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3 years ago

3. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation

2 answers:

34kurt3 years ago

3 0

<span>When it hit the ground
y = 0
y = –0.04x^2 + 8.3x + 4.3
0.04x^2 - 8.3x - 4.3 = 0
x = [-(-8.3) +/- √{(-8.3)^2 - 4 x 0.04 x (-4.3)}]/[2 x 0.04]
x = [8.3 +/- 8.34]/[0.08]
x = 208.02 m
A is right option
hope this helps</span>

babunello [35]3 years ago

3 0

Answer:

A) 208.02 meter

Step-by-step explanation:

We are given,

The path of the rocket is modeled by $y = -0.04x^2+8.3x+4.3$ .

We have that the when the rocket lands, the value of y = 0.

Substituting y= 0, we get,

$y=-0.04x^2+8.3x+4.3\\\\0=-0.04x^2+8.3x+4.3\\\\0.04x^2-8.3x-4.3=0\\\\(x+0.517)(x-208.02)=0\\\\x=-0.517,\ x=208.02$

As 'x' represents the horizontal distance, it cannot be negative.

So, we have x= 208.02 meter.

<h3>That is, the rocket will land after 208.02 meters.</h3>

Thus, option A is correct.

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See attachment for the new figure.

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