Question

find two consecutive whole numbers that \\(sqrt{101}\\) lies between.

Explanation:

Step1: Find square of 10

$10^2 = 100$

Step2: Find square of 11

$11^2 = 121$

Step3: Compare with 101

$100 < 101 < 121$, so $\sqrt{100} < \sqrt{101} < \sqrt{121}$

Answer:

10 and 11