91Ó°ÊÓ

We are given the following information: - The slope of the plane's path is 0.22. - The plane gains an altitude of 1000 ft during a certain period.

The slope is given by the formula: slope = \( \frac{rise}{run} \) In this problem, the slope is 0.22 and the rise (change in altitude) is 1000 ft. We can plug these values into the formula to find the run (change in horizontal distance): 0.22 = \( \frac{1000}{run} \)

We can now solve for the run (horizontal distance) by multiplying both sides of the equation by the run: 0.22 * run = 1000 To find the run, divide both sides by 0.22: run = \( \frac{1000}{0.22} \approx 4545.45 \) ft So the horizontal distance covered by the plane during that period is approximately 4545.45 ft.

To make sure that our answer is correct, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the two other sides: \(c^2 = a^2 + b^2\) In this problem, we can consider 'a' as the rise (1000 ft), 'b' as the horizontal distance we found (run = 4545.45 ft), and 'c' as the actual distance the plane travelled along its path. Using our values: \(c^2 = (1000)^2 + (4545.45)^2 = 1000000 + 20640425.2 = 21640425.2\) Now, calculate c: \(c = \sqrt{21640425.2} \approx 4651.94\) ft Since the actual distance the plane travelled (approximately 4651.94 ft) is greater than the horizontal distance we found (approximately 4545.45 ft), our answer is reasonable and the plane covered a horizontal distance of approximately 4545.45 ft over the given period of time.