document.write( "Question 42046: you have at least $30.00in change consisting of dimes and quaters. Write an inequality that shows the differant number of coins \n" ); document.write( "

Algebra.Com's Answer #27206 by josmiceli(19441) $\"\"$ $\"About$

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You have at least $30.00 in change consisting of dimes and quaters.
\n" ); document.write( "What's the least number of coins you can have? If you could have all quarters,
\n" ); document.write( "that would be the least number, but you must have some dimes.
\n" ); document.write( "All quarters would be $30/.25 = 120 coins.
\n" ); document.write( "You can't take away 1 quarter to get change in dimes. You must take away 2
\n" ); document.write( "quarters, and get 5 dimes change.
\n" ); document.write( "120 coins - 2 quarters + 5 dimes = 123
\n" ); document.write( "That's the low end of the inequality.
\n" ); document.write( "The most coins you could have would be all dimes.
\n" ); document.write( "That would be $30/.10 = 300, but you must have quarters. Dimes don't
\n" ); document.write( "divide evenly into 1 quarter, but there are 5 dimes in 2 quarters.
\n" ); document.write( "300 coins - 5 dimes + 2 quarters = 297
\n" ); document.write( " $\"123+%3C=+x+%3C=+297\"$
\n" ); document.write( "for exactly $30 worth of dimes and quarters \n" ); document.write( "

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