How do you simplify $(16q^0 r^-6) /( 4q^-3 r^-7)$ ?

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Answer 1 · 2016-01-24T17:48:43

$(16q^0r^(-6))/(4q^(-3)r^(-7))=4q^3r$

with exclusions $q!=0$ and $r!=0$

If $a != 0$ and $b$ and $c$ are integers, then $a^b/a^c = a^(b-c)$

So:

$(16q^0r^(-6))/(4q^(-3)r^(-7))$

$=(16/4)(q^0/q^(-3))(r^(-6)/r^(-7))$

$=4 q^(0-(-3)) r^((-6)-(-7))$

$=4q^3r^1$

$=4q^3r$

with exclusions $q!=0$ and $r!=0$

The exclusions are required since if $q=0$ or $r=0$ then at least one of $q^(-3)$ , $r^(-6)$ , $r^(-7)$ is undefined.

How do you simplify (16q^0 r^-6) /( 4q^-3 r^-7) (16q^0 r^-6) /( 4q^-3 r^-7)?