A practice soccer field has a length of \( 98 \frac{1}{2} \) meters and a width of \( 78 \frac{5}{12} \) meters. Is the perimeter of the field greater than 350 meters? Explain your thinking.

To find out if the perimeter of the soccer field is greater than 350 meters, we first need to calculate the perimeter. The formula for the perimeter \( P \) of a rectangle is \( P = 2 \times (\text{length} + \text{width}) \). First, we convert the mixed numbers into improper fractions for easier calculations: Length: \( 98 \frac{1}{2} = 98 + \frac{1}{2} = \frac{196}{2} + \frac{1}{2} = \frac{197}{2} \) meters. Width: \( 78 \frac{5}{12} = 78 + \frac{5}{12} = \frac{936}{12} + \frac{5}{12} = \frac{941}{12} \) meters. Now, we need to find the sum of the length and width: \[ \text{Length} + \text{Width} = \frac{197}{2} + \frac{941}{12} \] To add these fractions, we must first find a common denominator, which is 12. Converting \( \frac{197}{2} \) to twelfths gives us: \[ \frac{197}{2} = \frac{197 \times 6}{2 \times 6} = \frac{1182}{12} \] Now, we can add: \[ \frac{1182}{12} + \frac{941}{12} = \frac{1182 + 941}{12} = \frac{2123}{12} \] Next, we compute the perimeter: \[ P = 2 \times \left(\frac{2123}{12}\right) = \frac{4246}{12} = 353.83 \text{ meters (approximately)} \] Since \( 353.83 \) is greater than \( 350 \), we conclude that the perimeter of the field is indeed greater than \( 350 \) meters!