Let number [tex]c=\sqrt{45}.[/tex]
Square this number:
[tex]c^2=45.[/tex]
Since
[tex]36<45<49,[/tex]
you can state that
[tex]\sqrt{36}<\sqrt{45}<\sqrt{49},\\ \\6<\sqrt{45}<7.[/tex]
Also [tex]\sqrt{45}\approx 6.708[/tex] and [tex]6.5<x_C<7[/tex], then point C is the best approximation of [tex]\sqrt{45}.[/tex]
Answer: option C.
