What is #(16x^3)/(5y^9) * (x^3y^7)/(80xy^2)#?

1 Answer
ProfLayton · H Dev
Apr 17, 2016

#=(x^5)/(25y^4)#

Explanation:

#(16x^3)/(5y^9)*(x^3y^7)/(80xy^2)#

You can cross cancel 16 and 80 because both are multiples of 16. So #16 div 16 = 1# and #80 div 16 = 5#.
#(x^3)/(5y^9) times (x^3y^7)/(5xy^2)#

Cancel out terms. You will need to know this: #(a^m)/(a^n)=a^(m-n)#

#=(x^5)/(25y^4)#

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