How do you rewrite #(2x^-2 y^2 )/ y^-3# using a positive exponent?

1 Answer
Oct 22, 2015

Answer:

#(2x^(-2)y^2)/(y^(-3)) = (2y^5)/(x^2)#

Explanation:

In general
#color(white)("XXX")s^(-k) = 1/s^k#
and
#color(white)("XXX")1(/t^-q) = t^q#

#(2x^(-2)y^2)/(y^(-3))#

#color(white)("XXX")=(2y^2)/1 * (x^-2)/1 * 1/(y^(-3))#

#color(white)("XXX")=(2y^2)/1*1/x^2*y^3/1#

#color(white)("XXX")=(2y^5)/(x^2)#