A tank contains 210 liters of fluid in which 30 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the rate at which the number of grams of salt in the tank is changing at time t.

Answers

Answer 1

Answer:

[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min

Step-by-step explanation:

The mass flow rate dA(t)/dt = mass flowing in - mass flowing out

Since water is pumped in at a rate of 3 L/min, and it contains no salt, the concentration in is thus 0 g/L. the mass flow in is thus 0 g/L × 3 L/min = 0 g/min.

Let A(t) be the mass present at any time, t. The concentration at any time ,t is thus A(t)/volume = A(t)/210. Since water flows out at a rate of 3 L/min, the mass flow out is thus, A(t)/210 g/L × 3 L/min = A(t)/70 g/min.

So, dA(t)/dt = mass flowing in - mass flowing out

dA(t)/dt = 0 g/min - A(t)/70 g/min

dA(t)/dt =  - A(t)/70 g/min

 Since the tank initially contains 30 g of salt, the initial mass of salt in the tank is 30 g. So A(0) = 30

So, the initial value problem is thus  

dA(t)/dt =  - A(t)/70 , A(0) = 30

Separating the variables, we have

dA(t)/A(t) = -dt/70

Integrating both sides, we have

∫dA(t)/A(t) = ∫-dt/70

㏑A(t) = -t/70 + C

taking exponents of both sides, we have

A(t) = exp(-t/70 + C)

A(t) = exp(-t/70)expC

[tex]A(t) = e^{-t/70}e^{C}\\A(t) = Ce^{-t/70} whereC = e^{C}[/tex]

A(0) = 30, So

[tex]A(0) = Ce^{-0/70}\\30 = Ce^{0}\\C = 30[/tex]

[tex]A(t) = 30e^{-t/70}[/tex]

The rate at which the number of grams of salt in the tank is changing at time t is

[tex]dA(t)/dt = \frac{d30e^{-t/70} }{dt} \\dA(t)/dt = -\frac{30e^{-t/70} }{70}\\dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex]

So, the rate at which the number of grams of salt in the tank is changing at time t is

[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min


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Step-by-step explanation:

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First let's reduce the feet to miles

there are 5280 feet in a mile therefore

26400 feet=5 miles

31680 feet=6 miles

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We can compare these as fractions to see which is steeper. This can be viewed as slope and the origin (0,0) is the airport.

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1/20=?=5/96

Now we know that 5/100 =1/20 so 5/96 must be bigger than 5/100 because you are dividing by a smaller number.

so 1/20<5/96

So Jet B is descending steeper than Jet A.

As for linear model, I don't exactly know what your teacher means but I think I actually used the linear model when I'm thinking of steepness as slope in the coordinate plane, I will include a picture.

In this extremely zoomed out graph, you can see the blue line is just slighly higher than the red line(slope as in explanation is way easier to tell) this could be seen as the linear model) :) Hope it helped!

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Answers

Answer:

-75 x^4 + 85 x^5 - 32 x^6 + 4 x^7

Step-by-step explanation:

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Expand and get

-75 x^4 + 85 x^5 - 32 x^6 + 4 x^7

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The given expression is

[tex](x-\dfrac{5}{2})^2(x-1)^2(x-0)^4=0[/tex]

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[tex]X^4(x^4-7x^3+12x^2-\dfrac{35x}{2} +\dfrac{25x^2}{4} +\dfrac{25}{4} =0[/tex]

Thus the standard form will be  

[tex]X^4(x^4-7x^3+12x^2-\dfrac{35x}{2} +\dfrac{25x^2}{4} +\dfrac{25}{4} =0[/tex]

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Answer:

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Answer:

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Answers

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Step-by-step explanation:

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Answers

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Answers

Answer:

Step-by-step explanation:

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Answers

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‍♀️‍♀️

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Answers

x=the number
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Answer:

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editing for clarity.

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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