How do you simplify $p (q^{- 2}) (r^{- 3})$ $p(q^-2) (r^-3)$ ?

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Answer 1 · 2015-05-03T11:54:48

We know that $a^{- m} = \frac{1}{a^{m}}$ $color(blue)(a^-m = 1/a^m$ ( $a \neq 0$ $a!=0$ )

Hence $p (q^{- 2}) (r^{- 3}) = p \cdot (\frac{1}{q^{2}}) \cdot (\frac{1}{r^{3}})$ $p(q^-2) (r^-3) = p*(1/q^2)*(1/r^3)$

$= \frac{p}{q^{2} \cdot r^{3}}$ $=color(green)( p/(q^2*r^3)$

Answer 2 · 2015-05-03T11:55:55

In general $b^{- m} = \frac{1}{b^{m}}$ $b^(-m) = 1/(b^m)$

So $p (q^{- 2}) (r^{- 3})$ $p(q^(-2))(r^(-3))$

$= \frac{p}{q^{2} r^{3}}$ $= p/(q^2r^3)$

How do you simplify p(q^-2) (r^-3)p(q−2)(r−3)p(q^-2) (r^-3)?