Respuesta :

Answer:

3 and 4

Step-by-step explanation:

Consider squares on either side of 15, that is 9 and 16, so

[tex]\sqrt{9}[/tex] < [tex]\sqrt{15}[/tex] < [tex]\sqrt{16}[/tex], that is

3 < [tex]\sqrt{15}[/tex] < 4

Answer:

[tex]3<\sqrt{15}<4[/tex]

Step-by-step explanation:

We are asked to find the integers such that square root of 15 lies between them.

We know that perfect square less than 15 is 9 and perfect square greater than 15 is 16.

We can represent this information in an inequality as:

[tex]\sqrt{9}<\sqrt{15}<\sqrt{16}[/tex]

[tex]\sqrt{3^2}<\sqrt{15}<\sqrt{4^2}[/tex]

[tex]3<\sqrt{15}<4[/tex]

Therefore, the square root of 15 lies 3 and 4.

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