$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{ k}}}{x}
\end{array}\\\\
-----------------------------\\\\
\textit{p is inversely proportional to the cube of (q-2)}\implies p=\cfrac{k}{(q-2)^3}
\\\\\\
now \quad 
\begin{cases}
p=6\\
q=3
\end{cases}\implies 6=\cfrac{k}{(3-2)^3}$

solve for "k", to find k or the "constant of variation"

then plug k's value back to $\bf p=\cfrac{k}{(q-2)^3}$

now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is

4 0

3 years ago

Is it scaling isosceles or equilateral and what is the answer

garik1379 [7]

Answer:

$scalene$