1. The problem asks which products of fractions Renee could be modeling as the area of a rectangle inside a unit square divided into 6 columns and 3 rows (total 18 small rectangles). 2. The unit square is divided into $6 \times 3 = 18$ small rectangles, so any fractional side lengths must correspond to fractions with denominators 6 and 3 or factors thereof. 3. To model the product as an area inside this grid, the product must be representable as a fraction with denominator 18 or reducible to such, since the total area is 1. 4. Let's analyze each product: - $\frac{1}{6} \times \frac{1}{3} = \frac{1}{18}$: This matches exactly one small rectangle in the grid. So, Yes. - $\frac{4}{3} \times \frac{3}{6} = \frac{4}{3} \times \frac{1}{2} = \frac{4}{6} = \frac{2}{3}$: The length $\frac{4}{3}$ is greater than 1, which cannot fit inside the unit square. So, No. - $\frac{2}{3} \times \frac{1}{12} = \frac{2}{36} = \frac{1}{18}$: $\frac{1}{12}$ is not aligned with the grid divisions (denominator 12 vs 6 and 3), so it cannot be modeled exactly. So, No. - $\frac{2}{3} \times \frac{5}{6} = \frac{10}{18} = \frac{5}{9}$: Both denominators 3 and 6 fit the grid, so Yes. - $\frac{2}{6} \times \frac{1}{3} = \frac{1}{3} \times \frac{1}{3} = \frac{2}{18} = \frac{1}{9}$: $\frac{2}{6}$ simplifies to $\frac{1}{3}$, so Yes. - $\frac{3}{6} \times \frac{1}{18} = \frac{1}{2} \times \frac{1}{18} = \frac{1}{36}$: $\frac{1}{18}$ is smaller than the smallest grid unit $\frac{1}{18}$, but $\frac{1}{18}$ is the size of one small rectangle, so $\frac{1}{36}$ is smaller than a grid unit and cannot be modeled exactly. So, No. Final answers: $\frac{1}{6} \times \frac{1}{3}$: Yes $\frac{4}{3} \times \frac{3}{6}$: No $\frac{2}{3} \times \frac{1}{12}$: No $\frac{2}{3} \times \frac{5}{6}$: Yes $\frac{2}{6} \times \frac{1}{3}$: Yes $\frac{3}{6} \times \frac{1}{18}$: No