Results 1 to 3 of 3

Math Help - how do i found the the height

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    23

    how do i found the the height

    The volume of a rectangular box is (x^3+6x^2+11x6). The
    box is (x+3) cm long and (x+2) cm wide. How high is the box?

    May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1

    Re: how do i found the the height

    Quote Originally Posted by waleedrabbani View Post
    The volume of a rectangular box is (x^3+6x^2+11x6). The
    box is (x+3) cm long and (x+2) cm wide. How high is the box?

    May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.
    Divide the volume by (x+3).
    You should get a reminder of zero.
    Take the quotient and divide by (x+2) .
    You should get a reminder of zero.
    That quotient is the height.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,570
    Thanks
    1428

    Re: how do i found the the height

    Quote Originally Posted by waleedrabbani View Post
    The volume of a rectangular box is (x^3+6x^2+11x6). The
    box is (x+3) cm long and (x+2) cm wide. How high is the box?

    May I also add that this is in the long division section of the text book so i am assuming i will have to apply long division in it somewhere, i actually do not understand where to start, I am not looking for an answer, just ways that i can tackle the problem. thank you.
    As an alternative, for any prism I always remember that the volume is equal to the cross-sectional area multiplied by the height.

    You are told the length is \displaystyle \begin{align*} x + 3 \end{align*} and the width is \displaystyle \begin{align*} x + 2 \end{align*}, so the cross sectional area is \displaystyle \begin{align*} (x + 3)(x + 2) = x^2 + 5x + 6 \end{align*}.

    Therefore, you have

    \displaystyle \begin{align*} V &= AH \\ x^3 + 6x^2 + 11x + 6 &= \left(x^2 + 5x + 6\right)H \\ H &= \frac{x^3 + 6x^2 + 11x + 6}{x^2 + 5x + 6}  \end{align*}

    Now perform this long division to simplify the height
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. So pleased I found you all
    Posted in the New Users Forum
    Replies: 1
    Last Post: April 13th 2012, 04:04 PM
  2. How was this result found
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 3rd 2011, 06:56 AM
  3. [SOLVED] Almost found limit of sum
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 30th 2010, 05:21 AM
  4. Just found this website
    Posted in the Math Forum
    Replies: 0
    Last Post: July 1st 2010, 07:19 AM
  5. Weird prob that i found
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 16th 2008, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum