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Mathematics 20 Online
OpenStudy (vshiroky):

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OpenStudy (vshiroky):

OpenStudy (vshiroky):

72?

OpenStudy (vshiroky):

@agent0smith

OpenStudy (vshiroky):

@jdoe0001

OpenStudy (vshiroky):

Is my 72 right?

OpenStudy (jdoe0001):

ohh ahemm \(\bf \begin{array}{cccllll} \textit{something }&\textit{varies directly to }&\textit{something else}\\ \quad \\ \textit{something }&={\color{red}{ \textit{some value }}}&\textit{something else}\\ \quad \\ y&={\color{red}{ n}}&x&\implies y={\color{red}{ n}}x \end{array}\\ \quad \\ x=60\qquad y=24\qquad 60={\color{red}{ n}}(24)\) solve for "n" to find the "constant of variation" is not 72 though

OpenStudy (vshiroky):

ummmmm

OpenStudy (jdoe0001):

hmm lemme recheck that... maybe I got my units mixed up =(

OpenStudy (jdoe0001):

... nope, I think they're ok

OpenStudy (vshiroky):

I still don't understand. That's not the equation I was given and when I did mine I got 72

OpenStudy (vshiroky):

is it 144?

OpenStudy (jdoe0001):

nope... I'm afraid 72 makes sense to me since 3 * 60 is 180 thus 3 * 24 will be 72 making the constant of variation = 3 I'm thinking I may have the output as the input up there

OpenStudy (vshiroky):

So it is 72 then

OpenStudy (jdoe0001):

it makes logical sense.... trying to see ... what's off in the procedure

OpenStudy (vshiroky):

Thank you.

OpenStudy (jdoe0001):

anyhow... , I was using "x" as the given in the function, thus but the function above is correct

OpenStudy (jdoe0001):

\(\bf y=\cfrac{5}{2}x\qquad y=180\implies 180=\cfrac{5}{2}x\implies 72=x\)

OpenStudy (vshiroky):

Thanks.

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