6. chris is mailing his friend a poster that has been rolled up in a long tube. he has a box that measures 20 inches by 8 inches by 4 inches. what is the maximum length the rolled poster can be?

Question

Accepted Answer

# Explanation:
## Step1: Encontrar la diagonal de la base
La base del cuadro tiene dimensiones $20$ pulgadas y $8$ pulgadas. Usando el teorema de Pitágoras $d_{base}=\sqrt{20^{2}+8^{2}}$.
$$
\begin{align*}
d_{base}&=\sqrt{400 + 64}\
&=\sqrt{464}
\end{align*}
$$
## Step2: Encontrar la diagonal del cubo
Ahora, consideramos la diagonal de la base ($\sqrt{464}$) y la altura del cubo ($4$ pulgadas). Usando el teorema de Pitágoras de nuevo $d=\sqrt{(\sqrt{464})^{2}+4^{2}}$.
$$
\begin{align*}
d&=\sqrt{464+16}\
&=\sqrt{480}\
&=\sqrt{16\times30}\
& = 4\sqrt{30}\approx4\times5.477\
&=21.908
\end{align*}
$$

# Answer:
La longitud máxima del póster enrollado es aproximadamente $21.91$ pulgadas.

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QUESTION IMAGE

Question

Explanation:

Step1: Encontrar la diagonal de la base

Step2: Encontrar la diagonal del cubo

Answer: