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Plane A leaves Albany at 2:00 p.m., averaging 400 mph and flying in a northerly direction. Plane B leaves Albany at 2:30 p.m., averaging 325 mph and flying due east. At 5:00 p.m., how far apart will the planes be? A. 1,449 B. 1,649 C. 1,849

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meerkat18
From 2:00 pm to 5:00 pm, Plane A traveled for 3 hours. The total distance it had traveled is the product of its speed and the number of hours which is,
                        400 mph x 3 hours = 1200 miles
For Plane B, from 2:30 pm to 5:00 pm, it traveled for 2.5 hours and the total distance it had traveled is 812.5 miles. The unknown distance is the hypotenuse of the right triangle such that,
                              x² = (1200 miles)² + (812.5 miles)²
The value of x from the equation is 1449.19 miles. Thus, the answer is letter A. 
Plane A traveled for 3 hours and total distance it had traveled is: 
400 mph x 3 hours = 1200 miles    ( using D = V x t )
Plane B traveled for 2.5 hours and the total distance  traveled is 812.5 miles
Using Pythagoras theorem ;
 x² = (1200)² + (812.5)²
 x  = 1449.19 miles
Hence,the best answer is option A. 1,449

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