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[ x^2  + 4x ]  / [ -12x - 2 ]

This graph will  have a vertical asymptote   where  -12x - 2  =  0   ⇒   at  x  =  -1/6

We won't have any "holes"....the denominator would have to be a factor of the numerator for this to happen

Being confused is quite natural. I am not going to say that my method is completely sound, but it seemed to get the job done. 

I broke up each fraction into individual terms, and I knew that each one would, on average, be \(\frac{1}{2}\) since \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\), so my goal was to make every fraction as close as possible to the value 1/2. If I can do that, then I should be reasonably close to 2. 

I filled in the first a/b with 1/2 because that is exactly 1/2, so I have already solved one fraction, but I have used up the 1 and the 2. 

Now, a=1 and b=2. That seems pretty good. I then tried this with the other fractions. c/d=3/6 and e/f equals 4/8. I have one fraction left. We now have a few variables defined. They are the following:
 

a=1

b=2

c=3

d=6

e=4

f=8

The numbers I have left are 5,7, and 9, and there is unfortunately no way of making a fraction that is less than 1/2, so I had to backtrack. 

I then realized that I had to make the fractions' average to 1/2, so I decided to take a different approach. I still left a/b=1/2, so a=1 and b=2. 

This means that I should have denominators 7,8, and 9. This way, I do not affect the average as much. Let me explain. If I have a fraction with a denominator of 4, for example, then the halfway point is 2. However, the closest possibilities such as 3 and 1 affect the fraction's value by 25%, so I realized that, the larger the denominator, the less the affect is if I increment the numerator by 1.
 

a=1

b=2

I have now reduced the possibilities to \(\frac{1}{2}+\frac{c}{7}+\frac{e}{8}+\frac{g}{9}\). I knew that none of the fractions should have a 6 because that would be too large for the average to work, so I just tested a few possibilities and \(\frac{1}{2}+\frac{3}{7}+\frac{4}{9}+\frac{5}{8}\) is the one that worked the best.

There are 12^2  =  144 possible outcomes

Outcome        Freq

2                      1

3                      2

4                      3

5                      4

6                      5

7                      6

8                      7

9                      8

10                    9

11                   10

12                   11

Sum of            66

Freq

So.....the sum of the frequencies for rolling   a number > 12  = 144 - 66 = 78

So...the probability of rolling a number 12 or greater  = 78/144  = 13/24  ≈ 54.16%  

It is also possible to use knowledge of difference and sum of squares and cubes to find all the possible solutions. 

Let's consolidate those roots together.

\(x_{1,2}=\pm 2\\ x_{3,4}=1\pm i\sqrt{3}\\ x_{5,6}=-1\pm i\sqrt{3}\)

Thank you! Could you explain how you got there though? I'm a little confused.

Using this online generator :

http://keisan.casio.com/exec/system/14059932254941

-56.80x^2 + 437.91x - 65.5

x^6  =  64

64  =  64 + 0i

The sixth root divides the unit  circle into six even parts  and the beginning angle is  

arctan (0 / 64)  =  0°      

We have

⁶√64  [  cos (0) + i sin (0)]  =   2 [ 1 + 0 i ]  =  2 + 0i

⁶√64 [  cos (pi/ 3)  + i  sin (pi /3)  = 2 [ 1/2 + √3/2 i ]  =  1 + √3 i

⁶√64 [  cos (2pi/ 3) + i sin (2pi/ 3)  ]  = 2 [ -1/2  +  √3 i ]  =   - 1 + √3 i

⁶ √64 [  cos (pi)  + i sin (pi) ]   =     2 [ - 1 + 0 i ]  =  2[-1]  =  -2 + 0i

⁶√64  [ cos (4pi/3 + i sin (4pi/3) ]  =  2 [ -1/2  - √3 / 2 i ]  =  - 1 - √3 i

6√64 [ cos (5 pi / 3) + i sin (5pi/ 3) ]  = 2 [ 1/2  - i √3 /2 ]  =  1 - √3 i

Solve for x:

x^6 = 64

Take 6^th roots of both sides.

Taking 6^th roots gives 2 times the 6^th roots of unity:

x = -2    or x = 2     or x = -2 (-1)^(1/3)     or x = 2 (-1)^(1/3)    or x = -2 (-1)^(2/3)    or x = 2 (-1)^(2/3)

Note that to put in the first post and return, she walks  2 * 2 m  =  4m = 4 (1) m

And  the total distance she walks to put in the second post is    2 * 4m  =  8m  = 4 (2) m

And the distance she walks to put in the third post is  2 * 6m  = 12m  = 4 (3) m

So....to put in  N posits...it appears that the total distance waked is

4 ( 1  + 2 + 3 + ......+  N )  m

And the sum of the first positive N  integers can be expressed  as    (N /2) * (N + 1)

So...the total distance she walks (in meters)   for 200 posts is :

4 [ (200/2) (201) ]   =  2 (200)(201) = 80, 400 m

Converting this to kilometers, she walks   80,400 / 1000  =

80.4  km !!!      that's a lot of walking  !!!

None are prime....as the guest said....

To see this more clearly....note that we can write

43!  + n         where       n is an integer that  varies from 2 to 43

Note that in every case, n  will divide both terms evenly  because n will be a factor of both terms

Thus  

43! + n      will be divisible by 2,3,4,5,6......41, 42, 43

So.....it can't be prime because it is diviible by some integer other that 1 (and itself)

Yes, that is what I meant. I am sorry about the confusing. It has no holes?

Wow!! 👍

Bravo, X^2 !!!!.

After thinking about this one for quite some time, I think that I have found the magical conjugate: \(3^\frac{1}{3}-1\).

First of all, I will convert all the radicals to exponential form. 

\(\frac{2*9^\frac{1}{3}}{1+3^\frac{1}{3}+9^\frac{1}{3}}\)

Now, let's use the magical conjugate. 

That's a glorious realization to make, huh? The magical conjugate has transformed such a complex denominator into something extraordinarily simple. Let's calculate the numerator now. 

None of them is prime !! They are 43 consecutive composite numbers by definition, since all the added numbers[2, 3, 4, 5..........43] are already factors of 43!

Notice, rosala, that if n= 2, we have...

1! + 2!  =  1 + 2  =  3   which is, of course, divisible by 3

If n   ≥ 3  we can write

1!  +  2!  +  3!   +  4! +   .........+ (n - 2)! + (n - 1)!  +  n!

Note that  every term in  red  will have 3   as a factor because we can write

( 1! + 2!)  +  (1 * 2 * 3) +  (1 * 2 * 3 * 4) + ......+ (1 * 2 * 3 *...* (n - 2) )  + 

(1 * 2 * 3 * ... * (n - 1) )  +  ( 1 * 2 * 3 * ..... n )

So  the first term  = 3 and the sum of the rest of the terms can be written  as  3a   since  3 is a factor of each

So....the sum becomes   3  +  3a  =  3 (1  + a) which is divisible by 3

There are 15 ways.

Melody, CPhill, Hectictar: The above reduces to: 3 - 3^(2/3), but cannot get that far !! See if you can!

\(\dfrac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}\\ =\dfrac{2*(3^{1/3})^2}{1 + (3^{1/3}) + (3^{1/3})^2}\\ \)

there is not much you can do with this.  

A bag contains some sticks with lengths three feet, four feet, five feet, six feet, and seven feet.  Three sticks are chosen at random.

a) What is the probability that a triangle can be formed with the three chosen sticks?

3,4,5,6,7   there are 5C3 ways to chose three different ones  5C3=10  that is not many

If 3 are the same then there are 5*4C2 ways = 30 ways      

If they are all the same then there are 5 ways 

There are 45 ways to chose the 3 sticks

Also the sticks will from a triangle is the sum of the 2 little ones is longer than the big one. 

So

3,3,6

3,3,7

3,4,7 

are the only three selections that will not form a triangle

So

P(a triangle can be formed)= 42/45

b) What is the probability a right triangle can be formed with the three sticks?

The three sticks will from a right angle if the lengths of the sides are a pythagorean triad.  3,4,5 is the only one. 

so

P(forming a right angled triangle) = 1/45

P(forming a right angled triangle given that a triangle is able to be formed) = 1/42

Omi is right ... 

f(x)= x^2+4x-12x-2

has no holes.

A sinusoidal function whose frequency is 16π, the maximum value is 12, the minimum value is  −6   has a y-intercept of 3.

Do you mean  f(x)= x^2+4x/(-12x-2) ???

Between the 5th and 7th years, the tree grows approximately 7 ft.