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Tanya [424]
3 years ago
9

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

per gallon of water is pumped into the tank at the rate of 5 liters per minute, and the well-stirred mixture is pumped out at the same rate. Let A (t) represent the amount of salt in the tank at time t. The amount of salt in the tank at time t is:
Mathematics
1 answer:
goblinko [34]3 years ago
7 0

Answer:

A(t)=600-300e^t^/^3^0^0

Step-by-step explanation:

let A(t) represent the amount of salt in the tank at time t. ThenA(t)=300 \ at \ t=0 and

\frac{dA}{dt}=rate \ in- \ rate \ out\\\frac{dA}{dt}=2-\frac{A}{300}\\\\\frac{dA}{dt}+\frac{A}{300}=2\\\\e^t^/^3^0^0 \frac{dA}{dt}+\frac{e^t^/^3^0^0}{300}A=2e^t^/^3^0^0\\\\e^t^/^3^0^0A=\int2e^t^/^3^0^0dt=600e^t^/^3^0^0+C\\\\A(t)=600+Ce^t^/^3^0^0#To find C we use  A=300, \ when \ t=0

Hence:

300=600+Ce^t^/^3^0^0\\C=-300

#The amount of salt in tank at time time is expressed as:

A(t)=600-300e^t^/^3^0^0

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Given:

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Solution:

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Sum of first n terms of a GP is

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S_2=\dfrac{a(1-r^2)}{1-r}

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Putting n=4, we get

S_4=\dfrac{a(1-r^4)}{1-r}

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Divide both sides by 6.

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\dfrac{5-4}{4}=r^2

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Taking square root on both sides, we get

\pm \sqrt{\dfrac{1}{4}}=r

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6=a(1+0.5)  

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S_6=7.875

Case 2: If r is negative, then using (ii) we get

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Therefore, the sum of the first six terms is 7.875.

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