A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 7 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.

Answers

Answer 1

Answer:

The number A(t) of pounds of salt in the tank at time 't' is;

[tex]{A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100[/tex]

Step-by-step explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank,  = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

[tex]The \ rate \ of \ change \ in \ A(t) \ with \ time = \dfrac{dA}{dt}[/tex]

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

[tex]\dfrac{dA}{dt} = 21 - \dfrac{A(t)}{100}[/tex]

Therefore;

[tex]\dfrac{dA}{dt} + \dfrac{A(t)}{100}= 21[/tex]

The integrating factor is [tex]e^{\int\limits {\frac{1}{100} } \, dx } = e^{\frac{x}{100} }[/tex]

[tex]e^{\dfrac{t}{100} } \times \dfrac{dA}{dt} + e^{\dfrac{t}{100} } \times\dfrac{A(t)}{100}= e^{\dfrac{t}{100} } \times21[/tex]

[tex]\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21[/tex]

Using an online tool, we get;

[tex]{A(t)}= c_1 \times e^{\dfrac{-t}{100} } + 2,100[/tex]

At time t = 0, A(t) = 0

We get;

[tex]0= c_1 \times e^{\dfrac{-0}{100} } + 2,100[/tex]

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

[tex]{A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100[/tex]


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Answer:

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Step-by-step explanation:

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Answer:

The 20-count batteries have a lower unit price of $0.76 per battery

Step-by-step explanation:

Unit price means how much does it cost if you buy one of the item. To calculate this take the amount of money it cost and divide by the amount of object you have.

In the case of this question we know that you can buy 8 batteries for $6.16 and 20 batteries for $15.20. Let's find the unit cost (how much for one battery) for each of these packs. To calculate that we can take $6.16/8 batteries = $0.77 per battery for the 8 pack and $15.20/20 = $0.76 per battery for the 20 pack. Therefore, the 20-count batteries have a lower unit price of $0.76 per battery.

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Answers

Answer:

X= 67

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

if you're in k12

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Answers

Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides. 

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We have,

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A rhombus is a quadrilateral with all four sides of equal length.

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The two diagonals of a rhombus are of equal length. If we label the endpoints of the diagonals as E, F, G, and H, then EF = GH.

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All rhombuses have equal sides and diagonals that bisect each other at right angles.

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Answers

Answer:

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Step-by-step explanation:

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Answer:

B

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

Binomial distribution

Step-by-step explanation:

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Answers

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Step-by-step explanation:

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11

-

 4

____

7

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Answers

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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