9514 1404 393
Answer:
136
Explanation:
The x-value at one end of the beam will be where f(x) = 0.
0 = -x²/272 +17
x² = 17(272) = 4624
x = ±√4624 = ±68
The beam extends from x=-68 to x=68, so has a width of ...
68 -(-68) = 136
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Note that the x in the equation for f(x) is not the same as the x in the picture corresponding to width.