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qwelly [4]
3 years ago
6

The term hydrosphere describes one of the Earth's spheres and it includes all of the ________. The term hydrosphere describes on

e of the Earth's spheres and it includes all of the ________. fresh water on the planet liquid water on the Earth minus ice waters on the surface of the continents water on and below the continents water in the Earth system
Geography
1 answer:
Amanda [17]3 years ago
6 0

Answer:

water in the Earth system.

Explanation:

Earth crust refers to the outermost layer or thin shell which makes up the earth such as sedimentary, metamorphic and igneous rocks.

The Earth crust is typically made up of two (2) main types;

1. Oceanic crust: this crust has a thickness ranging from 5 km to 10 km and comprises of denser rocks such as diabase, gabbro and basalt.

2. Continental crust: this crust has a thickness ranging from 30 km to 60 km and comprises of minerals such as quartz, silicate, feldspar, aluminum and less dense rocks such as granite

The term hydrosphere describes one of the Earth's spheres and it includes all of the water in the Earth system. It makes up about 86% of the earth surface.

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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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