CPhill

9.8 is the largest decimal that can be formed given the constraints  and 2.1 is the smallest....so....their difference will be the greatest

Freq  =  1  / Period

Freq =  1 / 0.00024   =   4166.66 Hz  = 4167 Hz

A reflection across the y axis just changes the sign on the x coordinate....the y coordinate remains the same

So 

A = (2, 5)   and   A'  =  (-2, 5)

And

B = ( 6, 3)  and  B' = (-6, 3)

Sorry...I mis-read that as a multiplication....I think you wanted decimal values

The largest possible difference is

9.8  -  1.2   = 

8.6

D. frequency

a*b will be largest  when  a = 9  and b = 8  [or  vice-versa]

c*d  will be smallest when  c =2 and d =1  [ or vice-versa] 

So

The largest possible difference is

9 * 8  - 2*1  =

72  -  2  =

70

B. rotation

Using the sum of an infinite series, we have :

a1  /  [ 1 - r ]  = the sum

Where a1 is the first term  and  r is the common difference.....so we have...

a1  / [ 1 - (-1/2) ] =  45     simplify

a1 /  [ 3/2]  =  45

a1 =  45 (3/2)

a1  =  67.5

OK.....maybe someone else can shed some light on this....!!!

Here's how I view this....

5^1 / 8 = (8 - 3)^1 / 8   

Every term in the expansion is a multiple of 8 except the last  which is

(-3)^1 / 8  = (-3)^1  mod 8  = 5

Similarly

5^2 /8  =  (8 - 3)^2 / 8

Again we only need consider the last term which is

(3)^2/8  =  (3)^2 mod 8  =   1

etc....

So...we only need to consider  the last term in each subsequent power expansion and note that

(3)^n mod 8 =  1      and

(-3)^n mod 8 =  5

So

5^137 / 8    =  (8 - 3)^137 / 8      will  have (-3)^137 / 8  as a last term

And the remainder is  (-3)^137 mod 8  =   5