A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = -0.04x^2 + 8.3x + 4.3, 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? Round your answer to the nearest hundredth meter.
A. 208.02 m.
B. 416.03 m.
C. .52 m.
D. 208.19 m.
But it is not. Try again. It is A.
just solve for x when y=0.
be careful, as A and D are almost the same
I think D would be the answer.
Ohhh yea.
hoping this is right.. thx if it is
To find the horizontal distance that the rocket will land, we need to find the x-value where the height, y, is equal to zero. This represents the point at which the rocket lands on the ground.
First, let's set the equation equal to zero:
-0.04x^2 + 8.3x + 4.3 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -0.04, b = 8.3, and c = 4.3.
Plugging in these values into the quadratic formula, we get:
x = (-8.3 ± √(8.3^2 - 4(-0.04)(4.3))) / 2(-0.04)
Simplifying further:
x = (-8.3 ± √(68.89 + 0.688)) / -0.08
x = (-8.3 ± √(69.578)) / -0.08
x = (-8.3 ± 8.34) / -0.08
Now we have two possible values for x:
x = (-8.3 + 8.34) / -0.08 ≈ -0.52
x = (-8.3 - 8.34) / -0.08 ≈ -415.31
Since we're looking for the horizontal distance, we can ignore the negative value.
Therefore, the rocket will land approximately 415.31 meters horizontally from its starting point.
Rounding this to the nearest hundredth, the answer is:
B. 416.03 m.