Consider this solution, if it is possible, check the result:
1) if to divide polynomial 'x³-3x²-6x+8' on 'x+2' result is: x²-5x+4, then x³-3x²-6x+8=(x+2)(x²-5x+4);
2) x²-5x+4=(x-1)(x-4), then
3) x³-3x²-6x+8=(x+2)(x-1)(x-4).
4) zeros are: -2;1;4. Other zeros are: 1 and 4.
Answer: 1;4.
Answer:
ill help when you give the problem lol
Step-by-step explanation:
well, the recursive rule of aₙ = aₙ₊₁ + 7, where a₁ = 15, is simply saying that
we start of at 15, and the next term is obtained by simply adding 7, and so on.
well, that's the recursive rule.
so then let's use that common difference and first term for the explicit rule.

Answer:
x = 9
Step-by-step explanation:
x−(3)(3=x+−9 )
Answer:

Step-by-step explanation:
Given


Required [Missing from the question]
G(T(x))
We have:

This implies that:

Substitute: 
![G(T(x)) = 3[9(x + 6.9)]](https://tex.z-dn.net/?f=G%28T%28x%29%29%20%3D%203%5B9%28x%20%2B%206.9%29%5D)
Open bracket
