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 )
Consider, pls, this option.
Please, change the design according to local requirements.
She is incorrect because 14 multiplied by 2 equals 28, which means she wouldn’t have any money left over.
The answer in simplest form would be 1/6
Use the formula of arc length that is
s = rθ
where s = arc length, r = radius, θ = angle (unit must be in radian)
Given r = 5 in and θ = 48°( π /180°) = (4/15)* π = 0.2667 π,
s = 5 in (0.2667π) = 1.333 π in
