Question Number 2968 by Karting7442 last updated on 01/Dec/15
$${Powers}\:{of}\:{Monomials}\:\:\:\:\:\:\:{Alg}. \\ $$$$ \\ $$$$\left(\mathrm{0}.\mathrm{6}{p}^{\mathrm{5}} \right)^{\mathrm{3}} \\ $$
Answered by Filup last updated on 02/Dec/15
$$=\mathrm{0}.\mathrm{6}^{\mathrm{3}} {p}^{\mathrm{15}} \\ $$$$ \\ $$$$\mathrm{0}.\mathrm{6}=\frac{\mathrm{6}}{\mathrm{10}}\Rightarrow\mathrm{0}.\mathrm{6}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{100}}=\mathrm{0}.\mathrm{36} \\ $$$$ \\ $$$$=\mathrm{0}.\mathrm{36}{p}^{\mathrm{15}} \\ $$
Commented by Yozzi last updated on 02/Dec/15
$$\mathrm{0}.\mathrm{6}^{\mathrm{3}} =\mathrm{0}.\mathrm{36}×\mathrm{0}.\mathrm{6}=\mathrm{0}.\mathrm{216} \\ $$
Commented by Filup last updated on 03/Dec/15
$$\mathrm{oops}\:\mathrm{I}\:\mathrm{missread} \\ $$