Question Number 2968 by Karting7442 last updated on 01/Dec/15
![Powers of Monomials Alg. (0.6p^5 )^3](https://www.tinkutara.com/question/Q2968.png)
$${Powers}\:{of}\:{Monomials}\:\:\:\:\:\:\:{Alg}. \\ $$$$ \\ $$$$\left(\mathrm{0}.\mathrm{6}{p}^{\mathrm{5}} \right)^{\mathrm{3}} \\ $$
Answered by Filup last updated on 02/Dec/15
![=0.6^3 p^(15) 0.6=(6/(10))⇒0.6^2 =((36)/(100))=0.36 =0.36p^(15)](https://www.tinkutara.com/question/Q2975.png)
$$=\mathrm{0}.\mathrm{6}^{\mathrm{3}} {p}^{\mathrm{15}} \\ $$$$ \\ $$$$\mathrm{0}.\mathrm{6}=\frac{\mathrm{6}}{\mathrm{10}}\Rightarrow\mathrm{0}.\mathrm{6}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{100}}=\mathrm{0}.\mathrm{36} \\ $$$$ \\ $$$$=\mathrm{0}.\mathrm{36}{p}^{\mathrm{15}} \\ $$
Commented by Yozzi last updated on 02/Dec/15
![0.6^3 =0.36×0.6=0.216](https://www.tinkutara.com/question/Q3012.png)
$$\mathrm{0}.\mathrm{6}^{\mathrm{3}} =\mathrm{0}.\mathrm{36}×\mathrm{0}.\mathrm{6}=\mathrm{0}.\mathrm{216} \\ $$
Commented by Filup last updated on 03/Dec/15
![oops I missread](https://www.tinkutara.com/question/Q3030.png)
$$\mathrm{oops}\:\mathrm{I}\:\mathrm{missread} \\ $$