Question

drag \\(sqrt{53}\\) based on your estimate above:
to two decimal places, \\(sqrt{53}\\) must lie between \\(square\\) and \\(square\\)

Explanation:

Step1: Find lower bound square

We know that \(7.2^2 = 7.2\times7.2 = 51.84\) and \(7.3^2 = 7.3\times7.3 = 53.29\). Since \(51.84<53<53.29\), so \(\sqrt{51.84}<\sqrt{53}<\sqrt{53.29}\), which means \(7.2 < \sqrt{53}<7.3\).

Step2: Check decimal places

We need to find two - decimal - place bounds. Let's check \(7.28^2\) and \(7.29^2\).
Calculate \(7.28^2=7.28\times7.28 = 52.9984\)
Calculate \(7.29^2 = 7.29\times7.29=53.1441\)
Since \(52.9984 < 53<53.1441\), we have \(7.28<\sqrt{53}<7.29\)

Answer:

\(7.28\) and \(7.29\)