Answer:
The with of the tunnel at the ground level is 20 feet
Step-by-step explanation:
The equation that model the entrance of the tunnel is y = -11/50(x - 4)(x - 24)
Where the dimensions of 'x' and 'y' are in feet and the x-axis represent the ground level
The y-axis represent the height, and the equation is a quadratic equation with the shape of an upside down parabola to represent the shape of the entrance to the tunnel
The two coordinates of the parabola at the ground level (where y = 0)
of the tunnel entrance from which we can find the width by subtraction are given when y = 0 as follows;
y = 0 = -11/50(x - 4)·(x - 24)
∴ (x - 4)·(x - 24) = 0/(-11/50) = 0
When (x - 4)·(x - 24) = 0, x = 4, or x = 24
The coordinates at the ground corners of the tunnel entrance are therefore;
(4, 0), and (24, 0)
The with of the tunnel at the ground level is equal to the width of the parabola at the x-axis, which is given as follows;
The width of the parabola at the x-axis = Distance between points (4, 0), and (24, 0)
∴ The width of the parabola at the x-axis = √((0 - 0)² + (24 - 4)²) = √(24 - 4)² = 24 - 4 = 20 (distance between points having the same y-coordinates)
The width of the parabola at the x-axis = 20 ft. = The with of the tunnel at the ground level
∴ The with of the tunnel at the ground level = 20 feet.
