Answer c
Step-by-step explanation:
they have to be equivalent
Answer:

Step-by-step explanation:
Given :-
The sum of two numbers is 1 .
The product of the nos . is 12 .
And we need to find out the numbers. So let us take ,
First number be x
Second number be 1-x .
According to first condition :-

Hence the numbers are 4 and -3
Answer:
- (-16x² +10x -3) +(4x² -29x -2)
- (2x² -11x -9) -(14x² +8x -4)
- 2(x -1) -3(4x² +7x +1)
Step-by-step explanation:
I find it takes less work if I can eliminate obviously wrong answers. Toward that end, we can consider the constant terms only:
- -3 +(-2) = -5 . . . . possible equivalent
- -10 -5 = -15 . . . . NOT equivalent
- 3(-5) -2(5) = -25 . . . . NOT equivalent
- -9 -(-4) = -5 . . . . possible equivalent
- -7 -(-5) = -2 . . . . NOT equivalent
- 2(-1) -3(1) = -5 . . possible equivalent
Now, we can go back and check the other terms in the candidate expressions we have identified.
1. (-16x² +10x -3) +(4x² -29x -2) = (-16+4)x² +(10-29)x -5 = -12x² -19x -5 . . . OK
4. (2x² -11x -9) -(14x² +8x -4) = (2-14)x² +(-11-8)x -5 = -12x² -19x -5 . . . OK
6. 2(x -1) -3(4x² +7x +1) = -12x² +(2 -3·7)x -5 = -12x² -19x -5 . . . OK
All three of the "possible equivalent" expressions we identified on the first pass are fully equivalent to the target expression. These are your answer choices.
Answer:
The answer is "
"
Step-by-step explanation:
Given:

Find critical points:

differentiate the value with respect of x:
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 