document.write( "Question 1061875: An aircraft flying 3,500 feet above the ground at 11:05 A.M. descends to 1,000 feet at 11:10 A.M. What is the aircraft's average rate of change in feet per minute? \n" ); document.write( "
Algebra.Com's Answer #676646 by Theo(13342)\"\" \"About 
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he starts off at 3500 feet at 11:05 am.
\n" ); document.write( "he descends to 1000 feet5 at 11:10 am.\r
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\n" ); document.write( "\n" ); document.write( "he has descended 2500 feet in 5 minutes.\r
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\n" ); document.write( "\n" ); document.write( "divide 2500 by 5 and you get an average descent of 500 feet per minute.\r
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\n" ); document.write( "\n" ); document.write( "the rate of change would be -500 feet per minute.\r
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\n" ); document.write( "\n" ); document.write( "you can use the general equaion of rate * time = distance.\r
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\n" ); document.write( "\n" ); document.write( "the time is 5 minutes.\r
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\n" ); document.write( "\n" ); document.write( "the distance is -2500 feet.\r
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\n" ); document.write( "\n" ); document.write( "rate * time = distance becomes rate * 5 = -2500\r
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\n" ); document.write( "\n" ); document.write( "solve for rate to get rate = -2500 / 5 = -500 feet per minute.\r
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