Given the two functions:
![\begin{gathered} R(x)=2\sqrt[]{x} \\ S(x)=\sqrt[]{x} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20R%28x%29%3D2%5Csqrt%5B%5D%7Bx%7D%20%5C%5C%20S%28x%29%3D%5Csqrt%5B%5D%7Bx%7D%20%5Cend%7Bgathered%7D)
We need to find (RoS)(4). THis is the functional composition. We take S(x) and put it into R(x) and then put "4" into that composed function. Shown below is the process:
![(RoS)(x)=2\sqrt[]{\sqrt[]{x}}](https://tex.z-dn.net/?f=%28RoS%29%28x%29%3D2%5Csqrt%5B%5D%7B%5Csqrt%5B%5D%7Bx%7D%7D)
When we plug in "4", into "x", we have:
![\begin{gathered} (RoS)(x)=2\sqrt[]{\sqrt[]{x}} \\ (RoS)(4)=2\sqrt[]{\sqrt[]{4}} \\ =2\sqrt[]{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28RoS%29%28x%29%3D2%5Csqrt%5B%5D%7B%5Csqrt%5B%5D%7Bx%7D%7D%20%5C%5C%20%28RoS%29%284%29%3D2%5Csqrt%5B%5D%7B%5Csqrt%5B%5D%7B4%7D%7D%20%5C%5C%20%3D2%5Csqrt%5B%5D%7B2%7D%20%5Cend%7Bgathered%7D)
I'm pretty sure that this one is the correct factorization of the polynomial above: <span>a) (3x+4)(9x^2-12x+16)</span>
5 over and 4 ^. not sure if this is correct but that’s what i got
<em>Answer:</em>
n²+3
<em>Explanation:</em>
The differences between the terms are not the same, so this is not "linear". Knowing that the sequence may have started with a 1, you can try subtracting the first number with a number to get 1, and use that number to subtract the rest.
4 - 3 = 1
7 - 3 = 4
12 - 3 = 9
19 - 3 = 16
28 - 3 = 25
In this case, subtracting 3 to all the numbers gave us perfect squares! So this means the nth term has to do with squaring the number and adding three afterward! This can be checked.
√1 = 1
√4 = 2
√9 = 3
√16 = 4
√25 = 5
As we found the values of these terms by subtracting three first and then finding its square root, the nth term will be the opposite; squaring and then adding three! Again, this can be checked!
1² + 3 = 4
2² + 3 = 7
3² + 3 = 12
4² + 3 = 19
5² + 3 = 28
Hope this helps !! :D
