The plane flies a distance of approximately 10.536 kilometers in <em>straight</em> line and with a bearing of approximately 035°.
A plane that travels a distance 
, in kilometers, with a bearing of 
sexagesimal degrees can be represented in standard position by means of the following expression:
(1)
We can obtain the resulting vector (
) by the principle of superposition:
![\vec R = \Sigma_{i=1}^{n} [r_{i}\cdot (\sin \theta_{i}, \cos \theta_{i})]](https://tex.z-dn.net/?f=%5Cvec%20R%20%3D%20%5CSigma_%7Bi%3D1%7D%5E%7Bn%7D%20%5Br_%7Bi%7D%5Ccdot%20%28%5Csin%20%5Ctheta_%7Bi%7D%2C%20%5Ccos%20%5Ctheta_%7Bi%7D%29%5D)
(2)
If we know that 
, 
, 
, 
, 
and 
, then the resulting vector is:
![\vec R = (5\sqrt{3}, 6) \,[km]](https://tex.z-dn.net/?f=%5Cvec%20R%20%3D%20%285%5Csqrt%7B3%7D%2C%206%29%20%5C%2C%5Bkm%5D)
The magnitude of the resultant is found by Pythagorean theorem:
And the bearing is determined by the following <em>inverse</em> trigonometric relationship:
(3)
If we know that 
and 
, then the magnitude and the bearing of the resultant is:
The plane flies a distance of approximately 10.536 kilometers in <em>straight</em> line and with a bearing of approximately 035°.
To learn more on vectors, we kindly invite to check this verified question: brainly.com/question/21925479
Hi! For the area of a parallelogram, we just need to multiply the base and the height. It may sound simple but the tricky part in this problem is determining which one is the height.
For a parallelogram, the height is defined as the measurement PERPENDICULAR to the base. Therefore, for this problem, we need to multiply 14 and 4.5 instead of 14 and 6.
ANSWER: B. 63 square meters.
Answer: 7^5 or 7 to the fifth power.
Step-by-step explanation: This is an exponent meaning you count the 7’s and in the first picture there’s 5 7’s. This means that the number your putting to a power is the yellow box so that’s 7 and the exponent is 5 not 4 because if you did 7 to the first power it’s still 7 so it would be 7 to the 5th power or 7^5
