First to answer gets brainlest <33

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$

Recall that when t=0, A(t)=20 (our initial condition)

$20=240+Ce^{-\frac{0}{40}}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-\frac{t}{40}}$

3 0

3 years ago

Two Friends met up for coffee and dessert. Each friend got a medium coffee and two cupcakes, and one friend bought a mug for $12

marishachu [46]

Answer: each cupcake was 6.75

Step-by-step explanation: 12.00 + 3.50 =15.50 29.00- 15.50 is 13.50 divide by 2 boom 6.75 is each cupcake

3 0

3 years ago

Someone help me with this question please I don’t know

ra1l [238]

The magic of the internet will help!

6 0

3 years ago

Select the correct answer.

mariarad [96]

The answer to this Question is A

What is Absolute Value Function or so called Modulus Function or Mod function?

It is a function that takes all real values as input and returns same values but +ve in nature eg if you give input as 2, output will be 2 and if you give input as -2, you will get output 2 again

Solution:

We have expression Ix-5I + 2

we know that I anything I , called as mod of anything, is always greater than or equal to zero so if we take its minimum value as 0 and we get 0 + 2 = 2

means this expression cant be less than 2 hence it will be greater than or equal to 2

because minimum value of mod is 0 and we are adding 2 to it hence overall the expression will be greater than 2 only

To learn more about Absolute Value Function click the link:

brainly.com/question/28478005

#SPJ9

4 0

1 year ago

Graph the functions y=.05x

lisabon 2012 [21]

The slope is .05
the y intercept is 0

6 0

3 years ago