Here is you're answer:
In order to solve this equation you must first solve the equation then make a model.
Draw three circles then divide 15 between the 3 circles to equal 5.
Therefore you're answer is "5."
Hope this helps!
If the parallelogram is dilated by a factor V what is the new area
2 answers:
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<u>Given</u>:
The exterior angle P is 74°
The measure of ∠PRQ is 51°
We need to determine the measure of ∠PQR
<u>Measure of ∠QPR:</u>
From the figure, it is obvious that P is the intersection of the two lines.
The angle 74° and ∠QPR are vertically opposite angles.
Since, vertically opposite angles are always equal, then the measure of ∠QPR is 74°
Thus, the measure of ∠QPR is 74°
<u>Measure of ∠PQR:</u>
The measure of ∠PQR can be determined using the triangle sum property.
Thus, we have;
Substituting the values, we get;
Thus, the measure of ∠PQR is 55°
Hence, Option B is the correct answer.
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: