Answer:
![P=M(1-e^{-kt})](https://tex.z-dn.net/?f=P%3DM%281-e%5E%7B-kt%7D%29)
Step-by-step explanation:
The relation between the variables is given by
![\frac{dP}{dt} = k(M- P)](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%20%3D%20k%28M-%20P%29)
This is a separable differential equation. Rearranging terms:
![\frac{dP}{(M- P)} = kdt](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7B%28M-%20P%29%7D%20%3D%20kdt)
Multiplying by -1
![\frac{dP}{(P- M)} = -kdt](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7B%28P-%20M%29%7D%20%3D%20-kdt)
Integrating
![ln(P-M)=-kt+D](https://tex.z-dn.net/?f=ln%28P-M%29%3D-kt%2BD)
Where D is a constant. Applying expoentials
![P-M=e^{-kt+D}=Ce^{-kt}](https://tex.z-dn.net/?f=P-M%3De%5E%7B-kt%2BD%7D%3DCe%5E%7B-kt%7D)
Where
, another constant
Solving for P
![P=M+Ce^{-kt}](https://tex.z-dn.net/?f=P%3DM%2BCe%5E%7B-kt%7D)
With the initial condition P=0 when t=0
![0=M+Ce^{-k(0)}](https://tex.z-dn.net/?f=0%3DM%2BCe%5E%7B-k%280%29%7D)
We get C=-M. The final expression for P is
![P=M-Me^{-kt}](https://tex.z-dn.net/?f=P%3DM-Me%5E%7B-kt%7D)
![P=M(1-e^{-kt})](https://tex.z-dn.net/?f=P%3DM%281-e%5E%7B-kt%7D%29)
Keywords: performance , learning , skill , training , differential equation
Answer:
Number of jars filled is 3
(approximately 4).
Step-by-step explanation:
Initial amount of salt = 1/2 cup.
Amount of salt left = 1/9 cup.
Amount of salt used = 1/2 - 1/9
= ![\frac{9-2}{18}](https://tex.z-dn.net/?f=%5Cfrac%7B9-2%7D%7B18%7D)
= 7/18
The size of each jar = 1/10 cup.
Amount of salt used = size of the each jar x number of jars filled
7/18 = 1/10 x number of jars filled
number of jars filled = 7/18 ÷ 1/10
= 7/18 x 10/1
= 70/18
= 3 ![\frac{8}{9}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B9%7D)
= 3.9
The number of jars filled is approximately 4.
Answer: (0,10)
Step-by-step explanation: y=mx+b
y=12.5x+10
0=12.5x+10
10 is y because its on the outside
Answer:
Area = 140 in2
Step-by-step explanation:
Area = (15 +15+4+6)*7 /2
= 140 in2
216 yards.
Hope this helped!