Asked by Amber
At noon a private plane left Austin for Los Angeles, 2100 km away, flying at 500km/h. One hour later a jet left Los Angeles for Austin at 700 km/h. At what time did they pass each other?
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Answered by
Writeacher
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Answered by
bobpursley
The fly a combined 2100km
2100km=500km/hr*time1 + 700km/hr*time2
but time2=time1-1
2100=500*time1+700(time1-1)
now solve for time 1
2100km=500km/hr*time1 + 700km/hr*time2
but time2=time1-1
2100=500*time1+700(time1-1)
now solve for time 1
Answered by
Timothy
The equation to simply put words into numbers would be....."500x + 700x - 700 = 2100".....
Then if you solved for "x", you would get x=7/3.
So the plane and jet will meet in 7/3 hours. If yoy add that to 12:00 PM being noon, you would get 2:20, so the plane and the jet will meet at 2:20 PM.
Then if you solved for "x", you would get x=7/3.
So the plane and jet will meet in 7/3 hours. If yoy add that to 12:00 PM being noon, you would get 2:20, so the plane and the jet will meet at 2:20 PM.
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