CPhill

Magnitude  of  a ±  bi =   

sqrt  [ a^2  + b^2 ] 

Note that triangles DXE  and  EYF  have the same height as triangle  DEF  and their combined bases = DF

So.....their total areas must be the same as  the area of  triangle  DEF

Since rectangle  DXYF  =  40......then it is composed of two equal areas .....

 So.......[ DEF  ]   =  40   / 2    =  20

1/ x^(-3/6)  =   1 / x^(-1/2)  =   x^(1/2)    =  sqrt (x)

   { Remember that    1/a^(-n)   =  a^n }

By the Triangle Inequality :

25  +  shortest side  > the remaining side

If shortest  side =  1      remaining side =  34    not possible

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If shortest side =  5     remaining side =  30      not possible

If shortest side  =   6   remaining side =  29

The shortest possible side = 6  inches

Using an exponent  law, we can write P  as

5 ^ ( 1/5 + 1/25 + 1/125  .....)

The  sum  of the exponents is the sum of an infinite series = 

(1/5)  / ( 1 - 1/5)  =  (1/5) / (4/5)   =   1/4

So P  can be expressed as

5^(1/4)

a = 5        b  =   1        c   = 4

a + b + c   =   10

C(5,1)   =   5 different amts

C(5,2)  =  10 different amts

C(5,3)  = 10 different amts

C(5,4) = 5 different amts

C(5,5)  =  1 amt

So

31  different amts

M of first earthquake =   log 10^5.1  =   5.1

M of second earthquake  = log 10^7.7   = 7.7

Ratio of M's   =    7.7  / 5.1 =   1.509     =    about 2 times as strong based on their Richter scale magnitudes

Let f(x)   = ax^2 + bx +c

g(f(x))  =   (ax^2 + bx + c)^2   - 11(ax^2 + bx + c)  + 30

g (f(x))  = a^2 x^4 + 2 a b x^3 + 2 a c x^2 + b^2 x^2 + 2 b c x + c^2 - 11ax^2 - 11bx - 11c  + 30

g(f(x)) =  a^2x^4    + 2abx^3  + (2ac + b^2 - 11a)x^2 + (2bc - 11b)x + (c^2 - 11c + 30)

And  g (f(x))  = x^4    -8x^3  + 11x^2   + 20x +  6 

Equating coefficients

a  = 1  or  -1

b = - 4 or 4

Let  a  =1   and b  =  -4

(2ac + b^2 - 11a)  = 11   →   2c + 16 - 11  = 11  →  c = 3

So

f(x)  =   x^2  - 4x  +  3

And

f(7) =   7^2  - 4(7) + 3   =     24

a/b  = 2      b/c   =  4/3

Note that   (a/b)   *  ( b/c)  =  ( ab ) / (bc)    =  a / c   =      2 * 4/3  =   8/3  

Here's what I get

A)   Assuming that the chairs  of each color are identical

You have calculated the  the number of possible arrangements correctly

If two white chairs  are together  they could occupy  positions 

1,2    and the other white chair can occupy positions  4,5,6,7 or 8

2,3    and the other white chair can occupy positions 5,6,7 or 8

3,4    and the other white chair can occupy positions 1, 6, 7 or 8

4,5    and the other white chair can occupy  positions 1,2, 7 or 8

5,6   and  the other white chair can occupy  positions  1,2,3 or 8

6,7   and the othe white chair can occupy  positions  1,2,3 or 4

7,8   and the other white chair can occupy positions 1,2,3,4,5

So.....we have   30  arrangements here

Three white chairs together  can occupy   any of 6 positions

So......the total number of excluded arragements =     36  

So....the total number of  acceptable arrangememts =   5040   -  36   =   5004