How do you divide #(m^2n^3)/p^3# by #(mp)/n^2#?

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Answer 1 · 2016-04-19T06:04:50

#=(mn^5)/p^4#

As you will recall, to if you want to divide a fraction, you'll flip it, then multiply it e.g. #2 divide 1/2 = 2 * 2/1 = 2*2 = 4#

In this case, we'll flip #(mp)/n^2#:
#(m^2n^3)/p^3 divide (mp)/n^2=(m^2n^3)/p^3*n^2/(mp)#
#=((m^2n^3)(n^2))/((p^3)(mp))#

Now there is a rule you'll need to know to go further: #a^m*a^n=a^(m+n)#

#=(m^2n^5)/(p^4m)#

And there's another rule you need to know: #a^m/a^n=a^(m-n)#

#=(mn^5)/p^4#

Note: If you wish for me to explain those rules and why they work, just say so