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3 years ago

The image shows a hydrothermal vent what would geologists expect to find around this vent?

2 answers:

7 0

Hydrothermal Vents occur along divergent plate boundaries on the ocean floor. Hydrothermal vent systems give off plumes of hot, mineral-laden water with temperatures as high as 350 degrees Celcius. The water is ejected through a spout known as black smokers. Metal-rich minerals precipitate out of the circling sea water and forms ores of copper, zinc and iron.

sertanlavr [38]3 years ago

4 0

metal ore deposits is correct for plato

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Which method of showing Earth's surface is the most accurate?

Alexxandr [17]

Answer:

I think option B would be more accurate aND effective and easy for everyone to show Earth surface. No need for any other stuffs globe will do everything fine.

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3 years ago

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Which of the following terms is not a name for a disadvantaged squatter settlement in a city?

Alenkasestr [34]

I believe the answer is either A or D. However, I think D sounds the most fitting but like i said, it could be either one. :)

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3 years ago

What is the value of X6? <br><br> Show the solution.

wariber [46]

Answer : The value of $x_6$ is $\sqrt{7}$ .

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of $x_1$ .

Using Pythagoras theorem in triangle 1 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_1)^2=(1)^2+(1)^2$

$x_1=\sqrt{(1)^2+(1)^2}$

$x_1=\sqrt{2}$

Now we have to calculate the value of $x_2$ .

Using Pythagoras theorem in triangle 2 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_2)^2=(1)^2+(X_1)^2$

$(x_2)^2=(1)^2+(\sqrt{2})^2$

$x_2=\sqrt{(1)^2+(\sqrt{2})^2}$

$x_2=\sqrt{3}$

Now we have to calculate the value of $x_3$ .

Using Pythagoras theorem in triangle 3 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_3)^2=(1)^2+(X_2)^2$

$(x_3)^2=(1)^2+(\sqrt{3})^2$

$x_3=\sqrt{(1)^2+(\sqrt{3})^2}$

$x_3=\sqrt{4}$

Now we have to calculate the value of $x_4$ .

Using Pythagoras theorem in triangle 4 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_4)^2=(1)^2+(X_3)^2$

$(x_4)^2=(1)^2+(\sqrt{4})^2$

$x_4=\sqrt{(1)^2+(\sqrt{4})^2}$

$x_4=\sqrt{5}$

Now we have to calculate the value of $x_5$ .

Using Pythagoras theorem in triangle 5 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_5)^2=(1)^2+(X_4)^2$

$(x_5)^2=(1)^2+(\sqrt{5})^2$

$x_5=\sqrt{(1)^2+(\sqrt{5})^2}$

$x_5=\sqrt{6}$

Now we have to calculate the value of $x_6$ .

Using Pythagoras theorem in triangle 6 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_6)^2=(1)^2+(X_5)^2$

$(x_6)^2=(1)^2+(\sqrt{6})^2$

$x_6=\sqrt{(1)^2+(\sqrt{6})^2}$

$x_6=\sqrt{7}$

Therefore, the value of $x_6$ is $\sqrt{7}$ .

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3 years ago

How do nearby bodies of water help to keep the climate of certain areas mild?

GarryVolchara [31]

How do nearby bodies of water help to keep the climate of certain areas mild?
A Land and water absorb and store heat at similar rates.
B Water and land temperatures reach extremes of hot and cold.
C Water temperatures change more slowly than land temperatures.
D Land temperatures change more slowly than water temperatures.

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3 years ago

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Most of the land in Algeria, Libya, and Egypt is in the Sahara desert. Which best

julsineya [31]

Answer:

Correct answer is 4. Under 2.5 people per sq. mi. (VERY low density)

Explanation:

Option four is the correct because areas such as Sahara have a low density and are not populated. The conditions of living are not allowing people in these type of areas to create some larger permanent settlements. The similar case is for example in Australia where 3/4 is covered with desert, which is practically uninhabited. Or we can also created a parallel with Greenland which also has a low density due to climate factor.

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3 years ago