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That's an elegant solution, Melody. Definitely better than bashing with incircle coordinates 

1. Oh my god I answered on the wrong question 

Using algebra, we can see after subtracting a few numbers, we notice there are 10 numbers, therefore there are 5 pairs and one number.

$1-2, 4-8, 16-32, 64-128, 256-512, 1024$

using normal subtraction 

We get 

683

Using algebra, we can see after subtracting a few numbers, we notice there are 10 numbers, therefore there are 5 pairs and one number.

$ -1, -4, -16, -64, -256, and 1024$

$1-2, 4-8, 16-32, 64-128, 256-512, 1024

using normal subtraction 

We get 

$683$

We can bring the y^-2 down to the denominator and divide by the x to get 3x^2/y^5

:D

Mmmmm......let's  see....maybe  ???

We  have  63 + 18 =   81 total spots

18 / 81      

 2 /  9

2 : 9

There are 11 terms and the common ratio is -2. Using the geometric series calculator, the answer is 683.

:D

OK......glad  to   help   !!!

[ My Calculus is rusty  ....LOL!!!   ]

If we  inscribed  the  dodecagon  in a circle  we  can  find   the radius  as

1 = (1/2)R^2  ( sin 360 /12)

1  =  (1/2) R^2  ( sin 30°)

1  =  (1/4) R^2

4  = R^2

2   = R  of  circle

Diagonal of  square =  4

Side of square =   4/sqrt (2)

Area of square =   (4/sqrt (2))^2   =  16/2   =   8

The answer to the question is 1.02040816327 you can use a calculator and see for yourself hope this helps

looks like i got the first one correct! Which is awesome!

The second one however, I see where i made my mistake, very simple one that i should have caught! I will take note.

Thank you CPhill!

x^2  - ax  +  2016 +  5  =  0

factors of 2021 =   1, 43, 47, 2021

x^2   - ax  + 2021  =   0       factor as

(x  - 43) ( x  - 47)  =  0

Min  value  of a  = -(43 + 47)  =  -90

2. 

∫ 3  /  sqrt ( 1  - x^2)  dx   =    3arcsin x  + C

2 =  3 arcsin (0)  +  C            (arcsin 0    =  0  )

(0, 2)

2  = 0  + C

2   = C

So

3arcsin x  +  2

1.    ∫ 3^x  dx  =   3^x/ ln 3  +  constant ,  (C)

So

4  =   3^x / ln 3  +  C

4 =  3¹ / ln 3  +  C

4 - 3 / ln 3  =   C   ≈ 1.27

So

3^x/ln 3  + 1.27

Recursive rule for the arithmetic  sequence

next term =  previous term +  2

a_n+1   = a_n + 2       where  a₁  = 1

Explicit rule for  above  =

an = first term  +  d ( n -1)       where a_n  is the  nth term  and  d  =the common difference between terms = 2

So

a_n  = 1 +  2 ( n -1)

The second  ISN'T  a geometric series  because  the common ratio between  the terms is not constant

4/2  =2   but  6/4   =3/2

We can still write  rules for  it

Recursive rule

a_{n+ 1}  = an + 2  where a₁  =2

Explicit rule =     a_n   = first term* (2)n     where n is the nth term ....so...

a_n =  2 n

I have sent you a private message. 

Certainly if AoPS is written into the question or into a link address then it is an AoPS question.

There are other more subtle ways to tell as well.

Oh I forgot, like I kinda know like for example, the explicit form of the arithmetic sequence is like a_n=1+2(n-1)

But my teacher showed things like with exponents, so I am completely lost there.

90*9=9*9*10=81*10=810

6*14=6*2*7=12*7=84

2.  Similar to  1

Reverse the coordinates of the  center   = (3,-3)

The "new" coordinates   =  [(x  + 3 , (y - 3)  ]

Apply the scale factor   = ( (1/3)x + 1, (1/3)y - 1)

Add  back  the coordinates of the center  

[ (1/3)x + 1 - 3  ,  (1/3)y  - 1 +3 ]  =

[ x/3 - 2 , y/3 - 2 ]    =dilation rule

You should be able to finish this one  by applying the dilation rule to  L,M,N,P

Also....find the  coordinates of the parallelogram  at the bottom and use the  same proceedure to find the  dilation rule  for the  first one

For the second one....just apply the scale factor to  the  computed  points  you got  in  1

(1)    E = (0,0)   F(1,5)    G(7,2)    center of dilation =  (5,2)   scale factor  = 2

Dilation rule   is

Reverse the signs on the  coordinates of the  center =  ( -5 , -2)

The "new " coordinates  are  ( x - 5, y - 2)

Apply the scale factor  =  (2x - 10, 2y - 4)

Add  back the center coordinates  = ( 5, 2)

 [ (2x - 10 + 5). ( 2y - 4 + 2) ] =

   ( 2x  - 5   ,  2y   - 2)

E' = [ (2(0) - 5, 2(0)-2)  ]  = (-5, -2  )

F'  =   [ ( 2(1)  - 5 , 2(5) -2)  ] =  ( -3, 8)

G' =    [ 2(7)  -  5, 2(2) - 2 ] =  ( 9, 2) 

See here  :

LOL

See the following :

Let A  = (0,0)   B = (37,0)  

And  the equation of the  circle  is x^2 + y^2  = 33^2

It's clear  that   the greatest distance  from B to D  will  be when T  =(-33,0)

And this distance  is  just   37 + 33   = 70

We can see  that if we construct a  circle with a radius of  70 centered at B....it will  only intersect  the circle with a radius of AC at   D  =(-33,0)

S = a₁/(1-r)      r is  1/8

   = (1)/7/8    = 8/7

(3^1/2 )^sqrt2   *  (3^1/2)^-sqrt2      =   3 ^{(sqrt2)/2  - sqrt(2)/2}    =3⁰ = 1