Asked by joe mawna

9. Using rational approximations, what is an inequality statement comparing √64 and √59.
The inequality is: v64 √59

Answers

Answered by joe mawna
wat iz da anwsr?
Answered by GPT-4o mini
To compare \(\sqrt{64}\) and \(\sqrt{59}\) using rational approximations, we can start by calculating the values of these square roots.

1. We know that:
\[
\sqrt{64} = 8
\]

2. For \(\sqrt{59}\), we can use a rational approximation:
- We know \(\sqrt{49} = 7\) and \(\sqrt{64} = 8\), so \(\sqrt{59}\) is between 7 and 8.
- To get a more accurate approximation, we can estimate by checking perfect squares around 59:
- \( 7.5^2 = 56.25 \)
- \( 7.6^2 = 57.76 \)
- \( 7.7^2 = 59.29 \)

Since \( 7.6^2 < 59 < 7.7^2 \), we can conclude that:
\[
7.6 < \sqrt{59} < 7.7
\]

3. Hence, we can place \(\sqrt{59}\) approximately between 7.6 and 7.7.

Now we can compare:
\[
\sqrt{64} = 8 \quad \text{and} \quad 7.6 < \sqrt{59} < 7.7
\]

Thus, the inequality statement comparing \(\sqrt{64}\) and \(\sqrt{59}\) is:
\[
\sqrt{64} > \sqrt{59}
\]

The final inequality statement is:
\[
8 > \sqrt{59}
\]
Answered by GPT-4o mini
The answer is:

\(\sqrt{64} > \sqrt{59}\)

This means that \(\sqrt{64}\) (which is 8) is greater than \(\sqrt{59}\).

Related Questions