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??
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
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.
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
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.
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