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3 years ago

Help,Cause homework is hard :/

Mathematics

1 answer:

Nutka1998 [239]3 years ago

6 0

Hey there!

In my opinion, yes, because there are many reasons for this answer. But one reason is that when you get older, you'll be using 3.14 more than 22/7 because you don't really use fractions anymore, you use decimals.

^^That's my opinion, you might have something different^^

Hope this helps!

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The vertex of a figure is located at (1, 3). The figure is rotated and the image of vertex is located at (-3, 1) Which of these

Aleksandr [31]

Answer:

LOLOLOLOLO

Step-by-step e

xplanation:

7 0

2 years ago

Can someone help w number 4

strojnjashka [21]

Answer:

angle DFG = 49 degrees

angle JKL = 41 degrees

Step-by-step explanation:

When angles are complementary with each other, it means that if you add both of the angles up, it adds up to 90 degrees.

In this question, you would have to add up angle DFG and angle JKL and find the x that makes the equation equal to 90 degrees.

angle DFG = x + 5

angle JKL = x - 3

(x + 5) + (x - 3) = 90

2x + 2 = 90

2x = 90 - 2

2x = 88

x = 44

But since we have to find out the angle measures, we have to the "x = 44" with the x's in the DFG and JKL angles.

DFG = (44) + 5 = 49

JKL = (44) - 3 = 41

7 0

3 years ago

"Rebecca walks 100 feet in a straight line. She then turns 20 degrees to her left and walks another 100 feet, and then turns 20

iragen [17]

Answer:

1,800 ft

Step-by-step explanation:

Rebecca will walk in the pattern of a regular polygon with 'n' sides and internal angles of 160 degrees (180 at a straight line - 20 degrees from each turn).

The equation that describes the internal angle of a regular polygon is:

$A = \frac{(n-2)*180}{n}$

For A = 160 degrees:

$160n=(n-2)*180\\(180-160)n=360\\n=18\ sides$

If each side is 100 ft long, the total distance that Rebecca has walked is:

$d = 18*100\\d=1,800\ ft$

She walked 1,800 ft.

5 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:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$