CPhill

1 divides   5  other integers

2 dvides    2  other integers

3 divides   1  other integer

We have  C(6, 2)   =  15 possible outcomes

The probability =   8  / 15

2/9  =10/ x    also means that

9/2  = x /10   multiply both sides by 10

x  =  10 (9/2)  =  45

We have   6^3   =  216 possible outcomes

Ways of rolling 12

4 , 4, 4  =  1

4, 5 , 3  = 3! =  6

4 , 6 , 2  = 3!  =  6

5, 6 ,1  = 3! =  6

5 , 5 , 2 =  3

6 , 3 , 3  =  3

                 ___

                  25

P (12)  =   25  / 216

Let x be the side of the triangle

Using the Pythagorean Theorem  we have this  relationship

sqrt  ( x^2  - 9)  + sqrt (x^2 - 16)  = sqrt (x^2  -1)      square both sides

x^2  -9 + 2 sqrt  [ (x^2 - 9)(x^2  -16) ]  + x^2  -16   =  x^2   -1

-25  + 2x^2  + 2sqrt  [ x^4 - 25x^2 + 144 ]   =  x^2   -1

2sqrt [ x^4  - 25x^2 + 144 ]  =  24 - x^2            square both sides  again

4[ x^4 - 25x^2 + 144]  =  x^4 - 48x^2  + 576

4x^4 - 100x^2 + 576  =  x^4  - 48x^2 + 576     simplify

3x^4 = 52x^2

x^2   ( 3x^2  - 52)   = 0

x^2  = 52/3

x =  sqrt  (52)  /sqrt (3)  =   2sqrt (13)/ sqrt (3)  =  (2/3)sqrt (39) ≈ 4.163

3y + 2x  =  7 - x + 8y    rearrange as

3x - 5y  = 7

In the form    Ax + By =  C, the slope is  given by      -A/B

A = 3    B  = -5

So.....

-A / B  =  -3 / -5  = 3/5

To find  m   we have

(3/5) *m    =  -1

m = -1 (5/3)

m  =   -5/3

This pentagon is  irregular

See the image 

We can break this up into  3 triangles  AED , ADC and ABC

I'm going to  describe how to find the area of triangle  AED.....based on  this you should be able to  figure the areas of the  other two triangles  ( I'm going to express roots  as decimals to  ease the computations.....because of this your  computation will be a close approximation of the true  area )

Find the side lengths  using the distance  formula

AE  = sqrt [ (1 - -3)^2  + (7-4)^2 ]  = 5

DE =  sqrt [ (-1- -3)^2 + (4  - - 1)^2 ]  =  sqrt  [ 29 ]  ≈ 5.39

AD  = sqrt [  ( 1- -1)^2  + (7 - -1)^2 ] =  sqrt  [80 ]  ≈  8.25

Add the sides  and  divide by 2   =     [ 5 + 5.39 + 8.25]  / 2  =    9.32

Using Heron's Formula  to  calculate  the area  we have

sqrt   [s * ( s -AE)(s - DE) (s - AD)  ]   =  sqrt  [ 9.32 * (9.32  - 5) (9.32 -5.39) (9.32 - 8.25) ]  ≈   13   units^2

Follow the same  procedure for  the other two triangles .....

Then....add the ares you get to find the area of the pentagon

arc length  =   r * (theta in  radians)

So

arc length   =   7.2  ( 5pi /2)  =   36 pi / 2  =  18 pi ≈  56.5 cm

The area of the square  =  3^2  =  9

The small unshaded right triangles have bases of 1 and heights of 2.....

Their combined areas =  4 (1/2)(1) (2)  =  4

So....the pink region has an area of   9 - 4   = 5

A little hard to visualize this   (I did it with a piece of paper   - hope I didn't mess it up  !!!)

When we make the fold  EF = EA  = 9  and  EB = 6

And triangle EBF is right  so    BF  = sqrt  [ 9^2 - 6^2 ]  = sqrt  45  =  3sqrt 5

And FD =  BC

And triangle FDC is right  with   DC = 9    FC  = BC - BF =   BC - 3sqrt5  = BC - sqrt 45

So

FC =  sqrt  [ FD^2 -  DC^2 ]

(BC - sqrt 45)  =  sqrt  [ BC^2  - 81) 

(BC - sqrt 45)^2   = BC^2 - 81

BC^2 - 2*BC*sqrt45 + 45  = BC^2 - 81

-2sqrt (45) BC =  -81 -  45

-2 sqrt (45) BC =  -126

BC  =    126 / ( 2 sqrt 45)  =   126 / (2 * 3 * sqrt 5)  =   21 / sqrt 5  =  (21/5)sqrt 5

Look at the graph here :   https://www.desmos.com/calculator/em32xqiof4

We can split the area into two triangles....one with a base and  height of 4

Area =  (1/2)4*4  =  8

And the second one  with a base of 16 and a  height of 8

Area  = (1/2)(16)(8)  = 64

The area  we are interested in  =    64  - 8   =  56