Question

kaylib’s eye-level height is 48 ft above sea level, and addison’s 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.\
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$2\sqrt{2}$ mi\
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$28\sqrt{2}$ mi\
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$\sqrt{2}$ mi\
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$14\sqrt{2}$ mi

Explanation:

Step1: Calcular \( d \) para Kaylib

Sustituir \( h = 48 \) en \( d=\sqrt{\frac{3h}{2}} \):
\( d_{Kaylib}=\sqrt{\frac{3\times48}{2}}=\sqrt{72}=\sqrt{36\times2}=6\sqrt{2} \)? Espera, no, \( \frac{3\times48}{2}=72 \)? No, \( 3\times48 = 144 \), \( 144\div2 = 72 \)? Wait, no: \( 3\times48 = 144 \), \( 144/2 = 72 \)? Wait, no, \( 3h/2 \) cuando \( h=48 \): \( 3*48=144 \), \( 144/2=72 \), así que \( d_{Kaylib}=\sqrt{72}=6\sqrt{2} \)? No, espera, \( 85\frac{1}{3} \) es \( \frac{256}{3} \). Vamos a corregir:

Step1: Calcular \( d \) para Kaylib

\( h_{Kaylib}=48 \), entonces \( d_{K}=\sqrt{\frac{3\times48}{2}}=\sqrt{\frac{144}{2}}=\sqrt{72}=6\sqrt{2} \)? No, \( 144\div2=72 \), \( \sqrt{72}=6\sqrt{2} \)? Wait, no, \( 72 = 36\times2 \), así que \( \sqrt{72}=6\sqrt{2} \).

Step2: Calcular \( d \) para Addison

\( h_{Addison}=85\frac{1}{3}=\frac{256}{3} \), entonces \( d_{A}=\sqrt{\frac{3\times\frac{256}{3}}{2}}=\sqrt{\frac{256}{2}}=\sqrt{128}=8\sqrt{2} \).

Step3: Calcular la diferencia

\( d_{A}-d_{K}=8\sqrt{2}-6\sqrt{2}=2\sqrt{2} \)? Wait, no, me equivoqué en \( h_{Kaylib} \). Espera, el problema es "how much farther", así que debemos calcular \( d_{Addison} - d_{Kaylib} \).

Wait, volvamos:

Para Kaylib: \( h = 48 \), \( d_K = \sqrt{\frac{3*48}{2}} = \sqrt{\frac{144}{2}} = \sqrt{72} = 6\sqrt{2} \)? No, \( 144/2 = 72 \), \( \sqrt{72} = 6\sqrt{2} \).

Para Addison: \( h = 85\frac{1}{3} = \frac{256}{3} \), entonces \( \frac{3h}{2} = \frac{3*(256/3)}{2} = \frac{256}{2} = 128 \), así que \( d_A = \sqrt{128} = 8\sqrt{2} \).

Entonces la diferencia es \( 8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2} \)? Wait, pero las opciones incluyen \( 2\sqrt{2} \) mi. Wait, pero en el problema, quizás me equivoqué en \( h_{Kaylib} \). Wait, \( 48 \) ft: \( 3*48/2 = 72 \), \( \sqrt{72}=6\sqrt{2} \). \( 85\frac{1}{3} \) ft: \( 3*(256/3)/2 = 256/2 = 128 \), \( \sqrt{128}=8\sqrt{2} \). Entonces \( 8\sqrt{2} - 6\sqrt{2} = 2\sqrt{2} \).

Answer:

\( 2\sqrt{2} \) mi (la opción "2√2 mi")