Answer:
a

b
![x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B-%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
c

Step-by-step explanation:
From the question we are told that

Now integrating both sides

Now taking the exponent of both sides

=> 
Let 
So

Now from the question we are told that

Hence

=> 
So

From the question we are told that

substituting for y

=> 
Now integrating both sides

Now taking the exponent of both sides

=> 
Let 
=> 
Now from the question we are told that

So

=> 
divide both side by 
=> 
So

=> 
=> ![x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
Generally as t tends to infinity ,
tends to zero
so

Answer:
4
Step-by-step explanation:
Answer:
ok
Step-by-step explanation:
Answer:
The Answer is B
Ive hade this test before
Laim descended 5,050.25 feet and his elevation changed by -5,050.25.