Certainly! Let's find the product of the fractions [tex]\(\frac{7}{16}\)[/tex], [tex]\(\frac{4}{3}\)[/tex], and [tex]\(\frac{1}{2}\)[/tex] step by step.
1. First, multiply [tex]\(\frac{7}{16}\)[/tex] by [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[
\frac{7}{16} \times \frac{4}{3} = \frac{7 \times 4}{16 \times 3} = \frac{28}{48}
\][/tex]
2. Next, simplify [tex]\(\frac{28}{48}\)[/tex]:
Both [tex]\(28\)[/tex] and [tex]\(48\)[/tex] are divisible by 4.
[tex]\[
\frac{28}{48} = \frac{28 \div 4}{48 \div 4} = \frac{7}{12}
\][/tex]
So, the intermediate result of [tex]\(\frac{7}{16} \times \frac{4}{3}\)[/tex] is [tex]\(\frac{7}{12}\)[/tex].
3. Now, multiply this result by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[
\frac{7}{12} \times \frac{1}{2} = \frac{7 \times 1}{12 \times 2} = \frac{7}{24}
\][/tex]
Thus, the product of [tex]\(\frac{7}{16}\)[/tex], [tex]\(\frac{4}{3}\)[/tex], and [tex]\(\frac{1}{2}\)[/tex] is [tex]\(\frac{7}{24}\)[/tex].
Therefore, the correct answer is:
D. [tex]\(\frac{7}{24}\)[/tex]
