Question 1117530
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*[illustration ovalTrack1.jpg]


The length of the track is just 2 times the length of the field plus the circumference of the circle formed by putting the two semicircles together.



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2(96)\ +\ \pi(59)]


I'll let you do the arithmetic.


By the way, NOT rounding this answer to the nearest whole meter is absolutely, positively, 100% <i><b>wrong</b></i> because the least precise measurements given are given to the nearest whole meter.  When you say something measures 96 meters, you are only guaranteeing that the true measurement is somewhere in the interval *[tex \Large 95.5\text{m}\ \leq\ m\ <\ 96.5\text{m}].  Given that you are doing this calculation with <i>two</i> measurements that are equally imprecise, you have absolutely no idea about the value of the first decimal place in the answer, let alone the second.  And this is true regardless of how precisely you state your estimate for *[tex \Large \pi].  You might share with your instructor that had s/he wanted two decimal place precision in the answer, s/he should have specified the measurements of the field as 96.00 meters by 59.00 meters.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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