CPhill

Nice, AD  !!!!

Probably an easier way, but.....

Height of triangle EAD =10

(1/2) the base = 5

Slope of line through AE = 10/5 = 2

So....Let the line through AE  have  the  equation  y =2(x - 10)   =  2x -20

Slope of line through CD =  [ 10 -0 ] / [0 -20 ]  =  -1/2

So....Let the line through CD  have the equation    y = (-1/2)x + 10

Find the  intersection of these lines = x coordinate of N

2x -20  = (-1/2)x + 10

(5/2)x = 30

x =  12

The y coordinate of A  is   y = (-1/2)(10) + 10  = 5  = base of triangle formed by shaded region

 (12 -10)  = 2  =  height of triangle formed by the shaded region

Shaded region    =  (1/2) (5) * 2  =   5

1/2 +  1/3  +  1/8  +  1/30  +   1/144  + 1/840 +  ........

Progressive sums

5/6,   23/24,   119/120,    719 / 720,    5039 / 5040    + ......  =  (n! - 1) /n!  =    n! / n !  +  1 / n!

Note that   n! / n!   = 1     (for  integer n   ≥ 3)

And as n gets very large, 1 / n!  ≈  0

So....the sum  =   1

r =  .12 / .4  =  12 / 40 =   3/10

Sum  = 

First term / [ 1 - r ]  =

.4 / [ 1 - 3/10]  =

.4 / .7   = 

4 /  7

Here's a similar problem

https://web2.0calc.com/questions/pls-help-asap_78#r1

See if you can take it  from  here

If not, let me  know and  I'll help you

All are true

The highest power term in each polynomial  determines the end behavior

-2x^3    goes to inf as x approaches -infinity    (x^3 will be negative but the -2 makes it positive)

2x^4  goes to inf as x approaches infinity  

-9x^5 goes to -inf as  x approacjes  infinity

Coordinate Geometry

Let the center  of the hexagon = (0,0)

Let ED  = 2

D= (1, sqrt 3)

F = (-2,0)

A= (-1 , -sqrt 3)

N =  (-3/2 , -sqrt (3) / 2 )

M = (0, sqrt (3) )

B = (1 , -sqrt (3))

Slope of line containing  segment  (DN) =    [ sqrt (3) + sqrt (3) / 2 ] / [5/2]  =  3sqrt (3) / 5 

Equation of this  line

y = [ (3/5)sqrt 3 ]  ( x - 1) + sqrt (3)         (1)

Slope of line  containing segment  MB  =  [ 2sqrt (3) ] / -1 =  -2sqrt (3)

Equation of this  line

y = [ -2sqrt (3) ] (x ) + sqrt (3)         (2)

Set  (1) = (2)   to find the x coordinate of P

[ (3/5) sqrt (3) ] (x -1) + sqrt (3)  =  [-2sqrt (3)]x + sqrt (3)

(3/5)sqrt (3) x - (3/5)sqrt (3) = -2sqrt (3)x

(13/5) sqrt (3) x  =  (3/5)sqrt (3)

(13/5)x = (3/5)

x = (3/5)(5/13)  = 15/65  =  3/13

y =  -2sqrt (3) (3/13) + sqrt (3)

y =  (7/13)sqrt (3)

DP^2  = ( 1-3/13)^2 + ( sqrt 3 - (7/13)sqrt 3)^2  

DP^2  = (10/13)^2  + [(6/13) sqrt 3]^2

DP^2  =  16/13

PN^2  = ( 3/13 + 3/2)^2 + [ (7/13)sqrt 3  + sqrt (3)  / 2)^2

PN^2   = (45/26)^2  +  (27sqrt (3) / 26)^2

PN^2  =   81/13

DP^2 / PN^2  =   16 / 81

DP  / PN  =   4 / 9

Circumradius  = product of the sides /  [ 4 * area ]  =

a^3  / [ sqrt (3) / 4   * a^2 ] =

[4/sqrt (3) ] a

Inradius =  area  /semi-perimeter

Area =  (1/2)a^2 * sqrt (3) / 2  =  (sqrt (3) / 4) a^2

Seimi-perimeter =  (3/2)a

Inradius =  [ sqrt (3) / 2 ] a^2  / [ (3/2)a] =  a / sqrt 3