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xenn [34]
3 years ago
11

- 3 + 3x = 2x – 13 I have to use more characters...

Mathematics
1 answer:
DENIUS [597]3 years ago
6 0

- 3 + 3x = 2x – 13

Bring -3 to the right side of the equation by adding 3 to both sides

(- 3 + 3) + 3x = 2x – 13 + 3

0 + 3x = 2x - 10

3x = 2x - 10

Bring 2x to the other side by subtracting 2x to both sides

3x - 2x = 2x - 2x - 10

x = -10

Check:

-3 + 3(-10) = 2(-10) - 13

-3 + (-30) = -20 - 13

-33 = -33

Hope this helped!

~Just a girl in love with Shawn Mendes

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A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo
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(b) Rate at which the chlorine is entering the pool.

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= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}

Taking the integral of both sides

\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}

Recall that when t=0, A(t)=3000 (our initial condition)

3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}

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0.9871 - 0.1587 = 0.8284, thus:

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A similar problem is given at brainly.com/question/25151638

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