Answer:
[tex]\displaystyle \large{\frac{11}{3} < \sqrt{17} < \sqrt{20}}[/tex]
Step-by-step explanation:
( 1 ) 11/3
[tex]\displaystyle \large{\frac{11}{3}}[/tex] can be converted to mixed fractions to [tex]\displaystyle \large{3\frac{2}{3}}[/tex]
Therefore, 11/3 is at least around 3, almost 4. It could be more than 3.5 by approximation but not exactly 4.
( 2 ) √20
[tex]\displaystyle \large{\sqrt{20}=\sqrt{5\cdot 4}}\\\displaystyle \large{\sqrt{20}=\sqrt{5\cdot 2\cdot 2}}\\\displaystyle \large{\sqrt{20}=2\sqrt{5}}[/tex]
We know that [tex]\displaystyle \large{\sqrt{4}=2}[/tex] so [tex]\displaystyle \large{\sqrt{5} > \sqrt{4}}[/tex] which means approximately, [tex]\displaystyle \large{\sqrt{20}}[/tex] can be either at least 4.3 up to 4.4
( 3 ) √17
We know that [tex]\displaystyle \large{\sqrt{16} = 4}[/tex] so [tex]\displaystyle \large{\sqrt{17}}[/tex] can be at least 4.1 up to 4.2, but it’s obvious that the square root of 17 is less than square root of 20 since [tex]\displaystyle \large{\sqrt{a} > \sqrt{b}}[/tex] for a > b
Hence, from least to greatest:
[tex]\displaystyle \large{\frac{11}{3} < \sqrt{17} < \sqrt{20}}[/tex]
