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Note that     x^3 - x^2 -21x + 45 =  (x + 5) (x - 3) (x - 3)^2

Multiply through by this factorization and we  have

x^2 - 21x -3 = A(x - 3)^2  + B( x + 5) ( x-3) + C( x + 5)

x^2 -21x - 3 =  A( x^2 - 6x + 9) + B( x^2 + 2x -15) + C(x + 5)

Equate coefficients on each  side

1 =  A + B    →    B =  1 - A      (1)

-21 = -6A + 2B + C     (2)

-3 = 9A -15B + 5C     (3)

Sub (1) into (2) , (3)

-21 = -6A + 2 ( 1 - A) + C     →  -21 = - 6A + 2 - 2A + C →         -8A + C  =  -23      (4)

-3  = 9A - 15 ( 1 - A) + 5C  →  -3  =  9A - 15 + 15A  + 5C   →      24A + 5C =   12      (5)

Multiply  (4)  through by (-5)    add to  (5)

40A - 5C = 115

24A + 5C =  12

____________

64 A      =  127

A = 127/64

This simplifies to   ( x + 2x - 2) / ( x -1)  =  (3x -2) / ( x -1)

We have a horizontal asymptote  at  y= (3x/ x)  =  3

The range is  [(-inf , 3) U ( 3 , inf)

I like your thinking  !!!

Note that  we can    first  add  f(1) + f(2005)  and we  get

log (1) / ( log (1) + log (2005) )  +  log (2005) / ( log (2005) + log (1) )  =   

[ log (1) + log (2005) ]  / [ log(1) +Log (2005)] =   1

Next  we pair  f(2) + f(2004) and we  have

log (2) / ( log 2 + log 2004)  + log (2004) /( log (2004) + log (2) ) = (log 2 + log 2004) /(log 2 + log 2004) = 1

Continuing this pattern....We end up with   (2005 - 1 ) / 2 =  1002  pairs that each  evalute to 1

And we  have one evaluation  which is    n = (2005 + 1) / 2 =  1003

The evaluation of   this  is      log (1003) /( log 1003 + log (2006 -1003))  = 

log (1003) / ( log 1003  + log 1003)  =  1* log (1003) / 2 *log ( 1003 )  =  1/2 =  .5

So.....the sum is   1002 + 1/2  =   1002.5

There are \(n^n\) possible outcomes. Since there are n runners, and there are n amount of finishing positions, the total number of possible outcomes are \(n^n\). This should work if you consider some of the tied positions to be different (eg: tied for second place with two runners and tied for first place with two runners).

\([f(x) = \frac{2-x}{\log (2 - x)}\)

Only the denominator matters

We can't take the  log of a  non-zero  number

Therefore  the domain is  ( -inf , 2)   

Alternatively you could solve for a polynomial whose roots are \(p+3\) and \(q+3\) and then use Vieta's. 

The roots  are  5 and  -9/2

Can you take it from  there ????

We can construct 12  congruent  isosceles triangles....with sides of  1 and  apex angles  of 30°

The area =   12* (1/2) ( side)^2 * sin (30°)  = 12* (1/2) (1^2) * sin (30°)  = 6 ( 1 / 2)  = 3   units^2

No you don't you want us to do it for you. You try to give the illusion that you have solved it while simply copy-pasting your stupid homework question and giving us no information on your retarded thought-process whatsoever.

To solve the inequality \(x(x + 6) > 16\), follow these steps:

1. Expand the left side of the inequality:
   \(x^2 + 6x > 16\)

2. Move all terms to one side of the inequality:
   \(x^2 + 6x - 16 > 0\)

3. Factor the quadratic expression:
   \((x + 8)(x - 2) < 0\)

Now we have the factors of the quadratic expression. To determine the intervals where the inequality is satisfied, you can use a sign chart or test points within the intervals. The inequality is satisfied when the expression is greater than 0. Let's analyze the sign of the expression within different intervals:

- When \(x < -8\):
  Both factors \((x + 8)\) and \((x - 2)\) are negative. Their product is positive. So, the inequality holds in this interval.

- When \(-8 < x < 2\):
  \((x + 8)\) is positive, and \((x - 2)\) is negative. Their product is negative. So, the inequality holds in this interval.

- When \(x > 2\):
  Both factors \((x + 8)\) and \((x - 2)\) are positive. Their product is negative. So, the inequality does not hold in this interval.

Now, we can write the solution in interval notation:

Solution: \((- \infty, -8) \cup (-8,2)\)

This indicates that the inequality \(x(x + 6) > 16\) is satisfied when \(x\) is in the intervals from negative infinity to -8 (excluding -8), and from -8 (excluding -8) to 2 (excluding 2).

ab - 6a + 5b  =  37      subtract 30 from both sides   (the product of -6 * 5)        

ab - 6a + 5b - 30  =  37 - 30       factor the left side, simplify the  right

(a + 5) ( b - 6)  =  7

Let a = 2  and b = 7

abs ( a - b)  =  abs ( 2 - 7)  =  abs (-5)  =   5  is the  smallest  value

P.S.  .......check  to  see that  both of these  values work in  the  original  equation !!!!

Greetings, SpectraSynth!

You made a great attempt to solve this inequality, but I noticed you made an algebraic error along the way. You subtracted the quantity \(6x-2\) incorrectly from both sides of the inequality, which lead your answer astray.

If you manipulate the inequality by completing the square, the x-values that satisfy and do not satisfy this inequality become clear rather quickly.

\(2x^2 + 8x < -6 + 6x + 4 \\ 2x^2 + 8x < 6x - 2 \\ 2x^2 + 2x < -2 \\ 2(x^2 + x) < -2 \\ 2\left(x^2 + x + \frac{1}{4}\right) < -2 + 2 * \frac{1}{4} \\ 2\left(x + \frac{1}{2}\right)^2 < -\frac{3}{2} \)

Now, analyze the inequality carefully. I noticed that the left-hand side is the product of two nonnegative quantities. 2 is positive, and \(\left(x + \frac{1}{2}\right)^2\) has a square, which guarantees that the result is either 0 or positive in nature. If we multiply 2 by a quantity that is either 0 or positive, then it is impossible for that quantity to be less than \(-\frac{3}{2}\). This indicates that there is no solution to this inequality. In interval notation, that is typically denoted as \(\emptyset\).

Area = 36a^8b^4

Height = ab

Area =  (1/2) AC ( height)

2 *Area / (height ) =  AC

2 (36a^8b^4) / (ab) =  AC

72 a^7 b^3  =  AC

To graph the function \(f(x) = -3x^3 - 6x^2 + 3x + 6\) and identify its x-intercepts, we can follow these steps:

1. Find the x-intercepts by setting \(f(x)\) to 0 and solving for \(x\):

   \[f(x) = -3x^3 - 6x^2 + 3x + 6 = 0\]

2. Once we find the x-intercepts, we can plot them on the graph.

Let's proceed with these steps:

Step 1: Finding x-intercepts

To find the x-intercepts, we'll solve the equation \(-3x^3 - 6x^2 + 3x + 6 = 0\). This equation can be factored to some extent:

\(-3x^3 - 6x^2 + 3x + 6 = -3(x^3 + 2x^2 - x - 2)\)

Now, let's use a tool like a graphing calculator or a computer algebra system to find the roots of the polynomial \(x^3 + 2x^2 - x - 2\). The roots are approximately -1, 1, and 2.

So, the x-intercepts are: \(x = -1\), \(x = 1\), and \(x = 2\).

Step 2: Graphing the function

Now that we have the x-intercepts, let's proceed to graph the function. Here's a rough sketch of the graph:

```
   ^
   |                 *
   |               *
   |            *  
   |          *      
   |       *         
   |      *           * *
   |    *          *
   | *            *
   |*___________*______________>
            -2    -1   1   2
```

In this graph, the x-intercepts are marked with asterisks. The function \(f(x)\) approaches negative infinity as \(x\) moves towards negative infinity and approaches positive infinity as \(x\) moves towards positive infinity. The graph shows a general shape of the cubic polynomial, with its behavior around the x-intercepts.

Please note that this is a rough sketch and not to scale. For a more accurate and detailed graph, you can use graphing software or a graphing calculator.

Alright, thank you so much:))

This is a  little  difficult to  see

Here's a graph :

Notice  that   0   is  not  in the range......

maximum, this is a good-faith effort towards solving this problem, but you forgot to consider the right-hand side of the equation. Also, your attempt at factorization is not correct.

If you check your work, you will discover that your claims are not accurate.\(2(x + 10)(x + 3) = 2(x^2 + 10x + 3x + 30) = 2(x^2 + 13x + 30) \neq 2(x^2 + 12x + 30)\).

SpectraSynth, this is a good-faith effort towards solving this problem, but why did the line \(2x^2 + 24x - 60 - x^2 - 13x = 0\) turn into \(x^2 + 4x - 60 = 0\)? Notice how \(24x - 13x \neq 4x\).