![]() |
Sign Up FREE! |
Sign In |
Classroom Setup |
Common Core Alignment ![]() |
![]() |
#1 |
Join Date: Sep 2012
Posts: 1
|
![]() Help!!
A large box has a volume of 180ft^3. Its length is 9ft greater than its height and its width is 4ft less than its height. What are the dimensions of the box? I have 180=(h+9)(h-4)(h) and I keep coming up with h=4 which would mean the height is 0...not possible. Help!! |
![]() |
![]() |
#2 | |
![]() ![]() ![]() Join Date: Dec 2008
Posts: 249
|
![]() Quote:
(h^2 + 9h)(h - 4) = 180 h^3 - 4h^2 + 9h^2 - 36h = 180 h^3 + 5h^2 - 36h = 180 h^3 + 5h^2 - 36h - 180 = 0 Let's try 6 as a solution. (6)^3 + 5(6)^2 - 36(6) - 180 = 0 216 + 180 - 216 - 180 = 0 0 = 0 It turns out the other 2 roots are -6 and -5 (you can check by plugging them into the equation) so we will use h = 6 ft for the height. Then the length is 6 + 9 = 15 ft and width is 6 - 4 = 2 ft. Hope this helps. |
|
![]() |
Thread Tools | |
Display Modes | |
|
|