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Including this exercise into training sessions for your client or athlete is a MUST if you want to produce RESULTS!
Why do you think so many professional sports teams, athletes, high-level trainers, coaches and the military include BURPEES in their workouts?
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It is also one of the best exercises to build conditioning, and the strength to help prevent injuries and perform at optimal level.
The problem is, this exercise done over and over again, will get boring, monotonous and soon become the movement your clients or athletes LOVE to HATE, if they don't already.
3. The ever so popular comment that fly through the air "Burpees, do we have to? I hate Burpees!
You know how effective burpees are for you clients, but the challenge is how can you continually include them into your training sessions without your clients suffering from BURN OUT or worse leaving you for another trainer.
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Yes I know…you can always increase the reps or have them do burpees longer to make it more challenging, but that's just not fun.
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Variety - Doing the same old burpees over and over again can get boring and soon enough burn out can occur. Most people stop a fitness program because they become BORED.
Implementing a variety of new burpee variations will help break up the monotony and make your session even more exciting. Even if you change your workouts every day, which I'm sure you do, there is still a chance that they will feel "burned out". More variety you incorporate into your session the better the chance of clients returning on a reoccurring basis and new clients signing up.
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All workout videos, eBooks and information is secured on the Member-Only website. This will ensure that only those that have purchased the program can access the information.
What do you get when you combine the ultimate exercise with the most popular 4-minute workout ever? Tabata Finishers "Attack of the Burpees" program. Use these 24 Tabata bodyweight burpee circuits with an additional 40+ new burpee variations to continue to keep things new and exciting for your clients while getting results fast!
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FTC Legal Disclaimer: Results may vary, and testimonials are not claimed to represent typical results. All testimonials are real, and all the men and women pictured transformed their body through using Funk Roberts Fitness programs. However, these results are meant as a showcase of what the best, most motivated clients have done and should not be taken as average or typical results.
Funk Roberts is not a doctor, and his advice is not a substitute for medical advice. Consult your physician before beginning any exercise or nutrition program.


Space and place has become a popular study in recent years. It can be found in many titles of seminar papers, monographs, articles, and museum/library exhibitions. However, as an analytical category 'space' is a recent innovation for the History discipline, falling behind many other social sciences who took the category up much earlier.
In this paper Stuart Minson surveys the changing concepts and variety of thoughts about 'space' with particular mention of the early modern urban space. Even the meaning of the word varies considerably. China and the West have very different views of what space constitutes.
Minson looks at the meanings given to space, whether these are practical, spiritual or invisible. He asks what relationships humans have to spaces and how these influence and control actions. Certain spaces, for example encourage certain actions over others (you would act differently in a work place than a home or a café for example). These spaces are always open to reinterpretation, although there are always some – either official or unofficial – who attempt to retain the status quo when this happens. Space therefore can be a place where a variety of social interests are played out.
In the second half of the paper, Minson turns to the question of where this concept of space came from. This is a look at Max Weber, Carl Friedrich Gauss and others as well as an examination of the term in regards specifically to urban landscape studies which itself is routed in cultural geography. Most historical work, Minson argues, still continues to follow traditional and in some cases outdated models from geography and anthropology, when it should be increasingly focusing on synthesising the two way relationship between people and space. Thus, for Minson, the study of space and place in the History discipline still needs much work.
on 23 October 2012 (Digital History Seminar).
Just a quick reminder of tonight's live streamed seminar from the Digital History seminar. Tonight we will be hearing from Luke Blaxill from King's College London on the topic of 'Quantifying the Language of British Politics, 1880-1914'. As usual this talk will focus on the digital side of the research.
If you wish to follow on the live stream you will also have the opportunity to take part in the post-paper discussion to ask questions of the speaker either through Twitter (hashtag #dhist) or through History SPOT's chat function.
The Festival of Britain held in 1951 is generally described as a rather insular affair focused on Britain without much consideration beyond the borders. Most of the landmarks created for the festival were soon afterwards destroyed (apparently for ideological reasons by Churchill's government), although the Royal Festival Hall survived, as did the Miss World competition (then properly entitled the Festival Bikini-girl competition). It is not to these aspects, however, that Iain Wilton looks. Instead he examines the generally unmentioned role of sport during the festival and how such a study suggests that the festival was not so inward looking as is generally claimed.
Although there is barely any mention of sport in the official festival report (written soon after the end of the festival) there was nevertheless a variety of matches and events played across the country as part of the festival. Some of these, such as cricket and woman's netball were indeed insular affairs, focusing only upon England, Scotland, Wales, and Ireland, perhaps with some input from Australia. Other games, however, were more international. Women's Hockey included several European nations outside of the Empire, while cycling events included seven other nations. Football included the local derby's' but also held English club matches against foreign clubs, as well as truly international matches. In the late 1940s England had time and again shown to be a lacklustre force in international football, largely because of its focus on playing European teams. However, the festival offered them an opportunity to put themselves against teams further afield. For example, 1951 saw the very first encounter between England and Argentina (Wilton doesn't mention the final score, although I'm sure many football fans will already know that England won 2-1). Scotland too played various matches and performed well.
The Argentina game might also have been arranged in-part for political relations. 1950s Britain was still a rationed nation whilst the Argentine market promised cheap meat exports. Iain Wilton suggests that the match was therefore partially organised to build relations. In all, Wilton argues that sport is the missing piece of the 1951 Festival of Britain that truly makes it a national event and suggests that not all elements of the festival were as inward looking as is often suggested.
The focus of this seminar by Stephen Duane Dean jr is to address the impact of firearms, such as the Queen Anne musket on inter-communal violence in Armagh and the rest of Ireland, and factors which catalysed their growing ownership.
a quickly expanding population and little peaceful interaction between the Catholic and Protestant communities.
Firearms were a catalyst for violence in Irish communities, despite laws preventing Catholics from owning them. The law was not being enforced from the top except sporadic ineffective clampdowns, and as a result was attempted to be done locally which fractured communities. Furthermore checks for firearms by locals were mainly used as a cover for looting Catholic houses, which led to Catholics attempting to defend themselves and their property.
Was Armagh special? Stephen uses this case study to analyse the impact of firearms on an area with a pre-existing tradition of faction fighting between religious groups. Industry and employment led to increased expansion of populations of rival religious groups with overlapping religious days. Prosperity added to conflict as when the poor had money leftover they spent more time and money on alcohol and weaponry.
Please join us on the 23rd October for the second seminar from the Digital History seminar. As per usual the live stream will include options to Tweet (using the #dhist tag) and chat on the chat feature. Both options allow you to ask questions of the speaker. The History SPOT live stream is now only available as pop outs from the podcasts page. Please let us know what you think of this change and if there are any ways we can improve things for you.
This paper explores the power, potential, and challenges of studying historical political speeches using a specially constructed multi-million word corpus via quantitative computer software. The techniques used – inspired particularly by Corpus Linguists – are almost entirely novel in the field of political history, an area where research into language is conducted nearly exclusively qualitatively. The paper argues that a corpus gives us the crucial ability to investigate matters of historical interest (e.g. the political rhetoric of imperialism, Ireland, and class) in a more empirical and systematic manner, giving us the capacity to measure scope, typicality, and power in a massive text like a national general election campaign which it would be impossible to read in entirety.
Since the 1980s house size has increased by around 50% in Sydney, Australia. This is a large increase and brings with it a rapid change of lifestyle. Robyn Dowling examines what happens in these enlarged spaces and what makes them a home. She also investigates materiality and the impact on the family unit that greater space provokes.
The study is based around interviews carried out in the early 2000s and focuses on a study of 26 Sydney mass-produced houses. These mass-produced houses are increasingly popular in Australia and in most cases are designed with the nuclear family in mind. The interviewees were found to be largely in their thirties with children under the age of twelve. Most were middle class and natural born Australians.
So what did Dowling discover? Well the most significant change is in the fracturing of the household. With increased room to create individual spaces the needs of each family member is given greater focus over that of the whole. The expectation has shifted from a family-focused unit to individual privacy, with specific areas designated for quality family time.
This transformation of the family unit brings Dowling to look into methods of parenting in these enlarged spaces and the perception surrounding homeliness (such as notions of comfort and relaxation). A home is made up from a mixture of standard possessions and personal touches including family heirlooms. Identification of status and class (including demonstrating wealth through material accumulation and display) is also still highly sought in the design of rooms. As a final point, Dowling notes that despite modern lifestyle, women still remain the centre of the home.


Send this cute, loving greeting to someone you care about. Say what it is easily forgotten with a delicious surprise of sweets and refreshing bubbly, and Molly, our new 15-cm teddy bear. The gift we deliver will resemble as closely as possible the one in the picture. The base may vary depending on the product range of the individual shop. This also applies to the varieties of flowers.


text:
Q: How to solve query janusgraph by gremlin with Java takes too much time? I have 500 vertices and 400 edges in remote graph created with janusgraph and I use tinkerpop.gremlin to search. But opening a GraphTraversalSource and closing it tends to be time consuming.For example, if a neo4j's query takes 200ms, the corresponding operation in janusgraph takes about 4 minutes.
I've searched Google and official documentations but did not find anything that could solve the issue I face.
Here is a sample code:
public class ALittleTest {
    public static void main(String[] args) throws Exception{

        double startTime =  System.currentTimeMillis();

        String objectID = "xxxxxxxxxxxxxxx";

        // open graph
        Graph graph = EmptyGraph.instance();
        GraphTraversalSource g = graph.traversal().withRemote("remote-graph.properties");

        double makeGraphTime =  System.currentTimeMillis();
        // query
        Vertex rootVertex = g.V().has("ID", objectID).next();
        double queryGraphTime =  System.currentTimeMillis();

        // close graph
        g.close();
        double endTime =  System.currentTimeMillis();

        double queryTime = (queryGraphTime - makeGraphTime)/1000.00;
        double usedTime = (endTime - startTime)/1000.00;

        System.out.println(queryTime);
        System.out.println(usedTime);
    }
}

the result of code above is:
"queryTime":0.538,"usedTime":4.307.

It seems that open the connection with gremlin-server and close it takes a lot of time. How can we improve the performance?

A: Creation and destruction of the connection to the server is expensive. I don't think there is anything you can do to specifically speed up either of those processes. Of course, you really shouldn't be creating and destroying connections with any regularity. You should create and re-use the connection you make to the server for the life of your application. That said, four minutes even with opening and closing the connection seems outside of what I would normally expect for execution time, so there may be other issues at play.

redpajama_set_name: RedPajamaStackExchange


text:
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    <item android:id="@android:id/background">
        <shape>
            <solid android:color="@color/lightest"></solid>
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    </item>
    <item android:id="@android:id/progress">
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            <shape>
                <solid android:color="@color/primary"></solid>
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redpajama_set_name: RedPajamaGithub


text:
Huguette Bourgeois est une poétesse acadienne née en 1949 à Rogersville, au Nouveau-Brunswick (Canada).

Biographie
Huguette Bourgeois naît le  à Rogersville, au Nouveau-Brunswick, et grandit à Saint-Louis-de-Kent. Elle s'inscrit à l'Université de Moncton, où elle obtient successivement un baccalauréat en littérature française, un autre en pédagogie et finalement une maîtrise en littérature française. Elle part en Colombie-Britannique, où elle enseigne le français durant plusieurs années. Elle est ensuite propriétaire d'une entreprise de services linguistique au même endroit. Elle revient au Nouveau-Brunswick, où elle est embauchée comme conseillère pédagogique au campus de Shippagan de l'Université de Moncton. Elle devient coordonnatrice des services linguistique de l'Université d'Ottawa en 2006.

Dans ses poèmes, Huguette Bourgeois recherche la musicalité et l'expression des sentiments. Au fil du temps, elle tend vers l'essentiel en ce qui a trait au sujet. Elle publie Les Rumeurs de l'amour en 1984, dans lequel la confrontation entre l'amour et la mort fait naître la tristesse. L'Enfant-fleur (1987) évoque l'enfance, au moyen de courts poèmes faisant apparaître la quotidienneté. Le recueil remporte le prix France-Acadie. Espaces libres poursuit la quête de simplicité d'Huguette Bourgeois.

Notes et références

Voir aussi

Bibliographie

Naissance en 1949
Naissance au Nouveau-Brunswick
Personnalité liée au comté de Kent (Nouveau-Brunswick)
Étudiant de l'Université de Moncton
Professeur acadien
Franco-colombien
Poétesse acadienne
Poétesse néo-brunswickoise

redpajama_set_name: RedPajamaWikipedia


How were these Silicon Valley VCs so wrong about Facebook's IPO?
Before Friday, almost every Silicon Valley insider was saying Facebook stock would finish its first day between $50 and $60 per share.
In reality, the share price at the closing bell was a pathetic $38.37.
How were these masters of the universe — some of them the most powerful investors on the West Coast — so woefully inaccurate in their predictions?
Facebook closing price predictions were coming in fast and furious over the past several weeks, but as Facebook's pre-trading pop fizzled, estimates dwindled lower and lower.
Still, few thought the price would dip as low as it did — almost a full $4 below the $42 opening price and just 37 cents higher than Facebook's offering price.
Here's a graph we made to name and shame (just kidding … sorta) those who aimed way, way too high on Facebook's value in the open market.
Interestingly, in the week before the IPO, all the Wall Street analyst types we talked to were not nearly as optimistic as their tech-obsessed counterparts on the West Coast. As the opening bell rang, we published their conservative estimates — low $40s, maybe $44 by the end of the year.
Of all the Valley dudes and dudettes we called, however, only one man gave an accurate estimate of Facebook's closing price. Ladies and gentlemen, the winner of this guess-how-many-jellybeans-are-in-Facebook's-jar contest is … Menlo Ventures managing director Mark Siegel!
When we floated the $60 figure by him, he practically snurffled his coffee in surprise, saying there was "no way, not a chance" Facebook's numbers would get that high any time within the next year or so.
Siegel bought his Facebook shares today, as did fellow Menlo Ventures VC Shervin Pishevar (who guessed $63) and Chris Sacca of Lowercase Capital (who guessed $56).
But for those of you who didn't hit your closing-price target Friday, don't dismay. Facebook still has a long way to go, and we're just at the beginning in almost every way.


Stoneheadbikes came into existence for the sheer passion for traveling in a snazzy way. What's better than hitting the road on your own terms at affordable prices? Whether you are a leisure traveler, a roadie or a sightseer for whom capturing breathtaking moments wouldn't go amiss, Stoneheadbikes has got your back !
Since its inception in August 2009, Stoneheadbikes has made a prominent name in motorbike hire in India. We love spreading the passion for traveling and as a bunch of explorers ourselves we are fixated with wheels and know how much having an astounding experience matter to you. Whether you love to jump on the bikes with your backpacks or hop into a Self driven four wheeler with your spouse and toddlers or choose to paddle around with a bunch of mates on a bicycle, you will be spoilt for choices at Stoneheadbikes. We are specialized in customized India bike tours, bike self guided, tailor made and escorted tours & self driven car rentals to enkindle more exhilarating and more rewarding travel experiences.
At Stoneheadbikes, we do not have old junk heaps, but regularly maintained and conditioned fleet that doesn't leave you stranded halfway. With more than 165 bikes, 20 bicycles and 10 cars with a whole gamut of models, you can choose what fits you the best or let us help you make a wise decision for an unprecedented and comfy traveling experience. No matter if you are looking for an individual motorbike rental or corporate leasing, we are just a call or click away.
We are a sought after bike rental and car rental enterprise pan operational for bike rental in India, majorly in Delhi, Mumbai, Pune, Chandigarh, Dehradun, Jaipur, Leh, Bengaluru and Manali. Stonehead bikes renders easy and swift booking and payment options with pick up and door-step delivery making traversing a breeze for you. If you are new to the region and want us to guide you through, Stoneheadbikes folks will be more than happy to help you with custom itineraries, bike tours and all other assistance you need before you head off.
Count on us for the most competitive pricing for bike rental in India and guided motorbike tours.
Custom & daily India bike tours designed especially for you and your mates. Just drop us a line, call or drop by.
Hassle-free hotel bookings at competitive prices ensuring a lavish and memorable stay. And yes the coffee is on us !
24X7 customer support and experienced & pleasant staff at your back and call at all times.
We are simply fond of our fleet. As a result you have regularly serviced and well maintained bikes and cars at your disposal at all times. We did not become one of the most sought after bike rental companies in India.
Tailor made itineraries just for you. Unleash the explorer in you and don't forget to take along that camera. Gear up for guided motorbike tours.


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text:
\section{Introduction}
In differential geometry, the Ricci flow is an intrinsic geometric flow. It is a process that deforms the metric of a Riemannian manifold in a way formally analogous to the diffusion of heat.

In 1982, R. Hamilton introduced a nonlinear evolution equation for Riemannian metrics with the aim of finding canonical metrics on manifolds (see \cite{SB} or \cite{Ha}). This evolution equation is known as the Ricci flow, and it has since been used widely and with great success, most notably in Perelman's solution of the Poincaré conjecture. Furthermore, several convergence theorems have been established.

Bryant \cite{BRYANT} proved that there exists a
complete, steady, gradient Ricci soliton that is spherically symmetric for any
$n\geq 3$, which is known as Bryant's soliton. In the bi-dimensional case, this solution was obtained explicitly and it is known as the Hamilton cigar.

Recently, Cao-Chen \cite{CAOCHEN} showed that any complete, steady, gradient Ricci soliton that is locally conformally flat up to homothety, is either flat or isometric to Bryant's soliton. Complete, conformally flat, shrinking gradient solitons have been characterized as being quotients of $\mathbb{R}^n$, $S^n$ or $R\times S^{n-1}$ (see \cite{FG}).

Motivated by the notion of Ricci solitons on a semi-Riemannian manifold $(M^{n},g)$, $n\geq3$, it is natural to consider geometric flows of the following type:

\begin{equation}\label{Bouguignon}
\frac{\partial}{\partial t}g(t)=-2(Ric-\rho Kg)
\end{equation}
for $\rho\in\mathbb{R}$, $\rho\neq0$, as in \cite{CA}. We call these the  Ricci-Bourguignon flows. We notice that  short time existence for the geometric flows described in \eqref{Bouguignon} is provided
in (\cite{CCD}).  Associated to the flows, we have the following notion of gradient $\rho$-Einstein solitons, which generate self-similar solutions:

\begin{definition} Let $\left(M^{n},g\right), n\geq3$, be a Riemannian manifold and let $\rho\in \mathbb{R},\rho\neq0$. We say that $(M^{n},g)$ is a gradient $\rho-$Einstein soliton if there exists a smooth function $h:M\longrightarrow\mathbb{R}$, such that the metric $g$ satisfies the equation
	\begin{equation}\label{def. 1}
	Ric_{g}+Hess_{g}h=\rho K_{g}g +\lambda g
	\end{equation}
	for some constant $\lambda\in\mathbb{R}$, where $K_{g}$ is the scalar curvature of the metric $g$.
\end{definition}

A $\rho$-Einstein soliton is said to be shrinking, steady, or expanding  if $\lambda>0$, $\lambda=0$, or $\lambda<0$, respectively. Furthermore, a $\rho$-Einstein soliton is said to be a gradient Einstein soliton, gradient traceless Ricci soliton, and gradient Schouten soliton if $\rho=\frac{1}{2}$, $\rho=\frac{1}{n}$, and $\rho=\frac{1}{2(n-1)}$, respectively.

In \cite{CA}, the authors studied $\rho-$Einstein solitons and obtained important rigidity results, proving that every compact gradient Einstein, Schouten, or traceless Ricci soliton, is trivial.

In \cite{IL}, the authors considered $\rho$-Einstein solitons that are conformal to a
pseudo-Euclidean space and invariant under the action of the pseudo-orthogonal group. We proved all the solutions for the gradient Schouten soliton case. Moreover, we proved that if a gradient Schouten soliton is both complete, conformal to a Euclidean metric, and rotationally symmetric, then it is isometric to $\mathbb{S}^{n-1}\times\mathbb{R}$.

The objective of this paper is to generalize the results obtained in \cite{IL}, considering  a $\rho$-Einstein soliton as being product manifolds of type $M= \left(B^n, g^{*}\right), \times \left(F^m,g_F\right)$, where $\left(B^n,g^{*}\right)$ is conformal to a pseudo-Euclidean space and invariant under the action of the pseudo-orthogonal group, $\left(F^m,g_{F}\right)$ is Einstein manifold. In addition, we will consider that the potential function is defined only in $B^n$.

More precisely, let $(\mathbb{R}^{n},g)$ be the standard pseudo-Euclidean space with metric $g$ and coordinates $(x_{1},...,x_{n})$, with $g_{ij}=\delta_{ij}\varepsilon_{i}$, $1\leq i,j \leq n$, where $\delta_{ij}$ is the Kronecker delta, and $\varepsilon_{i}=\pm1$, with at least one $\varepsilon_{i}$ equal to one. Let $r=\sum_{i=1}^{n}\varepsilon_{i}x_{i}^{2}$ be a basic invariant for an $(n-1)-$dimensional
pseudo-orthogonal group. Let us found the necessary and sufficient conditions for the existence of differentiable functions $\psi(r)$ and $h(r)$ such that the metric $\tilde{g}=g^{*}+g_{F}$, where $g^{*}=\frac{g}{\psi^{2}}$ satisfies the equation
\begin{align}\label{0000}
Ric_{\tilde{g}}+Hess_{\tilde{g}}h=\rho K_{\tilde{g}}\tilde{g} +\tilde{\lambda} \tilde{g}
\end{align}
with $\rho=1/2(n-1)$.

In the proof of the results, we will consider product manifolds of type $\left(\mathbb{R}^n,g^{*}\right)\times F^m$ as being a particular case of the warped product $\left(\mathbb{R}^n,g^{*}\right)\times_{f} F^m$ considering $f \equiv 1$. For more details, see \cite{O'neil}.

We initially find a system of differential equations, which functions $h$ and $\psi$ must satisfy, so that the metric $\tilde{g}=g^{*}+g_{F}$ satisfies \eqref{0000} (see Theorem \ref{theorem 1}).

Note that if the solutions are invariant under the action of the pseudo-orthogonal group, the system of partial differential equations given in Theorem \ref{theorem 1} can be transformed into a system of ordinary differential equations (see Corollary \ref{Theorem 2}).

Making $\rho=\frac{1}{2(n-1)}$, in Lemma \ref{lema2} we found conditions necessary on $\psi(r)$
so that the problem allows for a solution. Finally, we obtain the necessary and sufficient conditions on $h(r)$ and $\psi(r)$ for the existence of one gradient Schouten soliton. In this case, all solutions are given explicitly. Moreover, in the Riemannian case, we prove that if 	$\left(\left(B^{n}, g^{*}\right)\times F^m\right)$, $n\geq3$ is a complete gradient Schouten soliton, where $\left(B^n,g^{*}\right)$ is conformal to a Euclidean metric and rotationally symmetric where $g^{*}_{ij}=\delta_{ij}/\psi^{2}$, and $F^m$ is an Einstein Riemannian manifold. Then $\left(B^{n},g^{*}\right)$ is isometric to $\mathbb{S}^{n-1}\times \mathbb{R}$, $F^m$ is a compact manifold.

In what follows, we state our main results. We denote the second order derivative of $\psi$ and $h$ by $\psi_{,x_{i}x_{j}}$ and $h_{,x_{i}x_{j}}$, respectively, with respect to $x_{i}x_{j}$.
\section{Main results}
\begin{theorem} \label{theorem 1}
	Let $(\mathbb{R}^n,g)$ be a pseudo-Euclidean space, $n\geq 3$ with coordinates $x=(x_1,\cdots, x_n)$ and $g_{ij}=\delta_{ij}\varepsilon_i$. Consider the product manifold $M = (\mathbb{R}^{n}, g^{*})\times F^{m}$ with metric tensor  $\widetilde{g} = g^{*} + g_{F}$, where $\displaystyle g^{*} =
	\dfrac{1}{\psi^{2}}g$,  $F$ is an Einstein semi-Riemannian manifold with constant Ricci curvature $\lambda_{F}$, $m\geq 1$, $\psi, h:\mathbb{R}^{n}\rightarrow \mathbb{R}$, are smooth functions. Then $M$ is a gradient $\rho$--Einstein soliton with
	\begin{equation}\label{0}
	 Ric_{\widetilde{g}} +Hess_{\widetilde{g}}h = \left(\rho K_{\tilde{g}}+\tilde{\lambda}\right) \widetilde{g}, \ \ \ \ \ \rho\in\mathbb{R},
	 \end{equation}
	if, and only if, the functions $\psi$ and $h$ satisty

	\begin{equation} \label{dpe1}
	\begin{array}{l}
	(n-2) \psi_{x_ix_j}+\psi h_{,x_ix_j}+\psi_{,x_i}h_{,x_j}+\psi_{,x_j}h_{,x_i}=0,\hspace{0.5cm} \forall  i\neq j,
	\end{array}
	\end{equation}
	\begin{equation} \label{dpe2}
	\begin{array}{l}
	 \varepsilon_i\sum\limits_{k=1}^{n}\varepsilon_k\left[\psi\psi_{,x_kx_k}-(n-1)\left(\psi_{,x_k}\right)^2-\psi\psi_{,x_k}h_{,x_k}-2(n-1)\rho\psi\psi_{,x_kx_k}+(n-1)n\rho \left(\psi_{,x_k}\right)^2\right]\\
	\displaystyle
	+\left[(n-2)\psi\psi,_{x_ix_i}+\psi^2 h,_{x_ix_i}+2\psi\psi_{,x_i} h_{,x_i}\right]=\left(\lambda_F m\rho+\tilde{\lambda}\right)\varepsilon_i, \hspace{0.5cm}
	\forall\hspace{0.5cm} i=1,\ldots,n,
	\end{array}
	\end{equation}
	\begin{equation} \label{dpe3}
	\begin{array}{l}
	 \sum\limits_{k=1}^n\varepsilon_k\left((n-1)n\rho\left(\psi_{,x_k}\right)^2-2(n-1)\rho \psi\psi_{,x_kx_k}\right)=\lambda_{F}\left(m\rho-1\right)+\tilde{\lambda}.
	 \end{array}
	\end{equation}
\end{theorem}
\begin{proof} Suppose initially that $M=\left(\mathbb{R}^n,g^{*}\right)\times F^{m}$ is a gradient $\rho$--Einstein soliton with potential function $h$, that is,
 \begin{equation}\label{14}
 Ric_{\widetilde{g}} +Hess_{\widetilde{g}}h = \left(\rho K_{\tilde{g}}+\tilde{\lambda}\right) \widetilde{g}, \ \ \ \ \ \rho\in\mathbb{R}.
\end{equation}

Let $X_{1}, X_{2},\ldots, X_{n}\in{\mathcal{L}}(\mathbb{R}^{n})$ and $Y_{1}, Y_{2}, \ldots, Y_{m}\in{\mathcal{L}}(F)$, where ${\mathcal{L}}(\mathbb{R}^{n})$ and ${\mathcal{L}}(F)$  are respectively the spaces of lifts of a vector field on $\mathbb{R}^{n}$ and $F$ to ($\mathbb{R}^{n}\times F^m$ ), then
\begin{equation} \label{Ricprodsemi}
\begin{cases}
Ric_{\widetilde{g}}(X_i,X_j) = \dfrac{(n-2)\psi_{,x_ix_j}}{\psi}, \ \forall \hspace{0.2cm}i\neq j,\hspace{0.2cm} i,j=1,\ldots,n\\
Ric_{\widetilde{g}}(X_i,X_i)=\dfrac{(n-2)\psi_{,x_ix_i}+\varepsilon_i\sum\limits_{k=1}^{n}\varepsilon_k\psi_{,x_kx_k}}{\psi}-(n-1)\varepsilon_i\sum\limits_{k=1}^n\left(\dfrac{\psi_{,x_k}}{\psi}\right)^2, \ \forall\hspace{0.2cm}\hspace{0.2cm} i=1,\ldots,n\\
Ric_{\widetilde{g}}(X_i,Y_j)=0, \ \forall i=1,\ldots,n, j=1,\ldots,m\\
Ric_{\widetilde{g}}(Y_i,Y_j)=\lambda_{F}\widetilde{g}(Y_i,Y_j), \ \forall\hspace{0.2cm} i,j=1,\ldots,m.
\end{cases}
\end{equation}

As the potential function $h:\mathbb{R}^n\longrightarrow\mathbb{R}$ is defined only on the basis, we have
\begin{equation*}
Hess_{\tilde{g}}(h)\left(X_i,X_j\right)=Hess_{g^{*}}(h)\left(X_i,X_j\right)\,\hspace{0.2cm} \forall\hspace{0.2cm} i,j=1,\ldots,n.
\end{equation*}

Therefore,
\begin{equation}\label{Hessprodsemi}
\begin{cases}
Hess_{\widetilde{g}}h(X_i,X_j) = h_{,x_{i}x_{j}}+\frac{\psi_{,x_{j}}h_{,x_{i}}}{\psi}+\frac{\psi_{,x_{i}}h_{,x_{j}}}{\psi}, \ \forall\hspace{0.2cm}i\neq j,\hspace{0.2cm} i,j=1,\ldots,n\\
Hess_{\widetilde{g}}(X_i,X_i)=h_{,,x_{i}x_{i}}+\frac{2\psi_{,x_{i}}h_{,x_{i}}}{\psi}-\varepsilon_{i}\sum_{k=1}^{n}\varepsilon_{k}\frac{\psi_{,x_{k}}h_{,x_{k}}}{\psi}, \ \hspace{0.2cm}\forall\hspace{0.2cm} \hspace{0.2cm} i=1,\ldots,n.
\end{cases}
\end{equation}

We split the proof into two cases. In the first case, we will consider the vector fields on the base and in the second one, on the fiber.

If $X_{1}, X_{2},\ldots, X_{n}\in{\mathcal{L}}(\mathbb{R}^{n})$, then to $i\neq j$, from equation \eqref{14} together with equations \eqref{Ricprodsemi} and \eqref{Hessprodsemi}, follows that
\begin{equation*}
\dfrac{(n-2)\psi_{,x_ix_j}}{\psi}+h_{,x_ix_j}+\dfrac{\psi_{,x_i}h_{,x_j}}{\psi}+\dfrac{\psi_{,x_j}h_{,x_i}}{\psi}=0.
\end{equation*}
Therefore,
\begin{equation*}
\begin{array}{l}
(n-2) \psi_{x_ix_j}+\psi h_{,x_ix_j}+\psi_{,x_i}h_{,x_j}+\psi_{,x_j}h_{,x_i}=0.
\end{array}
\end{equation*}

For $i=j$ we will need the following facts. It is well known that (see, e.g., \cite{Paul})
\begin{equation}\label{scalar}
K_{\tilde{g}}=K_{g^{*}}+K_{F}
\end{equation}
where $K_{\tilde{g}}$, $K_{g^{*}}$ and $K_{F}$ represent the scalar curvatures of $M=\left(\mathbb{R}^n,g^{*}\right)\times F^{m}$,  $\left(\mathbb{R}^n,g^{*}\right)$ and  $F^m$, respectively. Note that
\begin{equation*}
K_{g^{*}}=\sum\limits_{k=1}^n\varepsilon_k\left[2(n-1)\psi\psi_{,x_kx_k}-(n-1)n\left(\psi_{,x_k}\right)^2\right],\hspace{0.3cm} K_{F}=\lambda_{F}m.
\end{equation*}

Substituting this expression into \eqref{scalar}, we get
\begin{equation*}
K_{\tilde{g}}=\sum\limits_{k=1}^n\varepsilon_k\left[2(n-1)\psi\psi_{,x_kx_k}-(n-1)n\left(\psi_{,x_k}\right)^2\right]+\lambda_{F}m
\end{equation*}

Multiplying by $\rho$ on both sides of the above equality, we have

\begin{equation}\label{8}
\rho \left(K_{g^{*}}+K_{F}\right)=\rho\lambda_{F}m+\rho\sum\limits_{k=1}^n\varepsilon_k\left[2(n-1)\psi\psi_{,x_kx_k}-(n-1)n\left(\psi_{,x_k}\right)^2\right].
\end{equation}

Replacing the expressions found in \eqref{Ricprodsemi}, \eqref{Hessprodsemi} and \eqref{scalar} in \eqref{14}, we have
\begin{equation}\label{9}
\dfrac{(n-2)\psi_{,x_ix_i}+\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\psi_{,x_kx_k}}{\psi}-(n-1)\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\left(\dfrac{\psi_{,x_k}}{\psi}\right)^2+h_{,x_ix_i}+2\dfrac{\psi_{,x_i}h_{,x_i}}{\psi}-\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\dfrac{\psi_{,x_k}h_{,x_k}}{\psi}
\end{equation}
\begin{equation*}
=\left(\rho \left(K_{g^{*}}+K_{F}\right)+\tilde{\lambda}\right)g^{*}\left(X_i,X_i\right).
\end{equation*}

Substituting the equation \eqref{8} into \eqref{9}
 and multipying by $\psi^2$, we yields
\begin{equation*}
\begin{array}{l}
 \varepsilon_i\sum\limits_{k=1}^{n}\varepsilon_k\left[\psi\psi_{,x_kx_k}-(n-1)\left(\psi_{,x_k}\right)^2-\psi\psi_{,x_k}h_{,x_k}-2(n-1)\rho\psi\psi_{,x_kx_k}+(n-1)n\rho \left(\psi_{,x_k}\right)^2\right]\\
\displaystyle
+\left[(n-2)\psi\psi,_{x_ix_i}+\psi^2 h,_{x_ix_i}+2\psi\psi_{,x_i} h_{,x_i}\right]=\left(\lambda_F m\rho+\tilde{\lambda}\right)\varepsilon_i.
\end{array}
\end{equation*}

\item If  $Y_{1}, Y_{2}, \ldots, Y_{m}\in{\mathcal{L}}(F)$, we have

\begin{equation}\label{10}
Hess_{\tilde{g}}(h)\left(Y_i,Y_j\right)=0
\end{equation}

Substituting the equations given by \eqref{Ricprodsemi}, \eqref{8} and \eqref{10} into \eqref{14}, we obtain

\begin{equation}\label{11}
\left(\rho\lambda_{F}m+\rho\sum\limits_{k=1}^n\varepsilon_k\left(2(n-1)\psi\psi_{,x_kx_k}-(n-1)n\left(\psi_{,x_k}\right)^2\right)+\tilde{\lambda}\right)g_{F}\left(Y_i,Y_j\right)=\lambda_{F}g_{F}\left(Y_i,Y_j\right),
\end{equation}
equivalently,
\begin{equation*}
\rho\lambda_{F}m+\rho\sum\limits_{k=1}^n\varepsilon_k\left(2(n-1)\psi\psi_{,x_kx_k}-(n-1)n\left(\psi_{,x_k}\right)^2\right)+\tilde{\lambda}=\lambda_{F},
\end{equation*}
which implies
\begin{equation*}
\sum\limits_{k=1}^n\varepsilon_k\left((n-1)n\rho\left(\psi_{,x_k}\right)^2-2(n-1)\rho \psi\psi_{,x_kx_k}\right)=\lambda_{F}\left(m\rho-1\right)+\tilde{\lambda}.
\end{equation*}
\end{proof}

We want to find solutions to the system of equations (\ref{dpe1}),
(\ref{dpe2}) and (\ref{dpe3})
of the form $\psi(r)$ and $h(r)$, where   $r=\sum\limits_{k=1}^n\varepsilon_kx_{k}^2$. The following theorem reduces the system of partial differential equations (\ref{dpe1}),
(\ref{dpe2}) and (\ref{dpe3}) into a system of ordinary differential equations that must be satisfied by such solutions.

\begin{corollary}\label{Theorem 2}
	Let $( \mathbb{R}^n, g)$ be a
	pseudo-Euclidean space, $n\geq 3$, with coordinates
	$x=(x_1,\cdots, x_n)$ and $g_{ij}=\delta_{ij}\varepsilon_i$.
	Consider $M = (\mathbb{R}^{n}, g^{*})\times F^{m}$,
	where $\displaystyle g^{*} = \frac{1}{\psi^{2}}g$, $F^{m}$ an Einstein semi--Riemannian  manifold
	with constant Ricci curvature $\lambda_{F}$ and smooth
	functions $\psi(r)$ and $h(r)$, where
	$r=\sum\limits_{k=1}^n\varepsilon_kx_{k}^2$. Then $M$ is a
	gradient $\rho$--Einstein soliton with $h$ as a potential function if, and only if,
	the functions  $h$ and $\psi$ satisfy:
\begin{equation}\label{15}
(n-2)\psi''+\psi h''+2\psi' h'=0,
\end{equation}

\begin{equation}\label{16}
2\psi \left[2(n-1)(1-n\rho)\psi'+\psi h'\right]+
\end{equation}
\begin{equation*}
4r\left[\left(1-2(n-1)\rho\right)\psi\psi''+(n-1)(n\rho-1)\left(\psi'\right)^2-\psi\psi'h'\right]=\left(\lambda_{F}m\rho+\tilde{\lambda}\right).
\end{equation*}
and
\begin{equation}\label{17}
-4n(n-1)\rho\psi\psi'+4r\left[(n-1)n\rho\left(\psi'\right)^2-2(n-1)\rho\psi \psi''\right]=\lambda_{F}\left(m\rho-1\right)+\tilde{\lambda}.
\end{equation}
\end{corollary}
\begin{proof}Let $g^{*}=\psi^{-2}g$ be a conformal metric of $g$. We are assuming that $\psi(r)$
	and $h(r)$ are functions of $r$, where $r=\sum_{k=1}^{n}\varepsilon_{k}x_{k}^{2}$. Hence, we have

	\begin{equation*}
	\psi_{,x_{i}}=2\varepsilon_{i}x_{i}\psi',\hspace{0,5cm} \psi_{,x_{i}x_{i}}=4x_{i}^{2}\psi''+2\varepsilon_{i}\psi', \hspace{0,5cm} \psi_{,x_{i}x_{j}}=4\varepsilon_{i}\varepsilon_{j}x_{i}x_{j}\psi'',
	\end{equation*}
	and
	\begin{equation*}
	h_{,x_{i}}=2\varepsilon_{i}x_{i}h',\hspace{0,5cm} h_{,x_{i}x_{i}}=4x_{i}^{2}h''+2\varepsilon_{i}h', \hspace{0,5cm} h_{,x_{i}x_{j}}=4\varepsilon_{i}\varepsilon_{j}x_{i}x_{j}h''.
	\end{equation*}
Substituting these expressions into the first equation of Theorem \ref{theorem 1}, we get
\begin{equation*}
(n-2)\left(4\varepsilon_i\varepsilon_j x_ix_j\psi''\right)+\psi\left(4\varepsilon_i\varepsilon_j x_ix_jh''\right)
+\left(2\varepsilon_ix_i\psi'\right)\left(2\varepsilon_jx_jh'\right)+\left(2\varepsilon_jx_j\psi'\right)\left(2\varepsilon_ix_ih'\right)=0,
\end{equation*}
equivalently,
\begin{equation*}
4\varepsilon_i\varepsilon_j\left[(n-2)\psi''+\psi h''+2\psi' h'\right]x_ix_j=0.
\end{equation*}

Since there exists $i\neq j$, such that $x_ix_j\neq 0$, it follows that
\begin{equation*}
(n-2)\psi''+\psi h''+2\psi' h'=0
\end{equation*}

Similarly, considering the second equation of Theorem \ref{theorem 1}, we obtain
\begin{equation*}
\psi \left[(n-2)\left(4x_i^2\psi''+2\varepsilon_i\psi'\right)+\psi \left(4x_i^2h''+2\varepsilon_ih'\right)+2\left(2\varepsilon_ix_i\psi'\right)\left(2\varepsilon_ix_ih'\right)\right]+
\end{equation*}
\begin{equation*}
\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\left[\left(1-2(n-1)\rho\right)\psi\left(4x_k^2\psi''+2\varepsilon_k\psi'\right)+(n-1)(n\rho-1)\left(2\varepsilon_kx_k\psi'\right)^2\right]+
\end{equation*}
\begin{equation*}
\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\left[-\psi\left(2\varepsilon_kx_k\psi'\right)\left(2\varepsilon_kx_kh'\right)\right]
=\left(\lambda_{F}m\rho+\tilde{\lambda}\right)\varepsilon_i,
\end{equation*}
equivalently,
\begin{equation*}
4\varepsilon_i\sum\limits_{k=1}^n\varepsilon_kx_k^2\left[\left(1-2(n-1)\rho\right)\psi\psi''+(n-1)(n\rho-1)\left(\psi'\right)^2-\psi\psi'h'\right]+
\end{equation*}
\begin{equation*}
4\psi \left[(n-2)\psi''+\psi h''+2\psi'h'\right]x_i^2+2\psi \left[(n-2)\psi'+\psi h'\right]\varepsilon_i+
\end{equation*}
\begin{equation*}
2\varepsilon_i\sum\limits_{k=1}^n\varepsilon_k\left[\left(1-2(n-1)\rho\right)\psi\psi'\varepsilon_k\right]=\left(\lambda_{F}m\rho+\tilde{\lambda}\right)\varepsilon_i.
\end{equation*}
From equation \eqref{15}, it followos that
\begin{equation*}
2\psi \left[(n-2)\psi'+\psi h'+\left(1-2(n-1)\rho\right)n\psi'\right]+
\end{equation*}
\begin{equation*}
+4r\left[\left(1-2(n-1)\rho\right)\psi\psi''+(n-1)(n\rho-1)\left(\psi'\right)^2-\psi\psi'h'\right]=\left(\lambda_{F}m\rho+\tilde{\lambda}\right),
\end{equation*}
which implies,
\begin{equation*}
2\psi \left[2(n-1)(1-n\rho)\psi'+\psi h'\right]+
\end{equation*}
\begin{equation*}
4r\left[\left(1-2(n-1)\rho\right)\psi\psi''+(n-1)(n\rho-1)\left(\psi'\right)^2-\psi\psi'h'\right]=\left(\lambda_{F}m\rho+\tilde{\lambda}\right).
\end{equation*}

Finally, from the last equation of Theorem \ref{theorem 1}, we have
\begin{equation*}
\sum\limits_{k=1}^n\varepsilon_k\left[-2(n-1)\rho\psi \left(4x_k^2\psi''+2\varepsilon_k\psi'\right)+(n-1)n\rho\left(2\varepsilon_kx_k\psi'\right)^2\right]=\lambda_{F}(m\rho-1)+\tilde{\lambda},
\end{equation*}
equivalently,

\begin{equation*}
-4n(n-1)\rho\psi\psi'+4r\left[(n-1)n\rho\left(\psi'\right)^2-2(n-1)\rho\psi \psi''\right]=\lambda_{F}\left(m\rho-1\right)+\tilde{\lambda},
\end{equation*}
therefore, the proof is done.
\end{proof}

When $\rho=\frac{1}{2(n-1)}$, we obtain the following result:
\begin{corollary}\label{lemma1}
	Let $( \mathbb{R}^n, g)$ be a
pseudo-Euclidean space, $n\geq 3$, with coordinates
$x=(x_1,\cdots, x_n)$ and $g_{ij}=\delta_{ij}\varepsilon_i$.
Consider $M = (\mathbb{R}^{n}, g^{*})\times F^{m}$,
where $\displaystyle g^{*} = \frac{1}{\psi^{2}}g$, $F^{m}$ is an Einstein semi--Riemannian  manifold
with constant Ricci curvature $\lambda_{F}$ and smooth
functions $\psi(r)$ and $h(r)$, where
$r=\sum\limits_{k=1}^n\varepsilon_kx_{k}^2$. Then $M$ is a
	gradient Schouten soliton with $h$ as a potential function if, and only if,
	the functions $h$ and $\psi$ satisfy:
	\vspace{12pt}
	\begin{equation}\label{18}
		(n-2)\psi''+\psi h''+2\psi'h'=0,
	\end{equation}
	and
	\begin{center}
		\begin{equation}\label{19}
			\psi\left[\left(n-2\right)\psi'+\psi h'\right]
			+r\left[\left(2-n\right)(\psi')^{2}-2\psi \psi'h'\right]=\frac{\lambda_{F}m}{4(n-1)}+\frac{\tilde{\lambda}}{2},
		\end{equation}
	\end{center}
	and
	\begin{center}
		\begin{equation}\label{20}
		-n\psi\psi'+2r\left[\frac{n}{2}\left(\psi'\right)^2-\psi\psi''\right]=\frac{\lambda_{F}}{4(n-1)}\left(m-2n+2\right)+\frac{\tilde{\lambda}}{2}.
		\end{equation}
	\end{center}
\end{corollary}
\begin{proof} By simply inserting $\rho=\frac{1}{2(n-1)}$ into equations \eqref{19} and \eqref{20}, the result follows.
\end{proof}

The next lemma provides us with the necessary conditions for the existence of gradient Schouten solitons.

\begin{lemma}\label{lema2}
	Consider smooth functions $\psi(r)$ and $h(r)$ with $r=\sum_{i=1}^{n}\varepsilon_{i}x_{i}^{2}$. If $\psi(r)$ and $h(r)$ satisfy \eqref{18} and \eqref{19}, then $\psi(r)$ satisfies the following ordinary differential equation:
	\begin{equation}
	r\psi\psi''=r\left(\psi'\right)^{2}-\psi\psi'.
	\end{equation}
\end{lemma}
\begin{proof} Differentiating both sides of equation \eqref{19}, we get
	\begin{align*}
	\psi'\left[(n-2)\psi'+\psi h'\right]+\psi\left[(n-2)\psi''+\psi' h'+\psi h''\right]+\left[(2-n)\left(\psi'\right)^{2}-2\psi\psi'h'\right]+\\+r\left\{2(2-n)\psi'\psi''-2\left[\left(\psi'\right)^{2}h'+\psi\psi''h'+\psi\psi'h''\right]\right\}=0.
	\end{align*}

	This is equivalent to
	\begin{equation}\label{21}
	(n-2)\left(\psi'\right)^{2}+\psi\psi'h'+\left(n-2\right)\psi\psi''
	+\psi\psi'h'+\psi^{2}h''+\left(2-n\right)\left(\psi'\right)^{2}-2\psi\psi'h'+\\
	\end{equation}
	\begin{equation*}
	+2\left(2-n\right)r\psi\psi''-2r\left(\psi'\right)^{2}h'-2r\psi\psi''h'-2r\psi\psi'h''=0.
	\end{equation*}

	From \eqref{18} we obtain
	\begin{equation}\label{22}
	\psi h''=-\left[(n-2)\psi''+2\psi'h'\right].
	\end{equation}

	Substituting \eqref{22} into \eqref{21}, we obtain
	\begin{align*}
	(n-2)\psi\psi''-\psi\left[(n-2)\psi''+2\psi'h'\right]+2(2-n)r\psi'\psi''-2r\left(\psi'\right)^{2}h'-\\-2r\psi\psi''h'+2r\psi'\left[(n-2)\psi''+2\psi'h'\right]=0,
	\end{align*}
	which is equivalent to
	\begin{align*}
	2\psi\psi'h'-2r\left(\psi'\right)^{2}h'+2r\psi\psi''h'=0.
	\end{align*}

	How $h ' \ne 0$, we have that
	\begin{align*}
	r\psi\psi''=r(\psi')^{2}-\psi \psi'.
	\end{align*}

	This concludes the proof.
\end{proof}

As a consequence of Lemma \ref{lema2} we are going to prove a classification theorem for gradient schouten sotitons of type $M = (\mathbb{R}^{n}, g^{*})\times F^{m}$, where  $(\mathbb{R}^{n}, g^{*})$ is  conformal to a pseudo-Euclidean space and invariant by the action of the pseudo-orthogonal group and $F^m$ is an Einstein manifold.

\begin{theorem}\label{theorem 4}
	Let $( \mathbb{R}^n, g)$ be a
	pseudo-Euclidean space, $n\geq 3$, with coordinates
	$x=(x_1,\cdots, x_n)$ and $g_{ij}=\delta_{ij}\varepsilon_i$.
	Consider $M = (\mathbb{R}^{n}, g^{*})\times F^{m}$ a semi--Riemannian manifold product,
	where $\displaystyle g^{*} = \frac{1}{\psi^{2}}g$, $F^{m}$ is an  Einstein semi--Riemannian  manifold
	with $m\geq2$, constant Ricci curvature $\lambda_{F}$ and smooth
	functions $\psi(r)$ and $h(r)$, where
	$r=\sum\limits_{k=1}^n\varepsilon_kx_{k}^2$. Then $M$ is a
	gradient Schouten soliton with $h$ as a potential function if, and only if,
	the functions $h$ and $\psi$ satisfy:

	\begin{equation}\label{23}
	\left\{\begin{array}{lcl}
	\psi(r)=k_2r,\\
	h(r)=\frac{\lambda_{F}}{2k_2^2}r^{-1}+k_1,\\
	\tilde{\lambda}=-\frac{\lambda_{F}}{2(n-1)}\left(m-2n+2\right),
	\end{array}\right.
	\end{equation}
	and
	\begin{equation*}
	\left\{\begin{array}{lcl}
	\psi(r)=k_2r^{\frac{1}{2}},\\
	h(r)=\frac{(n-2)}{8}\left(\ln r\right)^{2}+c\ln r + c_1,\\
	\tilde{\lambda}=-\frac{(m-n+1)(n-2)}{2(n-1)}k_2^2,\\
	\lambda_{F}=(n-2)k_2^2,
	\end{array}\right.
	\end{equation*}
where $c$, $c_{1}$ and $k_1\in \mathbb{R}$ and $k_2\in\mathbb{R}_{+}^{*}$.
\end{theorem}
\begin{proof} From Lemma \ref{lema2} we know that $\psi(r)=k_2r^s$ with $ k_2 >0$. Since $\psi(r)$ must satisfy equation \eqref{20}, it follows that
	\begin{equation*}
	-nk_2r^sk_2sr^{s-1}+nrk_2^2s^2r^{2s-2}-2rk_2r^sk_2s(s-1)r^{s-2}=\frac{\lambda_{F}}{4(n-1)}\left(m-2n+2\right)+\frac{\tilde{\lambda}}{2},
	\end{equation*}
	equivalently,
	\begin{equation*}
	-nk_2^2sr^{2s-1}+nk_2^2s^2r^{2s-1}-2k_2^2s(s-1)r^{2s-1}=\frac{\lambda_{F}}{4(n-1)}\left(m-2n+2\right)+\frac{\tilde{\lambda}}{2},
	\end{equation*}
which implies
	\begin{equation}\label{27}
	s(s-1)(n-2)k_2^2r^{2s-1}=\frac{\lambda_{F}}{4(n-1)}\left(m-2n+2\right)+\frac{\tilde{\lambda}}{2}.
	\end{equation}

Since the right side must be constant, we have either $s=1$ or $s=\frac{1}{2}$.

If $s=1$, we have $\tilde{\lambda}=-\frac{\lambda_{F}}{2(n-1)}\left(m-2n+2\right)$, by the fact that $\psi(r)$ and $h$ satisfy equation \eqref{19}. Note that
\begin{equation*}
\psi(r)=k_2r,\hspace{0.5cm}\psi'(r)=k_2,\hspace{0.5cm}\psi''(r)=0.
\end{equation*}
Consequently
\begin{equation*}
	(n-2)k_2^2r+k_2^2r^2h'+(2-n)k_2^2r-2k_2^2r^2h'=\frac{\lambda_{F}m}{4(n-1)}+\frac{\tilde{\lambda}}{2},
\end{equation*}
which implies,
\begin{equation*}
h(r)=\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-1}+k_1,
\end{equation*}
thus we have,
\begin{equation*}
h(r)=\frac{\lambda_{F}}{2k_2^2}r^{-1}+k_1.
\end{equation*}

Reciprocally, the function $\psi(r)$ and $h(r)$ given in \eqref{23} satisfy \eqref{18}, \eqref{19} and
\eqref{20}. Since

\begin{equation*}
h'=-\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-2},\hspace{0.5cm}h''=2\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-3}.
\end{equation*}

In this case
\begin{equation*}
(n-2)\psi''+\psi h''+2\psi'h'= 2\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)rr^{-3} +2 \left(-\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-2}\right),
\end{equation*}
equivalently,
\begin{equation*}
(n-2)\psi''+\psi h''+2\psi'h'=2\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-2}-2\left(\frac{\lambda_{F}m}{4(n-1)k_2^2}+\frac{\tilde{\lambda}}{2k_2^2}\right)r^{-2}.
\end{equation*}
Therefore $(n-2)\psi''+\psi h''+2\psi'h'=0$.

Clearly, in this case the second and third equation hold.

If $s=\frac{1}{2}$, we have by \eqref{19} and \eqref{20} that  $\frac{\lambda_{F}}{4(n-1)}\left(m-2n+2\right)+\frac{\tilde{\lambda}}{2}=-\frac{(n-2)k_2^2}{4}$ and $\frac{\lambda_{F}m}{4(n-1)}+\frac{\tilde{\lambda}}{2}=\frac{(n-2)k_2^2}{4}$.
 Consequently we obtain 	$\tilde{\lambda}=-\frac{(m-n+1)(n-2)}{2(n-1)}k_2^2$ and
 $\lambda_{F}=(n-2)k_2^2$.
  Substituting $\psi(r)=k_2r^{\frac{1}{2}}$ into \eqref{18}, we obtain $h(r)=\frac{(n-2)}{8}\left(\ln r\right)^{2}+c\ln r + c_1$, where $c, c_1\in \mathbb{R}$.

This concludes the proof of Theorem \ref{theorem 4}.
\end{proof}

As a direct consequence of  Theorem \ref{theorem 4} we get,
in the Riemannian case,  the following result of rigidity.

\begin{corollary}\label{coro1}
	Let $\left(\left(M ^{n}, g^{*}\right)\times F^m\right)$, $n\geq3$ and $m\geq 2$ be a complete gradient Schouten soliton, where $(M^n,g^{*})$ is conformal to an Euclidean metric and rotationally symmetric where $g^{*}_{ij}=\delta_{ij}/\psi^{2}$ and $F^m$ is a Einstein Riemannian manifold. Then $\left(M^{n},g^{*}\right)$ is isometric to $\mathbb{S}^{n-1}\times \mathbb{R}$, $F^m$ is a compact manifold with ricci curvature equal to $\lambda_{F}=(n-2)$ and 	$\tilde{\lambda}=-\frac{(m-n+1)(n-2)}{2(n-1)}$,\\

\end{corollary}
\begin{proof}

Except for homothety we can consider $ k_2 = 1 $ in Theorem \ref{theorem 4}.	It has been proven in \cite{IL} that $(R^n,g^{*})$,
	 $g^{*}_{ij}=\delta_{ij}/\psi^{2}$,  with $\psi(r) = r^s$, is complete only for $ s= \frac{1}{2}$. Moreover,  $\left(\mathbb{R}^{n}\setminus\{0\},g^{*}=\frac{1}{r}g_{0}\right)$ is isometric to $\mathbb{R}\times \mathbb{S}^{n-1}$. Since $\lambda_{F}=(n-2)>0$, it follows from the Bonnet--Myers Theorem that $F$ is compact.

\end{proof}

Note that depending on the choice of $m$ and $n$, $\tilde{\lambda}$ can be negative, zero or positive. In this case, it is possible to build examples of complete gradient  schouten soliton, either expanding, steady or  shrinking. We provide three examples  of gradient Schouten solitons.
\begin{example}
	The product manifold
	 $\left(\mathbb{S}^{2}\times\mathbb{R}\right)\times\left(\mathbb{S}^{2}\times\mathbb{S}^{2}\right)$, with the usual product metric, is a  complete expanding gradient Schouten soliton with $\tilde{\lambda}=-\frac{1}{2}$.
	 In fact, considering $m = 4$ and $n = 3$, and since $S^2 \times S^2$ is an  Eisntein manifold with com $\lambda_F$= 1, it meets the conditions of Theorem \ref{theorem 4}.
\end{example}

\begin{example}
The product manifold
$\left(\mathbb{S}^{n-1}\times\mathbb{R}\right)\times\mathbb{S}^{n-1}$, with the usual product metric, is a  complete steady gradient Schouten soliton with $\tilde{\lambda}=0$. In fact, considering $m =n-1 $ and since the unitary sphere $(n-1)$ - dimensional $S^{n-1}$ is an Eisntein manifold with  $\lambda_F= n-2$, it meets the conditions of Theorem \ref{theorem 4}.
\end{example}

\begin{example}
	The product manifold
	$\left(\mathbb{S}^{3}\times\mathbb{R}\right)\times\mathbb{S_R}^{2}$, with the usual product metric, is a  complete shrinking gradient Schouten soliton with $\tilde{\lambda}= \frac{1}{3}$. In fact, considering $m =2 $ and $n=4$,  and since the sphere $S_R^{2}$, with $R= \frac{\sqrt{2}}{2}$, is a  Eisntein manifold with  $\lambda_F= 2$, it meets the conditions of Theorem \ref{theorem 4}.
\end{example}

\begin{example}
	As a consequence of Corollaly \ref{coro1}  we can build examples of complete gradient Schouten  solitons that are not locally flat. To do this, just consider $((\mathbb{S}^{n-1}\times \mathbb{R}) \times F^m)$ whre $F^m$  is a  non- trivial compact Einstein manifold with ricci curvature equal to $\lambda_{F}=(n-2)$.

\end{example}

\vspace{.2in}

redpajama_set_name: RedPajamaArXiv


This ring sparkles like a vampires blood and shines like the sun, each time I see it on my finger I'm taken aback by its beauty. I got a size 6 which is larger than expected but it does fit on my pointer finger. Photo shows only part of the beauty.


I am gratified by St. Paul Mayor Melvin Carter's commitment to directly address the root causes of poverty and inequity in our community by "Making Saint Paul a City that Works for All." As a city council member representing people living in poverty on St. Paul's East Side, I welcome his goal to tackle poverty head-on in our community.
As a member of the board of directors of the non-profit Community Action Partnership of Ramsey & Washington Counties, one of the largest and oldest continually-operating poverty-fighting organizations in the Eastern Metro, I particularly appreciate the Carter administration's intentions to coordinate the work of city government with the efforts of local service providers from both the for-profit and non-profit sectors.
Mayor Carter recently unveiled plans to create an Office of Financial Empowerment that will connect our city's low-income residents to financial education and counseling, help them establish savings and build credit, and provide them with tools to achieve financial stability. The launch of the OFE, along with the initiation of a college savings account program and the establishment of a $10 million affordable housing trust fund, form a multi-faceted approach to reversing poverty in our city.
Mayor Carter's goals align perfectly with the work of Community Action, so we, the staff and board led by Executive Director Dr. Clarence Hightower, celebrate the Mayor's vision and stand ready to assist these efforts by leveraging our organization's resources. During five decades of work Community Action has learned from the experts themselves — people living in poverty — what interventions work, what services are effective, where barriers to getting ahead persist, and what is broken in our systems.
Community Action staff members have honed their knowledge and skills by serving nearly 60,000 people each year. We have created an infrastructure designed to serve. And we have been entrusted with financial resources from government agencies, foundations, and private individuals with the explicit purpose that they be invested in addressing the needs of low-income people. We look forward to supporting Mayor Carter's vital and game-changing agenda for the City of St. Paul.
Jane Prince is a member of the St. Paul City Council.


The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS is a new wide-angle zoom lens for the Sony full-frame FE-mount cameras (A7/A7R/A7S). It features Optical SteadyShot image stabilisation, has five aspherical elements including a large diameter AA (advanced aspherical) element and three ED (extra-low dispersion) elements, is dust- and moisture-resistant, and has a minimum focusing distance of 28cm with a maximum magnification of 0.19x. It has a near-circular 7 blade diaphragm which creates an attractive blur to the out-of-focus areas of the image, while an internal focusing mechanism means the lens barrel doesn't move, and a ZEISS T* anti-reflective coating minimises flare. The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens is available for around £1289 / $1350 in the UK and the US, respectively.
Weighing in at 518grams, the metal-bodied Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS feels very well-constructed. It's overall size and weight makes it feel well-balanced on the Sony A7R body that we tested it with, as shown in the photos below. Note that the length of the lens does increase slightly by 2cm when you set the focal length to 16mm.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS has a sealed dust and moisture resistant design and is therefore a natural partner for the A7/A7R/A7S cameras.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens boasts built-in Optical SteadyShot image stabilisation to help ensure clearer, sharper handheld shots. This is important as the A7/A7R/A7S bodies don't have a built-in image stabilisation system. In practice we found that the Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS reliably offered around 3-stops of compensation.
The wide and ridged zoom ring and the narrower manual focus ring are the only external controls of note. Manual focusing is possible via the textured focus ring when set on the specific camera body. The lens utilizes double linear motors to produce quiet and smooth focusing, making it well-suited to shooting video.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens has a metal lens mount. It accepts 72mm filters via plastic threads.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens is commendably supplied with both a soft case and a fairly good quality plastic petal-shaped lens hood.
At the 35mm focal length the angle of view is 44 degrees.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens has quite a wide focus ring. There are no hard stops at either end of the range, making it a little more difficult to set focus at infinity. Polariser users should be pleased that the 72mm filter thread doesn't rotate on focus.
When it comes to auto-focusing, the Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS zoom is a quiet and fairly quick performer on the Sony A7R that we tested it with, taking about 0.25 seconds to lock onto the subject.
Chromatic aberrations, typically seen as blue or purple fringes along contrasty edges, were almost completely absent from our test shots. The examples below show the worst-case scenario.
With the Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens set to its maximum aperture of f/4, there is very obvious light fall-off in the corners, requiring you to stop down by at least 3 f-stops to prevent it. There's quite a lot of barrel distortion evident at 16mm, and at 35mm there's some obvious pin-cushioning too.
The Sony Vario-Tessar T* FE 16-35 mm F4 ZA OSS isn't claimed to be a macro lens, offering a minimum focusing distance of 28cm at 16mm, with a maximum magnification of 0.19x. The following example demonstrates how close you can get to your subject at 35mm, in this case a Compact Flash memory card.
Bokeh is a word used for the out-of-focus areas of a photograph, and is usually described in qualitative terms, such as smooth / creamy / harsh etc. In the Vario-Tessar T* FE 16-35 mm F4 ZA OSS lens, Sony employed an iris diaphragm with seven rounded blades, which has resulted in quite appealing bokeh in our view. We do realise, however, that bokeh evaluation is subjective, so we've included several 100% crops for your perusal.


As part of the termination, Connaught Plaza Restaurant Ltd is ceased to use McDonalds's name, system, trademark, designs and its associated intellectual property, among others within 15 days of the termination notice.
CPRL was a 50:50 joint venture between Mr Bakshi and McDonald's India.
The popular food outlet, known for leading the expansion of American fast food in India, has seen its fortunes declining since past few years over a tussle with Mr Bakshi for the management of CPRL, after he was ousted from the post of managing director of the McDonald's franchisee in 2014.


text:
Q: How can I add a Finder Service that would use a single selected file, perform a shell script on it then display it as a dialog window? How can I add a Finder Service that would use a single selected file, perform a shell script on it then display it as a dialog window?
I currently have Automator receiving files and folders within Finder then running an Applescript with the following:
on run {input, parameters}
    tell application "Finder"

        set filename to selection
        set filename to quoted form of POSIX file filename
        set theCMD to "/usr/local/bin/exiftool "
        set theScript to theCMD & filename
        set theResult to do shell script theScript
        display dialog theResult
    end tell
end run

I continuously receive errors.  The intention is to show a dialog window with Exif metadata information for a single selected file within a Finder window.  I used brew to install exiftool to retrieve the data.
I'm new to applescript and can't figure out how to get this to work.  Any help is appreciated.

A: An Automator service passes input items to the workflow, so you don't need to do anything else to get the selection.  In a Run AppleScript action the input parameter is a list, so you should pick a specific item or just loop though all the items:
on run {input, parameters}
    repeat with anItem in the input
        set anItem to quoted form of POSIX path of anItem
        set theResult to (do shell script "/usr/local/bin/exiftool " & anItem)
        display dialog theResult
    end repeat
    # return input
end run


text:
Home»Movies»Movies by Decade»Movies - 1970s
Movies - 1970s Movies - D 2 Mins Read
Debbie Does Dallas (1978)
Debbie Benton (Bambi Woods) is the captain of her high school cheerleading squad and has been accepted to try out for the Texas Cowgirls. Her parents disapprove, though, and refuse to pay the fare to Texas.
In a bid to help Debbie, her cheerleader friends Lisa (Georgette Sanders), Roberta (Misty Winter), Tammy (Arcadia Lake), Pat (Kasey Rodgers), and Annie (Jenny Cole) decide to accompany her to Texas. They have two weeks to raise the money and form a company called Teen Services to help.
Debbie gets a job at a sports store run by Mr Greenfield (Richard Balla), Tammy takes a job in the local record store run by Tony (Tony Mansfield), Roberta convinces Mr Hardwick (Eric Edwards) to give her a job at the candle store with Mrs Hardwick (Robyn Bird), while Rikki (Sherri Tart) and Annie agree to wash Mr Bradly's car.
While working for Mr Greenfield at the sports store, Debbie is talked into allowing him to see her breasts for $10, fondle her breasts for another $10, then she lets him suck them for an additional $20.
Realising that they will not be able to raise enough money by legitimate means, Debbie convinces the other girls to engage in sexual activities for money. Cue Mr Greenfield – dressed in his old football uniform – indulging in his most carnal fetishes with Debbie on the floor of his store . . .
Debbie Does Dallas was highly successful, selling 50,000 copies when it was released on videotape.
It is regarded as one of the most important releases during the so-called "Golden Age of Porn", and remains one of the best-known pornographic films.
The Dallas Cowboys filed suit against this film for using its cheerleader uniform in the film without permission. The sequels used a different uniform.
Debbie Benton
Bambi Woods
Mr Greenfield
Richard Balla (Robert Kerman)
Georgette Sanders
Arcadia Lake
Misty Winter (Christie Ford)
Kasey Rodgers
Jenny Cole
Rikki 'Sherri Tart' O'Neal
Mr Hardwick
Eric Edwards
Mrs Hardwick
Robin Byrd
Tony Mansfield
Mr Bradley
David Pierce
Merril Townsend
Mr Biddle
Jake Teague
Paul Hughs
Peter Lerman
Ben Pierce
David Buckley
Magic (1978) Failing magician Charles "Corky" Withers (Anthony Hopkins) suddenly hits the…
Frightmare (1974) In the 1950s, Dorothy Yates (an iconic performance from Sheila…
Class of Miss MacMichael, The (1978) Selkirk is an "Approved School" where young miscreants are sent…
Railway Children, The (1970) London, Christmas Day 1904. Two men call on Charles Waterbury…
Logan's Run (1976) Logan's Run is a great example of what passed for…
Getaway, The (1972) Carter "Doc" McCoy (Steve McQueen) is about as tough as…
American Graffiti (1973) Four California teenagers get a final, nostalgic glimpse of innocence…
Home Before Midnight (1979) Mike Beresford (James Aubrey) is a successful 28-year-old lyricist for…
Previous ArticleDeath Wish (1974)
Next Article Decoy For Terror (1967)
Launched in the UK in 1953, Babycham was the first alcoholic […]
24 Hours to Kill (1965)
Tarzan star Lex Barker and Hollywood legend Mickey Rooney head a […]
1 9 5 7 – 1 9 5 9 (USA) 52 x 30 minute episodes Three […]
Dr Syn Alias the Scarecrow (1963)
Dr Syn, the vicar of Dymchurch (Patrick McGoohan), is also a […]
Kiss Me, Stupid (1964)
Dean Martin, portraying himself, is stranded in Climax, Nevada, […]
Atari's Warlords got Medieval on the old Breakout formula. […]

redpajama_set_name: RedPajamaCommonCrawl


text:
Хипобарична комора Квест (), је хипобарична комора Међународне свемирске станице (МСС) која се највише користи. У буквалном преводу са енглеског језика Quest значи трагање, потрага. Ова комора конструисана је тако да се из ње могу спроводити свемирске шетње и у америчким и у руским Орлан свемирским оделима. Комора је лансирана мисијом СТС-104 14. јула 2001. године. Пре уградње овог модула, руске шетње могле су се спроводити само из сервисног модула Звезда, док су шетње у америчким свемирским оделима могле да се спроводе само када је са станицом спојен орбитер спејс-шатла, из хипобаричне коморе коју он поседује. Касније исте године, 16. септембра, лансиран је и руски модул Пирс, који се поред за спајање летелица Сојуз и Прогрес може користити и као хипобарична комора.

Дизајн

Хипобарична комора Квест састоји се из два сегмента – први се назива "преграда за опрему" () и у њему су складиштена свемирска одела и остала опрема коју астронаути користе приликом свемирских шетњи, а у истој прегради пред саму шетњу астронаути облаче та одела и врше провере свих система; други сегмент назива се "преграда за посаду" () која је у ствари хипобарична комора у којој се атмосферски притисак може контролисати од вакуума па све до нормалног притиска који влада унутар станице, и у овај сегмент астронаути улазе након што обуку одела, затим се затвори поклопац између ова два сегмента и притисак се лагано спушта све до вакуума, након чега астронаути могу изаћи "напоље" (у отворени свемир). Као модел при дизајнирању ове коморе послужила је хипобарична комора орбитера спејс-шатла, али је тај дизајн значајно унапређен тако да се потроши знатно мање ваздуха приликом употребе. Комора шатла није била ефикасна јер су мисије трајале релативно кратко, до две седмице, тако да није био проблем ако се потроши више ваздуха; међутим, станица је константно у орбити, тако да је веома битно да се уштеди и рециклира где год је то могуће како би што мање залиха морало да се шаље са Земље. По лансирању и спајању орбитера Атлантис са станицом, Квест је уграђена на десни отвор модула Јунити. На свом спољашњем ободу има носаче за четири резервоара под високим притиском, два која садрже кисеоник и два која садрже азот, која служе за допуњавање ваздухом америчког дела МСС, тачније, из ових резервоара се надокнађује губитак притиска који настаје приликом отварања поклопца при спровођењу свемирских шетњи. На преградним вратима (поклопцу), који дели сегмент за опрему и сегмент за посаду, уграђен је мали прозор кроз који астронаути који остају унутар станице током шетње могу пратити да ли све пролази како треба током декомпресије и пред излазак астронаута у свемир, или на крају шетње да провере да ли је сва опрема и алат коју су астронаути изнели на свом месту.

На спољашњости коморе постоји велики број рукохвата које астронаути користе приликом свемирских шетњи за кретање. Један број тих рукохвата уграђен је на Земљи, док је остатак уграђен након што је модул спојен са МСС. Врата (поклопац) који се налази на прегради за посаду и кроз која астронаути излазе у свемир поседује додатан термални поклопац, односно изолациони материјал у облику поклопца који штити главни поклопац од удара микро-метеорида, али и као заштита од Сунца док се астронаути не навикну у случају да у свемир излазе у време када је станица осветљена Сунчевим зрацима.

Комора Квест била је неопходна јер америчка свемирска одела не могу да се провуку кроз врата руских хипобаричних комора, а поред тога поседују различите компоненте, начин напајања и конекторе у односу на одела Орлан. Комора Квест је дизајнирана тако да може да подржи и једна и друга свемирска одела, међутим, тренутно се из ње спороводе изласци у отворени свемир само у америчким оделима јер опрема која је потребна за руска одела Орлан није још лансирана. Ово је велика мана у слуачају да нешто није у реду са једним или другим типом свемирских одела. Када је дошло до проблема са америчким оделима, током Експедиције 9, космонаути су у руским свемирским оделима типа Орлан морали да користе хипобаричну комору на руском делу станице, а затим пређу скоро целу дужину станице како би стигли до места на којем је требало да раде.

Процедура "камповање"
Комора Квест обезбеђује окружење у којем астронаути могу да "кампују" пре свемирске шетње у атмосфери са сниженим нивоом азота како би избацили што више овог гаса из крвотока и избегли декомпресиону болест у свемирским оделима сниженог притиска (5 -{psi}-; 34 kPa) и у којима је уместо ваздуха чист кисеоник. Пре ове методе астронаути су морали неколико сати пред излазак у отворени свемир да носе маске којима су из малих боца удисали чист кисеоник и тако избацивали вишак азота из крвотока. У априлу 2006. године чланови Експедиције 12, командант Бил Макартур и летачки инжењер Џефри Вилијамс, тестирали су нови метод припрема за свемирску шетњу "кампујући" у комори Квест ноћ пре шетње.. Астронаути су се увече затворили у "преграду за опрему" у којој је затим атмосферски притисак снижен са нормалних 101 на 70 -{kPa}- (14,7 на 10,2 psi). Четири сата након што су чланови експедиције отишли на спавање огласио се аларм због којег су контролори мисије морали да прекину процедуру "камповања", али је тест проглашен успешним. Од тада се пред сваки излазак у свемир из америчког сегмента станице користи процедура "камповања".

Резервоари високог притиска

На спољашњи део модула Квест инсталирана су четири резервоара високог притиска, два са кисеоником и два са азотом. Ови резервоари се могу допуњати, и обезбеђују залихе гаса систему за регулацију атмосфере МСС (), као и кисеоник под притиском од 6,2 -{MPa}- за допуну свемирских одела.

Ови резервоари допуњавани су из орбитера спејс-шатла док је он био у употреби. По спајању орбитера са станицом, аутоматски би се оствариле везе са цевима за проток кисеоника, која су спроведена по спољашњости станице од адаптера за спајање () до коморе Квест. Након пензионисања спејс-шатла 2011. године ови резервоари се допуњавају комерцијалним бродовима за снабдевање Драгон и Сигнус.

Конструкција
Сама хипобарична комора и системи резервоара изграђени су од алуминијума и тестирани у центру за свемирске летове Маршал, у Хантсвилу (Алабама). Предузимач је била компанија Боинг.

Спецификације коморе
 Материјал: алуминијум
 Дужина: 5,5 m
 Пречник: 4 m
 Маса: 6.064 kg
 Запремина: 34 m3
 Цена: 164.000.000 $, укључујући резервоаре

Галерија

Референце

Спољашње везе
 Излазак у отворени свемир из коморе Квест, снимљено 2010. (YouTube)
 Обука на Земљи за свемирску шетњу из модула Квест, (YouTube)

Међународна свемирска станица


it's fading everywhere. What is friendship?
that a hopeless kind of vanity even for us?
Derek's poems have appeared in Portland Review, The Boiler, and Radar Poetry. His chapbook, What the Dying Man Asked Me, is available from ELJ Publications. He lives in Oklahoma.


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Putting a Price Tag on the Decision to Opt-Out of the Medicaid Expansion
Brad Wright June 28, 2013 June 28, 2013
Conservative opposition to the Affordable Care Act is old news. Even after the Supreme Court ruling last year, it was quickly established that–given the option–many traditionally Republican states would not be participating in the Medicaid expansion. I've written before about the irony that the very states choosing to opt-out of the Medicaid expansion are the ones who would benefit from it the most, because a higher proportion of their population is uninsured, their residents' health is generally worse, and the amount of federal matching funds they'd bring into their states is higher.
Conservative opposition to the Affordable Care Act is old news. Even after the Supreme Court ruling last year, it was quickly established that–given the option–many traditionally Republican states would not be participating in the Medicaid expansion. I've written before about the irony that the very states choosing to opt-out of the Medicaid expansion are the ones who would benefit from it the most, because a higher proportion of their population is uninsured, their residents' health is generally worse, and the amount of federal matching funds they'd bring into their states is higher. But new research from the RAND Corporation puts a price tag on the states' decision to opt-out and it has me wondering if what we're seeing isn't just farsighted politics at its best. Let me explain.
In a recent Health Affairs article, Carter Price and Christine Eibner report the results of a microsimulation model to predict insurance coverage and federal and state spending under several different scenarios tied to the Medicaid expansion. They find that if the 14 states currently vowing to opt-out of the Medicaid expansion keep their word, there will be a number of consequences compared to a scenario in which all states participate in the expansion. First, there will be an additional 3.6 million persons who remain uninsured. Second, the Federal government will save about $8.4 billion. That figure comes about because the Feds will spend $21.3 billion less on Medicaid (remember, fewer people are enrolled), but $11.2 billion more on subsidizing premiums for individuals who would have been eligible for Medicaid but instead ended up enrolling in the health insurance exchange, and $1.7 billion more on uncompensated care. Third, and finally, states that opt-out will lose out on $1 billion, because while their Medicaid costs decrease slightly, they will also have to pick up the tab for uncompensated care when Federal disproportionate share hospital payments are cut.
If you assume that these effects are spread evenly over the 14 non-participating states (they're not, of course), it works out to the following: $71,428,571 per state in additional state spending for 257,143 fewer persons with insurance. In other words, each state is willing to pay about $278 for a person not to be insured. According to the Kaiser Family Foundation, in 2009, the average amount paid per Medicaid enrollee per year was $5,527. Thus, by 2020, when states would be responsible for 10% of these costs, that would equal $553. The conclusion: states would rather limit their total spending to $278 per person and get nothing in return, than pay an additional $275 so that each person could have health insurance, and face the possibility that that amount could increase over time.
But I think there's a bit more to it than that. As the authors write: "The decision to expand Medicaid would not have substantial effects on state budgets prior to 2017." What we are seeing is less about math, and more about politics. First, with the mid-term elections in 2014, the GOP is hoping that the health insurance exchanges that will begin enrollment in October of this year will be a disaster and that they will pick up seats in Congress as a result. Then, with the Presidential election in 2016, the GOP is hoping that their resistance to the Medicaid expansion will have caused further disruptions to the implementation of "Obamacare" and they will be able to leverage that to win the White House. If they are successful on both counts, they may find that they are closer to repealing or seriously dismantling the Affordable Care Act. That matters to them. A lot. And they're willing to gamble away billions of dollars in Federal matching funds in the hope that doing so will reclaim the Senate and the Presidency for Republicans. If at least one of those things doesn't happen, I think you'll see at least some states have a change of heart in 2017, and opt-in to the Medicaid expansion at that later date.
image: USHealth/shutterstock
Person-Centered HealthCare: How Caregivers are Different Than Other Health Seekers
Welcome to the Affordable Care Act Call Center, Please Press One for Healthcare, Two for….
Diana Hope - January 19, 2020


text:
require 'spec_helper'
include ValidationMacros

describe Conflict do
  let(:conflict) { FactoryGirl.build(:conflict, :without_rider) }
  let(:attrs) { [ :rider_id, :start, :end, :period ] }

  subject { conflict }

  it "should respond to all attributes" do
    check_attributes conflict, attrs
  end

  describe "validation" do

    it { should be_valid }
    it "should be invalid without required attribtues" do
      check_required_attributes conflict, attrs
    end

    describe "with end before start" do
      before do
        conflict.end = 5.years.ago
      end
      it { should_not be_valid }
    end
  end

  describe "after save" do
    before { conflict.save }
    its(:period){ should_not be nil }
  end

  it "should respond to association messages" do
    check_associations conflict, [ :rider ]
  end
end


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