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## 4meisu Group Title t = 2π √l/g Re-arrange to make g the subject. one year ago one year ago

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1. zepdrix Group Title

$\large t=2\pi \frac{\sqrt \ell}{g}$Is that formatted correctly? With the way it's written, it's a little hard to tell whether or not you have the g placed in the square root.

2. zepdrix Group Title

If this is correct, here are the steps we would take: ~Multiply both sides by g,$\large g t=2\pi \frac{\sqrt \ell}{\cancel g}\cancel g$The g's will cancel out on the right, ~Divide both sides by t,$\large \frac{g \cancel t}{\cancel t}=2\pi \frac{\sqrt \ell}{t}$The t's will cancel on the left, giving us,$\large \color{brown}{g=2\pi \frac{\sqrt \ell}{t}}$

3. zepdrix Group Title

Alternatively, if you meant to write,$\large t=2\pi \sqrt{\frac{\ell}{g}}$We would start by writing it like this,$\large t=2\pi \dfrac{\sqrt \ell}{\sqrt g}$Then we would apply similar steps to the last method, ~Multiply both sides by $$\sqrt g$$, $\large t \sqrt g=2\pi \frac{\sqrt \ell}{\cancel{\sqrt g}}\cancel{\sqrt g} \qquad \rightarrow \qquad t \sqrt g=2\pi \sqrt \ell$~Then divide both sides by t, $\large \frac{\cancel t \sqrt g}{\cancel t}=2\pi\frac{\sqrt \ell}{t} \qquad \rightarrow \qquad \sqrt g=2\pi \frac{\sqrt \ell}{t}$ ~Then we'll finish up by squaring both sides,$\large \left(\sqrt g\right)^2=\left(2\pi \frac{\sqrt \ell}{t}\right)^2 \qquad \rightarrow \qquad g=2^2 \pi^2\frac{(\sqrt \ell)^2}{t^2}$When we square a term, we square every part of it as I have done above. Which gives us,$\large \color{brown}{g=4\pi^2\frac{\ell}{t^2}}$

4. 4meisu Group Title

Thank you, that was very helpful and easy to understand :)