A lemur weighs 4.2 kilograms. How many grams does the lemur weigh? Use the number of zeros of the product when multiplying a num
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Answer:
The top option is false.
Step-by-step explanation:
Both segments have a <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>] of ⅔. It just that their ratios have unique qualities:
Greatest Common Factor: 2
___ ___
<em>BC</em><em> </em>is at a 4⁄6 slope, and <em>AB</em><em> </em>is at a ⅔ slope. Although their quantities are unique, they have the exact same value.
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B.
C.
D.
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C.
D.
please mark as brainliest.
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;
Therefore;
Which gives;
Similarly, we can get;
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: