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anonymous

  • 5 years ago

A boat can go 25 mph in still water. It takes as long to go 160 miles upstream as it does to go downstream 240 miles. How fast is the current?

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  1. anonymous
    • 5 years ago
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    Formula please!

  2. anonymous
    • 5 years ago
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    \[T= \frac{D}{R}=\frac{160}{25+x}=\frac{240}{25-x}\]

  3. anonymous
    • 5 years ago
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    if x = current than rate going up stream is 25-x and rate going down stream is 25+x

  4. anonymous
    • 5 years ago
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    since \[T=\frac{D}{R} \] and these times are the same set them equal. then solve as a ratio \[160(25-x)=240(25+x)\]

  5. anonymous
    • 5 years ago
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    divide both sides by 80 to make life easier \[2(25-x)=3(25+x)\] \[20-2x=75+3x\] \[5x=-50\] \[x=-10\] so i probably messed up and x = 10

  6. anonymous
    • 5 years ago
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    oh yes of course. it is \[\frac{160}{25-x}=\frac{240}{25+x}\] i had it backwards sorry

  7. anonymous
    • 5 years ago
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    smaller distance over slower rate = bigger distance over bigger rate. my mistake. lets solve this one

  8. anonymous
    • 5 years ago
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    \[160(25+x)=240(25-x)\] \[2(25+x)=3(25-x)\] \[20+2x=75-3x\] \[5x=50\] \[x=10\]

  9. anonymous
    • 5 years ago
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    now we check. if the current is 10 mpr then the rate going up stream is 25-10=15mph . \[\frac{160}{15}=10\tfrac{2}{3}\] hours

  10. anonymous
    • 5 years ago
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    isn't it 5x=25?

  11. anonymous
    • 5 years ago
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    yea i screwed up twice. must be early

  12. anonymous
    • 5 years ago
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    \[2(25+x)=3(25-x)\] \[50+2x=75-3x\] \[5x=25\] \[x=5\]

  13. anonymous
    • 5 years ago
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    sorry. good eye!

  14. anonymous
    • 5 years ago
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    can you do one more for me?

  15. anonymous
    • 5 years ago
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    whew. \[\frac{160}{20}=\frac{240}{30}=8\] so trip is 8 hours in either case

  16. anonymous
    • 5 years ago
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    I just need a formula

  17. anonymous
    • 5 years ago
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    sure

  18. anonymous
    • 5 years ago
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    yeah since clearly i am unable to solve a linear equation

  19. anonymous
    • 5 years ago
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    A boat can go 150 miles downstream in the same time it can go 100 miles upstream. The speed of the current is 6 miles per hour.

  20. anonymous
    • 5 years ago
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    Find the speed of the boat in still water.

  21. anonymous
    • 5 years ago
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    same exact eqution as before, only this time the variable will be speed, not current

  22. anonymous
    • 5 years ago
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    again \[T=\frac{D}{R}\]

  23. anonymous
    • 5 years ago
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    if you put x = speed of the boat then you know that the speed going downstream is x + 6 whereas the speed going up is x - 6

  24. anonymous
    • 5 years ago
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    so time going down is \[T=\frac{150}{x+6}\] and time going up is \[T=\frac{100}{x-6}\]

  25. anonymous
    • 5 years ago
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    set them equal and solve. \[\frac{160}{x+6}=\frac{100}{x-6}\]

  26. anonymous
    • 5 years ago
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    you can do this better than i can

  27. anonymous
    • 5 years ago
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    typo \[\frac{150}{x+6}=\frac{100}{x-6}\]

  28. anonymous
    • 5 years ago
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    Thanks!!!!!

  29. anonymous
    • 5 years ago
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    welcome

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