Question

Simplify 81p6q2÷3p2q5 HELPPP​

Answer 1

Answer:

$\displaystyle \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} = \frac{27\, p^{4}}{q^{3}}$.

Step-by-step explanation:

Make use of the fact that for any $x \ne 0$ and integer $n$:

$\displaystyle \frac{1}{x^{n}} = x^{-n}$.

For example, in this question:

$\displaystyle \frac{1}{p^{2}} = p^{-2}$.

$\displaystyle \frac{1}{q^{5}} = q^{-5}$.

Thus, the original expression would be equivalent to:

$\begin{aligned}& \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} \\ =\; & \frac{81}{3}\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\ =\; & 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5})\end{aligned}$.

Also make use of the fact that for any $x \ne 0$, integer $m$, and integer $n$:

$x^{m}\, x^{n} = x^{m + n}$.

Thus:

$\begin{aligned}& 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\=\; & 27\, p^{6 + (-2)}\, q^{2 + (-5)} \\ =\; & 27\, p^{4}\, q^{-3} \\ =\; & \frac{27\, p^{4}}{q^{3}}\end{aligned}$.