Two ships leave a harbor entrance at the same time. The first ship is traveling at a constant 10 miles per hour, while the second is traveling at a constant 22 miles per hour. If the angle between their courses is 100°, how far apart are they after 2 hours? (Round your answer to the nearest whole number.)

Billy plus billy equals Tim, but since dong dong minus butthole equals marijuana the distance between Egor and Tryrone is 35 degrees minus the turd from marijuanas bong, so if 70 degrees is between Egor and his mom then 2,234.69 bong rips is how many it takes to get to the center of a tootsie pop.

To find the distance between the two ships after 2 hours, we can use the law of cosines.

The formula for the law of cosines is:

c^2 = a^2 + b^2 - 2ab * cos(C)

where c is the side opposite the angle C, and a and b are the other two sides of the triangle.

In this case, we want to find the distance between the two ships, which is the side opposite the angle of 100 degrees.

Let's call this distance "d".

Using the law of cosines, we have:

d^2 = (10*2)^2 + (22*2)^2 - 2(10*2)(22*2) * cos(100°)

Simplifying this equation:

d^2 = 40^2 + 88^2 - 2(40)(88) * cos(100°)
d^2 = 1600 + 7744 - 7040 * cos(100°)
d^2 = 9344 - 7040 * cos(100°)

To find the value of cos(100°), we can use a calculator or trigonometric table.
cos(100°) ≈ -0.1736

Substituting this value into the equation:
d^2 = 9344 - 7040 * (-0.1736)
d^2 = 9344 + 1222.144
d^2 ≈ 10566.144
d ≈ √10566.144
d ≈ 102.8 (rounded to one decimal place)

Therefore, the distance between the two ships after 2 hours is approximately 103 miles.

To find how far apart the two ships are after 2 hours, we can use the concept of vector addition.

Let's break down the motion of each ship into its horizontal and vertical components.

The first ship is traveling at a constant speed of 10 miles per hour. After 2 hours, it would have traveled a distance of 2 * 10 = 20 miles.

For the second ship, we need to find its horizontal and vertical components. We can use trigonometry to determine these components.

The angle between their courses is 100°. The horizontal component of the second ship's velocity is given by 22 * cos(100°), while the vertical component is given by 22 * sin(100°).

Now, we can find the total horizontal and vertical distances traveled by the second ship after 2 hours by multiplying these components by 2:

Horizontal distance = 2 * 22 * cos(100°)
Vertical distance = 2 * 22 * sin(100°)

To find the total distance between the two ships after 2 hours, we can use the Pythagorean theorem:

Distance = √(20^2 + (2 * 22 * sin(100°))^2)

Calculating this distance will give you the answer, rounded to the nearest whole number.

recall the law of cosines. The distance z will be

z^2 = 20^2 + 44^2 - 2*20*44*cos100°