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.
Answer:Its B,C, and E man
Step-by-step explanation:
Composite numbers more than 2 numbers can go into it like 4 numbers that can go into 4 are 1,2,4
Answer:
82
Step-by-step explanation:
<u>Vertical rectangle:</u>
A = bh
A = (5)(10)
A = 50
<u>Horizontal rectangle:</u>
h = 10 - (3 + 3)
h = 4
A = bh
A = (8)(4)
A = 32
<u>Combined:</u>
50 + 32
82
N<6 is the answer I think
