2. Which product below is equivalent to the fraction \( \frac{20 n^{5}}{35 n^{2}} \) ? \( \begin{array}{ll}\text { (1) } \frac{5 n^{5}}{5 n^{5}} \cdot \frac{4}{7 n^{3}} & \text { (3) } \frac{10 n^{2}}{10 n^{2}} \cdot \frac{2 n^{3}}{25} \\ \text { (2) } \frac{10 n}{10 n} \cdot \frac{2 n^{4}}{25 n} & \text { (4) } \frac{5 n^{2}}{5 n^{2}} \cdot \frac{4 n^{3}}{7}\end{array} \)

To simplify the fraction \( \frac{20 n^{5}}{35 n^{2}} \), we can divide both the numerator and the denominator by 5, which gives us \( \frac{4 n^{5}}{7 n^{2}} \). We can further simplify this by canceling out \( n^{2} \) from both the numerator and the denominator, resulting in \( \frac{4 n^{3}}{7} \). Now, let's check which option yields the same result: 1. Evaluating option (1): \( \frac{5 n^{5}}{5 n^{5}} \cdot \frac{4}{7 n^{3}} = 1 \cdot \frac{4}{7 n^{3}} = \frac{4}{7 n^{3}} \) - Not equivalent. 2. Evaluating option (2): \( \frac{10 n}{10 n} \cdot \frac{2 n^{4}}{25 n} = 1 \cdot \frac{2 n^{4}}{25 n} = \frac{2 n^{3}}{25} \) - Not equivalent. 3. Evaluating option (3): \( \frac{10 n^{2}}{10 n^{2}} \cdot \frac{2 n^{3}}{25} = 1 \cdot \frac{2 n^{3}}{25} = \frac{2 n^{3}}{25} \) - Not equivalent. 4. Evaluating option (4): \( \frac{5 n^{2}}{5 n^{2}} \cdot \frac{4 n^{3}}{7} = 1 \cdot \frac{4 n^{3}}{7} = \frac{4 n^{3}}{7} \) - This is equivalent to the simplified fraction. Thus, the equivalent product is option (4).