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3 years ago

13

Where is the eastern Mediterranean

1 answer:

RideAnS [48]3 years ago

7 0

Eastern Mediterranean denotes the religion in the east of the Mediterranean Sea (the Levantine sea) in Western Asia. It includes the region of Syria(in the levant), the island of Cyprus, Anatolia, and Egypt

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What is the measure of C ? 38° 76° 90° 152°

SVETLANKA909090 [29]

Answer:

Well, I'm not sure because degrees Fahrenheit doesn't show. The degrees Celcius depends on Fahrenheit.

Explanation:

4 0

3 years ago

What words best describe the Baltic landscape? Select all that apply.

Leto [7]

Marshland
mountainous
tropical

8 0

3 years ago

Read 2 more answers

What city does the meridian pass directly through

tamaranim1 [39]

Hello,

Here is your answer:

The proper answer to this question is "Greenwich,England".

Here is how:

The prime minister thought that the longitude was observatory which is why he decided the pass all the way through.

Your answer is Greenwich,England!

If you need anymore help feel free to ask me!

Hope this help!

7 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

Simplify: (3x2 − 8xy + y2) − (7x2 − 4xy − 5y2)

kramer

-4x^2-4xy+6y^2 hope this helps

8 0

3 years ago