1. ## make L subject

$
t=2(pi)\sqrt\frac{100-L}{g}
$

i have to make L subject. and find L when t=20 g =2

i have $g(\frac{t}{2pi})^2-100 = L$

not sure on this one? do i remove the sqrt first or bring the 2pi across first?
thanks

2. Originally Posted by decoy808
$
t=2(pi)\sqrt\frac{100-L}{g}
$

i have to make L subject. and find L when t=20 g =2

i have $g(\frac{t}{2pi})^2-100 = L$

not sure on this one? do i remove the sqrt first or bring the 2pi across first?
thanks
$T = 2\pi \sqrt{\frac{100-L}{g}}
$

$\frac{T}{2\pi} = \sqrt{\frac{100-L}{g}}$

$\left(\frac{T}{2\pi}\right)^2 = \frac{100-L}{g}$

$g \cdot \left(\frac{T}{2\pi}\right)^2 = 100-L$

$L = 100 - g \cdot \left(\frac{T}{2\pi}\right)^2$

3. L=992.9475997 i get an error when i try put back into original equation. as i should be get in t=20

4. Originally Posted by decoy808
L=992.9475997 i get an error when i try put back into original equation. as i should be get in t=20
$L = 100 - 2\left(\frac{20^2}{4\pi^2}\right) \approx 79.7
$