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:
30
Step-by-step explanation:
If you plug in 5 for x and 3 for y then the equation would be 3(5)+5(3) or 15+15.
The width and height of the tv as well as the length and width of the back of your car
