Answer:
2x^2x(x^+3x-2)
Step-by-step explanation:
F ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
Answer:
I say that the answer is A, B, and C. I'm not positive but I think that is the answer.
Step-by-step explanation:
Tell me if I'm wrong it might just be B and C.
297 because 289 rounds to 290 plus 7 is 297
The last one does not represent a function because for it to be a function none of the x’s can repeat and in the last one the 0 repeats.
