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3 years ago

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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Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A

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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

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

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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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3 years ago

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igor_vitrenko [27]

Answer:

B) Mean

Step-by-step explanation:

Standard Deviation is the measure of the amount of variation or dispersion of a set of values. Standard Deviation is represented by lower case Greek alphabet sigma σ. Standard Deviation is the square root of variance. Variance is the average of the squared differences or variation or dispersion from the Mean. Therefore, to compute standard deviation, the mean of the given data must be known.

Standard Deviation, σ = $\sqrt{Variance}$