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What is the length of the longest item that will fit inside the shipping box? Enter the correct answer by replacing the values of [tex]\(m\)[/tex] and [tex]\(n\)[/tex].

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$\frac{D}{D}$ & $\sqrt{a}$ & $|0|$ & $\sqrt{0}$ & $0^0$ & + & - & $x$ & = & $\ \textless \ $ & $\ \textgreater \ $ & $\leq$ & $z$ & $\pi$ & $\beta$ & $\mu$ & $r$ & $\rho$ & 9 & $\varphi$ & $\sin$ & csC & $\cos$ & $\sec$ & $\tan$ & cot & $\sin^{-1}$ & $\log$ \\
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\end{tabular}\][/tex]


Sagot :

Given the dimensions of the shipping box with a length of 20 units, a width of 15 units, and a height of 10 units, we are to find the length of the longest item that will fit inside the box.

To determine this length, we can use the 3D Pythagorean theorem, which allows us to calculate the diagonal of a rectangular prism. The formula for the diagonal [tex]\(d\)[/tex] of such a box is:

[tex]\[ d = \sqrt{l^2 + w^2 + h^2} \][/tex]

where [tex]\( l \)[/tex] is the length, [tex]\( w \)[/tex] is the width, and [tex]\( h \)[/tex] is the height of the box.

Substituting the given dimensions into this formula:
[tex]\[ l = 20, \quad w = 15, \quad h = 10 \][/tex]

Calculating the square of each dimension:
[tex]\[ l^2 = 20^2 = 400 \][/tex]
[tex]\[ w^2 = 15^2 = 225 \][/tex]
[tex]\[ h^2 = 10^2 = 100 \][/tex]

Now, summing these squares gives:
[tex]\[ l^2 + w^2 + h^2 = 400 + 225 + 100 = 725 \][/tex]

Finally, taking the square root of this sum to find the diagonal:
[tex]\[ d = \sqrt{725} \approx 26.92582403567252 \][/tex]

Thus, the length of the longest item that will fit inside the shipping box is approximately [tex]\( 26.93 \)[/tex] units (rounded to two decimal places).

Therefore, the answer is:

[tex]\[ 26.93 \][/tex]