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.

Explanation:

Step1: Recall perfect squares

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

Step2: Use linear approximation or trial - and - error

Let's try values between \(6\) and \(7\). Let's start with \(6.7\): \(6.7^2=6.7\times6.7 = 44.89\).
Then \(6.8^2 = 6.8\times6.8=46.24\). Since \(44.89<46<46.24\), \(\sqrt{46}\) is between \(6.7\) and \(6.8\).
Now, let's try \(6.78\): \(6.78^2=6.78\times6.78=(6 + 0.78)\times(6 + 0.78)=6^2+2\times6\times0.78 + 0.78^2=36+9.36+0.6084 = 45.9684\).
\(6.79^2=6.79\times6.79=(6.78 + 0.01)\times(6.78+0.01)=6.78^2+2\times6.78\times0.01+0.01^2=45.9684 + 0.1356+0.0001 = 46.1041\).
Since \(45.9684<46<46.1041\), and we want to approximate to the hundredths place. We can see that \(6.78^2 = 45.9684\) and \(6.79^2=46.1041\). The actual value of \(\sqrt{46}\approx6.782\), so to the hundredths place, it is \(6.78\).

Answer:

\(6.78\)