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ss7ja [257]
2 years ago
9

Can someone pls help me...

Mathematics
1 answer:
joja [24]2 years ago
5 0
It is not a function and that is becuase a vertical line would pass through it twice
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A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t
nignag [31]
If A(t) is the amount of salt in the tank at time t, then the rate at which the amount of salt in the tank changes is given by

\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}
\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}

Let's drop the units for now. We have

\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24
e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}
\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}
e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt
e^{t/100}A(t)=2400e^{t/100}+C
A(t)=2400+Ce^{-t/100}

We're given that the water is pure at the start, so A(0)=0, giving

A(0)=0=2400+Ce^{-0/100}\implies C=-2400

So the amount of salt in the tank (in lbs) at time t is

A(t)=2400\left(1-e^{-t/100}\right)
4 0
2 years ago
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