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galina1969 [7]
3 years ago
12

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

; it mixes with the solution there, and then the mixture is pumped out at a rate of 5 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal. A(t) = Incorrect: Your answer is incorrect. lb Without actually graphing, conjecture what the solution curve of the IVP should look like. Then use a graphing utility to plot the graph of the solution on the interval [0, 300].
Mathematics
1 answer:
Dmitry [639]3 years ago
6 0

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

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a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r
galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ---->  the area of the second square painting

D -----> the difference of the areas

we have

A1=81x^{2}-90x-25

A2=25x^{2}+30x+9

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

D=(25x^{2}+30x+9)-(81x^{2}-90x-25)

D=(-56x^{2}+120x+34)

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

D=(81x^{2}-90x-25)-(25x^{2}+30x+9)

D=(56x^{2}-120x-34)

4 0
3 years ago
NEED HELP!!!
AnnyKZ [126]

Answer:

see explanation

Step-by-step explanation:

(1)

\frac{6}{2} = \frac{4}{p} ( cross- multiply )

6p = 8 ( divide both sides by 6 )

p = \frac{8}{6} = \frac{4}{3}

(2)

\frac{n}{4} = \frac{8}{7} ( cross- multiply )

7n = 32 ( divide both sides by 7 )

n = \frac{32}{7}

(3)

\frac{5}{3} = \frac{x}{4} ( cross- multiply )

3x = 20 ( divide both sides by 3 )

x = \frac{20}{3}

6 0
3 years ago
What is the solution set of {x | x < 2} {x | x ≥ 2}? all numbers less than -5 and greater than 5 the empty set the numbers be
juin [17]

Answer:

The solution set is the empty set.

Step-by-step explanation:

{x | x < 2} is the set of all numbers less than 2. This means x can take values such as 1.99, 0, -2000 and so on. That is, all values less than 2.

{x | x ≥ 2} is the set of all numbers equal to or greater than 2. This means x can take values such as 2, 2.1, 5000, and so on. That is, 2 or any value greater than 2.

Since there is no sign between the two sets, the question is asking for the intersection between these two sets. That is, what elements are common to these two sets? As we can see, the two sets don't have any common element. Hence, their intersection is the empty set.

(Note that the union of these two sets would be the set of all real numbers as that includes all elements from either set).

3 0
2 years ago
Adam is going to cook a turkey for 14 people and wants to allow ¾ lb of turkey for each person.
ArbitrLikvidat [17]
I did (450*3/4=337.5) to get how much each person would eat, then I did (337.5*14=4725g) to see how big the turkey should be. I then converted grams to kilos (4725*0.001=4.725 kg), and I got how big the turkey would be. It turns out the appropriate size of the turkey would be medium, so then what I did was that I looked for the medium price of the turkey per kg, and found that it was £5.99 per kg. I did (4.725*5.99) and I got about £28.30. It would cost £28.30 for 14 people.
8 0
3 years ago
Read 2 more answers
Show all work to solve 3x^2 + 5x − 2 = 0
Ratling [72]

Answer:

x=1/3 or x=−2

if i can be brainliest that would be great

Step-by-step explanation:

Step 1: Add 2 to both sides.

3x^2+5x−2+2=0+2

3x^2+5x=2

Step 2: Since the coefficient of 3x^2 is 3, divide both sides by 3.

3x^2+5x/3=2/3

x^2+5/3x=2/3

Step 3: The coefficient of 5/3x is 5/3. Let b=5/3.

Then we need to add (b/2)^2=25/36 to both sides to complete the square.

Add 25/36 to both sides.

x^2+5/3x+25/36=2/3+25/36

x^2+5/3x+25/36=49/36

Step 4: Factor left side.

(x+5/6)^2=49/36

Step 5: Take square root.

x+5/6=±√49/36

Step 6: Add (-5)/6 to both sides.

x+5/6+ −5/6=

−5/6±√49/36x=−5/6±√49/36x=

1/3 or x=−2

3 0
3 years ago
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