Answer:
The property of polynomial addition that says that the sum of two polynomial is always a polynomial is called closure property of addition or under addition.
Polynomials are closed under addition because when you add polynomials the letters and their exponents do no change, you just add the coefficients of the like terms (those with same letters raised to the same exponents), so the result will be other polynomial of the same kind, except for the terms that cancel (positive with negative) which does not change the fact that the result is still a polynomial.
In mathematics the closure property means that the result of an operation over a kind of "number" will result in a "number" of the same kind.
Answer: 
<u>Step-by-step explanation:</u>
Convert everything to "sin" and "cos" and then cancel out the common factors.
![\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)](https://tex.z-dn.net/?f=%5Cdfrac%7Bcot%28x%29%2Bcsc%28x%29%7D%7Bsin%28x%29%2Btan%28x%29%7D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%5B%5Cdfrac%7Bsin%28x%29%7D%7B1%7D%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%5D%5C%5C%5C%5C%5C%5C%5Cbigg%28%5Cdfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%2B%5Cdfrac%7B1%7D%7Bsin%28x%29%7D%5Cbigg%29%5Cdiv%5Cbigg%28%5Cdfrac%7Bsin%28x%29cos%28x%29%7D%7Bcos%28x%29%7D%2B%5Cdfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5Cbigg%29)


Answer:
Can't be determined
Step-by-step explanation:
You just don't have enough info!
It goes by 8!
If u see, 4 + 8= 12 +(another) 8 = 20! It skips 8 every time!
If u were to continue it, 20 + 8 = 28 and so on!
The difference is the sum of the number totalled.
Hope this helps!
Problem
evaluate the function
f(-x) = f(x)
plug in the points (2,1) then solve
Solution
For this case we can conclude that:
f(2) =1
So then we satisfy that:
f(-2) =1