You can put this solution on YOUR website!
A plane takes off 200 miles east of its destination with a 50 mph wind from the south. If the plane’s velocity is 250 mph and its heading is always toward its destination, its path can be described by the equation below. If the location of the initial airport is (0,0) and the location of the destination airport is (200, 0), y represents the perpendicular distance from the straight line path between initial location and the destination location. When x = 150 miles, how far from the straight line path is the plane? That is, what is y?
y=0.5x^0.8-.5x^1.2 where x is greater than or equal to zero and less than or equal to 200.
y = 0.5(150)^0.8 - 0.5(150)^1.2
y = 27.5323 - 204.3052
y = -176.77 miles
Comment: That doesn't make any sense.