create a complete social media manager business Page for $20

create a complete social media manager business Page

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Complete Facebook business Page

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路 Setup and Optimize Impressive Facebook page.

路 Fan Page, Coach Page, ecommerce, Product. Online shop

路 Configure Tabs Button

路 Facebook.businessSetUp

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路 Website Integration

路 SEO friendly Page

路 SEO friendly Profile image and Cover photo.

路 Suitable Template Related Your Niche

路 Setup Facebook business page Shop and add the products

路 Call to Action Button: linked to a destination website for specific action required by users (ex:

Book Now, Contact Us, Use App, Shop Now, etc.)

路 Auto Response Setup

路 Facebook Business Page Messenger Setup

路 Including All Information About your company

路 Linkup With all other Social Platform

路 Install a location map of your business page

Moreover!!!

100% Money Back Guarantee.

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Full customer support even when the gig is over.

Note: – Feel Free to drop me a message, we will discuss your project thoroughly. Thank You for your attention.

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complexity theory – What is wrong with this argument that if A is NP Complete, but B is in P, then AB is NP Complete and BA is NP Complete as well?

The following seems to me to be relevant to this question, but to me is an interesting exercise, especially since I have not formally worked with complexity before, but I want to learn more:

Suppose that A is NP Complete, but B is in P. I claim that AB is NP Complete and BA is NP Complete as well. To see this, assume first that AB is in P, and let X and Y be polynomial time algorithms for B and for AB, respectively. “Concatenating” X and Y as follows yields an algorithm Z for A:

Given L, test L using X;
if X outputs “yes”, test using Y;
if Y yields “yes”, output “no” and stop;
if X yields “no”, output “no” and stop; output “yes” otherwise and stop.

This algorithm Z runs in polynomial time, because if the (polynomial time) complexity exponent of X is k and the (polynomial time) complexity exponent of Y is n, then this algorithm clearly has (polynomial time) complexity exponent m=max(k,n). This would provide a proof that P=NP, so AB is NP Complete.

Now suppose that BA is in P. This time, let Y’ be a polynomial time algorithm for BA and let X be as above. We construct an algorithm Z’ for A, as follows:

Given L, test L using X;
if X outputs “yes”, test using Y’;
if Y yields “no”, output “yes” and stop;
if X yields “no”, test using Y’; if Y yields “no”, output “yes” and stop;
output “no” otherwise and stop.

This yields a polynomial time algorithm for A, and so again, this would entail that P=NP, so BA also is NP Complete.

+++++++++End of Example++++++++++++

While I don’t see anything wrong with the above at the moment, perhaps I have a mistake or complexity miscalculation? …because for a while as I was writing the second algorithm, I began to think it was odd and perhaps impossible that I can be right about BA also being NP Complete…

Like I said, I’m somewhat new to this area, so feedback would be appreciated.

gm techniques – As a GM, how do I build a complete group of players that fit my style and preferences?

I’ve been GMing for some years now. During that time, I feel like I’ve developed somewhat of a style/preference with the kinds of games I like to run. While I feel my games can be enjoyed by a variety of players, I find that certain types of players mesh better with my style/preference. Selfishly, this sort of meshing really improves my satisfaction as a GM and I’d really like to build an entire group of only players like this, even if it was just to fulfill a vision I have for a single, particular type of campaign.

Have other GMs attempted and/or succeeded in this sort of utopian endeavor? What are some strategies/tactics I could/should consider?

I’d be looking to build a group of 3-4 players, maybe 5. As an example of where my mind is currently at, I’m considering running a series of 4-session campaigns for different groups of players (a complete story arc for each group) and using these mini-campaigns as incognito recruiting events. Having a pre-defined end point after 4 sessions would (hopefully) allow me to gracefully cut ties with the players that don’t fit the profile I’m looking for, without creating any hard feelings — though this is all still theoretical.

gm techniques – As a GM, how do I build a complete group of players that mesh well with my style and preferences?

I’ve been GMing for some years now. During that time, I feel like I’ve developed somewhat of a style/preference with the kinds of games I like to run. While I feel my games can be enjoyed by a variety of players, I find that certain types of players mesh better with my style/preference. Selfishly, this sort of meshing really improves my satisfaction as a GM and I’d really like to build a group around such players, even if it was just to fulfill a vision I have for a single, particular type of campaign.

Have other GMs attempted and/or succeeded in this sort of utopian endeavor? What are some strategies/tactics I could/should consider?

I’d be looking to build a group of 3-4 players, maybe 5. As an example of where my mind is currently at, I’m considering running a series of 4-session campaigns for different groups of players (a complete story arc for each group) and using these mini-campaigns as incognito recruiting events. Having a pre-defined end point after 4 sessions would (hopefully) allow me to gracefully cut ties with the players that don’t fit the profile I’m looking for, without creating any hard feelings — though this is all still theoretical.

np complete – Reducing subsetsum to { | G is a weighted graph that has a spanning tree with weight between l and u}

Let $X = { x_1, x_2, dots, x_n }$ be the (multi-)set of elements in your subset-sum instance and let $t$ be the target value.

Create the undirected graph $G=(V,E)$ where $V={a,b} cup X$ and $E = ( {a,b} times X) cup {(a,b)}$.
The weight of edge $(a, x_i)$, for $x_i in X$, is $x_i$. The weight of edge $(b, z)$ for $z in {a} cup X$ is $0$. Finally, pick $l=u=t$.

There is some $Y subset X$ such that $sum_{x_i in Y} x_i = t$ if and only if there is a spanning tree $T$ of $G$ of total weight between $l$ and $u$.

Indeed, given $Y$, you can select $T$ as the tree induced by the edges in ${(a,x_i) mid x_i in Y} cup {(b,z) mid z in V setminus Y}$.
Moreover, given a tree $T=(V,F)$ of total weight between $l$ and $u$, you can select $Y = { x_i mid (a,x_i) in F }$.

np complete – Reducing subsetsum to $L = { langle G, l, urangle | G text{ is a weighted graph that has a spanning tree with weight between $l$ and $u$}}$

Let $X = { x_1, x_2, dots, x_n }$ be the (multi-)set of elements in your subset-sum instance and let $t>0$ be the target value.

Create the undirected graph $G=(V,E)$ where $V={a,b} cup X$ and $E = {a} times X$.
The weight of edge $(a, x_i)$ is $x_i$. The weight of edge $(b, x_i)$ is $0$. Finally, pick $l=u=t$.

There is some $Y subset X$ such that $sum_{x in Y} x = t$ if and only if there is a spanning tree $T$ of $G$ of total weight between $l$ and $u$.

Indeed, given $Y$, let $y$ be an arbitrary element from $Y$. You can select $T$ as the tree induced by the edges in ${(a,x) mid x in Y} cup {(b,x) mid x in X setminus Y} cup { (b,y) }$.

Moreover, given a tree $T=(V,F)$ of of total weight between $l$ and $u$, you can select $Y = { x mid (a,x) in F }$.

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uk – When is “2 weeks from the vaccination dose” complete for entry to France?

Obviously, it’s safer to use the later date. By inductive reasoning I would argue that the day of the vaccination should not be counted, but the same line of reasoning also leads to the conclusion that the relevant date is August 4th.

First, assume that “two weeks” is fourteen days (perhaps questionable for France, where they often use the expression quinze jours, “fifteen days,” to mean two weeks, but I’ll assume for now that the French-language rule is deux semaines).

Next, consider that if the requirement were “one day after the second dose” then the relevant date after a vaccination on 21 July would be 22 July. Therefore, “two days” would denote 23 July, “three days” 24 July, and so on, until you arrive at “ten days,” denoting 31 July, “11 days,” denoting 1 August, and eventually “14 days,” which denotes 4 August.

After a quick search, I found a page on the French government site that uses “2 semaines,” so there should be no need to worry about quinze jours.


However, as noted in a comment, it seems that at least Germany, for whatever reason, considers that one is fully vaccinated only as of the beginning of the fifteenth day after vaccination. I suppose that this is likely because during the earlier part of the fourteenth day after the vaccination, before the time at which you were vaccinated, you aren’t considered fully vaccinated, and it is easier just to wait until midnight than to try to measure the 14 days down to the second.

Now that this wrinkle has been introduced, and without being able to find anything explicit about how France approaches this question, I would assume that they approach it as Germany does, and use August 5th.

data structures – What will be The Complete code for It ? Please Help me

data structures – What will be The Complete code for It ? Please Help me – Computer Science Stack Exchange

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