document.write( "Question 661215: (FLIGHT CONDITIONS) In stable air, the air temperature drops about 5 °(F) for each 1,000-foot rise in altitude.\r
\n" ); document.write( "\n" ); document.write( "(A) If the temperature at sea level is 100 °F and a commercial pilot reports a temperature of -30 °F at 26000 feet, write a linear equation that expresses temperature T in terms of altitude A.
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Algebra.Com's Answer #411561 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Plot altitude on the horizontal and temp on the vertical
\n" ); document.write( "so, \"+T+=+f%28A%29+\"
\n" ); document.write( "The slope of the line will be \"+-5+%2F+1000+\" degrees/feet
\n" ); document.write( "You are given 2 points on the line,
\n" ); document.write( "( 0, 100 ) the sea level temp, and
\n" ); document.write( "( 26000, -30 )
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\n" ); document.write( "slope = \"+%28+-30+-+100+%29+%2F+%28+26000+-+0+%29+\"
\n" ); document.write( "slope = \"+-130+%2F+26000+\"
\n" ); document.write( "slope = \"-+.005+=+-5%2F1000+\"
\n" ); document.write( "This agrees with the given slope
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\n" ); document.write( "Using the point-slope formula:
\n" ); document.write( "\"+%28+T+-+100+%29+%2F+%28+A+-+0+%29+=+-.005+\"
\n" ); document.write( "\"+T+-+100+=+-.005A+\"
\n" ); document.write( "\"+T+=+-.005A+%2B+100+\" answer
\n" ); document.write( "check:
\n" ); document.write( "does it go through (26000, -30) ?
\n" ); document.write( "\"+T+=+-.005A+%2B+100+\"
\n" ); document.write( "\"+-30+=+-.005%2A26000+%2B+100+\"
\n" ); document.write( "\"+-30+=+-130+%2B+100+\"
\n" ); document.write( "\"+-30+=+-30+\"
\n" ); document.write( "OK
\n" ); document.write( "Here's the plot with the horizontal axis expressed
\n" ); document.write( "in thousands of feet
\n" ); document.write( "\"+graph%28+400%2C+400%2C+-10%2C+50%2C+-50%2C+200%2C+-5x+%2B+100+%29+\"\r
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