Y is between points p and b. pt= 15 and tb = 10. what is pb

1 answer:

4 0

PT = 15, and TB = 10

Take the 15 from PT, and add the 10 from TB. Which equals 25.

PB = 25

So, 25 is your answer.

Hope this helps! ☺

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The circles are identical. What is the circumference of each circle? Circle 1: radius of 2x; Circle 2: diameter of 2x+3. A. 3/2

Delicious77 [7]

Answer:

C=6π

Step-by-step explanation:

It is given that, two circles are identical. The radius of first circle is 2x and the diameter of other circle is 2x+3.

If identical, it means that the radius of both circles are same i.e.

$2x=\dfrac{2x+3}{2}\ \ (\because\ r=\dfrac{D}{2}, d=\text{diameter})\\\\4x=2x+3\\\\2x=3\\\\x=1.5$

Radius of both circles is 2(1.5)=3

The circumference of a circle is given by :

$C=2\pi r\\\\C=2\pi (3)\\\\C=6\pi$

So, the circumference of each circle is 6π.

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2 years ago

Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently

garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

$\small \sum_{i = 1}^{k} L_i= 2\pi$

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.

we get :

$\small \sum_{i = 1}^{k} C_i.l = 2\pi$

where C(i) is a constant coefficient obviously between 0 and 1.

$\small \sum_{i = 1}^{k} C_i= 2\pi/l$

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]

So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.

We already know the sum so it is easy to compute the average :

$\small L_{Exp} = \frac{2\pi}{k}$