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aalyn [17]
3 years ago
6

What is 3-2i divided by 5i

Mathematics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer:

-2/5 - 3/5 i

Step-by-step explanation:

3-2i           i                 3i - 2i^2            3i  + 2          -(2+3i)

--------- *  ----    =       -----------            = --------------=   -----------

5i             i                   5 i^2                     -5                 5

multiply by i to get rid of the complex number in the denominator

remember i^2 = -1

we need to rewrite this in the form a+bi

-2/5 - 3/5 i

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A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
Maggie recently took a road trip. She bought 15 gallons of gasoline for $1.68 per gallon and 12 gallons at a price of $1.76 per
photoshop1234 [79]
The numbers of gallons bought every stop are extraneous...you can ignore them. Add 1.68, 1.76 and 1.65 together and divide the total by three. The average is 1.69
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Subtract 8 from n then divide m by the result
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What exactly is n? I need to know n so I can figure out m.

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7 0
3 years ago
I don’t get this explain pls
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Im not 100% sure but RT and US are the same distance so theyd be the same...
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