Answer:
[tex]\large\boxed{\large\boxed{\sqrt{3} <\sqrt{5} <\pi <2\sqrt{3}}}[/tex]
Explanation:
You are comparing irrational numbers.
By inspection, i.e. at first sight you can only compare [tex]\sqrt{3} \text{ }and\text{ } 2\sqrt{3}[/tex] because they have the same radicand.
You can order: [tex]\sqrt{3} <2\sqrt{3}[/tex]
You can introduce the 2 inside the radical by squaring it:
[tex]2\sqrt{3}=\sqrt{2^2\times3}=\sqrt{12}[/tex]
Since 5 is between 3 and 12, you can order:
- [tex]\sqrt{3} <\sqrt{5} <\sqrt{12}[/tex]
Which is:
- [tex]\sqrt{3} <\sqrt{5} <2\sqrt{3}[/tex]
You must know that π ≈ 3.14.
5 is less than 9 and the square root of 9 is 3; hence, [tex]\sqrt{5} <3[/tex] and [tex]\sqrt{5} <\pi[/tex]
Now you must determine whether π is less than or greater than [tex]\sqrt{12}[/tex]
Using a calculator or probing numbers between 3 and 4 you get [tex]\sqrt{12} \approx3.46[/tex]
Hence, the complete order is:
- [tex]\sqrt{3} <\sqrt{5} <\pi <2\sqrt{3}[/tex]
