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poizon [28]
3 years ago
7

Geographically, people in western Russia have tended to identify with Europe more than Asia because

Geography
1 answer:
Sholpan [36]3 years ago
3 0
Answer:  [D]:  "T<span>he Ural Mountains seal them off from the rest of Asia</span>."
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Why do countries trade goods and services
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Countries trade with each other when, on their own, they do not have the resources, or capacity to satisfy their own needs and wants. By developing and exploiting their domestic scarce resources, countries can produce a surplus, and trade this for the resources they need.

3 0
3 years ago
Who would most likely have a pro-slavery point of view? A. A slave b. A White plantation owner c. A free Black d. An abolitionis
guajiro [1.7K]
The answer would be b
5 0
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According to the article about coral reefs, which of the following problems is threatening the world’s reefs today?
igor_vitrenko [27]
C. Rising sea temperatures
5 0
3 years ago
Were gladiators usually slaves or usually free
Rasek [7]
A gladiator<span> was a professional fighter in the </span>Roman Empire. Since<span> fights were most likely to the death they had a short life expectancy and the majority of fighters were slaves or former slaves.</span>
3 0
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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