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The equation you provided:

 

`T=2pi\sqrt(L/9.8) `

is the equation for the period of a simple pendulum.

Since L is the only variable, then T is a function of L.

1. `T(L) = 2pi\sqrt(L/9.8)`

2. The domain is `{LinR|Lgt=0}` or  `[0,+oo)`

3. The range is `{TinR|Tgt=0}` or `[0,+oo)`

4. This is a sideways-parabola graph (a square root graph).

T = 8.98 s at L = 20 m

5. Let T = 10 s, Solve for L

The above equation is rearranged:

`T=2pi\sqrt(L/9.8)`

`T/(2pi)=\sqrt(L/9.8)`

`(T/(2pi))^2=L/9.8`

`9.8*(T^2/(4pi^2))=L`

Now we can plug in T=10

`L=9.8*((10)^2/(4pi^2))`

`L=24.824 m` long to the nearest 0.001 m

6. The mass has no affect on the pendulum according to this equation.

 

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