What did the Egyptians believe about spirits? What role did they believe that the Gods played in their world?

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

6 0

Answer:

I only know the first one well. I don't want to answer the others and maybe give false info.

Step-by-step explanation:

Egyptians were very well known to be very interested in Spirits and Gods.

Their whole dynasty was built on the dawn of their Gods. There were many, like the Greek and Roman Gods. One of the sun, one of the underworld....etc..

The pharaohs believed themselves to be god. (Due to their immense power)

Egyptians believed that after death there was an Afterlife. In that afterlife, there was games, other people (spirits), homes (graves). This might sound weird but in the graves especially of Pharaohs they placed beds, tables, chairs, games..etc..

<h3>Hope this Helps!</h3>

You might be interested in

A. Analyze the number of dark triangles. Describe the pattern.

Kamila [148]

We need a picture to answer this...

3 0

4 years ago

If i raked 5/12 of grass and someone else raked 4/9 how much more needs to be raked

pashok25 [27]

To solve this problem, you can subtract the sum of 5/12 and 4/9 from 1. To find the sum of 5/12 and 4/9, you need to make them have common denominators.
5/12*3/3=15/36
4/9*4/4=16/36

Now add the two fractions.
15/36+16/36=31/36

Lastly, subtract the sum from 1.
1-31/36=5/36.

5 0

3 years ago

The vertices of ΔMNO are M (1, 3), N (4, 9), and O (7, 3). The vertices of ΔPQR are P (3, 0), Q (4, 2), and R (5, 0). Which conc

vodka [1.7K]

Answer:

They are similar by the definition of similarity in terms of a dilation.

Step-by-step explanation:

Given:

The vertices of ΔMNO are M (1, 3), N (4, 9), and O (7, 3). The vertices of ΔPQR are P (3, 0), Q (4, 2), and R (5, 0).

To choose: the correct option

Solution:

According to distance formula distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For ΔMNO,

$MN=\sqrt{(4-1)^2+(9-3)^2}=\sqrt{(3)^2+(6)^2}=\sqrt{45}\\NO=\sqrt{(7-4)^2+(3-9)^2}=\sqrt{(3)^2+(-6)^2}=\sqrt{45}\\\\MO=\sqrt{(7-1)^2+(3-3)^2}=\sqrt{(6)^2+(0)^2}=\sqrt{36}=6\\$

For ΔPQR,

$PQ=\sqrt{(4-3)^2+(2-0)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{5}\\QR=\sqrt{(5-4)^2+(0-2)^2}=\sqrt{(1)^2+(-2)^2}=\sqrt{5}\\\\PR=\sqrt{(5-3)^2+(0-0)^2}=\sqrt{(2)^2+(0)^2}=2\\$