MATH SOLVE

3 months ago

Q:
# z varies jointly with y and the square of x. If x=4 when y=β7 and z=β336, find x when z=36 and y=3.

Accepted Solution

A:

[tex]\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}\\\\
-------------------------------\\\\[/tex]

[tex]\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[/tex]

[tex]\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[/tex]