A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter
1 answer:
Answer:
Step-by-step explanation:
Let x be the track straights lengths
Let y be the track ends diameter and the other rectangle side lengths.
1800 = 2x + πy
y = (1800 - 2x) / π
A = xy
A = x((1800 - 2x) / π
A = (1/π)(1800x - 2x²)
dA/dx = (1/π)(1800 - 4x)
0 = (1/π)(1800 - 4x)
0 = 1800 - 4x
4x = 1800
x = 450 m
y = (1800 - 2(450)) / π
y = 900/π or approximately 286.5 m
You might be interested in
Answer:
Yes, 35 feet is enough
The angle of elevation is 4.1º
It is £21 but I have to be 20 characters long to post this comment but that’s the answer
Answer:
2 sqrt(15)
Step-by-step explanation:
sqrt(60) = sqrt(4*15) = 2 sqrt(15)
A it’s a I think ok because I’m new
Answer:
BEC = C. 50
BCD = A. 120
Step-by-step explanation:
BEC
65 + 65 = 130
180 - 130 = 50
BCD = 180 - 60 = 120
BEC is 50
