hmmm from the table, let's just pick one point since we know "y" is inversely proportional, so hmmm let's say hmm the point when x = 4 and y = 9/16
a)
[tex]\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{"y" inversely proportional to }x^2}{ {\LARGE \begin{array}{llll} y = \cfrac{k}{x^2} \end{array}}}\qquad \textit{we also know that} \begin{cases} x=4\\ y=\frac{9}{16} \end{cases} \\\\\\ \cfrac{9}{16}=\cfrac{k}{(4)^2}\implies \cfrac{9}{16}=\cfrac{k}{(16)}\implies \cfrac{9(16)}{16}=k\implies 9=k~\hfill \boxed{y=\cfrac{9}{x^2}}[/tex]
b)
when y = 16, what's "x"?
[tex]16=\cfrac{9}{x^2}\implies x^2=\cfrac{9}{16}\implies x=\pm\sqrt{\cfrac{9}{16}}\implies x=\pm \cfrac{\sqrt{9}}{\sqrt{16}}\implies \stackrel{positive~value}{x=+\cfrac{3}{4}}[/tex]
