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I'm sorry, I'm just a bit confused. When I did it, I got 16. Where does the other 2 come from? (Also, thank you very much for your help!)

Thank you, CPhill! 

Um... I'm only in 7th grade so I just going to solve WITHOUT the complex conjugate and Zbar thingy just for the fun of it.

1. Sqaure both sides.

9z^2-4z^2=1+8

9z^2-4z^2=9

5z^2=9

z^2=9/5

\(\boxed{z=\frac{\sqrt{9}}{5}}\)

Let a/2  = B      so a  = 2B

And we have the identity that

cos 2B  = 2cos^2 B - 1        add 1 to both sides

cos 2B + 1  = 2 cos^2 B        divide both sides by 2

cos 2B + 1

________    =  cos^2 B       take both roots

     2

±√ [ ( cos 2B + 1 ) / 2 ]  = cos B

So.....

True  !!!

Thank you so much CPhill.

Ok... This is the shape (green dots)

I will give you a strategy on how to solve this, quite complicated though.

1. Use the distance formula to find the distance between all the points to find each side length of the five-sided polygon.

2. Divide the shape into three triangles.

3. Use Heron's formula to solve for the area of each triangle (using the side lengths you got from step 1)

4. Add up each triangle's area to find the total area of the shape.

This is definitely not the best way to solve this, and might be very tedious. But it still works.

It is about -1.88 and about 7.12

using the quadratic formula

"Can I have a non-trig solution?"

The goal here...is to write the expression as a sum of squares plus a number.

Try this, and see what you get.

(x,y)=(2,-3)

Plug the values into the expression, and get -1?

Thank you!! Awesome as always! 

wow, I totally messed up the exponent rules!!!

Thanks, I thought this would be a hard problem.

cos 2a  =  cos a * cos a -  sin a  * sin a   =   

cos^2a - sin^2a  = (1 - sin^2 a) - sin^2 a  = 1 - 2sin^2 a

So...

True   !!!!

Are you trying to solve for one variable? If so, please say it.

Find the intersection points

2x^2  = 12 - x^2

3x^2  = 12

x^2  = 4

x  = - 2  and x  = 2

So we have, using symmetry

    2                                              2                                                 2

2 * ∫  12 - x^2 - 2x^2 dx  =   2    ∫  12 - 3x^2  dx   =  2 [ 12x - x^3]      =  2 [ 12(2) - (2)^3 ]  =

    0                                              0                                                 0

2 [ 24 - 8]  =  

2 * 16  =

32 units^2

The form of the circle is

(x - 5)^2 + (y - 15)^2 = 130

Since the x coordinate is twice the y coordinate, we have

(2y - 5)^2 + (y - 15)^2  = 130      simplify

4y^2 - 20y + 25  + y^2 - 30y + 225  = 130

5y^2 - 50y  + 120  = 0   

y^2 - 10y + 24  = 0

(y - 6) (y - 4)  = 0      set each factor to 0 and solve for y  and we get that  y = 6  and y  = 4

So....the max value for x  = 2y  =  2(6)  =  12

  7

Andy ate  4/7  = 84/168  pie

Barry ate 7/12  =  98/168 pie

Craig ate 5/8  = 105/168 pies

Andy's pie remaining = (1 - 4/7)  = 3/7

Barry's pie remaining = (1 - 7/12)  = 5/12

Craig's pie remaining = (1 - 5/8)  =  3/8

So  Fido  ate   3/7 + 5/12 + 3/8   =  205/168  pies

So Fido ate   205/168 - 105/168  = 100/168  = 25 / 42  more pies than the winner

sorry for the confusion , hopefully  i explained it better where im confused

We have that either

x + 4 + 3x > x + 9         or      3x + x + 9 > x + 4        or      x + 4 + x + 9 > 3x

4x + 4 > x + 9                           4x + 9 > x + 4                       2x + 13 > 3x

3x > 5                                        3x > -5                                    13 > x        

x > 5/3                                        x > -5/3                                   x < 13

Taking the most restrictive interval

5/3 < x < 13

So

n - m      =    13  - 5/3   =   39/3  - 5/3   =    34 / 3

Well actually this is a sample , I was unsure how they found out the period was 12 , why it was pi/6 , as well as why they changed 25 to negatitive  in order to be that equation from the table above.

Nice, tertre!!!!!

No clue, Jenny....did your instructor point you to any technology that might be used to do this???.....

x^2 - 6x  = 8      complete the square on x

x^2 - 6x + 9  =  8 + 9

(x - 3)^2  =  17     take both roots

x -  3  =   ±√17      add 3 to both sides

x  = 3 ±√17

WolframAlpha shows the last to be an identity.....but.....I don't know how to verify it..[it may be lengthy]...maybe someone else does

(sin x - cos x)^2  =   sin^2x  - 2sinxcosx  + cos^2x  =  sin^2x + cos^2x  - 2sinxcosx   =  1 - sin 2x 

Not an identity

The third one is an identity  because

sin (8x)  =  sin (2 * 4x)  =  2sin4xcos4x