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Answer this........wrong answers Will be reported and deleted.... |
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Answer» is ur solutionspeed of aircraft= xDistance = 2800 kmTime = Distance / Speed = 2800/x Given: The speed has reduced by 100 and time is reduced by 30 minNew speed = (x – 100)That is, 30 MIN = 30/60 HR = (1/2) hr Time taken= 2800/(x – 100) hr=2800/(x – 100) = (2800/x) + (1/2)2800/(x-100) -` (2800/x) = 1/2x² - 100x - 560000 = 0x² - 800x + 700x - 560000 = 0X(x - 800) + 700(x - 800) = 0(x + 700) (x - 800) = 0SO,x = - 700 or 800Thus ORIGINAL duration of the flight = 2800/800 = 7/2 hoursOR = 3 hr 30 MINS^_^ mark as brainliest ^_^ |
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