<span>A. (-3, 5)
the equation is xy=-15
or x=-15/y
put y=5
x=-15/5
x=-3</span>
I'm sorry I'm too lazy to read all of that. However, I did get the main point. The river was going at 3 mph. Since he wanted to travel 10 miles upstream and downstream, you would have to subtract 20 by 3. This would mean he was going at a speed of 17 mph.
Sub x = 2-y^2 to Q, we get:
Q = 3(2-y^2)*y^2
let y^2 = k
Q = 3(2-k)k = 3(2k-k^2)
2k-k^2 has a max when k = 1
Then y^2 = 1 -> y = 1 or -1
Answer:
19/20 is the correct answer
