f you just want to see the really short way, just skip down to AAAAAAAAAAAA
so, here is the long explanation
exponential properties
\(x^{-m}= \frac{1}{x^m}\)
don't forget pemdas
2x^2=2(x^2)
so
\(\frac{2x^{-4}}{3xy}\)=\(\frac{2 \frac{1}{x^4} }{3xy}= \frac{2}{3x^5y}\)
AAAAAAAAAAAAAAAAAAAAAAAA
so we see
the original equatio is
\(\frac{2x^{-4}}{3xy}\)
remember
\(\frac{ab}{cd} =( \frac{a}{c})( \frac{b}{d})\)
so we can seperate the constants
\(\frac{ab}{cd} =( \frac{2}{3})( \frac{x^{-4}}{xy})\)
we know that the placeholders cannot affet the position of the constants unless they are grouped together which they are not
terfor the answer must have 2/3 in it
the only one that hsa that is A
ANSWER IS A
