Answer:
The answer is a
Step-by-step explanation:
23.95×12=287.4+135=422.4
39,500-422.4=39,077.6
28. You do 2x + (4x + 12) = 180. That gets you 6x + 12 = 180. Minus the 12 from both sides. 6x = 168. Divide both sides by 6 then you get 28.
Answer:
96.46
Step-by-step explanation:
Volume of the Gallon =140 gallons
At time t=0, V(0)=40 gallons
![\frac{dV}{dt}=Rate \:In - Rate \:Out \\\frac{dV}{dt}=3\frac{gal}{min} - 1\frac{gal}{min} =2\frac{gal}{min}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3DRate%20%5C%3AIn%20-%20Rate%20%5C%3AOut%20%5C%5C%5Cfrac%7BdV%7D%7Bdt%7D%3D3%5Cfrac%7Bgal%7D%7Bmin%7D%20%20-%201%5Cfrac%7Bgal%7D%7Bmin%7D%20%20%3D2%5Cfrac%7Bgal%7D%7Bmin%7D)
V(t)=2t+k, Where K is a constant.
Since V(0)=40
V(0)=2(0)+k=40
k=40
V(t)=40+2t
The Change in Salt Concentration
![\frac{dC}{dt}=Rate \:In - Rate \:Out \\\frac{dC}{dt}=1\frac{lb}{gallon}*3\frac{gallon}{minute}-1\frac{lb}{gallon}*\frac{C}{40+2t} \frac{gallon}{minute}\\\frac{dC}{dt}=3-\frac{C}{40+2t}](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%3DRate%20%5C%3AIn%20-%20Rate%20%5C%3AOut%20%5C%5C%5Cfrac%7BdC%7D%7Bdt%7D%3D1%5Cfrac%7Blb%7D%7Bgallon%7D%2A3%5Cfrac%7Bgallon%7D%7Bminute%7D-1%5Cfrac%7Blb%7D%7Bgallon%7D%2A%5Cfrac%7BC%7D%7B40%2B2t%7D%20%5Cfrac%7Bgallon%7D%7Bminute%7D%5C%5C%5Cfrac%7BdC%7D%7Bdt%7D%3D3-%5Cfrac%7BC%7D%7B40%2B2t%7D)
Rearranging
![\frac{dC}{dt}+\frac{C}{40+2t}=3](https://tex.z-dn.net/?f=%5Cfrac%7BdC%7D%7Bdt%7D%2B%5Cfrac%7BC%7D%7B40%2B2t%7D%3D3)
Using Integrating Factor: ![e^{\int\frac{1}{40+2t} dt} =e^{ln|40+2t|}=40+2t](https://tex.z-dn.net/?f=e%5E%7B%5Cint%5Cfrac%7B1%7D%7B40%2B2t%7D%20dt%7D%20%3De%5E%7Bln%7C40%2B2t%7C%7D%3D40%2B2t)
![\frac{dC(40+2t)}{dt}=3(40+2t)](https://tex.z-dn.net/?f=%5Cfrac%7BdC%2840%2B2t%29%7D%7Bdt%7D%3D3%2840%2B2t%29)
![\int d[C(40+2t)]=\int(120+6t)dt\\C(40+2t)=120t+3t^2+k_0\\C(t)=\frac{120t+3t^2}{40+2t}+\frac{k_0}{40+2t}](https://tex.z-dn.net/?f=%5Cint%20d%5BC%2840%2B2t%29%5D%3D%5Cint%28120%2B6t%29dt%5C%5CC%2840%2B2t%29%3D120t%2B3t%5E2%2Bk_0%5C%5CC%28t%29%3D%5Cfrac%7B120t%2B3t%5E2%7D%7B40%2B2t%7D%2B%5Cfrac%7Bk_0%7D%7B40%2B2t%7D)
At t=0, C(0)=0.125, k_0=5
![C(t)=\frac{120t+3t^2}{40+2t}+\frac{5}{40+2t}\\C(t)=\frac{120t+3t^2+5}{40+2t}](https://tex.z-dn.net/?f=C%28t%29%3D%5Cfrac%7B120t%2B3t%5E2%7D%7B40%2B2t%7D%2B%5Cfrac%7B5%7D%7B40%2B2t%7D%5C%5CC%28t%29%3D%5Cfrac%7B120t%2B3t%5E2%2B5%7D%7B40%2B2t%7D)
At V(t)=140
V(t)=40+2t=140
t=50
![C(50)=96.46](https://tex.z-dn.net/?f=C%2850%29%3D96.46)
Just the tank overflows, C(t)=96.46
Answer:yes
Step-by-step explanation:
Answer:
1a) C = 600 + 5n
1b) R = 6n
Step-by-step explanation: