PLEAASE HELP
Kaylib’s eye-level height is 48 ft above sea level, and Addison’s eye-level height is ft above sea level. How much farther can Addison see to the horizon? Use the formula , with d being the distance they can see in miles and h being their eye-level height in feet. FIRST ONE IS QUESTION THE OTHER ONE IS THE FORMULA

well you are given the equation so let's plug in for kaylib and see how many miles she can see
distance = sqrt [(3 * height) / 2]
d = sqrt [(3 *48) / 2]
d = sqrt (144 / 2)
d = sqrt (72)
d = sqrt (3 * 3 * 2 * 2 * 2)
d = 6 * sqrt (2)

You you did not list Addisons height but I will say she is at x feet above sea level. we plug in x for height:
d = sqrt [(3x) / 2]

It it says how much farther for Addison which means she can see farther. to find difference we just subtract kaylibs distance from Addison. so:
sqrt [(3x) / 2] - 6 * sqrt (2)

plug in your x and use a calculator to get a decimal approximation

eudora eudora · Answer 2 · 2019-05-19T11:43:06+02:00

Answer:

2√2 miles is the difference in the distance which Addison can see more than Kaylib.

Step-by-step explanation:

Kaylib's eye-level height is 48 ft above the sea level and Addison's eye -level height is [tex]85\frac{1}{3} feet[/tex]

Formula to calculate the distance they can see has been given as

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

where d is the distance they can see in miles and h is their eye-level height in feet.

For Kaylib

[tex]D_{1}=\sqrt{\frac{(3)(48)}{2}}[/tex]

[tex]D_{1}=\sqrt{72}[/tex]

[tex]D_{1}=6\sqrt{2}[/tex]

For Addison

[tex]D_{2}=\sqrt{\frac{(3)(256)}{(2)(3)}}[/tex]

[tex]D_{2}=\sqrt{128}[/tex]

[tex]D_{2}=8\sqrt{2}[/tex]

Now difference in [tex]D_{2}-D_{1}[/tex] will be

[tex]D_{2}-D_{1}=8\sqrt{2}-6\sqrt{2}[/tex]

= 2√2 miles