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.
n(A-B) denotes elements which are in A but not in B
n(Au B) denotes elements in A and B
n(AnB) denotes elements that are common in A and B
Now I will add one more set
n(B-A) which denotes elements in B but not in A
So, n(AuB) = n(A-B) + n( B-A) +n(AnB)
70 = 18 +n(B-A) + 25
70 = 43 + n(B-A)
n(B-A) = 70-43
n(B-A) = 27
So, n(B) = n( B-A) + n( AnB)
= 27+25
= 52
F(x) = -x² + 25
g(x) = x + 5
(f ÷ g)(x) = ⁻ˣ²⁺²⁵/ₓ₊₅
(f ÷ g)(x) = ⁻¹⁽ˣ⁺⁵⁾⁽ˣ⁻⁵⁾/ₓ₊₅
(f ÷ g)(x) = -x + 5
(-4,0) (3,-5) I don't know if they mean two ordered pairs so I'm pretty sure this is right