Z varies jointly with y and the square of x. If x=4 when y=−7 and z=−336, find x when z=36 and y=3.

2 answers:

6 0

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{z varies jointly with y and the square of x}\qquad z=kyx^2
\\\\\\
\textit{we also know that }
\begin{cases}
x=4\\
y=-7\\
z=-336
\end{cases}\implies -336=k(-7)(4)
\\\\\\
\cfrac{-336}{-28}=k\implies 12=k\qquad \qquad \boxed{z=12yx^2}
\\\\\\
\textit{when z=36 and y=3, what is \underline{x}?}\qquad 36=12(3)x^2
\\\\\\
\cfrac{36}{12\cdot 3}=x^2\implies \sqrt{\cfrac{36}{12\cdot 3}}=x$

disa [49]3 years ago

4 0

Answer:

$\bf \textit{z varies jointly with y and the square of x}\qquad z=kyx^2 \\\\\\ \textit{we also know that } \begin{cases} x=4\\ y=-7\\ z=-336 \end{cases}\implies -336=k(-7)(4) \\\\\\ \cfrac{-336}{-28}=k\implies 12=k\qquad \qquad \boxed{z=12yx^2} \\\\\\ \textit{when z=36 and y=3, what is \underline{x}?}\qquad 36=12(3)x^2 \\\\\\ \cfrac{36}{12\cdot 3}=x^2\implies \sqrt{\cfrac{36}{12\cdot 3}}=x$