Answer:the plane in still air is 904 km/h
the rate of the wind is 104 km/h
Step-by-step explanation:
Let x represent the rate or speed of the plane in still air.
Let y represent the rate or speed of the wind.
Flying against the wind, an airplane travels 4000 kilometers in 5 hours. This means that the total speed of the airplane would be (x - y) km/h
Distance = speed × time
Therefore
4000 = 5(x - y)
4000 = 5x - 5y - - - - - - - - -1
Flying with the wind, the same plane travels 5040 kilometers in 4 hours. This means that the total speed of the airplane would be (x + y) km/h. Therefore,
5040 = 4(x + y)
5040 = 4x + 4y - - - - - - - - - 2
Multiplying equation 1 by 4 and equation 2 by 5, it becomes
16000 = 20x - 20y
20160 = 20x + 20y
Subtracting, it becomes
- 4160 = - 40y
y = - 4160/-40
y = 104
Substituting y = 104 into equation 1, it becomes
4000 = 5x - 5 × 104
4000 = 5x - 520
5x = 4000 + 520
5x = 4520
x = 4520/5 = 904
Answer:
1/4y
Step-by-step explanation:
(-x^-1)^0/(4y^-1*y^2)
anything to the power of 0 gives you 1,
1/(4y^-1*y^2)
-1+2=1
1/4y^1=1/4y
