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By the quadratic formula, the roots are -1/6*i*sqrt(59) - 5/6 and i/6*i*sqrt(59) + 5/6.  Then

u/v + v/u = (1/6*i*sqrt(59) - 5/6)/(i/6*i*sqrt(59) + 5/6) + (1/6*i*sqrt(59) - 5/6)/(i/6*i*sqrt(59) + 5/6)

u/v + v/u = 3/4

x = -1/2.

The largest value of N is 234.

Im pretty sure it is 1.05$ if not private message lCPhill

expanding:

(2x+1/6)(2x+1/6)

4x^2+x/3+x/3+1/36

combining like terms:

3x^2+2x/3+1/36

i hope this helped! please correct me if i am wrong:)

x  + y  = 1       square both sides

x^2  + y^2 + 2xy =  1

2  + 2xy  =  1

2 -1  = -2xy

1  =  -2xy  

-(1/2)  = xy

1/2  = -xy

x^3 + y^3  factors as  a  sum of  cubes =    (x + y)  ( x^2  - xy + y^2)  =

( x + y)  ( x^2 + y^2    - xy )     =

(1)  (  2  +  1/2)   =

5/2

Is it just Web 2.0 or other things as well,  perhaps it is everything?

We  have  a transversal   CA  cutting parallel lines....this makes  alternate  interior angles    equal

So  angle  2  = angle  DCA   = 53°

Let the  number of red pens  = R  and blue pens =  B

So

B / R  = 3  /4   ⇒    B =  (3/4) R

In the second scenario we  have  that

(R - 1)  / ( B - 2)  = 3 /2

Sub in for B

(R  -1)  / ( (3/4)R -2 ) =  3/2      cross-multiply

2 ( R  -1)  = 3  ( (3/4) R  - 2)      simplify

2R  -2  =  (9/4)R   - 6        rearrange  as

6  - 2   = (9/4)R  - 2R

4  =   (9/4)R - (8/4)R

4 =   (1/4) R      multiply through  by 4

16 =   R

Check 

Initially  :  R / B   =    16   / (3/4)R =    16  /12  =  4 /3

Then  :    (16 - 1) / ( 12 -2)   =     15  /10     =    3   /  2

I plotted this is Desmos, and got that the range is [0,inf).

That is incorrect.

360 and here is a link to the same prob

If  we choose  the  one who is fine either way to be a spiker....we  have  to  choose  one of 6 spikers and one of  4 setters  =   6 * 4  =   24 ways

If we choose  the  one who is fine either way to be a setter  we have to choose one of 5 spikers and  one of 5  setters =  5 * 5  = 25 ways

So    25  +  24 =     49 ways

1.  The sum of the angles surrounding the common vertex is $360^\circ$. The right angle from the square corner is $90^\circ$ of that, and the interior angle of the hexagon is another $120^\circ$. Thus, the remaining angle is $360^\circ-90^\circ-120^\circ=150^\circ$. Since $150^{\circ}$ is an interior angle of the $n$-gon, each exterior angle measures $30^{\circ}$. Thus, the number of vertices is $\dfrac{360^{\circ}}{30^{\circ}}=\boxed{12}$.

2.  The largest and smallest exterior angles measure $44^{\circ}$ and $4^{\circ}$. Since the exterior angles are in an arithmetic progression, the average of all of them is equal to the average of the largest and smallest. Since we have $k$ angles in total, and their sum is $360^{\circ},$ this means $$\dfrac{360^{\circ}}{k}=\dfrac{44^{\circ}+4^{\circ}}{2}=24^{\circ},$$ so $k=\dfrac{360^{\circ}}{24^{\circ}}=\boxed{15}$.

I think everybody would know it was just a typeo.  :-)  

r = 6 sin  θ

In the boxes

f(x)   =6sin x

r

Not sure about this  one

circle with a radius of  3

the sum of the interior angles of a triangle is 180 degrees, so:

3x+1+2x-3+9x = 180

simplifying,,

14x-2 = 180

14x = 182

x = 13

the obvious smallest angle is the 2x-3 one, so plugging our x in:

2(13) - 3 = 26 - 3 = 23

so, the answer is 23 degrees. 

hope this helped! please correct me if i am wrong:)

Sorry.....just  a typo   !!!.....corrected  !!!