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3 years ago

PLAS ANSWER ASAP!!!Johnny can read 10 pages of a book in 15 minutes.

Mathematics

1 answer:

Finger [1]3 years ago

5 0

The answer to this is b- 8

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HELP ASAP!!

viktelen [127]

Answer:

The answer is terminating!

Step-by-step explanation:

repeating never ends

4 0

3 years ago

Read 2 more answers

Help???????????!!!!!!!!!!!

Butoxors [25]

Salutations!

12 $\frac{2}{3}$

- 9 $\frac{3}{4}$

= $\frac{38}{3}$

Then,

- 9 $\frac{3}{4}$

- $\frac{39}{4}$

Next, our final answer:

= $\frac{35}{12}$

= 2 $\frac{11}{12}$

Our decimal would be 2.916667

Hopefully This Helps!!!

3 0

3 years ago

6x+9y=15<br> -12x-18y=-30<br> Solve system by elimination <br> Giving 10,pts

larisa [96]

Answer:

dssfgffdddfggrrghygfdedfhyjjujtg

7 0

2 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

3 years ago

13. 11 = 2b +17 14. 5 = 4a -7

dedylja [7]

$13.\ 2b+17=11\qquad|-17\\\\2b=-6\qquad|:2\\\\\boxed{b=-3}\\\\14.\ 4a-7=5\qquad|+7\\\\4a=12\qquad|:4\\\\\boxed{a=3}\\\\16.\ 2g-3=-19\qquad|+3\\\\2g=-16\qquad|:2\\\\\boxed{g=-8}\\\\17.\ 5x-9=16\qquad|+9\\\\5x=25\qquad|:5\\\\\boxed{x=5}\\\\19.\ 5+\dfrac{y}{8}=-3\qquad|-5\\\\\dfrac{y}{8}=-8\qquad|\cdot8\\\\\boxed{y=-64}$

3 0

3 years ago