CPhill

a / (1 - r)   = 2

a  = 2(1 - r)   →    a^3  = 8 ( 1 - r)^3         (1)

And

a^3  + (ar)^3  + (ar^2)^3 + (ar^3)^3  + ......  =   4

a^3  / ( 1 - r^3)  = 4

a^3  = 4 ( 1 - r^3)    (2)

Equate (1), (2)

8 ( 1 - r)^3  = 4 (1 - r^3)

8 ( 1 - r)^3  = 4 ( 1 - r) (1 + r + r^2)

8 (1 -r)^2  = 4 ( 1 + r + r^2)

8 (r^2 - 2r + 1) =  4 ( 1 + r + t^2)

2r^2 - 4r + 2  = 1 + r + r^2

r^2 - 5r + 1  =  0

r  =  [ 5 -  sqrt [ 25 - 4] ] /  2   = [   5 - sqrt (21) ]  / 2

f(x) =  72

(x - 1) ( x - 3 ) ( x + 5)  +  72   =  72

(x -1) ( x -3) ( x +5) =  0

x =  1, 3,  -5

Sum =  -1

. 

https://web2.0calc.com/questions/bisector

A + n = 10 (B - n)

A - n = 4 (B + n)

A +  n   = 10B -10n

A - n  = 4B + 4n              add these   and  subtract them

2A = 14B - 6n

2n = 6B - 14n

A = 7B - 6n       (1)

n = 3B - 7n  →   8n  = 3B  →  n = (3/8)B      (2)

Sub (2)  into (1)

A = 7B - 6(3/8)B

A = 7B - (18/8)B

A  = 7B  - (9/4)B

A = ( 28 - 9) / 4    * B

A =  (19 / 4)B

A / B  =   (19 / 4)

i^1 + i^2 + i^3 + i^4  = 

i + -1   - i  + 1  =  0

And this  pattern will repeat  

So....sum  of   i^1  tp  i^100 inclusive =  0

i^(-1)  = 1 / i  =   i / i^2  =  -i

i^(-2) = 1/i^2  = 1/ -1 = -1

i^(-3)  =  1/^2  * 1/i  =  -1 * 1/i =  -1 *-i  =  i

i^(-4) = 1/I^2  * 1/i^2 =  -1 * -1  = 1

And this pattern  will repeat

So....sum of i^(-100)  to i^(-1)  inclusive  = 0 

The only term not evaluated is i^0   =  1

So....the sum will be  1

1.

B              

       3        E

3            1

C    1   D                           A

For convenience, let BC = 3  and the radius of the semi-circle =  1

BC = BE

Let D be the center of the semi-circle   and let E  be the point where the semi-circle touches AB

Triangle EAD  similar to Triangle CAB

DE / EA = BC /  CA

1 / (BA - 3)  = 3 / CA

CA =  3 (BA - 3)

AC = 3BA - 9

BA^2  = 3^2 + (3BA - 9)^2

BA^2  = 9 + 9BA^2 - 54Ba + 81

BA^2  = 9BA^2 -54BA + 90

8BA^2  - 54BA + 90 =  0

Solving this produces .....BA =  15/4

AC = 3(15/4)  - 9  =  9/4

AC /  BC  =   (9/4) / 3  =  9 / 12  =   3 / 4

Note that  the sequence is

1000 , 1000(-1/3)  , 1000(-1/3)^2 , 1000(-1/3)^3   ........

We only need to  consider the terms that are positive  (these will be the odd terms)

So

1000 (1/3)^(n -1)  >  1         where n is  the nth term

(1/3)^(n -1)  >  1/1000    take the log of  both sides

log (1/3)^(n -1) >  log (1/1000)       and we can write

(n -1) log (1/3) > log (.001)      

Divide  both sides by log (1/3)....since this is  negative, we need to reverse the inequality sign

n - 1   <  log (.001) / log (1/3)

n <    1 +  log (.001) / log (1/3)   ≈  7.28

Note  that  the 7th term    =  1.37

Note that the 9th term = .152

So  only  terms   1 , 3 , 5  and 7   are >  1

x( x + 2)  + x ( x+ 1) =  kx (x + 1) (x + 2)

x^2 + 2x  + x^2 + x  =  kx ( x^2 + 3x + 2)

2x^2 + 3x  = kx^3 + 3kx^2 +  2kx

kx^3 + (3k-2)x^2 + (2k - 3)x  =  0

x [ kx^2  + (3k -2)x + (2k -3)  = 0

(3k - 2)^2  - 4 [k *(2k -3)]  < 0

9k^2 - 12k + 4 - 8k^2 + 12k  < 0

k^2 + 4  <  0

k^2  <   -4

k < 2i      and  k > -2i

k = (-2i, 2i) 

Something like this  

Law of Cosines   {   cos (angle BDA)  =  -cos (angle CDA)  }

BA^2   = BD^2  + DA^2   -2 (BD * DA) cos (BDA)

AC^2   = DC^2  + DA^2  - 2 (BD * DA) (CDA)

(8 + r)^2  = 2^2  + (10 -r)^2  -2 [ 2 (10-r) ] (-cos (CDA))

(2 + r^2)  = 8^2  + (10 - r)^2 - 2 [ 8 (10-r)] (cos (CDA))

(8 + r)^2  = 2^2  + (10 -r)^2  +2 [ 2 (10-r) ] (cos (CDA))

(2 + r^2)  = 8^2  + (10 - r)^2 - 2 [ 8 (10-r)] (cos (CDA))

Equating  Cosines

[(8 + r)^2  - 2^2 -(10 - r)^2] / [ 4 (10 -r))] = [ (2 + r)^2  - 8^2 - (10-r)^2] / [ -16 (10 -r) ]

[ r^2 + 16r + 64 - 4 - r^2 + 20r -100 ] / 4  =  [ r^2 + 4r + 4 - 64 - r^2 + 20r - 100] / -16

[ 36r -40 ] / 4 =  [ 24r -160 ] / -16

- 4 [ 36r -40 ] =  24r - 160

-144r  + 160  = 24r - 160

24r + 144r  =  320

r [ 168 ] = 320

r = 320 / 168   =  40/21  ≈ 1.9

          A

                       D

                 4

          E            8           C       

 triangle  ADE  similar to triangle  AEC

AE / DE  = AC / EC

h / 4  = sqrt [h^2 + 8^2] / 8  

8h  =  4sqrt [ h^2 + 64]

2h  = sqrt [ h^2 + 64 ]       square both sides

4h^2  = h^2  +  64

3h^2    =  64

h^2  =  64 / 3

h = 8 / sqrt 3  ≈   4.6