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 + 6.5x + 6.7, 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. (Hint: When does y = 0)
To find the horizontal distance at which the rocket will land, we need to determine the value of x when y equals 0.
We are given the equation y = 0.04x + 6.5x + 6.7. To find the value of x when y = 0, we substitute 0 for y and solve for x:
0 = 0.04x + 6.5x + 6.7
To simplify the equation, we combine like terms:
0 = 6.54x + 6.7
Next, we isolate the variable x by subtracting 6.7 from both sides of the equation:
-6.7 = 6.54x
Now, we divide both sides of the equation by 6.54 to solve for x:
x = -6.7 / 6.54
Using a calculator, we find that x is approximately -1.02 (rounded to two decimal places).
Since the distance cannot be negative, we take the absolute value of -1.02, which gives us 1.02.
Therefore, the rocket will land approximately 1.02 meters horizontally from its starting point.
Try typing instead of copying.
You surely meant
y = 0.04x^2 + 6.5x + 6.7
Even that is wrong, since it would indicate that the rocket never came back down. That 0.04x^2 is clearly wrong. Whatever. Fix your equation as necessary, and then just solve the quadratic in the usual way - use the quadratic formula.