Answer:
C
Step-by-step explanation:
3/5 + 1/4 = 0.85
0.85 as a fraction is 17/20
So the answer is C
Answer:
Step-by-step explanation:
![\frac{(x + 6)}{(x - 4)} = - 16 \\ \\ x + 6 = - 16(x - 4) \\ x + 6 = - 16x + 64 \\ x + 16x = 64 - 6 \\ 17x = 58 \\ x = \frac{58}{17} \\ x = 3\frac{7}{17} \\](https://tex.z-dn.net/?f=%20%5Cfrac%7B%28x%20%2B%206%29%7D%7B%28x%20-%204%29%7D%20%20%3D%20%20-%2016%20%5C%5C%20%20%5C%5C%20x%20%2B%206%20%3D%20%20-%2016%28x%20-%204%29%20%5C%5C%20x%20%2B%206%20%3D%20%20-%2016x%20%2B%2064%20%5C%5C%20x%20%2B%2016x%20%3D%2064%20-%206%20%5C%5C%2017x%20%3D%2058%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B58%7D%7B17%7D%20%20%5C%5C%20x%20%3D%20%203%5Cfrac%7B7%7D%7B17%7D%20%20%5C%5C)
Step-by-step explanation:
2 1/2 + 1 1/16
Adding whole number parts and fraction parts together,
(2+1) + (1/2 + 1/16)
= 3 + (1*8/2*8 + 1/16)
= 3 + (8/16 + 1/16)
= 3 + 9/16
= 3 and 9/16
At any time
(min), the volume of solution in the tank is
![500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}](https://tex.z-dn.net/?f=500%5C%2C%5Cmathrm%7Bgal%7D%2B%5Cleft%285%5Cdfrac%7B%5Crm%20gal%7D%7B%5Crm%20min%7D-5%5Cdfrac%7B%5Crm%20gal%7D%7B%5Crm%20min%7D%5Cright%29t%3D500%5C%2C%5Cmathrm%7Bgal%7D)
If
is the amount of salt in the tank at any time
, then the solution has a concentration of
.
The net rate of change of the amount of salt in the solution,
, is the difference between the amount flowing in and the amount getting pumped out:
![A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)](https://tex.z-dn.net/?f=A%27%28t%29%3D%5Cleft%285%5Cdfrac%7B%5Crm%20gal%7D%7B%5Crm%20min%7D%5Cright%29%5Cleft%28%5Cleft%282%2B%5Csin%5Cdfrac%20t4%5Cright%29%5Cdfrac%7B%5Crm%20lb%7D%7B%5Crm%20gal%7D%5Cright%29-%5Cleft%285%5Cdfrac%7B%5Crm%20gal%7D%7B%5Crm%20min%7D%5Cright%29%5Cleft%28%5Cdfrac%7BA%28t%29%7D%7B50%7D%5Cdfrac%7B%5Crm%20lb%7D%7B%5Crm%20gal%7D%5Cright%29)
Dropping the units and simplifying, we get the linear ODE
![A'=10+5\sin\dfrac t4-\dfrac A{10}](https://tex.z-dn.net/?f=A%27%3D10%2B5%5Csin%5Cdfrac%20t4-%5Cdfrac%20A%7B10%7D)
![10A'+A=100+50\sin\dfrac t4](https://tex.z-dn.net/?f=10A%27%2BA%3D100%2B50%5Csin%5Cdfrac%20t4)
Multiplying both sides by
allows us to identify the left side as a derivative of a product:
![10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}](https://tex.z-dn.net/?f=10e%5E%7B10t%7DA%27%2Be%5E%7B10t%7DA%3D%5Cleft%28100%2B50%5Csin%5Cdfrac%20t4%5Cright%29e%5E%7B10t%7D)
![\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}](https://tex.z-dn.net/?f=%5Cleft%28e%5E%7B10%7DtA%5Cright%29%27%3D%5Cleft%28100%2B50%5Csin%5Cdfrac%20t4%5Cright%29e%5E%7B10t%7D)
![e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt](https://tex.z-dn.net/?f=e%5E%7B10t%7DA%3D%5Cdisplaystyle%5Cint%5Cleft%28100%2B50%5Csin%5Cdfrac%20t4%5Cright%29e%5E%7B10t%7D%5C%2C%5Cmathrm%20dt)
Integrate and divide both sides by
to get
![A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}](https://tex.z-dn.net/?f=A%28t%29%3D10-%5Cdfrac%7B200%7D%7B1601%7D%5Ccos%5Cdfrac%20t4%2B%5Cdfrac%7B8000%7D%7B1601%7D%5Csin%5Cdfrac%20t4%2BCe%5E%7B-10t%7D)
The tanks starts off with 30 lb of salt, so
and we can solve for
to get a particular solution of
![A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}](https://tex.z-dn.net/?f=A%28t%29%3D10-%5Cdfrac%7B200%7D%7B1601%7D%5Ccos%5Cdfrac%20t4%2B%5Cdfrac%7B8000%7D%7B1601%7D%5Csin%5Cdfrac%20t4%2B%5Cdfrac%7B32%2C220%7D%7B1601%7De%5E%7B-10t%7D)