How do you multiply $(m+3)/(4m) div (m^2-9)/(32m^2(m-3))$ ?

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Answer 1 · 2015-05-08T08:38:58

we are given :
$(m+3)/(4m)$ / $(m^2-9)/(32m^2(m-3)$

for multiplying we take the reciprocal of the second term:

$= (m+3)/(4m)$ $xx$ $32m^2xx(m-3)/(m^2-9)$

we can write
$m^2-9$ as $(m+3)(m-3)$

$= (m+3)/(4m)$ $xx$ $32m^2xx(m-3)/((m+3)(m-3)$

$= cancel(m+3)/(4m)$ $xx$ $32m^2xxcancel(m-3)/(cancel(m+3)cancel(m-3)$

$= (32m^2) / (4m)$

$= 8m$

How do you multiply (m+3)/(4m) div (m^2-9)/(32m^2(m-3))(m+3)/(4m) div (m^2-9)/(32m^2(m-3))?