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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...

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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.

Approved by eNotes Editorial Team