Answer
$\frac{7}{x^{3}z^{5}}$
Work Step by Step
We are given the expression $\frac{14x^{-2}yz^{-4}}{2xyz}$.
To simplify, we can separate like terms and then use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers).
$(\frac{14}{2})\times(x^{-2-1})\times(y^{1-1})\times(z^{-4-1})=7\times x^{-3}\times y^{0}\times z^{-5}$
Note that $y^{0}=1$.
$7\times x^{-3}\times z^{-5}$
To simplify this into only positive exponents, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer).
$7\times x^{-3}\times z^{-5}=\frac{7}{x^{3}z^{5}}$