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in still air a bird flies 36 miles per hour with the wind it travels 6 miles farther in 3 hours then it flies in 6 hours against the wind, in miles per hour? round your answer to the nearest tenth

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frika

Let v be the speed of wind.

1. With the wind a bird flies 3 hours. The speed of bird with the wind is 36+v mi/h.

Then it flies the distance

[tex](36+v)\cdot 3[/tex] miles.

2. Against the wind a bird flies 6 hours with the speed 36-v, then it flies the distance

[tex](36-v)\cdot 6[/tex] miles.

3. If with the wind a bird travels 6 miles farther than against wind, you have an equation

[tex](36+v)\cdot 3=(36-v)\cdot 6+6.[/tex]

4. Solve it

[tex]108+3v=216-6v+6,\\ \\9v=222-108,\\ \\v=\dfrac{114}{9}=\dfrac{38}{3}\approx 12.7[/tex]

5. You find the speed of the wind [tex]v=\dfrac{38}{3}\approx 12.7[/tex] mi/h. Then a bird:

  • has speed [tex]36-12.7=23.3[/tex] mi/h against the wind;
  • has speed [tex]36+12.7=48.7[/tex] mi/h with the wind;
  • flies distance [tex](36+v)\cdot 3=\left(36+\dfrac{38}{3}\right)\cdot 3=108+38=146[/tex] miles with the wind;
  • flies distance [tex](36+v)\cdot 3=\left(36-\dfrac{38}{3}\right)\cdot 6=216-76=140[/tex] miles against the wind.

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