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3 years ago

1 and 2 are examples but I still need help , I don’t understand n this is due today

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

Answer:

A rational number is a number that can be express as the ratio of two integers. A number that cannot be expressed that way is irrational. ... However, numbers like √2 are irrational because it is impossible to express √2 as a ratio of two integers.

The perfect squares are the squares of the whole numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 … Here are the square roots of all the perfect squares from 1 to 100. 1.

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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

2 years ago

Use the distributive parenthesis: 7(m + 9)<br> will give brainliest if answered correctly!

Phantasy [73]

Answer:

The simplified answer would be 7m+63. You would have to solve for the variable.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

1. The triangles above are similar. What is the ratio (larger to smaller) of the perimeters? (1 point)

eimsori [14]

Answer:

Ratio = 5/3

The ratio (larger to smaller) of the perimeters is 5/3

Step-by-step explanation:

Attached is an image of the two triangles.

Since both triangles are similar, the ratio of their perimeter is equal to the ratio of each similar sides.

Ratio = P1/P2 = S1/S2

Similar side for triangle 1 S1 = 15ft

Similar side for triangle 2 S2 = 9ft

Substituting the values;

Ratio = 15ft/9ft = 5/3

Ratio = 5/3

The ratio (larger to smaller) of the perimeters is 5/3

7 0

3 years ago

56 out of 154 in simplest form of a fraction

marissa [1.9K]

We start with 56/154. Dividing each number by two, we have:

28/77

Dividing each number by seven, we have:

4/11

As the greatest common factor of 4 and 11 is 1, this fraction is in simplest form. Thus, 56/154 is equal to 4/11.

4 0

3 years ago

Read 2 more answers

Solve the inequality. –8 ≤ 2x – 4 < 4

Nostrana [21]

First;
-8≤2x-4
-4≤2x
-2≤x

Then;
2x-4<4
2x<8
x<4

-2≤x<4
C is the correct answer.

4 0

2 years ago