Answer:
2 touchdowns, 3 field goals
Step-by-step explanation:
The number of touchdowns cannot be more than 3, so it is relatively easy to find the solution by trial and error.
23 is not divisible by 3, so 0 touchdowns is not a solution
23 -7 = 16 is not divisible by 3, so 1 touchdown is not a solution
23 -14 = 9 is divisible by 3, so 2 touchdowns and 3 field goals is a solution
21 -21 = 2 is not divisible by 3, so there is only one solution.
Step-by-step explanation:
You have to do first and second derivatives :
y = a sin 3x + b cos 3x
dy/dx = 3a cos 3x - 3b sin 3x
d²y/dx² = - 9a sin 3x - 9b cos 3x
Next, you have to compare it :
let d²y/dx² = ky,
-9a sin 3x - 9b cos 3x = k(a sin 3x + b cos 3x)
-9(a sin 3x + b cos 3x) = k(a sin 3x + b cos 3x)
Therefore, k is equals to 9.