CPhill

(2 - i)(2/5 + 1/5i) =     factor 1/5 from the second term

(1/5) (2 - i)  (2 + i)  =

(1/5)  (4  - i^2)  =

(1/5)  (  4 -  -1)  =

(1/5) (5)   =  

1

cos(2t)   =   -1/10

arccos ( -.1)   = 2t   ≈  1.67 rads     

So    

2t  =  1.67  rads

t  ≈  .835 rads  = .84 rads

sin(4x)cos(2x)-sin(2x)cos(4x)

Using 

sin ( A - B) =   sinAcosB - sinBcosA   where  A  = 4x   and B  = 2x, we have

sin ( 4x  - 2x)  =

sin (2x)   →    "A"

 (4y^3+2y^2-3y-4) / y  =

(4y^3 / y)  + (2y^2 / y  - (3y) / y  - (4) / y  =

4y^2     + 2y     -    3      -     4 / y

y  = 2x + 5

y  = 5x + 2     

We have two lines with different slopes...........they will intersect at some point giving us one solution.....thus......this is a consistent system

x^2  = x + 20    rearrange as

x^2  - x - 20   = 0      fractor

(x - 5)  ( x + 4)  = 0

Setting both factors to 0 and solving for x, the two solutions are x = 5  and x  = -4

So    a^2  + b^2  +  a  +  b   =     (5)^2  + (-4)^2  + 5  +  - 4   =     42

250 pi :  1024 pi   =

250 : 1024  =

125 :  512  =

5^3  :  8^3        the similarity ratio (scale ratio )  is the cube root of both of these  = 

5 :  8

Note

2!  =  3! / 3

1!  =  2! / 2

0!  =  1!  / 1      = 1

cos^2(t)+2cos(t)+1=0       factor

(cos (t) + 1)  (cos (t) + 1)  = 0

(cos (t) + 1 ) ^2  = 0         take the square root of both sides

cos (t)  + 1   =  0             subtract 1 from both sides

cos (t)  = -1           and this occurs at   t = 180°

We want to choose 2 of the 9 men  and 2 of the 7 women.....so

C(9, 2) * C(7, 2)   =  756  ways

4sin (x)  = -1      divide both sides by 4

sin  (x)   = -1 / 4        take the sine inverse

arcsin ( -1 / 4)   = x  ≈  -14.48°

sin B / b  =  sin A / a

sin B  /  .6   =   sin (30) / 1.3       multiply by .6 on both sides

sin B   =  (.6 / 1.3) sin (30)

sin B  =  (.6 / 1.3) (1/2)

sin B  =  (.3 / 1.3)       take the sine inverse

arcsin (.3 / 1.3)  = B   ≈  13.34°