Value/depreciating problem..Need help please

[adevon2122]

The value of a computer t years after purchase is given by v(t)=2000e^-0.35t


a) What is the value of the computer after 3 years?

b)At what rate is the computer's value depreciating after 3 years?

c)When will the value of the computer be $800?

 

[wjm11]

The value of a computer t years after purchase is given by v(t)=2000e^-0.35t


a) What is the value of the computer after 3 years?

b)At what rate is the computer's value depreciating after 3 years?

c)When will the value of the computer be $800?

For a) simply plug in t = 3.
For b) the word “rate” is the key. You’re looking for the slope of the function, so use the derivative to find the slope at t=3.
For c) plug in 800 for v(t) and solve for t.

 

[xceler8ari1984]

Ok, this is what I have figured out

a) v(t)=(2000)e^-.35(3)
=(2000)e^-1.05
=$699.87


b) v'(3)=-.35*v(t)
=-.35(699.87)=-244.95


c) 800=(2000)e^-.35t
800/2000=e^-.35t
ln8/20=-.35t
.916=-.35t


Is this correct?
t=2.61

 

[xceler8ari1984]

correct?

--------------------------------------------------------------------------------

Ok, this is what I have figured out

a) v(t)=(2000)e^-.35(3)
=(2000)e^-1.05
=$699.87


b) v'(3)=-.35*v(t)
=-.35(699.87)=-244.95


c) 800=(2000)e^-.35t
800/2000=e^-.35t
ln8/20=-.35t
.916=-.35t
t=2.61

 

[wjm11]

a) v(t)=(2000)e^-.35(3)
=(2000)e^-1.05
=$699.87


b) v'(3)=-.35*v(t)
=-.35(699.87)=-244.95


c) 800=(2000)e^-.35t
800/2000=e^-.35t
ln8/20=-.35t
.916=-.35t
t=2.61

a) and c) are correct. b) looks okay, too.