Question

I do not know how to solve this problem or know what to do with the exponents

Answer 1

K, remember
(ab)/(cd)=(a/c)(b/d) or whatever
also
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}= \frac{1}{x^m}$
and
$x^ \frac{m}{n}= \sqrt[n]{x^m}$
and
$( \frac{x}{y} )^m= \frac{x^m}{y^m}$
and
$(x^m)^n=x^{mn}$
and
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)


so
$( \frac{-7x^ \frac{3}{2} }{5y^4} )^{-2}$=
$( \frac{-7}{5} )^{-2}( \frac{x^ \frac{3}{2} }{y^4} )^{-2}$=
$( \frac{(-7)^{-2}}{5^{-2}} )( \frac{(x^ \frac{3}{2})^{-2} }{(y^4)^{-2}} )$=
$( \frac{ \frac{1}{(-7)^2} }{ \frac{1}{5^2} } )( \frac{x^ \frac{-6}{2} }{y^{-8}} )$=
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{(x^{-3}) }{\frac{1}{y^8}} )$
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{\frac{1}{x^3} }{\frac{1}{y^8}} )$=
$(\frac{25}{49} )( \frac{y^8}{x^3}$=
$\frac{25y^8}{49x^3}$