document.write( "Question 162699: Q: A rectangle has an area of 6 yards squared. If the width of the rectangle is 2/5 yards, what is the length of the rectangle?\r
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\n" ); document.write( "\n" ); document.write( "I assume that to get the 6yds sq. It was LxW assuming L=3 & W=2 but I do not no where to go from there?
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Algebra.Com's Answer #119917 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Not quite sure how you arrived at L = 3 & W = 2, but that is not correct.
\n" ); document.write( "Here's how we would find the length from the information given in the problem:
\n" ); document.write( "\"A+=+L%2AW\" You were correct here.
\n" ); document.write( "\"W+=+2%2F5\" yds. and the area, A, = 6 sq.yds. So, we make the appropriate substitutions into the formula for the area.
\n" ); document.write( "\"6+=+L%2A%282%2F5%29\" Multiply both sides by the multiplicative inverse of \"2%2F5\" which is \"5%2F2\":
\n" ); document.write( "\"%285%2F2%29%2A6+=+L%2A%282%2F5%29%285%2F2%29\" Evaluate this.
\n" ); document.write( "\"15+=+L\"
\n" ); document.write( "So the length of the rectangle is 15 yards.\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "\"A+=+L%2AW\" Substitute L = 15 and W = 2/5
\n" ); document.write( "\"A+=+15%282%2F5%29\"
\n" ); document.write( "\"A+=+6\" It checks!
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