The stopping distance for a car traveling 25 mph is 69.2 feet, and for a car traveling 40 mph it is 112 feet. The stopping distance in feet can be described by the equation \( \mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx} \), where x is the speed in mph. (a) Find the values of a and b . (b) Use the answers from part (a) to find the stopping distance for a car traveling 60 mph . (a) Write the augmented matrix that will be used to find the values of a and b . \( [\square \square 69.2 \) \( \square \) (Type whole numbers.) (Ty

To find the values of \( a \) and \( b \) in the equation \( y = ax^2 + bx \), we can use the data provided. First, we can set up the equations based on the given stopping distances: 1. For the speed of 25 mph: \( 69.2 = a(25^2) + b(25) \) 2. For the speed of 40 mph: \( 112 = a(40^2) + b(40) \) This leads us to: 1. \( 69.2 = 625a + 25b \) 2. \( 112 = 1600a + 40b \) Next, we'll write the augmented matrix for this system: \[ \begin{bmatrix} 625 & 25 & | & 69.2 \\ 1600 & 40 & | & 112 \end{bmatrix} \] To find \( a \) and \( b \), you would solve this augmented matrix using methods such as row reduction or substitution. Now, solving this system will give you the values of \( a \) and \( b \). Once you have \( a \) and \( b \), you can substitute \( x = 60 \) into the equation \( y = ax^2 + bx \) to find the stopping distance for a car traveling at that speed. Get ready to dust off your algebra skills because the equations will reveal the hidden secrets behind those stopping distances! And trust me, if you think stopping at 60 mph is exhilarating, wait till you see the calculations in action! 🏎️💨