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)
Answer:
No solution.
Step-by-step explanation:
Given
Required
Find all possible solutions
If 
, then the triangle is right-angled and the hypotenuse is at b
Given that 
This implies that a > b
In a right-angled triangle, the hypotenuse (side b) is the longest.
Since this is not true for the given sides, then the triangle has no solution.
To get better at 12's:
Write down on your paper your 1's facts in column skip 5 and 11 going to 14 (a vertical line - line that goes up and down). To the right of that column, write your two's facts 0 to 8 and repeat again. Then you will have your 12's! Should look as follows
12's:
0 0 = 12 x0
1 2 = 12 x1
2 4 = 12 x2
3 6 = 12 x3
4 8 = 12 x4
6 0 = 12 x5
7 2 = 12 x6
8 4 = 12 x7
9 6 = 12 x8
10 8 = 12 x9
12 0 = 12 x10
13 2 = 12 x 11
14 4 = 12 x 12
It would be A.) 7. hope this helps
