Since
[tex]6^2 = 36,\quad 7^2 = 49[/tex]
Then we have
[tex]6 = \sqrt{36}<\sqrt{45}<\sqrt{49}=7[/tex]
So, the square root of 45 is somewhere between 6 and 7.
Answer:
C
Step-by-step explanation:
Note that
[tex]\sqrt{36}[/tex] < [tex]\sqrt{45}[/tex] < [tex]\sqrt{49}[/tex], that is
6 < [tex]\sqrt{45}[/tex] < 7
The square root lies between 6 and 7 → point C