Question
Simplify:$\frac{32\text{m}^2\text{n}^3\text{p}^2}{4\text{mnp}}$
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Answer
$\frac{32\text{m}^2\text{n}^3\text{p}^2}{4\text{mnp}}$ $=\frac{32\times{\text{m}\times\text{m}\times\text{n}\times\text{n}\times\text{n}\times\text{p}\times\text{p}}}{4\times\text{m}\times\text{n}\times\text{p}}$ $=8\text{m}^{(2-1)}\text{n}^{(3-1)}\text{p}^{(2-1)}$ $=8\text{mn}^2\text{p}$
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