Kaylib's eye-level height is 48 ft above sea level, and Addison's eye-level height is [tex]$85 \frac{1}{3}$[/tex] ft above sea level. How much farther can Addison see to the horizon?

Use the formula [tex]d = \sqrt{\frac{3h}{2}}[/tex], with [tex]d[/tex] being the distance they can see in miles and [tex]h[/tex] being their eye-level height in feet.

A. [tex]\sqrt{2}[/tex] mi
B. [tex]2 \sqrt{2}[/tex] mi
C. [tex]14 \sqrt{2}[/tex] mi
D. [tex]28 \sqrt{2}[/tex] mi



Answer :

To determine how much farther Addison can see to the horizon than Kaylib, let's follow the given formula [tex]\(d = \sqrt{\frac{3h}{2}}\)[/tex], where [tex]\(d\)[/tex] represents the distance to the horizon in miles and [tex]\(h\)[/tex] represents the eye-level height in feet.

First, let's calculate the distance for each of them individually.

### Kaylib's Distance to the Horizon:
Kaylib's eye-level height [tex]\(h_kaylib\)[/tex] is 48 feet.

Using the formula [tex]\(d = \sqrt{\frac{3h}{2}}\)[/tex], we get:
[tex]\[ d_{kaylib} = \sqrt{\frac{3 \cdot 48}{2}} \][/tex]

[tex]\[ d_{kaylib} = \sqrt{\frac{144}{2}} \][/tex]

[tex]\[ d_{kaylib} = \sqrt{72} \][/tex]

[tex]\[ d_{kaylib} = 8.485 \text{ miles} \][/tex]

### Addison's Distance to the Horizon:
Addison's eye-level height [tex]\(h_addison\)[/tex] is [tex]\(85 \frac{1}{3}\)[/tex] feet, which is equal to [tex]\(\frac{256}{3}\)[/tex] feet.

Using the formula [tex]\(d = \sqrt{\frac{3h}{2}}\)[/tex], we get:
[tex]\[ d_{addison} = \sqrt{\frac{3 \cdot 85 \frac{1}{3}}{2}} \][/tex]

[tex]\[ d_{addison} = \sqrt{\frac{3 \cdot 256/3}{2}} \][/tex]

[tex]\[ d_{addison} = \sqrt{\frac{256}{2}} \][/tex]

[tex]\[ d_{addison} = \sqrt{128} \][/tex]

[tex]\[ d_{addison} = 11.314 \text{ miles} \][/tex]

### Difference in Distance:
Now, the difference between the distances they can see to the horizon is:

[tex]\[ \text{Difference} = d_{addison} - d_{kaylib} \][/tex]

[tex]\[ \text{Difference} = 11.314 - 8.485 \][/tex]

[tex]\[ \text{Difference} = 2.829 \text{ miles} \][/tex]

### Simplified to Options:
Given the list of options:

- [tex]\(\sqrt{2}\)[/tex] miles
- [tex]\(2\sqrt{2}\)[/tex] miles
- [tex]\(14\sqrt{2}\)[/tex] miles
- [tex]\(28\sqrt{2}\)[/tex] miles

We need to express [tex]\(2.829\)[/tex] miles in terms of [tex]\(\sqrt{2}\)[/tex]. We know:

[tex]\[ \sqrt{2} \approx 1.414 \][/tex]

So,

[tex]\[ 2 \sqrt{2} \approx 2 \times 1.414 = 2.828 \][/tex]

Thus, [tex]\(2.828 \approx 2.829\)[/tex], implying:

[tex]\[ 2.829 \approx 2\sqrt{2} \][/tex]

Hence, Addison can see approximately [tex]\(2 \sqrt{2}\)[/tex] miles farther than Kaylib. Therefore, the correct answer is:

[tex]\[ \boxed{2\sqrt{2} \text{ miles}} \][/tex]

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