Your answers for (a) and (c) are correct.
(b) Salt flows into the tank at a rate of
![\left(0.025 \dfrac{\rm kg}{\rm L}\right) \left(5 \dfrac{\rm L}{\rm min}\right) = 0.125 \dfrac{\rm kg}{\rm min} = \dfrac18 \dfrac{\rm kg}{\rm min}](https://tex.z-dn.net/?f=%5Cleft%280.025%20%5Cdfrac%7B%5Crm%20kg%7D%7B%5Crm%20L%7D%5Cright%29%20%5Cleft%285%20%5Cdfrac%7B%5Crm%20L%7D%7B%5Crm%20min%7D%5Cright%29%20%3D%200.125%20%5Cdfrac%7B%5Crm%20kg%7D%7B%5Crm%20min%7D%20%3D%20%5Cdfrac18%20%5Cdfrac%7B%5Crm%20kg%7D%7B%5Crm%20min%7D)
If
is the amount of salt (in kg) in the tank at time
(in min), then the salt flows out of the tank at a rate of
![\left(\dfrac{A(t)}{1000+(5-5)t} \dfrac{\rm kg}{\rm L}\right) \left(5 \dfrac{\rm L}{\rm min}\right) = \dfrac{A(t)}{200} \dfrac{\rm kg}{\rm min}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7BA%28t%29%7D%7B1000%2B%285-5%29t%7D%20%5Cdfrac%7B%5Crm%20kg%7D%7B%5Crm%20L%7D%5Cright%29%20%5Cleft%285%20%5Cdfrac%7B%5Crm%20L%7D%7B%5Crm%20min%7D%5Cright%29%20%3D%20%5Cdfrac%7BA%28t%29%7D%7B200%7D%20%5Cdfrac%7B%5Crm%20kg%7D%7B%5Crm%20min%7D)
The net rate of change in the amount of salt in the tank at any time is then governed by the linear differential equation
![\dfrac{dA}{dt} = \dfrac18 - \dfrac{A(t)}{200}](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdt%7D%20%3D%20%5Cdfrac18%20-%20%5Cdfrac%7BA%28t%29%7D%7B200%7D)
![\dfrac{dA}{dt} + \dfrac{A(t)}{200} = \dfrac18](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdt%7D%20%2B%20%5Cdfrac%7BA%28t%29%7D%7B200%7D%20%3D%20%5Cdfrac18)
I'll solve this with the integrating factor method. The I.F. is
![\mu = \exp\left(\displaystyle \int \frac{dt}{200}\right) = e^{t/200}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cexp%5Cleft%28%5Cdisplaystyle%20%5Cint%20%5Cfrac%7Bdt%7D%7B200%7D%5Cright%29%20%3D%20e%5E%7Bt%2F200%7D)
Distributing
on both sides of the ODE gives
![e^{t/200} \dfrac{dA}{dt} + \dfrac1{200} e^{t/200} A(t) = \dfrac18 e^{t/200}](https://tex.z-dn.net/?f=e%5E%7Bt%2F200%7D%20%5Cdfrac%7BdA%7D%7Bdt%7D%20%2B%20%5Cdfrac1%7B200%7D%20e%5E%7Bt%2F200%7D%20A%28t%29%20%3D%20%5Cdfrac18%20e%5E%7Bt%2F200%7D)
![\dfrac d{dt} \left(e^{t/200} A(t)\right) = \dfrac18 e^{t/200}](https://tex.z-dn.net/?f=%5Cdfrac%20d%7Bdt%7D%20%5Cleft%28e%5E%7Bt%2F200%7D%20A%28t%29%5Cright%29%20%3D%20%5Cdfrac18%20e%5E%7Bt%2F200%7D)
Integrate both sides.
![\displaystyle \int \frac d{dt} \left(e^{t/200} A(t)\right) \, dt = \frac18 \int e^{t/200} \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%20d%7Bdt%7D%20%5Cleft%28e%5E%7Bt%2F200%7D%20A%28t%29%5Cright%29%20%5C%2C%20dt%20%3D%20%5Cfrac18%20%5Cint%20e%5E%7Bt%2F200%7D%20%5C%2C%20dt)
![e^{t/200} A(t) = \dfrac{200}8 e^{t/200} + C](https://tex.z-dn.net/?f=e%5E%7Bt%2F200%7D%20A%28t%29%20%3D%20%5Cdfrac%7B200%7D8%20e%5E%7Bt%2F200%7D%20%2B%20C)
![A(t) = 25 + Ce^{-t/200}](https://tex.z-dn.net/?f=A%28t%29%20%3D%2025%20%2B%20Ce%5E%7B-t%2F200%7D)
Given that
, we find
![50 = 25 + Ce^0 \implies C = 25](https://tex.z-dn.net/?f=50%20%3D%2025%20%2B%20Ce%5E0%20%5Cimplies%20C%20%3D%2025)
so that
![A(t) = 25 + 25e^{-t/200}](https://tex.z-dn.net/?f=A%28t%29%20%3D%2025%20%2B%2025e%5E%7B-t%2F200%7D)
Then the amount of salt in the tank after 1 hr = 60 min is
![A(60) = 25 + 25e^{-60/200} = \boxed{25 \left(1 + e^{-3/10}\right)}](https://tex.z-dn.net/?f=A%2860%29%20%3D%2025%20%2B%2025e%5E%7B-60%2F200%7D%20%3D%20%5Cboxed%7B25%20%5Cleft%281%20%2B%20e%5E%7B-3%2F10%7D%5Cright%29%7D)
Answer:
0.08333333333 or 1/2
Step-by-step explanation:
Answer:
9 plates
Step-by-step explanation:
The number of plates Kyle can make is the GCF of 9 and 18. Since 9 is a factor of 18, that GCF is 9.
Kyle can make 9 plates, each containing 1 avocado roll and 2 cucumber rolls.
Answer:
0.185
Step-by-step explanation:
Please mark as brainliest