document.write( "Question 1043519: A farmer wants to fence in a field that has a length of 12 2/5 yards and a width of 9 3/8 yards. How many yards of fence is needed to fence in the field? \n" ); document.write( "

Algebra.Com's Answer #658643 by Boreal(15235) $\"\"$ $\"About$

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A field has two lengths and two widths.
\n" ); document.write( "L=12 2/5 and twice that is 24 and 4/5 yards
\n" ); document.write( "W=9 3/8 and twice that is 18 6/8 yards
\n" ); document.write( "There are different ways to add them, but I like to add the 24 and 18 to get 42 yards and then add the fractions 4/5 and 3/4 (reduced).
\n" ); document.write( "the common denominator is 20, and 4/5=16/20, whereas 3/4=(15/20)
\n" ); document.write( "They make (31/20, which is 1 (11/20) yards
\n" ); document.write( "Remembering the 42 above, the answer is 42 + 1 (11/20)=43 (11/20) yards.
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\n" ); document.write( "using improper fractions
\n" ); document.write( "L=(62/5) and twice that is (124/5)
\n" ); document.write( "W=(75/8) and twice that is (150/8) or (75/4) reduced
\n" ); document.write( "(124/5)+((75/4) has a common denominator of 20
\n" ); document.write( "(496/20)+(375/20)=(871/20)
\n" ); document.write( "This is 43 (11/20) yards, because 20 goes into 100 5 times, into 800 40 times, and into 860 43 times, leaving 11 left over. \n" ); document.write( "

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