CPhill

Using  partial fraction decomposition :

1/ [n * (n + 3)] = A / n  + B / (n +3)             multiply through by n *(n + 3)

1 = A (n + 3)  + Bn

1 = (A + B)n  + 3A

3A  = 1

A = 1/3

A+B  = 0

B = -1/3

So    

1 / [1x 4]  =   (1/3)/1 - (1/3)/4 =   1/3 - 1/12

1/ [ 4 x 7] =   (1/3)/4 - (1/3)/7  =  1/12 - 1/21

1/[7 * 10] =   (1/3)/7 - (1/3)/10 =  1/21 - 1/30

......

1/(34 *37)  = ((1/3)/34 - (1/3)/37  = 1/ 102 - 1/111

1 / (37 * 40) =  (1/3)/(37) - (1/3)/40  =  1/111  - 1/120

All the intermediate terms  "cancel"     and  we  are left with

1/3  -  1/120   =

[ 120 - 3 ] / 360   =

117 / 360    =

13 /  40

Diagonal^2  =  20'2 + 20^2 -2 ( 20 *20) * cos (144°)

Diagonal^2  =  800 - 800 *(-1 - sqrt 5) / 4

Diagonal  =   sqrt [ 800 + 200 ( 1 + sqrt 5) ] =  10 sqrt [ 10 + 2sqrt 5 ] = 10 [ 2 (5 + sqrt 5)] ≈ 38.04

https://web2.0calc.com/questions/altitudes_39

From A,B  draw  AE  and BF perpendicular to BC

Let FC = x

By the PythagoreanTheorem

sqrt [24^2 - x^2 ]  =  height of  trapezoid

And DE = 104 - (78 + x)  =  26  - x

And, again,

sqrt [ 10^2  - ( 26 - x)^2 ] = height of trapezoid

So

sqrt [ 24^2 - x^2 ] = sqrt [ 10^2 - (26-x)^2 ]    square both sides

24^2 - x^2  = 10^2 - (26 - x)^2

576 - x^2  = 100  - x^2 + 52x - 676

52x  =  1152

x =  1152 / 52   = 288/13

height of trapezoid = sqrt [ 24^2  - (288/13)^2 ]    =  120 /  13

Area =  (1/2) (120 / 13) ( 104 + 78)  =  (1/2) (120/13) ( 182)   =  (91) (120 / 13)  =   840

Without  explanation.....the  side of the square = [ product of leg lengths ]  / [ sum of the leg lengths ]  =

 [6 * 6 ]  /  [ 6 + 6 ] = 36 / 12  =  3

I can show you how I got this  if you want me to.......

J / S =  3 / 2

J = (3/2)S

[J + 12 ] / [ S - 1 ]  =  2 / 1

[ (3/2)S + 12 ]  / [ S - 1 ] = 2 / 1

(3/2)S + 12  = 2  [S -1]

(3/2)S + 12 = 2S  - 2

2S - (3/2)S = 12 + 2

(1/2)S  = 14

S = 28

J = (3/2)S = (3/2) (28)  = 42

After the changes  there are  54 J   and  27 S  =    81  students total

https://web2.0calc.com/questions/triangle_4602

\( a + \frac{1}{a} + a^2 + \frac{1}{a^2} = 0 \)

[a^2 + 1] / a  +  [ a^4 + 1] / a^2  =   0          multiply through by a^2

a^3 + a + a^4 + 1   =   0

a^4 + a^3  +  a + 1  = 0

a^3 ( a + 1)  + (a + 1)  =  0

(a + 1) ( a^3 + 1)  =   0

a + 1 =  0               or       a^3 + 1 =  0 

a =  -1                             a^3  =  -1

                                       a =  -1

a^5  = (-1)^5 =   -1

      A 30

12  D

      C 90                          B

Angle  B  = 60

So angle BCD =  30

Side  BC =   12/sqrt 3  =  4sqrt 3

So...in triangle  BDC,, DC  =  BC / sqrt 3 =  4sqrt 3 /sqrt 3 =  4

Area of BDC = (1/2) BC * DC =  (1/2) (4sqrt 3) (4)  =  8 sqrt 3

Area of ABC  =  (1/2) (AC)(BC)  = (1/2)(12)(4sqrt 3)  =  24sqrt 3

So....  [ ABD ]  =   [ ABC ] - [ BDC ] =  24sqrt 3 - 8sqrt 3 =  16 sqrt 3

CORRECTED 

https://web2.0calc.com/questions/triangles_35925