We're given 2 equations for the stopping distance on dry pavement and wet pavement. We're asked to find what the stopping distance for a car traveling 65 mph would be. So literally we just stick in 65 for the X's in both equations and compute it. So we get 331 approximately and 522. The directions tell us to round in the nearest whole number. Part B says based on your answers, which rectangular coordinate graph shows stopping distance on Dr. pavement and which shows stopping distance on wet pavement. So if we come up here and we look at our graph at 65, we should know that the one with the smaller value is going to be the dry payment because drive payment was 331 and the one with the bigger value is going to be the wet pavement or 522. So the dry pavement is equation B and the wet pavement is the equation A. The next part of this problem asks us to how well do the answers model the actual stopping distance shown in the graph at the right. So if we think about the values that we had for 65, they're very, very close. We have 335 and 520, and the answers a minute ago were 331 and 522. So they're very close right here. And these two, so they're very close. So our answer is the model values and the actual values are practically the same. Part D says determine the speed on dry pavement. So we're going to use the F of X equation requiring stopping distances that exceed the length of 1 1/2 football fields including the end zones or 540 feet. So now we're going to put 540 in for the F of X because that's how long or how far the stopping distance we're going to solve for the X. So we're going to start by subtracting 540 and doing the quadratic formula. We're going to get X is approximately 85 mph. So if you're going 85 mph, it's going to take you 540 feet to stop, or 1 1/2 football fields.