Question 845256
A pilot of a downed airplane fires the emergency flare into the sky. 
The path of the flare is modeled by the equation h= -0.096(d-25)^2 + 60, where h is the height of the flare in metres when its horizontal distance from where it was propelled is d metres.
 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
:
Rewrite the first equation using x and y
y = -.096(x-25)^2 + 60
Write the 2nd equation in the slope intercept form
y = .9x + 1.4
:
a) solve the system and give answers rounded to two decimal places.
-.096(x-25)^2 + 60 = .9x + 1.4
-.096(x^2 - 50x + 625) + 60
-.096x^2 + 4.8x - 60 + 60 = .9x + 1.4
-.096x^2 + 4.8x - .9x - 1.4 = 0
-.096x^2 + 3.9x - 1.4 = 0
using the quadratic formula, obtained two solutions
x = 40.26 m, horizontal distance
x = .36 m  "  "
:
b) the line of sight from the binoculars spots the flare twice.
 How high was the flare closest to the ground when it was spotted the first time?
Replace x with .36 in the easiest equation y = .9x + 1.4
y = .9(.36) + 1.4
y = 1.72 m above the ground, when spotted the 1st time