Question

reduce to lowest terms
(38x^2yz^2)/(-19xy^2z^3)

Answer 1

38/-19=-2
\[\frac{x^2}{x}=x\]
\[\frac{y}{y^2}=\frac{1}{y}\]
\[\frac{z^2}{z^3}=\frac{1}{z}\]
multiplying together, we get
\[\frac{-2x}{yz}\]