A plane takes off from City A and goes towards city B at noon. After arriving in city B it has a 5 hour layover before returning. During the return flight the plane runs into a headwind that causes it to go 20 mph slower than the original flight and makes the flight back one more take one extra hour more than the original time. Write an equation that represents the total time traveled with clearly labeled variables

5y² + 2xy – 20x - 100

Step-by-step explanation:

Assume that the distance between City A and City B is x miles,

& the speed of the plane from City A to B is y mph;

The return trip is 20 mph slower so the plane's speed is (y – 20) mph

Since time = distance/speed; The time taken by the plane from City A to B is x / y hrs.

The time taken in return flight is;

x / (y - 20) hrs

The layover was 5hrs long. Therefore the total time taken on the who return trip journey is;

x/y  +  x/(y-20)  + 5

multiply each individual fraction with the LCM of the denominators

x(y-20) + xy + 5 y(y-20)

xy -20x +xy + 5y² – 100

5y² + 2xy – 20x - 100

the correct answer to your question is a. 4^4

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the answer is simply just 6n add the two together as they have an n