Results 1 to 7 of 7

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

  1. #1
    Newbie
    Joined
    Apr 2018
    From
    hamilton
    Posts
    4

    how can i solve for h in this fraction?

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

    thanks!!820(1+h6400)2<10
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,181
    Thanks
    2635

    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

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

    $57600 < h$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,860
    Thanks
    3071

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

    If h< 0 then $\displaystyle \frac{h}{6400}< -9$, $\displaystyle h< -9(6400)= -57600$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,906
    Thanks
    352

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

    Quote Originally Posted by romsek View Post
    $\dfrac{820}{1+\left(\frac{h}{6400}\right)^2} < 10$
    Shouldn't denominator be (1 + h/6400)^2 ???
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,880
    Thanks
    2889
    Awards
    1

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

    Quote Originally Posted by DenisB View Post
    Shouldn't denominator be (1 + h/6400)^2 ???
    YES! The OP is $\dfrac{820}{\left(1+\frac{h}{6400}\right)^2}<10$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,860
    Thanks
    3071

    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 $\displaystyle h> 6400(\sqrt{82}- 1)$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,906
    Thanks
    352
    Last edited by DenisB; Apr 18th 2018 at 05:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solve the complex fraction
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 26th 2016, 05:04 AM
  2. how to solve this algebra fraction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 10th 2013, 03:20 AM
  3. Fraction & Inequality: How to solve for x?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 23rd 2009, 07:56 AM
  4. Replies: 2
    Last Post: Oct 5th 2009, 10:53 AM
  5. How do I solve when A fraction is involved?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 6th 2008, 08:31 AM

Search Tags


/mathhelpforum @mathhelpforum