A running track in the shape of an oval is shown. The ends of the track form semicircles. What is the perimeter of the inside of the track? (π = 3.14) 392.00 m 372.22 m 444.44 m 588.88 m

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A running track in the shape of an oval is shown. The ends of the track form semicircles. What is the perimeter of the inside of the track? (π = 3.14) 392.00 m 372.22 m 444.44 m 588.88 m

Mathematics
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get the perimeter of the 2 semicircles add them up add the rectangle in the middle
do u know the answer i thought it was 372.22 but idk

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Other answers:

how'd you get that?
i did the work what u said but idk i have a feeling its eitheir that or 444.44
so the perimeter of a circle is \(2\pi r\) the perimeter of a semicircle is half that or \(\pi r\) we have 2 semi circles.|dw:1435687589686:dw|
Yeah
given that the whole side is 48, we want the radius which is half of that or 24 then we can get the perimeters semicircle 1: 24pi semicircle 2: 24pi never mind the rectangle area, I was mistaken
I keep getting 450.72 hmm
do u think its 444.44
oh my bad, that's 46 meters
23pi+23pi+150+150=444.44

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