Question

Drag the tiles to the table. The tiles can be used more than once, but not all tiles will be used.

On John's flight to New York City, the plane flies 740 miles with a tailwind, which increases its speed. On the flight back
home, the plane travels against the same wind, which decreases its speed. Assume the airplane cruises at a speed of
495 miles/hour without any wind. Let x represent the wind speed. Use this information to complete the table.
Distance
Rate
Time
Flight to NYC
Flight Home
740
495 + x
495 - X
740
740
х
495 + x
495 - X
495 +
740
495 - X
740

Answer 1

Explanation:

Answer 2

Answer: flight to NYC

distance: 740 rate: 495+x time: 740/495+x

flight home:

distance: 740 rate: 495-x time: 740/495-x

Step-by-step explanation: the plane is flying with the wind when flying to NYC so the planes speed, or rate, can be modeled as 495+x

on the flight back home plane is flying against the wind so the planes speed, or rate, can be modeled as 495-x

multiply across the table to find the time each journey will take by using the equation time=distance/rate. Hope this helped :)