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vichka [17]
3 years ago
7

Find an angle between 0 and 2pi that is coterminal with the following angle. 32pi/9

Mathematics
1 answer:
never [62]3 years ago
5 0
<span>To do these you will be adding or subtracting  2pi (or integer multiples of .

Since the given angles are in fraction form, it will help to have 2pi  in fraction form, 2pi=10/5=6pi/3=4pi/2=18pi/9     NOTE: this>(/) stands for over like 1 over 2 EX. 1/2

too, so the addition/subtraction is easier.

Hint: When deciding if you have a number between 0 and 2pi, compare it to the fraction version of 2pi that you've been adding/subtracting.


For 17pi/5...
First we can see that 17pi/5 is more than 10pi/5 (aka 2pi). So we need to start subtracting: 17pi/5 - 10pi/5 = 7pi/5

Now we have a number between 0 and 10pi/5. So 7pi/5 is the co-terminal angle between 0 and 2pi.

I'll leave the others for you to do. Just remember that you might have to add or subtract  multiple times before you get a number between 0 and 2pi.
P.S don't add or subtract at all if the number starts out between 0 and 2pi.</span>
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Answer:

<h2><em>4</em><em>1</em></h2>

<em>Sol</em><em>ution</em><em>,</em>

<em>a</em><em>=</em><em>1</em><em>4</em>

<em>b</em><em>=</em><em>2</em><em>9</em>

<em>c</em><em>=</em><em>1</em>

<em>D=</em><em>1</em>

<em>Now</em><em>,</em>

<em>a + b - c - d \\  = 14 + 29 - 1 - 1 \\  = 43 - 1 - 1 \\  =42 - 1 \\  = 41</em>

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An airplane is at a location 800 miles due west of city X. Another airplane is at a distance of 1,200 miles southwest of city X.
andrezito [222]

Answer:

The distance between the two airplanes (to the nearest mile) is 1058 miles.

Step-by-step explanation:

An airplane A is at a location 800 miles due west of city X. So AX = 800 miles.

Another airplane is at a distance of 1,200 miles southwest of city X. So BX = 1200 miles.

The angle at city X created by the paths of the two planes moving away from city X measures 60°. So angle ∠AXB = 60°.

In triangle ΔAXB, AX = 800 miles, BX = 1200 miles, ∠AXB = 60°.

Using law of cosines:-

AB² = AX² + BX² - 2 * AX * BX * cos(∠AXB).

AB² = 800² + 1200² - 2 * 800 * 1200 * cos(60°).

AB² = 640000 + 1440000 - 2 * 960000 * 1/2

AB² = 2080000 - 960000

AB² = 1120000

AB = √(1120000) = 1058.300524

Hence, the distance between the two airplanes (to the nearest mile) is 1058 miles.

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Hey There!

Remember, first find the gcf of 4,3 in this problem it is 12.

Steps to solve

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So, The answer is <span>2\frac{11}{12}</span>

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