math-calculus

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a plane flew from red deer to winnipeg, flying distance of 1260 km. on the return journey, due to a strong head wind, the plane travelled 1200 km in the same time it took to complete the outward journey. on the outward journey, the plane be able to maintain an average speed 20km/hr greater than on the return journey.

if average speed of plane from red deer to winnipeg is xkm/hr, state an expression for the average speed of the plane from winnipeg to red deer in km/hr.

i wrote (x-20)km/hr

calculate average speed of the plane from winnipeg to red deer.

i also not real get how to do this. i go speed = distance/time?

calculate total flying time for round trip.

  • math-calculus -

    Time to W = 1260/x

    Time back if had gone all the way = (1260/1200)* time to W = 1260^2/1200x

    speed back =(x-20)

    so
    1260 = (x-20)(1260^2/1200x)
    1200 x = (x-20)(1260)
    1200 x = 1260 x - 20(1260)
    60 x = 20(1260)
    x = 420
    x-20 = 400

  • math-calculus -

    thanks very much damon :)

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