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2 years ago

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Item 3 The Hampton family used 17,158 kilowatts of electricity last year. They used about the same amount of electricity each we

ek. About how many kilowatts of electricity did the Hampton family use each week? (There are 52 weeks in a year).

1 answer:

katen-ka-za [31]2 years ago

7 0

17,158/52= 329.96 rounded to the nearest tenth is 330

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I want to buy a car that cost 36,718 I also have a 9% sales tax to keep in mind

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Answer:

40,022.62

Step-by-step explanation:

9% of 36,718=3304.62

36,718+3304.62=40,022.62

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3 years ago

In 2004 pin header Marie walk 680 miles to the south pole the trip took 58 days suppose they travel the same distance each day w

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Answer: 12 miles

Step-by-step explanation:

From the question, we are informed that Marie walk 680 miles to the south pole and that the trip took 58 days and they travel the same distance each day.

The number of miles that they travel each day to the nearest whole number will be gotten by dividing 680 miles by 58. This will be:

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2 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

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$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

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Allushta [10]

23,718<span>

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3 years ago

A flagpole flying the Ohio state flag is 9/10 the height of a 30-foot-tall flagpole that is flying the U.S. Flag. What is the he

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The answer to this problem is 27

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