Answer:
the answer is c, less than or equal to 4
Answer:
no it is not the same because you get different answers
(f+g)(3)=f(3)+g(3)
f(x)=7 so f(3)=7
g(x)=x+5 so g(3)=3+5=8
(f+g)(3)=f(3)+g(3)=7+8=15
Answer:
x = -9
Step-by-step explanation:
You are solving for x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.
First, subtract 7 from both sides:
2x + 7 (-7) = -11 (-7)
2x = -11 - 7
2x = -18
Next, divide 2 from both sides:
(2x)/2 = (-18)/2
x = -18/2
x = -9
x = -9 is your answer.
~
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 
