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inessss [21]
4 years ago
10

A plumber works 80 hours. He charges $60 plus $50 per hour. Which equation shows this situation plz help

Mathematics
1 answer:
Marianna [84]4 years ago
6 0
60+50(80)=Y


Y= Money Made
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Caroline took a math and got 19 out of 25 questions correct. What percent of the question did she answer correctly? What percent
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Answer:

Step-by-step explanation:

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What is the slope of the line?
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Draw a conclusion as to why zero students had a 54-inch shoelace
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A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and
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and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of A(t) is governed by the ODE,

\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}

or

A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}

Multiply both sides by the integrating factor e^{7t/1080} to condense the left side into the derivative of a product:

e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}

\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}

Integrate both sides:

e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt

e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C

Solve for A(t):

A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}

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0=\dfrac{378}5+C\implies C=-\dfrac{378}5

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\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5

or 75.6 kg of sugar.

7 0
3 years ago
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