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( ^ω^ ) thanks!

The divisor pairs of 6 are 1 and 6, and 2 and 3. Summing the factors of 6 excluding 6 gets 1+2+3=6, so no matter how many times you apply the function to 6, the answer will always be 6.

my friend lost my dog and took my lunch

Hmmmm....  ALL 12 of my triangles have the equal side lengths of 1.....and 4 have bases 0f 3    I guess there is only two that can be found?

  The question says nothing about rotating/flipping being the same triangle.....   I still think there is 16.....reflected / rotated / flipped makes it another triangle in the diagram  .....   IMO  

Very nice, heureka   !!!!

I liked this problem.....

Totally agree with Melody.....two  real assets    !!!

y  = 6x^2 (x + 1)^3

y  =  6x^2 ( x^3 + 3x^2 + 3x  + 1)

y  =  6x^5 + 18x^4  + 18x^3  + 6x^2

y'   =  30x^4  + 72x^3 + 54x^2  + 12x

Set the derivative to 0  ......solve for x

30x^4 + 72x^3 + 54x^2 + 12x   =   0

x ( 5x^3 + 12x^2 + 9x + 2 )  =  0

x (x^2 + 1)^2  (5x + 2)  =  0

The two critical  numbers  are    x  = 0     and x  = -2/5  = -0.4

Since we can use any method.....look at the graph, here  :

https://www.desmos.com/calculator/ehkvynz2gf

A relative max occurs at  ( -0.4, .20736) 

A realtive  min occurs at  (0, 0)

2^4 * 3^2 * 5^1    = 720

The number of divisors   =

(4 +1) (2 + 1) + (1 + 1)  =

(5) (3) (2)  =

15   *  2   =

30

a +  b  =  7

a^2 + 2ab + b^2  =  49

a^2 + b^2  =  49 - 2ab       (1)

a^3 + b^3  =  42

(a + b)(a^2 - ab + b^2)  = 42

 (7) (a^2 - ab + b^2)  =  42

a^2 - ab + b^2  =  6     

a^2 + b^2 - ab   =  6       (2)

Sub (1) into (2)  and we have that

49 - 2ab - ab  =  6

43 = 3ab

43/3  = ab

So

1/a  + 1/b  =

(a + b)

______    =

    ab

    7

______  =

 (43/3)

 21

 __

 43

Thrown horizontally....how long before it his the ground?

x = x₀ + v_i t+ 1/2 a t²     x = 0     xo = 36.7 m   vi =0   a = -9.81 m/s²

0=36.7 + 1/2 (-9.81) t²         solve for t = 2.735 seconds

v_f = vo + at

v_f = 0 -9.81(2.735) = -26.83 m/s   (negative because it is in downward direction)    if ya just want the magnitude: 26.83 m/s

Can you show me an isosceles triangle in the diagram with base lenght two?  ...or three?   Remember....two sides need to be of the sam length....

Solve for z:
z^2 = 21 - 20 i

Take the square root of both sides:
z = sqrt(21 - 20 i) or z = -sqrt(21 - 20 i)

21 - 20 i = 25 - 20 i - 4 = 25 - 20 i + 4 i^2 = (-2 i + 5)^2:
z = 5 - 2 i or z = -sqrt(21 - 20 i)

21 - 20 i = 25 - 20 i - 4 = 25 - 20 i + 4 i^2 = (-2 i + 5)^2:
 
z = 5 - 2 i       or       z = -5 - 2 i

6! =720 has exactly 30 divisors.

I like that some people get the topic that I was wondering about to, I was wondering if you wouldent mind me reposting this, or adding it to my welcome message. I will put your name in it also for credit if you want, If you dont mind me doing so. You might want to change it up a bit for something that is going to be there for awhile.

---Killersteve27

The smallest a, b, c that I got are:

a=2, b=2, c=2

So: a + b + c =2 + 2 + 2 =6

Is 209 the last term? If so, the sum of the sequence =21,945.

21,945 has this many consecutive sequences:

1  - 10972  to 10973 = 21945
2  - 7314  to 7316 = 21945
3  - 4387  to 4391 = 21945
4  - 3655  to 3660 = 21945
5  - 3132  to 3138 = 21945
6  - 2190  to 2199 = 21945
7  - 1990  to 2000 = 21945
8  - 1561  to 1574 = 21945
9  - 1456  to 1470 = 21945
10  - 1146  to 1164 = 21945
11  - 1035  to 1055 = 21945
12  - 987  to 1008 = 21945
13  - 717  to 746 = 21945
14  - 649  to 681 = 21945
15  - 610  to 644 = 21945
16  - 559  to 596 = 21945
17  - 502  to 543 = 21945
18  - 372  to 426 = 21945
19  - 357  to 413 = 21945
20  - 300  to 365 = 21945
21  - 279  to 348 = 21945
22  - 247  to 323 = 21945
23  - 184  to 278 = 21945
24  - 157  to 261 = 21945
25  - 145  to 254 = 21945
26  - 136  to 249 = 21945
27  - 99  to 231 = 21945
28  - 66  to 219 = 21945
29  - 51  to 215 = 21945
30  - 21  to 210 = 21945
31  - 1  to 209 = 21945

Note: let me know if I misunderstood your question.

Guest, Why would I want to solve it from anywhere?

If there are any errors in my solution they are only careless ones.

I was only answering for your benefit not for mine. 

Do you realize how rude your comment sounds?

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Thanks for the point Chris.  

Let P(x) = x^2.  Then P(1) = 1, p(2) = 4, p(3) = 9, so p(n) is always a perfect square.  You can do this for any polynomial that factors, like P(x) = x^3 and P(x) = x(x + 1).

? I don't really notice that, I mean, there's not much spam on the website, it's mostly just random stuff and the occasional nasty Guest...

LOL, I love it!

I see you CPhill answering this question:

Let us see if you agree with I got!!.

Longer side (a):    21

Shorter side (b):   9

Legs (c):                10

Diagonal (d):       17

Height (h)             8

Central mediam:   15

Circumcircle radius (rc): 10.625

Overlap (g):               6

Perimeter (p):           50

Area (A):                    120

Acute angle (α):       53.13

Obtuse angle (β):    126.87

thank you so much!!!!