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.
In factored form the answer would be <span>(<span><span>3x</span>−2</span>)</span><span>(<span><span>2x</span>+5)...
Hope This Helped
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The diameter is 12 cm and the radius is alsays half of the diameter so it will be 6 cm
