document.write( "Question 813685: a small motorboat travels 10mph in still water it takes 3 hours longer to travel 68 miles going upsteam than it does going downstream. find the rate of the current \n" ); document.write( "

Algebra.Com's Answer #489891 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+c+\"$ = the rate of the current in mi/hr
\n" ); document.write( " $\"+10+%2B+c+\"$ = speed of the boat going downstream
\n" ); document.write( " $\"+10+-+c+\"$ = speed of the boat going upstream
\n" ); document.write( "Let $\"+t+\"$ = time in hrs for going downstream
\n" ); document.write( "---------------------------------------
\n" ); document.write( "Equation for going upstream:
\n" ); document.write( "(1) $\"+68+=+%28+10+-+c+%29%2A%28+t+%2B+3+%29+\"$
\n" ); document.write( "Equation for going downstream:
\n" ); document.write( "(2) $\"+68+=+%28+10+%2B+c+%29%2At+\"$
\n" ); document.write( "-------------------
\n" ); document.write( "(1) $\"+68+=+10t+-+c%2At+%2B+30+-+3c+\"$
\n" ); document.write( "(1) $\"+68+=+%28+10+-+c+%29%2At+%2B+30+-+3c+\"$
\n" ); document.write( "(1) $\"+%28+10+-+c+%29%2At+=+3c+%2B+38+\"$
\n" ); document.write( "(1) $\"+t+=+%28+3c+%2B+38+%29+%2F+%28+10+-+c+%29+\"$
\n" ); document.write( "--------------------------
\n" ); document.write( "By substitution:
\n" ); document.write( "(2) $\"+68+=+%28+10+%2B+c+%29%2A%28+3c+%2B+38+%29+%2F+%28+10+-+c+%29+\"$
\n" ); document.write( "(2) $\"+68%2A%28+10+-+c+%29+=+%28+10+%2B+c+%29%2A%28+3c+%2B+38+%29+\"$
\n" ); document.write( "(2) $\"+680+-+68c+=+30c+%2B+3c%5E2+%2B+380+%2B+38c+\"$
\n" ); document.write( "(2) $\"+3C%5E2+%2B+136C+-+300+=+0+\"$
\n" ); document.write( "Use the quadratic formula
\n" ); document.write( " $\"+C+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$ ( note: C and not c is the solution )
\n" ); document.write( " $\"+a+=+3+\"$
\n" ); document.write( " $\"+b+=+136+\"$
\n" ); document.write( " $\"+c+=+-300+\"$
\n" ); document.write( "-----------
\n" ); document.write( " $\"+C+=+%28-136+%2B-+sqrt%28+136%5E2-4%2A3%2A%28-300%29+%29%29%2F%282%2A3%29+\"$
\n" ); document.write( " $\"+C+=+%28-136+%2B-+sqrt%28+18496+%2B+3600+%29%29%2F6+\"$
\n" ); document.write( " $\"+C+=+%28+-136+%2B+sqrt%28+22096+%29+%29+%2F+6+\"$
\n" ); document.write( " $\"+C+=+%28+-136+%2B+148.65+%29+%2F+6+\"$
\n" ); document.write( " $\"+C+=+2.108+\"$
\n" ); document.write( "The rate of the current is 2.108 mi/hr
\n" ); document.write( "-----------
\n" ); document.write( "check:
\n" ); document.write( "(2) $\"+68+=+%28+10+%2B+c+%29%2At+\"$
\n" ); document.write( "(2) $\"+68+=+%28+10+%2B+2.108+%29%2At+\"$
\n" ); document.write( "(2) $\"+t+=+68+%2F+12.108+\"$
\n" ); document.write( "(2) $\"+t+=+5.616+\"$ hrs
\n" ); document.write( "---------------
\n" ); document.write( "(1) $\"+68+=+%28+10+-+c+%29%2A%28+t+%2B+3+%29+\"$
\n" ); document.write( "(1) $\"+68+=+%28+10+-+2.108+%29%2A%28+5.616+%2B+3+%29+\"$
\n" ); document.write( "(2) $\"+68+=+7.892%2A8.616+\"$
\n" ); document.write( "(2) $\"+68+=+67.997+\"$
\n" ); document.write( "pretty close
\n" ); document.write( "Hope I got it \n" ); document.write( "

\n" );