A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = mc006-1.jpg0.04xmc006-2.jpg+ 5.8x + 4.9, 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?
Also not trying to be pushy, but I need this answer tonight and no one in my family knows the answer so can someone please help me?
Just type the equation normally using your normal keyboard
I can't make out what "mc006-1.jpg0.04xmc006-2.jpg" is
To find how far horizontally from its starting point the rocket will land, we need to determine the value of x when the height of the rocket (y) is equal to zero.
Given the equation y = -0.04x^2 + 5.8x + 4.9, we can set y = 0 and solve for x.
0 = -0.04x^2 + 5.8x + 4.9
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Since factoring is not straightforward in this case, let's use the quadratic formula.
The quadratic formula is given by x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation (in this case, a = -0.04, b = 5.8, and c = 4.9).
Substituting these values into the formula, we get:
x = (-5.8 ± sqrt(5.8^2 - 4(-0.04)(4.9))) / (2(-0.04))
Simplifying this equation will give us the two possible values of x. Since the rocket can only land once, we need to choose the appropriate value.
I'll calculate this using a calculator.
(Solving the equation using the quadratic formula)
x = (-5.8 ± sqrt(33.64 - (-0.784))) / (-0.08)
x = (-5.8 ± sqrt(34.424)) / (-0.08)
x ≈ (-5.8 ± 5.87) / -0.08
Now we evaluate the two possible values of x:
x ≈ 1.33 or x ≈ 71.42
Therefore, the rocket will land either approximately 1.33 meters or 71.42 meters horizontally from its starting point.