CPhill

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%  

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

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)

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

7x-(x-5) ≥ 14     simplify

7x - x + 5 ≥  14     combine like terms

6x + 5 ≥ 14     subtract  5 from both sides

6x ≥ 9        divide both sides by 6

x ≥  9/6  ⇒   x  ≥ 3/2

OK.....I'm gonna' answer this.....you just can't trust a procrastinator like that darn hectictar....(HEHEHEHE!!)

(10, 2)  and (8, 7)

Slope

(1) Subtract the y coordinates in some order  ⇒ 7 - 2   =  5        ....    (1)

(2) Subtract the x coordinates in the SAME order ⇒   8 - 10 =   - 2   ...    (2)

Put   (1)  over (2)

5 / -2     =   -5 / 2        and that's the slope

Oops...you're late, hectictar....it's already 12:07 !!!  LOL!!!!!!!

See???.....you already got a point....!!!