Answer:
see below
Step-by-step explanation:
y' = -2e⁻ˣcos(x) -sin(x)2e⁻ˣ +e⁻ˣsin(x)-cos(x)e⁻ˣ = e⁻ˣ(-2cos(x)-2sin(x)+sin(x) -cos(x))
=e⁻ˣ(-3cos(x)-sin(x))
y'' = -e⁻ˣ(-3cos(x)-sin(x)) + (3sin(x)-cos(x))e⁻ˣ
y'' = e⁻ˣ[3cos(x)+sin(x) + 3sin(x)-cos(x)] = e⁻ˣ[2cos(x)+4sin(x)]
y''+2y'+2y = 0
e⁻ˣ[2cos(x)+4sin(x)] + 2[e⁻ˣ(-3cosx-sinx)] + 2[2e⁻ˣ cos(x)−e⁻ˣ sin(x)]
e⁻ˣ[2cos(x)+4sin(x)-6cos(x)-2sin(x)+4cos(x)-2sin(x)]
e⁻ˣ[-4cos(x)+2sin(x)+4cos(x)-2sin(x)]
e⁻ˣ[0cos(x)+0sin(x)] = 0
Answer:
C, 2.15
Step-by-step explanation:
Arabian Plate: 1.29x10^-4 = .000129
African Plate: 6x10^-5 = .000006
Once you've converted both into standard form, divide them to find the difference.
.000129/.000006 = 2.15
The cubic function is f(x) = x^3
You need to perform three transformations to the cubic function to obtain
f(x) = - (x + 2)^3 - 5.
Those transfformations are:
1) Shift f(x) = x^3, 2 units leftward to obtain f(x) = (x + 2)^3
2) reflect f(x) = (x + 2)^3 across the x-axis to obtain f(x) = - (x + 2)^3
3) shift f(x) = - ( x + 2) ^3, 5 units downward to obtain f(x) = - (x + 2)^3 - 5
The answer is going to be c
Answer is in the attachment below.
