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3 years ago

Can someone explain this geometry problem?? Will give Brainliest!!

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

Answer:

Area is square units so if length is tripled (times 3), area is times 3^2 or 9

so 600•9 = 5400 cm^2

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6 0

4 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

81-1 1/5x < 55

Norma-Jean [14]

Answer:

A. I guess

Step-by-step explanation:

not really sure

8 0

2 years ago

The large bottle of nasal spray is 9.46 centimeters tall. The small bottle is 5.29 centimeters tall. How much shorter is the sma

posledela

Answer:

4.17 centimeters

Step-by-step explanation:

The large bottle of nasal spray is 9.46 centimeters tall. Assuming that the large bottle of nasal spray has the shape of cylinder, then the height of the cylinder is 9.46 centimeters.

The small bottle is 5.29 centimeters tall. Assuming that the small bottle of nasal spray has the shape of a cylinder too, then the height of the cylinder is 9.46 centimeters.

We want to know how much shorter the smaller bottle is than the big bottle. This is calculated by

Height of big bottle - height of small bottle

= 9.46 centimeters. - 5.29 centimeters

= 4.17 centimeters

5 0

3 years ago

Chloe is painting on a stretched canvas that has an area of 108 square inches. The length of the canvas is 12 inches. The width

Stolb23 [73]

Answer: Width of canvas = 9 inches

Area left for painting after painting the border = 70 square inches.

Explanation :

Since we have given that

Area of painting = 108 square inches

Length of canvas = 12 inches

As we know that

$\text{Area of rectangle} = length\times breadth$

$108=12\times breadth\\\\\frac{108}{12}=breadth\\\\9=breadth$

So, breadth of canvas = 9 inches .

Now, Chloe paints a 1 inch wide blue border on the canvas.

$\text{So, remaining length left for painting after painting the border} = 9-1-1=9-2=7\text{ inches}$

$\text{ Remaining breadth left for painting after painting the border }= 12-1-1=12-2= 10\text{ inches}$

$\text{So , area of remaining portion }= length\times breadth \\\\ \text{ area of remaining portion }=7\times 10\\\\\text{ area of remaining portion }=70 \text{ square inches}$

Hence, remaining area = 70 square inches .

6 0

3 years ago