@1e2d0806: Interesting math problem! To solve this, let's denote the speed of the plane in still air as "x" (in miles per hour).

When the plane flies against the headwind, its effective speed is reduced by the wind speed. So, the plane's speed against the wind is (x - 30) miles per hour.

On the return trip, the plane's speed is increased by the wind speed, so its speed with the wind is (x + 30) miles per hour.

We know that the total time for the round trip is 10 hours.

Using the formula: time = distance / speed, we can set up two equations:

720 / (x - 30) + 720 / (x + 30) = 10

Now, we can solve this equation to find the speed of the plane in still air. Can you give it a try?

Ah, a math word problem! Let's see if we can solve it together.

We have a plane flying against a steady headwind for 720 miles, and then it returns to the same point with the wind. We're given that the entire trip takes 10 hours.

To solve this problem, let's first find out how long it takes for the plane to fly against the headwind.

Let's call the speed of the plane in still air "p" and the speed of the wind "w". Since distance equals speed multiplied by time (d = s * t), we can set up two equations based on this information:

Equation 1: (p - w) * t1 = 720 (representing time and speed against headwind)

Equation 2: (p + w) * t2 = 720 (representing time and speed with tailwind)

Since we don't know either "t1" or "t2," let's solve for them in terms of p and w:

t1 = 720 / (p - w)

t2 = 720 / (p + w)

Now, from what is given in the problem, we know that:

t1 + t2 = 10

Substituting our expressions for t1 and t2 into this equation:

(720 / (p - w)) + (720 / (p + w)) = 10

To make things clearer, let's multiply both