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Mathematics 8 Online
OpenStudy (anonymous):

can someone help me isolate g?

OpenStudy (anonymous):

\[T=2\pi \sqrt{L/g}\]

OpenStudy (anonymous):

\[\frac{T}{2\pi}=\sqrt{\frac{L}{g}}\] is a start then square both sides

OpenStudy (anonymous):

i did that then you have L/g what is next

OpenStudy (anonymous):

flip it

OpenStudy (anonymous):

\[\frac{T^2}{4\pi^2}=\frac{L}{g}\] \[\frac{4\pi^2}{T^2}=\frac{g}{L}\]

OpenStudy (anonymous):

then multiply both sides by \(L\)

OpenStudy (anonymous):

are you sure you can just flip it?

OpenStudy (anonymous):

i can if i want

OpenStudy (doc.brown):

Yes you can.

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

or you can solve always\[\frac{a}{b}=\frac{c}{x}\] for \(x\) vial \[x=\frac{bc}{a}\]

OpenStudy (doc.brown):

They are still equal \[\frac{1}{2}=\frac{2}{4}\]\[\frac{2}{1}=\frac{4}{2}\]

OpenStudy (anonymous):

ok ill try it thanks

OpenStudy (anonymous):

\[((2\pi)^{2}/T ^{2})L=g\]

OpenStudy (anonymous):

is this right

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