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can someone help me isolate g?
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\[T=2\pi \sqrt{L/g}\]
\[\frac{T}{2\pi}=\sqrt{\frac{L}{g}}\] is a start then square both sides
i did that then you have L/g what is next
flip it
\[\frac{T^2}{4\pi^2}=\frac{L}{g}\] \[\frac{4\pi^2}{T^2}=\frac{g}{L}\]
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then multiply both sides by \(L\)
are you sure you can just flip it?
i can if i want
Yes you can.
thanks
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or you can solve always\[\frac{a}{b}=\frac{c}{x}\] for \(x\) vial \[x=\frac{bc}{a}\]
They are still equal \[\frac{1}{2}=\frac{2}{4}\]\[\frac{2}{1}=\frac{4}{2}\]
ok ill try it thanks
\[((2\pi)^{2}/T ^{2})L=g\]
is this right
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