Respuesta :
[tex]\bf \begin{array}{cccccclllll}
\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}\\\\
-----------------------------\\\\[/tex]
[tex]\bf % 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]
now... let's see
"m varies directly as the square root of y, inversely as p², and directly as n"
[tex]\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}[/tex]
now, let's quadruple "y", multiply "n" by 5 and "p" by 3
[tex]\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}\qquad \begin{cases} y=4y\\ p=3p\\ n=5n \end{cases}\implies m=\cfrac{k\sqrt{4y}\cdot 5n}{(3p)^2} \\\\\\ m=\cfrac{k\sqrt{2^2y}\cdot 5n}{(3^2p^2)}\implies m=\cfrac{k2\sqrt{y}\cdot 5n}{(9p^2)}\implies m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{2\cdot 5}{9} \\\\\\ m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{10}{9}\implies \cfrac{9}{10}\cdot m=\cfrac{k\sqrt{y}\cdot n}{p^2}[/tex]
[tex]\bf % 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]
now... let's see
"m varies directly as the square root of y, inversely as p², and directly as n"
[tex]\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}[/tex]
now, let's quadruple "y", multiply "n" by 5 and "p" by 3
[tex]\bf m=\cfrac{k\sqrt{y}\cdot n}{p^2}\qquad \begin{cases} y=4y\\ p=3p\\ n=5n \end{cases}\implies m=\cfrac{k\sqrt{4y}\cdot 5n}{(3p)^2} \\\\\\ m=\cfrac{k\sqrt{2^2y}\cdot 5n}{(3^2p^2)}\implies m=\cfrac{k2\sqrt{y}\cdot 5n}{(9p^2)}\implies m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{2\cdot 5}{9} \\\\\\ m=\cfrac{k\sqrt{y}\cdot n}{p^2}\cdot \cfrac{10}{9}\implies \cfrac{9}{10}\cdot m=\cfrac{k\sqrt{y}\cdot n}{p^2}[/tex]
