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\begin{tabular}{|c|c|c|}
\hline
& Rational & Irrational \\
\hline
[tex]$-\sqrt{36}$[/tex] & 0 & 0 \\
\hline
[tex]$\frac{9}{14}$[/tex] & 0 & 0 \\
\hline
[tex]$4.\overline{32}$[/tex] & 0 & 0 \\
\hline
[tex]$11\pi$[/tex] & 0 & 0 \\
\hline
[tex]$\sqrt{14}$[/tex] & 0 & 0 \\
\hline
\end{tabular}


Sagot :

To categorize each of the given values as rational or irrational, let's break down each step-by-step:

1. Analyzing [tex]\(-\sqrt{36}\)[/tex]:
- The value [tex]\(\sqrt{36}\)[/tex] equals 6. Therefore, [tex]\(-\sqrt{36}\)[/tex] equals -6.
- Since -6 is an integer, it is also a rational number.

2. Analyzing [tex]\(\frac{9}{14}\)[/tex]:
- The fraction [tex]\(\frac{9}{14}\)[/tex] represents a ratio of two integers.
- Fractions where both the numerator and denominator are integers (and the denominator is not zero) are considered rational numbers.

3. Analyzing [tex]\(4.\overline{32}\)[/tex]:
- The notation [tex]\(4.\overline{32}\)[/tex] signifies a repeating decimal where 32 repeats indefinitely.
- Repeating decimals are representations of rational numbers.

4. Analyzing [tex]\(11\pi\)[/tex]:
- Pi ([tex]\(\pi\)[/tex]) is known to be an irrational number. Any non-zero multiple of an irrational number is also irrational.
- Therefore, [tex]\(11\pi\)[/tex] is irrational.

5. Analyzing [tex]\(\sqrt{14}\)[/tex]:
- The number 14 is not a perfect square, hence [tex]\(\sqrt{14}\)[/tex] is not an integer.
- The square root of any non-perfect square is an irrational number.

Now, filling in the table with these results:

[tex]\[ \begin{tabular}{|c|c|c|} \hline & \text{rational} & \text{irrational} \\ \hline -\sqrt{36} & \text{1} & \text{0} \\ \hline \frac{9}{14} & \text{1} & \text{0} \\ \hline 4 . \overline{32} & \text{1} & \text{0} \\ \hline 11\pi & \text{0} & \text{1} \\ \hline \sqrt{14} & \text{0} & \text{1} \\ \hline \end{tabular} \][/tex]

The correct answer with the values filled in:

[tex]\[ \begin{tabular}{|c|c|c|} \hline & \text{rational} & \text{irrational} \\ \hline -\sqrt{36} & 1 & 0 \\ \hline \frac{9}{14} & 1 & 0 \\ \hline 4 . \overline{32} & 1 & 0 \\ \hline 11\pi & 0 & 1 \\ \hline \sqrt{14} & 0 & 1 \\ \hline \end{tabular} \][/tex]