The simplification of square root will be option A; 6.
How to find the closest integer less than and greater than the square root of a number?
The function is strictly increasing function for x ≥ 0.
Assuming that all the values of x are ≥ 1, whenever there is increment in the value of 'x', the value of y increases too.
We have given an expression;
[tex]\dfrac{3\sqrt{20} }{\sqrt{5} }[/tex]
So,
[tex]\dfrac{3\sqrt{5 \times 2 \times 2} }{\sqrt{5} } \\\\\\\dfrac{3 \times 2 \sqrt{5 } }{\sqrt{5} } \\\\3 \times 2 = 6[/tex]
Hence, the simplification of square root will be option A; 6.
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