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Side length of the square  =   [ product of the legs  / [ sum of the legs ] =  (1*1) / ( 1 + 1)  = 1/2

                                           Q

                                   8               X      5

                           P                      Y       R   

                                          1

Impossible.....PQ  >  QR + PR   violates the triangle inequality

There are an infinite number of solutions to the question you asked - see below.  Perhaps you meant to add another condition.

x - 1  =   x (x +x^3 + x^5 +.....)   -  1(x + x^3  + x^5 + .....)

x - 1 =  (x -1) (x + x^3 + x^5 + .......)

(x -1)   / ( x -1)  =   ( x + x^3 + x^5  +  ......)

1  =  ( x + x^3 + x^5 +  .......)

-1 =  -x - x^3 - x^5 +  .........

So

x = 1 - 1  + x^2 +x^4 + x^6+ ........

x = x^2 + x^4 + x^6 + .......

Infinite sum 

x^2 / ( 1 - x^2)  =  x

x^2  =  x - x^3

x^3 + x^2   - x  = 0

x ( x^2 + x - 1)   = 0

x = 0   reject

x^2 + x - 1   = 0

x^2 + x  = 1

x^2 + x + 1/4 =  1 + 1/4

(x + 1/2)^2  = 5/4

x + 1/2  = sqrt (5)  / 2          or    x + 1/2  =  -sqrt (5) / 2

x = [ sqrt (5) - 1 ]  2            or  x =  [ -sqrt (5)  -1 ] / 2

x =  "phi"      or    x =  "-Phi "

https://web2.0calc.com/questions/angle-bisector_13

https://web2.0calc.com/questions/angle-bisectors_37

a = 1

6a = a2 + a3

6 = a2 +a3

6 = ar + ar^2

6 = 1r + 1r^2

6 = r + r^2

r^2  + r  -  6 =  0

(r - 2) ( r + 3)  =  0

r - 2 =  0 

r =  2

So

S4  =   1 + 2 + 2^2 + 2^3  =   15

a5 + a16 =  20

a1 + 4d  +  a1 + 15d     =  20

2a1  + 19d  =  20

(a1)  + ( a1 +  19d)  =  20

(a1)  + (a20)  =  20             

S20 =   [ a20 +a1 ] [20 ] / 2    =  [20] [10]   =  200

Simplify as

x^2 - 4x  + y^2 + 8y =  25             complete the square on x, y

x^2 - 4x +4  +  y^2 + 8y + 16 =  25 + 4 + 16

(x - 2)^2  + ( y + 4)^2  =  45

This is a  circle  centered at  ( 2 , -4)   with a radius of sqrt (45)  = 3sqrt (5)

Largest value of x =   2 + 3sqrt (5)

S10  = a ( 1 - r^10) / ( 1 -r) =  10

S30 =  a ( 1 - r^30) / (1- r)  =   70

a / (1 -r )  =  10 / (1 -r^10)

a/(1- r)   =   70 / (1-r^30)

10 / ( (1 -r^10)  =  70 / ( 1 -r^30)

10 ( 1 - r^30) = 70 (1 -r^10)

10  - 10r^30  = 70 - 70r^10

70r^10 - 10r^30   =  60

7r^10 - r^30  = 6

r^30 - 7r^10  + 6 =  0       

Let   r^30  = x^3 ,  7r^10  = 7x

So we have 

x^3 - 7x + 6   =  0

1 is a root

Use synthetic division  to find the other roots

1  [ 1  0   -7     6]

          1    1    -6

     ____________

      1   1   -6    0

The remaining polynomial  is

x^2 + x  -  6   =  0 

(x + 3) ( x -2) = 0

x = 2  is the only feasible  solution

So

r^30  = x^3

r ^10  =  x

r^10  = 2

r = 2^(1/10)

a ( 1 - r*10) / (1 - r)  = 10

a  ( 1 - 2) / ( 1 - 2^(1/10) )  =  10

a  = - 10 ( 1 - 2^(1/10))   

a =  10 [ 2^(1/10) - 1)]

S40  =    10 [ 2^(1/10) - 1 ] [ 1 - 2^4 ]  / [ 1 - 2^(1/10) ] = 

-10 [ 1 - 2^4 ]  =

-10 (-15)  =

150

"Can you solve these?"<- I think you misspelled "solve these problems for me so I can please the answer box." Quite an egregious error if you ask me!

The only thing you are "absolutely destroying" is your own education as you 'outsource' your homework questions .

This is a nice question , kind of challenging.

So then here is my explanation: 

To find the length of segment AS in triangle AST.

You have information about the lengths of AC, AT, and BT.

Using the Pythagorean theorem, you calculate BC as √(AC² - AB²), which is √(35² - 20²) = √825.

Then, you find CT by adding BC and TB:

CT² = BC² + TB²

= 825 + 9²

= 906.

Now, you use the Pythagorean theorem for triangle CST to get CS² + ST² = CT² = 906.

And for triangle AST, you have AS² + ST² = AT², which is 121.

Subtracting equation 1 from equation 2, you get

AS² - CS² = -785.

Rearrange to get :

AS² - (35 - AS)² = -785.

Further simplify to AS² - AS² + 70AS - 35² = -785.

Now, solve for AS, and you get AS ≈ 6.28.

So, the length of segment AS is approximately 6.28 units.

I think there might be some things missing but I heard of the question, and I think it is:

In the figure, angles ABC and AST are right angles. If AC = 35, AT = 11, and BT = 9, then what is AS?

Um, is there an image to this question, becuase I have no idea without any image. 

I know you can find all the terms from a1 to a17 and add them up, but can someone find a more simple way to get the answer.

And I don't even get this question, so 

Yes, in some point, the planets will line up in a line that is not in the original starting line.

3.

2* area ABC =  AD * BC                  2* area ABC =  BE * AC

2* area ABC =  12 * BC                   2* area ABC  = 16 * AC

area ABC = 6 BC                             area ABC  =  8 AC

6 BC =  8 AC

AC = (8/6) BC = (4/3) BC

Triangle inequality

AB + BC >  AC

AB + BC >  (4/3)BC

AB  >  (4/3)BC - BC

AB > (1/3)BC

Then

2 [ ABC ]  =   BC  *  AD

2[ ABC ]   = BC * 12

2 [ ABC ] = (1/3)BC *  36

Since AB > (1/3)BC

2[ ABC ]  < AB *  36

2[ ABC ] < AB * CF

Then CF  < 36

So

Max integer value for CF =  35

Good Question, 

83pi*r^2/360 = 60pi

83pi*r^2 = 60pi*360

cancel pi on both sides: 83*r^2 = 21600

r^2 = 21600/83

r would be (rounded to the nearest hundreth): 16.13