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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?

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how did you calculate that? are you sure?? Thanks!!!
@satellite73 ?? do you know how to solve this??

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Other answers:

and 1680 i think is too far
i set up (5p+28)=6(p-28)..
close, i think it should be \[5(p+28)=6(p-28)\]
if you call "p" the rate
yeah what you said at first
Yeah but then when you simplify it's 28 and thats not an answer..
oh i could be wrong, i jumped the gun maybe 1680 is the right answer
but how do you get that?!?!?!
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
to find the distance replace the rate of 308 in either expression for example \[5(308+28)=5\times 336=1680\] as stated above
THANK YOU!! but that is the rate in still air as well?
@Anyssah was right , i jumped the gun
the rate in still air is 308 mph, yes. the distance is therefore 1680
Its not asking for the distance thought, its the rate it traveled in still air.
ok, then 308 mph is the right answer to that question
Thank you guys :)

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