Question

Approximate 11√31, correct to one decimal place, by using a sequence of values that gets closer and closer to √31
.(1 point)

Answer 1

To approximate \( 11\sqrt{31} \) by using a sequence of values that converges to \( \sqrt{31} \), we need to first find a good approximation for \( \sqrt{31} \).

We know that \( \sqrt{25} = 5 \) and \( \sqrt{36} = 6 \), so \( \sqrt{31} \) is between 5 and 6.

We can refine this approximation by checking the squares of numbers between 5 and 6. Let's try 5.5:

\[
5.5^2 = 30.25
\]

Next, let's check 5.6:

\[
5.6^2 = 31.36
\]

Now we see that \( \sqrt{31} \) is between 5.5 and 5.6. Let's average these to refine our guess further:

\[
\text{Midpoint} = \frac{5.5 + 5.6}{2} = 5.55
\]

Now we calculate \( 5.55^2 \):

\[
5.55^2 = 30.8025
\]

Next, check 5.57:

\[
5.57^2 = 31.0449
\]

Now we know \( \sqrt{31} \) is between 5.55 and 5.57. Let's try 5.56:

\[
5.56^2 = 30.9136
\]

Now we know that \( \sqrt{31} \) is between 5.56 and 5.57. We can conclude that:

\[
\sqrt{31} \approx 5.57
\]

Now, multiply this approximation by 11 to find \( 11\sqrt{31} \):

\[
11\sqrt{31} \approx 11 \times 5.57 = 61.27
\]

Rounding to one decimal place, we get:

\[
\boxed{61.3}
\]