Answer
$= \frac{2z^{3}}{3x^{2}y}$
Work Step by Step
Simplify $\frac{-32}{-48}$
$= \frac{32}{48}$
$= \frac{16}{24}$
$= \frac{8}{12}$
$= \frac{4}{6}$
$= \frac{2}{3}$
Simplify$ \frac{xy^{2}z^{4}}{x^{3}y^{3}z}$
1) Cancel $x$ from the numerator and denominator to result in $1$ in the numerator and $x^{2}$ in the denominator.
2) Cancel $y^{2}$ from the numerator and denominator to result in $1$ in the numerator and $y$ in the denominator.
3) Cancel $z$ in the denominator and numerator to result in $z^{3}$ in the numerator and $1$ in the denominator.
$= \frac{(1)y^{2}z^{4}}{x^{2}y^{3}z}$
$= \frac{(1)(1)z^{4}}{x^{2}yz}$
$= \frac{(1)(1)z^{3}}{x^{2}y(1)}$
$= \frac{z^{3}}{x^{2}y}$
Multiply the two components to get the result
$= \frac{2}{3}(\frac{z^{3}}{x^{2}y})$
$= \frac{2z^{3}}{3x^{2}y}$
