Question 975102
(5x^0y^-7/2x^-2y^4)^-2;;; 5x^0=1    ;;;y^(-7/2)*(-2)=y^7;;;;  (-2)^-2=(1/(-2)^2) or (1/4);;;(y^4)^-2=y^-8

1* y^7*(1/4)y^-8
y^7/(1/4)y^8

(4/y)