Question

P^-4q^3r^-7 over p^-2q^3p^-2 simplify

Answer 1

$\bf \cfrac{p^{-4}~~\begin{matrix} q^3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ r^{-7}}{p^{-2}~~\begin{matrix} q^3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ p^{-2}}\implies \cfrac{1}{p^{4}p^{-2}p^{-2}r^7}\implies \cfrac{1}{p^{4-2-2}r^7}\implies \cfrac{1}{p^0r^7}\implies \cfrac{1}{r^7}$

Answer 2

Answer:

$\large\boxed{r^{-7}=\dfrac{1}{r^7}}$

Step-by-step explanation:

$\dfrac{p^{-4}q^3r^{-7}}{p^{-2}q^3p^{-2}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=p^{-4-(-2)-(-2)}q^{3-3}r^{-7}\\\\=p^{-4+2+2}q^0r^{-7}\\\\=p^0q^0r^{-7}\\\\=r^{-7}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{1}{r^7}$