Answer: D. (z2-3z+14)
Step-by-step explanation:
took the test
The possible x-values of the equation are options C and F. These are the + and - values that make the equation true.
Let

. Then

, and differentiating both sides with respect to

gives


Now, when

, you get

You have

, so

and

. So
Answer:
Step-by-step explanation:
3x+6=-5-2x-6
+2x. +2x
5x+6= -5-6
5x+6= -11
-6. -6
5x= -17
divide by 5
x= -3.4