CPhill

Very nice, Max  !!!

abcd  =  8!   =  40320

ab  +  a  + b  =  524  (1)

bc + b + c  =  146   (2)

cd + c +  d  =  104    (3)

Manipulating  (1)  and (2)  we have that

(a + 1) (b + 1)  =  525   ⇒  (b + 1)  =  525 / (a + 1)   

(b + 1) (c + 1)  =  147  ⇒  (b + 1)  =  147 / (c + 1)   

Which implies that    

525 / (a + 1)  =  147/ (c + 1)

525/147  = 25/7 =  (a + 1) / ( c + 1)

Which implies that a  = 24   and c  =  6

Which implies that

b + 1  =   525/25  ⇒  b  =   525/25 - 1  =    21 - 1  =  20

And  manipulating (3), we have

(c + 1) (d + 1)  =  105

(7) (d + 1)  =  105

d + 1  =  15

d  = 14

So

a  -  d  =    24  -  14   =  10

( 3√5, d + 3)  =   (√45 , d + 3 )

So we have that

3d  = √  [  (√ 45)^2  +  (d + 3)^2 ]

3d  =  √ [  45  +  d^2 + 6d + 9 ]

3d  =  √ [ 54 + d^2 + 6d ]       square both sides

9d^2   = d^2 + 6d + 54      rearrange as

8d^2 - 6d - 54  = 0

4d^2  - 3d  -  27  =  0      factor

(4d + 9) (d - 3)  = 0

Setting both factors to 0  and solving for d, we have that

d  = -9/4     or  d  =  3

The first value  gives the  distance  of 3(-9/4)  =  -27/4 units.....but distance is  a positive quantity

So......d  = 3   is the correct value for d  and  3d  =  9 units

R + Y  + O  + W   =  35

R/2  =  Y - 2   ⇒  Y  = R/2 + 2    (1)

R/2 =  O/3  ⇒  O  =  (3/2)R   (2)

R/2  =  (W + 3) / 3  ⇒  (3/2)R  = W + 3  ⇒  (3/2)R - 3  = W    (3)

Sub (1), (2), (3)  into  the first equation

R  + R/2 + 2  + (3/2)R  + (3/2)R - 3  = 35

4R + R/2   =  36

8R + R  =  72

9R  = 72

R  = 8  = red marbles

And

Y = 8/2 + 2  =  6

O  = (3/2)8   = 12

W  = (3/2)(8) - 3  =  9

2x  =  5y     ⇒    y  =  (2/5)x      (1)

7y = 10z     (2)

Sub  (1)  into (2)

7 [(2/5)x]  = 10z

[14/5] x  =  10z

z / x  =   14 / 50   =   7 / 25

No prob....I still gave you some points....!!!

Notice that  one possible  "Pythagorean Triple" right triangle is  one with the sides  5 - 12 - 13

Any any  multiple  of these will also be a Pythagorean Triple...so....this implies that

5(3)  -  12(3)  - 13(3)   =

15  -  36   -  39   will also be a Pythagorean Triple

So....the shorter leg  =  15 units

I think the "i" is supposed to be there, MapleTheory

So....

The quantity  under the radical  must be  = -15

So......we can solve this...

7^2  - 4(4)k  = - 15

49  -  16k   =   -15

-16k  =  - 64

k  =  4

First.....we can solve

[ 4x + 1 ] / 3  =  1

4x  + 1   =   3

4x  =  2

x  =  2/4  =  1/2

So  ( 1/2, 1)  is on the orginal graph

So..... (1, 1/2)  is on the inverse

So....

  [  f ^-1  (1) ] ^-1  =   [ (1/2) ]^-1  =    1 / (1/2)  =   2

Thanks, ant!!!!