Situated a stone's throw from anywhere you need to be, Paddington has everything the modern professional needs.
Anyone with an eye for retail or the buzz of a thriving central area will be intrigued by the prospect of being able to rent property in Paddington. It's a high-end central area that has boutique retail opportunities at every turn, trendy eateries, and a wealth of property to satiate the most affluent of tastes.
In short, it's the city-dwellers paradise, and a real jewel for those seeking to rent property in the eastern suburbs.
Going by 2011 census data produced by the Australian Bureau of Statistics, Paddington is a haven for young professionals. More than a quarter of the suburb's population is aged between 25 and 34, and a whopping 63 per cent work 40 or more hours per week. Compared to the 45.8 per cent who do this on a statewide level, it is clear that those in Paddington want disposable income, and it shows.
Census data also shows that for families with and without children in Paddington, median income is significantly higher than state and national averages. A total of 46.7 per cent of residents work as 'professionals', further cementing the luxury lifestyle that many live.
Paddington is also established as a fashionable part of town. Emerging trends can be found here on any given weekend, be they in street food, high fashion, local wine or music. The Paddington Markets are hailed by Lonely Planet as the place to find crafted jewellery and clothing, although considerably more on the mainstream and high-end side than in days gone by – this is where many of the big names in fashion got their start.
Victorian-era townhouses dominate the landscape, and there is also a considerable amount of apartments – ideal for those who want to rent property in Paddington, or investors seeking property management in Paddington.
Placed further towards the centre of Sydney than some other areas you might rent property in eastern suburbs, Paddington is also a haven for those who need quick access to amenities, shops and their place of work. A considerable number of people don't even need to drive, preferring to take the bus or even walk to their job.
To rent property in Paddington is to move into a an area that effortlessly combines vintage style, high fashion and trending food with a classic Australian work hard play hard attitude. This blend of the old and the new means property management in eastern suburbs will always have a client base seeking to rent property in Paddington.


Relax back in a comfortable clean-lined chair whose size suits so many locations and whose beautiful solid wood frame and neutral upholstery look good in almost every interior style, from antique through eclectic to mid-century modern. Part of our popular Jasmeen collection, the upholstered armchair is handcrafted from sheesham hardwood and, employing the best in traditional Indian furniture making skills, comes from a Rajasthan workshop with whom we've enjoyed a long fair trade relationship. The chair has a slatted back and base, and piped stone-grey coloured cotton seat and back cushions. Cushion covers remove for machine washing.
Sheesham hardwood (Indian Rosewood) with a hand waxed finish.
The chair cushion upholstery is natural cotton, removable so you can upholster to match your decor should you wish. Wax the chair occasionally to enhance the grain.


Bright shopping mails, standing out against the background of the surrounding bedroom districts and industrial estates, have dramatically changed the city's face. Today they are not only a part of consumer culture, but a symbol of the first decade of the 2000s.
Built from short-term construction materials from the shambles of the Soviet past, countless shopping mails today metaphorically resemble a computer glitch brought about by the mismatch between post-Soviet social realities and the market relations of a nascent capitalism.
Otdelnov looks into the shopping mail as an architectural phenomenon, which is linked to the aesthetics of2000-s. The characteristics of this architecture are its temporality, ephemerality and the juxtaposition to the surrounding landscape.
The Mall. Glitch project is a kind of a monument to the "colourful hangars" epoch, the period of stability. Today in many cities you can already find abandoned or unfinished trade centers.
The exhibition presents a 3D visualization of the object.
The book of Konstantin Zatsepin "The space of the sight. Art of 2000–2010. Articles"–Samara, ООО Книжное издательство, 2016, 120p.


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Omicron: New Covid variant 'of concern' named by WHO
Omicron carries a higher risk of reinfection than previous strains, WHO says
Bevan Hurley,John Bowden
Saturday 27 November 2021 00:09
Fears growing around new COVID variant
The World Health Organisation has designated the new Covid-19 variant Omicron.
The variant, which originated in southern Africa, is of growing concern to public health officials who fear it may be more transmissible than previous strains and vaccine-resistant.
The United States announced on Friday afternoon it was joining European countries in closing its borders to eight African countries where the variant has been detected.
The US said it would be restricting travel to South Africa, Botswana, Eswatini, Lesotho, Mozambique, Namibia, Malawi and Zimbabwe from Monday.
Fauci says US may block travel from countries again if new Covid variant evades vaccine
CDC and White House non-committal on US travel restrictions over new Covid-19 variant
Omicron Covid variant – live: EU to halt flights as first European case reported
"President Biden has promised to take every measure necessary to keep Americans safe and defeat the pandemic, and this was a step recommended by U.S. government medical experts and the COVID-19 Response Team," a White House official said in a statement.
The ban does not apply to American citizens and lawful permanent residents, the official said.
The variant, previously known as B.1.1.529, has been identified in Israel, Hong Kong and Europe.
Belgium confirmed Europe's first case just hours after South Africa had identified the new variant, and the European Commission has now proposed that all EU member states suspend travel from countries where it has been detected.
The United Kingdom imposed travel restrictions on flights from six countries in southern Africa from midday Friday (7am EST).
"More data is needed but we're taking precautions now," UK Health Secretary Sajid Javid wrote on Twitter, adding that flights will be temporarily banned and UK travellers must quarantine.
European Commission President Ursula von der Leyen said it was vital for the continent to act "swiftly, decisively and united", and called for EU citizens to get vaccinated.
International passengers walk through the arrivals area at Terminal 5 at Heathrow Airport on 26 November where flights to six African countries have been suspended
She added: "All air travel to these countries should be suspended until we have a clearer understanding about the danger posed by this new variant."
Belgium health officials said a case had been detected in a traveller returning from Egypt on 11 November.
In the United States, the White House's chief medical adviser Anthony Fauci told CNN US public health officials were "rushing" to get data from their counterparts in South Africa to assess whether vaccines are effective against the new strain.
"We want to find out, scientist-to-scientist, exactly what is going on," said Dr Fauci, director of the National Institute of Allergy and Infectious Diseases.
Dr Fauci said the variant may already be spreading in the United States.
"Of course anything is possible, we don't know that. There is no indication that it is right now," he told CNN.
Scientists in the US were holding emergency meetings with health officials in South Africa on Friday to learn more abut the strain, Dr Fauci said.
The US could reimpose travel bans to affected countries once it learns more about the variant, Dr Fauci said.
Drs Rochelle Walensky and Anthony Fauci at a Senate hearing
Financial markets across the world tumbled on Friday on news of the new strain.
The Dow Jones plunged 2.5 per cent, it's worst day since October 2020. The S&P 500 closed down 2.3 per cent, while the Nasdaq fell 2.2 per cent.
In the UK, $96 million was wiped from the FTSE100, the index of largest 100 companies.
The European Stoxx 600 closed down almost 3.7 per cent.
Spain's IBEX fell by 5.0 per cent, France's CAC index dropped by 4.75 per cent, Germany's DAX slipped by 4.15 per cent.
The price of oil also plummeted by 9 per cent, to a new two-month low.
The pathogen is thought to have developed in an immunocompromised person whose immune system was unable to purge itself of the virus, allowing it to mutate time after time before transmission to somebody else.
In a statement, the WHO said preliminary evidence suggests the variant carries a higher risk of reinfection than other strains.
The WHO said it would take weeks to determine how effective vaccines were against the new strain, which was first identified this week.
Dr Eric Feigl-Ding, an epidemiologist and senior fellow at the Federation of American Scientists (FAS), tweeted that the variant has more than double the number of "bad spike mutations" compared with the highly transmissible Delta variant.
⚠My god—the new #B11259 variant being possibly ~500% more competitively infectious is the most staggering stat yet. Also, #NuVariant has more than >2x the number of bad spike mutations than Delta. Here's an updated 🧵👇
Model by @JPWeiland matches up with graph by @jburnmurdoch pic.twitter.com/SFvFEbD7QO
— Eric Feigl-Ding (@DrEricDing) November 26, 2021
Cases of the Delta variant were already spiking in many European countries.
New daily Covid cases surged to 50,000 in the UK, the highest one-day tally in more than a month.
In the US, cases are creeping up again to the highest levels since mid-October.
According to the Centers for Disease Control and Prevention, there were 100,000 new cases on 23 November, and the 7-day average of new cases reached 94,000.
New York Governor Kathy Hochul said the state was on "high alert".
At least 600,000 people cancelled December flights from Heathrow
What rules the public were following when a No 10 aide organised a lockdown party
Asian stocks mostly lower, eyeing Fed, China omicron cases
"We will continue to actively monitor virus samples and will continue working with the CDC and WHO to closely follow developments."
Previous Covid-19 variants have included Delta, Omega and Alpha, and take their names from letters in the Greek alphabet.
CovidCovid variantOmicron
1/3New Covid-19 variant is named Omicron by WHO
New Covid-19 variant is named Omicron by WHO
Covid variant named

redpajama_set_name: RedPajamaCommonCrawl


The book, co-authored by Dr. John Hillen and Dr. Mark Nevins, discusses what happens when an organization's growth outpaces its leaders' skills — and how to reinvent themselves.
The event was organized by WashingtonExec and held at the Tower Club in Tysons Corner. A special thanks goes to presenting sponsor MacAulay-Brown, Inc. (MacB), silver sponsor SOS International LLC, book sponsor Array Information Technology and beverage sponsor RSM.
Hillen said nearly each of the leaders at the event contributed to the book in some way.
Sid Fuchs, President and CEO of MacB, said it was wonderful to see so many of Hillen's friends and colleagues at the event. Fuchs was also the emcee for the evening and introduced Hillen to the crowd.
"If you're a leader, you can engineer your way through many of the challenges of complexity, but you'll need to reinvent yourself personally if you have lagged in developing new skills, behaviors and mindsets for a changed organization and higher level of performance," according to a synopsis of the book.
The book details seven points at which leaders often stall and how to navigate toward personal growth and organization success. Hillen is the executive in residence and professor of practice in the School of Business at George Mason University, where he and Array President and CEO Sumeet Shrivastava were connected as early drivers of the university's GovCon initiative.
Hillen, a decorated combat veteran and business leader in the national security community, has been honored with numerous business and military awards. He moderates leadership seminars for The Aspen Institute, writes a monthly column for Washington Technology and has authored several critically acclaimed books and many articles.


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Home Armenian New Eelement in Periodic Table Named After Armenian Scientist Yuri Oganessian
New Eelement in Periodic Table Named After Armenian Scientist Yuri Oganessian
Karabakh Self-Determination 'Not Negotiable' for Yerevan
ZURICH — The International Union of Pure and Applied Chemistry (IUPAC) approved the name and symbols for four elements: nihonium (Nh), moscovium (Mc), tennessine (Ts), and oganesson (Og), respectively for element 113, 115, 117, and 118. Oganesson was named in honor of Armenian scientist Yuri Oganessian.
In concordance with and following the earlier reports that the claims for discovery of these elements have been fulfilled, the discoverers have been invited to propose names. Keeping with tradition, the newly discovered elements have been named after a place or geographical region, or a scientist. The ending of the names also reflects and maintains historical and chemical consistency: "-ium" for elements 113 and 115 and as for all new elements of groups 1 to 16, "-ine" for element 117 and belonging to group 17 and "-on" for element 118 element belonging to group 18.
The name nihonium with the symbol Nh for element 113 was proposed by the discoverers at RIKEN Nishina Center for Accelerator-Based Science (Japan); the name came from Nihon which is one of the two ways to say "Japan" in Japanese, and literally mean "the Land of Rising Sun".
Moscovium with the symbol Mc for element 115 and tennessine with the symbol Ts for element 117 were proposed by the discoverers at the Joint Institute for Nuclear Research, Dubna (Russia), Oak Ridge National Laboratory (USA), Vanderbilt University (USA) and Lawrence Livermore National Laboratory (USA). Both are in line with tradition honoring a place or geographical region. Moscovium is in recognition of the Moscow region and honors the ancient Russian land that is the home of the Joint Institute for Nuclear Research, where the discovery experiments were conducted using the Dubna Gas-Filled Recoil Separator in combination with the heavy ion accelerator capabilities of the Flerov Laboratory of Nuclear Reactions. Tennessine is in recognition of the contribution of the Tennessee region of the United States, including Oak Ridge National Laboratory, Vanderbilt University, and the University of Tennessee at Knoxville, to superheavy element research.
Lastly, and in line with the tradition of honoring a scientist, the name oganesson and symbol Og for element 118 was proposed by the collaborating teams of discoverers at the Joint Institute for Nuclear Research, Dubna (Russia) and Lawrence Livermore National Laboratory (USA) and recognizes Professor Yuri Oganessian (born 1933) for his pioneering contributions to transactinoid elements research. His many achievements include the discovery of superheavy elements and significant advances in the nuclear physics of superheavy nuclei including experimental evidence for the "island of stability".
European Court: Armenian Genocide Denial No Crime, Constitutes Freedom of Expression
STRASBOURG — The European Court of Human Rights (ECHR) ruled Dec. 17…
Hrant Dink Assassination Records Erased by Ankara Police
ISTANBUL — A chief police inspector has claimed that phone recordings of…
FM Nalbandian: Nothing Related to Nagorno-Karabakh Can be Settled Without the Consent of its People
YEREVAN — No question related to the Karabakh conflict settlement can be…
French Senate Passes Bill Criminalizing Denial of Armenian Genocide
PARIS — The French Senate approved on Friday a new bill criminalizing…


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Home » Danish Sheikh » 20 Feb|Fri|6 PM|Conversations Around a Comprehensive Anti-Discrimination Law
20 Feb|Fri|6 PM|Conversations Around a Comprehensive Anti-Discrimination Law
Conversations Around a Comprehensive Anti-Discrimination Law
Danish Sheikh and Sumathi Murthy
Friday, 6 pm, Alternative Law Forum
About the discussion
The landscape of discrimination law in the country is deeply fragmented and lacks insight into intersectional experiences across recognized identity markers. The manner in which the constitutional guarantees under Article 15 of the Constitution have been operationalized has been extremely inadequate.
To facilitate conversations on bringing together multiple strands of discrimination under a common legal framework, ALF organized a consultation on anti-discrimination laws at the National Law School, Bangalore in December 2014. The consultation brought together lawyers, activists and academics to engage in discussions around comparative discrimination law regimes; providing for intersectionality in the law; lived realities of direct and indirect discrimination; and the possible ways to further these conversations.
The consultation was part of a broader project on anti-discrimination laws that ALF has been working on alongside Sumathi Murthy and Sunil Mohan. While ALF has been engaged in doctrinal research on different strands of discrimination law, Sumathi and Sunil have been conducting interviews with activists and lawyers across South India, on the different experiences of discrimination and gauging their responses to different legal frameworks.
We invited you to a discussion intended to take this work forward. Danish Sheikh will present on some of the key issues raised at the National Law School conference, while Sumathi Murthy will discuss her and Sunil's experiences on the field. A report of the conference proceedings can be found on our website:
http://altlawforum.org/campaigns/consultation-on-antidiscrimination-laws/
About the Discussants
Danish graduated with a bachelor's degree in law from NALSAR, Hyderabad and a master's degree from the University of Michigan Law School. He primarily works on research and advocacy relating to Gender and Sexuality law, along with Free Speech and Law and Literature. He has taught courses at NLSIU and St. Joseph's in Bangalore; NALSAR, Hyderabad; NUJS, Kolkata and Jamia Millia Islamia, New Delhi.
Sumathi Murthy started working with women, contract workers' union and marginalized genders and sexualities from 1999. She composes music and has also done work on sociological understanding of music and gender and written a play on the Raga Kalyani called Sanchari.


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1968 Lamborghini Espada Rat Rod Failed To Sell At $100K
Jeffrey N. Ross
motorious January 11, 2020
1968 Lamborghini Espada Rat Rod
Lamborghini's 2+2 Grand Touring coupe as you've never seen it before!
Mecum's 2020 Kissimmee auction definitely wasn't lacking in terms of amazing collector cars, but one of the coolest and most indescribably unique vehicles to cross the auction block was this one-off rat rod that started life as an ultra-rare 1968 Lamborghini Espada. Despite its incredible style, this world-class custom failed to find a new home with a high bid of just $100,000.
Sitting a staggering 8.2 feet wide, this Lamborghini rat rod was built by Danton Art Kustoms in France to celebrate the 50th anniversary since the launch of the Espada, which was built from 1968 through 1978. Despite the heavy customization, the Espada's unique body is still evident with its large side window openings and the Kammback rear end design. The front axle has been move forward considerably resulting in an extreme wheelbase, and the track itself has been widened by moving the rear wheels outside of the body. To help tie the lines of the body in with the widened track, this Espada features the wide side door scoops of a Lamborghini Reventon.
The custom interior has even more of the rat rod look with plenty of exposed rivets and sheet metal. Just like it did from the factory, this Espada is still a four-seater with custom fabricated seats. The most incredible part of this build is that it used the Espada's original body, chassis and engine, and speaking of the engine, most of the front clip was removed – in true hot rod fashion – to fully expose monster 3.9-liter V12. And don't worry, if you're the winning bidder, this Lamborghini Espada Rat Rod will be street legal and sold with a U.S. title. You'll definitely be the only one on the planet with such a car!
Continue the conversation on this Espada rat rod over on Lamborghini Chat.
More From 2020 Mecum Kissimmee...
1967 Ford Mustang Hero Car From 'Gone In 60 Seconds' Up For Grabs
Mecum Reveals Bullitt Mustang To Be Auctioned At Kissimmee 2020
Own A Different Kind Of Hemi Challenger
How to create a guaranteed income into retirement years
How negative rates will impact the global economy in the long-term
Zulily launches price-comparing feature to compete with Amazon, Walmart


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Tag Archives: Crash Bandicoot
WEEKLY NEWS UPDATE FOR 14TH – 20TH JULY
July 20, 2014 Darren Cooke	1 Comment
Microsoft is looking to lay off 18,000 staff over the next six months; the largest turnaround in workforce the company has seen in 39 years. It is expected that over $800 million will be paid in benefits and severance packages, and although it is expected that many of the cuts will be taken from the recently acquired Nokia branch, the Xbox division will also be affected with Microsoft looking to close down Xbox Entertainment Studios.
XES, which was formed only as recently as February 2013 will continue to produce the Halo: Nightfall series and a document about Atari, having already premiered football documentary Every Street United which coincided with the World Cup. The shutdown of the studio means however that progress on over a dozen series will be cancelled, and heavily calls into question Microsoft's strategy going forward as original Xbox content was one of the original selling points for the console (existing TV deals with the likes of NFL will continue), alongside the recently jettisoned Kinect peripheral. The move also flies in the face of a recent trend for streaming services providing original content with Netflix proving to be the most successful, and Amazon & Sony amongst others having projects waiting in the wings.
Recent changes do seems to suggest that after a years indecision and u-turns, Microsoft are once again looking to focus the Xbox One on the gaming audience. You can read the internal e-mail sent by Xbox chief Phil Spencer to staff about XES via Game Informer. Continue reading WEEKLY NEWS UPDATE FOR 14TH – 20TH JULY →
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Best VGM 399 – Crash Bandicoot – Pinstripe
Sounds odd without the gunshots and maniacal laughter in the background doesn't it?
Crash BandicootMark MothersbaughPinstripe PotorooVideo Game Music
Best VGM 301 – Crash Bandicoot – Boulders
Part IV of the Best VGM List (301-400) begins today! And we start this new installment by returning to the adventure that began my love of video games – didgeridoos at the ready…
BouldersCrash BandicootPlaystationVideo Game Music
Playstation All Stars Battle Royale – Who Should Be On The Roster?
July 7, 2012 Darren Cooke	1 Comment
Yesterday, Evolution 2012 kicked off in all its glory at Caesar's Palace in Las Vegas. For those of you who may be unaware, Evolution is an annual tournament bringing together the greatest players of fighting games in the world, who contest over a variety of titles for big cash prizes. In amongst footage of some scarily good Street Fighter 4 players, there was a demonstration of Sony's upcoming Playstation All Stars Battle Royale, which is now set for release on October 26th 2012. The headline news from this demonstration was the announcement of two new characters for the roster – Heihachi (from the Tekken series) and Toro (Sony's mascot cat which is most popular in Japan). This brings the total announced roster thus far to 10 characters; and while there is still time for further announcements, some people found last nights reveals to be a bit underwhelming. This begs the question then – who is left to make the cut?
To answer this question, I have once again asked for the help of a guest writer. Therefore taking the reins for a record breaking second time is Mr Stephen Jackman, a fellow part time games jorunalist and all round fan of all things Sony and fighting. Enjoy!
Playstation All Stars Battle Royale Remaining Roster Speculation
With leaks of characters and reveals popping up here, there and everywhere, it's hard not to speculate about the cast of characters that are going to appear in a big crossover fighter such as Playstation All Stars Battle Royale. Maybe we are leading ourselves to disappointment or forgetting some memorable names leading to genuine surprise and excitement. Regardless, the process of putting together lists with friends has been fun, and my own views on the matter can be seen below. The way I have structured my character choices puts them into a 'one or the other' scenario; I see it possible in some situations that some if not all the characters are included but I think at least one will make it in to represent that field. There are ten fields that I believe will be filled by one of each at least (a cheeky way of avoiding narrowing it down to a top ten list). It is a sort of top ten list but clearly not. Continue reading Playstation All Stars Battle Royale – Who Should Be On The Roster? →
Cloud StrifeCrash BandicootDanteLara CroftPlaystation All Stars Battle RoyaleSolid SnakeSonySpyro The DragonTop TenVideo Game News
Best VGM 1 – Crash Bandicoot – N-Sanity Beach
January 24, 2012 Darren Cooke	Leave a comment
In what I see to be a fitting tribute, the first piece of game music to make my list is…the first piece of game music I ever heard. It remains legendary to this very day.
Crash BandicootN-Sanity BeachVideo Game Music
Features And News, Top Ten Lists
Top Ten…Most Important Games On Sony Consoles
January 23, 2012 Darren Cooke	2 Comments
Today, It's part two of my initiative to get other people writing on this blog, as we chronicle some of the most important games of all time. I covered the efforts of Microsoft in this article, and today it is the turn of Stephen Jackman to present his case for Sony. Stephen is a good friend of mine from Lincoln University, where he holds the enviable position of being the reigning champion at the Gaming Society I help to run (trust me – its a big deal). So, take it away…
Top 10…Most Important Games On Sony Consoles
I just want to thank Mr Cooke for allowing me to post on his blog as I always find it enjoyable to write about games, since most of the time they are pretty awesome. As someone who grew up with Sony and the Playstation console line, starting back in 1997 with a package deal along with the not so stellar title of Bubsy 3D (one of the times when games aren't awesome), up to the present day with the Playstation 3 and my acquisition of Uncharted 3: Drake's Deception, I have seen a selection of great titles come and go with some leaving a large impact on the games industry and becoming a huge benefit, if not a major means of success, for Sony and their endeavours into the games industry.
Whereas the first console played a lot of copycat with the N64, for example the introductions of analogue sticks and the controller rumble feature, it had some strong titles that allowed it to reach the level of success that it did. The Playstation 2 built on what the original foundations the Playstation set out and ended up having one of the largest selections of titles any console ever supported. The most current iteration of the console, the Playstation 3, is a powerhouse of a machine and the most powerful console on the market at the moment, bringing high definition with blue ray among many other features.
From each Playstation console I have had a chance to play a fair amount of the games. From when the older titles were new on shelves and not yet nostalgic memories, as well as the newer releases, I feel I am able to look back at Sony's history and find a top ten of important titles. As with Darren my aim was to give preference to console exclusives over multi platform releases, or at the very least the console the game launched on and what it was well known for. Continue reading Top Ten…Most Important Games On Sony Consoles →
Crash BandicootDevil May CryFinal Fantasy 7God Of WarGran TurismoGTA 3Heavy RainLittle Big PlanetMetal Gear Solid 3PlaystationPlaystation 2Playstation 3SonyUncharted 2
Playstation, Video Game Reviews
The Best Karting Game – Crash Team Racing Review
July 29, 2011 Darren Cooke	3 Comments
CRASH TEAM RACING (1999) – NAUGHTY DOG
I'll make no excuses; Crash Team Racing is a Mario Kart 64 clone, pure and simple. When the Italian plumber took his karts to the Nintendo 64 in 1997, it set the bar for everything else, and most of the efforts from developers in the two years afterwards (with the notable exception of Diddy Kong Racing) neatly smashed into it. Crash Team Racing differed from all of the other imitators however, and I personally feel that it did the unthinkable and actually surpassed Mario Kart 64, despite borrowing so heavily from it. Now I am aware of the heavy bias I carry into this review (being a massive Crash fan and all), but keep reading and you will hopefully realise why the marsupial fits the kart better than his moustachioed rival. Continue reading The Best Karting Game – Crash Team Racing Review →
Crash BandicootCrash Team RacingKart RacingMario Kart 64PlaystationVideo Game Reviews
Top Ten…Levels In Gaming
Sometimes when you play a game, you have to stand back and just simply admire the game world that you are playing in. It can be a variety of things that can trigger this off; an epic vista, a serious confrontation, a challenging enviorment, a funny situation, a serious situation etc. The point remains: some locations in games simply get lodged in our heads and we never forget them.
My purpose in writing this list is to share an insight into the sorts of gaming locations that have had an effect on me and become great favourites of mine. This list is not ranked as some of my others have been; quite frankly I would find it difficult to rank these memories on merit, and so they are simply listed in the order they came to my head. I also admit that the title of the article is a bit loose; some of the locations mentioned aren't 'levels' in the platforming sense of the word, but I couldn't think of a better term so it will do. And so without further ado; my favourite levels/places/locations in my gaming history. Continue reading Top Ten…Levels In Gaming →
Crash BandicootLevelsMarioRatchet And ClankSpyro The DragonTop Ten
Time To Get Warped: Crash Bandicoot 3 Review
(First published on Gamepad Magazine, 5th March 2011)
CRASH BANDICOOT 3: WARPED – Naughty Dog – 1998
If only I had a Apple Launching Bazooka…
Back In 1997, if you wanted a platform game for the Playstation, the choice was simple; you bought the Crash Bandicoot series. It was the best of the genre on the console, and nothing surpassed it. Come into 1998 however, and we find that Crash's reign at the top was not as solid as it may have seemed. The Playstation's great nemesis, the Nintendo 64, had been pumping out quality such as Banjo Kazooie, and Crash now had a rival on his own console, in the form of the massively popular Spyro The Dragon. Naughty Dog had to take action, and in the space of just 10 ½ months they managed to produce the third game in the series just in time for Christmas. Did the rush affect the quality? Happily, it didn't. It raised the bar again. Continue reading Time To Get Warped: Crash Bandicoot 3 Review →
Crash BandicootPlaystationVideo Game Reviews
'I Come From A Land Down Under' – Crash Bandicoot Review
March 19, 2011 Darren Cooke	2 Comments
(First published on Gamepad Magazine, 7th November 2010)
CRASH BANDICOOT (1996) – NAUGHTY DOG STUDIOS
Ah, how the memories come flooding back
I should make it aware at this point that Crash Bandicoot was the first video game I ever played. Therefore it seems only fitting that 13 years later, the adventures of the spinning orange marsupial forms the basis of my first review.
Crash Bandicoot was released at a strange time for platformers. The king of the hill at the time was undoubtedly Super Mario 64 on the N64 which had revolutionised platforming in 3D, and Crash was Sony's attempt to at least emulate the success of the Italian plumber's adventures. It was also unique because the only competition it had on its own console was Bubsy 3D, meaning that it was almost guaranteed sales.
Cue the mayhem… Continue reading 'I Come From A Land Down Under' – Crash Bandicoot Review →
Crash BandicootNaughty DogPlaystationVideo Game Reviews


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UTEP, Lockheed Martin…
Lockheed Martin Missiles and Fire Control Storefront announcement and MOU signing with UTEP, Wednesday, October 9, 2019, in El Paso, Texas. Photo by Ivan Pierre Aguirre/UTEP Communications
UTEP, Lockheed Martin sign agreement to employ students in El Paso
News Regional News Tech
The University of Texas at El Paso (UTEP) President Heather Wilson and Lockheed Martin Missiles and Fire Control Vice President of Engineering & Technology Travis Coomer signed a Memorandum of Agreement Wednesday that will employ UTEP students as interns with Lockheed Martin in El Paso.
"Lockheed Martin is a strong partner," said UTEP President Heather Wilson. "Through this agreement, we are taking the next step in our relationship and bringing more engineering and business jobs to El Paso. We educate students from the region for jobs in industry, but UTEP will also work with community leaders to attract companies willing to employ people here in El Paso."
In addition to Wilson and Coomer, U.S. Rep. Veronica Escobar, El Paso Mayor Dee Margo and El Paso County Judge Ricardo A. Samaniego participated in today's announcement and signing ceremony.
Over the past 10 years, Lockheed Martin has supported research, infrastructure and program development at UTEP in the College of Engineering, leading to successful collaborations on dozens of professional and educational projects.
The company also has supported the College of Business Administration, where the Lockheed Martin Enterprise Resource Planning Academy has offered students an application-oriented view of planning systems since 2017.
As part of the new agreement, Lockheed Martin will open a location in El Paso near the UTEP campus where they will employ UTEP undergraduate students as interns. In addition, Lockheed Martin will hire UTEP doctoral students and postdoctoral researchers to serve as supervisors.
"UTEP is an invaluable asset to El Paso and key in training the next generation of workforce and promoting our binational region's economic development," Congresswoman Escobar said. "This partnership with Lockheed Martin will support growth and innovation in the aerospace and defense industries and open the doors of opportunity to a vibrant community of talented, highly trained, highly skilled students."
The Lockheed Martin location may also provide an opportunity for existing Lockheed Martin employees from El Paso to return to their hometown while continuing to work for the company.
UTEP is one of only nine top tier research universities in Texas. It educates 35% more undergraduate engineers than the University of New Mexico, and is the largest university between Tucson, Arizona; the Colorado border; and Lubbock, Texas.
For Lockheed Martin, the partnership provides opportunities to recruit a highly educated and diverse workforce while continuing to work closely with other high-tech UTEP facilities, including the W.M. Keck Center for 3D Innovation and the NASA MIRO Center for Space Exploration and Technology Research (cSETR).
Lockheed Martin is one of the most active and successful recruiters of UTEP graduates. Its F-35 manufacturing plant in Fort Worth, Texas, employs more than 100 UTEP alumni.
"Lockheed Martin and UTEP have a strong history of partnership," Coomer said. "With the establishment of this program, we see great potential for the UTEP interns to make invaluable contributions to the engineering community and to the future of our business."
Leaders and policymakers from across the Paso del Norte region said they're encouraged by the prospects this latest addition to the partnership between UTEP and Lockheed Martin presents.
"The City of El Paso fully supports UTEP's commitment to economic development efforts that harness the talent of this community's students," El Paso Mayor Dee Margo said. "Lockheed Martin's increased presence in El Paso will bolster our city's reputation as a place where Fortune 100 companies, like Lockheed Martin, can grow their businesses by seizing on the opportunity that our homegrown, dedicated workforce presents."
"The County of El Paso stands with UTEP in its efforts to create local jobs for a highly educated and competitive workforce that continues to grow in our region," El Paso County Judge Ricardo A. Samaniego said. "Both the County and UTEP take great pride in the preparation of our future leaders who, through their vision and leadership, will create growth and prosperity in our region. We are pleased to have Lockheed Martin as a partner in this endeavor."
UTEP is a comprehensive public research university that is increasing access to excellent higher education. It advances discovery of public value and positively impacts the health, culture, education and economy of the community it serves.
Headquartered in Bethesda, Maryland, Lockheed Martin is a global security and aerospace company that employs approximately 105,000 people worldwide and is principally engaged in the research, design, development, manufacture, integration and sustainment of advanced technology systems, products and services.
Tags: lockheed martin utep
Special Promotions Set For UTEP Men's And...
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BUILD YOUR OWN TERRARIUM FROM START TO FINISH. WE HAVE EVERYTHING YOU NEED AND MORE. OUR TRAINED STAFF WILL GUIDE YOU, NO APPOINTMENT NECESSARY. CHOOSE FROM OUR VAST SELECTION OF GLASS AND POTS OR BRING YOUR OWN CONTAINER.
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text:
Ceratosanthes es un género  de plantas perteneciente a la familia Cucurbitaceae.

Taxonomía
El género fue descrito por Michel Adanson y publicado en Familles des Plantes 2: 139, 535. 1763. La especie tipo es: Ceratosanthes palmata (L.) Urb.

Especies aceptadas
A continuación se brinda un listado de las especies del género Ceratosanthes aceptadas hasta noviembre de 2014, ordenadas alfabéticamente. Para cada una se indica el nombre binomial seguido del autor, abreviado según las convenciones y usos.

Referencias

Cucurbitaceae

redpajama_set_name: RedPajamaWikipedia


Un-PreCompose is a plug-in for After Effects that allows you to select a pre-comp (nested composition) and extract all of its layers into the parent composition while maintaining each layers effects and other properties. You are given various options for maintaining or deleting properties from the pre-comp and the precomp itself.
This option will help you see which layers came from the pre-comp. Whatever you put into this textbox will be added to the beginning of each layer that is un-precomposed. If no prefix is desired, just leave the box empty.
If you have key frames, effects, etc. applied to your pre-comp and want those attribute to be maintained, select this option. Each layer inside will then receive the effects from its parent and will be made a child of a null object that contains the transform attributes. The null object will have the name ||CONTROLLER||.
option will get rid of any attributes that are applied to your pre-comp. Layers inside the pre-comp will maintain their current positions and attributes. You may notice a change in position if your precomp is a different resolution then its parent composition.
If this option is selected, once the pre-comp has been un-precomposed, it will be deleted from the project. This option will not be available if the pre-comp is being used elsewhere in the project. This option will be disabled if the 'Keep Pre-Comp in Composition' option is selected.
For the moment, masks applied to your pre-comp will not be copied over to the extracted layers. This may change in the future, so keep a look out for any updates to the plug-in.
If your pre-comp is a 3D layer that contains particle layers, you may get some issues on running the plug-in. Particle layers are usually meant to be applied to 2D layers, but if the parent comp is 3D, every layer inside will be changed to a 3D layer, so you might get some strange results.
For now, the, any transform attributes of the pre-comp are applied to the null object, which then becomes the parent of each extracted layer. If a layer already has a parent, that layer will not be affected.
Parenting is maintained within the pre-comp, however if any layers are parented to the pre-comp, that connection will be broken.
If layers inside your pre-comp contain expressions that use random numbers, you may notice a slight change due to random numbers being based on the layer's index, which may be changed when running an Un-PreCompose.


Recently, Korean S/W developers attended at the world renowned industry conferences and started to actively engage in foreign businesses. Korean ERP consulting company, LAPHA & COMPANY, attended at SAP annual Sapphire conference last month. Following it, an APM solution leader, Jennifersoft(formerly known as JSC) also attended at 'Java One' conference which is the largest event among Java developers, held in San Francisco on 11th, May.
Jennifersoft set company booth to publicize its solution 'Jennifer' Only Jennifersoft was Korean developer among many world leading S/W companies at the exhibition and laid a foundation for its overseas business. Especially, it's noticeable because a Korean company attended at Java One conference in 4 years.
Established in 2005, Jennifersoft draws huge attention in APM market by launching Jennifer APM solution. Jennifersoft has overtaken other rival companies like CA and Symantec and now occupies more than 50% of domestic market share. Based on product quality and business experience, it opened a branch office in Japan and established partnerships with famous IT companies such as FUJITSU, HITACHI and MARUBENI. Through the experience at the conference, Jennifersoft plans to establish a branch office in this year and start its business in America. At '2007 Java One Conference', all employees of Jennifersoft attended and enthusiastically publicized their company and APM solution-'Jennifer'.


text:
Q: Netsuite, Filtering the Items allowed to sell based on a customer classification I am wondering if it is possible in NetSuite to filter items available to sell based on a customer.
The requirement is that the company restricts selling certain items to some selected customers. Just wanted to know if this is possible in NetSuite.
I am aware that it is possible in sales order form to have an 'Item saved search' as the filtering criteria. But this does not allow filtering items based on selected customer fields. Seems 'Item saved search' that is assigned for the form can only use static fields or some variables from the 'Item work flow'.
Other way I tried is by restricting the 'Custom' price level on the sales order form. This way I am able stop selling anything not assigned to the customer price level. But I cannot stop those items being listed in the drop-down and search list.
Really appreciate any help on this.

A: *

*You can use clientscript with validateLine function

*Create script parameter with item list record type and select item which you want to restrict.

*Within validateLine function restrict item which is selected within a script parameter record.

A: I was successful in filtering class based on a particular department and the following approach was followed:
var selectedCustom = 'T';
var v_form = type + '_form';
var formCustom = document.forms[v_form];
var fldCustom = getFormElement(formCustom, fldnamCustom.toLowerCase());

Using the above approach the following function was able to filter the list of values
addSelectOption(document, ldCustom, v_class, v_class_id, selectedCustom);

redpajama_set_name: RedPajamaStackExchange


text:
Q: oracle string replace odd number of slash at the end of string In Oracle pl sql how to add one SLASH if a field ends with ODD NUMBER OF SLASH?
For example for a column c I want:
    AAA\ converts to AAA\\
    AAA do nothing
    AAA\\ do nothing
    AAA\\\ converts to AAA\\\\
    AAA\\\\ do nothing
    AAA\\\\\ converts to AAA\\\\\\

A: You can use REGEXP_SUBSTR, LENGTH and MOD functions as follows:
SELECT YOUR_COLUMN,
       CASE WHEN MOD(LENGTH(REGEXP_SUBSTR(YOUR_COLUMN,'[\]+$')),2) = 1
            THEN YOUR_COLUMN || '\'
       ELSE YOUR_COLUMN
       END AS RESULT
FROM YOUR_TABLE;


text:
Latest in Marco Polo
'Marco Polo' Produer Files Amended Sexual Harassment Suit Against Harvey Weinstein, TWC – Updated
UPDATED, 12:39 PM: An associate producer on the former Weinstein Company series Marco Polo has filed an amended complaint charging sexual and physical abuse by Harvey Weinstein. Alexandra Canosa's filing Monday in New York Supreme Court (read it here) details further abuse at the hands of the disgraced mogul, charging…
By Dominic Patten
Harvey Weinstein & TWC Hit With Sexual Battery & Assault Suit By Gloria Allred For New Accuser
As the NYPD, the LAPD and police in London and Beverly Hills investigate sexual assault complaints against Harvey Weinstein, the disgraced producer and The Weinstein Company were hit with a sexual battery and assault lawsuit today by a new and anonymous accuser. "Prior to the incidents involving Plaintiff, the…
'Marco Polo' Canceled After 2 Seasons On Netflix
Marco Polo has come to an end. Netflix has opted not to renew the drama series about the famous 13th century merchant and explorer for a third season. Marco Polo, which launched in 2014, was the first Netflix original series developed in the US for the global market. Created by John Fusco, it was inspired by Marco…
'Marco Polo' Creator: Netflix Series "Breaking Barriers" On Diversity – Guest Column
Editors note: While diversity in casting is increasingly front and center, there's evidence of a reward in greater viewership. Take Netflix's Marco Polo, the John Fusco-created historical drama that just bowed its Season 2. Netflix doesn't provide ratings figures, but its makers cite Parrot Analytics which places…
By John Fusco
'Marco Polo' Season 2 Trailer: Kublai Khan Took The Wall, But Can He Keep His Power?
"Marco Polo, I conscript you into the order of the Mongol Knights," says Kublai Khan (Benedict Wong) teeing off Netflix's second season of Marco Polo as we see that the famed explorer (Lorenzo Richelmy) has become a trusted part of the Middle Kingdom lord's 13th Century court. "This foreigner saved my life," retorts…
Netflix Sets Premiere Dates For 'The Crown' & Season 2 Of 'Marco Polo'
Netflix has set the worldwide premiere of its Stephen Daldry-directed drama The Crown for November 4 and Season 2 of international adventure series Marco Polo for July 1. Reuniting Daldry with his The Queen team of writer Peter Morgan and producer Andy Morgan, The Crown stars Claire Foy as Queen Elizabeth II, a…
Michelle Yeoh Joins Netflix's 'Marco Polo' For Season 2
Michelle Yeoh is joining the Season 2 cast of Netflix drama Marco Polo in a recurring role. Based on the famed 13th Century explorer, the story follow's Marco Polo's (Lorenzo Richelmy) adventures, as he fights greed, betrayal and rivalries in Kublai Khan's court. Yeoh will play a mysterious figure from Hundred Eyes…
Ron Yuan Joins Netflix's 'Marco Polo'; Jolene Purdy In 'Orange Is The New Black'
Ron Yuan (Fast & Furious) has joined the cast of Netflix drama Marco Polo as a series regular. Based on the famed 13th Century explorer, the story follow's Marco Polo's (Lorenzo Richelmy) adventures, as he fights greed, betrayal and rivalries in Kublai Khan's court. Yuan will play Prince Nayan in the second season…
Tax Incentives Not "Free Money" Says TWC's David Glasser – Produced By
"Size and certainty, that's what matters," said David Glasser today on a panel at the Produced By Conference about California's new $330 million a year film and TV tax incentive program. "As it gets bigger we'll be more confident we can make something here," The Weinstein Co. COO and president told a packed house at…
Netflix Renews 'Marco Polo', Sets Premiere Dates For 'Daredevil', 'Kimmy Schmidt', 'Bloodline', 'Grace & Frankie' – TCA
At the top of Netflix's executive session, Chief Content Officer Ted Sarandos made a slew of announcements. On the heels of the debut of the streaming service's latest original series, Marco Polo, Netflix has renewed the adventure drama, from the Weinstein Co. and Electus, for a 10-episode second season. Daredevil,…
By Nellie Andreeva, Lisa de Moraes
Jan 7, 2015 9:53 am
'Marco Polo' Review: Dominic Patten On Netflix's Fabulous New Epic
Empires, betrayal, bloodshed and seduction on an epic scale. That's the semi-historical account of Marco Polo coming from Netflix on Dec. 12. And, as my video review above says, while it may not be the most original show out there on streaming-video platforms or otherwise, it is a feast to be savored. The $90 million…
'Marco Polo' Teaser: First Look At The Netflix-Weinstein Company Drama
Here's our first look at Italian newcomer Lorenzo Richelmy exploring 13th century China. Marco Polo, which bows December 12 on Netflix, also stars Benedict Wong, Joan Chen, Chin Han, Zhu Zhu, Olivia Cheng, Claudia Kim, Mahesh Jadu, Tom Wu, Remy Hii, Uli Latukefu and Rick Yune. Keeping with its international flavor…


text:
Whitford Promotes WH's 'Big Block of Cheese Day'
White House Holds Third Annual Social Media Engagement Event
Jan. 13 marks the third year in a row that the White House will be making top officials, including the President, Vice President and first Lady, available during a daylong (10 a.m. to 6:30 p.m.) social media opportunity to ask questions via Twitter, Facebook, Instagram and Tumblr and ex-West Wing and current Transparent actor Bradley Whitford was trying to help the White House drive Web traffic.
The day has been dubbed the virtual Big Block of Cheese Day, a reference to both an 1837 Andrew Jackson public open house featuring a 1,400-pound block of cheese, and to NBC's West wing, which referred the event and whose cast has been helping promote the event (https://www.youtube.com/watch?v=ex_jJlv5HKk).
In an e-mail from the White House account Whitford, who played Josh Lyman in West Wing, pointed out that "in my gig on the show we hosted our very own day like this, so I've been around the cheese block a few times." He and urged those with questions about Tuesday night's State of the Union address to log on and weigh in at hashtag #BigBlockOfCheeseDay.
Whitford said he would be online for the event (@WhitfordBradley) as would the real-life version, White House Chief of Staff Denis McDonough (@Denis44), of his boss on the show.
Bradley WhitfordobamaSOTU
Obama in SOTU: We've Protected an Open Internet
Also Urges Passage of TPP, Campaign Finance Reform
Title II Fans Celebrate Big Victory
Say Grassroots Efforts Helped New Rules Take Hold
Wheeler Defends Title II Decision in Senate
Says Communications With White House Were Not Unusual
Sen. Markey: I Will Not Block STELAR
Bill passed Wednesday with bipartisan vote in House
SOTU: President Intends to Protect Open Internet
But Doesn't Talk About High-Speed Broadband Vision
House Oversight Committee Grills Wheeler
Dems Deem White House Communications Par for Course
Jerry Lewis Remembered
White House cites "incredible charity work"
President To Promote Broadband in Iowa Speech
To Visit 1-Gig Broadband Network Ahead of 'SOU'


text:
A mildly insane idea for disabling the coronavirus
November 28, 2020 [email protected]_84 coronavirus, disabling, Idea, insane, mildly
Enlarge / Diagram of the structure of the virus' spike protein.
When the COVID-19 pandemic was first recognized for the threat that it is, researchers scrambled to find anything that might block the virus' spread. While vaccines have grabbed much of the attention lately, there was also the hope that we could develop a therapy that would block the worst effects of the virus. Most of these have been extremely practical: identify enzymes that are essential for the virus to replicate, and test drugs that block similar enzymes from other viruses. These drugs are designed to be relatively easy to store and administer and, in some cases, have already been tested for safety in humans, making them reasonable choices for getting something ready for use quickly.
But the tools we've developed in biotechnology allow us to do some far less practical things, and a paper released today describes how they can be put to use to inactivate SARS-CoV-2. This is in no way a route to a practical therapy, but it does provide a fantastic window into what we can accomplish by manipulating biology.
Throw it in the trash
The whole effort described in the new paper is focused on a simple idea: if you figure out how to wreck one of the virus' key proteins, it won't be able to infect anything. And, conveniently, our cells have a system for destroying proteins, since that's often a useful thing to do. In some cases, the proteins that are destroyed are damaged; in others, the proteins are made and destroyed at elevated paces to allow the cell to respond to changing conditions rapidly. In a few cases, changes in the environment or the activation of signaling pathways can trigger widespread protein destruction, allowing the cell to quickly alter its behavior.
This system relies on a small protein called "ubiquitin." When a protein is to be targeted for destruction, enzymes called ubiquitin ligases chemically link a chain of ubiquitins to it. These serve as a tag that is recognized by enzymes that digest any proteins with ubiquitin attached to them.
So, the idea behind the new work is to identify a key viral protein, and figure out how to attach ubiquitin to it. The cell would then take care of the rest, digesting the viral protein and thus blocking the production of any useful viruses in that cell. In this case, the researchers decided to target the spike protein that sits on the surface of coronaviruses and allows them to attach to and infect new cells.
Unfortunately, there are no proteins that attach ubiquitin to the viral spike protein. Or, rather, there were no proteins that fit that description.
But a team at Harvard has now produced one.
The team's method of doing so started with the fact that we do know something that sticks to the viral spike protein: the cellular protein it latches on to in order to enter the cell. This is called the angiotensin-converting enzyme 2,


text:
\section{Introduction}

An expression $G=HK$ of a group $G$ as the product of subgroups $H$ and $K$ is called a \emph{factorization} of $G$, where $H$ and $K$ are called \emph{factors}. A group $G$ is said to be \emph{almost simple} if $S\leqslant G\leqslant\Aut(S)$ for some nonabelian simple group $S$, where $S=\Soc(G)$ is the \emph{socle} of $G$. In this paper, by a factorization of an almost simple group we mean that none its factors contains the socle. The main aim of this paper is to solve the long-standing open problem:

\begin{problem}\label{PrbXia1}
Classify factorizations of finite almost simple groups.
\end{problem}

Determining all factorizations of almost simple groups is a fundamental problem in the theory of simple groups, which was proposed by Wielandt~\cite[6(e)]{Wielandt1979} in 1979 at The Santa Cruz Conference on Finite Groups. It also has numerous applications to other branches of mathematics such as combinatorics and number theory, and has attracted considerable attention in the literature.
In what follows, all groups are assumed to be finite if there is no special instruction.

The factorizations of almost simple groups of exceptional Lie type were classified by Hering, Liebeck and Saxl~\cite{HLS1987}\footnote{In part~(b) of Theorem~2 in~\cite{HLS1987}, $A_0$ can also be $\G_2(2)$, $\SU_3(3)\times2$, $\SL_3(4).2$ or $\SL_3(4).2^2$ besides $\G_2(2)\times2$.} in 1987.
For the other families of almost simple groups, a landmark result was achieved by Liebeck, Praeger and Saxl~\cite{LPS1990} thirty years ago, which classifies the maximal factorizations of almost simple groups. (A factorization is said to be \emph{maximal} if both the factors are maximal subgroups.)
Then factorizations of alternating and symmetric groups are classified in~\cite{LPS1990}, and factorizations of sporadic almost simple groups are classified in~\cite{Giudici2006}.
This reduces Problem~\ref{PrbXia1} to the problem on classical groups of Lie type.
Recently, factorizations of almost simple groups with a factor having at least two nonsolvable composition factors are classified in~\cite{LX2019}\footnote{In Table~1 of~\cite{LX2019}, the triple $(L,H\cap L,K\cap L)=(\Sp_{6}(4),(\Sp_2(4)\times\Sp_{2}(16)).2,\G_2(4))$ is missing, and for the first two rows $R.2$ should be $R.P$ with $P\leqslant2$.}, and those with a factor being solvable are described in~\cite{LX} and~\cite{BL2021}.

As usual, for a finite group $G$, we denote by $G^{(\infty)}$ the smallest normal subgroup of $X$ such that $G/G^{(\infty)}$ is solvable.
For factorizations $G=HK$ with nonsolvable factors $H$ and $K$ such that $L=\Soc(G)$ is a classical group of Lie type, the triple $(L,H^{(\infty)},K^{(\infty)})$ is described in~\cite{LWX}. Based on this work, in the present paper we characterize the triples $(G,H,K)$ such that $G=HK$ with $H$ and $K$ nonsolvable, and $G$ is an orthogonal group of minus type.

For groups $H,K,X,Y$, we say that $(H,K)$ contains $(X,Y)$ if $H\geqslant X$ and $K\geqslant Y$, and that $(H,K)$ \emph{tightly contains} $(X,Y)$ if in addition $H^{(\infty)}=X^{(\infty)}$ and $K^{(\infty)}=Y^{(\infty)}$.
Our main result is the following Theorem~\ref{ThmOmegaMinus}.
Note that it is elementary to determine the factorizations of $G/L$ as this group has relatively simple structure (and in particular is solvable).

\begin{theorem}\label{ThmOmegaMinus}
Let $G$ be an almost simple group with socle $L=\POm_{2m}^-(q)$, where $m\geqslant4$, and let $H$ and $K$ be nonsolvable subgroups of $G$ not containing $L$. Then $G=HK$ if and only if (with $H$ and $K$ possibly interchanged) $G/L=(HL/L)(KL/L)$ and $(H,K)$ tightly contains $(X^\alpha,Y^\alpha)$ for some $(X,Y)$ in Table~$\ref{TabOmegaMinus}$ and $\alpha\in\Aut(L)$.
\end{theorem}

\begin{remark}
Here are some remarks on Tables~\ref{TabOmegaMinus}:
\begin{enumerate}[{\rm(I)}]
\item The column $Z$ gives the smallest almost simple group with socle $L$ that contains $X$ and $Y$. In other words, $Z=\langle L,X,Y\rangle$.
It turns out that $Z=XY$ for all pairs $(X,Y)$.
\item The groups $X$, $Y$ and $Z$ are described in the corresponding lemmas whose labels are displayed in the last column.
\item The description of groups $X$ and $Y$ are up to conjugations in $Z$ (see Lemma~\ref{LemXia04}(b) and Lemma~\ref{LemXia03}).
\end{enumerate}
\end{remark}

\begin{table}[htbp]
\captionsetup{justification=centering}
\caption{$(X,Y)$ for orthogonal groups of minus type}\label{TabOmegaMinus}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Row & $Z$ & $X$ & $Y$ & Remark & Lemma\\
\hline
1 & $\POm_{2m}^-(q)$ & $\Omega_{2m-1}(q)$ & $\SU_m(q)$ & $m$ odd & \ref{LemOmegaMinus01}\\
2 & $\POm_{2m}^-(q)$ & $q^{2m-2}{:}\Omega_{2m-2}^-(q)$ & $\SU_m(q)$ & $m$ odd & \ref{LemOmegaMinus02}\\
\hline
3 & $\Omega_{2m}^-(2)$ & $\Omega_{2m-2}^-(2).2$ & $\SU_m(2)$ & $m$ odd & \ref{LemOmegaMinus03}\\
4 & $\mathrm{O}_{2m}^-(2)$ & $\Omega_{2m-2}^-(2).2$ & $\SU_m(2).2$ & $m$ odd & \ref{LemOmegaMinus04}\\
5 & $\GaO_{2m}^-(4)$ & $\Omega_{2m-2}^-(4).4$ & $\SU_m(4).4$ & $m$ odd & \ref{LemOmegaMinus04}\\
\hline
6 & $\mathrm{O}_{2m}^-(2)$ & $\GaO_m^-(4)$ & $\Omega_{2m-1}(2).2$ & $m$ even & \ref{LemOmegaMinus05}\\
7 & $\GaO_{2m}^-(4)$ & $\GaO_m^-(16)$ & $\Omega_{2m-1}(4).4$ & $m$ even & \ref{LemOmegaMinus05}\\
\hline
8 & $\mathrm{O}_{2m}^-(2)$ & $\SU_{m/2}(4).4$ & $\Omega_{2m-1}(2).2$ & $m/2$ odd & \ref{LemOmegaMinus06}\\
9 & $\GaO_{2m}^-(4)$ & $\SU_{m/2}(16).8$ & $\Omega_{2m-1}(4).4$ & $m/2$ odd & \ref{LemOmegaMinus06}\\
\hline
10 & $\Omega_{10}^-(2)$ & $\A_{12}$, $\M_{12}$ & $2^8{:}\Omega_8^-(2)$ & & \ref{LemOmegaMinus07}\\
11 & $\Omega_{18}^-(2)$ & $3.\J_3$ & $2^{16}{:}\Omega_{16}^-(2)$ & & \ref{LemOmegaMinus08}\\
\hline
\end{tabular}
\vspace{3mm}
\end{table}

\section{Preliminaries}

In this section we collect some elementary facts regarding group factorizations.

\begin{lemma}\label{LemXia01}
Let $G$ be a group, let $H$ and $K$ be subgroups of $G$, and let $N$ be a normal subgroup of $G$. Then $G=HK$ if and only if $HK\supseteq N$ and $G/N=(HN/N)(KN/N)$.
\end{lemma}

\begin{proof}
If $G=HK$, then $HK\supseteq N$, and taking the quotient modulo $N$ we obtain
\[
G/N=(HN/N)(KN/N).
\]
Conversely, suppose that $HK\supseteq N$ and $G/N=(HN/N)(KN/N)$. Then
\[
G=(HN)(KN)=HNK
\]
as $N$ is normal in $G$. Since $N\subseteq HK$, it follows that $G=HNK\subseteq H(HK)K=HK$, which implies $G=HK$.
\end{proof}

Let $L$ be a nonabelian simple group. We say that $(H,K)$ is a \emph{factor pair} of $L$ if $H$ and $K$ are subgroups of $\Aut(L)$ such that $HK\supseteq L$. For an almost simple group $G$ with socle $L$ and subgroups $H$ and $K$ of $G$, Lemma~\ref{LemXia01} shows that $G=HK$ if and only if $G/L=(HL/L)(KL/L)$ and $(H,K)$ is a factor pair.
As the group $G/L$ has a simple structure (and in particular is solvable), it is elementary to determine the factorizations of $G/L$.
Thus to know all the factorizations of $G$ is to know all the factor pairs of $L$.
Note that, if $(H,K)$ is a factor pair of $L$, then any pair of subgroups of $\Aut(L)$ containing $(H,K)$ is also a factor pair of $L$.
Hence we have the following:

\begin{lemma}\label{LemXia02}
Let $G$ be an almost simple group with socle $L$, and let $H$ and $K$ be subgroups of $G$ such that $(H,K)$ contains some factor pair of $L$. Then $G=HK$ if and only if $G/L=(HL/L)(KL/L)$.
\end{lemma}

In light of Lemma~\ref{LemXia02}, the key to determine the factorizations of $G$ with nonsolvable factors is to determine the minimal ones (with respect to the containment) among factor pairs of $L$ with nonsolvable subgroups.

\begin{lemma}\label{LemXia03}
Let $L$ be a nonabelian simple group, and let $(H,K)$ be a factor pair of $L$.
Then $(H^\alpha,K^\alpha)$ and $(H^x,K^y)$ are factor pairs of $L$ for all $\alpha\in\Aut(L)$ and $x,y\in L$.
\end{lemma}

\begin{proof}
It is evident that $H^\alpha K^\alpha=(HK)^\alpha\supseteq L^\alpha=L$. Hence $(H^\alpha,K^\alpha)$ is a factor pair.
Since $xy^{-1}\in L\subseteq HK$, there exist $h\in H$ and $k\in K$ such that $xy^{-1}=hk$. Therefore,
\[
H^xK^y=x^{-1}Hxy^{-1}Ky=x^{-1}HhkKy=x^{-1}HKy\supseteq x^{-1}Ly=L,
\]
which means that $(H^x,K^y)$ is a factor pair.
\end{proof}

The next lemma is~\cite[Lemma~2(i)]{LPS1996}.

\begin{lemma}\label{LemXia05}
Let $G$ be an almost simple group with socle $L$, and let $H$ and $K$ be subgroups of $G$ not containing $L$. If $G=HK$, then $HL\cap KL=(H\cap KL)(K\cap HL)$.
\end{lemma}

The following lemma implies that we may consider specific representatives of a conjugacy class of subgroups when studying factorizations of a group.

\begin{lemma}\label{LemXia04}
Let $G=HK$ be a factorization. Then for all $x,y\in G$ we have $G=H^xK^y$ with $H^x\cap K^y\cong H\cap K$.
\end{lemma}

\begin{proof}
As $xy^{-1}\in G=HK$, there exists $h\in H$ and $k\in K$ such that $xy^{-1}=hk$. Thus
\[
H^xK^y=x^{-1}Hxy^{-1}Ky=x^{-1}HhkKy=x^{-1}HKy=x^{-1}Gy=G,
\]
and
\[
H^x\cap K^y=(H^{xy^{-1}}\cap K)^y\cong H^{xy^{-1}}\cap K=H^{hk}\cap K=H^k\cap K=(H\cap K)^k\cong H\cap K.\qedhere
\]
\end{proof}

\section{Notation}

Throughout this paper, let $q=p^f$ be a power of a prime $p$, let $m\geqslant4$ be an integer, let $\,\overline{\phantom{\varphi}}\,$ be the homomorphism from $\GaO_{2m}^-(q)$ to $\mathrm{P\Gamma O}_{2m}^-(q)$ modulo scalars, let $V$ be a vector space of dimension $2m$ over $\bbF_q$ equipped with a nondegenerate quadratic form $Q$, whose associated bilinear form is $\beta$, let $V_\sharp$ be a vector space of dimension $m$ over $\bbF_{q^2}$ with the same underlying set as $V$, and let $\Tr$ be the trace of the field extension $\bbF_{q^2}/\bbF_q$.
For $w\in V$ with $Q(w)\neq0$, let $r_w$ be the reflection in $w$ defined by
\[
r_w\colon V\to V,\quad v\mapsto v-\frac{\beta(v,w)}{Q(w)}w.
\]

We fix the notation for some field extension subgroups of $\mathrm{O}(V)$ as follows, according to the parity of $m$.

First assume that $m=2\ell+1$ is odd. In this case we may equip $V_\sharp$ with a nondegenerate unitary form $\beta_\sharp$ such that $Q(v)=\beta_\sharp(v,v)$ for all $v\in V$, and thus $\GU(V_\sharp)<\mathrm{O}(V)$. Take a standard $\bbF_{q^2}$-basis $E_1,F_1,\dots,E_\ell,F_\ell,D$ for $V_\sharp$, so that
\[
\beta_\sharp(E_i,E_j)=\beta_\sharp(F_i,F_j)=\beta_\sharp(E_i,D)=\beta_\sharp(F_i,D)=0,\quad\beta_\sharp(E_i,F_j)=\delta_{i,j},\quad\beta_\sharp(D,D)=1
\]
for all $i,j\in\{1,\dots,\ell\}$. Let $\psi\in\GaU(V_\sharp)$ such that
\[
\psi\colon a_1E_1+b_1F_1+\dots+a_\ell E_\ell+b_\ell F_\ell+cD\mapsto a_1^pE_1+b_1^pF_1+\dots+a_\ell^pE_\ell+b_\ell^pF_\ell+c^pD
\]
for $a_1,b_1\dots,a_\ell,b_\ell,c\in\bbF_{q^2}$, and let $\lambda\in\bbF_{q^2}$ such that $\lambda+\lambda^q=1$. Then we have
\begin{align}
&Q(\lambda E_1)=\beta_\sharp(\lambda E_1,\lambda E_1)=0,\label{EqnOmegaMinus01}\\
&Q(F_1)=\beta_\sharp(F_1,F_1)=0,\label{EqnOmegaMinus02}\\
&Q(\lambda E_1+F_1)=\beta_\sharp(\lambda E_1+F_1,\lambda E_1+F_1)=\lambda+\lambda^q=1,\nonumber
\end{align}
and hence
\begin{equation}\label{EqnOmegaMinus03}
\beta(\lambda E_1,F_1)=Q(\lambda E_1+F_1)-Q(\lambda E_1)-Q(F_1)=1.
\end{equation}
From~\eqref{EqnOmegaMinus01}--\eqref{EqnOmegaMinus03} we see that $(\lambda E_1,F_1)$ is a hyperbolic pair with respect to $Q$.
Thus there exists a standard basis $e_1,f_1,\dots,e_{m-1},f_{m-1},d,d'$ for $V$ as in~\cite[2.2.3]{LPS1990} such that
\[
e_1=\lambda E_1\ \text{ and }\ f_1=F_1.
\]
Then $Q(e_1+f_1)=\beta_\sharp(e_1+f_1,e_1+f_1)=1$, which implies that $e_1+f_1$ is a nonsingular vector in both $V$ (with respect to $Q$) and $V_\sharp$ (with respect to $\beta_\sharp$). Let $\phi\in\GaO(V)$ be as defined in~\cite[\S2.8]{KL1990} with respect to the basis $e_1,f_1,\dots,e_{m-1},f_{m-1},d,d'$.
Then $\phi$ fixes $e_1,f_1,\dots,e_{m-1},f_{m-1},d$, and commutes with $r_{e_1+f_1}$.

Next assume that $m=2\ell$ is even. In this case we may equip $V_\sharp$ with a nondegenerate quadratic form $Q_\sharp$ of minus type such that $Q(v)=\Tr(Q_\sharp(v))$ for all $v\in V$, and thus $\mathrm{O}(V_\sharp)<\mathrm{O}(V)$. Let $\beta_\sharp$ be the associated symmetric bilinear linear form of $Q_\sharp$, and take a standard $\bbF_{q^2}$-basis $E_1,F_1,\dots,E_{\ell-1},F_{\ell-1},D,D'$ for $V_\sharp$, so that
\begin{align*}
&Q_\sharp(E_i)=Q_\sharp(F_i)=0,\quad Q_\sharp(D)=1=\beta_\sharp(D,D'),\quad Q_\sharp(D')=\mu,\quad\beta_\sharp(E_i,F_j)=\delta_{i,j},\\
&\beta_\sharp(E_i,E_j)=\beta_\sharp(F_i,F_j)=\beta_\sharp(E_i,D)=\beta_\sharp(F_i,D)=\beta_\sharp(E_i,D')=\beta_\sharp(F_i,D')=0
\end{align*}
for all $i,j\in\{1,\dots,\ell-1\}$, where $\mu\in\bbF_{q^2}$ such that $x^2+x+\mu$ is irreducible over $\bbF_{q^2}$.

\section{Infinite families of $(X,Y)$ in Table~\ref{TabOmegaMinus}}\label{SecOmegaMinus01}

In the first lemma we construct the factor pair $(X,Y)$ in Row~1 of Table~\ref{TabOmegaMinus}.

\begin{lemma}\label{LemOmegaMinus01}
Let $G=\Omega(V)=\Omega_{2m}^-(q)$ with odd $m$, let $H=G_{e_1+f_1}$, let $K=\SU(V_\sharp)$, let $Z=\overline{G}$, let $X=\overline{H}$, and let $Y=\overline{K}$. Then $H\cap K=\SU_{m-1}(q)$, and $Z=XY$ with $Z=\POm_{2m}^-(q)$, $X\cong H=\Omega_{2m-1}(q)$ and $Y\cong K=\SU_m(q)$.
\end{lemma}

\begin{proof}
It is clear that $Z=\POm_{2m}^-(q)$, and $Y\cong K=\SU_m(q)$ as $m$ is odd. Since $e_1+f_1$ is a nonsingular vector in both $V$ (with respect to $Q$) and $V_\sharp$ (with respect to $\beta_\sharp$), we have $X\cong H=G_{e_1+f_1}=\Omega_{2m-1}(q)$ and
\[
H\cap K=G_{e_1+f_1}\cap\SU(V_\sharp)=\SU(V_\sharp)_{e_1+f_1}=\SU_{m-1}(q).
\]
It follows that
\[
\frac{|K|}{|H\cap K|}=\frac{|\SU_m(q)|}{|\SU_{m-1}(q)|}=q^{m-1}(q^m+1)=\frac{|\Omega_{2m}^-(q)|}{|\Omega_{2m-1}(q)|}=\frac{|G|}{|H|},
\]
and so $G=HK$. Hence $Z=\overline{G}=\overline{H}\,\overline{K}=XY$.
\end{proof}

The second lemma is on the factor pair $(X,Y)$ in Row~2 of Table~\ref{TabOmegaMinus}.

\begin{lemma}\label{LemOmegaMinus02}
Let $G=\Omega(V)=\Omega_{2m}^-(q)$ with odd $m$, let $H=G_{E_1}$, let $K=\SU(V_\sharp)$, let $Z=\overline{G}$, let $X=\overline{H}$, and let $Y=\overline{K}$. Then $H\cap K=(q.q^{2m-4}){:}\SU_{m-2}(q)$, and $Z=XY$ with $Z=\POm_{2m}^-(q)$, $X\cong H=q^{2m-2}{:}\Omega_{2m-2}^-(q)$ and $Y\cong K=\SU_m(q)$.
\end{lemma}

\begin{proof}
It is clear that $Z=\POm_{2m}^-(q)$. Since $m$ is odd, we have $Y\cong K=\SU_m(q)$. Since $Q(E_1)=\beta_\sharp(E_1,E_1)=0$, we see that $E_1$ is a singular vector in both $V$ (with respect to $Q$) and $V_\sharp$ (with respect to $\beta_\sharp$). Thus $X\cong H=G_{E_1}=q^{2m-2}{:}\Omega_{2m-2}^-(q)$ and
\[
H\cap K=G_{E_1}\cap\SU(V_\sharp)=\SU(V_\sharp)_{E_1}=(q.q^{2m-4}){:}\SU_{m-2}(q).
\]
It follows that
\[
\frac{|K|}{|H\cap K|}=\frac{|\SU_m(q)|}{|(q.q^{2m-4}){:}\SU_{m-2}(q)|}=(q^m+1)(q^{m-1}-1)=\frac{|\Omega_{2m}^-(q)|}{|q^{2m-2}{:}\Omega_{2m-2}^-(q)|}=\frac{|G|}{|H|},
\]
and so $G=HK$. Therefore, $Z=\overline{G}=\overline{H}\,\overline{K}=XY$.
\end{proof}

In the following two lemmas we construct the factor pairs $(X,Y)$ in Rows~3--5 of Table~\ref{TabOmegaMinus}.
Recall the notation of the field extension subgroups of $\mathrm{O}(V)$ for odd $m$ given at the start of this section.

\begin{lemma}\label{LemOmegaMinus03}
Let $Z=\Omega(V)=\Omega_{2m}^-(2)$ with $q=2$ and $m$ odd, let $X=Z_{\{e_1,f_1\}}$, and let $Y=\SU(V_\sharp)$. Then $Z=XY$ with $X=\Omega_{2m-2}^-(2).2$, $Y=\SU_m(2)$ and $X\cap Y=\SU_{m-2}(2)$.
\end{lemma}

\begin{proof}
Clearly, $Y=\SU_m(2)$. Since $(e_1,f_1)$ is a hyperbolic pair with respect to $Q$, we have $Z_{e_1,f_1}=\Omega_{2m-2}^-(q)$ and $Z_{\{e_1,f_1\}}=Z_{e_1,f_1}{:}\langle s\rangle$ for some $s$ swapping $e_1$ and $f_1$. In particular, $X=\Omega_{2m-2}^-(2).2$.

Suppose that $Y_{\{e_1,f_1\}}\neq Y_{e_1,f_1}$, that is, there exists $t\in Y$ swapping $e_1=\lambda E_1$ and $f_1=F_1$. Then as $t\in Y=\SU(V_\sharp)$, we obtain
\[
\lambda=\beta_\sharp(\lambda E_1,F_1)=\beta_\sharp((\lambda E_1)^t,F_1^t)=\beta_\sharp(F_1,\lambda E_1)=\lambda^q,
\]
contradicting the condition $\lambda+\lambda^q=1$.

Thus we conclude that $Y_{\{e_1,f_1\}}=Y_{e_1,f_1}$. Consequently,
\[
X\cap Y=Z_{\{e_1,f_1\}}\cap Y=Y_{\{e_1,f_1\}}=Y_{e_1,f_1}=\SU(V_\sharp)_{e_1,f_1}=\SU(V_\sharp)_{E_1,F_1}=\SU_{m-2}(2).
\]
As $q=2$, we then derive that
\[
\frac{|Y|}{|X\cap Y|}=\frac{|\SU_m(2)|}{|\SU_{m-2}(2)|}=2^{2m-3}(2^m+1)(2^{m-1}-1)=\frac{|\Omega_{2m}^-(2)|}{|\Omega^-_{2m-2}(2).2|}=\frac{|Z|}{|X|},
\]
which yields $Z=XY$.
\end{proof}

\begin{lemma}\label{LemOmegaMinus04}
Let $Z=\Omega(V){:}\langle\rho\rangle$ with $q\in\{2,4\}$, $m$ odd and $\rho\in\{\phi,(r_{e_1+f_1})^{2/f}\phi\}$, let $X=\Omega(V)_{e_1,f_1}{:}\langle\rho\rangle$, and let $Y=\SU(V_\sharp){:}\langle\psi\rangle$. Then $Z=XY$ with $Z=\GaO_{2m}^-(q)$, $X=\Omega_{2m-2}^-(q).(2f)$, $Y=\SU_m(q).(2f)$ and $X\cap Y=\SU_{m-2}(q)$.
\end{lemma}

\begin{proof}
Clearly, $Z=\GaO_{2m}^-(q)$, $X=\Omega_{2m-2}^-(q).(2f)$, and $Y=\SU_m(q).(2f)$.

Suppose that $X\cap Y\nleqslant\Omega(V)$. Then there exist $s\in\Omega(V)_{e_1,f_1}$ and $t\in\SU(V_\sharp)$ such that $\phi^fs=\psi^ft$. Since $\phi$ and $s$ both fix $e_1$ and $f_1$, it follows that $\phi^fs$ fixes $e_1=\lambda E_1$ and $f_1=F_1$. This together with $t\in\SU(V_\sharp)$ implies that
\begin{align*}
\lambda=\beta_\sharp(\lambda E_1,F_1)&=\beta_\sharp((\lambda E_1)^{\phi^fs},F_1^{\phi^fs})\\
&=\beta_\sharp((\lambda E_1)^{\psi^ft},F_1^{\psi^ft})=\beta_\sharp((\lambda^qE_1)^t,F_1^t)=\beta_\sharp(\lambda^qE_1,F_1)=\lambda^q,
\end{align*}
contradicting the condition $\lambda+\lambda^q=1$.

Thus we conclude that $X\cap Y\leqslant\Omega(V)$. Accordingly,
\[
X\cap Y=(X\cap\Omega(V))\cap Y=\Omega(V)_{e_1,f_1}\cap Y=\SU(V_\sharp)_{e_1,f_1}=\SU(V_\sharp)_{E_1,F_1}=\SU_{m-2}(q).
\]
Observe $q=2f$ as $q\in\{2,4\}$. We then derive that
\[
\frac{|Y|}{|X\cap Y|}=\frac{|\SU_m(q).(2f)|}{|\SU_{m-2}(q)|}=2fq^{2m-3}(q^m+1)(q^{m-1}-1)=\frac{|\GaO^-_{2m}(q)|}{|\Omega_{2m-2}^-(q).(2f)|}=\frac{|Z|}{|X|}.
\]
This yields $Z=XY$.
\end{proof}

In the next lemma we construct the factor pairs $(X,Y)$ in Rows~6--7 of Table~\ref{TabOmegaMinus}.
Recall the notation of the field extension subgroups of $\mathrm{O}(V)$ for even $m$ given at the start of this section.

\begin{lemma}\label{LemOmegaMinus05}
Let $Z=\GaO(V)=\GaO_{2m}^-(q)$ with $q\in\{2,4\}$ and $m$ even, let $X=\GaO(V_\sharp)$, and let $Y=Z_{D'}$. Then $Z=XY$ with $X=\GaO_m^-(q^2)$, $Y=\Omega_{2m-1}(q).(2f)$ and $X\cap Y=\Omega_{m-1}(q^2).2$.
\end{lemma}

\begin{proof}
It is clear that $X=\GaO_m^-(q^2)$ is maximal in $Z$. Since $x^2+x+\mu$ is irreducible, we have $\mu\notin\bbF_q$, that is, $\mu\neq\mu^q$. Then as $q$ is even, we obtain
\[
Q(D')=\Tr(Q_\sharp(D'))=\Tr(\mu)=\mu+\mu^q\neq0.
\]
It follows that an element in $Z$ stabilizes $\langle D'\rangle_{\bbF_q}$ if and only if it fixes $D'$. Hence
\[
Y=Z_{D'}=Z_{\langle D'\rangle_{\bbF_q}}=\N_1[Z]=\Omega_{2m-1}(q).(2f)
\]
is maximal in $Z$. Then~\cite[Theorem~A]{LPS1990} asserts that $Z=XY$, which implies
\[
|X\cap Y|=\frac{|X|}{|Z|/|Y|}=\frac{|\GaO_m^-(q^2)|}{|Z|/|\N_1[Z]|}=\frac{|\GaO_m^-(q^2)|}{|\Omega_{2m}^-(q)|/|\Omega_{2m-1}(q)|}=\frac{|\GaO_m^-(q^2)|}{q^{m-1}(q^m+1)}=2|\Omega_{m-1}(q)|
\]
as $q=2f$. Let $r$ be the reflection in $V_\sharp$ with respect to the vector $D'$. Then $r\in\mathrm{O}(V_\sharp)\setminus\Omega(V_\sharp)$, and $r$ fixes $D'$. This implies that
\[
X\cap Y=\GaO(V_\sharp)\cap Z_{D'}\geqslant\Omega(V_\sharp){:}\langle r\rangle=\Omega_{m-1}(q^2).2,
\]
which together with $|X\cap Y|=2|\Omega_{m-1}(q)|$ leads to $X\cap Y=\Omega(V_\sharp){:}\langle r\rangle=\Omega_{m-1}(q^2).2$.
\end{proof}

Now we give the factor pairs $(X,Y)$ in Rows~8--9 of Table~\ref{TabOmegaMinus}.

\begin{lemma}\label{LemOmegaMinus06}
Let $Z=\GaO(V)=\GaO_{2m}^-(q)$ with $q\in\{2,4\}$ and $m=2\ell$ for some odd $\ell$, let $X=\SU_\ell(q^2){:}(4f)<\GaO_m^-(q^2)<Z$, and let $Y=Z_{D'}$. Then $Z=XY$ with $Y=\Omega_{2m-1}(q).(2f)$ and $X\cap Y=\SU_{\ell-1}(q^2).2$.
\end{lemma}

\begin{proof}
Similarly as in the proof of Lemma~\ref{LemOmegaMinus05} we have $Y=\Omega_{2m-1}(q).(2f)$. Let $A=\GaO_m^-(q^2)$ be a maximal subgroup of $Z$ containing $X$. Then Lemma~\ref{LemOmegaMinus05} asserts that $Z=AY$ with $A\cap Y=\Omega_{m-1}(q^2).2$. Moreover, from Lemma~\ref{LemOmegaMinus01} we see that $X\cap Y=X\cap(A\cap Y)$ has a normal subgroup $\SU_{\ell-1}(q^2)$, and $A=X(A\cap Y)$. Thus $Z=X(A\cap Y)Y=XY$ and
\[
|X\cap Y|=|X\cap(A\cap Y)|=\frac{|X||A\cap Y|}{|A|}=\frac{|\SU_\ell(q^2){:}(4f)||\Omega_{m-1}(q^2).2|}{|\GaO_m^-(q^2)|}=2|\SU_{\ell-1}(q^2)|.
\]
Consequently, $X\cap Y=\SU_{\ell-1}(q^2).2$.
\end{proof}

\section{Sporadic cases of $(X,Y)$ in Table~\ref{TabOmegaMinus}}\label{SecOmegaMinus02}

In this section, we give the sporadic pairs $(X,Y)$ in Table~\ref{TabOmegaMinus}, namely, the pairs $(X,Y)$ in Rows~10--11 of Table~\ref{TabOmegaMinus}.

\begin{lemma}\label{LemOmegaMinus07}
Let $Z=\Omega_{10}^-(2)$, let $X$ be a subgroup of $Z$ isomorphic to $\A_{12}$ or $\M_{12}$ (there is a unique conjugacy class of such subgroups $X$ in each case), and let $Y=\Pa_1[Z]=2^8{:}\Omega_8^-(2)$. Then $Z=XY$ with
\[
X\cap Y=
\begin{cases}
(\A_4\times\A_8).2&\textup{if }X=\A_{12}\\
2_+^{1+4}.\Sy_3&\textup{if }X=\M_{12}.
\end{cases}
\]
\end{lemma}

\begin{lemma}\label{LemOmegaMinus08}
Let $Z=\Omega_{18}^-(2)$, let $X=3.\J_3<\SU_9(2)<Z$, and let $Y=\Pa_1[Z]=2^{16}{:}\Omega_{16}^-(2)$. Then $Z=XY$ with $X\cap Y=2^{2+4}.(3\times\Sy_3)$.
\end{lemma}

\begin{proof}
Let $M=\SU_9(2)$ be a subgroup of $Z$ containing $X$. By Lemma~\ref{LemOmegaMinus02} we have $M\cap Y=2^{1+14}{:}\SU_7(2)$ with $M\cap Y\ngeqslant\Z(M)=3$.
Then it follows from Lemma~\ref{LemUnitary18} that
\[
X\cap Y=X\cap(M\cap Y)\cong(X/\Z(M))\cap((M\cap Y)\Z(M)/\Z(M))=2^{2+4}.(3\times\Sy_3).
\]
Therefore,
\[
\frac{|X|}{|X\cap Y|}=\frac{|3.\J_3|}{|2^{2+4}.(3\times\Sy_3)|}=(2^9+1)(2^8-1)=\frac{|\Omega_{18}^-(2)|}{|2^{16}{:}\Omega_{16}^-(2)|}=\frac{|Z|}{|Y|},
\]
and so $Z=XY$.
\end{proof}

\section{Proof of Theorem~\ref{ThmOmegaMinus}}

Let $G$ be an almost simple group with socle $L=\POm_{2m}^-(q)$, and let $H$ and $K$ be nonsolvable subgroups of $G$ not containing $L$.
In Subsections~\ref{SecOmegaMinus01} and~\ref{SecOmegaMinus02} it is shown that all pairs $(X,Y)$ in Table~\ref{TabOmegaMinus} are factor pairs of $L$.
Hence by Lemma~\ref{LemXia02} we only need to prove that, if $G=HK$, then $(H,K)$ tightly contains $(X^\alpha,Y^\alpha)$ for some $(X,Y)$ in Table~\ref{TabOmegaMinus} and $\alpha\in\Aut(L)$. Suppose that
\[
G=HK.
\]
Then by~\cite[Theorem~7.1]{LWX} the triple $(L,H^{(\infty)},K^{(\infty)})$ lies in Table~\ref{TabInftyOmegaMinus}.

\begin{table}[htbp]
\captionsetup{justification=centering}
\caption{$(L,H^{(\infty)},K^{(\infty)})$ for orthogonal groups of minus type}\label{TabInftyOmegaMinus}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Row & $L$ & $H^{(\infty)}$ & $K^{(\infty)}$ & Conditions\\
\hline
1 & $\POm_{2m}^-(q)$ & $\Omega_{2m-1}(q)$, $\Omega^-_{2m-2}(q)$, $q^{2m-2}{:}\Omega^-_{2m-2}(q)$ & $\SU_m(q)$ & $m$ odd\\
2 & $\Omega_{2m}^-(2)$ & $\SU_{m/2}(4)$ ($m/2$ odd), $\Omega_m^-(4)$, & $\Sp_{2m-2}(2)$ & \\
 & & $\SU_{m/4}(16)$ ($m/4$ odd), $\Omega_{m/2}^-(16)$ & & \\
3 & $\Omega_{2m}^-(4)$ & $\SU_{m/2}(16)$ ($m/2$ odd), $\Omega_m^-(16)$ & $\Sp_{2m-2}(4)$ & \\
\hline
4 & $\Omega_{10}^-(2)$ & $\A_{12}$, $\M_{12}$ & $2^8{:}\Omega_8^-(2)$ & \\
5 & $\Omega_{18}^-(2)$ & $3.\J_3$ & $2^{16}{:}\Omega_{16}^-(2)$ & \\
\hline
\end{tabular}
\vspace{3mm}
\end{table}

\begin{lemma}\label{LemOmegaMinus09}
Suppose that $H^{(\infty)}=\Omega_{2m-2}^-(q)$ and $K^{(\infty)}=\SU_m(q)$ with $m$ odd. Then $q\in\{2,4\}$, and $(H,K)$ tightly contains some pair $(X,Y)$ in Rows~\emph{3--5} of Table~$\ref{TabOmegaMinus}$.
\end{lemma}

\begin{proof}
Let $M=\Nor_G(H^{(\infty)})$. Then $H\leqslant M=\N_2^+[G]$, and so $G=MK$. Note from~\cite[Table~3.5.F]{KL1990} that $M$ is maximal in $G$ for $q\geqslant4$. Then by~\cite[Theorem~A]{LPS1990} we have $q\leqslant4$. Without loss of generality, we may assume $H^{(\infty)}=\Omega(V)_{e_1,f_1}$ and $K^{(\infty)}=\SU(V_\sharp)$. Then
\[
H\cap K\geqslant H^{(\infty)}\cap K^{(\infty)}=\SU(V_\sharp)_{e_1,f_1}=\SU(V_\sharp)_{E_1,F_1}=\SU_{m-2}(q).
\]
Let $\calO=G/L$, $\calO_1=H/H^{(\infty)}$ and $\calO_2=K/K^{(\infty)}$. It follows that
\[
|H^{(\infty)}|_p|\calO_1|_p|K^{(\infty)}|_p|\calO_2|_p=|H|_p|K|_p=|G|_p|H\cap K|_p\geqslant|L|_p|\calO|_p|\SU_{m-2}(q)|_p,
\]
and thus
\begin{equation}\label{EqnOmegaMinus04}
|\calO_1|_p|\calO_2|_p\geqslant\frac{|L|_p|\SU_{m-2}(q)|_p|\calO|_p}{|H^{(\infty)}|_p|K^{(\infty)}|_p}
=\frac{|\Omega_{2m}^-(q)|_p|\SU_{m-2}(q)|_p|\calO|_p}{|\Omega_{2m}^-(q)|_p|\SU_m(q)|_p}=q|\calO|_p.
\end{equation}
Note from~\cite[Proposition~4.3.18]{KL1990} that $|K\cap L|_p=|K^{(\infty)}|_p$.
Hence $|\calO|_p\geqslant|KL/L|_p=|K/K\cap L|_p=|\calO_2|_p$, and so~\eqref{EqnOmegaMinus04} yields $|\calO_1|_p\geqslant q$.

First assume $q=2$. Then $|\calO_1|_2\geqslant2$, and so $H\geqslant X$ for some $X=\Omega_{2m-2}^-(2).2$.
Applying this conclusion to the factorization $HL\cap KL=(H\cap KL)(K\cap HL)$ we also obtain $H\cap KL\geqslant\Omega_{2m-2}^-(2).2$.
If $X\leqslant L$, then $(H,K)$ tightly contains the pair $(X,Y)$ with $Y=K^{(\infty)}$, as in Row~3 of Table~\ref{TabOmegaMinus}.
If $X\nleqslant L$ and $K\leqslant L$, then $H\cap L=H\cap KL\geqslant\Omega_{2m-2}^-(2).2$ and so we are led to the previous case where $X\leqslant L$.
If $X\nleqslant L$ and $K\nleqslant L$, then $(H,K)$ tightly contains the pair $(X,Y)$ with $Y=\SU_m(2).2$, as in Row~4 of Table~\ref{TabOmegaMinus}.

Next assume $q=3$. Then $|H|_3\leqslant|M|_3=|\Omega_{2m-2}^-(3)|_3=|H^{(\infty)}|_3$, which implies that $|\calO_1|_3=1$, contradicting the conclusion $|\calO_1|_p\geqslant q$.

Finally assume $q=4$. By~\cite[Theorem~A]{LPS1990} we have $G=\Aut(L)=\Omega_{2m-2}^-(4).4$. This conclusion applied to the factorization $HL\cap KL=(H\cap KL)(K\cap HL)$ leads to $HL=KL=\Omega_{2m-2}^-(4).4$. It follows that $H=(H\cap L).4$ and $K=(K\cap L).4$. Hence $(H,K)$ tightly contains the pair $(X,Y)$ in Row~5 of Table~\ref{TabOmegaMinus}.
\end{proof}

By virtue of Lemma~\ref{LemOmegaMinus09}, if $(L,H^{(\infty)},K^{(\infty)})$ lies in Row~1 of Table~\ref{TabInftyOmegaMinus}, then $(H,K)$ tightly contains some pair $(X,Y)$ in Rows~1--5 of Table~\ref{TabOmegaMinus}.

If $(L,H^{(\infty)},K^{(\infty)})$ lies in Row~2 or~3 of Table~\ref{TabInftyOmegaMinus}, then the next two lemmas show that $(H,K)$ tightly contains $(X,Y)$ in Rows~6--9 of Table~\ref{TabOmegaMinus}.

\begin{lemma}
Suppose that $K^{(\infty)}=\Sp_{2m-2}(q)$ with $q\in\{2,4\}$ and either $H^{(\infty)}=\SU_{m/2}(q^2)$ with $m/2$ odd or $H^{(\infty)}=\Omega_m^-(q^2)$. Then $(H,K)$ tightly contains some pair $(X,Y)$ in Rows~\emph{6--9} of Table~$\ref{TabOmegaMinus}$.
\end{lemma}

\begin{proof}
Since $K^{(\infty)}=\Sp_{2m-2}(q)$, we have $K\leqslant B$ for some $B=\N_1[G]$. Since either $H^{(\infty)}=\SU_{m/2}(q^2)$ with $m/2$ odd or $H^{(\infty)}=\Omega_m^-(q^2)$, there exists a maximal subgroup $A$ of $G$ such that $A^{(\infty)}=\Omega_m^-(q^2)$. From $G=HK$ we deduce $G=AB$. Then~\cite[Theorem~A]{LPS1990} implies $G=\Aut(L)=\mathrm{O}_{2m}^-(q)$. Applying this conclusion to the factorization $HL\cap KL=(H\cap KL)(K\cap HL)$ we obtain $HL=KL=\mathrm{O}_{2m}^-(q)$. Hence $H/H\cap L\cong K/K\cap L=2f$, and so $(H,K)$ tightly contains some pair $(X,Y)$ in Rows~6--9 of Table~\ref{TabOmegaMinus}.
\end{proof}

\begin{lemma}
If $q=2$ and either $H^{(\infty)}=\SU_{m/4}(16)$ with $m/4$ odd or $H^{(\infty)}=\Omega_{m/2}^-(16)$, then $K^{(\infty)}\ncong\Sp_{2m-2}(2)$.
\end{lemma}

\begin{proof}
Suppose for a contradiction that $K^{(\infty)}\cong\Sp_{2m-2}(2)$. Then $K\leqslant B$ for some $B=\N_1[G]$. Since either $H^{(\infty)}=\SU_{m/4}(16)$ with $m/4$ odd or $H^{(\infty)}=\Omega_{m/2}^-(16)$, there exist subgroups $M$ and $A$ of $G$ such that $H\leqslant M<A$ such that $A$ is a maximal subgroup of $G$ with $A^{(\infty)}=\Omega_m^-(4)$ and $M$ is a maximal subgroup of $A$ with $M^{(\infty)}=\Omega_{m/2}^-(16)$. From $G=HK$ we deduce $A=M(A\cap B)$ and $G=AB$. Then~\cite[Theorem~A]{LPS1990} implies $G=\Aut(L)=\mathrm{O}_{2m}^-(2)$. It follows that $A=\GaO_m^-(4)$ and $B=\Omega_{2m-1}(2).2$. Hence $A\cap B=\Omega_{m-1}(4).2$ as Lemma~\ref{LemOmegaMinus05} asserts. Let $N=(A\cap B)\Soc(A)$. Then since $A/\Soc(A)=4$, it follows that $N<A$. From $A=M(A\cap B)$ and $A\cap B\leqslant N$ we deduce
\[
N=(M\cap N)(A\cap B).
\]
Moreover, $\Soc(N)=\Soc(A)=\Omega_m^-(4)$, $(M\cap N)^{(\infty)}=\Omega_{m/2}^-(16)$, and $(A\cap B)^{(\infty)}=\Omega_{m-1}(4)$. Thus~\cite[Theorem~A]{LPS1990} implies $N=\GaO_m^-(4)$, contradicting the conclusion that $N<A$.
\end{proof}

Finally, if $(L,H^{(\infty)},K^{(\infty)})$ lies in Row~4 or~5 of Table~\ref{TabInftyOmegaMinus}, then the pair $(H,K)$ tightly contains $(X,Y)=(H^{(\infty)},K^{(\infty)})$ in Row~10 or~11, respectively, of Table~\ref{TabOmegaMinus}. This completes the proof of Theorem~\ref{ThmOmegaMinus}.

\section*{Acknowledgments}
The first author acknowledges the support of NNSFC grants no.~11771200 and no.~11931005. The second author acknowledges the support of NNSFC grant no.~12061083.

redpajama_set_name: RedPajamaArXiv


text:
Das Webertief ist ein Meerestief im Westteil des Pazifischen Ozeans (Pazifik) und mit 7440 m Meerestiefe die tiefste Stelle der Bandasee. Es befindet sich im Osten der Bandasee, nordwestlich der Tanimbarinseln.

Das Meerestief wurde nach Max Wilhelm Carl Weber benannt, der auf der Siboga-Expedition zwischen 1899 und 1900 diese Gewässer detailliert hydrografisch erkundete.

Meerestief
Bandasee


text:
Ross Stores Reports Record Second Quarter 2009 Results, With Earnings Per Share Up 52%, Raises Third and Fourth Quarter 2009 Sales and EPS Targets
PLEASANTON, Calif., Aug. 20 /PRNewswire-FirstCall/ -- Ross Stores, Inc. (Nasdaq: ROST) today reported that earnings per share for the 13 weeks ended August 1, 2009 rose 52% to $.82, from $.54 for the 13 weeks ended August 2, 2008. Net earnings for the second quarter of 2009 grew 45% to a record $103.4 million, from $71.3 million in the second quarter of 2008. Sales for the 13 weeks ended August 1, 2009 increased 8% to $1.769 billion, with comparable store sales up 3% on top of a strong 6% gain in the prior year.
For the six months ended August 1, 2009, earnings per share rose 37% to $1.55, from $1.13 for the six months ended August 2, 2008. Net earnings for the six months ended August 1, 2009 grew 29% to a record $194.8 million, compared to $150.8 million in the prior year period. Sales for the first six months of 2009 increased 8% to $3.460 billion, with comparable store sales up 3% on top of a 5% gain last year.
Michael Balmuth, Vice Chairman, President and Chief Executive Officer, commented, "We are extremely pleased with our healthy sales and robust earnings gains in the second quarter and first six months, both of which were well ahead of plan. This outstanding performance is even more notable considering the ongoing difficult macro economic and retail climate. Our ability to deliver compelling bargains, while operating our business on much lower inventories, remains the primary driver of our strong results. The best performing merchandise categories during the second quarter and year to date periods were Dresses and Shoes, while the strongest regions were the Mid-Atlantic and Southeast."
Mr. Balmuth continued, "Second quarter operating margin grew about 260 basis points to 9.7%, driven by a 240 basis point improvement in gross margin and a 20 basis point decrease in selling, general and administrative costs versus the prior year. The gross margin improvement was mainly due to substantial gains in merchandise margin and much lower freight costs as a percent of sales."
Mr. Balmuth also noted, "As we ended the first half of the year, our balance sheet and cash flows remained healthy. We continued to return capital to stockholders through our stock repurchase and dividend programs. During the first six months of fiscal 2009, we repurchased 4.2 million shares of common stock for an aggregate purchase price of $154 million. We are on track to complete the remaining $146 million stock repurchase authorization by the end of fiscal 2009."
Looking ahead, Mr. Balmuth said, "We are optimistic about the important back-to-school and holiday periods for a number of reasons. We delivered exceptional sales and earnings growth for the first six months on top of strong results the prior year and are up against much easier comparisons in the second half. More importantly, we are well positioned in the value retailing sector and excited about our merchandise offerings and the availability of great product as we enter the fall season. As a result, we are now forecasting same store sales gains of 5% to 6% for both the third and fourth quarters, up from our previous guidance for a 2% to 3% increase."
Based on these updated sales targets, the Company is projecting third and fourth quarter earnings per share to be in the ranges of $.57 to $.63 and $.88 to $.94, respectively. These 2009 ranges compare to earnings per share of $.44 and $.76 in the 2008 third and fourth quarters, respectively. For the fiscal year ending January 30, 2010, the Company now projects earnings per share to increase 29% to 34% to $3.00 to $3.12, up from $2.33 in fiscal 2008.
The Company will provide additional details concerning its second quarter results and management's outlook for the third and fourth quarters on a conference call to be held Thursday, August 20, 2009 at 11:00 a.m. Eastern time. Participants may listen to a real time audio webcast of the conference call by visiting the Company's website located at www.rossstores.com. A recorded version of the call will also be available until the end of October at the website address and via a telephone recording through 8:00 p.m. Eastern time on Thursday, August 27, 2009 at (706) 645-9291, PIN # 86370053.
Forward-Looking Statements: This press release and the recorded comments and transcript on our corporate website contain forward-looking statements regarding expected sales and earnings levels in future periods that are subject to risks and uncertainties which could cause our actual results to differ materially from management's current expectations. The words "plan," "expect," "target," "anticipate," "estimate," "believe," "forecast," "projected," "guidance," "looking ahead" and similar expressions identify forward-looking statements. Risk factors for Ross Dress for Less ("Ross") and dd's DISCOUNTS include, without limitation, competitive pressures in the apparel or home-related merchandise industry; changes in the level of consumer spending on or preferences for apparel or home-related merchandise, including the potential impact from uncertainty in financial and credit markets and the severity and duration of the current recession; changes in geopolitical and general economic conditions; unseasonable weather trends; disruptions in supply chain; lower than planned gross margin, including higher than planned markdowns and higher than expected inventory shortage; greater than planned operating costs; our ability to continue to purchase attractive brand-name merchandise at desirable discounts; our ability to attract and retain personnel with the retail talent necessary to execute our strategies; our ability to effectively operate our various supply chain, core merchandising and other information systems; our ability to improve our merchandising capabilities through the development and implementation of new processes and systems enhancements; achieving and maintaining targeted levels of productivity and efficiency in our distribution centers; and obtaining acceptable new store locations. Other risk factors are detailed in our SEC filings including, without limitation, the Form 10-K for fiscal 2008 and Form 10-Q and 8-K's for fiscal 2009. The factors underlying our forecasts are dynamic and subject to change. As a result, our forecasts speak only as of the date they are given and do not necessarily reflect our outlook at any other point in time. We do not undertake to update or revise these forward-looking statements.
Ross Stores, Inc., a Fortune 500 and Nasdaq 100 (ROST) company headquartered in Pleasanton, California, is the nation's second largest off-price retailer with fiscal 2008 revenues of $6.5 billion. As of August 1, 2009 the Company operated 939 Ross Dress for Less ("Ross") stores and 51 dd's DISCOUNTS locations, compared to 888 Ross and 55 dd's DISCOUNTS locations at the end of the same period last year. Ross offers first-quality, in-season, name brand and designer apparel, accessories, footwear and home fashions for the entire family at everyday savings of 20 to 60 percent off department and specialty store regular prices. dd's DISCOUNTS features a more moderately-priced assortment of first-quality, in-season, name brand apparel, accessories, footwear and home fashions for the entire family at everyday savings of 20 to 70 percent off moderate department and discount store regular prices. Additional information is available at www.rossstores.com.
Three Months Ended      Six Months Ended
---------------------   ------------------
($000, except stores and per  August 1,  August 2,   August 1,  August 2,
share data, unaudited)      2009       2008        2009       2008
----------------------------    ----       ----        ----       ----
Sales                        $1,768,636 $1,640,412  $3,460,235 $3,196,740
Costs of goods sold         1,311,136  1,255,222   2,579,845  2,436,779
Selling, general and
administrative               286,158    268,839     558,188    516,511
Interest expense (income),
net                            1,390     (1,052)      3,046     (2,673)
-----     ------       -----     ------
Total costs and expenses  1,598,684  1,523,009   3,141,079  2,950,617
Earnings before taxes           169,952    117,403     319,156    246,123
Provision for taxes on
earnings                        66,545     46,104     124,362     95,339
------     ------     -------     ------
Net earnings                   $103,407    $71,299    $194,794   $150,784
========    =======    ========   ========
Basic                           $0.84      $0.55       $1.57      $1.15
Diluted                         $0.82      $0.54       $1.55      $1.13
Weighted average shares
outstanding (000)
Basic                         123,467    130,110     124,080    130,714
Diluted                       125,658    132,433     126,063    132,914
Cash dividends declared per
share                          $0.11      $0.10       $0.11      $0.10
Stores open at end of period        990        943         990        943
August 1,     August 2,
($000, unaudited)                       2009          2008
-----------------                       ----          ----
Cash and cash equivalents           $520,424      $309,554
Short-term investments                 1,135         2,821
Accounts receivable                   49,375        49,423
Merchandise inventory                926,244     1,018,726
Prepaid expenses and other            63,926        63,223
Deferred income taxes                 13,669        20,883
------        ------
Total current assets             1,574,773     1,464,630
Property and equipment, net            942,745       906,533
Long-term investments                   21,752        44,176
Other long-term assets                  61,379        63,078
Total assets                        $2,600,649    $2,478,417
==========    ==========
Accounts payable                    $702,977      $682,565
Accrued expenses and other           219,479       234,423
Accrued payroll and benefits         172,913       153,077
Income taxes payable                  11,268             -
------           ---
Total current liabilities        1,106,637     1,070,065
Long-term debt                         150,000       150,000
Other long-term liabilities            168,558       168,814
Deferred income taxes                  106,032        83,418
Stockholders' Equity                 1,069,422     1,006,120
---------     ---------
Total liabilities and
stockholders' equity               $2,600,649    $2,478,417
August 1,  August 2,
($000, unaudited)                                  2009       2008
-----------------                                  ----       ----
Net earnings                                     $194,794   $150,784
Adjustments to reconcile net earnings
to net cash provided by operating activities:
Depreciation and amortization                    75,502     65,866
Stock-based compensation                         13,017     11,330
Deferred income taxes                             9,400      3,275
Tax benefit from equity issuance                  5,256      6,608
Excess tax benefit from stock-based
compensation                                    (4,008)    (4,714)
Merchandise inventory                         (45,186)     6,569
Other current assets                          (16,890)   (23,257)
Accounts payable                              180,240     58,145
Other current liabilities                        (678)    23,200
Other long-term, net                            2,521      8,750
-----      -----
Net cash provided by operating activities     413,968    306,556
-------    -------
Additions to property and equipment             (80,731)  (113,472)
Proceeds from sales of property and equipment        10        117
Net proceeds from (purchases of) investments     16,811     (1,950)
------     ------
Net cash used in investing activities         (63,910)  (115,305)
-------   --------
compensation                                     4,008      4,714
Issuance of common stock related to stock
plans                                           31,745     36,470
Treasury stock purchased                         (4,546)    (2,907)
Repurchase of common stock                     (154,371)  (152,631)
Dividends paid                                  (27,825)   (24,923)
Net cash used in financing activities        (150,989)  (139,277)
--------   --------
Net increase in cash and cash equivalents         199,069     51,974
Cash and cash equivalents:
Beginning of period                             321,355    257,580
End of period                                  $520,424   $309,554
========   ========
Interest paid                                    $4,834     $4,834
Income taxes paid                              $105,012   $109,099
Non-Cash Investing Activities
Increase (decrease) in fair value of
investment securities                             $886    $(1,817)
John G. Call,
Chief Financial Officer,
+1-925-965-4315,
Bobbi Chaville,
Senior Director,
Investor Relations,
bobbi.chaville@ros.com,
both of Ross Stores, Inc.
Web Site: http://www.rossstores.com