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scilark posted at 18:02 UTC
artisci posted at 18:26 UTC
stupiscied posted at 18:28 UTC
scirocket.com
@anon: nice idea, but then it's not chronological.
scift.com
@anon: y? because time flies.
spience.com
scithused.com posted at 18:06:25
sciflyer.com posted at 18:12:18
scite.com posted at 18:13:29
sci77.com posted at 18:15:09
sciencium.com
3762
check with division. 376/99=3R79 792/99=8, hence 3762/99 = 38, and post # 38.
21307
check it by division. 21/13=1R8 83/13=6R5 50/13=3R11 117/13=9, so 21307/13=1639 and post #1639.
does this work for your holiday??
1 gives to 4 and 8.
2 gives to 5 and 9.
3 gives to 6 and A.
4 gives to 7 and B.
5 gives to 8 and 1.
6 gives to 9 and 2.
7 gives to A and 3.
8 gives to B and 4.
9 gives to 1 and 5.
A gives to 2 and 6.
B gives to 3 and 7.
How's dat??
Now i will try to list the two sides that are not adjacent...
03
04
05
06
07
14
15
16
17
18
25
26
27
28
29
36
37
38
39
XX30 is already listed above (alphabetical numerical order neglect repeats)
47
48
49
XX neglect repeats
58
59
69
Okay just count 'em up and add to 70 for answer...
Zave a byeutizfull daze.
number the sides: 0 to 9: (0 1 2 3 4 5 6 7 8 9).
Okay... now you can do the adjacent sides first with the point going out to meet across somewhere, and then add that set to the ones where the two sides are not adjacent.
Now letter the vertices and number the sides as follows:
0Z1A2B3C4D5E6F7G8H9Y
Okay, now we can talk of the first set where two adjacent polygon sides are used like this list of them:
90A
90B
90C
90D
90E
90F
90G
then rotate thse around to get more of them...
for example later you rotate to
45F
45G
45H
45Y
45Z
45A
45B
This is the first set, partially shown, but only 2 tenths of the set is shown above.
So seven times ten is 70, so the first adjacent sided set has 70 four-sided figures.
See next post for the other set...
You said ". i need to prove that d2f/dx2 + d2f/dv2 = 0.".
I see f, x, f, v, but no u, why no u ?
i also guess C based on 12 X 3.141_.
Maybe it's a board game, but I don't know the rules to it yet. Here's a hint though, examine the green numbers. These numbers might ring a bell in your memory banks.
Why do you think these numbers are arranged like this?
[solar system] --> sassy; artsy; Taylor; metro; Troy; stream; moss; meal; Leary; slater; lossy; stray; May; Ma; a
#4468.)
Also, you may utilize the utility called "uuencode" and post the alphanumerics for someone else to "uudecode".
It is a free software utility easily obtainable on the web.
You might be able to embed you excel spreadsheet into a non-compressed *.png file and post the pic and then a diligent observer could disentangle the pixels into the excel bytes. Just a thought.
Also some web sites allow temporary sharing of binaries but I don't know where off hand...
thanks for the binary dots in a square, very interesting...
Here's an awkward way to avoid absolute value signs:
Abs(X) = (x**2)**(1/2) = (x^2)^(1/2)
Basically square root the square of the interested value.
max value of a and b = (a + b + |a - b|) / 2
minValue(C,D)= (C + D - |C - D|) / 2
Put it all together in one large equation and get...
minValue(maxValue(dTop,dBottom),maxValue(dLeft,dRight))
Now just expand this with you trigonometric equations you have
mentioned in your original post and I believe it is possible.
I hope you are allowed to use the absolute value operator though!!!!
Have a super week...
Hi my friend. If you are allowed to use the absolute value operator, I do believe an elegant solution will arrive. For starters, let's couple together the dbottom with the dtop in an expression to find the one that is positive. And also let's couple together the dleft with the dright in an expression to find the one that is positive meaning non-negative. In these two cases, we simply need a "maximum finder" equation that uses the absolute value signs. After these two terms are resolved, they can be embedded in the larger single equation in which the "minimum of two values" is found, and that is the answer. Now let me give you an idea as to how one would find the max value of two numbers with the absolute value sign. See next post for more details...