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prohojiy [21]
3 years ago
14

In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The average speed of the trip was decreased

by 200 km/hr and thus the time of flight increased by 30 minutes. Find the average speed of the aircraft originally. __________ Please don't spam or your answer will be reported :) Thank you :)
Mathematics
1 answer:
UkoKoshka [18]3 years ago
5 0

The average speed would be 600 km/hr.

Step-by-step explanation:

Let the original speed be 'x'

Let the decreased speed be 'x-200'

Distance = 600 km

Time increased by 30 minutes

According to question, it becomes,

\dfrac{600}{x-200}-\dfrac{600}{x}=\dfrac{30}{60}\\\\\dfrac{x-(x-200)}{x^2-200x}=\dfrac{1}{2\times 600}\\\\\dfrac{200}{x^2-200x}=\dfrac{1}{1200}\\\\240000=x^2-200x\\\\x^2-200x-240000=0\\\\x=600,-400

So, The average speed would be 600 km/hr.

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Answer:

<h3>            x = -9,  y = -13 </h3><h3>    or    x = 13,   y = 9</h3><h3>    or    x = -13,  y = -9</h3><h3>    or     x = 9,   y = 13</h3>

Step-by-step explanation:

x^2+y^2=250\\\\x^2-2xy+y^2+2xy=250\\\\(x-y)^2=250-2xy\\\\(x-y)^2=250-2\cdot117\\\\ (x-y)^2=16\\\\x-y=4\qquad\qquad\vee\qquad \qquad  x-y=-4\\\\x=4+y \qquad\qquad \vee\qquad\qquad x=-4+y\\\\(y+4)y=117\qquad\vee\qquad\quad (y-4)y=117\\\\y^2+4y-117=0\qquad\vee\qquad y^2-4y-117=0\\\\y=\dfrac{-4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\qquad\vee\qquad y=\dfrac{4\pm\sqrt{4^2-4(-117)}}{2\cdot1}\\\\y=\dfrac{-4\pm\sqrt{16+468}}{2}\qquad\ \ \vee\qquad y=\dfrac{4\pm\sqrt{16+468}}{2}

y_1=\dfrac{-4-22}{2}\ ,\quad y_2=\dfrac{-4+22}{2}\ ,\quad y_3=\dfrac{4-22}{2}\ ,\quad y_4=\dfrac{4+22}{2}\\\\y_1=-13\ ,\qquad y_2=9\ ,\qquad\quad\qquad\ y_3=-9\ ,\qquad y_4=13\\\\x_{1,2}=4+y_{1,2}\qquad\qquad\qquad\qquad\qquad x_{3,4}=-4+y_{3,4}\\\\x_1=-9\ ,\qquad x_2=13\ ,\qquad\quad\qquad x_3=-13\ ,\qquad x_4=9

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