Question 76491
{{{ (5x)^0y^-2 }}}
Anything to the zero = 1
{{{ 1y^-2 }}}
Remember that anything over 1 = itself
{{{ (1y^-2)/1 }}}
Anything with a negative exponent that is in the numerator, drops to the denominator, and the exponent becomes positive
{{{ 1/y^2 }}}
done