document.write( "Question 113995This question is from textbook Introductory and Intermediate Algebra
\n" ); document.write( ": The Hudson River flows at a rate of 3mph. A patrol boat travels 60 miles upriver and returns in a total time of 9 hr. What is the speed of the boat in still water?\r
\n" ); document.write( "\n" ); document.write( "D= rt upriver - 9r = 60 still water - 3rt = 60
\n" ); document.write( "Have I set this equation up correctly? Where do I go from here? Please help me.
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Algebra.Com's Answer #82960 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let r=speed of the boat in still water\r
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\n" ); document.write( "\n" ); document.write( "When the boat goes upstream, we subtract 3mph from r to get (the boat is slowed by 3mph)\r
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\n" ); document.write( "\n" ); document.write( "\"60=%28r-3%29t%5B1%5D\"\r
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\n" ); document.write( "\n" ); document.write( "When the boat goes downstream, we can add 3mph to r to get (the boat is sped up by 3mph)\r
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\n" ); document.write( "\n" ); document.write( "\"60=%28r%2B3%29t%5B2%5D\"\r
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\n" ); document.write( "\n" ); document.write( "Now since the distance is the same we can set the two equations equal to one another to get\r
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\n" ); document.write( "\n" ); document.write( "\"60%2F%28r-3%29=t%5B1%5D\" Now solve for \"t%5B1%5D\" in the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"60%2F%28r%2B3%29=t%5B2%5D\" Now solve for \"t%5B2%5D\" in the second equation\r
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\n" ); document.write( "\n" ); document.write( "Since the total time was 9 hr, this means \"t%5B1%5D%2Bt%5B2%5D=9\"\r
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\n" ); document.write( "\n" ); document.write( "\"60%2F%28r-3%29%2B60%2F%28r%2B3%29=9\" Plug in \"t%5B1%5D=60%2F%28r-3%29\" and \"t%5B1%5D=60%2F%28r%2B3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28r-3%29%28r%2B3%29%2860%2F%28r-3%29%2B60%2F%28r%2B3%29%29=%28r-3%29%28r%2B3%29%289%29\" multiply by the LCD \"%28r-3%29%28r%2B3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"60%28r%2B3%29%2B60%28r-3%29=%28r-3%29%28r%2B3%29%289%29\" Distribute\r
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\n" ); document.write( "\n" ); document.write( "\"60%28r%2B3%29%2B60%28r-3%29=%28r%5E2-9%29%289%29\" Foil\r
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\n" ); document.write( "\n" ); document.write( "\"60r%2B180%2B60r-180=9r%5E2-81\" Distribute again\r
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\n" ); document.write( "\n" ); document.write( "\"120r=9r%5E2-81\" Combine like terms\r
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\n" ); document.write( "\n" ); document.write( "\"0=9r%5E2-81-120r\" Subtract 120r from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"0=9r%5E2-120r-81\" Rearrange the terms\r
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\n" ); document.write( "\n" ); document.write( "Now let's use the quadratic formula to solve for r:\r
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\n" ); document.write( "\n" ); document.write( "Starting with the general quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"ar%5E2%2Bbr%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "the general solution using the quadratic equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29\"\r
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\n" ); document.write( "\n" ); document.write( "So lets solve \"9%2Ar%5E2-120%2Ar-81=0\" ( notice \"a=9\", \"b=-120\", and \"c=-81\")\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28--120+%2B-+sqrt%28+%28-120%29%5E2-4%2A9%2A-81+%29%29%2F%282%2A9%29\" Plug in a=9, b=-120, and c=-81\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+sqrt%28+%28-120%29%5E2-4%2A9%2A-81+%29%29%2F%282%2A9%29\" Negate -120 to get 120\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+sqrt%28+14400-4%2A9%2A-81+%29%29%2F%282%2A9%29\" Square -120 to get 14400 (note: remember when you square -120, you must square the negative as well. This is because \"%28-120%29%5E2=-120%2A-120=14400\".)\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+sqrt%28+14400%2B2916+%29%29%2F%282%2A9%29\" Multiply \"-4%2A-81%2A9\" to get \"2916\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+sqrt%28+17316+%29%29%2F%282%2A9%29\" Combine like terms in the radicand (everything under the square root)\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+6%2Asqrt%28481%29%29%2F%282%2A9%29\" Simplify the square root (note: If you need help with simplifying the square root, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B-+6%2Asqrt%28481%29%29%2F18\" Multiply 2 and 9 to get 18\r
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\n" ); document.write( "\n" ); document.write( "So now the expression breaks down into two parts\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28120+%2B+6%2Asqrt%28481%29%29%2F18\" or \"r+=+%28120+-+6%2Asqrt%28481%29%29%2F18\"\r
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\n" ); document.write( "\n" ); document.write( "Now break up the fraction\r
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\n" ); document.write( "\n" ); document.write( "\"r=%2B120%2F18%2B6%2Asqrt%28481%29%2F18\" or \"r=%2B120%2F18-6%2Asqrt%28481%29%2F18\"\r
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\n" ); document.write( "\n" ); document.write( "Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"r=20+%2F+3%2Bsqrt%28481%29%2F3\" or \"r=20+%2F+3-sqrt%28481%29%2F3\"\r
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\n" ); document.write( "\n" ); document.write( "So these expressions approximate to\r
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\n" ); document.write( "\n" ); document.write( "\"r=13.9772373998204\" or \"r=-0.643904066487102\"\r
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\n" ); document.write( "\n" ); document.write( "So our possible solutions are:\r
\n" ); document.write( "\n" ); document.write( "\"r=13.9772373998204\" or \"r=-0.643904066487102\"\r
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\n" ); document.write( "\n" ); document.write( "However, we cannot have a negative speed, so our only solution is \"r=13.9772373998204\"\r
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\n" ); document.write( "\n" ); document.write( "So the speed of the boat in still water is about 13.977 mph
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