anonymous
  • anonymous
A tortoise crawling at a rate of .1mi/h passes a resting hare. The hare wants to rest another 30 minutes before chasing the tortoise at a rate of .5mi/h. How many feet must the hare run to catch the tortoise?
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
we start with an easy question in 30 minutes is what portion of an hour?
anonymous
  • anonymous
half
anonymous
  • anonymous
k good so in half an hour at a rate of \(.1\) mph how far did the tortoise go?

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anonymous
  • anonymous
In ft, he went 264ft
anonymous
  • anonymous
lets keep it in miles, since that is the unit given other wise we are going to get all messed up with the units
anonymous
  • anonymous
Well, then he would have traveled .05 miles
anonymous
  • anonymous
k good now we use "distance equals rate times time"
anonymous
  • anonymous
the tortoise has rate \(.1\) and a \(.05\) mile head start, so his distance is going to be \[.05+.1t\] where \(t\) is time (in hours)
anonymous
  • anonymous
the hare has rate \(.5\) so his distance is going to be \(.5t\)
anonymous
  • anonymous
and since evidently they meet at the same distance, you can put \[.05+.1t=.5t\] and solve for \(t\)
anonymous
  • anonymous
t= -1?
anonymous
  • anonymous
not too hard with these decimals, or we can work with whole numbers by writing \[5+10t=50t\] either way, you get \[5=40t\] so \[t=\frac{5}{40}=\frac{1}{8}\]
anonymous
  • anonymous
or \[.05+.1t=.5t\\.05=.4t\\ \frac{.05}{.4}=t\]still get \(t=\frac{5}{40}=\frac{1}{8}\)
anonymous
  • anonymous
or if you like decimals \(t=.125\) now we find the distance it is \(.5\times .125=.0625\) miles convert to feet by multiplying \[5280\times .0625\]
anonymous
  • anonymous
So, the hare has to travel 330 ft?

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