LaToya throws a ball from the top of a bridge. Her throw is modeled by the equation y=-0.6x^2+4x+11, and the bridge is modeled by the equation y=-0.1x+8. About how far does the ball travel horizontally before its first bounce?

The horizontal location, from the starting point at which the ball first bounces = 7.5

Step-by-step explanation:

The equation representing LaToya's throw y = -0.6·x² + 4·x + 11

The equation that represent the path of the bridge y = -0.1·x + 8

Where;

x = The horizontal location

y = The vertical location

The location where the ball bounces is given by the common solution of both equations as follows;

-0.6·x² + 4·x + 11 = -0.1·x + 8

-0.6·x² + 4·x + 11 - (-0.1·x + 8) = 0

-0.6·x² + 4·x + 11 + 0.1·x - 8 = 0

-0.6·x² + 4.1·x + 3 = 0

Which gives;

-0.6/(-0.6)·x² + 4.1/(-0.6)·x + 3/(-0.6) = 0

Which gives;

x² - 41/6·x - 5 = 0

By the quadratic formula, we have;

x = (41/6 ± √((-41/6)² - 4× 1 × (-5)))/(2 × 1)

x = 7.5 or x = -2/3

Therefore;

The horizontal location, from the starting point at which the ball first bounces, x = 7.5.

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60