again, the quarterback throws the football, but this time the model y = -16x² + 35x + 6.5 describes the…

again, the quarterback throws the football, but this time the model y = -16x² + 35x + 6.5 describes the height (y), in feet, of the football x seconds after it is thrown. if the ball is not caught, how long will it be in the air? 0 = -16x² + 35x + 6.5 x = \\frac{-35\\pm\\sqrt{35² - 4(-16)(6.5)}}{2(-16)} rounded to the nearest tenth, the solutions of the equation are -0.2, 2.4. why can you eliminate the solution of -0.2 in the context of this problem? check all that apply. the height of the ball must be positive. it does not make sense for time to be negative. the ball cannot hit the ground before it is thrown. to the nearest tenth of a second, the ball is in the air for s.

again, the quarterback throws the football, but this time the model y = -16x² + 35x + 6.5 describes the height (y), in feet, of the football x seconds after it is thrown. if the ball is not caught, how long will it be in the air? 0 = -16x² + 35x + 6.5 x = \\frac{-35\\pm\\sqrt{35² - 4(-16)(6.5)}}{2(-16)} rounded to the nearest tenth, the solutions of the equation are -0.2, 2.4. why can you eliminate the solution of -0.2 in the context of this problem? check all that apply. the height of the ball must be positive. it does not make sense for time to be negative. the ball cannot hit the ground before it is thrown. to the nearest tenth of a second, the ball is in the air for s.

Answer

Explanation:

Step1: Identify valid solution

We are looking for the time the ball is in the air. Since time cannot be negative in this context (the ball cannot hit the ground before it is thrown), we take the positive solution of the quadratic - equation.

Step2: Determine time in air

The positive solution of the quadratic equation $0=-16x^{2}+35x + 6.5$ is $x = 2.4$. This represents the time the ball is in the air.

Answer:

$2.4$