Answer:
The proof is detailed below.
Step-by-step explanation:
We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.
Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.
It is a rule that alternate angles are equal, so using this fact, we can say that:
3x + 17 = x + 53
- x
2x + 17 = 53
- 17
2x = 36
÷ 2
x = 18
Which means that your answer would be B. x = 18
I hope this helps!
The answer is the first option. 12m-12n
Angle Y would be 40 degrees
5/3 is the answer because if you do 5/1 multiplied by 1/3 you get 5/3
Hope this helps(:
