Question

which point on the number line shows the position of \\(\sqrt{96}\\)?

Explanation:

Step1: Find nearby perfect squares

$9^2=81$, $10^2=100$

Step2: Locate $\sqrt{96}$ between integers

Since $81<96<100$, $\sqrt{81}<\sqrt{96}<\sqrt{100}$, so $9<\sqrt{96}<10$

Step3: Narrow down position

96 is closer to 100 than 81, so $\sqrt{96}$ is closer to 10.

Answer:

point B