a plane flies due north at 300 mi/h for 3 h then due west at 250 mi/h. If x represents the number of hours the plane is flying due west, determine for what values of x the plane will be more than 3700 mi from it's original point.
(250x)^2 + 900^2 > 3700^2
x^2 > 206.08
x > 14.3555
To determine for what values of x the plane will be more than 3700 mi from its original point, we need to calculate the distance traveled both due north and due west separately.
Let's start by calculating the distance traveled due north. The plane flies at 300 mi/h for 3 hours, so the distance traveled due north is 300 mi/h * 3 h = 900 miles.
Next, let's calculate the distance traveled due west. The plane flies at 250 mi/h for x hours, so the distance traveled due west is 250 mi/h * x h = 250x miles.
To find the total distance from the original point, we can use the Pythagorean theorem, which states that for a right triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In this case, the distance traveled due north (900 miles) represents one side of the triangle (a), and the distance traveled due west (250x miles) represents the other side (b). The distance from the original point (c) represents the hypotenuse.
Using the Pythagorean theorem, we can write the equation:
c^2 = a^2 + b^2
c^2 = 900^2 + (250x)^2
c^2 = 810,000 + 62,500x^2
To be more than 3700 miles from the original point, the distance c needs to be greater than 3700. Therefore, we can set up the following inequality:
c^2 > 3700^2
810,000 + 62,500x^2 > 13,690,000
62,500x^2 > 12,880,000
x^2 > 206.08
To solve for x, we take the square root of both sides:
x > sqrt(206.08)
Now, we can approximate the square root of 206.08 using a calculator:
x > 14.35
Therefore, for values of x greater than 14.35, the plane will be more than 3700 miles from its original point.