calculus

First we use the volume of a cone equation to find the volume of the conical tank when the water level is 2 meters: Volume = (π/3)*h*(r^2+r*R+R^2) Substituting 2 m for h, 6 m for R, and (2/3)x4 m for r, we get Volume = (π/3)*2*((2/3)x4^2+2/3x4x6+6^2) Volume = (π/3)*2*(16+24+36) Volume = (π/3)*2*76 ≈ 502.65 m^3 We can now use a conversion rate to convert cm^3 to m^3 by multiplying by (1/1,000,000), so the rate at which water is leaking out is 10,000 cm^3/min x (1/1,000,000) = 0.01 m^3/min. We know that the rate at which water is going in must match the rate at which is it going out, meaning we need to calculate the rate at which water is being pumped into the tank as 0.01 m^3/min. This is equal to the rate of 10,000 cm^3/min. The rate at which water is being