Question

drag \\(\sqrt{82}\\) based on your estimate above:

to one decimal place, \\(\sqrt{82}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Find perfect squares around 82

We know that \(9^2 = 81\) and \(10^2=100\). So \(\sqrt{81} = 9\) and \(\sqrt{100}=10\). So \(\sqrt{82}\) is between 9 and 10.

Step2: Estimate to one decimal place

Let's check \(9.0^2 = 81\), \(9.1^2=82.81\). Since \(81<82<82.81\), so \(\sqrt{82}\) is between 9.0 and 9.1? Wait, no, wait \(9.0^2 = 81\), \(9.1^2=82.81\), but 82 is between 81 and 82.81, so \(\sqrt{82}\) is between 9.0 and 9.1? Wait, no, wait \(9.0^2 = 81\), \(9.1^2 = 82.81\), so \(9.0<\sqrt{82}<9.1\)? Wait, no, 82 - 81 = 1, 82.81 - 81 = 1.81, so \(\sqrt{82}\) is closer to 9.1? Wait, no, let's calculate \(9.0^2=81\), \(9.1^2 = 82.81\). So 82 is between 81 (9.0²) and 82.81 (9.1²). So to one decimal place, \(\sqrt{82}\) is between 9.0 and 9.1? Wait, no, wait 9.0 squared is 81, 9.1 squared is 82.81. So 82 is between 81 and 82.81, so \(\sqrt{82}\) is between 9.0 and 9.1. Wait, but let's check 9.0 and 9.1. So the lower bound is 9.0 and upper bound is 9.1.

Answer:

9.0 and 9.1