wyatts eye - level height is 120 ft above sea level, and shawns eye - level height is 270 ft above sea…

wyatts eye - level height is 120 ft above sea level, and shawns eye - level height is 270 ft above sea level. how much farther can shawn see to the horizon? use the formula $d = sqrt{\frac{3h}{2}}$, $hgeq0$, with $d$ being the distance they can see in miles and $h$ being their eye - level height in feet.\n$sqrt{5}$ mi\n$3sqrt{5}$ mi\n$15sqrt{5}$ mi\n$45sqrt{5}$ mi

wyatts eye - level height is 120 ft above sea level, and shawns eye - level height is 270 ft above sea level. how much farther can shawn see to the horizon? use the formula $d = sqrt{\frac{3h}{2}}$, $hgeq0$, with $d$ being the distance they can see in miles and $h$ being their eye - level height in feet.\n$sqrt{5}$ mi\n$3sqrt{5}$ mi\n$15sqrt{5}$ mi\n$45sqrt{5}$ mi

Answer

Answer:

B. $3\sqrt{5}$ mi

Explanation:

Step1: Calculate Wyatt's distance

Substitute $h = 120$ into $d=\sqrt{\frac{3h}{2}}$. $d_{1}=\sqrt{\frac{3\times120}{2}}=\sqrt{180}=6\sqrt{5}$

Step2: Calculate Shawn's distance

Substitute $h = 270$ into $d=\sqrt{\frac{3h}{2}}$. $d_{2}=\sqrt{\frac{3\times270}{2}}=\sqrt{405}=9\sqrt{5}$

Step3: Find the difference

$d_{2}-d_{1}=9\sqrt{5}-6\sqrt{5}=3\sqrt{5}$