Help please. Order from greatest (left) to least.

Mathematics

1 answer:

Anna35 [415]3 years ago

8 0

Answer:

$\pi +3 >2\pi -\sqrt{9}>\frac{\pi }{2}$

Step-by-step explanation:

$Lets\ take\ \pi 's\ approximate\ value\ as\ 3.14159\\Hence,\\ \pi +3\\=3.14159+3\\=6.14159\\2\pi -\sqrt{9} \\=2*3.14159-3\\=6.28318-3\\=3.28318\\\frac{\pi }{2} \\=1.570795\\$

$As\ we\ can\ see\ that,\\6.14159>3.28318>1.570795,\\Their\ corresponding\ values\ are\ also\ in\ the same\ order:\\\pi +3 >2\pi -\sqrt{9}>\frac{\pi }{2}$

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