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:
Deal B is better
Step-by-step explanation:
over a year in payment Deal b will be less
You would need to multiply the monthly cost, $14, by the number of months and then that would be added to the cost to join, $25
For 6 months it would cost 14 x 6 = 84 + 25 = $109
The answer would be C.
Use elimination and subtitution method to solve the problem.
First, eliminate x and you'll find the value of y
x - 4y = 12
x - y = 0
--------------- - (substract)
-3y = 12
y = 12/-3
y = -4
Second, subtitute -4 as y and you'll find the value of x
x - y = 0
x- (-4) = 0
x + 4 = 0
x = -4
The solution
x,y = -4,-4
