If y varies inversely as x, and y = 10 when x = 7, find y for the x-value of 10.
\(\bf \begin{array}{cccllll} \textit{something }&\textit{varies inversely to }&\textit{something else}\\ \quad \\ \textit{something }&=\cfrac{{\color{red}{ \textit{some value }}}}{\textit{something else}}\\ \quad \\ y&=\cfrac{{\color{red}{ n}}}{x} &&\implies y=\cfrac{{\color{red}{ n}}}{x} \end{array} \\ \quad \\ y=10\qquad x=7\qquad 10=\cfrac{{\color{red}{ n}}}{7}\) solve for "n" to find the "constant of variation" to find "y" when x = 10 just plug in those values in the original equation, along with the found "n" value to get "y"
so it's 7?
yeap
I need help with one more question
hmmm seems to be 2 pieces of the graph to the left of 0, is increasing thus is a direct variation from 0 to the right is decreasing, so is an inverse variation on that side
but if you were to consider only the integer value for "x" regardless of its sign then yes, it'd be an inverse variation
so it would be inverse
so is that the answer?
well, you'd be correct, is an inverse variation, if only the integer value is used
keep in mind that an inverse variation means, when "x" increases, moves to the right you could say, "y" decreases, goes down towards 0
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