Crypto-mining – Bonus – Proof Of Payment

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Compensation amount is from 0.01 cents up to $10. Roi: R1 Server: 12.5 – 8125 Gh/s Minimum amount of Hash purchases : 12.5 ( only 2 USD ) ( 1 GH/s 0.16 USD ) Maximum amount of Hash purchase: 8125 Term of Server lease 300 days Profit is accrued: from Monday to Friday Weekends are not paid: Saturday, Sunday Profit per day: from Monday to Friday Profit per month: 46% Profit for the entire lease term: 460% The minimum withdrawal amount: 0.1 USD R2 Server 625 – 18750 Gh/s Minimum amount of Hash purchases : 625 Maximum amount of Hash purchase: 18750 Term of Server lease 300 days Profit is accrued: from Monday to Friday Weekends are not paid: Saturday, Sunday Profit per day: from Monday to Friday Profit per month: 54% Profit for the entire lease term: 540% The minimum withdrawal amount: 0.1 USD R3 R4 R5 R6 Server…. [b]R7 Server 156250 – 3125000 Gh/s Minimum amount of Hash purchases : 156250 Maximum amount of Hash purchase: 3125000 Term of Server lease 240 days Profit is accrued: from Monday to Friday Weekends are not paid: Saturday, Sunday Profit per day: from Monday to Friday Profit per month: 90% Profit for the entire lease term: 720% The minimum withdrawal amount: 0.1 USD Deposit Processor: Bitcoin, Perfect Money, Payeer, ADVcash Withdraw Processor: Bitcoin, Perfect Money, Payeer, ADVcash Minimum / Maximum Lim.: 0.10 USD – 10 000.00 USD REGISTER HERE LINK Payment of Proof: [/b] Exclusive explanation of the site russiamoney + proof of withdrawal of the amount of 1179 rubles Peace be upon you The progress continues to prove a new draw of 56 rubles, for those who have been exhausted and not taken advantage of The opportunity after the total draw is currently 1179.89 rubles a good amount of capital Simple means that if there is substantial capital, the result would be beautiful but now The one asks God the bles Exclusive explanation of the site russiamoney + proof of withdrawal of the amount of 1179 rubles Do I need Proof of onward travel when entering India on one-year multiple entry tourist visa as Lithuanian (EU Citizen)? I am Lithuanian (EU Citizen) and got one-year multiple entry tourist visas for my family. My entry point will be Kochi (COK) airport from Singapore. I am getting conflicting information in regards proof of onward travel when entering India. I would find it somewhat unreasonable to ask for plane tickets one year in advance. Do I need Proof of onward travel when entering India on one-year multiple entry tourist visa? Question regarding proof of Tychonoff’s theorem On Wikipedia it states that a space $$X$$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff’s theorem, and I don’t quite see how. In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which is a convergent subsequence. This is obviously not true, since $$\{0,1\}^\mathbb{R}$$ is not sequentially compact, but it is compact by Tychonoff’s theorem. A Deeper Explanation of this Proof (or Alternative Proof) On the Maximality of Valuation Rings I’m trying to see if there is a proof of the following theorem using Galois theory, or number theory of some kind. Specifically, I’m looking at (iii) implies (i) below. Theorem: Let $$K$$ be a field and let $$\mathcal{O}$$ be a subring. Let $$l$$ be an algebraically closed field, and let $$\phi : \mathcal{O} \rightarrow l$$ be a map into $$l$$. Then, for any $$a \in K^\times$$, there is either an extension of $$\phi$$ from $$\mathcal{O}[a]$$ into $$l$$ or an extension of $$\mathcal{O}[a^{-1}]$$ into $$l$$. My intuition for the theorem is as follows. For any subring $$\mathcal{O}$$ of $$K$$, we have a partially ordered abelian group $$K^\times / \mathcal{O}^\times$$, where the partial order on $$K^\times$$ is given by declaring $$x \leq y$$ when there is $$a \in \mathcal{O}$$ such that $$x a = y$$, and the partial order on $$K^\times / \mathcal{O}^\times$$ is given by declaring $$xU \leq yU$$ when there exists $$a \in A$$ such that $$xa \mathcal{O}^\times = y \mathcal{O}^\times$$. This theorem is sort of saying that we can add at least one of the relations $$0 \leq a$$ and $$0 \geq a$$ in $$K^\times / \mathcal{O}^\times$$ (mod out by the convex subgroup they generate), collapsing it so that it now corresponds to the ring $$\mathcal{O}[a]$$ or $$\mathcal{O}[a^{-1}]$$. A corollary is that we can get $$K^\times / \mathcal{O}^\times$$ to be totally ordered this way. What is going on in this proof? Maybe it could be clarified with some higher number theory or Galois theory. UK standard Visa – bank statement type, proof of vacation reqest I am super thankful to you guy for helping. So, I want to go with my boyfriend to London and he will be paying for the trip mostly (9 days). We want to show them his savings account (bank statement). Is this possible or it has to be the bank statement from the account you get salary om? I have a permanent contract and pay slips but my paycheck is not even medium amount. And so, does having an approved vacation proof is very very important? Thank you!!! Leona Proof of heuristic function in A star algorithm Maybe I am missing something very easy and obvious. But, I don’t understand why estimate cost of source vertex is subtracted from the overall estimate cost, if heuristic function $$h$$ is monotonic: $$d′(x,y)=d(x,y)+h(y)−h(x)$$ What I currently know: A* algorithm can be used as an extension for Dijkstra’s algorithm. At each iteration of its main loop, it chooses the vertex with the minimum of estimation cost plus cost of the path to this vertex: For vertex $$u$$ and its successor $$v$$, overall cost is calculated with $$f(u, v) = d(u, v) + h(v)$$ using some heuristic function $$h$$. Where: • $$d(u,v)$$ cost of the path from $$u$$ to $$v$$ • $$h(v)$$ estimate cost from $$v$$ to the target vertex $$t$$ If for any adjacent vertices $$u$$ and $$v$$, it is true that $$h(u) <= d(u, v) + h(v)$$ then $$h$$ is a monotonic. In other words, graph holds triangle inequality property. It is stated in Wiki page of A* algorithm: If the heuristic h satisfies the additional condition $$h(x) ≤ d(x, y) + h(y)$$ for every edge $$(x, y)$$ of the graph (where d denotes the length of that edge), then h is called monotone, or consistent. In such a case, A* can be implemented more efficiently—roughly speaking, no node needs to be processed more than once (see closed set below)—and A* is equivalent to running Dijkstra’s algorithm with the reduced cost $$d'(x, y) = d(x, y) + h(y) − h(x)$$. My questions are: and A* is equivalent to running Dijkstra’s algorithm with the reduced cost $$d'(x, y) = d(x, y) + h(y) − h(x)$$. Any proof for this equivalence ? It is clear for me that $$0 <= d(x, y) + h(y) – h(x)$$, and it is feasible. But: • Why this formula is chosen as a new distance function ? • Is there any formal proof that it works ? • Why it is not enough to run Dijkstra with $$d'(x, y) = d(x, y) + h(y)$$ ? • What is the math behind it ? Proof of QuickSort algorithm correctness Recently I’ve studied QuickSort and understood its general idea. Basically, we do the following: 1. Pick an element from the array (no matter which one and how in this context) 2. Rearrange elements in the array so that elements less than the pivot are placed prior to it (in the left part of the array), and ones that are greater than the pivot are placed after it (in the right part). 3. Apply these procedures to the left and right parts of the array. Everything looks quite clear, but my question is: why does it eventually sorts the entire array? I’ve read lots of articles about the QuickSort, but none of them provides a clear proof that algorithm does really work in a way they describe. So I’m looking for such a proof. I realize that: • Array of one element (i.e with length 1) is already sorted. • After any partition the pivot is placed on it’s final position. So what’s the proof of the fact that after applying QuickSort procedure to a given array of length N its elements will be ordered after a procedure completes? P.S. I’ve seen several proofs based on mathematical induction. Unfortunately, I’m not familiar with it yet, so I’m primarily looking for non-induction proof. Proof$SEQ_{CFG}\$ = {⟨G,H⟩ | G,H are CFGs and L(G) ⊆ L(H)} is decidable

How can I proof that $$SEQ_{CFG}$$ = {⟨G,H⟩ | G,H are CFGs and L(G) ⊆ L(H)} is decidable ?

ps: I know $$EQ_{CFG}$$ = {⟨G, H⟩ | G and H are CFGs and L(G) = L(H)} is not.

Trouble understanding proof regarding elements of an integral domain which are not product of irreductible elements.

I’m having trouble understanding the proof of a proposition regarding elements which are not product of irreductible elements in integral domains. The proposition is the following:

Let $$A$$ be an integral domain and $$a\in A$$ different of 0 and not a unit. If $$a$$ is not product of irreductible elements then there exists a sequence $$\{a_n\}_{n\in N}$$ of elements of $$A$$ such that $$a_{n+1}$$ is a proper divisor of $$a_n$$ for every natural $$n$$.

The proof that has been given to me is this:

Let’s build inductively a sequence with this property. Let $$a_0 = a$$ and suppose $$a_0,\dots,a_n$$, $$n\geq0$$ are already built. Since $$a_n$$ is not product of irreductibles it must be composite, this $$a_n = bc$$ with $$b,c$$ proper divisors of $$a_n$$. Clearly at least one of the two factors must not be product of irreductibles. We define $$a_{n+1}$$ as this factor.

The problem I have understanding this proof is that I don’t get why not being product of irreductibles implies it is composite. Wouldn’t such elements be irreductibles itself? I mean if it is composite, couldn’t we keep decomposing its factors until all of them are irreductible?