Answer: Since the second differences are constant, there is a quadratic relationship between and .
Step-by-step explanation:
Answer: 6x squared minus 17x minus 14
8=2*2*2=2^3
32=2*2*2*2*2=2^5
<span>[8^(2/3)] / [32^(2/5)] = [</span>(2^3)^(2/3)] / [(2^5)^(2/5)] =
= [2^(3*2/3)] / [2^(5*2/5)]= [2^2] / [2^2] = 4/4 = 1
Answer: 1
Solve for the first variable in one of the equations, then substitute the result into the other equation.
3. (1/2, -1,2)
4. (2, -4)
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Answer:
The answer is "
"
Step-by-step explanation:
Given:
Find critical points:
differentiate the value with respect of x:
![\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]](https://tex.z-dn.net/?f=%5Cto%20g%27%28x%29%3D%20%28x-e%29%5E3%20%5Cfrac%7Bd%7D%7Bdx%7De%5E%7Be-r%7D%20%2Be%5E%7Be-r%7D%20%20%5Cfrac%7Bd%7D%7Bdx%7D%28x-e%29%5E3%3D%28x-e%29%5E2%20e%5E%7B%28e-x%29%7D%20%5B-x%2Be%2B3%5D)
critical points 
![\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3](https://tex.z-dn.net/?f=%5Cto%20%28x-e%29%5E2%20e%5E%7B%28e-x%29%7D%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20e%5E%7B%28e-x%29%7D%5Cneq%200%20%5C%5C%5C%5C%5Cto%20%28x-e%29%5E2%3D0%5C%5C%5C%5C%20%5Cto%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20x%3De%5C%5C%5C%5C%5Cto%20x%3De%2B3%5C%5C%5C%5C%5Cto%20x%3D%20e%2Ce%2B3)
So,
The critical points of 
