Answer:
<em>The speed of the plane in still air is 590 mph</em>
<em>The speed of the wind is 20 mph</em>
Step-by-step explanation:
Let's call:
x = Speed of the plane in still air
y = Speed of the wind
The plane traveled d=4575 miles with the wind in t=7.5 h. The speed calculated with this data corresponds to the sum of the speed of the plane and the speed of the wind, thus:

x + y = 610 [1]
The plane traveled 4275 miles in 7.5 hours against the wind, thus the speed calculated is x - y:

x - y = 570 [2]
Adding [1] and [2]:
2x = 610 + 570 = 1180
x = 1180 / 2 = 590
From [1]:
y = 610 - 590 = 20
The speed of the plane in still air is 590 mph
The speed of the wind is 20 mph
Answer:
a = 9
Step-by-step explanation:
expand (y - 3)(y² + 3y + 9)
= y³ + 3y² + 9y - 3y² - 9y - 27
compare the coefficients of like terms with
y³ + 3y² + ay - 3y² - ay - 27
ay = 9y and - ay = - 9y ⇒ a = 9
The answer is -1 because, -2x+-6 = -4 is the same as -2x=2 which equals -1
I'm guessing that they took of in the same airport.
Each hour they will separate by 1260 miles or 21 miles each minute
2,000/21 = 95.23 minutes which is 1 hour and 35 minutes
Hope this helps :)