CPhill

Since all the boces are the same, this  boils down to partitions of  4

4            in any  one box

2 2         in any  two boxes

1 3         in any two boxes

1 2 1      in any three boxes

1 1 1 1   in any  four  boxes

5 ways

a)  We can choose any 2 of the 12  = C(12,2)  = 66 ways

b) Mark is a given....choose any 1 of the remaining 11 =   11 ways

c)  Mark is  out......choose any 2 of the  other 11  =  C(11,2)   = 55 ways

The points  (-4,0) and (0,2)  are on the graph

Slope =  (2 - 0)  / ( 0  - - 4 ) =   2/4  = 1/2

Using (0,2)  we can write the equation ofthe  line as

y = (1/2)x + 2     multiply throygh by 2

2y  = x + 4

x - 2y  =   -4

Here :

https://web2.0calc.com/questions/geometry_55509

Note

w^2 + l^2 + h^2 = the length we want

Sum of edges  = 

4( w + h + l)  =  44

w + h + l =  11      squaring both sides produces

(w^2 + l^2 + h^2) + (2 wh + 2 wl +  2 hl)  = 121

(w^2 + h^2 + l^2) + 2 (wh + wl + hl)   =121       (1)

Surface Area  =

2 (wh + wl + hl)  = 30               (2)

Sub (2)  into (1)

(w^2 + l^2 + h^2)  +  30  =  121

w^2 + l^2 + h^2 =  91

The horizontal distance from the center of one of the faces of the tetrahedron to the edge of the tetrahedron = (1/3)s, where s is the side length of the base of the tetrahedron

The  length of one  side  of the base of EFGH =   sqrt  [ (1/3 * s)^2  + (1/3 *s)^2 ] = 

sqrt [ (2/9) s^2 ] =   (sqrt (2) / 3)   s

So the area of the base of of EFGH = (2/9)s^2

And the height of the pyramid = (1/3) height of the  of the tetrahedron =  (1/3) h

Volume of tetrahedron  = (1/3)s^2 * h

18= (1/3) s^2 * h

s^2 * h  =  54

Volume of  pyramid  EFGH =  

(1/3) (2/9)s^2 * (1/3) h  = 

(2/81) s^2 * h  =

(2/81) (54)  =  2 (54 / 81)  =  2 (2/3) =   4/3

x^3 - ax^2 - b^2 x + ab^2  =

x^2 (x - a)  - b^2 ( x - a)  =

(x - a)  ( x^2 - b^2)  =

(x - a) (x + b) ( x - b)

(x- y)^2  - 3 ( y - x)  = 

(x - y)^2 + 3 ( x - y)   =

(x - y)  [  ( x + y) + 3  ]

sin A = 1/3

sin A   =  BD  / AB

BD / AB =  1/3

BD / 24 = 1/3

BD = (1/3)(24)  = 8

sin C = 1/2

arcsin (1/2)  = 30°  = C

BD =  8

So since triangle  CBD   = 30 -60 - 90

Then BC = 2 BD  =  2  *  8   = 16

The triangle is  isosceles....let the base = (sqrt 6   - sqrt (3)

cos ( base angle)  = ( [ sqrt  6 -sqrt 3  ] / 2 )  /  2 =    [sqrt 6 - sqrt 3 ]  / 4

arccos ( [ sqrt 6 -sqrt 3 ] / 4 )  ≈ 79.7° 

Apex angle =   180 - 2(79.7)   ≈  20.6°

Angles are    20.6° , 79.7° , 79.7°

Slope =  [ b^2   - a^2  ] /  [  b - a ]  =   [ (b -a) (b + a) ] / (b -a) =  b + a  =  2