How do you simplify (4/7m)^2(49m)(17p)(1/34p^5)?

Feb 27, 2017

See the entire solution process below:

Explanation:

First, we will use these rules for exponents to simplify the term on the left: $\textcolor{red}{{\left(\frac{4}{7} m\right)}^{2}}$: $a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

$\textcolor{red}{{\left(\frac{4}{7} m\right)}^{2}} \left(49 m\right) \left(17 p\right) \left(\frac{1}{34} {p}^{5}\right) =$

$\textcolor{red}{{\left({4}^{1} / {7}^{1} {m}^{1}\right)}^{\textcolor{b l u e}{2}}} \left(49 m\right) \left(17 p\right) \left(\frac{1}{34} {p}^{5}\right) =$

color(red)((4^(1xxcolor(blue)(2))/7^(1xxcolor(blue)(2))m^(1xxcolor(blue)(2)))(49m)(17p)(1/34p^5) =

$\left({4}^{2} / {7}^{2} {m}^{2}\right) \left(49 m\right) \left(17 p\right) \left(\frac{1}{34} {p}^{5}\right) =$

$\left(\frac{16}{49} {m}^{2}\right) \left(49 m\right) \left(17 p\right) \left(\frac{1}{34} {p}^{5}\right)$

Next, we can rewrite this expression as:

$\left(\frac{16}{49} {m}^{2} \times 49 m\right) \left(17 p \times \frac{1}{34} {p}^{5}\right)$

Then, factor and cancel the coefficients:

$\left(\frac{16}{\textcolor{red}{\cancel{\textcolor{b l a c k}{49}}}} {m}^{2} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{49}}} m\right) \left(\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{17}}} p \times \frac{1}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{17}}} 2} {p}^{5}\right) =$

$\left(16 {m}^{2} \times m\right) \left(p \times \frac{1}{2} {p}^{5}\right)$

We can then use this rule of exponents to simplify the variables:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\left(16 {m}^{\textcolor{red}{2}} \times {m}^{\textcolor{b l u e}{1}}\right) \left({p}^{\textcolor{red}{1}} \times \frac{1}{2} {p}^{\textcolor{b l u e}{5}}\right) =$

$\left(16 {m}^{\textcolor{red}{2} + \textcolor{b l u e}{1}}\right) \left(\frac{1}{2} {p}^{\textcolor{red}{1} + \textcolor{b l u e}{5}}\right) =$

$\left(16 {m}^{3}\right) \left(\frac{1}{2} {p}^{6}\right)$=#

$\frac{16 {m}^{3} {p}^{6}}{2} =$

$8 {m}^{3} {p}^{6}$