a reservoir is in the form of a frustrum of a cone with upper base of radius 9 ft and lower base of radius 4ft and altitute of 10 ft. the water in the reservoir is x ft deep. if the level of the water is increasing at 4 ft/min, how fast is the volume of the water in the reservoir increasing when its depth is 2 ft? NOTE: the volume of a frustrum of a cone of upper base raduis R and lower base raduis r and height h. V = 1/3 pi h (R^2 +r^2 + Rr)

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a reservoir is in the form of a frustrum of a cone with upper base of radius 9 ft and lower base of radius 4ft and altitute of 10 ft. the water in the reservoir is x ft deep. if the level of the water is increasing at 4 ft/min, how fast is the volume of the water in the reservoir increasing when its depth is 2 ft? NOTE: the volume of a frustrum of a cone of upper base raduis R and lower base raduis r and height h. V = 1/3 pi h (R^2 +r^2 + Rr)

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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calculus prob? did that formula come with the problem or did you just add it?
came from the problem

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