Let's determine which number, when multiplied by [tex]\(\frac{1}{4}\)[/tex], produces a rational number. We will analyze each option step-by-step:
A. [tex]\(\pi\)[/tex]
When multiplying [tex]\(\pi\)[/tex] by [tex]\(\frac{1}{4}\)[/tex], we get:
[tex]\[ \pi \times \frac{1}{4} \approx 0.7853981633974483 \][/tex]
Since [tex]\(\pi\)[/tex] is an irrational number and multiplying it by a rational number [tex]\(\frac{1}{4}\)[/tex] does not yield a rational number, the result of this multiplication is irrational.
B. [tex]\(1.7320508\ldots\)[/tex]
This number is an approximation of [tex]\(\sqrt{3}\)[/tex], which is an irrational number. When multiplying [tex]\(1.7320508\ldots\)[/tex] by [tex]\(\frac{1}{4}\)[/tex], we get:
[tex]\[ 1.7320508 \times \frac{1}{4} \approx 0.4330127 \][/tex]
Since [tex]\(1.7320508\ldots\)[/tex] is an approximation of an irrational number, the result of this multiplication is also irrational.
C. [tex]\(\sqrt{6}\)[/tex]
The number [tex]\(\sqrt{6}\)[/tex] is an irrational number. When multiplying [tex]\(\sqrt{6}\)[/tex] by [tex]\(\frac{1}{4}\)[/tex], we get:
[tex]\[ \sqrt{6} \times \frac{1}{4} \approx 0.612372436 \][/tex]
Since [tex]\(\sqrt{6}\)[/tex] is an irrational number, the result of this multiplication is irrational.
D. [tex]\(\frac{14}{11}\)[/tex]
[tex]\(\frac{14}{11}\)[/tex] is a rational number. When multiplying [tex]\(\frac{14}{11}\)[/tex] by [tex]\(\frac{1}{4}\)[/tex], we get:
[tex]\[ \frac{14}{11} \times \frac{1}{4} = \frac{14}{11} \times \frac{1}{4} = \frac{14 \cdot 1}{11 \cdot 4} = \frac{14}{44} = \frac{7}{22} \][/tex]
Since [tex]\(\frac{7}{22}\)[/tex] is a rational number, the result of this multiplication is rational.
So, the number that produces a rational number when multiplied by [tex]\(\frac{1}{4}\)[/tex] is [tex]\(\frac{14}{11}\)[/tex]. The answer is:
D. [tex]\(\frac{14}{11}\)[/tex]
