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\cdot (\sin 0^{\circ}, \cos 0^{\circ}) + 6\cdot (\sin 60^{\circ}, \cos 60^{\circ}) + 4\cdot (\sin 120^{\circ}, \cos 120^{\circ})](https://tex.z-dn.net/?f=%5Cvec%20R%20%3D%205%5Ccdot%20%28%5Csin%200%5E%7B%5Ccirc%7D%2C%20%5Ccos%200%5E%7B%5Ccirc%7D%29%20%2B%206%5Ccdot%20%28%5Csin%2060%5E%7B%5Ccirc%7D%2C%20%5Ccos%2060%5E%7B%5Ccirc%7D%29%20%2B%204%5Ccdot%20%28%5Csin%20120%5E%7B%5Ccirc%7D%2C%20%5Ccos%20120%5E%7B%5Ccirc%7D%29)
![\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:
![\|\vec R\| = \sqrt{R_{x}^{2}+R_{y}^{2}}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20R%5C%7C%20%3D%20%5Csqrt%7BR_%7Bx%7D%5E%7B2%7D%2BR_%7By%7D%5E%7B2%7D%7D)
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:
![\|\vec R\| = \sqrt{(5\sqrt{3})^{2}+6^{2}}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20R%5C%7C%20%3D%20%5Csqrt%7B%285%5Csqrt%7B3%7D%29%5E%7B2%7D%2B6%5E%7B2%7D%7D)
![\|\vec R\| \approx 10.536\,km](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20R%5C%7C%20%5Capprox%2010.536%5C%2Ckm)
![\theta_{R} = \tan^{-1} \left(\frac{6}{5\sqrt{3}} \right)](https://tex.z-dn.net/?f=%5Ctheta_%7BR%7D%20%3D%20%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B6%7D%7B5%5Csqrt%7B3%7D%7D%20%5Cright%29)
![\theta_{R} \approx 34.715^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_%7BR%7D%20%5Capprox%2034.715%5E%7B%5Ccirc%7D)
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
Answer:
The number of
Quarters = q = 58 quarters
Nickels = n = 32 nickels
Step-by-step explanation:
It is important to note that:
A quarter is worth 25 cents = 0.25q
A nickel is worth 5 cents = 0.05n
George has only Quarters and nickels in his piggy bank.
Let the number of
Quarters = q
Nickels = n
He has a total of 90 coins.
Hence,
q + n = 90.... Equation 1
q = 90 - n
The value of the coins is $16.10.
Hence,
0.25q + 0.05n = 16.10....Equation 2
We substitute 90 - n for q in Equation 2
0.25(90 - n) + 0.05n = 16.10
22.5 - 0.25n + 0.05n = 16.10
Collect like terms
- 0.25n + 0.05n = 16.10 - 22.5
-0.2n = - 6.4
n = -6.4/-0.2
n = 32 nickels
Solving for q
q = 90 - n
q = 90 - 32
q = 58 quarters
Therefore, the number of
Quarters = q = 58 quarters
Nickels = n = 32 nickels
The answer is 42 just simplify what's in the parentheses and then multiply the 7 with the answer in the parentheses