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confusedkarl · Jun 16, 2008, 12:02 PM

(R+ωL)(R-ωL)
= R^2-ωL

Z^2-R^2
= (Z+R)(Z-R)

h= -4.9t^2+55t+12
make ‘t’ subject

h= -4.9t^2+55t+12
h-12=-4.9t^2+55t+12
-12h=50.1t

-12h/50.1=t

Thank you

galactus · Jun 16, 2008, 04:48 PM

For the last one, you can't subtract a t term from a t^2 term:confused: .

You just broke the laws of algebra. :) :rolleyes: :p

You can easily check your answer by subbing it back in to the equation. If it works, you should get h. But you won't.:eek:

$-\frac{49}{10}t^{2}+55t+12=h$

Try completing the square. Can you do that?

Upon doing that you get $\frac{-(49t-275)^{2}}{490}+\frac{16301}{98}$

Multiply the right side by 5 to get the denominators the same, then we get:

$\frac{-(49t-275)^{2}+81505}{490}=h$

$(49t-275)^{2}-81505=-490h$

$(49t-275)^{2}=-490h+81505$

$49t-275=\sqrt{-490h+81505}$

$t=\frac{\sqrt{-490h+81505}+275}{49}, \;\ \frac{-\sqrt{-490h+81505}-275}{49}$

There, see what happened?