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Mama L [17]
3 years ago
13

How am I supposed to do this please help me

Mathematics
2 answers:
Temka [501]3 years ago
8 0
U add 1 1/4 in the left one
lisabon 2012 [21]3 years ago
6 0
You add them you times the two numbers that are next to eatch other :)
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The are 66 small wooden animals to sell at the craft fair . He displays in rows of 7 animals in a row. How many will not be in a
lord [1]

Answer:

9

Step-by-step explanation:

There would be 9 because 7 goes into 66 9 times

4 0
3 years ago
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
How do you figure up volume in cubic inches
Anna11 [10]

<em>Measure the length, width, and depth of an object in inches. ... </em>

<em>Write the length of your object. ... </em>

<em>Multiply the length by the width of your object. ... </em>

<em>Multiply your answer by the depth of your object. ... </em>

<em>Label your answer in cubic inches.</em>

3 0
3 years ago
If one of the roots of the quadratice quation x2+kx-12=0is4,what is the value of k?
uysha [10]
<h3>Given Equation:</h3>

\huge \purple {\rm { {x}^{2}  + kx - 12 = 0}}

<h3>Value:</h3>

\huge \purple {\rm {x = 4}}

<h3>To Find:</h3>

\huge \purple {\rm {The \: value \: of \: k.}}

<h3>Solution:</h3>

\huge \purple {\rm { {(4)}^{2}  + k(4) - 12 = 0}}

\huge \purple {\rm {or, \: 16 + 4 k - 12 = 0}}

\huge \purple {\rm {or, \: 4k =  - 4}}

\huge \purple {\rm {or, \: k =  \frac{ - 4}{4}  =  - 1}}

<h2>Answer:</h2>

\huge \purple {\sf {The \: value \: of \: k \: is \: -1}}

3 0
3 years ago
Read 2 more answers
Which number contains digits where the hundreds digit is 10 times as great as the tens digit? 7,988 , 7,898 , 7,889
grigory [225]
Answer : 7,889

Hundreds digit : 800
Tens digit : 80

800 ÷ 80 = 10 (800, the hundreds digit, is 10 times as great as the tens digit, 80)
5 0
3 years ago
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