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2 years ago

Need help?????????!!!!!!!

Mathematics

1 answer:

julsineya [31]2 years ago

3 0

Answer:

73 and -23 hope this helped hon :)

Step-by-step explanation:

y=-8x+9

plug in -8

y=-8(-8)+9

y=64+9

y=73

y=-8x+9

plug in 4

y=-8(4)+9

y=-32+9

y=-23

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2 years ago

What numbers multiply to -30 and add to -3

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Let the numbers be x and y. Then xy = -30 and x+y = -3.

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Multiply all 3 terms by x: x^2 - 30 = -3x, so x^2 + 3x - 10 = 0

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3 years ago

Divide 6 divided by 3/5 Answer only if u know please:)

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2 years ago

An Air Force plane flew to the

Free_Kalibri [48]

Answer:

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Step-by-step explanation:

Time taken for trip : t + 1

Time taken for return : t

Avg. speed (trip) : 256 km/h

Avg. speed (return) : 320 km/h

The distances covered on both trips are equal.

⇒ Distance (trip) = Distance (return)

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2 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

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:

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

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:

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

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:

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

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

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

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

The solution of this equation is:

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

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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3 years ago