Question

use the calculator below to approximate \\( \sqrt{46} \\) as a decimal to the hundredths place. note that you must do the approximation without using a square root button. your answer must be within a hundredth of the actual value. \\( \sqrt{46} \approx \square \\)

Explanation:

Step1: Identify perfect squares around 46

We know that \(6^2 = 36\) and \(7^2=49\). So \(\sqrt{46}\) is between 6 and 7.

Step2: Narrow down the range

Let's try \(6.7^2=6.7\times6.7 = 44.89\), \(6.8^2 = 6.8\times6.8=46.24\). So \(\sqrt{46}\) is between 6.7 and 6.8.

Step3: Refine to hundredths place

Now, let's check \(6.78^2=6.78\times6.78 = 45.9684\), \(6.79^2=6.79\times6.79 = 46.1041\). Since \(45.9684<46<46.1041\), and we need to approximate to the hundredths place. The difference between 46 and \(6.78^2 = 45.9684\) is \(46 - 45.9684=0.0316\). The difference between \(6.79^2 = 46.1041\) and 46 is \(46.1041 - 46 = 0.1041\). Since 0.0316 is smaller, but we can also check the midpoint or see the closeness. Alternatively, using linear approximation or just checking the decimal expansion, we know that \(\sqrt{46}\approx6.78\) (because \(6.78^2 = 45.9684\), \(6.78^2\) is 0.0316 less than 46, and \(6.79^2\) is 0.1041 more. The actual value of \(\sqrt{46}\approx6.7823\), so to the hundredths place, it's 6.78.

Answer:

\(6.78\)