Answer:
By continuing my education I increased my earning potential from $21,484 to $39,746 a year. That's a difference of $18262 a year.
If the additional education costs $18,000, then in one year it will pay for itself.
Question 1)
Given: F(x) = 3x^2 + 1, G(x) = 2x - 3, H(x) = x
F(G(x)) = 3(2x - 3)^2 + 1
F(G(x)) =3(4x^2 - 12x + 9) + 1
F(G(x)) = 12x^2 - 36x + 27 + 1
F(G(x)) =12x^2 - 36x + 28
Question 2)
Given: F(x) = 3x^2 + 1, G(x) = 2x - 3, H(x) = x
H -1 (x) = x (inverse)
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Let's say the first number is "a"
the next consecutive can just be "a+1" and the next, you guessed it, "a+2"
the two smaller ones are a and a+1, the larger one is a+2
now, the product of "a" and "a+1" equals 5 times "a+2" but less by 5
thus a(a + 1) = 5(a + 2) - 5 <----- solve for "a"
The Answer is x= -6 and x= -6
