## anonymous 3 years ago PLEASE HELP! A plane took 5 hours to get from Omaha to San Francisco with a tailwind of 28 mph. It took 6 hours to make the return trip with a headwind of the same speed, 28 mph. How fast was the plane traveling in still air?

1. anonymous

1680miles

2. anonymous

how did you calculate that? are you sure?? Thanks!!!

3. anonymous

@satellite73 ?? do you know how to solve this??

4. anonymous

yes

5. anonymous

and 1680 i think is too far

6. anonymous

i set up (5p+28)=6(p-28)..

7. anonymous

*5(p+28)

8. anonymous

close, i think it should be $5(p+28)=6(p-28)$

9. anonymous

if you call "p" the rate

10. anonymous

yeah what you said at first

11. anonymous

second

12. anonymous

Yeah but then when you simplify it's 28 and thats not an answer..

13. anonymous

oh i could be wrong, i jumped the gun maybe 1680 is the right answer

14. anonymous

but how do you get that?!?!?!

15. anonymous

lets solve $5(x+28)=6(x-28)$ $5x+140=6x-168$ $140=x-168$ $x=308$ so the rate is 308 miles per hour, but that is not the distance, that is the rate

16. anonymous

to find the distance replace the rate of 308 in either expression for example $5(308+28)=5\times 336=1680$ as stated above

17. anonymous

THANK YOU!! but that is the rate in still air as well?

18. anonymous

@Anyssah was right , i jumped the gun

19. anonymous

the rate in still air is 308 mph, yes. the distance is therefore 1680

20. anonymous

Its not asking for the distance thought, its the rate it traveled in still air.

21. anonymous

ok, then 308 mph is the right answer to that question

22. anonymous

Thank you guys :)