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3 years ago

What is the opposing plate involved in the formation of the volcanoes if the aegean sea?

Geography

1 answer:

eimsori [14]3 years ago

7 0

Answer:

Anatolian Plate

Explanation:

Aegean plate is a small tectonic plate located in the mediterranean sea to north is the eurasian plate which is the divergent boundary responsible for formation.

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What is the name of the earth layer that I'm standing in

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I believe it is the outer crust but i will do more reaserch for you :)

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3 years ago

Converting is a key skill when working with maps and scales. Tracking units and showing any conversion factors is an important p

alukav5142 [94]

Answer:

2.5 km = 250000 cm

500 ft = 15240 cm

Explanation:

part a

given data

length = 2.5 km

to find out

Convert into cm

solution

we know that 1 m = 100 cm ...............1

and also we know 1 km = 1000 m .................2

sop as that we can say from equation 1 and 2

1 km = 100000 cm

so for 2.5 km = 2.5 × 100000 cm

2.5 km = 250000 cm

and

part b

given data

length = 500 ft

to find out

Convert ft into cm

solution

we know here that

1 ft = 12 inches ..................1

and also we know 1 inch = 2.54 cm ...................2

so we can say

500 ft = 12 × 500 inches

500 ft = 6000 inches

and 500 ft = 6000 × 2.54 cm

500 ft = 15240 cm

4 0

3 years ago

his relatively large, symmetrical volcano contains interlayered lava flow, pyroclastic deposits, and volcanic mudflows. What kin

IrinaK [193]

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2 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

The most abundant type of rocks on earth are classified asa. igneous.

Sati [7]

<span>the most abundant type of rocks on earth are classified as - </span>a. igneous

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3 years ago