PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
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Chapter P.4, Problem 47E
Solution

The horizontal distance the airplane will fly to reach an altitude of 12000 ft above the takeoff point.

  32000 ft

Given:

A commercial jet airplane climbs at takeoff with slope m=38 .

Concept Used:

  • The slope-intercept form of a line is y=mx+b, where m is the slope of the line and b is the y -intercept of the line.
  • Slope of the line passing through the points x1,y1 and x2,y2 is given by m=y2y1x2x1
  • The slope of a line is defined as rise over run

Calculation:

Let the horizontal distance that airplane will fly be x .

Let the take off point be at origin 0,0 , then the terminal point at an altitude of 12000 ft would be x,12000 .

Since the airplane climbs with slope m=38 , use the slope formula with points 0,0 and x,12000 , and simplify further as shown below

  m=y2y1x2x138=120000x038=12000xx=12000×83x=32000

Thus, the airplane must fly 32000 ft in the horizontal direction.

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

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