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to two decimal places, \\(\\sqrt{96}\\) must lie between \\(\\square\\)…

Question

to two decimal places, \\(\sqrt{96}\\) must lie between \\(\square\\) and \\(\square\\)
answer attempt 1 out of 2

Explanation:

Step1: Calculate \(9.7^2\)

\(9.7^2 = 9.7\times9.7 = 94.09\)

Step2: Calculate \(9.8^2\)

\(9.8^2 = 9.8\times9.8 = 96.04\)

Step3: Analyze the value of \(\sqrt{96}\)

We know that \(9.7^2 = 94.09\) and \(9.8^2 = 96.04\). Since \(94.09<96<96.04\), taking square roots (and since square root is an increasing function for non - negative numbers), we have \(9.7<\sqrt{96}<9.8\).

Answer:

\(9.7\) and \(9.8\)