The answer is C: 3 books, 2.5 weeks.
The perimeter of a semicircle consists of two parts. (the curve and bottom)
That curve is half the distance around the circle, since it's been split in half.
The distance around a circle, the circumfrence, is equal to 2πr, where r is the radius of that circle. In this case, the circumfrence of the entire circle would be 16π. and so that curve would have a length of just 8π.
Using 3.14 for π, 8π = 8×3.14 = 25.12.
As for the flat part, that is the diameter (distance across) our circle.
The radius is the distance from the center of a circle to its edge, and always has half the length of the diameter. (you can break the diameter down into two radii)
If our radius is 8 meters, our diameter (the flat part of that semicircle) must be 16.
Now we add up the two parts of the perimeter...25.12 + 16 = 41.12.
Answer:
x=8
y=16
Step-by-step explanation:
2x+2y=48
3x+y=40
Ther is many ways to solve this, one of them is clearance:
WE CLEAR AN UNLOCKED (X) AND REPLACE IT IN THE OTHER
x= (40-y)/3
2[(40-y)/3] +2y=48
(80-2y)/3 +2y=48
(80-2y+6y)/3=48
80+4y=48(3)
4y=144
y= 144/4
y=16
x=(40-y)/3
x=(40-16)/3
x=24/3
x=8
Y = -0.01x^2 + 0.5x + 3
when the height y = 8
-0.01x^2 + 0.5x + 3 = 8
-0.01x^2 + 0.5x - 5 = 0
x = 13.82 and x = 36.18 so the ball is at least 8 ft high from 13.82 and 36.18 feet inclusive.- from the player.
If the player is 30 ft from the net then the ball is likely to go over the net.