There are 60 minutes in one hour. If we are trying to find how many minutes are in 4 hours and 30 minutes, we first need to find out how many minutes are in 4 hours.
To do that, we just need to multiply the amount of minutes in an hour by the amount of hours we have.
60 * 4 = 240
Then, we just add the remaining 30 minutes.
240 + 30 = 270
There are 270 minutes in 4 hours and 30 minutes.
Hope that helped! =)
Answer:
x = 1/2m
Step-by-step explanation:
Answer:
A. ![\frac{(a-b)^{2} +b^{2} }{a^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%28a-b%29%5E%7B2%7D%20%2Bb%5E%7B2%7D%20%7D%7Ba%5E%7B2%7D%7D)
Step-by-step explanation:
So to get the area of a square, we need to find the length of one side.
We know the length of the larger square is a, so the area of the larger cube is ![a^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D)
We can find the length of a side of the smaller square by using pythagoreans theorem to find the hypotenuse of the triangle formed in the bottom left corner. The length of one side along the x axis is a - b, and the length of the other side, along the y-axis, is b.
We can plug it into pythagoreans theorem to get
(C represents the length of one side of the smaller square, and the hypotenuse of the triangle)
![C = \sqrt{ (a-b)^{2} +b^{2}}](https://tex.z-dn.net/?f=C%20%3D%20%5Csqrt%7B%20%28a-b%29%5E%7B2%7D%20%2Bb%5E%7B2%7D%7D)
The area of the smaller triangle is C squared to the area of the smaller triangle is
![(a-b)^{2} +b^{2}](https://tex.z-dn.net/?f=%28a-b%29%5E%7B2%7D%20%2Bb%5E%7B2%7D)
To get the ratio of the smaller square in comparison to the larger square we divide the area of the smaller square by the area of the larger square.
So the ratio should be
![\frac{(a-b)^{2} +b^{2} }{a^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%28a-b%29%5E%7B2%7D%20%2Bb%5E%7B2%7D%20%7D%7Ba%5E%7B2%7D%7D)
Answer:
We are observing the galactic center as it was 27,000 years ago
Step-by-step explanation:
The Galactic Center, or Galactic Centre, is the rotational center of the Milky Way. It is 8,122 ± 31 parsecs (26,490 ± 100 ly) away from Earth in the direction of the constellations Sagittarius, Ophiuchus, and Scorpius where the Milky Way appears brightest. It coincides with the compact radio source Sagittarius A*