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8090 [49]
3 years ago
15

You will have 27 points if you can help me with this please!!!!

Mathematics
2 answers:
KatRina [158]3 years ago
5 0

Answer:1: P 2: d 3:a

Step-by-step explanation:

Alina [70]3 years ago
4 0

Answer:

P

D

A

Step-by-step explanation:

The center of the circle is the middle point.It is located in the middle of the diameter, in the middle of a central right angle, and all the points of a circle lie at an equal distance from the center.

You might be interested in
What’s the answer for 7+25•4
kotykmax [81]
Use Order of Operations!! Remember PEMDAS!
P: Parentheses first
E: Exponents (ie Powers and Square Roots, etc.)
MD: Multiplication and Division (left-to-right)
AS: Addition and Subtraction (left-to-right)

7+25•4 (multiply 25 by 4 first)
7+100 (then add 7 to 100)
107

So your final answer would be 107

Hope this helped and I hope you have an awesome day!! :D
3 0
3 years ago
Read 2 more answers
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
3 years ago
If you translated the point a (4,-3) left 2 units and down 6 units where is A
olga2289 [7]
It would be (-2, 1)
5 0
2 years ago
Now heres the tricky part​
loris [4]

13. Answer: n ≥ 108

<u>Step-by-step explanation:</u>

18 ≤ n ÷ 6

<u>×6 </u>  <u>    ×6 </u>

108 ≤ n       →   n ≥ 108

Note: The line under the inequality symbol represents a closed dot

Graph: 108 ·---------------→

************************************************************************************

14. Answer: x > -4

<u>Step-by-step explanation:</u>

 -3x <  12

<u>÷(-3) </u>  <u>÷(-3) </u>

    x > -4      <em>Flip the inequality sign when × or ÷ by a negative </em>

Note: No line under the inequality symbol represents an open dot

Graph: -4 o-----------→

4 0
3 years ago
PLEASE HELP
lozanna [386]
It’s 46 hope that helps I don’t know how to put a picture of the work I did on here but I ta 46
4 0
3 years ago
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