Question

(25h^3*jk^5)/(12h^2*k)

Probability and combinatorics
ANSWERED
asked 2021-06-04
\(\displaystyle\frac{{{25}{h}^{{3}}\cdot{j}{k}^{{5}}}}{{{12}{h}^{{2}}\cdot{k}}}\)

Answers (1)

2021-06-05

We are given: \(\displaystyle\frac{{{25}{h}^{{3}}\cdot{j}{k}^{{5}}}}{{{12}{h}^{{2}}\cdot{k}}}\)
Use the rule: \(\frac{a^m}{a^n}=a^{m-n} =\frac{(25h^{3-2})(jk^{5-1})}{12} =\frac{25hjk^4}{12}\)

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