Please help!!
Kaylib’s eye-level height is 48 ft above sea level, and Addison’s eye-level height is 85 1/3 ft above sea level. How much farther can Addison see to the horizon? Use the formula d=√3h/2, with d being the distance they can see in miles and h being their eye-level height in feet.
A. √2 mi
B. 2√2 mi
C. 14√2 mi
D. 28√2mi

For this case we have the following equation:
[tex]d = \sqrt{\frac{3h}{2}} [/tex]
Where,
d: the distance they can see in thousands
h: their eye-level height in feet
For Kaylib:
[tex]d = \sqrt{\frac{3(48)}{2}} [/tex]
[tex]d=\sqrt{3(24)}[/tex]
[tex]d=\sqrt{72}[/tex]
[tex]d=6\sqrt{2}[/tex]
For Addison:
[tex]d = \sqrt{\frac{3(85\frac{1}{3})}{2}}[/tex]
[tex]d = \sqrt{\frac{256}{2}} [/tex]
[tex]d=\sqrt{128}[/tex]
[tex]d=8\sqrt{2}[/tex]
Subtracting both distances we have:
[tex]8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2}[/tex]
Answer:
[tex]2\sqrt{2}[/tex]
B. 2√2 mi

joaobezerra joaobezerra · Answer 2 · 2022-01-31T17:19:24+01:00

Using the formula, it is found that the measure of how much farther can Addison see to the horizon is given by:

B. 2√2 mi

How we find the distance they can see?

We use the following formula:

[tex]d = \sqrt{\frac{3h}{2}}[/tex]

In which h is their eye-level height is feet.

Kaylib’s eye-level height is 48 ft above sea level, hence:

[tex]d_K = \sqrt{\frac{3(48)}{2}} = \sqrt{72} = \sqrt{9 \times 8} = 6\sqrt{2}[/tex]

Addison’s eye-level height is 85 1/3 ft above sea level, hence:

[tex]h = 85 + \frac{1}{3} = 85.3[/tex]

[tex]d_A = \sqrt{\frac{3(85.3)}{2}} = 11.29[/tex]

Hence, the difference is:

[tex]D = d_K - d_A = 11.29 - 6\sqrt{2} \approx 2.8[/tex]

Which is equivalent to approximately [tex]2\sqrt{2}[/tex], hence option B is correct.

A similar problem, in which a formula is also applied, is given at https://brainly.com/question/24648661