To determine which number produces an irrational number when multiplied by [tex]\(\frac{1}{4}\)[/tex], we'll examine each option one by one.
#### Option A: [tex]\(0.444444 \ldots\)[/tex]
This number is a repeating decimal, also commonly written as [tex]\(0.\overline{4}\)[/tex]. Repeating decimals are rational numbers because they can be expressed as a fraction. Let's see the result when we multiply it by [tex]\(\frac{1}{4}\)[/tex]:
[tex]\[ 0.444444 \ldots \times \frac{1}{4} = 0.111111 \ldots = 0.\overline{1} \][/tex]
Since [tex]\(0.\overline{1}\)[/tex] is also a repeating decimal, it is a rational number.
#### Option B: [tex]\(-\sqrt{36}\)[/tex]
First, simplify [tex]\(\sqrt{36}\)[/tex]:
[tex]\[ \sqrt{36} = 6 \][/tex]
Therefore, [tex]\(-\sqrt{36}\)[/tex] becomes [tex]\(-6\)[/tex]. Now, multiply this by [tex]\(\frac{1}{4}\)[/tex]:
[tex]\[ -6 \times \frac{1}{4} = -\frac{6}{4} = -1.5 \][/tex]
Since [tex]\(-1.5\)[/tex] can be expressed as a fraction [tex]\(\left(-\frac{3}{2}\right)\)[/tex], it is also a rational number.
#### Option C: [tex]\(\sqrt{12}\)[/tex]
To determine the nature of [tex]\(\sqrt{12}\)[/tex]:
[tex]\[ 12 = 4 \times 3 \implies \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \][/tex]
Now, multiply this by [tex]\(\frac{1}{4}\)[/tex]:
[tex]\[ 2\sqrt{3} \times \frac{1}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \][/tex]
Since [tex]\(\sqrt{3}\)[/tex] is an irrational number, [tex]\(\frac{\sqrt{3}}{2}\)[/tex] remains irrational.
#### Option D: [tex]\(\frac{4}{3}\)[/tex]
Finally, consider the rational number [tex]\(\frac{4}{3}\)[/tex]. Multiply it by [tex]\(\frac{1}{4}\)[/tex]:
[tex]\[ \frac{4}{3} \times \frac{1}{4} = \frac{4 \times 1}{3 \times 4} = \frac{1}{3} \][/tex]
Since [tex]\(\frac{1}{3}\)[/tex] can be expressed as a fraction, it is a rational number.
### Conclusion:
Only the multiplication involving Option C, [tex]\(\sqrt{12}\)[/tex], results in an irrational number. Therefore, the correct answer is:
[tex]\[ \text{C. } \sqrt{12} \][/tex]
