document.write( "Question 747998: A grocer mixes two kinds of coffee beans one worth $32.20 a kilogram and the other $18.40 a kilogram. If the mixture weights 78 lbs and sells for $ 23.00 a kilogram. How many lbs. of each kind does he use? \n" ); document.write( "
Algebra.Com's Answer #455276 by josgarithmetic(39630)\"\" \"About 
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H = $32.20* per kg
\n" ); document.write( "L = $18.40* per kg
\n" ); document.write( "T = $23.00 per kg
\n" ); document.write( "M = 78*454/1000 kg
\n" ); document.write( "u = amount of the \"L\" priced coffee
\n" ); document.write( "v = amount of the \"H\" priced coffee\r
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\n" ); document.write( "\n" ); document.write( "Note for M, we were given 78 POUNDS but we can do the analysis and calculations with kilograms instead; and then convert our results to pounds. \r
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\n" ); document.write( "\n" ); document.write( "Based on that intermediate conversion,
\n" ); document.write( "\"%28Lu%2BHv%29%2FM=T\" and \"u%2Bv=M\"
\n" ); document.write( "Solve for u and v. REMEMBER, convert them to pounds for the final answer. For some quantity h in kilograms, h*1000/454 is the number of pounds.\r
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\n" ); document.write( "\n" ); document.write( "You could if you wanted, convert the prices T, L, and H, into dollars per pound and then create equations for the system, and you would not need to later convert the results into pounds, since they would already have that conversion taken care of.
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