Question

answer this........

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Answer 1

Hii

Here is ur solution

speed of aircraft= x

Distance = 2800 km

Time = Distance / Speed = 2800/x 

Given:

The speed has reduced by 100 and time is reduced by 30 min

New 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)  =  1/2

x² - 100x - 560000  = 0

x² - 800x + 700x - 560000  =  0

x(x - 800) + 700(x - 800)  =  0

(x + 700) (x - 800)  =  0

SO,

x  =  - 700 or 800

Thus original duration of the flight =  2800/800 = 7/2 hours

OR = 3 hr 30 mins

^_^ mark as brainliest ^_^

Answer 2

let original speed be x kmph
distance =2800km
time=2800/x hr

new speed=x-100 kmph
distance =2800km
new time =2800/x-100 hr

difference in time =1/2hr

according to question

2800/×-100 -2800/x=1/2
2800x-2800x+280000/x^2-100x=1/2
280000/x^2-100x=1/2

x^2-100×=560000
x^2-100x-560000=0
x^2-800x+700x-560000=0
×(×-800)+ 700(×-800)
(x-800)(x+700)
x=800

original time=2800/800=3.5hr

HOPE IT HELP YOU
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