Y = t*e^(-t/2)
y' = t' [e^(-t/2)] + t [e^(-t/2)]' = e^(-t/2) + t[e^(-t/2)][-1/2]=
y' = [e^(-t/2)] [1 - t/2] = (1/2)[e^(-t/2)] [2 - t] = - (1/2) [e^-t/2)] [t -2]
Answer:
The rate at which the distance from the plane to the station is increasing is 331 miles per hour.
Step-by-step explanation:
We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:
a: is one side of the triangle = altitude of the plane = 3 miles
b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles
h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles
First, we need to find b:
(1)

Now, to find the rate we need to find the derivative of equation (1) with respect to time:
Since "da/dt" is constant (the altitude of the plane does not change with time), we have:
And knowing that the plane is moving at a speed of 500 mi/h (db/dt):
Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.
I hope it helps you!
Answer:
Step-by-step explanation:
I need help with same question pleas.
Answer:
x=6
Step-by-step explanation:
4x-1=2x+11
4x-2x-1=11
2x-1=11
2x=11+1
2x=12
x=12/2
x=6
Given:
f(x) = 4x² + 1
g(x) = x² - 5
Then
(f + g)(x) = 4x² + 1 + x² - 5
= 5x² - 4
Answer: 5x² - 4