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3 years ago

9

Please help! When water vapor gets cold it turns into..

2 answers:

Flura [38]3 years ago

8 0

I think the answer is

B liquid

Water vapor in the air reaches its dew point as it cools in the air around the can, forming liquid drops of water.

andre [41]3 years ago

4 0

B liquid.

When the water vapor cools enough, the attractions between the molecules bring them together. This causes the water vapor to change state and become tiny drops of liquid water. The process of changing from a gas to a liquid is called condensation.

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a group of 4 pupils can take 2 weeks to do a certain job.how long will 12 pupils take to complete the job if they work at the sa

Ksivusya [100]

Answer:

6 weeks

Step-by-step explanation:

Set up a proportion where x is the amount of time with 12 pupils:

$\frac{4}{2}$ = $\frac{12}{x}$

Cross multiply and solve for x:

4x = 24

x = 6

So, it will take 12 pupils 6 weeks to complete the job.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=32.5%20%3D%2030%20%2B%20%20%5Cfrac%7B20%20-%20%2840%20%2B%20x%29%7D%7B12%7D%20%20%5Ctimes%2010

erastova [34]

Answer:

$x = - 23$

Step-by-step explanation:

$\frac{20 - (40 + x)}{12} \times 10 = 32.5 - 30 = 2.5$

$= > \frac{20 - (40 + x)}{12} = 0.25$

$= > \frac{20 - (40 + x)}{12} = \frac{1}{4}$

$= > 20 - (40 + x) = \frac{12}{4} = 3$

$= > 40 + x = 20 - 3 = 17$

$= > x = 17 - 40 = - 23$

7 0

3 years ago

What could the answer be?

VashaNatasha [74]

Answer:

it 66

Step-by-step explanation:

8 0

3 years ago

This question has three parts. Answer the parts in order.

nalin [4]

Answer:

The area of the smallest section is $A_{1}=100yd^{2}$

The area of the largest section is $A_{2}=625yd^{2}$

The area of the remaining section is $A_{3}=250yd^{2}$

Step-by-step explanation:

Please see the picture below.

1. First we are going to name the side of the larger square as x.

As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:

- Area of the first section:

$A_{1}=10yd*10yd$

$A_{1}=100yd^{2}$

- Area of the second section:

$A_{2}=x^{2}$ (Eq.1)

- Area of the third section:

$A_{3}=width*length$

$A_{3}=10yd*x$ (Eq.2)

2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:

$A_{1}+A_{2}+A_{3}=975yd^{2}$

Replacing values:

$100+x^{2}+10x=975$

Solving for x:

$x^{2}+10x-875=0$

$x=\frac{-10+\sqrt{100+(4*875)}}{2}$

$x=\frac{-10+\sqrt{3600}}{2}$

$x=\frac{-10+60}{2}$

$x=25$

3. Replacing the value of x in Eq.1 and Eq.2:

- From Eq.1:

$A_{2}=25^{2}$

$A_{2}=625yd^{2}$

- From Eq.2:

$A_{3}=10*25$

$A_{3}=250yd^{2}$

3 0

3 years ago

Please Help me it would mean the world to me

alexdok [17]

Answer:

I Can't read it

Step-by-step explanation:

7 0

3 years ago