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Firdavs [7]
2 years ago
15

Suppose you mark n points on a circle, where n is a whole number greater than 1. The number of segments you can draw that connec

t these points is 1/2n^2 - 1/2n
How many segments can you draw if you mark 8 points on the circle?
Mathematics
1 answer:
leonid [27]2 years ago
5 0
The expression for the number of segments drawn out of the n number of points in the circle is,
                                n² / 2 - n / 2
Substituting directly to the expression the number of points, n, which is equal to 8,
                                8² / 2  - 8 / 2 = 28
Thus, there are 28 segments that can be drawn from the points. 
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I've solved most! Just need help with 9, 10, 14.
Ugo [173]
7. Correct

- - -

9. By the ratio test, the series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{b^{n+1}(x-a)^{n+1}}{\ln(n+1)}}{\frac{b^n(x-a)^n}{\ln n}}\right|

The limit reduces to

\displaystyle|b(x-a)|\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=b|x-a|

where |b|=b because b>0 is given. So the series converges when

b|x-a|

This means the radius of convergence is \dfrac1b.

- - -

10. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2}}{\frac{x^{2n}}{n(\ln n)^2}}\right|

The limit is

\displaystyle|x^2|

and so the radius of convergence is 1.

- - -

11. Incorrect. By the root test, the series converges for

\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\frac{(x-2)^n}{n^n}\right|}=\lim_{n\to\infty}\frac{|x-2|}n=0

which means the series converges for all x, and so the interval of convergence is (-\infty,\infty).

- - -

For 14 and 16, it'll probably be too late to edit this post by the time you see this. You can try posting the remaining problems in a new question.
7 0
2 years ago
Help m please!!!!!!!
liraira [26]
The Domain in #4 is (0,1,2,3)
The Range in #4 is (1,3,5,7)
5 0
3 years ago
What is the range of possible sizes for side x?
posledela

Answer:

x = 3.96

Step-by-step explanation:

apply the pitagoras theorem

x =

\sqrt{ {2.8}^{2}  +  {2.8 }^{2} }

4 0
2 years ago
The total deposit and rent collected by an apartment manager for one resident are given by the equation t = 350 + 700m, where m
Nat2105 [25]

B.......just took the test

5 0
3 years ago
What is the range for the following set of numbers: 199, 211, 63, 200, 544, 347, 727, 98, 110 *
LenKa [72]

Answer:

Minimum = 64

Maximum = 727

Medium = 200

Mean = 277.667

Mode = none

<em>s</em><em> </em>= 223.825

<em>Q</em><em> </em>= 211.024

<em>n</em><em> </em>= 9

8 0
3 years ago
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