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arlik [135]
3 years ago
12

A ship leaves port and travels due west for 30 knots, then changes course to S 30° W and travels 50 more knots. Find the bearing

from the port of departure

Mathematics
1 answer:
leva [86]3 years ago
5 0

Answer:

  232°

Step-by-step explanation:

There are a couple of ways to find the desired direction. Perhaps the most straightforward is to add up the coordinates of the travel vectors.

  30∠270° +50∠210° = 30(cos(270°), sin(270°)) +50(cos(210°), sin(210°))

  = (0, -30) +(-43.301, -25) = (-43.301, -55)

Then the angle from port is ...

  arctan(-55/-43.301) ≈ 231.79° . . . . . . . 3rd quadrant angle

The bearing of the ship from port is about 232°.

_____

<em>Comment on the problem statement</em>

The term "knot" is conventionally used to indicate a measure of speed (nautical mile per hour), not distance. It is derived from the use of a knotted rope to estimate speed. Knots on the rope were typically 47 ft 3 inches apart. As a measure of distance 30 knots is about 1417.5 feet.

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Step-by-step explanation:

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2 years ago
Let a and b be real numbers satisfying a^3 - 3ab^2 = 47 and b^3 - 3a^2 b = 52. Find a^2 + b^2.
boyakko [2]

Answer: a²+b² = -99/2

Step-by-step explanation:

Since we are given two equations, this equations will be solved simultaneously to get a² and b²

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From 1, a(a² - 3b²) = 47...3

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Adding 3 and 4, we have;

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Min-jun has a swimming pool that needs to be drained. His pool holds
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Read 2 more answers
The question is in the picture​
otez555 [7]

Answer:

x=27.8

Step-by-step explanation:

first, you need to find y. 2y-5=65. 65+5 is 60 and you are left with 2y=70. to get the 2 off of the y you divide everything by 2. 70/2 is 35 therefor y=35. Now you plug that into the other equation (2x+y) and get 2x+35 which is equal to segment 90.6 so 2x+35=90.6. subtract 35 on both sides and you have 2x=55.6. To get the 2 off the x, we divide everything by 2. all of that divided by 2 is x=27.8

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2 years ago
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