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Posts posted by Barcallica
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Please clarify right bottom corner
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How about the 2nd 9/20?
62
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Host: Molly Mae
1. Flamebirde - 2013
2. MikeD - 2013
3. DyalDragon
4. Kikacat123 - 2013
5. Marksmanjay
6.
7. Barcallica - 2013
8. Brainiac100
9.
10. araver - 2013
11.
12. dee_tot - 2013
13. TwoaDay
14.
15.Magic - 2013
16. Hirkala
17. Panther - 2013
18. Bmad - 2013
19. mboonAnd now for something completely different... Hope I can join in.
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what we are doing on the right side is what we do to convert from decimal to binary. So what we are trying to prove is "if n in decimal equals a
1a2....ak in binary then prove in decimal n=a1*2k-1+a2*2k-2......+ak-1*2+ak, which is another way we convert dec to bin. Maybe someone help me proving this conversion methods???this is one approach
i found it easier to prove it geometrically
Interesting, I have no idea how you could connect it to geometry.
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either 1.45 or 2.15 defending on interpretation of what which meant
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what we are doing on the right side is what we do to convert from decimal to binary. So what we are trying to prove is "if n in decimal equals a
1a2....ak in binary then prove in decimal n=a1*2k-1+a2*2k-2......+ak-1*2+ak, which is another way we convert dec to bin. Maybe someone help me proving this conversion methods??? -
Effectively, you are adding the numbers on the left. Now, all the numbers on the left are doubles (even) except the first number that you start with. So, irrespective of the number you have on the right, the addition on the left will be odd if you start with an odd number on the left.
Therefore, this algorithm is not correct.
That is not true.
if the first number number on the right is even then that row will be crossed, leaving sum of left column even not odd. So it works in every possible pair.
Actually this has nothing to do with the number on the left. And also has something to with the fact that every number can be put as sum of exponents of 2.
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Simplifying all equations without any solutions to make things easier:
nn-1/2
sqrt(n)-1
sqrt(n)+1
n2
(3n)!-n
(3(n!))!+n
2srqt(n)+1
n*1n (=n if n<infinity)
2+n
2n+1
2n+(n-1)!
2n+|n| (=3n, n>0, or =n, n<0)
2n*.n+.n (=.n*(2n+1))
2n
(n+1)/.n
2n-(sqrt(n)-1)!
2n+1
2nn+n
(n-1)!-n
2n!-1
n*sqrt((n-.n)/.n)
n(1+2/.n)
sqrt(.n-n)*n-sqrt(n)
(nn/n-n)!
n2
I thought it was more of a fill-in thing, that squares
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we will end up in 1, 4, 2, 1 cycle whatever n we choose?
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Can anyone tell me how Darvinists or evolutioists explain first living thing ever formed?
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For every round you have chance to get average 1/2$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1$ 5/56, to win 2$ is 1/56
Bonus: You guys will collectively lose 3$ every round to the house
How much does a player stand to lose per round they play?
I didnt quite understand the question. But Curtis Jackson??
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For every round you have chance to get average 1/2$. More detailed, you have chance to lose 5/8, chance to recover 15/56, chance to win 1$ 5/56, to win 2$ is 1/56
Bonus: You guys will collectively lose 3$ every round to the house -
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time: 6 days or 144 hours
distance: 133.33 miles
full speed: ~1.4 miles/hour
Full speed I think is 1.3888...miles/hour
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3535
care to explain your logic?
Let x be the page number. Then 2x/10+xa/7+303=x ---> x*(28-5a)=10605. On the other hand 10605=3*5*7*101 also x>303 and a<7. So 28-5a should be 3,5,7,15 or 21. Since a is natural a can only be 5 and x=3535
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26?
very good!
Failed students number would be 12+5+8-2-6-3+1=15 thus 26 didnt fail
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31
because 3193 has only one divider below 50, which is 31
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1. c*(c+3p)/(3c+p)
2. (c-p)/(3c-p)
3. Paul and peter will walk same distance regardless of c, p and how far Bob drops Paul. -
Let a1, a2, ....am be the sides of the cakes. we need to prove 4*(a1+a2+......+am)<=20*sqrt(mn) ------> (a1+a2+......+am)<=5*sqrt(mn) -------> (a1+a2+......+am)2<=25mn We know 25n>=a12+a22+.........+am2 so we need to prove m*(a12+a22+.........+am2)>=(a1+a2+......+am)2 which can be easily proven by writing equation a2+b2>=2ab; (m-1)*(m-2)/2 times.
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Every row can be put as [x^2-(2x-2)]+[x^2-(2x-1)]+......+(x^2-2)+(x^2-1)+x^2=(x-1)^3+x^3 which converts to {[x^2-(2x-2)]+x^2}/2=(x-1)^3+x^3 thus can be proved after some easy simplification.
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1. 367.92
2. 5.11
3. 72=8*9 then pretty easy -
The wish game
in Games
Posted
I wish someone close this tread