Answer:
20 meters
Step-by-step explanation:
The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.
So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.
The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be 
meters from the starting point.
C=2pr, r=c/(2p)
a=pr^2, using r found above we get:
a=p(c^2/(4p^2))
a=(c^2)/(4p), since c=106.76 and we approximate pi as 3.14
a=(106.76^2)/(4*3.14)
a=11612.2176/12.56 cm^2
a≈924.54 cm^2 (to nearest one-hundredth of a square cm)
We may draw a vertical that bisects the vertex angle and the shorter side of the triangle, to form a right-triangle with the measurements:
hypotenuse = 8
length = 1.5
We may then apply:
sin(∅) = P / H
sin(∅) = 1.5 / 8
∅ = 10.8 ≈ 11°
This is half of the vertex angle, so the vertex angle is:
11 * 2 = 22°
Multiply the second fraction by three.
4/9 + 3/9 = 7/9
Substitute the first equation into the second equation. You will get:
4x - (2x - 5) = 7
Distribute the negative sign into the parenthesis:
4x - 2x + 5 = 7
Simplify
2x + 5 = 7
Subtract 5 on both sides
2x = 2
x = 1
Now, substitute x = 1 into the first equation:
y = 2(1) - 5
y = 2 - 5
y = -3
The solution to the system of equations is (1, -3).
