How do you simplify # (14x^5y)/(2xy^ -2)#?

Question

1553 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-27T09:28:26

#=color(blue)(7x^4y^3#

#(14x^5y)/(2xy^-2#

By property
#color(blue)(1/a =a^-1# and,

#color(blue)(1/a^-1 =a#

#(14x^5y)/(2(color(blue)(xy^-2))) = (cancel14/cancel2) * (x^5y* color(blue)(x^-1 y^2))#

#=7 x^5y* color(blue)(x^-1 y^2)#

By property
#color(blue)(a^m*a^n = a^(m+n)#

#=7 x^color(blue)((5-1)) y^color(blue)((1+2))#

#=color(blue)(7x^4y^3#