Salt flows in at a rate of (5 g/L)*(3 L/min) = 15 g/min.
Salt flows out at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.
So the net flow rate of salt, given by
in grams, is governed by the differential equation,
![x'(t)=15-\dfrac{3x(t)}{10}](https://tex.z-dn.net/?f=x%27%28t%29%3D15-%5Cdfrac%7B3x%28t%29%7D%7B10%7D)
which is linear. Move the
term to the right side, then multiply both sides by
:
![e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}](https://tex.z-dn.net/?f=e%5E%7B3t%2F10%7Dx%27%2B%5Cdfrac%7B3e%5E%7Bt%2F10%7D%7D%7B10%7Dx%3D15e%5E%7B3t%2F10%7D)
![\implies\left(e^{3t/10}x\right)'=15e^{3t/10}](https://tex.z-dn.net/?f=%5Cimplies%5Cleft%28e%5E%7B3t%2F10%7Dx%5Cright%29%27%3D15e%5E%7B3t%2F10%7D)
Integrate both sides, then solve for
:
![e^{3t/10}x=50e^{3t/10}+C](https://tex.z-dn.net/?f=e%5E%7B3t%2F10%7Dx%3D50e%5E%7B3t%2F10%7D%2BC)
![\implies x(t)=50+Ce^{-3t/10}](https://tex.z-dn.net/?f=%5Cimplies%20x%28t%29%3D50%2BCe%5E%7B-3t%2F10%7D)
Since the tank starts with 5 g of salt at time
, we have
![5=50+C\implies C=-45](https://tex.z-dn.net/?f=5%3D50%2BC%5Cimplies%20C%3D-45)
![\implies\boxed{x(t)=50-45e^{-3t/10}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bx%28t%29%3D50-45e%5E%7B-3t%2F10%7D%7D)
The time it takes for the tank to hold 20 g of salt is
such that
![20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}](https://tex.z-dn.net/?f=20%3D50-45e%5E%7B-3t%2F10%7D%5Cimplies%20t%3D%5Cdfrac%7B20%7D3%5Cln%5Cdfrac32%5Capprox2.7031%5C%2C%5Cmathrm%7Bmin%7D)
T=ir/p i think this is correct
Solve the equation for k by finding a, b, and c of the quadratic then applying the quadratic formula.
Answer =
K= 1/2 , 2/5
Answer: x-7
Step-by-step explanation:
![f(x)+g(x)=-3x-5+4x-2=\boxed{x-7}](https://tex.z-dn.net/?f=f%28x%29%2Bg%28x%29%3D-3x-5%2B4x-2%3D%5Cboxed%7Bx-7%7D)
Answer:
Step-by-step explanation:
when s(t)=0
t^4-20t^2=0
t^2(t^2-20)=0
![t^{2} (t+2\sqrt{5} )(t-2\sqrt{5} )=0\\t=0,-2\sqrt{5} ,2\sqrt{5} \\so particle passes through origin when t=0,-2\sqrt{5} and 2\sqrt{5} \\\frac{x}{y} \frac{ds}{dt} =4t^3-40t\\when particle is motionless \frac{ds}{dt}=0\\4t^3-40t=0\\4t(t^2-10)=0\\t(t+\sqrt{10} )(t-\sqrt{10} )=0\\particle is motionless when t=0,-\sqrt{10} ~or~\sqrt{10}](https://tex.z-dn.net/?f=t%5E%7B2%7D%20%28t%2B2%5Csqrt%7B5%7D%20%29%28t-2%5Csqrt%7B5%7D%20%29%3D0%5C%5Ct%3D0%2C-2%5Csqrt%7B5%7D%20%2C2%5Csqrt%7B5%7D%20%5C%5Cso%20particle%20passes%20through%20origin%20when%20t%3D0%2C-2%5Csqrt%7B5%7D%20and%202%5Csqrt%7B5%7D%20%5C%5C%5Cfrac%7Bx%7D%7By%7D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%3D4t%5E3-40t%5C%5Cwhen%20particle%20is%20motionless%20%5Cfrac%7Bds%7D%7Bdt%7D%3D0%5C%5C4t%5E3-40t%3D0%5C%5C4t%28t%5E2-10%29%3D0%5C%5Ct%28t%2B%5Csqrt%7B10%7D%20%29%28t-%5Csqrt%7B10%7D%20%29%3D0%5C%5Cparticle%20is%20motionless%20when%20t%3D0%2C-%5Csqrt%7B10%7D%20~or~%5Csqrt%7B10%7D)