document.write( "Question 409077: When mineral deposits formed a coating 1mm thick on the inside of a pipe the area through wich fluid can flow was reduced by 20%. Find the original inside diameter of the pipe.
\n" ); document.write( "(Remeber: area of circle = pi(r^2) and diameter = 2r)
\n" ); document.write( "Thanks for the help =)
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Algebra.Com's Answer #288050 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
I think they are saying that the area of the pipe opening has been
\n" ); document.write( "reduced by 20%.
\n" ); document.write( "Let \"A\" = original area
\n" ); document.write( "Let \"r\" = original radius
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\n" ); document.write( "If the diameter is reduced by 1 mm, then the
\n" ); document.write( "radius is reduced by .5 mm.
\n" ); document.write( "given:
\n" ); document.write( "(1) \"A+=+pi%2Ar%5E2\"
\n" ); document.write( "(2) \"A+-+.2A+=+pi%2A%28r+-+.5%29%5E2\"
\n" ); document.write( "------------------------
\n" ); document.write( "(2) \".8A+=+pi%2A%28r+-+.5%29%5E2\"
\n" ); document.write( "substituting from (1):
\n" ); document.write( "(2) \".8%2Api%2Ar%5E2+=+pi%2A%28r%5E2+-+r+%2B+.25%29\"
\n" ); document.write( "(2) \".8%2Ar%5E2+=+r%5E2+-+r+%2B+.25\"
\n" ); document.write( "(2) \".2r%5E2+-+r+%2B+.25+=+0\"
\n" ); document.write( "(2) \"20r%5E2+-+100r+%2B+25+=+0\"
\n" ); document.write( "(2) \"4r%5E2+-+20r+%2B+5+=+0\"
\n" ); document.write( "Use quadratic formula
\n" ); document.write( "\"r+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "\"a+=+4\"
\n" ); document.write( "\"b+=+-20\"
\n" ); document.write( "\"c+=+5\"
\n" ); document.write( "\"r+=+%28-%28-20%29+%2B-+sqrt%28+%28-20%29%5E2-4%2A4%2A5+%29%29%2F%282%2A4%29+\"
\n" ); document.write( "\"r+=+%2820+%2B-+sqrt%28+400+-+80+%29%29%2F8+\"
\n" ); document.write( "\"r+=+%2820+%2B-+sqrt%28+320+%29%29%2F8+\"
\n" ); document.write( "\"r+=+%2820+%2B+8%2Asqrt%285%29%29%2F8\"
\n" ); document.write( "\"r+=+5%2F2+%2B+sqrt%285%29\"
\n" ); document.write( "\"r+=+4.736\" mm
\n" ); document.write( "The original diameter = 2r, or 9.472 mm
\n" ); document.write( "check answer:
\n" ); document.write( "(2) \"A+-+.2A+=+pi%2A%28r+-+.5%29%5E2\"
\n" ); document.write( "(2) \".8A+=+pi%2A%284.736+-+.5%29%5E2\"
\n" ); document.write( "(2) \".8A+=+pi%2A4.236%5E2\"
\n" ); document.write( "(2) \".8A+=+3.141%2A17.944\"
\n" ); document.write( "(2) \".8A+=+56.361\"
\n" ); document.write( "(2) \"A+=+70.451\" mm2
\n" ); document.write( "and
\n" ); document.write( "(1) \"A+=+pi%2Ar%5E2\"
\n" ); document.write( "(1) \"A+=+3.141%2A4.736%5E2\"
\n" ); document.write( "(1) \"A+=+70.451\" mm2
\n" ); document.write( "OK
\n" ); document.write( "Hope I got it\r
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