Answer:
Below!
Step-by-step explanation:
Given square root:
[tex]\sqrt41[/tex]
Let's note a few perfect square roots.
[tex]1 = \sqrt{1 \times 1} = \sqrt{1}[/tex]
[tex]2 = \sqrt{2 \times 2} = \sqrt{4}[/tex]
[tex]3 = \sqrt{3 \times 3} =\sqrt{9}[/tex]
[tex]4 = \sqrt{4 \times 4} =\sqrt{16}[/tex]
[tex]5 = \sqrt{5 \times 5} =\sqrt{25}[/tex]
[tex]6 = \sqrt{6 \times 6} =\sqrt{36}[/tex]
[tex]7 = \sqrt{7 \times 7} =\sqrt{49}[/tex]
[tex]8 = \sqrt{8 \times 8} =\sqrt{64}[/tex]
And so on...
When looking at the perfect square roots we identified, we can say that:
[tex]\sqrt{36} < \sqrt{41} < \sqrt{49}[/tex]
Therefore,
[tex]6 < \sqrt{41} < 7[/tex]
We can conclude that [tex]\sqrt{41}[/tex] is between 6 and 7.
