a cement walk of uniform width surrounds a rectangular pool that is 10m wide and 50m long. find the width of the walk if its area is 864 m squared.

can you help please??

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- April 19th 2007, 05:55 PMpieman396can anyone help wth this polynomial equation?
a cement walk of uniform width surrounds a rectangular pool that is 10m wide and 50m long. find the width of the walk if its area is 864 m squared.

can you help please?? - April 19th 2007, 05:57 PMJhevon

Umm... this is EXACTLY like the question i answered for you before, all you do is change the numbers up.

see http://www.mathhelpforum.com/math-he...ra-2-help.html

was there something you didn't understand about the method i used. you should say so if you don't get it - April 19th 2007, 06:03 PMpieman396my work for the problem.. can you tell me whats wrong??
yeah but i did it the same way but i ended up with

864=(2x+50)(2x+10)

864=4x^2 +20x + 100x +500

864=4x^2 + 120x +500...divided by 4

216= x^2 +30x +125

0= x^2 +30x -91

but it can't be broken up into two binomials. - April 19th 2007, 06:20 PMtopsquark
- April 19th 2007, 06:21 PMpieman396???
i have no idea what mean... i'm completely lost... bu ti really need help because i have a test on this stuff tomorrow..

- April 19th 2007, 06:21 PMJhevon
- April 19th 2007, 06:25 PMpieman396...
no i have no idea what the quadractic formula is....

- April 19th 2007, 06:28 PMtopsquark
It means this:

Compare the expressions:

ax^2 + bx + c = 0

x^2 + 30x - 91 = 0

So a = 1, b = 30, c = -91

Thus

x = [-b (+/-) sqrt{b^2 - 4ac}]/(2a)

x = [-(30) (+/-) sqrt{(30)^2 - 4(1)(-91)}]/(2(1))

x = [-30 (+/-) sqrt{900 + 364}]/2

x = [-30 (+/-) 35.5528]/2

x = (-30 + 35.5528)/2 = 2.77639

or

x = (-30 - 35.5528)/2 = -32.7764

If you haven't used the quadratic formula yet, then I suspect there is an error in deriving your equation.

-Dan - April 19th 2007, 06:30 PMJhevon
- April 19th 2007, 06:32 PMpieman396
thanks... but can help me with another problem now...

A rocket is launched from ground level with an initial vellcity of 83.3m/s. When will the ball reach a height of 294m??

i'm supposed to use polynomial equations to figure this out to but the only way i know how to do it is with physics. - April 20th 2007, 06:04 AMtopsquark
If you have a new (unrelated) question, you should post it in a new thread.

What's wrong with doing it by Physics?

Set an origin at the ground, where the ball-rocket started and set +y upward. Then y0 = 0m, v0 = +83.3 m/s and a = -9.8 m/s^2.

Thus

y = y0 + v0*t + (1/2)a*t^2 <-- Polynomial in t!!

y = 83.3t - 4.9t^2

So when is y = 294 m?

294 = 83.3t - 4.9t^2

4.9t^2 - 83.3t + 294 = 0

Since you've never seen the quadratic equation, let's multiply both sides of this by 10:

49t^2 - 833t + 2940 = 0

This looks like a terrible mess, but note that 833 and 2940 are multiples of 49. Thus we can factor a 49:

49(t^2 - 17t + 60) = 0

t^2 - 17t + 60 = 0

(t - 12)(t - 5) = 0

Thus

t = 12 s or t = 5 s.

-Dan