abbie1
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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Anyssah
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1680miles
abbie1
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how did you calculate that? are you sure?? Thanks!!!
abbie1
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@satellite73 ?? do you know how to solve this??
anonymous
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yes
anonymous
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and 1680 i think is too far
abbie1
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i set up (5p+28)=6(p-28)..
abbie1
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*5(p+28)
anonymous
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close, i think it should be
\[5(p+28)=6(p-28)\]
anonymous
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if you call "p" the rate
anonymous
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yeah what you said at first
anonymous
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second
abbie1
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Yeah but then when you simplify it's 28 and thats not an answer..
anonymous
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oh i could be wrong, i jumped the gun
maybe 1680 is the right answer
abbie1
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but how do you get that?!?!?!
anonymous
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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
anonymous
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to find the distance replace the rate of 308 in either expression
for example
\[5(308+28)=5\times 336=1680\] as stated above
abbie1
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THANK YOU!! but that is the rate in still air as well?
anonymous
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@Anyssah was right , i jumped the gun
anonymous
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the rate in still air is 308 mph, yes. the distance is therefore 1680
abbie1
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Its not asking for the distance thought, its the rate it traveled in still air.
anonymous
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ok, then 308 mph is the right answer to that question
abbie1
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Thank you guys :)