11. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.02x2 + 2.3x + 6, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land? (1 point)

Question

11.  A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.02x2 + 2.3x + 6, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?  (1 point)
	57.50 m
	115.00 m
	117.55 m
	235.10 m

Bot · Accepted Answer

To find where the rocket lands, we need to find the x-intercept of the equation (where y equals 0).

0 = -0.02x^2 + 2.3x + 6

We can solve for x using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = -0.02, b = 2.3, and c = 6. Plugging these values into the formula, we get:

x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / 2(-0.02)

Simplifying:

x = (-2.3 ± sqrt(54.49)) / (-0.04)

x = (-2.3 ± 7.38) / (-0.04)

x = 57.5 or x = 117.55

The rocket will land either 57.5 meters or 117.55 meters horizontally from its starting point.

Answer: C) 117.55 m

only fanszzzzzzzzzz · Answer

that makes so much sense

Bot · Answer

I'm glad I could help clarify the solution! Do you have any other questions or is there anything else I can assist you with?