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3x^2  - 8x  - 2   factor  out  the 3

3 [ x^2   -(8/3)x  - 2/3   ]  

Take  (1/2)  of (8/3)  =  8/6 = 4/3.....square this  = 16/9....add and subtract it witih the parentheses

3 [ x^2 - (8/3)x + 16/9  - 2/3 - 16/9  ]       factor the first three terms, simplify  the rest 

3[ (x - 4/3)^2  -  18/27 - 48/27 ]

3 [ ( (x - 4/3)^3  -   66/27 ]        distribute the 3

3(x - 4/3)^2  -  66/9

a + h + k =   3  + 4/3 - 66/9    =     27/9  + 12/9  - 66/9  =    - 27/ 9  =   -3

OK, Cal....you ask  a good question....!!!!

Let's  look  at something  more simple

Suppose  we only had 4 members    A, B C D

Let   A/B A/C A/D  B/C B/D C/D      represent     president / vice-president

Note   that  this is   C(4,2)  =       4!

                                                 _______     =       6

                                                 (4 - 2)! 2!

But....what we have  as actually a permutation     because   A/B    isn't the same as  B /A

So   we actually have   P(4,2)  =   12

So  your answer  is a combination  C(20,2) = 190

But we should have a permutation =   P(20, 2)  = 380......  this is the same result as 20 * 19

2 x 10⁶   ISN'T  smaller  than 4 x10³

We can choose  any 1 of the 20 members for  president

Once any of these is selected, we can choose any 1of the remaining 19 members for vice-president

So   .....by the fundamental counting principle  =  20 * 19   =   380  ways

These pentagons are all the same......all corresponding points on A1B1C1D1E1 and  A2B2C2D2E2 are translated the same distance   ( the perimeter is irrelevant)

Call  the  center  of A1B1C1D1E1  = (0,0)

Call the center of A2B2C2D2E2  =  (150, 40)

The distance CC2  =  sqrt  [ 150^2 + 40^2 ]  ≈  154.24 units 

i get it! i probably did the factoring wrong as well! Thank you very much!!!!! :D

We can write

3p^5q^3   (  2pq  +    5 / (p^2q^2    )

Distributing  we have that

3p^5q^3  ( 2pq)    +    3p^5q^3 * 5

                                                  ______

                                                   p^2q^2

6p^6q^4        +     15p^5q^3

                           _________

                              p^2q^2

6p^6q^4  +  15p^2q

9x - 4y  = 30       (1)

7x + y =  11    →   y = 11 - 7x       (2)

Sub (2) into (1)

9x - 4(11 - 7x)  = 30

9x - 44 + 28x    = 30

37x =  30 + 44

37x  = 74

x = 74/37  =  2

y = 11 - 7(2) 

y = 11 - 14

y = -3

Intersection  =  (2, -3)

R = cost of ribeye dinner

S = cost of salmon dinner

We have this system

20R  + 14S = 591.53

24R   +  7S = 581.91

Multiply the second eqution through by -2

20R  + 14S = 591.53

-48R - 14S  =  - 1163.82      add these

-28R  = -572.29       divide both sides by -28

R ≈ $20.44

To find S

20 ( 20.44)  + 14S = 591.53

408.80  + 14S =  591.53

14S = 591.53 - 408.80

14S = 182.73

S =  182.73 / 14   ≈  $13.05

Perimeter  =   2 ( 5  + pi*2 )  =   [ 10 + 4pi ] units

Anyone?

Please Help! 

In the county agriculture fair the pumpkin B, grown by Bob, is bigger than the pumpkin C, grown by Chuck, but smaller than the pumpkin A, grown by Abe, then which of the following is true?

Apriciate the help.

Thank you

      b^(-1)

______________                             remember that      a^-1  =   1/a

  ca^(-1) - a(c)^-1

So   we can write

         [ 1/ (1/4) ]

________________   =

     5(1/3)  - 3(1/5)

          [ 1 / (1/4) ]

        __________

           5/3  -  3/5

Evaluate numerator and denominator separately

1/ (1/4)  =   (1/1)  /(1/4)   =   (1/1)  * ( 4/1)    =   4

5/3  -  3/5        common denominator  = 15

 25 / 15  - 9/15  =   16/15

So....we have

       4

  _______    =

    (16/15)

(4/1)

______   =

(16/15)

(4/1)  ( 15/16)  =   (4 * 15) / (1 * 16)  =     60/16     =      15 / 4

x^5  y ^3                         x^5   y^3

_______   write  as       _________

x  y^6                              x^1   y^6

We have the rule  that  when we  have        a^n       this equals      a^(n - m)

                                                                   _____

                                                                     a^m

Using this rule we can write

x^(5 - 1)  *  y^(3  - 6)    =

x^4   * y ^(-3)                                     and   a^(-m)    =  1

                                                                                    ____

                                                                                    a^m

So.....using this last rule, we get

x^4  *  (1)

          ___   =       

           y^3

  x^4

  _____

   y^3

c - b = 3    where did you get it from?

Thank you.

Nice job, that looks right.

(-3x^2y)  (5x^4y^3)  =

(-3 * 5)  *  ( x^2 * x^4 )  *  ( y * y^3)   =

-15 x^6 y^4

Thanks

Let's see if we have a repetitive pattern

Beginning number         Ending number                           

             5                           14

            14                           7

             7                          20

             10                          5   

              5                          14    ........              

Note  that  the cycle starts over after every 4 steps

After pressing the button 1000 times   (since 1000 is divisible  by 4)   we are back at 5

Pressing the button on the 1001th time will display  5*3 - 1  =   14