SOLUTION The equation \( d=17 t \) gives the distance a raccoon runs in feet as a function of time in seconds. The graph shows the distance a squirrel runs over time. Does the raccoon or the squirrel run faster? Show your work.

To determine who runs faster, we can compare the rates of distance covered per unit of time for both the raccoon and the squirrel. The equation \( d = 17t \) tells us that the raccoon runs 17 feet for every second. This gives the raccoon a speed of 17 feet per second. Next, let's analyze the squirrel's distance over time from the graph. If we can extract the slope of the squirrel’s distance-time graph (which represents its speed), we can compare it to the raccoon's speed. For example, if the squirrel runs 34 feet in 2 seconds, its speed will be \( \frac{34 \text{ feet}}{2 \text{ seconds}} = 17 \text{ feet per second} \) as well. Once you have the squirrel's speed calculated, simply compare it to 17 feet per second. If it's higher, the squirrel is faster; if equal, they run at the same speed; and if lower, the raccoon is faster. Without the specific data from the graph, that's the general method!