CPhill

Assuming that the  venue can hold  14,000 people  ( not,  12,000 people)

(a)  Let x  be the number of incremental increases

So  the number of tickets sold  = (14000 - 20x)

(b)

We have that

Profit  =  Revenue  -    Cost of staging concert

Revenue   =  Number of people  *  Price 

Let x be the number of   incremental price increases......so....

Revenue  =  (14000 - 20x) ( 100 + x)      simplify

Revenue =  1400000 - 2000x + 14000x - 20x^2  =  -20x^2 + 12000x  + 1400000 

So.....the total profit is given by

Revenue - Expenses...so.....

y = -20x^2 + 12000x + 1400000 - 1000000

y= -20x^2 + 12000x  + 400000

(c)   See the graph, here : https://www.desmos.com/calculator/kqdbrpbuuz

The domain is  [0, ≈ 632 ]   

(d) The range of prices that  will guarantee a profit =

(100 + 0 , ≈ 100 + 632)  =    [100, ≈732 ]    (in dollars)

(e)  Looking at the graph again.......the max occurs at  (300, 2,200,000)

So  the price that will guarantee a max =  (100 + 300)  = $400

And the  number of tickets sold at this price =  14000 - 20(300)  = 8000 tickets

x^2 -3lxl  + 2   = 0        factor as

( l x l - 2 )  ( l x l  - 1 )   = 0

So either                           or             l x l - 1   = 0

lx l - 2  = 0                                         l x l    = 1

x = 2, -2                                             x = -1, 1         

Let the bottom  let  vertex of the  blue triangle  =A

Let the bottom right vertex of the  blue triangle  = B

Let the remaining vertex of the blue triangle  = C

Let the bottom right vertex of the figure  =  D

Let  DE  be  the edge  that is perpendicular to AD

And traingles 

ABC  and ADE   are similar triangles

So

BC / AB   =  DE / AD

BC / 12  =  12 / 16

16* BC  = 12 * 12

16* BC  = 144

BC  = 144/16  =  9

So  the dark  area is the area of  triangle ABC  =

(1/2) (AB) ( BC)  = 

(1/2)(12) (9)  =  

(1/2) 108  =

54 units^2

 tan(x)^2 - 4*tan(x) + 1   =  0      let tan (x)  = a

a^2  - 4a  +  1  =  0        complete the square on  a

a^2  - 4a  + 4  =   - 1  + 4

(a - 2)^2   =  3           take both roots

(a - 2)  =  ±√3

a  =  2  + √3                a  =  2 - √3

So   either

tan (x)  = 2 + √3            

arctan ( 2 + √3)  =  x  = 75°

So  sin (2 * 75°)   =  sin (150°)   =  1/2     

Or

tan (x)   =  2 - √3

arctan ( 2 - √3)  =  x  =  15°

So    sin (2 * 15°)   =    sin (30°)   =   1/2

a^2   =    sin^2(t) + 2sin(t )tan(t)  + tan^2 (t)

b^2    =   sin^2( t)  - 2sin (t)tan (t)  + tan^2( t)

a^2 - b^2 =   4sin(t)tan(t)   ⇒ (a^2 - b^2)^2  =  4sin^2(t) tan^2(t)

ab =  tan^2(t) -  sin^2 (t)

(a^2  - b^2)^2

_____________    = 

     ab 

4sin^2 (t) tan^2 (t)

________________       =

tan^2(t) - sin^2(t)

4 sin^2(t) sin^2(t) / cos^2(t)

_______________________________      =

[sin^2(t) - sin^2(t) cos^2(t)]  / cos^2(t)

4 sin^2(t) (sin^2(t)

____________________   =

sin^2 (t) ( 1 - cos^2(t) )

4 sIn^2(t) sin^2(t)

________________   =

sin^2(t) sin^2(t)

4

x  +  x^-1   = √2      multiply through by   x

x^2  + 1  =  √2x

x^2 - √2x  =   - 1       complete the square on x

x^2 - √2/x  + 1/2  =  -1 + 1/2

[ x - (1/√2)]^2   =  -1/2          take both roots

x - (1/√2 )  =   ±i/  √2

x  =   ( 1 + i)               x   =   ( 1 - i )

       ______     or               ________

             √2                                √2

x ^20  =  [  ( 1 + i)) ] ^20

             ____________

                     ( √2)^20

Note that  ( 1 + i)^2  =  2i  .....so we have

                         [ (1 + i)^2 ] ^10                            [ 2i ] ^10

x^20  =             _______________         =           ________   =    i^10  =   -1  

                              [( √2)^2]^10                                2^10

The same result  is obtained  for  the other solution for x raised to the 20th power

Not as hard as it seems

(1 + i)^2011

___________

( 1 - i)^2009

Note that   ( 1 + i)^2  =  2i

And ( 1 - i)^2  = -2i

So  we have

[ (1 + i)^2 ] ^1005  *  ( 1 + i)

_______________________  =

[  ( 1 - i)^2]^1004  *  ( 1 - i)

[2i ] ^1005  *  (1 + i)

________________  =

[ -2i]^ 1004  * ( 1 - i)

[ 2i] * [ 2i]^1004 * ( 1 + i)

_____________________

    [ 2i]^1004  * ( 1 - i)

2i ( 1 + i)

________

   ( 1 - i)

2i ( 1 + i) ( 1 + i)

_____________

 (1 - i) ( 1 + i)

2i ( 2i)

________

1 - i^2

4i^2

_______

1  - (-1)

-4

____

  2

-2

x + y  + sqrt (x +y)  = 12

x - y  + sqrt (x - y )  = 6        

Let   ( x +y)  =  a

Let   (x - y)  = b

So we have

a + sqrt (a)  =12          

a + sqrt (a) - 12  = 0       Factor as

(sqrt (a) + 4) ( sqrt (a) - 3)  =  0

sqrt (a) + 4  = 0                     sqrt (a) - 3  = 0

No real solutions exist           sqrt (a)  = 3

                                                 a  = 9

And

b + sqrt (b)  = 6

b +sqrt (b) - 6    = 0     Factor  as

(sqrt (b) + 3) (sqrt (b) - 2)  = 0

sqrt (b) + 3  = 0                               sqrt (b) - 2  = 0

No real solutions exist                     sqrt (b)  =  2

                                                               b   = 4 

So

x + y  = 9

x - y  = 4       add these

2x  = 13

x = 13/2

And

13/2 + y  =9

13/2 + y  = 18/2

y = 5/2

(x, y)   =  ( 13/2, 5/2)

Let  z  =  a + bi

Conjugate z  =  a - bi

So we have

2 (a + bi)  + 3 (a - bi)  =  -25  - 2i       simplify

5a - bi  =  -25  - 2i

Equate  real  and non-real   terms

5a  = -25            -bi  = -2i

a  = -5                  b  = 2

So

z =   -5 + 2i

CORRECT   !!!