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yawa3891 [41]
3 years ago
11

Find f'(x) if f(x)=3squareroot(x+2)using the definition of the derivative!

Mathematics
1 answer:
ycow [4]3 years ago
4 0
I hope this helps you



f (x)=(x+2)^3/2


f'(x)=3/2. (x+2)^3/2-1


f'(x)=3/2 (x+2)^1/2


f'(x)=3/2.square root of (x+2)
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PLEASE HELP! 50 Points!
gulaghasi [49]

The bearing of the plane is approximately 178.037°. \blacksquare

<h3>Procedure - Determination of the bearing of the plane</h3><h3 />

Let suppose that <em>bearing</em> angles are in the following <em>standard</em> position, whose vector formula is:

\vec r = r\cdot (\sin \theta, \cos \theta) (1)

Where:

  • r - Magnitude of the vector, in miles per hour.
  • \theta - Direction of the vector, in degrees.

That is, the line of reference is the +y semiaxis.

The <em>resulting</em> vector (\vec v), in miles per hour, is the sum of airspeed of the airplane (\vec v_{A}), in miles per hour, and the speed of the wind (\vec v_{W}), in miles per hour, that is:

\vec v = \vec v_{A} + \vec v_{W} (2)

If we know that v_{A} = 239\,\frac{mi}{h}, \theta_{A} = 180^{\circ}, v_{W} = 10\,\frac{m}{s} and \theta_{W} = 53^{\circ}, then the resulting vector is:

\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})

\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]

Now we determine the bearing of the plane (\theta), in degrees, by the following <em>trigonometric</em> expression:

\theta = \tan^{-1}\left(\frac{v_{x}}{v_{y}} \right) (3)

\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)

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The bearing of the plane is approximately 178.037°. \blacksquare

To learn more on bearing, we kindly invite to check this verified question: brainly.com/question/10649078

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\sin 58=\frac{|PQ|}{15}

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Answer:

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this is more of a statement then a question.

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