Two ships leave port at the same time one trip takes a bearing of 26° east of the north the other ship takes a bearing of 32° west of the north both ships are traveling at a speed of 30 mph how far apart to the nearest mile will they be at the end of three hours

Step-by-step explanation:

The diagram below shows the general situation.

One ship is heading NWbN at 30 mi/h. The other is heading NNE at the same speed.

They have been travelling for 3 h, so they have each covered 90 mi.

The angle between their tracks is

∠A = 32 + 26 = 58°

We can use the Law of Cosines to calculate a, the distance between the ships.

a² = b² + c² - 2bccosA

   = 90² + 90² - 2 × 90 × 90cos58

   = 8100 + 8100 – 16 200 × 0.5299

   = 16 200 - 8585

   = 7615

a = 87 mi

After 3 h, the distance between the ships will be

Hey!

Alright, let's take a look at this problem.
The first step would be to rearrange the equation. To do that we will group the like elements.

Original Equation :
7x - 8 - 3x - 4

New Equation {Changed by Grouping Like Elements} :
7x - 3x - 4 - 8

Next, we'll add both the 7x and 3x since they both contain variables.

Old Equation :
7x - 3x - 4 - 8

New Equation {Changed by Adding Similar Elements} :
4x - 4 - 8

Again, we'll add the other similar elements in the equation.

Old Equation :
4x - 4 - 8

New Equation {Changed by Adding Similar Elements} :
4x - 12

Our equation can no longer be simplified any farther, so we have our final answer.

The equation 7x - 8 - 3x - 4 simplified is  4x - 12.

Hope this helps!

- Lindsey Frazier ♥