CPhill

The remaining side  =  50 - 22 - 13   =  15

The semi-perimeter, S, of this triangle  =   50 / 2   = 25

Using Heron's Formula, the area, A,  = 

√ [ S (S - 15) ( S - 22) (S - 13)  ] =

√ [ 25 * ( 25 - 15) * ( 25 - 22) * (25 - 13) ] =

√ [ 25 * 10 * 3 *12  ]  =

 5 * √[3 *10 * 12 ]  =

5 * √ [ 36 * 10 ]  =

5*6 * √10

30 √ 10  units^2

This  would be similar to  the number of identifiable  "words" that can formed using  4 R's  and 6 B's

So we have

(4 + 6)!           10!                   3628800                    

_______ =   ________  =     __________ =  210  identifiable "words"

4!  * 6!          24 * 720               17280

7 are staying behind

So

7 / 30   ≈  .233  ≈  23.3%

Let H  be the number  of home series

Let A be the number of away series

Let number of home wins  = 3H

Let the number of home losses  = 1H

Let the number of away wins  = 2A

Let the numbe of away losses  = 1A

And we know that

3H + 2A  =  40  + 1H + 1A         rearrange as

2H + A  =  40      (1)

The number of home games  =  4H

The number of away games  =  3A

And we know that

4H =   3A

So

(4/3)H =  A      (2)

Sub  (2)  into (1)   and we have that

2H + (4/3)H  = 40 

(10/3) H  = 40        multiply both sides by  3/10

H  = 40 (3/10)

H =  120 / 10

H  = 12  =  12 home series

And

(4/3)H  = A  =  (4/3)(12)  =  16 away series

So....the total number of wins  = 3H + 2A  =  3(12) + 2(16)  =  36 + 32  =  68 wins

And the total number of losses   = 1H + 1A  =  1(12) + 1(16)  =  28 losses

9 ( 1.09)  =   9.81  = final price

Let x be one number  and  y the other  and we have this system

x = 3y + 2   (1)

x + 8  = 4y   (2)

Sub (1) into (2)  for  x  and we have

(3y + 2)  + 8  = 4y

3y + 10  = 4y       subtract 3y from both sides

10  =  y

And

x = 3(10) + 2

x =  32

Let   A be the price of an adult ticket  (in dollars)

Let S  be the price of a student's ticket  (in dollars)

We have this system

6A + 8S  =   130

2A + 12S   = 90   multiply this equation through by  - 3

6A  + 8S  = 130

-6A - 36S  = - 270         add these

-28S  = -140      divide both sides by -28

S  =  5

And  using   2A + 12S  = 90

2A   + 12 (5)  = 90

2A + 60  = 90     subtract 60 from both sides

2A  =30      divide both sides  by 2

A  =15

This series diverges  ( we have larger and larger negatives  as terms are added )

Only geometric series have common ratios......arithmetic series  have a common difference

The common difference  is   

15/2 - 16

15/2 - 32/2  =

-17/2

Tha absolute value  of this difference  >  1

Since the transverse axis is parallel to the y axis, we have this form

(y - k)^2         (x - h)^2

_______   -   ________  =   1

     a^2               b^2

(h, k)  is the center  (1,4)

a  =  20/2  = 10  ⇒  a^2  = 100

b =  16/2  =  8 ⇒ b^2  = 64.we have

(y - 4)^2        ( x - 1)^2

_______  -   ________   =    1

  100                  64

Here's a graph  :   https://www.desmos.com/calculator/mkgqzex2ft

Call x  the number of original girls

Call y the number of original boys

And we know that

x / y  =  5 /3   cross-multiply

3x =  5y

x  =  (5/3)y    (1)

And we also know that

x+ 3                 7

_______  =    ___          sub (1)  in for x

y + 5                5

(5/3)y + 3            7

________  =     ___       cross-multiply

y +  5                  5

5 ([ 5/3]y + 3 ]  =  7 [ y + 5]

(25/3)y + 15  =  7y   +  35       subtract  15, 7y from both sides

( 25/3  - 7) y  = 20

(25/3 - 21/3)y  = 20

(4/3)y   =  20         multiply both sides by 3/4

y = 20(3)/4  =   15   = original number of  boys

And 

3x  = 5y

3x  = 5 *15

3x  = 75     divide both sides by  3

x  = 25 =  original number of girls