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.
Let us denote the smaller even integer by

.
Then, the next even integer must be

.
To verify, set

as any even number, say

.
Then, the next even integer must be 2 more than this number, i.e.

.
The sum of these two consecutive even integers can be expressed as:

We are told that this sum divided by four is 189.5, i.e.

Multiplying both sides by 4 will reverse the division:

Subtracting 2 from each side will reverse the addition:

Dividing both sides by 2 will reverse the multiplication:

Therefore, the two consecutive integers must be
378 and
380.
Check this:
Maybe put them in order so I could possibly answer
Answer: 15/7
Step-by-step explanation: Simplify 6/2 to 3
8x-3=x+12
Add 3 to both sides
8x=x+15
Subtract x from both sides
7x=15
Divide both sides by 7
x=15/7
Answer:
2 x 2 x 3 x 3
Step-by-step explanation:
36
6 x 6
3 x 2 x 3 x 2