document.write( "Question 160708: Assume that the sales of a certain appliance dealer are approximated by a linear function. Suppose that sales were $6500 in 1982 and $64,500 in 1987. Let x =0 represent 1982. Find the equation giving yearly sales S(x).\r
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\n" ); document.write( "\n" ); document.write( "S(x)= 11,600x + 6,500
\n" ); document.write( "S(x)= 58,000x + 64,500
\n" ); document.write( "S(x) = 58,000x + 6,500
\n" ); document.write( "S(x)= 11,600x + 64,500 \n" ); document.write( "
Algebra.Com's Answer #118488 by gonzo(654)\"\" \"About 
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answer is 11,600*x + 6,500.
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\n" ); document.write( "here's why.
\n" ); document.write( "1982 = 6500
\n" ); document.write( "1987 = 64500
\n" ); document.write( "total increase is 64500 - 6500 = 58000
\n" ); document.write( "total number of years from 1982 to 1987 is 1987 - 1982 = 5.
\n" ); document.write( "yearly increase in sales = total increase in sales divided by total increase in years.
\n" ); document.write( "this becomes 58000 / 5 = 11600.
\n" ); document.write( "since you start at 6500, the equation has to be 6500 + yearly increase of 11600.
\n" ); document.write( "equation would be y = 6500 + 11600 * x.
\n" ); document.write( "when x = 0, y = 6500 + 0 * 11600 = 6500 + 0 = 6500 which is the same as the value in 1982.
\n" ); document.write( "when x = 5, y = 6500 + 5 * 11600 = 6500 + 58000 = 64500 which is the same as the value in 1987.
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\n" ); document.write( "answer is 11,600 * x + 6,500.
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