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2.) if the graph cannot be drawn without lines crossing each other, then you need another layer probably to get the 3-d effect, like knot theory sort of.
since a line segment or "edge" has two ends, then yes, 1/2 is the formula. v=dots d=ends of lines on a dot. e=#ofline segs
Pretty obvious really if you draw a few. but to prove it, ask Ricky.
I searched for Jane's formula but couldn't find it.
but i did find this weird discussion of mine and carlos.
http://www.mathisfunforum.com/viewtopic.php?id=2168
there is a formula for the length of a curve that might help. jane fairfax gave it to me on this forum a few years ago.
search for length of curve or something... then put that together with the height witha product i guess and maybe it'll work.
the half thing seems right because log(400) is about 2.6 and 2.6/2=1.3 and log (20) sorta is 1.301, so that's good...
If you wire the output of a logic gate to an input upstream from a bunch of logic gates, you create a sequential sort of circuit, depending what exactly you do.
I suggest reading about J-K flip-flops and D flip-flops and also thinking about
how these flip-flops are made out of logic gates and pursuing some thought
about making a sequential machine without using J-K or D flip-flops, but
simply by figuring out your own way of creating a sequential circuit.
I do not know how to make a mathematical expression for a sequential circuit.
For convenience, here is the image.
Click to enlarge.
Is
??(49/50)(500ml)(3) should work. The reason is that if 49 parts water out of 50 make 98% water, then that's 2% salt by weight.
so therefore you want to quadruple the water to go from 2 to 1 to 1/2 of a percent salt. So add 3 more times the 49 parts.
Am i right?
Venn diagrams are circular usually and Karnaugh maps are gridlike square-ish, but they are both one and the same except that the
Karnaugh map encloses the zero zero zero case in a box, whereas a venn diagram puts the zero zer zero case outside all the circles and does not
put a circle around it, though you could circle everything and that would make them more similar.
w h(w) j(w)
2 4 1
3 9 1.5
4 16 2
9 81 4.5
16 256 8
j(w) means takeHalfOf(w)
h(w) means square(w)
So j(h(2) = 2 because h(2) = 4 because 2 to the 2 power is 4 and j(4) is half of 4 or 2 back for the answer.
So j(j(j(j(32)=2 because j(32) = 32/2 = 16 and j(16) = 16/2 = 8, and j(8) = 8/2 or 4. and j(4) is 2.
log base e of 55 is too big for 4, should be 54.5981500331
as
2.718281828459045235 ^ 4 is 54.5981500331
1 = 1 | 0
0 = 0 | 0
| means OR.
now AND...
0 = 0 & X
X = 1 & X, where X can be 1 or 0 (boolean value)
now XOR...
(exclusive-OR)
1 = 1 ^ 0
1 = 0 ^ 1
0 = 1 ^ 1
0 = 0 ^ 0
basically the 1 is a toggle of the other value, toggle = flip from 0 to 1 or 1 to 0.
Given that Ab, Bb, Cb are boolean variables holding 0 or 1, we have the following:
The / sign is the negation or NOT symbol as / put in front of a variable.
(Ab & /Bb) | (Bb & /Ab) = A ^ B, where ^ means exclusive-or (XOR).
Also demorgans laws something like...
/(Ab & Bb) = /Ab + /Bb
Also see Karnaugh mapping for electronics ways...
Maurice Karnaugh of Bell Labs 1957 1953...
I think you are correct and the problem meant to say what is the probabil that the man has one (not none) of these accessori, in which case 3/4 would be he is wearing one of the things. Your iteration of the permutations seems correct. I would stick to your guns. Good Luck.
4005 (skipped) and 4095
4005 + post90 = 4095
90 x 91 / 2 = 4095
3916
3828 + post88 = 3916
post88 x 89 / 2 = 3916
dot org, rite? do u have to download the software to do this?
I'm sorry for any confusion, I just couldn't resist a little joke with the seasons greetings and so forth.
Actually, the seasons have little or nothing to do with the puzzle. I regret any wasted time this hath causeth.
#84 = Po, as in the november poisoning of the russian run-away with polonium-210 in 2006.
Okay, her4e's another try on the circular seasons number chart...
I guess 122K bytes photos are too large for our forum. I thought max was 64K at one point, but thought it was larger now.
This pic shows
that the order
of the numbers
belongs to the
seasons.
Happy seasons!!!!
This clue should give it away.
Try reading on a guy named, "Charles Janet".
Another clue: In 1929, these numbers were known such that this order could have been generated by a guy last name "Janet", though he did not put them in this particular order, but he had the knowledge to do it, based on what he did generate in 1929.
Aei => 333
Spcl => 24 Ways, So Make It 4444
Start List To Get Idea
posted at 18:43 UTC: (continued)
4444333
4443334
4433344
4333444
3334444
3344443
3344434
3344344
3343444
4334443
4334434
4334344
4433443
4433434
4443343
now do single 3 first then double 3 later...
3444433
3444334
3443344
3433444
4344433
4344334
4343344
4434433
4434334
4443433
now do all 3's apart and separated...
3444343
3443443
3434443
3443434
3434434
3434344
continue with 3 in second place...
4344343
4343443
4343434
now do 3 in third place.
4434343
that's it i think.
hope i didn't miss any.
count them all up and multiply by 24 for the answer to the question because the consonants can be arranged 24 ways = 4 x 3 x 2 x 1.
only thought that comes to mind is that route 5 minus 1 over 2 is golden ratio, and route 5 plus 1 over 2 is the other golden ratio.