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Rudiy27
3 years ago
9

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

; it mixes with the solution there, and then the mixture is pumped out at a rate of 3 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.
Mathematics
1 answer:
sesenic [268]3 years ago
3 0
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
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Volgvan

Given:

The graph of a proportional relationship.

To find:

The constant of proportionality, the value of y when x is 24 and the value of x when y is 108.

Solution:

If y is directly proportional to x, then

y\propto x

y=kx             ...(i)

Where, k is the constant of proportionality.

The graph of proportional relationship passes through the point (5,15).

Substituting x=5 and y=15 in (i), we get

15=k(5)

\dfrac{15}{5}=k

3=k

Therefore, the constant of proportionality is 3.

Substituting k=3 in (i) to get the equation of the proportional relationship.

y=3x              ...(ii)

Substituting x=24 in (ii), we get

y=3(24)

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Therefore, the value of y is 72 when x is 24.

Substituting y=108 in (ii), we get

108=3x

\dfrac{108}{3}=y

36=y

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3 years ago
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Murljashka [212]

Answer:

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Step-by-step explanation:

The question is badly formatted so cannot match the answer choice

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