kaylibs eye - level height is 48 ft above sea level, and addisons eye - level height is $85\frac{1}{3}$ ft…

kaylibs eye - level height is 48 ft above sea level, and addisons eye - level height is $85\frac{1}{3}$ ft above sea level. how much farther can addison see to the horizon? use the formula $d = sqrt{\frac{3h}{2}}$, with $d$ being the distance they can see in miles and $h$ being their eye - level height in feet.\n$sqrt{2}$ mi\n$2sqrt{2}$ mi\n$14sqrt{2}$ mi\n$28sqrt{2}$ mi

kaylibs eye - level height is 48 ft above sea level, and addisons eye - level height is $85\frac{1}{3}$ ft above sea level. how much farther can addison see to the horizon? use the formula $d = sqrt{\frac{3h}{2}}$, with $d$ being the distance they can see in miles and $h$ being their eye - level height in feet.\n$sqrt{2}$ mi\n$2sqrt{2}$ mi\n$14sqrt{2}$ mi\n$28sqrt{2}$ mi

Answer

Explanation:

Step1: Calculate Addison's distance

Substitute $h = 85\frac{1}{3}=\frac{256}{3}$ into $d=\sqrt{\frac{3h}{2}}$. $d_{Addison}=\sqrt{\frac{3\times\frac{256}{3}}{2}}=\sqrt{\frac{256}{2}}=\sqrt{128}=8\sqrt{2}$

Step2: Calculate Kaylib's distance

Substitute $h = 48$ into $d=\sqrt{\frac{3h}{2}}$. $d_{Kaylib}=\sqrt{\frac{3\times48}{2}}=\sqrt{72}=6\sqrt{2}$

Step3: Find the difference

Subtract Kaylib's distance from Addison's distance. $d_{Addison}-d_{Kaylib}=8\sqrt{2}- 6\sqrt{2}=2\sqrt{2}$

Answer:

B. $2\sqrt{2}$ mi