We pour so much water into a container that has the shape of an equilateral cone, the base of which has a radius r = 6 cm, that one-third of the volume of the cone is filled. How high will the water reach if we turn the cone upside down?
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Dear Sirs! Considering English is not my mother tongue, I hope I understood the word problem as well as the solution you have provided (since there's no image in solution I hope that I understood the labels of the variables that you've uses in the solution) . .. Anyway, I think there is an error in the given solution. .. When you turn the cone upside down, the shape of the cone part filled with water is a truncated cone with the height h2, so it's volume can not be calculated using formula that you have used, that is V1=1/3 * pi * r2 * h2 . .. This isn't the formula for the volume of a truncated cone. ..
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