Lighthouses Ranging From
lighthouses ranging from
Help Wwith Solving Linear Quadratic Systems Algebraically?
From a lighthouse, the range of visibility on a clear day is 40 km. On a coordinate system, where (0,0) represents the lighthouse, a ship is travelling on a course represented by y = 2x + 0. Between what tow points on the course can the ship be seen from the lighthouse?
i vaguely remember doing problems like this before but i haven't done math course since last year. can anyone plz explain how to solve this?
what 2 points*
sry sry its y=2x + 80 not 0
thx Kimoshu i completely understand it and remember how to do it now. i'll give u the best answer once i'm able to
let d = distance from lighthouse to tow points = 40 km
d = √((x - x1)² + (y - y1)²) = √((x - 0)² + (y - 0)²)
= √(x² + y²) , y = 2x + 80
= √(x² + (2x + 80)²)
= √(x² + 4x² + 320x + 6400)
= √(5x² + 320x + 6400) = 40 km
5x² + 320x + 6400 = 1600
5x² + 320x + 4800 = 0
use the quad formula,
x = [-b ± √(b² - 4ac)]/2a
= [-320 ± √(320² - 4(5)(4800))]/2(5)
= [-320 ± √(102400 - 96000)]/10
= [-320 ± √(6400)]/10
= [-320 ± 80]/10
= -32 ± 8
x = -40 and -24, so
(x,y) = (x,2x + 80) = (-40,0) and(-24,32), between these two points the ship is within 40 km from the lighthouse...
CHECK: if you choose any point -40 ≤ x ≤ -24, d ≤ 40 km,
and x < -40 or x > -24, d > 40 km.
∮κιмσ
