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Mathematics 10 Online
OpenStudy (anonymous):

If y varies inversely as x, and y = 10 when x = 7, find y for the x-value of 10.

OpenStudy (jdoe0001):

\(\bf \begin{array}{cccllll} \textit{something }&\textit{varies inversely to }&\textit{something else}\\ \quad \\ \textit{something }&=\cfrac{{\color{red}{ \textit{some value }}}}{\textit{something else}}\\ \quad \\ y&=\cfrac{{\color{red}{ n}}}{x} &&\implies y=\cfrac{{\color{red}{ n}}}{x} \end{array} \\ \quad \\ y=10\qquad x=7\qquad 10=\cfrac{{\color{red}{ n}}}{7}\) 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"

OpenStudy (anonymous):

so it's 7?

OpenStudy (jdoe0001):

yeap

OpenStudy (anonymous):

I need help with one more question

OpenStudy (anonymous):

OpenStudy (jdoe0001):

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

OpenStudy (jdoe0001):

but if you were to consider only the integer value for "x" regardless of its sign then yes, it'd be an inverse variation

OpenStudy (anonymous):

so it would be inverse

OpenStudy (anonymous):

so is that the answer?

OpenStudy (jdoe0001):

well, you'd be correct, is an inverse variation, if only the integer value is used

OpenStudy (jdoe0001):

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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