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Wyatt’s eye-level height is 120 ft above sea level, and Shawn’s eye-level height is 270 ft above sea level. How much farther can Shawn see to the horizon? Use the formula d=square root (3h/2), with d being the distance they can see in miles and h being their eye-level height in feet.

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kanest
The formula is defined in the question as follows:

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

h is the eye-level height. 

Wyatt's eye-level height is 120. Plug this value into the equation:

[tex]\sqrt{ \frac{3(120)}{2} } = \sqrt{ \frac{360}{2} } = \sqrt{180} = 13.4164[/tex]

Shawn's eye-level height is 270. Plug this value into the equation:

[tex]\sqrt{ \frac{3(270)}{2} } = \sqrt{ \frac{810}{2} } = \sqrt{405} = 20.1246[/tex]

Subtract Wyatt's viewing distance from Shawn's to find their difference:

[tex]20.1246 - 13.4164 = 6.7082[/tex]

Shawn can see 6.7082 miles farther than Wyatt.
Paris020605

Answer: 3√5 mi.

Step-by-step explanation:

The formula is: d = √(3h/2)

Wyatt:

h = 120 ft

d = √(3 * 120/2) = √180 = √(36 * 5) = √36 * √5 = 6√5 mi

Shawn:

h = 270 ft

d = √(3 * 270/2) = √405 = √(81 * 5) = √81 * √5 = 9√5 mi

How much farther can Shawn see to the horizon?

Shawn - Wyatt = 9√5 - 6√5 = 3√5 mi

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