How do you simplify #\frac { 72m ^ { 4} n ^ { 8} } { 8m ^ { 2} n ^ { 5} }#?

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Answer 1 · 2016-11-08T09:46:09

#(72m^4n^8)/(8m^2n^5)=9m^2n^3#

Using the properties that #a^x/a^y = a^(x-y)# and #a/b * c/d = (ac)/(bd)#, we have

#(72m^4n^8)/(8m^2n^5) = 72/8 * m^4/m^2 * n^8/n^5#

#=9 * m^(4-2) * n^(8-5)#

#=9m^2n^3#

Answer 2 · 2016-11-08T11:28:08

Just another way of presenting the same thing:

Write as:#" "(72xxm^2xxm^2xxn^5xxn^3)/(8xxm^2xxn^5)#

#72/8xxm^2/m^2xxm^2xxn^5/n^5xxn^3#

But #m^2/m^2" " &" " n^5/n^5=1#

#72/8xxm^2xxn^3#

#9m^2n^3 #

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Or if you prefer to use the 'cancel out method'.

#72/8xx(m^(cancel(4)^2)n^(cancel(8)^3))/(cancel(m^2)cancel(n^5)) = 72/8m^2n^3" "# and so on..