dV/dt = (1/3) * π * (3.0)^2 * dh/dt.

We know that dV/dt = rate at which water is being pumped into the tank, and dh/dt = rate at which the water level is rising.

Now we can plug in the given values:

dV/dt = (1/3) * π * (3.0)^2 * 27.0 cm/min.

Simplifying this expression:

dV/dt = 9π * 27.0 cm^3/min.

Now we need to find the value of 9π.

Using a calculator, we find that π ≈ 3.14159, so 9π ≈ 9 * 3.14159 ≈ 28.27433.

Finally, we can calculate the rate at which water is being pumped into the tank:

dV/dt = 28.27433 * 27.0 cm^3/min.

Calculating this expression:

dV/dt ≈ 763.39691 cm^3/min.

Therefore, the rate at which water is being pumped into the tank is approximately 763.39691 cm^3/min.