1. Some students solved the racing cars challenge with tables. Here is some data from a different racing dots scenario than the one you solved in the Desmos activity. \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Time (in \\ seconds) \end{tabular} & \begin{tabular}{c} Orange Car's Distance \\ (in feet) \end{tabular} & \begin{tabular}{c} Blue Car's Distance \\ (in feet) \end{tabular} \\ \hline 0 & 5 & 25 \\ \hline 2 & 11.4 & 29.8 \\ \hline 4 & 17.8 & 34.6 \\ \hline 6 & 24.2 & 39.4 \\ \hline 8 & 30.6 & 44.2 \\ \hline & & \\ \hline \end{tabular} When will the cars meet? Where will they be when they meet? Enter your answer in the last row of the table. How did you come up with your answer? Explain your thinking:

Step 1 ：First, let's calculate the rate of change for each car. The rate of change is the change in distance over the change in time. For the orange car, the rate of change is \((30.6 - 5) / (8 - 0) = 3.2\) feet per second. For the blue car, the rate of change is \((44.2 - 25) / (8 - 0) = 2.4\) feet per second.

Step 2 ：Next, we can set up the equations for the distances of the two cars. The distance of the orange car is \(5 + 3.2t\) and the distance of the blue car is \(25 + 2.4t\). We can set these two equations equal to each other to find the time at which the cars meet.

Step 3 ：Finally, we can substitute the time back into either of the equations to find the distance at which they meet.

Step 4 ：Let's solve this. We set up the equation \(3.2t + 5 = 2.4t + 25\). Solving for \(t\), we get \(t = 25.0000000000000\).

Step 5 ：Substituting \(t = 25.0000000000000\) back into either of the equations, we get the distance as \(85.0000000000000\).

Step 6 ：Final Answer: The cars will meet at time \(\boxed{25}\) seconds and they will be at a distance of \(\boxed{85}\) feet.