A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation
y = –0.04x2 + 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.
208.02 m
416.03 m
0.52 m
208.19 m
I think it is A...?
I am sure you meant
y = -.04x^2 + 8.3x+ 4.3
when it lands, y = 0
.04x^2 - 8.3x - 4.3=0
x = (8.3 ± √69.578)/.08
= 208.0167.. or a negative value of x
= 208.02 m
Thank you...
wow old question
Yea so old.
Ur mom
To find the horizontal distance from the starting point where the rocket will land, we need to determine the value of x when y equals zero. This is because the rocket will hit the ground when its height, y, is zero.
Given the equation y = -0.04x^2 + 8.3x + 4.3, set y to zero and solve for x:
0 = -0.04x^2 + 8.3x + 4.3
We can now solve this quadratic equation. One way to do this is by factoring or using the quadratic formula. However, in this case, the equation does not have simple integer factors, so we will use the quadratic formula.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
Where a, b, and c are the coefficients in the equation: ax^2 + bx + c = 0.
In our equation, a = -0.04, b = 8.3, and c = 4.3.
Substituting these values into the quadratic formula, we have:
x = (-8.3 ± √(8.3^2 - 4(-0.04)(4.3))) / (2(-0.04))
Simplifying:
x = (-8.3 ± √(68.9 + 0.688)) / (-0.08)
x = (-8.3 ± √69.588) / (-0.08)
Now we can calculate the two possible values for x:
x₁ = (-8.3 + √69.588) / (-0.08) ≈ 208.018
x₂ = (-8.3 - √69.588) / (-0.08) ≈ -0.518
We are interested in the positive value for x, as the distance cannot be negative in this context. Therefore, the rocket will land approximately 208.02 meters horizontally from its starting point.
So the correct answer is A) 208.02 m.