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The answer is 17*pi/32.

half of 12 is 6 and so a part of this will be  (x+6)

The 'a' will just be 1 since 1 is the coefficient of x^2

So it is going to be   (x+6)^2 + v

I will leave you to work out what v is. 

Hint:  Do the expansion.

\((x+2)(3x^2 + x + 5) \)

\(x(3x^2 + x + 5) + 2(3x^2 + x + 5)\)

\(x(3x^2 + x + 5) = 3x^3 + x^2 + 5x\)

\(2(3x^2 + x + 5) = 6x^2 + 2x + 10\)

\((6x^2 + 2x + 10) + (3x^3 + x^2 + 5x) = 3x^3 + 7x^2 + 7x + 10\)

\(3 + 7 - 7 - 10 = \color{brown}\boxed{-7}\)

wait, 55-49 = -6?

3x^2 -24x + 55 - x^2 - 49      simplify as

2x^2 - 24x  -  6

2 ( x^2 - 12x - 3)       complete the square on  x inside the parentheses

2 ( x^2 - 12x + 36 - 3 - 36)

2 (x - 6)^2    + 2 (-39)

2 (x - 6)^2  -  78

a +  d +  e     =     2 - 6  - 78  =    -82

4 galllons of the current  mixture is  ammonia  (80% of 5)

We need 3 gallons of ammonia at the  end

Let x be the number of gallons drained off and.8x the amount of  ammonia drained off

(4 - .8x) = .6 *5

4 - .8x  = 3

4 - 3  =  .8x

1 = .8x

1/.8  = x = 1.25

We need to drian off  1.25 gal

Proof

4 gallons of  ammmonia to start with

When we drain away 1.25  gallons, 80%  of this is ammonia.....so   4 - .8(1.25)  = 3 gallons of ammonia are left

Adding back 1.25 gallons  of  pure water  = 5 gallons  and 3 of  this is  ammonia...so....3/5=  .60  =  60% 

And 1.25 gallons =     5 qts

f ( x + 1)  = f ( 1 + 1)  = f(2)

So x = 1

So

f(2)  =  1/1    = 1

When x changes by  3, y changes by   (13 - a)

From  4 to 13   x changes by 9  so  y changes by  3 (13 - a)=   39 - 3a

And  notice that

39 - 3a =  b   rearrange   as

3a + b =  39 

Since x =  y ,  need to solve this

(x - 5)^2 + ( x - 15)^2  = 130

x^2 - 10x + 25 + x^2 - 30x + 225  = 130

2x^2 -40x + 120  =  0     divide through by 2

x^2 - 20x + 60  = 0

x^2 - 20x  =  - 60       complete the square on  x

x^2 -20x + 100 = -60 + 100

(x - 10)^2  = 40        take the positive root

x -10 =   sqrt (40)

x =  10 + sqrt (40) =   10 + 2sqrt (10) ≈  16.325 

Thanks Ginger.

My problem is that I do not understand the question. 

(And at present I am too unenthused to try and work it out from what you have said.)

Maybe I will become more enthused later...

I answered a different question.  

There is no casual solution to this problem; though the solution process is not ultra complex.

The use of the Inclusion-Exclusion principle will accelerate the process. Like usual, its application requires careful and (sometimes) laborious attention to avoid over-counting successes and failures.

The question resolves to a binary state of either an arrangement of eight (8) rooks occupying or attacking every square, or it does not. Analyzing the counts of unoccupied and un-attacked squares is not requires for this question.

(Permutations of the rooks among themselves are not necessary, as this will be done for both success and failure counts and will factor out.) 

Setup:

Vertical format.

 Starting with an 8 X 8 chessboard place eight rooks on the top row (row 1) and observe that every square is either occupied or attacked by a rook. Moving the rooks vertically (row-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically.

This same sequence is repeatable in a horizontal format.

Place the eight rooks in the left column (Column 1) observe that every square is either occupied or attacked by a rook. Moving the rooks horizontally (column-wise) still allows for every square to be either occupied or attacked by a rook. There are (8^8) unique positions when moving the rooks vertically. HOWEVER, in this horizontal format, there are two (2) cases where the positions of the rooks identically match the positions in the vertical format. These two (2) cases occur when the rooks align diagonal positions across the board.  This occurs twice: from the top-left to the bottom-right, and from the bottom-left to the top-right.  These two (2) cases are exclusions and need to be subtracted from the total. So at this point, the subtotal for success counts are (8^8) + [(8^8) – (2)].

At this point it’s (apparently) discernable that for every square to be either occupied or attacked by a rook, every column or every row must have at least one rook. Removing all rooks from a column leaves some (not all) row squares on that column un-attacked.  Like-wise, removing all rooks from a row leaves some (not all) column squares on that row un-attacked.   

If the above is true, then these counts (8^8) + [(8^8) – (2)] are the only successes.

Probability of a random arrangement of eight (8) rooks to occupy or attack every square on a chessboard:

\(\rho_{(s)} = \dfrac {(2*8^8)-2} {\binom {64}{8}} \approx 0.75809\%\)

GA

--. .-

Recall that: \({1 \over p} + {1 \over q} = {p \over pq} + {q \over pq} = {p + q \over pq} \)

We have: \({p + q \over 9} = 1\), so \(p + q = \color{brown}\boxed{9}\)

a^2 - b^2  + 2a   =

(a + b) (a - b)  + 2a  =                      {a + b  = -1 }

(-1) (a - b)  +  2a  =

-a + b + 2a = 

a + b  

And

a  + b  = -1

Using the equare of the distance formula

(4 -1)^2  + ( b - 2)^2  =  5^2

9 + (b -2)^2  = 25

(b -2)^2   =   25 - 9

(b - 2)^2 =  16      take  both roots

b  -  2 = 4                            or                        b -  2 =   -4

b =  6                                                                 b = -2

Substitute the first equation into the second: \(384b = 4068\), meaning \(b = 12\)

Now, plug this back into the first equation, and solving for a gives us \(a = \color{brown}\boxed{1 \over 54}\)

Convert vertex form to standard form: \(2(x-4)^2 + 8 = 2(x^2 - 8x + 16) + 8 = 2x^2- 16x + 40\)

This means that \(a = 2\), \(b = -16\), and \(c = 40\). So, the quadratic we need to convert to vertex form is \(6x^2 - 32x + 40\).

Note that in vertex form (which we need to convert it to), the vertex occurs at \((h,k)\).

This means we need to find the x-coordinate of the vertex, which occurs at \({-b \over 2a}={32 \over 12} = \color{brown}\boxed{8 \over 3}\). 

Rearrange as

2x^2  -2x  =  10   divide through by 2

x^2  - 2x   =   5        complete the square on  x

x^2 - 2x + 1 =  5 + 1

(x - 1)^2 =  6        take the positive root

x - 1 =  sqrt (6)

x =  1 + sqrt (6)

x =  (1 + sqrt (6)  )   / 1

a + b + c =      8

Start with numbers divisible by 16. The smallest number is 16 (16 x 1), and the largest number is 4000 (16 x 250). This makes for 250 integers. 

Now, we do the numbers divisible by 30. The smallest number is 30 (30 x 1), and the largest is 3990 (30 x 133). This makes for 133 integers. 

However, we double count for every multiple of 240 (lcm of 16 and 30). The smallest number we overcount for is 240 (240 x 1) and the largest is 2840 (240 x 16). This makes for 16 integers. 

Thus, there are \(250 + 133 - 16 = \color{brown}\boxed{367}\) integers.

The denominator  cannot  = 0

So

y^2  - 5 =  0

y^2  = 5

y =  ± 5 

The sum of these values =  0

Point A is 4 units to the right of the midpoint, so point B must be 4 units to the left of the midpoint, meaning its coordinates are \((4-4, 4) = (0,4) \).

Thus, the sum is \(0 + 4 = \color{brown}\boxed{4}\)

[17 + 10 + 9 + 14 + 16 + 8  + 10 + 10 + sum ] /  11 =  14

[ 94 + sum ] =  14 *11

94 + sum =   154

sum = 154 - 94   

sum = 60

Point B = (12,4), so the answer is 12 + 4 = 16.

Your answer is incorrect.  You should think about the problem some more.

The 200th term is    3 + 199 (5) =   998

The sum is 

[ 3 + 998 ]  * 100  =  

100,100

The possible rational roots come from :

All the factors of 45 =    ± [ 1, 3 , 5 , 9 , 15,  45 ] 

Divided by

All the  factors of  2 = ± [ 1, 2]

So we have

± [  1/2 , 3/2 5/2 , 9/2 , 15/2, 45/2 , 1, 3, 5 , 9, 15 ,45  ]