Question

Between which two consecutive whole numbers does √121 lie? ​

Answer 1

Answer:

$\sqrt{121} = \sqrt{ {11}^{2} } = \frac{ + }{} 11 \\ but \: it \: is \: whole \: number \\ so \\ number \: should \: be \: positive \\ = > 11 \\ two \: consecutive \: number \: in \: which \\ 11 \: lies \\ = 11 - 1 \: \: \: \: \: or \: 11 + 1 \\ 10 \: and \: 12$