Question

to one decimal place, \\(\sqrt{96}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Find the perfect squares around 96

We know that \(9^2 = 81\) and \(10^2 = 100\). Since \(81<96<100\), we have \(\sqrt{81}<\sqrt{96}<\sqrt{100}\), which simplifies to \(9 < \sqrt{96}<10\).

Step2: Check the tenths place

Now we calculate the square of \(9.7\) and \(9.8\) to get a better approximation.
\(9.7^2=(10 - 0.3)^2 = 10^2-2\times10\times0.3 + 0.3^2=100 - 6+0.09 = 94.09\)
\(9.8^2=(10 - 0.2)^2=10^2-2\times10\times0.2 + 0.2^2 = 100-4 + 0.04=96.04\)

Since \(9.7^2 = 94.09<96\) and \(9.8^2=96.04>96\), we know that \(9.7<\sqrt{96}<9.8\) when rounded to one decimal place.

Answer:

\(9.7\) and \(9.8\)