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mr Goodwill [35]

2 years ago

10

A scale drawing of a bedroom is shown below. The scale is 1 : 30. A rectangle is shown. The length of the rectangle is labeled 3

inches. The width of the rectangle is labeled 4 inches.
Show your work to determine the area of the room in square feet.

2 answers:

antoniya [11.8K]2 years ago

7 0

Here, scale is 1 : 30
We have length of scale = 3 in.
So, length of the room is 30 * 3 = 90 inch.

In the same manner, width of room = 30 * 4 = 120 in.

Now, Area = Length * Width
A = 90 * 120
A = 10800 In²
A = 10800 / 144 Ft² [ As 1 Ft² = 144 In² ]

In short, Your Answer would be 75 Ft²

Hope this helps!

Firdavs [7]2 years ago

3 0

Answer:

75 square feet.

Step-by-step explanation:

So all you're going to do is simply cross multiply.

1/30 = 4/x (we put x because we don't know what number goes there)

SO 1 x X = x= 4x 30+ 120 so, x=120 in.

Now we do the same to the other one.

1/30 = 3/x

1 x x = x= 3x 30 = 90 so, x=90 in.

Wait, we are not done yet. It also says to find the area of the room in sq ft.

We need to convert these numbers into feet. So divide 12 by the numbers.

90/ 12 = 7.5 120/12 = 10 Now, we can multiply to get the square feet.

7.5 x 10 = 75 square feet.

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son4ous [18]

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7 1/2 cups

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If Carol uses 1 1/4 cups of flour for 2 muffins, then she will need to repeat this conversion 6 times since she wants to make identical 12 muffins in all.

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7 0

2 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

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blondinia [14]

Answer:

1. 8/24

2. 9/27

3. 5/15

4. 10/30

5. 2/6

6.1/3

Step-by-step explanation:

Start with 1/3 then multiply that by the number that is given.

6 0

2 years ago

Please answer asap thanks

mylen [45]

its a or c not to sure

8 0

2 years ago