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3^(2c + 1)  = 28*3^c

3^(2c) * 3^1  =  28*3^c

(3^c)^2 * 3  = 28 *3^c      divide  both sids by  3^c

3^c * 3  =  28

3^c  =  28/3        take the log of each side

log 3^c =  log (28/3)

c * log 3  =  log (28/3)

c =  log (28/3)  / log (3)

I believe that each pair of complex roots are conjugates of each other, no?

But think about it, if they are conjugates, they are mirrors across the x axis. If the distance between two roots conjuagates of each other are different from the other pair of roots which are conjugates, then this would form a trapezoid. Could you see what I mean? Can you continue from here?

What if some fo the roots are real, however? If the roots are real, what does z have to be? Is there even a root that is real? I have to go real soon, so Ill try to post the solution when I am back. 

thanks

It was 60

4x + 5 = 19

4x  = 19 - 5

4x   = 14

So

[14 + 17 ]  /  [ 14 + 1 ] = 

31  /  15          { as pogfrog found  !!!!.....thx, pogfrog  !!!!  }

I ams orry that is wrong, try again 

i think its [-1,3/5]

(10 *8 * 6 * 4 *2)  / (9 *7 *5 * 3 *1)  =

10/5 * 6/3  * 2/1 *  ( 8/9) ( 4/7)  =

2^3  *  32 / 63  =

256 / 63

Angle YZX = 75 degrees.

The answer is 2*sqrt(2) - 2.

what

31/15

Let f(x) =  (x - 1)^2  - 2

Then

f ( 2x - 3)  =  [ (2x - 3) - 1]^2  - 2   =

[ 2x - 4 ]^2  - 2  =

[ 2 ( x - 2)]^2  - 2  =

4 ( x - 2)^2 - 2

The vertex is    ( 2 , -2)

See here :  https://www.desmos.com/calculator/yehdvvbzxe

Use Quadratic Formula to find roots \(a\) and \(b\) \( = - 3/2 +- \sqrt37 / 2\)

Use these values in\((2a-3)(4b-6)\)to find the answer .....

which is also BuilderBoi's answer! 

Assuming without replacement   the probability is

6  / 14  *    5 / 13     =    3/7 *5/13   =   15 / 91

look at builderboi's answer

The area of the rhombus is 2*sqrt(3).

We see that t + 8 is always the largest side.

The triangle inequality requires t+8 < t+1+t-1 to be true, which simplifies to t > 8. 

OHHHHHHH thank you! How come i didn't see!!!

Welcome aboard   !!!!

Use polynomial construction!
S_1 = 3

S_2 = 3

S_3 = 3

We can then construct a polyoimial where a, b, c are roots of this polynomial.

P(x) = x^3 - 3x^2 + 3x - 3

= (x - 1)^3 - 2

(x-1)^3 = 2

Since I believe a has to be real based on the fact that b = d + ei and c = d - ei , \(a = \sqrt[3]{2} + 1\)

a = 15/13, and b = 780, a+b = 781 2/13 or 10155/13

All real numbers.

(x-3)^2 - 1