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:
(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°.
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Answer:
<em>y = - </em>
<em> x + </em>
<em> </em>
Step-by-step explanation:
(6, 4)
(1, 8)
m =
= -
y - 4 = -
( x - 6 )
<em>y = - </em>
<em> x + </em>
⇔ y = - 0.8x + 8.8
The answer is 93.6
156x .6 = 93.6
Juan's time = x
Yumi's time = x + 2
Yumi's distance = 3 miles
Juan's distance = 2 miles
Rate = distance/time
Juan's rate = 2/x
Yumi's rate = 3/(x + 2)
The rates are equal so:
2/x = 3/(x + 2)
2x + 4 = 3x
x = 4 hours
Yumi's time = 4 + 2 = 6 hours
The answer is D.