anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Formula please!
anonymous
  • anonymous
\[T= \frac{D}{R}=\frac{160}{25+x}=\frac{240}{25-x}\]
anonymous
  • anonymous
if x = current than rate going up stream is 25-x and rate going down stream is 25+x

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More answers

anonymous
  • anonymous
since \[T=\frac{D}{R} \] and these times are the same set them equal. then solve as a ratio \[160(25-x)=240(25+x)\]
anonymous
  • anonymous
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
anonymous
  • anonymous
oh yes of course. it is \[\frac{160}{25-x}=\frac{240}{25+x}\] i had it backwards sorry
anonymous
  • anonymous
smaller distance over slower rate = bigger distance over bigger rate. my mistake. lets solve this one
anonymous
  • anonymous
\[160(25+x)=240(25-x)\] \[2(25+x)=3(25-x)\] \[20+2x=75-3x\] \[5x=50\] \[x=10\]
anonymous
  • anonymous
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
anonymous
  • anonymous
isn't it 5x=25?
anonymous
  • anonymous
yea i screwed up twice. must be early
anonymous
  • anonymous
\[2(25+x)=3(25-x)\] \[50+2x=75-3x\] \[5x=25\] \[x=5\]
anonymous
  • anonymous
sorry. good eye!
anonymous
  • anonymous
can you do one more for me?
anonymous
  • anonymous
whew. \[\frac{160}{20}=\frac{240}{30}=8\] so trip is 8 hours in either case
anonymous
  • anonymous
I just need a formula
anonymous
  • anonymous
sure
anonymous
  • anonymous
yeah since clearly i am unable to solve a linear equation
anonymous
  • anonymous
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.
anonymous
  • anonymous
Find the speed of the boat in still water.
anonymous
  • anonymous
same exact eqution as before, only this time the variable will be speed, not current
anonymous
  • anonymous
again \[T=\frac{D}{R}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
so time going down is \[T=\frac{150}{x+6}\] and time going up is \[T=\frac{100}{x-6}\]
anonymous
  • anonymous
set them equal and solve. \[\frac{160}{x+6}=\frac{100}{x-6}\]
anonymous
  • anonymous
you can do this better than i can
anonymous
  • anonymous
typo \[\frac{150}{x+6}=\frac{100}{x-6}\]
anonymous
  • anonymous
Thanks!!!!!
anonymous
  • anonymous
welcome

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