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help pls

2(a + bi)   - 3 ( a - bi)  =  -7  + 2i

-a + 5bi  =   -7  +  2i

a =  7

5bi  =  2i

5b = 2

b =  2/5 

z =  7 +  (2/5) i

(a^2  - 1)  (a^2 - 2)

Notice how each hexagon has at least 2 other hexagons around it. 

That means that once those 2 hexagons next to it are shaded, the other one has to be the other colour. 

I think the answer would be 2 because you can put yellow on the top and green on the bottom on the second column, or green on top and yellow on bottom. Once those 2 colours are put, the rest of the pattern can only be shaded in 1 way. 

=^._.^=

(6-2)^2 + (1-n)^2 = 5^2

(1-n)^2 = 9

1-n = 3, or 1-n = -3. 

n = -2, n = 4

4*-2 = -8

=^._.^=

Height  =   sqrt  (   8^2  -  2^2)   =  sqrt  (60)  =   2sqrt (15)

Area  =    (1/2)  (4)  ( 2 sqrt (15) )  =  4 sqrt (15)  ≈   15.49  

yeah ok, CPhill's solution is better lol

z  +  1/z  = 1

(z  + 1/z)^2  = 1^2                            

z^2  + 2  +  (1/z)^2   =  1                   

z^2   + 1/z^2   =   -1                                                                       

(z  +  1/z)^4  = 1^4

z^4  +  4z^2  + 6  + 4/z^2  + 1/z^4  = 1

z^4  + 4 (z^2 + 1/z^2)  + 6  + 1/z^4   = 1

z^4  + 1/z^4   -  4 +  6  =   1

z^4  +  1/z^4 =    -1

(z  + 1/z)^6  =  1^6

z^6  +  6z^4  + 15z^2  + 20  +  15/z^2  + 6/z^4  + 1/z^6    =  1

z^6  + 1/z^6   +  15 ( z^2 + 1/z^2)  + 6(z^4 + 1/z^4)  + 20   =1

z^6  +  1/z^6  + 15*-1  + 6*-1  + 20   =  1

z^6  + 1/z^6  -1  = 1

z^6  + 1/z^6  =  2

(z^6 + 1/z^6)  ( z^6  + 1/z^6)  =   2 * 2

z^12  +  2 ( z^6 * ( 1/z^6) )  +  1/z^12  =  4

z^12   +  2   +  1/z^12   =   4

z^12  + 1/z^12   =  2

Let ABCDEFGH be a rectangular prism, where EF = 4, EH = 15 (?), and EA = 6. Find the volume of pyramid CFAH.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Prism volume = (EF)(EH)(EA) = 360 u³

There are 5 pyramids of which 4 have a volume of 60 u³ each.

The volume of pyramid CFAH = 360 - (4 * 60) = 120 u³

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If EH = 5 then the volume of a pyramid CFAH = 40 u³

\(\begin{align*} x^8 - y^8 &= (x^4)^2 - (y^4)^2 \\ &= (x^4 - y^4)(x^4 + y^4) \\ &= ((x^2)^2 - (y^2)^2)(x^4 + y^4) \\ &= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \\ &= (x-y)(x+y)(x^2 + y^2)(x^4 + y^4)  \end{align*}\)

Can you take it from here?

Hint: Manipulate $x^2 + y^2$ into $(x+y)^2 - 2xy$ and do the same thing with $x^4 + y^4.$ You can substitute your equation for $x^2 + y^2$ and do some further manipulation and you are done. If you have degrees $>1,$ you can square root it.

1/4 = first term           1/3  = 5th term

So.....

1/4  +  4d   =    1/3

4d  =  1/3  - 1/4

4d  = 1/12

d  =1/48

1/4 + 1/48 =  13/48  =   2nd term

13/48  + 1/48   =  14/48  =  3rd term

14/48 +  1/48  =  15/48  =  4th term

Sum  =      (13 + 14 + 15)   / 48  =     42  / 48  =    7/8

This is the  absolute   value  graph......all  points  are in  the  form  ( -a , a)  or  ( a ,a)

f(x -3)  slides the graph to the right  3  units

So   (-1.5 ,1.5)  becomes   ( -1.5 + 3  ,  1.5)  =   ( 1.5 , 1.5)  = the intersection

Sum of  coordinates =  3   

Looking  at  Pascal's  Triangle,  it appears that   every   

n =  2, 6.10 , 14......  =   2  +  4(p - 1)   =  4p - 2 

and

n = 3, 7, 11 , 15 .....  =   3  +  4(q - 1) =     4q - 1

Will  have the  result  that   C(n, 2)  is  odd

So

4p - 2  = 98                                and           4q  -  1   =  99

4p =    100                                                  4q  =  100

p =  100 /  4  =  25                                        q = 100 / 4  =  25

So

p + q   =   25  + 25    =   50   values of n  will  have C( n , 2)  =   odd

x is an exterior angle, there is a rule that says

so x = 29 + 82 = 111

\(\:z=\frac{1}{2}-i\frac{\sqrt{3}}{2}\)

so \(z^{12}+\:\frac{1}{z^{12}}\) i belive is 2

Step by step:

\(5x+14=k\)

\(5x=k-14\) (subtract 14 from both sides)

\(x=\frac{k-14}{5}\) (divide both sides by 5)

idk if this is right

\(x=\frac{9+\sqrt{519}}{6},\:x=\frac{9-\sqrt{519}}{6}\)

We can solve this with proportions:

\(\frac{4}{10}=\frac{14}{x}\)

cross multiply: \(4x=10\cdot \:14\)

\(4x=140\)

\(x=35\)

35 dollars

The fraction of the swim team that are sophomores is 2/7.