Coplanar
do not intersect
answer is paralell lines
You can factor out the 4 and you would get....
![4(2b-1)](https://tex.z-dn.net/?f=4%282b-1%29)
Patel needs to solve the quadratic equation. The steps in the solution are shown below:
8x² + 16x +3 =0
Transpose 3, we have:
8x² + 16x = -3
Factor out 8, we have:
8 (x² + 2x) = -3
Divide both sides by 8, we have:
x² + 2x = -3/8
Add 2 in both sides, we have:
x² + 2x +2= -3/8 +2
(x+`1)² = -3/10
Answer:
![A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196](https://tex.z-dn.net/?f=%20A%2840%29%3D%20%5Cfrac%7B-200%7D%7B10%2B40%7D%20%2B4%20%2810%20%2B40%29%3D-4%2B200%20%3D%20196%20)
Step-by-step explanation:
For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by ![V= 10 +t](https://tex.z-dn.net/?f=%20V%3D%2010%20%2Bt)
Since the initial volume is 10 L and the volume increase at a rate of 1L/min.
For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:
![\frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}](https://tex.z-dn.net/?f=%20%5Cfrac%7BdA%7D%7Bdt%7D%3D%204%20%5Cfrac%7Bgr%7D%7BL%7D%20%2A2%20%5Cfrac%7BL%7D%7Bmin%7D%20-%20%5Cfrac%7BA%7D%7B10%2Bt%7D)
Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of ![\frac{A}{10+t}](https://tex.z-dn.net/?f=%20%5Cfrac%7BA%7D%7B10%2Bt%7D)
So then we can reorder the differential equation like this:
![\frac{dA}{dt} +\frac{A}{10+t} =8](https://tex.z-dn.net/?f=%20%5Cfrac%7BdA%7D%7Bdt%7D%20%2B%5Cfrac%7BA%7D%7B10%2Bt%7D%20%3D8)
We find the solution using the integration factor:
![\mu = -\int \frac{1}{10+t} dt = -ln(10+t)](https://tex.z-dn.net/?f=%20%5Cmu%20%3D%20-%5Cint%20%5Cfrac%7B1%7D%7B10%2Bt%7D%20dt%20%3D%20-ln%2810%2Bt%29)
And then the solution would be given by:
![A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})](https://tex.z-dn.net/?f=%20A%20%3D%20e%5E%7B-ln%20%2810%2Bt%29%7D%20%28%5Cint%20e%5E%7B%5Cint%20%5Cfrac%7B1%7D%7B10%2Bt%7D%20dt%7D%29)
And if we simplify this we got:
![A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)](https://tex.z-dn.net/?f=%20A%3D%20%5Cfrac%7B1%7D%7B10%2Bt%7D%20%28c%20%2B%20%5Cint%20%2810%20%2Bt%29%208%20dt%29)
And after do the integral we got:
![A= \frac{c}{10+t} +4 (10 +t)](https://tex.z-dn.net/?f=%20A%3D%20%5Cfrac%7Bc%7D%7B10%2Bt%7D%20%2B4%20%2810%20%2Bt%29%20)
And using the initial condition t=0 A= 20 we have this:
![20 = \frac{c}{10} +40](https://tex.z-dn.net/?f=%2020%20%3D%20%5Cfrac%7Bc%7D%7B10%7D%20%2B40)
![c= -200](https://tex.z-dn.net/?f=%20c%3D%20-200)
So then we have this function for the solution of A:
![A= \frac{-200}{10+t} +4 (10 +t)](https://tex.z-dn.net/?f=%20A%3D%20%5Cfrac%7B-200%7D%7B10%2Bt%7D%20%2B4%20%2810%20%2Bt%29%20)
And now replacinf t= 40 we got:
![A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196](https://tex.z-dn.net/?f=%20A%2840%29%3D%20%5Cfrac%7B-200%7D%7B10%2B40%7D%20%2B4%20%2810%20%2B40%29%3D-4%2B200%20%3D%20196%20)