Answer:
[tex]\pi +3 >2\pi -\sqrt{9}>\frac{\pi }{2}[/tex]
Step-by-step explanation:
[tex]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\\[/tex]
[tex]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}[/tex]
