Question 339860
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No can do.  You would be able to do this if the box were perfectly cubical, that is:  *[tex \Large l\ =\ w\ =\ h].  But since you only guaranteed that the base of the box was square, i.e. *[tex \Large l\ =\ w], any Volume function would have to include both *[tex \Large w] AND *[tex \Large h] as variables.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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