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Saw this question and have been unsure of how to solve it properly. Any help would be appreciated!

A pilot of a downed airplane fires the emergency flare into the sky. The path of the flare is modeled by the equation $y =-0.096(x-25)^2+60$, where $y$ is the height of the flare in meters when its horizontal distance from where it is propelled is $x$ meters. An emergency helicopter equipped with special binoculars has a line of sight to the spot where the flare was launched. The line of sight from the binocular is modeled by the equation $9x-10y=-14$.

Solve the system and give answers rounded to two decimal points.

The line of sight from the binoculars spots the flare twice. How high was the flare closest to the ground?

A.E
  • 2,473
  • You are looking for the intersection of the binocular's line of sight and the trajectory of the flare. So you should substitute $y=1.4+.9x$ into the equation for the flare. You will get two results, and I leave it to you to figure out which answer is correct. – A.E Jun 17 '13 at 16:18

1 Answers1

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Hint simplify and solve the following:

$$9x-10(-0.096(x-25)^2+60)=-14$$