Question

PLEAASE HELP

Answer 1

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

Answer 2

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 $85\frac{1}{3} feet$

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

$d=\sqrt{\frac{3h}{2} }$

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

For Kaylib

$D_{1}=\sqrt{\frac{(3)(48)}{2}}$

$D_{1}=\sqrt{72}$

$D_{1}=6\sqrt{2}$

For Addison

$D_{2}=\sqrt{\frac{(3)(256)}{(2)(3)}}$

$D_{2}=\sqrt{128}$

$D_{2}=8\sqrt{2}$

Now difference in $D_{2}-D_{1}$ will be

$D_{2}-D_{1}=8\sqrt{2}-6\sqrt{2}$

= 2√2 miles