water is being poured into a cylindrical tank at 9 ft/sec. If the tank has a base radius of 5ft and a height of 10ft, how fast is water being poured in when the height is at 6ft?

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water is being poured into a cylindrical tank at 9 ft/sec. If the tank has a base radius of 5ft and a height of 10ft, how fast is water being poured in when the height is at 6ft?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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V = pi * r^2 * h dV = pi * r^2 * dh dV/dt = pi * r^2 * dh/dt since dh/dt=9 , r=5, we subsititute these into that eqauation dV/dt = pi * 25 *9 = 225pi if you meant 9 ft^3 / sec then that would be dV/dt and you would be trying to find dh/dt so 9/25pi
I'm so sorry!! I meant tank shaped like a cone.
V = pi * r^2 * h/3 dV = pi *(2r * dr * h + r^2 *dh) also 5r = 10h so r = 2h and dr = 2dh dV = pi *(2(2h) * (2dh) * h + (2h)^2 *dh) divide both sides by dt and plug in 9for dh/dt, and 6 for h

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