Z varies directly with x and inversely with y. When x=6 and y=2, .z=15. Write function that models variation. Then, find z when x=4 and y=9. Write answer in a/b form if necessary.

Respuesta :

[tex]\bf \begin{array}{cccccclllll} \textit{direct proportional variation}\\\\\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\ \textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\ y&=&{{ k}}&\cdot&x \\ && y={{ k }}x \end{array}\\ \quad \\[/tex]


[tex]\bf \textit{inverse proportional variation}\\\\ \begin{array}{llllll} \textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\ \textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\ y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x} \\ &&y=\cfrac{{{ k}}}{x} \end{array}[/tex]


[tex]\bf \\\\ -------------------------------\\\\ \textit{z varies directly with x}\implies z=kx \\\\ \textit{and inversely with y}\implies z=\cfrac{kx}{y} \\\\\\ \textit{we also know that}\quad \begin{cases} x=6\\ y=2\\ z=15 \end{cases}\implies 15=\cfrac{k\cdot 6}{2}\implies 30=6k[/tex]

[tex]\bf \cfrac{30}{6}=k\implies \boxed{5=k}\impliedby \textit{constant of variation} \\\\\\ thus\implies z=\cfrac{5x}{y}\\\\ -------------------------------\\\\ \textit{what's "z" when } \begin{cases} x=4\\ y=9 \end{cases}\implies z=\cfrac{5\cdot 4}{9}[/tex]