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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
B2√2 mi
C14√2 mi
D28√2mi

Verified Answer :

carlosego

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

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:

\(d = \sqrt{\frac{3h}{2}}\)

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

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

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

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

\(h = 85 + \frac{1}{3} = 85.3\)

\(d_A = \sqrt{\frac{3(85.3)}{2}} = 11.29\)

Hence, the difference is:

\(D = d_K - d_A = 11.29 - 6\sqrt{2} \approx 2.8\)

Which is equivalent to approximately \(2\sqrt{2}\), hence option B is correct.

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

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