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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Hope there are y^(1/5) so than this mean 5 index radical of y
hope helped
Answer:
(a). 2.251 cm
, 2.349 cm, 2.295 cm
(b). 0.023 meters.
Step-by-step explanation:
(a). We are asked to to choose the options that shows the actual measurement of the rainfall.
Since the rounded figure is 2.3 cm, this means that either the tenths digit is 3 with hundredths digit less than 5, or tenths digit is 2 with hundredths digit greater than or equal to 5.
Let us check our given choices one by one.




Therefore, the correct choices are 2.251 cm
, 2.349 cm and 2.295 cm.
(b) We know that 1 cm equals 0.01 meters.
To convert 2.3 cm to meters, we will multiply 2.3 by 0.01.


Therefore, the rounded measurement of rain would be 0.023 meters.
Answer:
(-8,1)
Step-by-step explanation:
Imma assume you meant (x-2, y)
x coordinate becomes -6-2=-8
y coordinate stays the same
so the image is (-8,1)
5 x 7 because if 5 is the width is 25 then 18 minus that is 7 and 7x5 = 35