A + B + C = 476
A = 3B + 6
C = B + 45
now its just a matter of subbing..
A + B + C = 476
(3B + 6) + B + (B + 45) = 476...combine like terms
5B + 51 = 476
5B = 476 - 51
5B = 425
B = 425/5
B = 85 <== team B scored 85
A = 3B + 6
A = 3(85) + 6
A = 255 + 6
A = 261 <=== team A scored 261
C = B + 45
C = 85 + 45
C = 130 <=== team C scored 130
Answer:
The answer is "0.4125"
Step-by-step explanation:
33 divided by 3 is 11 plus 10 is 21 so your answer is 21:)
Answer:1002
Step-by-step explanation:
and 
and 
as 
Applying this we get
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D)
every 
there exist 
such that 
therefore
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D%3D1002)
A whole hour or revolution is 60 minutes and 360° so the area of a sector is:
A/πr^2=α/360
A=απr^2/360...now we need to find the angle α/360=5/60, α=30° so:
A=30πr^2/360
A=πr^2/12 and r=11.25ft so
A=11.25^2*π/12
A≈33.13 ft^2
