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Rudik [331]
3 years ago
5

Put in least to highest order 6.86,6.8,7,6.9,6.827 and 12.03,1.2,12.3,1.203,12.301

Mathematics
1 answer:
Anon25 [30]3 years ago
6 0
6.8,6.827,6.86,6.9

1.2,1.203,12.03, 12.301,12.31
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Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality \sin ^2\theta +\cos ^2\theta \geq 1

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

\sin \theta = side opposite to \theta/hypotenuse

\cos  \theta = side adjacent to \theta/hypotenuse

\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta

\cos  \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta

So, coordinates of A = \left ( \cos \theta ,\sin \theta  \right )

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by x^2+y^2=1

Put (x,y)=\left ( \cos \theta ,\sin \theta  \right )

\cos ^2\theta +\sin ^2\theta =1

So, (x,y)=\left ( \cos \theta ,\sin \theta  \right ) satisfies the equation x^2+y^2=1

For points  (x,y)=\left ( \cos \theta ,\sin \theta  \right ) inside the circle, \cos ^2\theta +\sin ^2\theta

For points  (x,y)=\left ( \cos \theta ,\sin \theta  \right ) outside the circle, \cos ^2\theta +\sin ^2\theta >1

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Answer:

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where

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X is the mean value of these observations, and

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