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4 years ago

Which segment is not a radius of circle C cq cp rc pq

Mathematics

2 answers:

attashe74 [19]4 years ago

7 0

The answer could be D- pq. It's not from the center of the circle.

zubka84 [21]4 years ago

4 0

Answer:

D. PQ.

Step-by-step explanation:

We have been given a circle. We are asked to find the segment which is not a radius of circle C.

We know that radius of a circle is a segment that connects any point on circle to center of the circle.

Upon looking at our given circle, we can see that CQ, CP and RC are radii of the circle C.

Since PQ is the diameter of our given circle, therefore, segment PQ is not radius of our given circle.

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3 years ago

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polet [3.4K]

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Step-by-step explanation:

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3 years ago

Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x

Tanzania [10]

Recall that

$\cosh^2(x) - \sinh^2(x) = 1$

Dividing both sides by cosh²(x) gives

$1 - \tanh^2(x) = \mathrm{sech}^2(x)$

Also, recall the identity

$\sinh(2x) = 2\sinh(x)\cosh(x)$

Then

$\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)$

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3 years ago