## What are the 2 kinds of links important for SEO?

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What are the 2 kinds of links important for SEO?

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## What Kinds of Leisure Chairs Are Available – Cryptocurrencies Corner

Nowadays people’s living standards are constantly improving, and they will enjoy life more and more. In order to meet this demand, many merchants will take them into account when producing products. Modern Leisure Chair are no exception. Now there is a leisure chair in the market that allows us relax and enjoy life while we are busy with our work.

Leisure Chair

The rattan chair is generally a bamboo chair frame. This type of leisure chair is very flexible and breathable. When choosing this kind of wicker chair, it is best to choose a backrest. You can then place an ottoman on the side so that your feet are flat.

Leisure chair folding chair

Speaking of folding chairs, you may not know the long history of this type of leisure chair. The development of the folding chair was about 2000-1500 BC, and the chair was the special seat of the army commander on the battlefield. This means that the folding chair is a symbol of power and status.

Lounge chair for leisure chairs

The recliner is no stranger to the Chinese. In the Qing Dynasty, the recliner was very popular. The appearance of the recliner is because of the continuous improvement of the craftsmanship and skills of the feudal society in China, and people pay more and more attention to the quality of life.

Leisure chair armrest lounge chair

Leisure chairs with armrests are often sold with coffee tables, usually with two chairs and a coffee table. Many of these chairs are made of wood. Sitting on a casual chair of this style, you can put your hands on the armrests and look particularly casual from the side. There is a charm that stops your hands and rests.

Leisure chair rocking chair

Rocking chair, according to its literal meaning, we can know that this chair can be shaken. Remember, it is a chair that swings back and forth. People can lie on top half, and the feeling is particularly comfortable as the chair sways gently.

As people life quality rise, the Modern Leisure Chair have not confined to the garden. Now in many public places such as: community, parks, entertainment plaza there are outdoor Leisure Chair . These outdoor leisure chair, beautiful appearance diversity. It has also become a lot of city beautiful elements, and brings great convenience to people. Modern Leisure Chair : https://www.insharefurniture.com/product/pp-leisure-chair/

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## dnd 5e – What kinds of training and research are possible within the paradises found within a Rod of Security?

The Rod of Security gives a person access to ‘any paradise one could imagine’, specifically:

‘… the rod then instantly transports you and up to 199 other willing creatures you can see to a paradise that exists in an extraplanar space. You choose the form that the paradise takes. It could be a tranquil garden, lovely glade, cheery tavern, immense palace, tropical island, fantastic carnival, or whatever else you can imagine.’

This ageless paradise has a 199 day limit, divided by the number of occupants staying within Example: a party of five gets ( 199 days / 5 people = ) 39.8 days maximum. Once out it takes ten days to regain this 199-day-persons capacity.

Plenty of rest, good times and amazing food (i.e. ‘possible weight-gain hazard’) – but so much free time… with incredible resources! Imagine gaining new tool proficiencies, catching up on some downtime stuff or whatever you like. But could one research old &/or new spells? Develop a formula / blueprint / pattern for almost any magic item? Research ancient lore in a paradise-library? Suddenly this rod is a very powerful item. Possibly derailing a campaign – or even presenting as a bit too powerful?

Given access to ‘any paradise’ – be that training &/or library-research – what limits exist for players’ gains in skills, abilities &/or knowledge?

## dnd 3.5e – How exactly do cylinder spells and effects “ignore obstructions” in their areas? Or what kinds of obstructions they can realistically ignore?

I think this is a poor description, because it is trying to describe a method of drawing a particular volume (by drawing a horizontal two-dimensional shape on the $$XY$$ plane and projecting it downward on the $$Z$$ axis), but then also using that as the description of what the spell itself is doing. The reason that’s a problem is because the combination of two factors here undercut any concept of the spell being “projected” at all. The first issue is that spells come into effect instantaneously; there is no time at which an effect is just the circle and then it travels downward. The other issue is the last quoted line, “A cylinder-shaped spell ignores any obstructions within its area,” which says we have to ignore anything blocking the projection on its way downward.

The sum of those two effects is that the spell isn’t “projected” at all, but simply comes into existence at all points within the volume simultaneously, without regard for what else is in that volume or where any point within is in relation to any other point in it. This is a valid way to imagine magic happening, of course, and probably appropriate for many spells. Moreover, even when it isn’t precisely appropriate, the top-down projection is very specific and definitely isn’t going to apply in a lot of cases, so it makes sense to avoid implementing that as a global default. All of which they could have easily just said, and relied on the “projection” description to talk solely about how you should draw the shape/know you have the right shape, but they didn’t do that. Instead, they talk about the “spell” doing that, and then need to apply the “hack” of having it ignore obstructions.

Anyway, yes, the long and short of it is, the official rule is that, by default, cylinder-shaped spells simply fill the entire volume without regard for what’s in it, or where any point is in relation to another or the existence of anything between them. Which is kind of nonsensical for some spells, but appropriate for others, so we should look to individual spell descriptions creating exceptions for those spells that actually want effects to come down from above. (Of course, I imagine, a great many authors of individual spells didn’t double-check the default rules and think through these concerns, and probably many spells that should have such an exception lack it—that’s where the DM comes in.)

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## closed form expressions – Why some kinds of elementary transcendental equations cannot be solved by elementary functions – Conjecture and proof

Are my conjecture and proof below correct and well formulated?

How can they be improved?

It’s a new conjecture. It’s an extension of Lin’s theorem in (Lin 1983) which is used in (Chow 1999) – see both references at the bottom of this page.

The proof is very very simple because the preconditions of my conjecture contain the preconditions of Lin’s theorem except irreducibility. The irreducibility follows directly from very simple arguments from elementary algebra. But I need qualified confirmations.

$$mathbb{L}$$ are the elementary numbers, or Liouvillian numbers:
http://mathworld.wolfram.com/LiouvillianNumber.html, https://en.wikipedia.org/wiki/Elementary_number, http://mathworld.wolfram.com/ElementaryNumber.html, (Chow 1999).


Theorem (Lin 1983):
If Schanuel’s conjecture is true and $$tilde{P}(X,Y)inoverline{mathbb{Q}}(X,Y)$$ is an irreducible polynomial involving both $$X$$ and $$Y$$ and $$tilde{P}left(alpha,e^{alpha}right)=0$$ for some non-zero $$alpha$$ in $$mathbb{C}$$, then $$alpha$$ is not in $$mathbb{L}$$.


Conjecture:
Let
$$f$$ be an elementary function,
$$ninmathbb{N}_{ge 1}$$,
$$P(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x)cupoverline{mathbb{Q}}(y))$$ irreducible over $$mathbb{C}$$,
$$Q(x,y)inoverline{mathbb{Q}}(x,y)$$ not zero,
$$p(x)inoverline{mathbb{Q}}(x)$$,
$$q(x)inoverline{mathbb{Q}}(x)$$ not zero,
$$R(x,y)=frac{P(x,y)}{Q(x,y)cdot q(x)^n}$$,
$$r(x)=frac{p(x)}{q(x)}$$ not constant
so that
$$P(x,y)$$ and $$Q(x,y)$$ coprime over $$mathbb{C}$$,
$$p(x)$$ and $$q(x)$$ coprime over $$mathbb{C}$$.
Suppose Schanuel’s conjecture is true.
If a $$tilde{P}(x,y)inoverline{mathbb{Q}}(x,y)setminus(overline{mathbb{Q}}(x) cup overline{mathbb{Q}}(y))$$ of $$degree_x=n$$ with $$frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$$ exists and $$R(f(z_0),e^{r(f(z_0))})=0$$ for $$z_0inmathbb{C}$$ and $$r(f(z_0))neq 0$$, then $$z_0$$ is not an elementary number.


Proof draft:
We will prove our conjecture using Lin’s theorem. Lin’s theorem makes statements about equations of the form $$tilde{P}(alpha,e^alpha)=0$$, wherein $$alphainmathbb{C}$$ and $$tilde{P}(x,y)$$ is irreducible over $$mathbb{C}$$. We consider the equation $$R(f(z_0),e^{r(f(z_0))})=0$$ from our conjecture and will show that under the preconditions of our conjecture there exists an equation equivalent to it in the form from Lin’s theorem.

Setting $$f(z_0)=x$$ and $$e^{r(f(z_0))}=y$$ gives the equation $$R(x, y)=0$$. We consider it in the form $$frac{P(x, y)}{Q(x, y)cdot q(x)^n}=0$$ from the preconditions of our conjecture. Because $$P(x,y)$$ and $$Q(x,y)$$ are coprime over $$mathbb{C}$$ by precondition, we can multiply both sides of the equation by $$Q(x,y)$$ and get the equation $$frac{P(x,y)}{q(x)^n}=0$$ equivalent to it. By precondition, $$frac{P(x,y)}{{q(x)}^n}=tilde{P}(r(x),y)$$ holds. So our equation becomes $$tilde{P}(r(x),y)=0$$. Backsubstitution of $$x$$ and $$y$$ gives $$tilde{P}(r(f(z_0)),e^r(f(z_0)))=0$$.
We have $$forall z_0inmathbb{C}colon r(f(z_0))inmathbb{C}$$. We can therefore set $$r(f(z_0))=alpha$$, which gives the form $$tilde{P}(alpha,e^alpha)=0$$ from Lin’s theorem. To fulfill all preconditions from Lin’s theorem, we still have to show that $$tilde{P}(x,y)$$ is irreducible over $$mathbb {C}$$.
If $$tilde{P}(x,y)$$ is reducible over $$mathbb{C}$$, then
$$n_1inmathbb{N}_{ge 1}$$,
$$n_2inmathbb{N}_0$$,
$$a_0,…,a_{n_1}inmathbb{C}(y)$$,
$$b_0,…,b_{n_2}inmathbb{C}(y)$$ not all constant
exist so that
$$a_{n_1},b_{n_2}neq 0$$,
$$tilde{P}(x,y)=(a_0+a_1x^1+…+a_{n_1}x^{n_1}) cdot(b_0+b_1x^1+…+b_{n_2}x^{n_2})$$.
The highest degree regarding $$x$$ of the polynomial product on the right-hand side of the equation is $$n_1+n_2$$. This must be equal to the highest degree regarding $$x$$ of the polynomial $$tilde{P}(x,y)$$ on the left-hand side of the equation. This was $$n$$ according to the preconditions of our conjecture. We have therefore $$n_1+n_2=n$$.
By substitution of $$x$$ by $$r(x)$$, we get

$$tilde{P}(r(x),y))=(a_0+a_1r(x)^1+…+a_{n_1}r(x)^{n_1})cdot(b_0+b_1r(x)^1+…+b_{n_2}r(x)^{n_2}).$$

Because, according to the preconditions of our conjecture, $$r(x)=frac{p(x)}{q(x)}$$, and $$p(x)$$ and $$q(x)$$ are coprime over $$mathbb{C}$$, we have

$$tilde{P}(r(x),y))=(a_0+a_1left(frac{p(x)}{q(x)}right)^1+…+a_{n_1}left(frac{p(x)}{q(x)}right)^{n_1})cdot(b_0+b_1left(frac{p(x)}{q(x)}right)^1+…+b_{n_2}left(frac{p(x)}{q(x)}right)^{n_2})$$

$$=(a_0+a_1frac{p(x)^1}{q(x)^1}+…+a_{n_1}frac{p(x)^{n_1}}{q(x)^{n_1}})cdot(b_0+b_1frac{p(x)^1}{q(x)^1}+…+b_{n_2}frac{p(x)^{n_2}}{q(x)^{n_2}})$$

$$=(a_0+frac{a_1p(x)^1}{q(x)^1}+…+frac{a_{n_1}p(x)^{n_1}}{q(x)^{n_1}})cdot(b_0+frac{b_1p(x)^1}{q(x)^1}+…+frac{b_{n_2}p(x)^{n_2}}{q(x)^{n_2}})$$

$$=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})}{q(x)^{n_1}}cdotfrac{(b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^{n_2}}$$

$$=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^{n_1+n_2}}.$$

Because, as stated above, $$n_1+n_2=n$$, we have

$$tilde{P}(r(x),y))=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

According to the preconditions of our conjecture, $$tilde{P}(r(x),y))=frac{P(x,y)}{q(x)^n}$$. We have therefore

$$frac{P(x,y)}{q(x)^n}=frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$

Multiplying by $$q(x)^n$$ yields

$$P(x,y)=(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+…+a_{n_1}p(x)^{n_1})cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+…+b_{n_2}p(x)^{n_2}).$$

If $$tilde{P}(x,y)$$ is reducible over $$mathbb{C}$$, then $$P(x,y)$$ is accordingly reducible over $$mathbb{C}$$. Because, according to the preconditions of our conjecture, $$P(x,y)$$ is irreduible over $$mathbb{C}$$, $$tilde{P}(x,y)$$ is irreducible over $$mathbb{C}$$.
Thereby all preconditions from Lin’s theorem are fulfilled and we can apply it to our equation. According to Lin’s theorem, $$alpha$$ is not in $$mathbb{L}$$. Because $$alpha=r(f(z_0))$$, $$r(f(z_0))notinmathbb{L}$$. Because $$r$$ and $$f$$ are elementary functions, $$r(f(z))inmathbb{L}$$ holds for all $$zcolon zin dom(rcirc f) land zinmathbb{L}$$. But because, as just stated, $$r(f(z_0))notinmathbb{L}$$, $$z_0notinmathbb{L}$$, so $$z_0$$ is not an elementary number.
q.e.d.


(Lin 1983) Ferng-Ching Lin: Schanuel’s Conjecture Implies Ritt’s Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
(Chow 1999) Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448