Ask question

Ask question

All categories

4 years ago

5

An mT air mass is best described as?

2 answers:

MrRa [10]4 years ago

7 0

Your answer is "warm and moist".

Varvara68 [4.7K]4 years ago

4 0

warm and wet is the gradpoint answer 100%

You might be interested in

According to the law of universal gravitation, what would happen to Earth if the Sun were somehow replaced by a black hole of th

Ilya [14]

If the black hole had the same gravitational effects as the sun then the black hole wouldn’t effect the earth, we wouldn’t have light and everyone would most likely die in 10 minuites

3 0

3 years ago

Which of the following is not a way in which redlining can increase economic problems in a city?

V125BC [204]

The answer is B. The citys tax base begins to increase because if it is redlines then they have to move somewhere else where it is harder to get money so they have to increase the taxes

5 0

3 years ago

Read 2 more answers

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

When all of your friends are fake and ignore you 24/7 <br> fracking mint

iren2701 [21]

Don't have fake friends then XD

People are wack frfr

7 0

4 years ago

Read 2 more answers

Explain the difference between underwater sinkholes and underwater caves? Select one:

kotykmax [81]

B..
Is the option.......

3 0

3 years ago

Read 2 more answers