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#