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The volume of a right regular pyramid is 1/3bh, where b is the pyramid's base and h is its height. If both the height and the base area are reduced by half, what is the new volume of the pyramid?
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1/3(h/2)(b/2) = V/4 isn't it ?
Volume Is \[\int\limits_{0}^{h}x ^{2}dx\] Where x^2 is the base and h is the height..
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