-x+y+z=-3,-2x-y+3z=2,3x-4y+z=1 me x, y, z ka man btaiye
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Answer:
x =2 y=6 z=4
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A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
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Let the usual speed of the plane = x km/hr
Distance to the destination = 1500 km
Increased speed = 100 km/hr
So, according to the question
1500/x - 1500(x+100) = 30/60
x2 + 100x - 300000 = 0
x2 + 600x - 500x - 300000 = 0
(x + 600)(x - 500) = 0
x = -600 or 500
Since, speed can not be negative, x = 500
Usual speed of plane is 500 k
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