Let's solve the question step-by-step:
1. Multiplying Fractions with Numerators and Denominators:
- [tex]\(\frac{1}{5} \times \frac{2}{7}\)[/tex]:
- Multiply the numerators: [tex]\(1 \times 2 = 2\)[/tex]
- Multiply the denominators: [tex]\(5 \times 7 = 35\)[/tex]
- The product is: [tex]\(\frac{2}{35}\)[/tex]
- [tex]\(\frac{3}{5} \times \frac{9}{8}\)[/tex]:
- Multiply the numerators: [tex]\(3 \times 9 = 27\)[/tex]
- Multiply the denominators: [tex]\(5 \times 8 = 40\)[/tex]
- The product is: [tex]\(\frac{27}{40}\)[/tex]
- [tex]\(\frac{2}{4} \times \frac{4}{9}\)[/tex]:
- Multiply the numerators: [tex]\(2 \times 4 = 8\)[/tex]
- Multiply the denominators: [tex]\(4 \times 9 = 36\)[/tex]
- The product is: [tex]\(\frac{8}{36}\)[/tex]
2. Completing the Table:
[tex]\[
\begin{tabular}{|l|l|l|}
\hline
\(\frac{2}{3} \times \frac{4}{5} = \frac{8}{15}\) & \(\frac{8}{15} < \frac{2}{3}, \frac{8}{15} < \frac{4}{5}\) & Product is less than each of the \\
\hline
\(\frac{1}{5} \times \frac{2}{7} = \frac{2}{35}\) & \(\frac{2}{35} < \frac{1}{5}, \frac{2}{35} < \frac{2}{7}\) & fractions involved. \\
\hline
\(\frac{3}{5} \times \frac{9}{8} = \frac{27}{40}\) & \(\frac{27}{40} < \frac{3}{5}, \frac{27}{40} < \frac{9}{8}\) & fractions involved. \\
\hline
\(\frac{2}{4} \times \frac{4}{9} = \frac{8}{36}\) & \(\frac{8}{36} < \frac{2}{4}, \frac{8}{36} < \frac{4}{9}\) & fractions involved. \\
\hline
\end{tabular}
\][/tex]
In our table, we have multiplied fractions and filled out both the product and the comparison between the product and the original fractions. Each product is indeed less than the fractions that were multiplied.
