Final Answer-Explanation:
[tex][/tex] To find where the footballs are at the same location, we need to set the two equations equal to each other and solve for d:
-0.25d² + 3d + 6 = -0.15d² + 2d + 6
Combining like terms gives:
-0.25d² + 3d = -0.15d² + 2d
Subtracting -0.15d² + 2d from both sides gives:
-0.1d² + d = 0
Now we can solve for d. Factoring is a bit tricky but we can do it by completing the square or using the quadratic formula. I will use the quadratic formula for simplicity:
d = (-b ± √(b² - 4ac)) / (2a)
Here, a = -0.1, b = 1, and c = 0. Plugging these values into the quadratic formula:
d = (-1 ± √(1 - 4(0)(-0.1))) / (2(-0.1))
d = (-1 ± √(1 + 0)) / (-0.2)
d = (-1 ± √1) / (-0.2)
So we have two potential solutions:
d = (-1 + 1) / (-0.2) = 0 / (-0.2) = 0
d = (-1 - 1) / (-0.2) = -2 / (-0.2) = 10
So the footballs are at the same location when they have traveled 0 yards and 10 yards.
To determine who threw the ball further, we need to analyze the maximum distances obtained from the two equations, which will occur at the vertex of the parabolas. We can use the formula for the x-coordinate of the vertex, which is given by -b/2a.
For quarterback A's equation h = -0.25d² + 3d + 6, the x-coordinate of the vertex is:
d = -3 / (2(-0.25)) = -3 / (-0.5) = 6 yards.
For quarterback B's equation h = -0.15d² + 2d + 6, the x-coordinate of the vertex is:
d = -2 / (2(-0.15)) = -2 / (-0.3) = 6.67 yards.
Thus, quarterback B threw the ball further, with the ball reaching the furthest at approximately 6.67 yards compared to 6 yards for quarterback A.
