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3 years ago

How many dogs do I have if I give two away

Mathematics

1 answer:

Helen [10]3 years ago

3 0

Answer:

Two less dogs

Your situation can be defined as x-2, where x was the number of dogs. Since the number of original dogs was not specified, two less dogs is the corrext answer

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Help<br> plz (A) 4.00<br> (B) 0.25<br> (C) 1.00<br> (D)0.10

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Answer:

B. 0.25

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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

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(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

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

By expanding the previous equation:

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

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

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

Since there is no accumulation within the tank, expression is simplified to this:

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

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$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

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

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

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