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andriy [413]
3 years ago
14

In Practing Partial Products Multiplictaion What im Going to do to 65 × 32 ?​

Mathematics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

2080

Step-by-step explanation:

<u>Step 1:  Multiply</u>

65 × 32

60*30 + 60*2 + 5*30 + 5*2

1800 + 120 + 150 + 10

2080

Answer:  2080

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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

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Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

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By expanding the previous equation:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)

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V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0

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c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

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The solution of this equation is:

c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})

The salt concentration after 8 minutes is:

c(8) = 0.166 \frac{pounds}{gallon}

The instantaneous amount of salt in the tank is:

m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds

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