Answer:
![x=9](https://tex.z-dn.net/?f=x%3D9)
Step-by-step explanation:
![\frac{x}{3}+5=8](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%2B5%3D8)
![\frac{x}{3}+5-5=8-5](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%2B5-5%3D8-5)
![\frac{x}{3}=3](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%3D3)
![\frac{3x}{3}=3\cdot \:3](https://tex.z-dn.net/?f=%5Cfrac%7B3x%7D%7B3%7D%3D3%5Ccdot%20%5C%3A3)
![x=9](https://tex.z-dn.net/?f=x%3D9)
Answer:
<h2>The distance to the Eath's Horizon from point P is 352.8 mi, approximately.</h2>
Step-by-step explanation:
You observe the problem from a graphical perspective with the image attached.
Notice that side
is tangent to the circle, which means is perpendicular to the radius which is equal to 3,959 mi.
We have a right triangle, that means we need to use the Pythagorean's Theorem, to find the distance to the Earth's Horizon from point P.
The hypothenuse is 3959 + 15.6 = 3974.6 mi.
![(3974.6)^{2}=x^{2} +(3959)^{2} \\x^{2} =15,797,445.16 - 15,673,681\\x=\sqrt{123,764.16} \approx 351.8](https://tex.z-dn.net/?f=%283974.6%29%5E%7B2%7D%3Dx%5E%7B2%7D%20%20%2B%283959%29%5E%7B2%7D%20%5C%5Cx%5E%7B2%7D%20%3D15%2C797%2C445.16%20-%2015%2C673%2C681%5C%5Cx%3D%5Csqrt%7B123%2C764.16%7D%20%5Capprox%20351.8)
Therefore, the distance to the Eath's Horizon from point P is 352.8 mi, approximately.
By the order of operation, we first do the operation in parentheses
( 6 * (2 + 15) ) /3 = (6* 17)/3 = 102 / 3
Then we do the division
Answer = 102 / 3 = 34
Hope that helps!
3 kilograms = 4.86
So one kilogram = 4.86/3 = 1.62
2 kilograms = 1.62 × 2 = 3.24