Instructions: For the following real-world problem, solve using any method. Use what you’ve learned to determine which method would be best. Put your answer in the context of the problem and determine the appropriate final answer.A rocket carrying fireworks is launched from a hill 400400 feet above a lake. The rocket will fall into the lake after exploding at its maximum height. The model of the trajectory of the rocket and when it hits the lake is given by the equation −10x2+180x+400=0−10𝑥2+180𝑥+400=0 where x𝑥 is the number of seconds that pass.What are the two solutions to this equation? x=𝑥= or In this context, which of these answers makes sense to be the number of seconds that have passed before the rocket hits the lake? x=𝑥=
Question
Instructions: For the following real-world problem, solve using any method. Use what you’ve learned to determine which method would be best. Put your answer in the context of the problem and determine the appropriate final answer.A rocket carrying fireworks is launched from a hill 400400 feet above a lake. The rocket will fall into the lake after exploding at its maximum height. The model of the trajectory of the rocket and when it hits the lake is given by the equation −10x2+180x+400=0−10𝑥2+180𝑥+400=0 where x𝑥 is the number of seconds that pass.What are the two solutions to this equation? x=𝑥= or In this context, which of these answers makes sense to be the number of seconds that have passed before the rocket hits the lake? x=𝑥=
Solution 1
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -10, b = 180, and c = 400.
We can solve this equation using the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / 2a.
Substituting the values of a, b, and c into the formula, we get:
x = [-180 ± sqrt((180)^2 - 4*(-10)400)] / 2(-10) x = [-180 ± sqrt(32400 + 16000)] / -20 x = [-180 ± sqrt(48400)] / -20 x = [-180 ± 220] / -20
This gives us two solutions:
x1 = (180 - 220) / -20 = 2 seconds x2 = (180 + 220) / -20 = -20 seconds
In the context of this problem, time cannot be negative. Therefore, the only solution that makes sense is x = 2 seconds. This means that the rocket will hit the lake 2 seconds after it has been launched.
Solution 2
The given equation is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -10, b = 180, and c = 400.
To find the solutions of this equation, we can use the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / (2a).
Substituting the values of a, b, and c into the formula, we get:
x = [-180 ± sqrt((180)^2 - 4*(-10)400)] / (2-10) x = [-180 ± sqrt(32400 + 16000)] / -20 x = [-180 ± sqrt(48400)] / -20 x = [-180 ± 220] / -20
This gives us two solutions:
x1 = (180 - 220) / -20 = 2 seconds x2 = (180 + 220) / -20 = -20 seconds
In the context of this problem, time cannot be negative. Therefore, the only solution that makes sense is x = 2 seconds. This means that the rocket will hit the lake 2 seconds after it has been launched.
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