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2 years ago

describe how axioms about addition in particular could be helpful in developing theorems about multiplication.

1 answer:

8 0

The axioms in addition help in developing theorems about multiplication because multiplication, in simple terms, repeated addition.
As an example,
This addition operation
3 + 3 + 3 + 3 = 12
can be expressed as a multiplication operation
3 x 4 = 12

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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

it rained 2.79 inches in july. What was the average daily rainfall in July? (Hint: July has 31 days.)

natali 33 [55]

Rainfall in July = 2.79 inches.

July has 31 days.

2.79 : 31 = x : 1 Set up the proportion

2.79 / 31 = x/1 Cross multiply and switch.

2.79*1 / 31 = x Do the division

0.09 inches = x Answer

3 0

3 years ago

Read 2 more answers

The percentage 21 is of 28 is equal to the percentage 30 is of 40 True? or false?

katovenus [111]

Answer:

False

Step-by-step explanation:

75%

28 5.88

21 x 28

100

= 5.88

x 100% = 75%33.3333333333%

28 - 21

x 100% = 33.3333333333%

5 0

2 years ago

In a rock-paper-scissors game, a winner gets 5 points and a loser gets 2 points. After a man played this game 20 times, his tota

aleksandr82 [10.1K]

Answer:

The man wins 9 times and looses 11 times in the rock - paper-scissor game.

Step-by-step explanation:

Let us assume that in the game:

Number of times man wins the games = x times

Number of times man loses the games = y times

Total number of attempts = 20

Points after winning = 5

Points after loosing = 2

Now, according to the question:

x + y =20

and 5 x + 2 y = 67

Now, from (1) y = 20-x, put in (2)

5x + 2y = 67 ⇒ 5x + 2(20 -x) = 67

or, 5x -2x = 67 - 40

or, 3x = 27

⇒ x = 27/3 = 9

then y = 20 - x = 20 - 9 = 11

⇒ x = 9 and y = 11 is the solution of the given system.

Hence, the man wins 9 times and looses 11 times in the rock - paper-scissor game.

3 0

3 years ago

Evaluate. (x + 4). 2 - x for x = 6 54 74 0 14

Scrat [10]

Answer: 74

Step-by-step explanation:

Substitute x=6 into the expression

(x+4) . 2^3 -6

Add the numbers

(6+4) . 2^3- 6

10 x 2^3 -6

Evaluate the power

10 x 2^3 -6

10 x 8-6

Multiply the numbers

10 x 8-6

80-6

Subtract the numbers

80-6

3 0

3 years ago