# Thread: how can i solve for h in this fraction?

1. ## how can i solve for h in this fraction?

820/(1+h/6400)2 <10

thanks!!820(1+h6400)2<10

2. ## Re: how can i solve for h in this fraction?

$\dfrac{820}{1+\left(\frac{h}{6400}\right)^2} < 10$

$1+\left(\frac{h}{6400}\right)^2 > 0$ so we can rearrange this as

$82 < 1+\left(\dfrac{h}{6400}\right)^2$

$81 < \left(\dfrac{h}{6400}\right)^2$

we'll assume that $h > 0$ so we have

$9 < \dfrac{h}{6400}$

$57600 < h$

3. ## Re: how can i solve for h in this fraction?

If h< 0 then $\displaystyle \frac{h}{6400}< -9$, $\displaystyle h< -9(6400)= -57600$.

4. ## Re: how can i solve for h in this fraction?

Originally Posted by romsek
$\dfrac{820}{1+\left(\frac{h}{6400}\right)^2} < 10$
Shouldn't denominator be (1 + h/6400)^2 ???

5. ## Re: how can i solve for h in this fraction?

Originally Posted by DenisB
Shouldn't denominator be (1 + h/6400)^2 ???
YES! The OP is $\dfrac{820}{\left(1+\frac{h}{6400}\right)^2}<10$

6. ## Re: how can i solve for h in this fraction?

Romsek had the square misplaced as DenisB and Plato point out. From
$\displaystyle \frac{820}{\left(1+ \frac{h}{6400}\right)^2}< 10$,
dividing both sides by 10 and multiplying both sides by $\displaystyle \left(1+ \frac{h}{6400}\right)^2$, both of which are positive, $\displaystyle 82< \left(1+ \frac{h}{6400}\right)^2$.

Taking the square root of both sides we have either
1) $\displaystyle -\sqrt{82}> 1+ \frac{h}{6400}$ or
2) $\displaystyle \sqrt{82}< 1+ \frac{h}{6400}$.

If (1), $\displaystyle \frac{h}{6400}< -1- \sqrt{82}$, so $\displaystyle h< -6400(1+ \sqrt{82})$.

If (2), $\displaystyle \frac{h}{6400}> \sqrt{82}- 1$, so $\displaystyle h> 6400(\sqrt{82}- 1)$.