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Ugo [173]
3 years ago
11

The age of one rock layer compared to the age of another rock layer is its __________.

Geography
2 answers:
jeka943 years ago
6 0
The correct answer is letter B. Relative Age. The age of one rock layer compared to the age of another rock layer is its relative age.
Maslowich3 years ago
4 0

the answer is B) relative age

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What is the fracturing of rock along curved lines that results when pressure is removed from bedrock?
mr_godi [17]
~Hello there! ^_^

Your question: What is the fracturing of rock along curved lines that results when pressure is removed from bedrock..?

Your answer: The fracturing of rock along curved lines that results when pressure is removed from bedrock is called fracking.

Happy Studying! =)



7 0
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Whaling was a major industry in the Northeast in the 1800’s.<br> true or false
arlik [135]

Indeed it was. So, yes, it's true.

8 0
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over time, limestone dissolves in the presence of acid rain on earths surface. relate this chemical property to the use of limes
dimaraw [331]
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What is the key funtion of geographic information systems
lara [203]

Answer:

the main function of a geographic information system is to present visual data of representation

Explanation:

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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}.

5 0
3 years ago
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