7. (Parabola/line) A cannon ball flies through the air following a parabola modeled by h = -0.002² +0.12d + 2, where h is the height of the cannon ball and d is the distance. The land slopes upward and is modeled by h = 0.15d. At what distance away from the cannon does the ball hit the ground?

[tex][/tex] To find the distance at which the cannonball hits the ground, we can set the equation for the cannonball's height (-0.002d² + 0.12d + 2) equal to the equation for the terrain (0.15d), and solve for the value of d.

-0.002d^2 + 0.12d + 2 = 0.15d

Rearranging the equation gives us:

-0.002d^2 + 0.12d - 0.15d + 2 = 0

Combining like terms:

-0.002d^2 - 0.03d + 2 = 0

This is a quadratic equation in the form ad^2 + bd + c = 0, where a = -0.002, b = -0.03, and c = 2. We can solve for d using the quadratic formula:

d = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:

d = (-(-0.03) ± √((-0.03)^2 - 4 * (-0.002) * 2)) / (2 * (-0.002))

d = (0.03 ± √(0.0009 + 0.016)) / -0.004

d = (0.03 ± √0.0169) / -0.004

d = (0.03 ± 0.1299) / -0.004

We get two potential solutions:

d = (0.03 + 0.1299) / -0.004 = 0.1599 / -0.004 = -39.975 (discarded as distance cannot be negative),

d = (0.03 - 0.1299) / -0.004 = -0.0999 / -0.004 = 24.975

Therefore, the cannonball hits the ground at a distance of approximately 24.975 units away from the cannon.

7. (Parabola/line) A cannon ball flies through the air following a parabola modeled by h = -0.002² +0.12d + 2, where h is the height of the cannon ball and d is the distance. The land slopes upward and is modeled by h = 0.15d. At what distance away from the cannon does the ball hit the ground? ​

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7. (Parabola/line) A cannon ball flies through the air following a parabola modeled by h = -0.002² +0.12d + 2, where h is the height of the cannon ball and d is the distance. The land slopes upward and is modeled by h = 0.15d. At what distance away from the cannon does the ball hit the ground?