A model rocket is launched from a roof into a large field. The path 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?

will some one please give the answers i’ve been sick for like a month and procrastinated math and did everything else but i learned nothing and i’m not even supposed to be in algebra and i need help

The correct answer is 208.02 m

the answer is 208.02 m

its 208.2

@Anonymous ...same bruh

yas

To find the horizontal distance at which the rocket will land, we need to determine the value of x when the rocket's height reaches the ground level (y = 0).

Given the equation of the rocket's path: y = -0.04x^2 + 8.3x + 4.3

To solve for x when y = 0, we substitute 0 for y in the equation:

0 = -0.04x^2 + 8.3x + 4.3

Now, we have a quadratic equation that we can solve. To do this, we can use factoring, completing the square, or the quadratic formula.

However, since factoring may not be straightforward in this case, let's use 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 case, a = -0.04, b = 8.3, and c = 4.3. Substituting these values into the quadratic formula:

x = (-(8.3) ± √((8.3)^2 - 4(-0.04)(4.3))) / (2(-0.04))

Simplifying further:

x = (-8.3 ± √(68.89 - (-0.86))) / (-0.08)
x = (-8.3 ± √(69.75)) / (-0.08)

Taking the square root of 69.75:

x = (-8.3 ± 8.35) / (-0.08)

There are two possible values of x:

x₁ = (-8.3 + 8.35) / (-0.08) ≈ -0.0625
x₂ = (-8.3 - 8.35) / (-0.08) ≈ 206.25

Since distance cannot be negative in this context, the rocket will land approximately 206.25 meters horizontally from its starting point on the roof.

my answer is 208.02 meters

Since you want the distance horizontally from the starting point, you want y to be 4.3m which is the initial height, thus

0 = -.04x^2 + 8.3x
x(.04x - 8.3) = 0
x = 0 , which would be our starting point and
x = 8.3/.04 or 207.5 m

your answer would be the distance from a height of 4.3 m to the ground, which would not be a horizontal line