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kirill115 [55]
3 years ago
11

What province is Gavilan, New Mexico?

Geography
2 answers:
snow_lady [41]3 years ago
5 0
A. Colorado Plateau Province
Ludmilka [50]3 years ago
3 0

Answer: Detail, Side Patio, Casa del Gavilan, near Cimarron,

The Casa del Gavilan is a historic bed and breakfast located six miles south of Cimarron, New Mexico. The property is at the base of Urraca Mesa and is surrounded by the property of Philmont Scout Ranch. The bed and breakfast has six guest rooms. The Casa del Gavilan is also referred to as The Nairn Place due to its original owners.

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Explain the reason for having an Equator and a Prime Meridian. How do these two lines help us to navigate our world?
EleoNora [17]

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By using the equator and prime meridian, we can divide the world into four hemispheres, north, south, east, and west. For instance, the United States is in the Western Hemisphere (because it is west of the prime meridian) and also in the Northern Hemisphere (because it is north of the equator).

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3 years ago
What is the value of X6? <br><br> Show the solution.
wariber [46]

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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

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

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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}

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Now we have to calculate the value of x_6.

Using Pythagoras theorem in triangle 6 :

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

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(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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