Not similar one is using a 20 in area and the other is using a 96 in area
The addison see to the horizon at 2 root 2mi.
We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.
We have to find the how much farther can addison see to the horizon
<h3>Which equation we get from the given condition?</h3>
![d=\sqrt{\frac{3h}{2} }](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5Cfrac%7B3h%7D%7B2%7D%20%7D)
Where, we have
d- the distance they can see in thousands
h- their eye-level height in feet
For Kaylib
![d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5Cfrac%7B3%5Ctimes%2048%7D%7B2%7D%20%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B%7B3%2824%29%7D%20%7D%5C%5C%5C%5C%5C%5Cd%3D%5Csqrt%7B72%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B36%5Ctimes%202%7D%5C%5C%5C%5C%5C%5Cd%3D6%5Csqrt%7B2%7D....%281%29)
For Addison h=85(1/3)
![d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5Cfrac%7B3%5Ctimes%2085%5Cfrac%7B1%7D%7B3%7D%20%7D%7B2%7D%20%7D%5C%5Cd%5Csqrt%7B%5Cfrac%7B256%7D%7B2%7D%20%7D%20%5C%5Cd%3D%5Csqrt%7B128%7D%20%5C%5Cd%3D8%5Csqrt%7B2%7D%20.....%282%29)
Subtracting both distances we get
![8\sqrt{2}-6\sqrt{2} =2\sqrt{2}](https://tex.z-dn.net/?f=8%5Csqrt%7B2%7D-6%5Csqrt%7B2%7D%20%20%3D2%5Csqrt%7B2%7D)
Therefore, the addison see to the horizon at 2 root 2mi.
To learn more about the eye level visit:
brainly.com/question/1392973
This is something I’ve never seen before I think you have to add the numbers.
Answer: 16/3 or 5.333333
Step-by-step explanation:
1/3 is 0.3333333 so 0.3333333 + 5 is 5.333333
Answer:
48=3b+v
Step-by-step explanation:
384=24b+8v
b standing for bus
v standing for vam
to simplify the equations you would divided by eight since all numbers are multiples
48=3b+v