enmobladatn

2022-07-25

A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude 25ft and base edge of 10ft. how much will the surface rise when 6,779 liters of water is added?

Izabelle Frost

Beginner2022-07-26Added 13 answers

$v=\frac{\sqrt{3}}{2}{s}^{2}h\phantom{\rule{0ex}{0ex}}6779liters\times 0.0353cu\frac{ft}{1}liter=239.299cuft\phantom{\rule{0ex}{0ex}}butbyproportion\frac{s}{h}=10/25\phantom{\rule{0ex}{0ex}}s=\frac{10}{25}h\phantom{\rule{0ex}{0ex}}andv=\frac{\sqrt{3}}{2}(\frac{10}{25}h{)}^{2}h\phantom{\rule{0ex}{0ex}}239.299=0.139{h}^{3}\phantom{\rule{0ex}{0ex}}h=(\frac{239.299}{0.139}{)}^{(\frac{1}{3})}=11.985ft$

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