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2018 LEAD AWARD WINNERS 21 and have them contribute while helping them discover how they matter to success, both of our team and the organization. It's tapping into their desire to contribute, to be valued and respected. I feel that everyone has valuable gifts to offer and as their leader I need to get them in a position where they can use those gifts and sometimes help them discover their gifts. We dedicate an inordinate amount of time to our jobs and when I can help #AWorldInspired them use their gifts and tap into their passion they are more personally fulfilled and engaged. My aim with a collaborative approach is that they develop ownership where it's not the company's plan or my plan, it's their plan. This creates alignment around purpose while also generating buy-in, commitment and enthusiasm for delivering on those commitments and the resiliency needed when obstacles do arise, which in life, they always do. Tell us a story of something you have done as a leader that has impacted or inspired others. It's been showing vulnerability as a leader. I've done this in several ways. With my team members, I've been open and honest regarding my personal development needs, how I am working to address them, and errors I've made. I've apologized directly to those involved, including my team members when my errors involved them. I've made it a practice to ask for feedback directly from them as well. The approach I've found most effective is my regular 1:1 meetings which I have on a recurring basis. I've used this avenue to solicit feedback on how I can better support them, the team, and their goals. I've also made sure that I've closed the loop with respect to acting on the feedback I've received— demonstrating I've not just heard their feedback, I've acted on it. I feel that by demonstrating openness and vulnerability it creates a safe space for them to do the same. This has been a powerful practice that has inspired my team members to be vulnerable with me and each other. This has enabled more effective coaching, guidance and feedback from me while strengthening relationships across the team. What are the lessons you've learned this year from your leadership experiences? As I think about this year, the lesson resonating with me is about servant leadership—the leader needs to serve the team and enable their success.
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GREEN LANTERN CORPS #10
David Andrews July 05, 2012 COMICS, DC, G.L. CORPS, PETE TOMASI
[ OVER-CHARGED? ]
John Stewart is really up sh*t creek. He does not know what to do or what to speak. Guy Gardner has an idea. Though it does come across quite queer. Plus inherently very-very Greek. Just like Writer: Peter J. Tomasi; Artist: Fernando Pasarin; and Publisher: DC Comics in May 2012. Kebab anyone?
Now there are many collective phrases which could easily sum up this solemn chapter of Alpha War, entitled 'Execution Song'. But to me, there is one phrase in particular which stands out from the crowd... 'Oh Shit!'.
Just imagine if you will what it must feel like to be a member of the Green Lantern Corps at this precise moment in time.
John Stewart has just been sentenced to death for the murder of Kirrt, and has asked to see the parents of the man he killed.
The Alpha Lanterns are making all the necessary arrangements to carry out their legal edict.
The rest of the Corp's are shocked and stunned by what is going on, just like Kirrt's wayward ring deep within space.
And after a quick snack, Guy Gardner and some of his colleges do something that is very in character for them to do.
JAIL BREAK!
I wonder what the Guardian's will think about that?
To be continued when the Alpha Lanterns give chase.
As much as I did enjoy Guy's phrase 'Boodicrap' very much, in the same vein I thought that John's play on life very pertinent when he said...
'Everybody plays Poker, Guy -- Everyday of their lives. They just don't use cards'.
Sum's up very well what Guy then does next, huh?
There's so much to like about this issue that I do not know where to begin. But I'll try my best: (1) The emotional core of this comic was just marvellous to digest, as it's melancholy spirit was something pure and dynamic to plough through. (2) I loved the fact that all the characters in this story were all depicted 'in character'. Sometimes a writer would 'fudge it' a bit, just for the sake of plot. But not this time, thank Tomasi. (3) One of the most charming aspects of this tale was the way that things plodded along at a very casual and easy going pace. Nothing rushed. All relaxed.
Now at a touch, the only slight gripe I had with this story was the way that the Guardians are starting to turn into 'Saturday Morning' cartoon-type villains. Maybe it is just me, but half of me expected them to gaze into the beyond and cackle mercilessly to themselves.
BWA-HA-HA-HA! For example.
STORY: The scene between John and Kirrt's parents was the stand out scene in my eyes. I do not think that I have ever read anything so wholesome in a comic book before. It had a nice tone. It was relatable. Plus I did love the way that this piece dissolved on a poignant note.
ART: The more that I follow Fernando Pasarins art, the more that I begin to appreciate what he is really doing on this title. Now being a bit of a film-buff myself, I can tell when a camera-man has the right style and pace for a piece comedy or an action scene. Fernando does too - I can see it in his layouts and his select choice of compositions. Class act.
BOODICRAP: OK, I know that she is smiling a tad too much in the picture provided, but trust me when I say that busty actress, Ashley Judd, can also do cold and frigid at the drop of a hat - just as she did in the movie, 'Heat'.
THE HORSE-FACED ALPHA-LANTERN: Although he is most probably dead by now, there is only one actor I know who can play this merciless mare like no nag on Earth - Mister Ed the talking horse.
You know it makes sense.
Guilty. Innocent. True. Or False. A Green Line Has Been Crossed And John Has To Pay.
OK, so this story is about the aftermath of a murder trial and how a friend tries his best to get his buddy out sing-sing. Oh! I know! What about comparing this comic to the Gene Wilder and Richard Pryor comedy 'Stir Crazy '?
Just reverse the temperament and it would be a nice match.
This is one of those books which has not disappointed me for quite some time now, and just continues month after month in delivering fresh and new takes on tried and tested ideas.
Well, have you seen a hero murder someone in a comic and then confront the parents just prior to him being sentenced to death? Also, have you seen a scene where a prisoner is reluctantly rescued from prison under his own duress in space?
No. Me neither. And that is one of the main reasons why I tell one and all to support this title, as I do not want it to take a nose-dive because the big-wigs at DC aren't draining it dry enough.
GREEN LANTERN CORPS #10 Reviewed by David Andrews on July 05, 2012 Rating: 5

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Mask of line in PIL I've got this function, which draws a bezier curve:
from random import randint
from PIL import Image

def plot_curve(image, px, py, steps=1000, color=(0, 0, 0)):
    def B(coord, i, j, t):
        if j == 0:
            return coord[i]
        return (B(coord, i, j - 1, t) * (1 - t) +
                B(coord, i + 1, j - 1, t) * t)

    img = image.load()
    for k in range(steps):
        t = float(k) / (steps - 1)
        x = int(B(px, 0, n - 1, t))
        y = int(B(py, 0, n - 1, t))
        try:
            img[x, y] = color
        except IndexError:
            pass

n = 4

for i in range(0,10):
    coord_x = [randint(0, 999), randint(0, 999), randint(0, 999), randint(0, 999)]
    coord_y = [randint(0, 999), randint(0, 999), randint(0, 999), randint(0, 999)]

    image = Image.new("RGBA", (1000, 1000))

    plot_curve(image, coord_x, coord_y)
    name = '{:03d}'.format(i) + "_double_curve" + ".png"
    image.save("./train/" + name)

how would I return the co-ordinate mask of this line, say 10px either side? I've tried to do the same bezier curve but with a thicker line, but I cannot see how to set the width of the line and then save the co-ordinates?

redpajama_set_name: RedPajamaStackExchange


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Laura Murray and complaints
Read online at https://workersliberty.org/node/33956
Submitted by AWL on 1 May, 2019 - 1:15 Author: Keith Road
The appointment (on 25 April) of Laura Murray to Head of Complaints for the Labour Party has real problems, whatever merits she may have.
Murray has been defended by Jon Lansman, chair of Momentum: "I can't think of anyone better for this role. She's developed wide knowledge about it [antisemitism], deeply wants to root it out, has engaged and worked with Jewish communities".
Jeremy Newmark, former chair of the Jewish Labour Movement, has described her as "responsible and fair", and defended her against charges that, intervening from a position in the Leader's Office, she pressed complaints staff to investigate but not suspend a Labour member who was subsequently arrested by police over antisemitic social media posts.
We remain concerned with comments Murray has made about ourselves which cast doubt on her propensity for dispassionate judgement and on whether she can really be said to have developed a view on antisemitism that fits with Lansman's description.
In a lengthy blog post in December 2016, Murray wrote about us as " a group with such extreme Trotskyist politics that they are almost a caricature of themselves ... Subtle support for imperialist wars, uncritical support for Israel and fanatical support for the European Union are amongst their policies".
We've been outside the Israeli Embassy many times criticising and protesting against Israeli government actions, and we've never seen Murray there. Such things as misinterpreting our opposition to destruction of Israel as "uncritical support" are the usual ploy of left antisemites.
Also, however much Murray may have privately improved, the Labour Party today needs to signal to the Jewish community and others that its complaints administration is visibly independent from the "Leader's Office", which is Stalinist-dominated.
The continued cycling into key jobs in the Labour "machine" of people from the same families, schools, and political networks is no way to do that.
Fight the Tories' Brexit coup plot
Boris Johnson is enough of an unprincipled opportunist that he might ditch his supporters...
New setback in USA
The conference of the US revolutionary socialist group Solidarity on the weekend 29-30...
Socialist Party calls a special conference on 21 July
The Socialist Party is not the party of Lenin and Trotsky, but rather a much-downgraded...
Who needs the "horseshoe" theory?
Significant sections of the left, or what considers itself the left, across Europe are...
Labour shifts on Brexit. Now clinch a victory!
Agitation and campaigning from the grassroots by groups like Labour for a Socialist Europe...
Solidarity 504, 1 May 2019Labour Party Momentum Left groups and people


text:
True Stories American Public Media Society Conversations
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Terrible, Thanks For Asking explicit
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系列首页•Feed
You know how when someone asks "How are you?" you just say "Fine," even if you're totally dying inside, so everyone can go about their day? "Terrible, Thanks For Asking" is the opposite of that. Nora McInerny asks real people to share their complicated and honest feelings about how they really are. It's sometimes sad, sometimes funny, and often both. From American Public Media.
类似 Terrible, Thanks For Asking 的节目
From the makers of the award-winning podcast Criminal, This is Love investigates life's most persistent mystery. Stories of sacrifice, obsession, and the ways in which we bet everything on each other.
Sincerely, X
Stories shared on the TED stage are seen and heard by millions around the world. But some ideas can't be shared in the open — there are stories that are too risky, painful, or controversial. Sincerely, X provides a space to share powerful ideas anonymously. Stories that deserve to be heard, from people who remain hidden.
Stories of the human heart. A candid, unscripted conversation between two people about what's really important in life: love, loss, family, friendship. When the world seems out of hand, tune in to StoryCorps and be reminded of the things that matter most.
The Mortified Podcast
The Mortified Podcast is a storytelling series where adults share the embarrassing things they created as kids—diaries, letters, lyrics & beyond—in front of total strangers. PS: It totally likes you. The Mortified Podcast is a proud member of Radiotopia, from PRX. Learn more at radiotopia.fm.
Extraordinary first person stories from around the world
The Kitchen Sisters Present
The Kitchen Sisters Present… Stories from the b-side of history. Lost recordings, hidden worlds, people possessed by a sound, a vision, a mission. The episodes tell deeply layered stories, lush with interviews, field recordings and music. From powerhouse producers The Kitchen Sisters (Hidden Kitchens, The Hidden World of Girls, The Sonic Memorial Project, Lost & Found Sound, Fugitive Waves and coming soon… The Keepers). "The Kitchen Sisters have done some of best radio stories ever broadcast ...
The Secret Room | True Stories
What's the one secret you've never told anybody? Welcome to The Secret Room, a podcast about the true stories no one ever tells. Hosted by Ben Hamm.
Beautiful Stories From Anonymous People
1 phone call. 1 hour. No names. No holds barred. Thats the premise behind Beautiful Stories from Anonymous People, hosted by comedian Chris Gethard (the Chris Gethard Show, Broad City, This American Life, and one of Time Outs 10 best comedians of 2015). Every week, Chris opens the phone line to one anonymous caller, and he cant hang up first, no matter what. From shocking confessions and family secrets to philosophical discussions and shameless self-promotion, anything can and will happen! T ...
Ear Hustle
Ear Hustle brings you the daily realities of life inside prison shared by those living it, and stories from the outside, post-incarceration. The podcast is a partnership between Nigel Poor, a Bay Area visual artist, and Earlonne Woods, formerly incarcerated at San Quentin State Prison, and was co-founded with former San Quentin resident Antwan Williams. The Ear Hustle team works in the San Francisco Bay Area, both in San Quentin State Prison's media lab and from offices on the outside, to pr ...
Radio Diaries
First-person diaries, sound portraits, and hidden chapters of history from Peabody Award-winning producer Joe Richman and the Radio Diaries team. From teenagers to octogenarians, prisoners to prison guards, bra saleswomen to lighthouse keepers. The extraordinary stories of ordinary life. Radio Diaries is a proud member of Radiotopia, from PRX. Learn more at radiotopia.fm


text:
For Native Americans, the Battle Against Voter Suppression in North Dakota Is Only the Beginning
Indigenous high schoolers are organizing to let their families vote in an election they argue is designed to block them from the polls.
Massoud Hayoun
In northern North Dakota's Turtle Mountain, Native American students are set to walk out of class and march to polling places with signs encouraging community members to vote.
(Photo: Stephen Morton/Getty Images)
North Dakota Native Americans say they're fired up to vote following a fraught battle over a voter identification measure that they say amounts to a thinly veiled attempt by the state's Republican officials to bar indigenous people from the polls. What's more, Native American leaders promise that their renewed energy for political engagement will last far beyond Tuesday, citing overwhelming efforts by the community's youth to get out the vote.
"This is far from the election of a candidate; it's more of an empowerment of the people," says O.J. Semans, the co-director of Four Directions, a Native American voting rights group. "I see it continuing and nothing but good happening from it hereafter."
"They're taking it upon themselves to become involved," Semans says, adding that the youth has engaged in "trickle-up" activism, inspiring their elders of voting age to travel what can be great distances and overcome technical obstacles posed by the state to vote. "That encourages us knowing the fact that we have future generations stepping up to ensure that we, the Native Americans are going to have people like Barb (Semans' partner and fellow organizer) and I generation after generation after generation. Regardless of what they try, our future generations will know how to stand up to it."
Semans attributes his optimism to the fact that the several setbacks in his battle against voter suppression seem only to have emboldened Native American voters and activists in their push toward the polls. More than ever before, community members of all stripes are encouraging their families and fellow tribal nation members to do what they can to participate in the electoral process, he says.
In their latest setback last week, North Dakota's Spirit Lake Tribe sued North Dakota Secretary of State Alvin Jaeger in a last-minute bid to block a measure requiring that North Dakotans submit identification with proof of address. That lawsuit cited several North Dakota Native Americans who had obtained the necessary identification listing a residential address but were still unable to vote, despite going to great lengths to comply with the state's rule. Many North Dakota Native Americans live in rural areas on tribal lands, typically without named and numbered roads.
The state has charged that the requirement is designed to stop voter fraud, a phenomenon that the George W. Bush administration study found to be virtually non-existent. Native Americans like Semans say that the measure is a poorly disguised attempt by North Dakota Republicans to give their candidate, Kevin Cramer, an advantage against Democratic Senator Heidi Heitkamp, a longtime champion of Native American rights. In 2012, Heitkamp won against Republican Rick Berg by just a few thousand votes.
The state had advised Native Americans to obtain addresses from local 911 coordinators, but the lawsuit filed last week cited circumstances in which Native Americans with those addresses were still blocked from voting.
On Thursday, a federal judge tossed that lawsuit, which sought a temporary injunction against the residential address requirement.
"Last week's ruling was another blow to the backbone of democracy," Semans says. But he is neither discouraged nor surprised. "Lately, rulings like that have been more common that that. We realize sometimes civil rights battles aren't won in court; it's won by movement to make change."
In October, the Supreme Court upheld a lower-court decision in favor of the state's measure to require proof of residential address. That lower-court decision acknowledged that the requirement would discourage Native American voters but held that the overwhelming concern was the prevention of voter fraud.
Jaeger's office did not respond to Pacific Standard's request for comment on the ruling last week and whether it could ensure that Native Americans would enjoy constitutionally guaranteed access to the polls.
The court decisions, which in one case seemed to admit that a ramification of the residential address requirement would be to suppress native votes, may have amounted to obstacles to the Native electorate. But they also inspired tireless work by Semans and others.
"The Supreme Court ruling and the last judge's ruling required us to do what takes six months in a matter of weeks. I'm confident we met the call but we'll see tomorrow," Semans says. "But I can tell you everyone I've talked to are all excited about voted. It's not the issue of who wins and loses; the issue is they're not going to steal their voice by creating nuisance laws." Semans' Four Directions has developed a system with California's Claremont Graduate University that allows tribe members to point to residences on electric maps and generate addresses. On-site tribal officials then printed identification that the voters use to cast ballots. There had been some question as to whether state officials would accept that identification, but the IDs were accepted at a recent early voting event in Standing Rock, North Dakota, Semans reports.
"That's a big victory for us," he says. Tribal representatives will be available at North Dakota polling stations to ensure that members aren't turned away from the polls, Semans says.
Whether they will accept ballots submitted with identification generated on site is about more than just the mid-terms—it's about the future of Native American human rights.
"It goes back to tribal sovereignty to be able to create addresses," he says. But he expects the support of non-Native North Dakotans in tomorrow's final battle against what some say is a discriminatory and suppressive voting law. "It shouldn't take a court to change [the voting law]. It should be the good people of North Dakota saying this isn't right. We're used to losing in court. But we're also used to coming back and winning. It ain't going to stop us. We're still going fight to empower tribes. That's just how it is off in Indian Country."
Native AmericansNorth DakotaPoliticsVoting RightsHeidi HeitkampMid-Terms 2018
Massoud Hayoun is a contributing writer to Pacific Standard.


Sky Zone Las Vegas will be hosting their first ever World 3-D Dodgeball Championship for adults ages 18 and over. Over 200 teams across 22 Sky Zone Indoor Trampoline Parks are expected to compete for the top prize of $25,000.
The Sky Zone Las Vegas Tournament will be held on September 28th and 29th. The regional winners will advance to the Championship Weekend at Sky Zone Torrance in Los Angeles, CA on November 9th to 11th.
The deadline to register for the Regional Qualifying Tournament at Sky Zone Las Vegas is September 27, 2012 and the entry fee per team (5-8 people) is $200. For more information and how to register visit www.World3DDodgeball.com.


This "moveable feast" of the state's literary heritage began in 1998, and brings together novelists, poets, historians, screenwriters, publishers, and more for one day of lectures, readings, and book exposition. The festival features authors from around the state of Georgia and is held in a different town every year.
The Georgia Literary Festival was last held on November 7, 2015 in Augusta.


Nearly 27,000 car crashes occurred in Colorado in 2017. Whether you're driving through wintery mountain passes, or crawling through rush hour traffic in Denver or Boulder, a car accident is a risk no matter how good of a driver you are. A car accident in Colorado can have specific terms and laws associated which help us understand your rights and access to compensation. Here are some local Colorado regulations which are important to understand.
If you are in a Colorado car accident you have approximately 3 years to file your case. This does not mean you should wait until the very last minute to file it, of course. It's in your best interest to file a claim as soon as or soon after the car accident.
Medical expenses, including medical bills, equipment or machines you might need to use in your home, hiring a caretaker, paying for rehabilitation, etc.
If your car needs to get fixed and you need a temporary car in the meantime you will need a rental car. This is not something you would want to pay for out of your own pocket, especially if you need it for weeks or months.
Damages usually fall under 2 types; economic and non-economic. An example of economic damage may include past and future medical expenses, loss of income, and other out of pocket expenses. Non-economic damages may include disabilities, disfigurement, injuries that are long-term, emotional pain and distress, etc.
If you've been injured in a Colorado car accident and you're wondering what steps you should take next, contact Sloat & Nicholson for a free consultation today by visiting their website or calling 1-800-873-3202.


Straight Teeth in 6 Months: Fast. Effective. Affordable.
You shouldn't have to wait 2 years for straight teeth. Six Month Smiles is a clear braces system designed to straighten your teeth in about 6 MONTHS.
Six Month Smiles is generally less expensive than traditional braces or aligners due to shorter treatment time. Most providers have payment plans suitable for every budget.
Six Month Smiles uses the latest in digital technology coupled with experienced orthodontic professionals. We use a modern twist on a reliable and safe method.
Six Month Smiles uses tried and true orthodontic techniques to provide excellent results. With an average treatment time of only 6 Months, you can have the straight smile you have always wanted before for your next cleaning.
What is Six Month Smile Braces?
What are Six Month Smiles Adult Braces?
With traditional braces, treatment can last between two and three years for results. In the meantime, you must live with the metal brackets and wires on your teeth that can be embarrassing, especially for adults. Metal braces also require constant adjustments and visits to the dentist or orthodontist during the long treatment period. This can be a hassle for a busy lifestyle. With Six Month Smiles, you can get the straight smile you want in a fraction of the time of traditional braces, making it more convenient for busy adults.
Will I have a mouth full of metal for six months?
Six Month Smiles are braces for adults, but they are designed to be discrete. The clear brackets are cleverly blended into your smile and the tooth-colored wires are difficult to notice. This is a great option for adults that don't want to advertise their choice to straighten their teeth every time they smile. The discrete design is a desirable attribute that many adults prefer.


Many food-handling facilities, lodging facilities, and institutions are regulated by state laws and rules. The Environmental Health Section of the NC Department Of Health and Human Services provides statewide regulatory oversight of these establishments.
The Food And Lodging section administers state regulatory programs for the following: restaurants and other food-handling establishments, meat markets, public swimming pools (more information), child care centers, school lunchrooms, school buildings, tattoo establishments, lodging establishments, bed and breakfast homes, bed and breakfast inns, hospitals, nursing and rest homes, summer camps, local confinement facility, and residential care facilities. Staff members are authorized agents of the State to enforce state laws and rules.
The regulatory program makes sanitation inspections of these facilities.
View Child Care Center regulatory information.
The State also oversees the regulation of shellfish commerce. Shellfish sanitation rules have been established for the protection of the public from shellfish and crustacea that could cause illness. Shellfish sanitation rules can be found by visiting the Department Of Environment And Natural Resources (DENR) web page for shellfish commerce.
New restaurants must obtain a permit from the local Health Department. As part of the permit application process, prospective owners/managers of new restaurants must submit a plan for review by the Brunswick County Health Department. View more information.
Food safety is a critical concern, whether we eat at home or at a restaurant. A basic premise of food safety is keep hot foods hot and cold foods cold. This Temperature Guide For Food Protection is a handy tool to print out and keep in your kitchen. For more information about food safety, visit foodsafety.gov.


Upsizing is one of the main reasons to sell your home.
A need arises to unlock capital.
Realising profit from an investment.
You can no longer afford to keep the property.
An opportunity arises that is too good to refuse.
Generally speaking, it's best not to sell a property during times of upheaval in your life or when you are unable to showcase the property in its best possible light. Before listing your property for sale, make sure that you are genuinely ready and willing to sell.
Which is the best season to sell?
When it comes to deciding which season in which to sell your property, it's useful to know how the property calendar works.
Traditionally, spring has been a favourite time to sell property, as gardens are looking their best, the weather is usually good and there are plenty of buyers wanting to move into their new homes in time for Christmas. This can also mean there are more properties on the market, meaning more competition as buyers have more to choose from.
This is the next-biggest selling season, again because of the reasonable weather, and also because buyers have possibly missed out on properties during spring.
Although these are generally quieter selling periods, some people prefer to sell at these times because of the reduced competition from other sellers and the belief that while there are fewer buyers looking, those who are tend to be more serious about purchasing.
Properties with gardens look their best in spring.
How does the property cycle work?
You may be familiar with the concept that the property market moves in a cycle – prices rise, fall, stabilise and then rise once more.
The value stage: Prices are flat, leading many people to believe it's a good time to buy.
The growth phase: Prices begin to rise, slowly at first, before picking up pace.
The peak: This marks the top of the market. Prices will have increased very rapidly – as much as 20 per cent year on year – but will have reached the highest point of the cycle.
The correction: Price moderation. People often equate a correction with a crash, but sometimes a correction is simply a long, slow period where prices stagnate.
Median property prices compared with annual growth rates shows the property cycle in action.
Together, the whole cycle generally takes about seven to 10 years, and ideally one full cycle will see the price of property double.
The factors that contribute to one phase will directly lead to the next phase developing. After a period of stability, people will regain confidence in the market and start investing, which leads to price growth. As prices grow, more and more people pile in, which quickens the pace of price rises. Eventually, affordability constraints will lead to prices peaking, before undergoing a correction to bring prices back from their dizzying heights. Once they hit this bottom, a period of some stability ensues, and the cycle begins over again.
The phases, though, don't transition purely of their own accord. There are a vast number of external factors that play a role.
The willingness of mortgage lenders to write new loans.
How much you can borrow to fund a purchase.
Broader economic outlook and unemployment rates.
The rate of property price growth/declines.
Vacancy rates for investment properties.
Whether there are available properties to suit your needs.
Additionally, property cycles are much more localised than most people recognise. For example, it is unusual for every Australian capital city to be in the same stage of the cycle simultaneously, as each market operates independently of the others. Within each city, each suburb can be subject to its own property cycle. One particular area may experience rapid growth thanks to a new shopping centre being constructed while the next suburb sees little to no growth.
The time between listing a property and it selling varies around Australia, but your agent should be able to offer an expected range of timings in your local area. A typical auction campaign runs for four weeks, while for private treaty sales the property will remain on the market until a buyer is found. You can also research your suburb to get an idea of how long it takes property to sell in your area.
Getting a home ready to sell will have an influence on the timing. If you have been thinking about selling and have been preparing a home for sale, you may be ready to get the property photographed and put on the market immediately. But if you need time to get it in shape, this can delay the process. Allow at least four weeks to declutter, make any repairs and spruce up a property with paint before having photos taken.
Allow a few weeks to spruce up your home before sale.
The price can also affect how long it takes a home to sell. If it is overpriced, it will most likely sit on the market for an extended period of time. Your agent will provide feedback from potential buyers and guide you about the price. The location and type of property also affect how quickly it can be sold, with properties in higher demand selling quicker, and vice versa.
Once the price has been agreed upon and the contract is signed, settlement can take anywhere from one to three months, but six weeks is the standard settlement period.
The process of selling a property can be streamlined if you carefully review the price, presentation, promotion, photography, placement and potential. The tips below may even help you achieve a higher price.
Compare your property with similar recent local sales to help set a realistic price, so potential buyers can see good value. Prices set too high will sit around for weeks until the price is reduced or a buyer is found. If selling by auction, you can set a 30-day auction campaign to create momentum in a set timeframe, leading to an enthusiastic response on auction day.
First impressions count, so making your home look great is vital to the success of your sale. For an instant facelift, declutter your house and make it sparkling clean. If the walls look tired, give them a quick refresh – a coat of paint can do wonders. The garden is another key area to focus on to attract buyers. Lastly, look at staging your house by hiring modern furniture and accessories to help create a warm ambience.
Attracting multiple interested buyers should create competition leading to offers in the higher range. Promotion is key to reaching out to these buyers. A strong digital presence and print advertising puts your house in front of potential buyers. Your agent can help you with these marketing strategies.
Renovated homes achieve a higher price, but be careful of overcapitalising.
Real estate photography gives buyers emotional connections to houses they are viewing. It is unwise to scrimp on professional photography as it will be used across your entire marketing campaign, both print and digital. Display the best features of your house including the view, setting, living spaces and unique features. Remember, with online advertising, your house may be viewed by potential buyers from not just the local area, but even around the world.
Low-maintenance houses could appeal more to an investor, so provide any rental history, strata levies and any other useful information to help progress a sale. If it is a family house, market its strengths to attract these buyers by mentioning a location close to parks, schools, shops and transport. First-home buyers are often motivated by price whereas empty-nesters desire security, level access and proximity to shops and cafes.
If you are thinking of renovating to increase the sale price, avoid overcapitalising as the return may not cover the outlay. As a general rule of thumb, it is best to complete any half-finished projects but avoid any polarising decorating choices to leave a blank canvas for the buyer. Your agent can advise you on possible renovations, and a budget to help you achieve the best return.
Next: What are the different ways to sell a property?
This article is part of Domain's Ultimate Guide To Selling Your Home, compiled by Domain Advice Editor Daniel Butkovich.


text:
pipHelp

service in module pipHelp.Service

Description

This service is provided an interface to manage the Help component.
It is available on the config and run application phases. On the both phases the interface is the same.
This module requires the 'pipState' module.

Dependencies

pipState

Methods

addPage(pageObj)

This method allows add new page into navigation menu. It accepts config object to define new state
with needed params.

View source

Parameters

Param	Type	Details

pageObj	Object

Configuration object contains settings for another page

pageObj.state	Object.<string>

Name of page state which is available via UI router

pageObj.title	Object.<string>

Page title in the navigation menu.

pageObj.access	Object.<boolean>

If it is true it will be available only for logged in users

pageObj.visible	Object.<boolean>

If it is true the page will be visible

pageObj.stateConfig	Object.<Object>

Configuration object in format like UI Router state

Example

// on the config phase
pipHelpProvider.addPage({
    state: 'test',
    title: 'Test help page',
    auth: true,
    stateConfig: {
        url: '/test',
        templateUrl: 'help/help_test1.html'
    }
});

getDefaultPage()

This method return name of the default state.

View source

Returns

string

Name of the state

Example

// on the config phase
pipHelpProvider.getDefaultPage();

getPages()

This method returns asset of all pages registered in the Help component.

View source

Returns

Array<Object>

List of registered states

Example

// on the config phase
pipHelpProvider.getPages();

setDefaultPage(name)

This method establishes passed state as default which is redirected at after transfer on abstract
state

View source

Parameters

Param	Type	Details

name	Object

Name of the state

Example

pipHelpProvider.setDefaultPage('test');

redpajama_set_name: RedPajamaGithub


The Personal Agent are pleased to present you with this attractive three-bedroom detached house, located on the periphery of Epsom Common. This property is immaculately presented and benefits from views of The Wells, Epsom and Ashtead Common.
The property has so much space and has recently been refurbished. The accommodation includes a beautiful living room with bay window, modern kitchen/dining room, three good sized bedrooms, and a family bathroom. This house also benefits from a stunning garden and garage. There is also potential to extend further.
We believe this property has everything ready for modern day family living. Call 01372 745 850 to book an appointment with The Personal Agent sales team.


No private land has been acquired for developing Karera Wildlife Sanctuary in Madhya Pradesh. The Karera Wildlife Sanctuary in Shivpuri District comprises both Government revenue land as well as private land and in view of the provisions of Wild Life (Protection) Act, 1972, as applicable to a Sanctuary, the villagers do face problems buying and selling land. The State Government has therefore, proposed to denotify the Sanctuary to resolve the problem of the villagers.
The proposal for denotification of any Sanctuary/National Park requires recommendation of the National Board for Wildlife as per the provisions of the Wild Life (Protection) Act, 1972 and thereafter, approval from Hon'ble Supreme Court also. The proposal for denotification of Karera Sanctuary has been recommended by the Standing Committee of the National Board for Wildlife. As informed by the State Government of Madhya Pradesh, the State Government has filed a petition before the Hon'ble Supreme Court for denotification of the sanctuary. No specific time limit for approval from Hon'ble Supreme Court can be indicated.
This information was given by the Minister of State for Environment and Forests (independent charge) Shrimati Jayanthi Natarajan in a written reply to a question by Shri Virendra Kumar, Shri Narendra Singh Tomar and Shri Shivraj Bhaiya in Lok Sabha today.


It is the critical element.
Every trip I've ever had, or ever will, will start with this vital component.
It gets harder and harder to find these days.
Work is humming and there are a thousand little things poking up that demand my attention.
There are the kids. They need their watering and care as well if they are to grow big and strong.
And the marriage itself also requires maintenance and nurturing.
My garage could use a good bit of straightening up.
When it works for the kids and the wife and work.
I check the calendar like a racetrack junkie checking the race results.
Is confronted by the round holes of modern life.
This entry was posted on Tuesday, December 5th, 2017 at 8:47 am	and is filed under All the other stuff. You can follow any comments to this entry through the RSS 2.0 feed. You can leave a comment, or trackback from your own site.
Your battery needs recharging so you can do all the other things with joy! Hope you make it happen soon. Agreed, good piece of prose. I'm sure the fly fishing community would like to see a Fly Paper III!


text:
Research Article| April 15 1986
Restriction fragment-length polymorphisms of class II gene sequences in mice expressing minor structural variants of I-Ak and I-Ap.
T J McConnell;
T J McConnell
B Darby;
B Darby
E K Wakeland
J Immunol (1986) 136 (8): 3076–3084.
https://doi.org/10.4049/jimmunol.136.8.3076
T J McConnell, B Darby, E K Wakeland; Restriction fragment-length polymorphisms of class II gene sequences in mice expressing minor structural variants of I-Ak and I-Ap.. J Immunol 15 April 1986; 136 (8): 3076–3084. https://doi.org/10.4049/jimmunol.136.8.3076
Serologic and structural analyses of the I-A molecules expressed among a large collection of wild mouse-derived H-2 haplotypes has led to the definition of "families" of I-A alleles which encode antigenically similar molecules that are identical in more than 90% of their tryptic peptides. Two of these families, denoted the I-Ak and I-Ap families, consist of 10 I-A alleles which encode I-A molecules whose structures are closely related to either I-Ap or I-Ak. The evolutionary relationships of the I-A alleles in these families were assessed by a molecular analysis of their genomic structures. The A alpha and A beta alleles within these I-A families were compared by analysis of restriction fragment-length polymorphisms (RFLP) detected at high stringency by Southern blot hybridization with DNA probes specific for either A alpha or A beta. The polymorphic restriction enzyme sites detected in this survey were distributed over more than 7 kb of genomic DNA surrounding each gene. Because both A alpha and A beta are encoded by about 700 bp of exon DNA, the majority of the restriction enzyme sites assayed by this RFLP analysis reflect polymorphisms in noncoding regions. The DNA sequence homologies of these alleles were estimated from the RFLP results with seven restriction endonucleases by calculating the fraction homologous value as defined previously. The results indicate that evolutionarily dissimilar I-A alleles can encode I-A molecules with very similar structures. The five I-A alleles in the I-Ak family could be divided into two discrete groups, denoted K1 and K2, on the basis of their restriction fragment (RF) genotypes. The RF genotypes of alleles within each group shared more than 80% of the restriction fragments for both A and A beta. In contrast, the RF genotypes of alleles in group K1 differed extensively from those in group K2, indicating that alleles in these separate groups may not be evolutionarily closely related. These observations suggest that gene conversion or intragenic recombinational events may have been involved in the evolution of groups K1 and K2 in the I-Ak family. The RF genotypes of alleles in the I-Ap family demonstrated a close evolutionary relationship among all but two of the alleles. These two alleles encoded I-A molecules whose structures were the least related to I-Ap of any of the alleles in the I-Ap family.(ABSTRACT TRUNCATED AT 400 WORDS)


text:
Heinz-Hermann Köhler [Paintings]
Heinz-Hermann Köhler grew up in a country village in Lower Saxony. In 1936 after completing elementary school, he moved in with relatives in nearby Hildesheim, a mid-sized provincial town, so he could attend the Agricultural School. In contrast to…	[more]
National Museum of African Art
Susan Douglass
The National Museum of African Art website is a colorful destination that showcases a dynamic mixture of historical, traditional, and contemporary…	[more]
Piñata [Object]
A piñata is a decorated container of paper or clay that contains sweets, small toys, fruits, and nuts. It is the object of a game played in Mexico at children's birthday parties and at Christmas celebrations, in which blindfolded children take turns…	[more]
Piegan Play Tipi [Photograph]
The photograph, dated 1928, from the well-known ethnographic collection The North American Indian by Edward Edward S. Curtis, is a portrait of a young Piegan girl standing in front of a painted tipi, or conical tent. The label on the photograph by…	[more]
Indus Valley Wheeled Ram Toy [Object]
The hand-modeled, terracotta, wheeled figurine of a bird with a ram's head was excavated from an archaeological site in the Indus Valley called Mohenjo-daro, a city that flourished between 2600 and 1900 BCE. Museum curators added the wooden axle…	[more]
Toys [Objects]
The material culture of early childhood in the 21st century is characterized by an emphasis on biological age and related levels of cognitive and motor skill development. All types of objects, including diapers, toys, food products, and clothing, are…	[more]
Recycled Wire Toy Truck, Uganda [Photograph]
The photograph shows a boy "driving" an ingeniously simple but functional toy truck made of wire, wood, and what are likely plastic bottlecap wheels. Purchased, often imported toys are very expensive and homemade toys are common in many poorer…	[more]
Tet Trung Thu festival, Vietnam [Photographs]
The photographs show children during the mid-Autumn Harvest Festival, or Tet Trung Thu in Vietnam, a children's festival associated with the full moon. Tet Trung Thu follows the harvest in the eighth lunar month, falling usually during September or…	[more]


After the huge success of the inaugural Bass Strait Masters last year, the tournament is set to be bigger and better again in 2017 as we are proud to announce Taylor Made as our major partner for the upcoming event.
Having the number one golf brand in the world on board, it has taken our event to a new level with up to $20,000 worth of prizes on offer over the 4 day tournament.
What better way to experience these amazing golf courses with Cape Wickham & Ocean Dunes on King Island and Barnbougle Dunes & Lost Farm on the North East coast of Tasmania recently rated the Top 4 public golf courses in Golf Australia Magazine.
The Bass Strait Masters is a 4 day all-inclusive golf tournament like you have never seen before.
Compete in the 90 person field playing a different world class course each day, whilst your daily score will tally up to announce an overall Bass Strait individual and team champion after arguably the 4 toughest days of golf you can compete in.
After each day's event, enjoy a fully catered function showcasing that region of Tasmania's finest produce with matching wines and listen to the daily winners being announced.
Breakfast and lunch packs are included each day, as well as all refreshments throughout the 4 days.
Cricketing legend and 'better than scratch' golfer Greg Blewett will be back to host the event and present the daily awards.
Vortex Air's professional team will again take care of all flights and on ground logistics to get you to your tee time, accommodation and dinner with ease.


text:
Get XtremeSorb spill materials here!
Topics Free Sites FAQ's Regulations Glossary Software Suppliers
Books Forum Poll Fun stuff Quiz Store
MSDS and safety supplies Search ALL our MSDS info
Title: 01/25/1995 - Guidance on whether an inert gas in the non-compressed state represent hazardous chemicals under the Hazard Communication Standard.
Record Type: Interpretation Standard Number: 1910.1200
Richard F. Andree, CSP, PE, Ph.D.
Vice President - Director of Safety and Health Services
Lovell Safety Management Company, Inc.
Dear Mr. Andree:
This is in further response to your letter of April 11, to the Occupational Safety and Health Administration (OSHA), requesting guidance on whether an inert gas in the non-compressed state represent hazardous chemicals under the Hazard Communication Standard (HCS), 29 CFR 1910.1200. Your letter specifically inquired about gases such as argon, nitrogen and helium at atmospheric pressure. Please accept my apology in the delay in this response.
OSHA agrees with your statement that inert gases are classified as a simple asphyxiants. Patty's Industrial Hygiene and Toxicology defines a simple asphyxiant as: "physiologically inert gases that can act principally by dilution of the atmospheric oxygen below partial pressure necessary to maintain an oxygen saturation in the blood sufficient for normal tissue respiration."
Inert gases, such as argon, nitrogen and helium, have the potential to create a hazardous atmosphere (less than 19.5% oxygen) and, where that potential exists, are considered by OSHA as hazardous chemicals presenting an acute health hazard and covered by the standard. The HCS would not apply where employees are not occupationally exposed, under normal conditions of use or in a reasonably foreseeable emergency, to the oxygen displacing hazard associated with inert gases. Obviously, argon, nitrogen and helium occur naturally in the atmosphere, and under those conditions they are not hazardous chemicals.
The HCS requires chemical manufacturers, or importers to assess the hazards of chemicals which they produce or import, and all employers to provide information to their employees about the hazardous chemicals to which they are exposed, by means of a hazard communication program, labels and other forms of warning, material data safety sheets (MSDSs), and information and training. Included in the HCS is paragraph (c) which contains a broad definition of the term "health hazard." Appendix A of the HCS also provides further definition and explanations of the scope of health hazards including acute health hazards. Enclosed for your reference is a copy of a February 9, Hazard Communication; Final Rule which contains a number of minor changes and technical amendments to clarify further the regulatory requirements.
You objected to the language in the March 4, 1993, letter to Matthew McFarland stating that toxicity has been established in inert gases, and that the manufacturer, importer, and/or distributor has the duty of performing a complete hazard determination. To elaborate on the underlined phrase we mean the toxic effects of excessive levels of inert bases has been well established. Hopefully, this clarifies the language and addresses your concern.
Get gas cylinder accessories such as tags, signs, and storage cabinets at Safety Emporium.
We hope this information provides you with the necessary guidance. Thank you for your interest in worker safety and health. Should you require additional information please contact the Office of Health Compliance Assistance at (202) 219-8036.
Joseph A. Dear
Assistant Secretary
Entry last updated: Saturday, June 11, 2016. This hypermarked and content-enhanced page is copyright 2001-2019 by ILPI, all rights reserved. Unauthorized duplication or posting on other web sites is expressly prohibited. For questions, comments and concerns, please contact us at our MSDS email address.
The official, public domain, OSHA version of this document is available at http://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=INTERPRETATIONS&p_id=21683&p_text_version=FALSE


Singer Sunitha in red designer plain shimmer saree with embroidery work paired with designer short sleeves saree blouse with round neck pattern.
Celebrities at 58th Filmfare Awards 2011.
Anushka ,Lakshami and Amala at Idea Filmfare Awards 2011.
Anushka in beautiful peach designer shimmer saree with velvet embroidery patch work , gold embellished border with studs paired with short sleeves saree blouse from designer Tarun Tahiliani at 58th idea Filmfare Awards South 2011.It is a Net Saree.
Lakshmi Prasanna in beautiful blue designer plain saree with gold embroidered and kundan work paired with designer rhinestone strap saree blouse at 58th Filmfare Awards 2011.
Amala in beautiful pink designer cotton saree with gold embellished border paired with halter neck saree blouse at 58th idae Filmfare awards 2011.
Pink Sarees at 58th Idea Filmfare Awards South 2011.
Kriti Kharbanda,Samantha ,Nikesha Patel and Richa Gangopadhyay at 58th idea Filmfare Awards 2011.
Kriti Kharbanda in beautiful pink designer saree with silver lace border paired with sleeveless saree blouse at 58th Filmfare Awards South 2011.
Samantha in beautiful pink designer shimmer saree with multicolor unique embroidery border work paired with sliver embroidery halter neck saree blouse at 58th idea Filmfare Awards South 2011.
Richa Gangopadhyay in pink designer embroidery saree with mirror,sequins and embroidered border work paired with designer rhinestone strap saree blouse at 58th Filmfare Awards 2011.
Nikesha Patel in pink and mustered dual color tissue saree with mirror,embroidery thread work and silver sequins buttes border paired with sleeveless saree blouse at 58th idea Filmfare Awards south 2011.It is a Net Saree.
Anushka in Tarun Tahiliani Sari .
South Indian actress Anushka in beautiful peach designer shimmer saree with velvet embroidery patch work , gold embellished border with studs paired with short sleeves saree blouse from designer Tarun Tahiliani at 58th idea Filmfare Awards South 2011.It is a Net Saree.
Samantha in Pink Sari at 58th idea Filmfare Awards South 2011.
South Indian actress Samantha in beautiful pink designer shimmer saree with multicolor unique embroidery border work paired with sliver embroidery halter neck saree blouse at 58th idea Filmfare Awards South 2011.
Nikesha Patel at 58th idea Filmfare Awards 2011.
South Indian actress Nikesha Patel in pink and mustered dual color tissue saree with mirror,embroidery thread work and silver sequins buttes border paired with sleeveless saree blouse at 58th idea Filmfare Awards south 2011.It is a Net Saree.
South Indian actress Kriti Kharbanda in beautiful pink designer saree with silver lace border paired with sleeveless saree blouse at 58th Filmfare Awards South 2011.


text:
\section{Two generalized order relations}

\medskip
We begin this section by recalling some facts about dual real numbers. For any two elements $(x_1,\, x_2)$  and $(y_1,\, y_2)$ from the $2$-dimensional real vector space
$\mathcal{R}^2=\mathcal{R}\times\mathcal{R}$, we  define their product according to the the fololowing rule called
{\bf dual number multiplication}:
$$(x_1,\, x_2)\,\cdot\,(y_1, \, y_2):= (x_1y_1,\, x_1y_2+x_2y_1).$$
The vector space $\mathcal{R}^2$ with respect to the dual number multiplication is a real associative algebra, which is called the  {\bf dual real number algebra} and denoted by $\mathcal{R}^{(2)}$. An element of $\mathcal{R}^2$ is called a dual real number. Clearly, the  dual real number algebra
$\mathcal{R}^{(2)}$ is both unital and commutative. We denote the multiplication identity $(1,\, 0)$ by $1$, and the element
$(0,\, 1)$ by $1^{\#}$. Then every dual real number  $(x_1,\, x_2)$ of $\mathcal{R}^{(2)}$ can be expressed in a unique way as a linear combination of $1$ and $1^{\#}$:
$$ x=(x_1,\, x_2)=x_1+x_2\,1^{\#}\qquad \mbox {for $x_1$, $x_2\in \mathcal{R}^{(2)}$},$$
where $Re\,x:=x_1$ and $Ze\,x:=x_2$ are called the {\bf real part} and the {\bf zero-divisor part} of $x$, respectively.

\medskip
The dual real number algebra $\mathcal{R}^{(2)}$ is not a field and has many zero-divisors.
In fact, if $0\ne x\in \mathcal{R}^2$, then $x$ is a zero-divisor if and only if $Re\,x=0$ and
$x$ is invertible if and only if $Re\,x\ne 0$. Moreover if $x$ is invertible,  then  the inverse $x^{-1}$ of $x$ is given by
$x^{-1}=\displaystyle\frac{1}{Re\, x}-\displaystyle\frac{Ze\, x}{(Re\, x)^2}\,1^{\#} .$

\medskip
\begin{definition}\label{def1.1} The real-valued function $||\quad||: \mathcal{R}^{(2)}\to \mathcal{R}$  defined by
\begin{equation}\label{eq1}
||x||:=\sqrt{2\,(Re\,x)^2+(Ze\,x)^2}\quad\mbox{for $x\in\mathcal{R}^{(2)}$}
\end{equation}
is called the {\bf norm} in $\mathcal{R}^{(2)}$
\end{definition}

\medskip
The dual real number algebra is a normed algebra with respect to the norm introduced in Definition \ref{def1.1} by the following proposition.

\medskip
\begin{proposition}\label{pr1.1} Let $x$, $y\in\mathcal{R}^{(2)}$ and $a\in\mathcal{R}$.
\begin{description}
\item[(i)] $||x||\ge 0$, with equality only when $x=0$.
\item[(ii)] $||ax||=|a|\,||x||$, where $|a|$ denotes the absolute value of the real number $a$.
\item[(iii)]  $||x+y||\le  ||x||+||y||$.
\item[(iv)]  $||xy||\le  ||x||\,||y||$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} a direct computation.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
Unlike the Cartesian product of finite many copies of the real number field and the complex field, there are two generalized order relations on $\mathcal{R}^{(2)}$  which are compatible with the multiplication in $\mathcal{R}^{(2)}$.

\begin{definition}\label{def1.2} Let $x$ and $y$ be two elements of $\mathcal{R}^{(2)}$.
\begin{description}
\item[(i)] We say that $x$ is {\bf type 1 greater than} $y$ ( or $y$ is {\bf type 1 less than} $x$) and we write $x\stackrel{1}{>} y$ (or $y\stackrel{1}{<} x$) if
$$
\mbox{either}\quad \left\{\begin{array}{c}
Re\,x>Re\,y\\ Ze\,x\ge Ze\,y\end{array}\right.\quad\mbox{or}\quad
\left\{\begin{array}{c}
Re\,x=Re\,y\\ Ze\,x> Ze\,y\end{array}\right.
$$
\item[(ii)] We say that $x$ is {\bf type 2 greater than} $y$ ( or $y$ is {\bf type 2 less than} $x$) and we write $x\stackrel{2}{>} y$ (or $y\stackrel{2}{<} x$) if
$$
\mbox{either}\quad \left\{\begin{array}{c}
Re\,x>Re\,y\\ Ze\,y\ge Ze\,x\end{array}\right.\quad\mbox{or}\quad
\left\{\begin{array}{c}
Re\,x=Re\,y\\ Ze\,y> Ze\,x\end{array}\right.
$$
\end{description}
\end{definition}

We use $x\stackrel{\theta}{\ge} y$ when $x\stackrel{\theta}{>} y$ or $x=y$ for $\theta =1,\, 2$.
By Definition 1.1, if $Re\,x=Re\,y$, then $x\stackrel{1}{>} y\Longleftrightarrow y\stackrel{2}{>} x$; if $Ze\,x=Ze\,y$, then $x\stackrel{1}{>} y\Longleftrightarrow x\stackrel{2}{>} y$.
The following proposition gives the basic properties of the two generalized order relations.

\begin{proposition}\label{pr1.2} Let $x$, $y$ and $z$ be elements of $\mathcal{R}^{(2)}$ and $\theta =1,\, 2$.
\begin{description}
\item[(i)] One of the following holds:
$$ x\stackrel{1}{>} y, \quad y\stackrel{1}{>} x, \quad x=y, \quad x\stackrel{2}{>} y, \quad y\stackrel{2}{>} x .$$
\item[(ii)] If $x\stackrel{\theta}{>} y$ and $y\stackrel{\theta}{>} z$, then $x\stackrel{\theta}{>} z$.
\item[(iii)] If $x\stackrel{\theta}{>} y$, then $x+z\stackrel{\theta}{>} y+z$.
\item[(iv)] If $x\stackrel{\theta}{>} 0$ and $y\stackrel{\theta}{>} 0$, then $xy\stackrel{\theta}{\ge} 0$.
\item[(v)] If $x\stackrel{\theta}{>} y$, then $-x\stackrel{\theta}{<} -y$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} Clear.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
\section{Differentiation}

\medskip
By Proposition \ref{pr1.1},  the dual real number algebra $\mathcal{R}^{(2)}$ is a metric space with the distance function $||\,\,||$.  If $c\in\mathcal{R}^{(2)}$ and $\epsilon\in \mathcal{R}$, we use $N(c;\,\epsilon)$  and $N^*(c;\,\epsilon)$ to denote the ordinary $\epsilon$-{\bf neighborhood} and {\bf deleted $\epsilon$-neighborhood} of $c$, respectively, i.e.,
$$N(c;\,\epsilon)=\{x\in \mathcal{R}^{(2)}\,|\, ||x-c||<\epsilon\}\quad\mbox{and}\quad
N^*(c;\,\epsilon):=N(c;\,\epsilon)\setminus\{c\}.$$
For $\theta\in\{1,\,2\}$, the  set
$
N_{\theta}(c;\,\epsilon):=\{x\in N(c;\,\epsilon)\,\big|\, \mbox{$x\stackrel{\theta}{\ge}c$ or
 $x\stackrel{\theta}{\le}c$}\}
$
is called the {\bf type $\theta$ $\epsilon$- neighborhood} of $c$ and the set
$
N_{\theta}^*(c;\,\epsilon):=N_{\theta}(c;\,\epsilon)\setminus\{c\}
$
is called the {\bf deleted type $\theta$ $\epsilon$- neighborhood} of $c$.
By Proposition\ref{pr1.2} (i), we have
$$
N(c;\,\epsilon)=N_{1}^*(c;\,\epsilon)\bigcup N_{2}^*(c;\,\epsilon).
$$

\medskip
We now introduce the differentiability in the following

\medskip
\begin{definition}\label{def2.1} Let $D$ be an open subset of $\mathcal{R}^{(2)}$ and let
$c\in D$.
\begin{description}
\item[(i)] We say that $f: D\to \mathcal{R}^{(2)}$ is {\bf type $\theta$ differentiable} at $c$ with $\theta\in\{1,\, 2\}$ if for each positive real number $\epsilon>0$ there exist a  positive real number $\delta>0$ and a dual real number $f_{\theta}'(c)\in \mathcal{R}^{(2)}$ such that
$$ x\in N_{\theta}^*(c;\,\delta)\subseteq D\Rightarrow
\frac{f(x)-f(c)-f_{\theta}'(c)(x-c)}{||x-c||}\in N(0;\,\epsilon).  $$
The dual real number $f_{\theta}'(c)$ is called the {\bf type $\theta$ derivative} of $f$ at $c$, which is also denoted by $\displaystyle\frac{df}{d_{\theta}x}(c)$.
\item[(ii)] We say that $f: D\to \mathcal{R}^{(2)}$ is {\bf differentiable} at $c$ if for each positive real number $\epsilon>0$ there exist a  positive real number $\delta>0$ and a dual real number $f'(c)\in \mathcal{R}^{(2)}$ such that
$$ x\in N^*(c;\,\delta)\subseteq D\Rightarrow
\frac{f(x)-f(c)-f'(c)(x-c)}{||x-c||}\in N(0;\,\epsilon).  $$
The dual real number $f'(c)$ is called the {\bf derivative} of $f$ at $c$, which is also denoted by $\displaystyle\frac{df}{dx}(c)$. If $f$ is  differentiable at each point of the open subset $D$, then $f$ is said to be {\bf differentiable} on $D$.
\end{description}
\end{definition}

\medskip
It is easy to check that if a dual real number-valued $f$ is  type $\theta$ differentiable at
$c\in \mathcal{R}^{(2)}$, then the type $\theta$ derivative of $f$ at $c$ is unique for $\theta\in\{1,\, 2\}$.

\bigskip
Let $S$ be a subset of $\mathcal{R}^{(2)}$. A function $f: S\to \mathcal{R}^{(2)}$ can be expressed as
$$f(x)=u(x)+v(x)\,1^{\#}\quad\mbox{for $x=x_1+x_2\,1^{\#}\in \mathcal{R}^{(2)}$},$$
where  $u(x):=u(x_1,\,x_2)$ and $v(x):=v(x_1,\,x_2)$ are two real-valued functions of two real variables $x_1$ and $x_2$, which are called the {\bf real component} and the {\bf zero-divisor component} of $f$, respectively. The following proposition provides an useful characterization of differentiability for  dual
real-valued functions in terms of their real and zero-divisor components.

\medskip
\begin{proposition}\label{pr2.1} Let $f: D\to \mathcal{R}^{(2)}$ be a  dual
real-valued function given by
$$f(x)=u(x_1,\,x_2)+v(x_1,\,x_2)\,1^{\#}\quad\mbox{for $x=x_1+x_2\,1^{\#}\in \mathcal{R}^{(2)}$},$$
where $D$ is an open subset of $\mathcal{R}^{(2)}$,  $u(x_1,\,x_2)$ and  $v(x_1,\,x_2)$ are  the real component and the zero-divisor component of $f$, respectively. Let $c=c_1+c_2\,1^{\#}\in D$ with
$c_1$, $c_2\in \mathcal{R}$.
\begin{description}
\item[(i)] If the first-order partial derivatives $u_{x_1}$, $u_{x_2}$, $v_{x_1}$ and $v_{x_2}$ exist at
$(c_1,\,c_2)$ and are continuous at $(c_1,\,c_2)$, and the following equations
\begin{equation}\label{eq2}
u_{x_1}=v_{x_2}, \qquad u_{x_2}=0
\end{equation}
hold at $(c_1,\,c_2)$, then $f$ is differentiable at $c$ and the  derivative $f'(c)$ of $f$ at $c$ is given by
\begin{equation}\label{eq3}
f'(c)=u_{x_1}(c)+v_{x_1}(c)\,1^{\#}.
\end{equation}
\item[(ii)] If $f$ is differentiable at $c$, then the equations in  (\ref{eq2}) hold at  $c=c_1+c_2\,1^{\#}$. In this case,  the  derivative $f'(c)$ of $f$ at $c$ is given by (\ref{eq3}).
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} The proof of Proposition \ref{pr2.1} is similar to the proof of the famous fact which characterizes the complex differentiability by using Cauch-Riemmann equations.

\medskip
Foe example, let us prove (ii). If $f$ is differentiable at $c$, then exists $L=L_1+L_2\,1^{\#}$ with $L_1$, $L_2\in  \mathcal{R}$ such that for every $\epsilon>0$, there exists a $\delta>0$ such that $N(c;\,\delta)\subseteq D$ and
\begin{equation}\label{eq4}
x\in N^*(c;\,\delta)\Rightarrow
\frac{\|f(x)-f(c)- L\,(x-c)\|}{\|x-c\|}<\epsilon .
\end{equation}

\medskip
By dual number multiplication, we have
\begin{eqnarray}\label{eq5}
&&f(x)-f(c)- L\,(x-c)=u(x_1,\,x_2)-u(c_1,\,c_2)-L_1(x_1-c_1)+\nonumber\\
&&\quad +[v(x_1,\,x_2)-v(c_1,\,c_2)-L_1(x_2-c_2)-L_2(x_1-c_1)]\,1^{\#}.
\end{eqnarray}

\medskip
Let $x_2=c_2$ and choose $x_1$ such that $0<|x_1-c_1|<\displaystyle\frac{\delta}{\sqrt{2}}$. Then
$$0<\|x-c\|=\|x_1-c_1\|=\sqrt{2} |x_1-c_1|<\delta ,$$
which implies that $x=x_1+c_2\,1^{\#}\in  N^*(c;\,\delta)$. By  (\ref{eq4}) and  (\ref{eq5}), we get
\begin{eqnarray}\label{eq6}
\epsilon&>&\frac{\|f(c_1+x_2\,1^{\#})-f(c)- L\,(c_1+x_2\,1^{\#}-c)\|}{\|c_1+x_2\,1^{\#}-c\|}\nonumber\\
&=&\frac{1}{\sqrt{2}}\Big\{2\Big[\frac{u(x_1,\,c_2)-u(c_1,\,c_2)}{x_1-c_1}-L_1\Big]^2+\nonumber\\
&&\quad +\Big[\frac{v(x_1,\,c_2)-v(c_1,\,c_2)}{x_1-c_1}-L_2\Big]^2\Big\}^{\frac12}.
\end{eqnarray}
It follows from  (\ref{eq6}) that for every $\epsilon>0$, there exists a $\delta>0$ such that
$$
\left|\frac{u(x_1,\,c_2)-u(c_1,\,c_2)}{x_1-c_1}-L_1\right|<\epsilon\quad\mbox{and}\quad
\left|\frac{v(x_1,\,c_2)-v(c_1,\,c_2)}{x_1-c_1}-L_2\right|<\epsilon\sqrt{2}
$$
whenever $0<|x_1-c_1|<\displaystyle\frac{\delta}{\sqrt{2}}$. This proves that
\begin{equation}\label{eq7}
u_{x_1}(c_1,\,c_2)=L_1\quad\mbox{and}\quad v_{x_1}(c_1,\,c_2)=L_2.
\end{equation}

\medskip
Similarly, let $x_1=c_1$ and choose $x_2$ such that $0<|x_2-c_2|<\delta$. Then
$$0<\|x-c\|=\|(x_2-c_2)\,1^{\#}\|= |x_2-c_2|<\delta ,$$
which implies that $x=c_1+x_2\,1^{\#}\in  N^*(c;\,\delta)$. By  (\ref{eq4}) and  (\ref{eq5}), we get
\begin{eqnarray*}
\epsilon&>&\frac{\|f(c_1+x_2\,1^{\#})-f(c)- L\,(c_1+x_2\,1^{\#}-c)\|}{\|c_1+x_2\,1^{\#}-c\|}\nonumber\\
&=&\left\{2\left[\frac{u(c_1,\,x_2)-u(c_1,\,c_2)}{x_2-c_2}\right]^2
+\left[\frac{v(c_1,\,x_2)-v(c_1,\,c_2)}{x_2-c_2}-L_1\right]^2\right\}^{\frac12}.
\end{eqnarray*}
which implies that for every $\epsilon>0$, there exists a $\delta>0$ such that
$$
\left|\frac{u(c_1,\,x_2)-u(c_1,\,c_2)}{x_2-c_2}\right|<
\frac{\epsilon}{\sqrt{2}
}\quad\mbox{and}\quad
\left|\frac{v(c_1,\,x_2)-v(c_1,\,c_2)}{x_2-c_2}-L_1\right|<\epsilon
$$
whenever $0<|x_2-c_2|<\delta$. This proves that
\begin{equation}\label{eq8}
u_{x_2}(c_1,\,c_2)=0\quad\mbox{and}\quad v_{x_2}(c_1,\,c_2)=L_1.
\end{equation}

By (\ref{eq7}) and (\ref{eq8}), (ii) holds.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
\section{Type $\theta$ Integrals}

\medskip
In the remaining of this paper, $\theta$ always denote an element in the set $\{1, 2\}$.
Let  $f: S\to \mathcal{R}^{(2)}$ be a function on a subset $S$ of  $\mathcal{R}^{(2)}$. For convenience, we will use $f_{Re}$ and $f_{Ze}$ to denote the  real component and the  zero-divisor component of a function $f: S\to \mathcal{R}^{(2)}$, respectively. Thus, we have
$$f(x)=f_{Re}(x)+f_{Ze}(x)\,1^{\#}\quad\mbox{for $x=x_1+x_2\,1^{\#}\in \mathcal{R}^{(2)}$}.$$
We say that the function $f: S\to \mathcal{R}^{(2)}$ is {\bf bounded on $S$} if both $f_{Re}$ and $f_{Ze}$ are bounded on $S$ ($\subseteq \mathcal{R}^2= \mathcal{R}\times  \mathcal{R}$).

\medskip
Let $a$, $b\in \mathcal{R}^{(2)}$ and $a\stackrel{\theta}{<}b$. The {\bf type $\theta$ closed interval} $[a, b]_{\theta}$ is defined by
$$
[a, b]_{\theta}:=\big\{\, x\in \mathcal{R}^{(2)}\,|\, a\stackrel{\theta}{\le}x\stackrel{\theta}{\le}b\,\big\}.
$$

\medskip
\begin{definition}\label{def3.1} Let $[a, b]_{\theta}$ be a type $\theta$ closed integral. A {\bf partition} $P$ of  $[a, b]_{\theta}$ is a finite set of points $\{x^{(0)}, x^{(1)}, \dots , x^{(n)}\}$ in  $[a, b]_{\theta}$ such
that
\begin{equation}\label{eq9}
 a=x^{(0)}\stackrel{\theta}{<} x^{(1)}\stackrel{\theta}{<} \dots \stackrel{\theta}{<} x^{(n)}=b.
\end{equation}
If $P$ and $Q$ are two partitions of $[a, b]_{\theta}$ with $P\subseteq Q$, then $Q$ is called a
{\bf refinement} of $P$.
\end{definition}

\medskip
Let  $[a, b]_{\theta}$ be a type $\theta$ closed interval. Suppose that
$f: [a, b]_{\theta}\to \mathcal{R}^{(2)}$ is bounded and $P=\{x^{(0)}, x^{(1)}, \dots , x^{(n)}\}$ is a partion of  $[a, b]_{\theta}$. For $1\le i\le n$, the {\bf length} $\Delta x^{(i)}$ of the $i$-th type $\theta$
subinterval $[x^{(i-1)},\,x^{(i)}]_{\theta}$ is defined by $\Delta x^{(i)}:=x^{(i)}-x^{(i-1)}$. Clearly,
 $\Delta x^{(i)}\stackrel{\theta}{>}0$, and  $\Delta x^{(i)}$ is  a real number if and only if
$Ze(x^{(i)})=Ze(x^{(i-1)})$ for $1\le i\le n$. Since both $f_{Re}$ and $f_{Ze}$ are bounded on
$[a, b]_{\theta}$, both $f_{Re}$ and $f_{Ze}$ are bounded on $[x^{(i-1)},\,x^{(i)}]_{\theta}$  for $1\le i\le n$. Hence,
both
$ \sup_if_{\clubsuit}:=\sup \big\{f_{\clubsuit}(x)\,|\, x\in [x^{(i-1)},\,x^{(i)}]\big\}$
and
$ \inf_if_{\clubsuit}:=\inf \big\{f_{\clubsuit}(x)\,|\, x\in [x^{(i-1)},\,x^{(i)}]\big\}$
exist as real numbers for $\clubsuit\in\{Re, \, Ze\}$ and  $1\le i\le n$. Based on these facts,
we define the {\bf type $\theta$ upper sum} $U_{\theta}(P,  f)$ of $f$ with respect to the partition $P$ to be
\begin{eqnarray*}
U_{\theta}(P,  f)&=&\left\{\begin{array}{c}
\displaystyle\sum_{i=1}^n\big(\sup_if_{Re}+1^{\#}\,\sup_if_{Ze}\big)\Delta x^{(i)}\quad\mbox{$\theta =1$;}\\  \\
\displaystyle\sum_{i=1}^n\big(\sup_if_{Re}+1^{\#}\,\inf_if_{Ze}\big)\Delta x^{(i)}\quad\mbox{$\theta =2$}
\end{array}\right.
\end{eqnarray*}
and the {\bf type $\theta$ lower sum} $L_{\theta}(P,  f)$ of $f$ with respect to the partition $P$ to be
\begin{eqnarray*}
L_{\theta}(P,  f)&=&\left\{\begin{array}{c}
\displaystyle\sum_{i=1}^n\big(\inf_if_{Re}+1^{\#}\,\inf_if_{Ze}\big)\Delta x^{(i)}\quad\mbox{$\theta =1$;}\\  \\
\displaystyle\sum_{i=1}^n\big(\inf_if_{Re}+1^{\#}\,\sup_if_{Ze}\big)\Delta x^{(i)}\quad\mbox{$\theta =2$}.
\end{array}\right.
\end{eqnarray*}
Then the following four sets
\begin{equation}\label{eq10}
\big\{\clubsuit U_{\theta}(P,  f)\,\big|\, P\in \mathcal{P}_{\theta}\big\},\quad
\big\{\clubsuit L_{\theta}(P,  f)\,\big|\, P\in \mathcal{P}_{\theta}\big\}\quad\mbox{with}\quad
\clubsuit\in\{Re, \, Ze\}
 \end{equation}\label
are bounded subsets of the real number field $\mathcal{R}$, where $\mathcal{P}_{\theta}$ is the set of all partitions of  $[a, b]_{\theta}$, i.e.
$\mathcal{P}_{\theta}:=\big\{\, P\,\big|\, \mbox{$P$ is a partition of $[a,\,b]_{\theta}$}\,\big\}$. Hence, the supremums and infimums of the four sets in (\ref{eq10}) exist. Using these facts, we introduce the {\bf type $\theta$ lower integral}
$\underline{\displaystyle\int_a^b} f(x)d_{\theta}x$ and  the {\bf type $\theta$ upper integral}
$\overline{\displaystyle\int_a^b} f(x)d_{\theta}x$ of $f(x)$ on $[a, b]_{\theta}$  in the following way:
$$\underline{\displaystyle\int_a^b} f(x)d_{1}x=
sup\big\{Re L_{1}(P,  f)\,\big|\, P\in \mathcal{P}_{1}\big\}+
1^{\#}\,sup\big\{Ze L_{1}(P,  f)\,\big|\, P\in \mathcal{P}_{1}\big\},
$$
$$
\underline{\displaystyle\int_a^b} f(x)d_{2}x=
sup\big\{Re L_{2}(P,  f)\,\big|\, P\in \mathcal{P}_{2}\big\}+
1^{\#}\,inf\big\{Ze L_{2}(P,  f)\,\big|\, P\in \mathcal{P}_{2}\big\},
$$
$$
\overline{\displaystyle\int_a^b} f(x)d_{1}x=
inf\big\{Re U_{1}(P,  f)\,\big|\, P\in \mathcal{P}_{1}\big\}+
1^{\#}\,inf\big\{Ze U_{1}(P,  f)\,\big|\, P\in \mathcal{P}_{1}\big\}
$$
and
$$
\overline{\displaystyle\int_a^b} f(x)d_{2}x=
inf\big\{Re U_{2}(P,  f)\,\big|\, P\in \mathcal{P}_{2}\big\}+
1^{\#}\,sup\big\{Ze U_{2}(P,  f)\,\big|\, P\in \mathcal{P}_{2}\big\}.
$$
If the  type $\theta$ lower integral and the type $\theta$ upper integral of $f(x)$ on $[a, b]_{\theta}$ are equal, i.e.,
if $\underline{\displaystyle\int_a^b} f(x)d_{\theta}x=\overline{\displaystyle\int_a^b} f(x)d_{\theta}x$, then
we say that $f$ is {\bf type $\theta$ integrable} on $[a, b]_{\theta}$, we denote their common value by
$\displaystyle\int_a^b f(x)d_{\theta}x$ which is called the {\bf type $\theta$ integral} of $f$  on
$[a, b]_{\theta}$.

\medskip
\begin{proposition}\label{pr3.1} Let $f: [a, b]_{\theta}\to \mathcal{R}^{(2)}$ is a bounded function, where $\theta\in\{1,\,2\}$.
\begin{description}
\item[(i)] If $P$  and $P^*$ are partitions of $[a, b]_{\theta}$ and $P^*$ is a refinement of $P$, then
$$
 L_{\theta}(P,  f)\stackrel{\theta}{\le} L_{\theta}(P^*,  f)\stackrel{\theta}{\le} U_{\theta}(P^*,  f)
\stackrel{\theta}{\le} U_{\theta}(P,  f).
$$
\item[(ii)] $
\underline{\displaystyle\int_a^b} f(x)d_{\theta}x \stackrel{\theta}{\le}
\overline{\displaystyle\int_a^b} f(x)d_{\theta}x .
$
\item[(iii)] $f$ is  type $\theta$ integrable iff for each $\varepsilon\in \mathcal{R}^{(2)}$ with
$\varepsilon\stackrel{\theta}{>}0$ and $(Re\,\varepsilon)(Ze\,\varepsilon)\ne 0$ there exists a  partition $P$ of
$[a, b]_{\theta}$ such that
$$ U_{\theta}(P,  f)-L_{\theta}(P,  f)\stackrel{\theta}{<}\varepsilon .
$$\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} The proof of Proposition \ref{pr3.1} follows from the definitions above and  the properties of the supremums and infimums. For example, let us prove (i) for $\theta=1$, i.e.,
\begin{equation}\label{eq11}
 L_{1}(P,  f)\stackrel{1}{\le} L_{1}(P^*,  f)\stackrel{1}{\le} U_{1}(P^*,  f)
\stackrel{1}{\le} U_{1}(P,  f).
\end{equation}

\medskip
The middle inequality  in (\ref{eq11}) follows directly from the definitions of type $1$ upper
and lower sums. Suppose that
$P=\{x^{(0)}, x^{(1)}, \dots , x^{(n)}\}$ and  consider the partition  $P^*$ formed by joining just  one
point $x^*$ to $P$, where $ x^{(k-1)}\stackrel{1}{\le} x^*\stackrel{1}{\le} x^{(k)}$ for some
$k$ with $1\le k\le n$. Let
$$
\alpha_1 (f_{\clubsuit}):=inf \big\{\, f_{\clubsuit}(x)\,\big|\, x\in [x^{(k-1)}, x^*]_1\,\big\},
$$
$$
\alpha_2 (f_{\clubsuit}):=inf \big\{\, f_{\clubsuit}(x)\,\big|\, x\in [x^*, x^{(k)}]_1\,\big\},
$$
where $\clubsuit\in\{Re,\, Ze\}$. The terms in $L_{1}(P^*,  f)$ and $L_{1}(P,  f)$ are all the
same except those over the subinterval $[x^{(k-1)},\,x^{(k)}]_1$. Thus we have
\begin{eqnarray}\label{eq12}
&&L_{1}(P^*,  f)-L_{1}(P,  f)\nonumber\\
&=&\big[(\alpha_1 (f_{Re})-\inf_kf_{Re})+1^{\#}(\alpha_1 (f_{Ze})-\inf_kf_{Ze})\big](x^*- x^{(k-1)})+
\nonumber\\
&&\quad +\big[(\alpha_2 (f_{Re})-\inf_kf_{Re})+1^{\#}(\alpha_2 (f_{Ze})-\inf_kf_{Ze})\big](x^{(k)}-x^*).
\end{eqnarray}
Since
$$
\alpha_j (f_{\clubsuit})\ge \inf_kf_{\clubsuit}\quad\mbox{for $ j\in\{1,\, 2\}$ and  $ \clubsuit\in\{Re,\, Ze\}$ },
$$
we get
\begin{equation}\label{eq13}
(\alpha_j (f_{Re})-\inf_kf_{Re})+1^{\#}(\alpha_j (f_{Ze})-\inf_kf_{Ze})\stackrel{1}{\ge}0
\quad\mbox{for $ j\in\{1,\, 2\}$ }.
\end{equation}

Using  (\ref{eq13}) and the facts:
$x^*- x^{(k-1)}\stackrel{1}{>}0$ and $x^{(k)}-x^*\stackrel{1}{>}0$,
we get from  (\ref{eq12}) that
$L_{1}(P^*,  f)-L_{1}(P,  f)\stackrel{1}{\ge}0$ or
$L_{1}(P^*,  f)\stackrel{1}{\ge}L_{1}(P,  f)$.

\medskip
Similarly,  let
$$
\beta_1 (f_{\clubsuit}):=sup \big\{\, f_{\clubsuit}(x)\,\big|\, x\in [x^{(k-1)}, x^*]_1\,\big\},
$$
$$
\beta_2 (f_{\clubsuit}):=sup \big\{\, f_{\clubsuit}(x)\,\big|\, x\in [x^*, x^{(k)}]_1\,\big\},
$$
where $\clubsuit\in\{Re,\, Ze\}$. Since the terms in $U_{1}(P^*,  f)$ and $U_{1}(P,  f)$ are all the
same except those over the subinterval $[x^{(k-1)},\,x^{(k)}]_1$, we have
\begin{eqnarray}\label{eq14}
&&U_{1}(P^*,  f)-U_{1}(P,  f)\nonumber\\
&=&\big[(\sup_kf_{Re}-\beta_2 (f_{Re}))+1^{\#}(\sup_kf_{Ze}-\beta_2 (f_{Ze}))\big](x^{(k)}-x^*)+
\nonumber\\
&&+\big[(\sup_kf_{Re}-\beta_1 (f_{Re}))+1^{\#}(\sup_kf_{Ze}-\beta_1 (f_{Ze}))\big](x^*- x^{(k-1)}).
\end{eqnarray}

Since
$$
\sup_kf_{\clubsuit}\ge \beta_j (f_{\clubsuit})\quad\mbox{for $ j\in\{1,\, 2\}$ and  $ \clubsuit\in\{Re,\, Ze\}$ },
$$
we have
\begin{equation}\label{eq15}
(\sup_kf_{Re}-\beta_j (f_{Re}))+1^{\#}(\sup_kf_{Ze}-\beta_j (f_{Ze}))\stackrel{1}{\ge}0
\quad\mbox{for $ j\in\{1,\, 2\}$}.
\end{equation}

It follows from  (\ref{eq14}) and  (\ref{eq15}) that
$U_{1}(P,  f)-U_{1}(P^*,  f)\stackrel{1}{\ge}0$ or $U_{1}(P,  f)\stackrel{1}{\ge}U_{1}(P^*,  f)$.

\medskip
This proves that (\ref{eq11}) holds.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
As a corollary of Proposition \ref{pr3.1} (iii), we have that if $f=f_{Re}+1^{\#}f_{Ze}$ is a function on a type $\theta$ closed interval  $[a, b]_{\theta}$ such that real-valued functions $f_{Re}$ and $f_{Ze}$ are continuous on the rectangle
$[Re\,a, Re\,b]\times [Ze\,2, \,Ze\,b]\subseteq \mathcal{R}^2=\mathcal{R}\times\mathcal{R}$, then $f$ is type $\theta$ integrable on $[a, b]_{\theta}$, where $\theta =1$ or $2$.

\medskip
The algebraic properties of the ordinary integral are still true for the type $\theta$ integrals.

\medskip
\begin{proposition}\label{pr3.2} Let $\theta =1$ or $2$ and let $k\in \mathcal{R}^{(2)}$ be a dual real number.
\begin{description}
\item[(i)] If $f$ and $g$ are  type $\theta$ integrable on $[a, b]_{\theta}$, then
$f+g$ and $kf$ are  type $\theta$ integrable on $[a, b]_{\theta}$  and
$$\displaystyle\int_a^b (f+g) d_{\theta}x=\displaystyle\int_a^b f d_{\theta}x+
\displaystyle\int_a^b g d_{\theta}x,\quad
\displaystyle\int_a^b kf d_{\theta}x=k\displaystyle\int_a^b f d_{\theta}x .$$
\item[(ii)] If $f$ is type $\theta$ integrable on both
$[a, c]_{\theta}$ and $[c, b]_{\theta}$, where $a\stackrel{\theta}{<}c\stackrel{\theta}{<}b$,
then $f$ is type $\theta$ integrable on both $[a, b]_{\theta}$ and
$$
\displaystyle\int_a^b f\,d_{\theta}x=\displaystyle\int_a^c f\,d_{\theta}x
+\displaystyle\int_c^b f\,d_{\theta}x.
$$
\item[(iii)] If $f$, $g: [a, b]_{\theta}\to \mathcal{R}^{(2)}$ are type $\theta$ integrable and $f(x)\stackrel{\theta}{\ge} g(x)$ for all $x\in [a, b]_{\theta}$, then
$\displaystyle\int_a^b f(x)\,d_{\theta}x\stackrel{\theta}{\ge}\
\displaystyle\int_a^b g(x)\,d_{\theta}x$.
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} Both (i) and (ii) are proved by using Proposition \ref{pr3.1}, and (iii) is proved by using the definitions of type $\theta$ inegrals and the properties of the two generalized order relations on the dual real number algebra.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\medskip
We finish this paper with the following counterpart of the odinary fundamental theorem of calculus in the context of dual real numbers.

\medskip
\begin{proposition}\label{pr3.3} Let $a$, $b\in \mathcal{R}^{(2)}$ and $a\stackrel{\theta}{<}b$, where $\theta=1$ or $2$.
\begin{description}
\item[(i)] If $f: [a, b]_{\theta}\to \mathcal{R}^{(2)}$ is  a function such that the real-valued functions  $f_{Re}$ and $f_{Ze}$ are continuous on the rectangle
$[Re\,a, \, Re\,b]\times [Ze\,a, \, Ze\,b]\subseteq \mathcal{R}^2$, then the function
$F(x): [a, b]_{\theta}\to \mathcal{R}^{(2)}$ defined by
$$
F(x):=\displaystyle\int_a^x f(t)\,d_{\theta}t\quad\mbox{for $x\in [a, b]_{\theta}$}
$$
is type $\theta$ diffrential at  each $c\in [a, b]_{\theta}$ and $F'_{\theta}(c)=f(c)$.
\item[(ii)] If $f(x): [a, b]_{\theta}\to  \mathcal{R}^{(2)}$ is differential on $[a, b]_{\theta}$ and the derivative $f'(x)$ of $f(x)$ is integrable on $[a, b]_{\theta}$, then
$$\displaystyle\int_a^b f'(x)\,d_{\theta}x=f(b)-f(a).$$
\end{description}
\end{proposition}

\medskip
\noindent
{\bf Proof} The way of proving Proposition \ref{pr3.3} comes from the application of the
algebraic properties of type $\theta$ integrals in Proposition \ref{pr3.2}. Let us prove (i) to explain the way of doing the proofs.

\medskip
By the definitions of type $\theta$ integrals, we have
\begin{equation}\label{eq16}
a\stackrel{\theta}{\le} b\Longrightarrow
\displaystyle\int_a^b \,d_{\theta}x=b-a\quad\mbox{for $\theta\in\{1,\, 2\}$}.
\end{equation}

\medskip
Clearly, (i) holds if we can prove that for each positive real number $\epsilon>0$ there exists a  positive real number $\delta>0$ such that
\begin{equation}\label{eq17}
x\in N_{\theta}^*(c;\,\delta)\cap [a, b]_{\theta}\Rightarrow
\frac{||F(x)-F(c)-f(c)(x-c)||}{||x-c||}<\epsilon ,
\end{equation}
where $\theta\in\{1,\, 2\}$. The proofs of (\ref{eq17}) for $\theta=1$ and $\theta=2$ are similar, so we prove (\ref{eq17}) for $\theta=1$. First,
we choose two positive real numbers $\varepsilon_{Re}$ and $\varepsilon_{Ze}$ such that
\begin{equation}\label{eq18}
 \mbox{$0<\varepsilon_{Re}<\displaystyle\frac{\epsilon}{3}$\quad and\quad
$0<\varepsilon_{Ze}<\displaystyle\frac{\epsilon -3\varepsilon_{Re}}{\sqrt{2}}$.}
\end{equation}

\medskip
Next, since both real-valued functions $f_{Re}$ and $f_{Ze}$ are continuous on the rectangle
$[Re\,a, \, Re\,b]\times [Ze\,a, \, Ze\,b]\subseteq \mathcal{R}^2$, both $f_{Re}$ and $f_{Ze}$ are uniformly
continuous on the rectangle $[Re\,a, \, Re\,b]\times [Ze\,a, \, Ze\,b]$. Hence, there exist a positive real number $\delta >0$ such that
\begin{equation}\label{eq19}
\sqrt{(t_1-s_1)^2+(t_2-s_2)^2}<\delta\nonumber\\
\Rightarrow |f_{\clubsuit}(t_1,\,t_2)-f_{\clubsuit}(s_1,\,s_2)|<\varepsilon_{\clubsuit}
\end{equation}
for all $(t_1,\,t_2)$, $(s_1,\,s_2)\in [a_1, b_1]\times [a_2, b_2]$ and
$\clubsuit\in\{Re,\, Ze\}$.

\medskip
Let $x\in N_{1}^*(c;\,\delta)\cap [a, b]_{\theta}$. Then $x\stackrel{1}{\ge}c$ and
$x\stackrel{1}{\le}c$.

\medskip
{\bf Case 1}: $x\stackrel{1}{\ge}c$, in which case, by
Proposition \ref{pr3.2} and (\ref{eq16}), we have
\begin{eqnarray}\label{eq20}
&&F(x)-F(c)-f(c)(x-c)=
\displaystyle\int_a^x f(t)\,d_{1}t-\displaystyle\int_a^c f(t)\,d_{1}t-f(c)(x-c)\nonumber\\
&=&\displaystyle\int_c^x f(t)\,d_{1}t-f(c)\displaystyle\int_c^x \,d_{1}t
=\displaystyle\int_c^x f(t)\,d_{1}t+\displaystyle\int_c^x \big(-f(c)\big)\,d_{1}t\nonumber\\
&=&\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t .
\end{eqnarray}

\medskip
For $c=c_1+c_21^{\#}\stackrel{1}{\le} t=t_1+t_21^{\#}\stackrel{1}{\le} x=x_1+x_21^{\#}$, where
$c_1$, $c_2$, $t_1$, $t_2$, $x_1$ and  $x_2\in\mathcal{R}$, we have
\begin{eqnarray*}
&&\sqrt{(t_1-c_1)^2+(t_2-c_2)^2}\le \sqrt{(x_1-c_1)^2+(x_2-c_2)^2}\\
&\le& \sqrt{2(x_1-c_1)^2+(x_2-c_2)^2}=||x-c||<\delta\\
&\Rightarrow& |f_{\clubsuit}(t_1,\,t_2)-f_{\clubsuit}(c_1,\,c_2)|<\varepsilon_{\clubsuit}
\quad\mbox{for $\clubsuit\in\{Re,\, Ze\}$,}
\end{eqnarray*}
which implies
\begin{equation}\label{eq21}
-\varepsilon=-\varepsilon_{Re}-\varepsilon_{Ze}1^{\#}\stackrel{1}{\le}f(t)-f(c)
\stackrel{1}{\le}\varepsilon_{Re}+\varepsilon_{Ze}1^{\#}=\varepsilon.
\end{equation}

It folows from (\ref{eq16}), (\ref{eq21}) and Proposition \ref{pr3.2} (iii) that
$$
-\varepsilon (x-c)=\displaystyle\int_c^x (-\varepsilon)\,d_{1}t\stackrel{1}{\le}
\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\stackrel{1}{\le}
\displaystyle\int_c^x \varepsilon\,d_{1}t=\varepsilon (x-c),
$$
which gives
\begin{equation}\label{eq22}
\Big|\clubsuit\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big|
\le \big|\clubsuit\big(\varepsilon (x-c)\big)\big|\quad\mbox{for $\clubsuit\in\{Re,\, Ze\}$}.
\end{equation}
Since
\begin{eqnarray*}
&&\varepsilon (x-c)=(\varepsilon_{Re}+\varepsilon_{Ze}1^{\#})[(x_1-c_1)+(x_2-c_2)1^{\#}]\\
&=&\varepsilon_{Re}(x_1-c_1)+[\varepsilon_{Re}(x_2-c_2)+\varepsilon_{Ze}(x_1-c_1)]1^{\#}),
\end{eqnarray*}
we have
\begin{equation}\label{eq23}
Re\big(\varepsilon (x-c)\big)=\varepsilon_{Re}(x_1-c_1)
\end{equation}
and
\begin{equation}\label{eq24}
Ze\big(\varepsilon (x-c)\big)=\varepsilon_{Re}(x_2-c_2)+\varepsilon_{Ze}(x_1-c_1).
\end{equation}

\medskip
By (\ref{eq22}), (\ref{eq23}) and (\ref{eq24}), we have
\begin{eqnarray}\label{eq25}
&&\Big|\Big|\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big|\Big|\nonumber\\
&=&\sqrt{2\Big[Re\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big]^2
+\Big[Ze\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big]^2}\nonumber\\
&\le&\sqrt{2}\Big|Re\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big|
+\Big|Ze\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big|\nonumber\\
&\le&\sqrt{2}\,\,\Big|\varepsilon_{Re}(x_1-c_1)\Big|+
\Big|\varepsilon_{Re}(x_2-c_2)+\varepsilon_{Ze}(x_1-c_1)\Big|\nonumber\\
&\le&\sqrt{2}\,\,\varepsilon_{Re}\big|x_1-c_1\big|+\varepsilon_{Re}\big|x_2-c_2\big|
+\varepsilon_{Ze}\big|x_1-c_1\big|
\end{eqnarray}

\medskip
It follows from (\ref{eq20}) and (\ref{eq25}) that
\begin{eqnarray}\label{eq26}
&&\frac{||F(x)-F(c)-f(c)(x-c)||}{||x-c||}
=\frac{\Big|\Big|\Big(\displaystyle\int_c^x \big[f(t)-f(c)\big]\,d_{1}t\Big)\Big|\Big|}
{||x-c||}\nonumber\\
&\le&
\frac{\sqrt{2}\,\,\varepsilon_{Re}\big|x_1-c_1\big|+\varepsilon_{Re}\big|x_2-c_2\big|+\varepsilon_{Ze}\big|x_1-c_1\big|}
{\sqrt{2(x_1-c_1)^2+(x_2-c_2)^2}}
\end{eqnarray}

\medskip
{\bf Case 11}: $x_1=c_1$, in which case, by (\ref{eq18}) and (\ref{eq26}), we get
\begin{equation}\label{eq27}
\frac{||F(x)-F(c)-f(c)(x-c)||}{||x-c||}
\le \frac{\varepsilon_{Re}\big|x_2-c_2\big|}{\sqrt{(x_2-c_2)^2}}=\varepsilon_{Re}<
\displaystyle\frac{\epsilon}{3}< \epsilon.
\end{equation}

\medskip
{\bf Case 12}: $x_1\ne c_1$ and
$\left|\displaystyle\frac{x_2-c_2}{x_1-c_1}\right|\le \displaystyle\frac{1}{\sqrt{2}}$, in which case, using (\ref{eq18}) and (\ref{eq26}) again, we get
\begin{eqnarray}\label{eq28}
&&\frac{||F(x)-F(c)-f(c)(x-c)||}{||x-c||}\nonumber\\
&\le&
\frac{\sqrt{2}\,\,\varepsilon_{Re}\big|x_1-c_1\big|+\varepsilon_{Re}\big|x_2-c_2\big|+\varepsilon_{Ze}\big|x_1-c_1\big|}
{\sqrt{2(x_1-c_1)^2+(x_2-c_2)^2}}\nonumber\\
&=&\frac{\sqrt{2}\,\,\varepsilon_{Re}+
\varepsilon_{Re}\left|\displaystyle\frac{x_2-c_2}{x_1-c_1}\right|+\varepsilon_{Ze}}
{\sqrt{2+\left(\displaystyle\frac{x_2-c_2}{x_1-c_1}\right)^2}}
<\sqrt{2}\,\,\varepsilon_{Re}+
\varepsilon_{Re}\left|\displaystyle\frac{x_2-c_2}{x_1-c_1}\right|+\varepsilon_{Ze}\nonumber\\
&<&\sqrt{2}\,\,\varepsilon_{Re}+\varepsilon_{Re}\displaystyle\frac{1}{\sqrt{2}}+\varepsilon_{Ze}
=\displaystyle\frac{3}{\sqrt{2}}\varepsilon_{Re}+\varepsilon_{Ze}
<\displaystyle\frac{1}{\sqrt{2}}\epsilon<\epsilon .
\end{eqnarray}

\medskip
{\bf Case 13}: $x_1\ne c_1$ and
$\left|\displaystyle\frac{x_2-c_2}{x_1-c_1}\right|\ge \displaystyle\frac{1}{\sqrt{2}}$. In this case, we have  $\left|\displaystyle\frac{x_1-c_1}{x_2-c_2}\right|\le \sqrt{2}$ and
\begin{eqnarray}\label{eq29}
&&\frac{||F(x)-F(c)-f(c)(x-c)||}{||x-c||}\nonumber\\
&\le&
\frac{\sqrt{2}\,\,\varepsilon_{Re}\big|x_1-c_1\big|+\varepsilon_{Re}\big|x_2-c_2\big|+\varepsilon_{Ze}\big|x_1-c_1\big|}
{\sqrt{2(x_1-c_1)^2+(x_2-c_2)^2}}\nonumber\\
&=&\frac{\sqrt{2}\,\,\varepsilon_{Re}\left|\displaystyle\frac{x_1-c_1}{x_2-c_2}\right|+
\varepsilon_{Re}+\varepsilon_{Ze}\left|\displaystyle\frac{x_1-c_1}{x_2-c_2}\right|}
{\sqrt{2\left(\displaystyle\frac{x_1-c_1}{x_2-c_2}\right)^2+1}}\nonumber\\
&\le& \sqrt{2}\,\,\varepsilon_{Re}\left|\displaystyle\frac{x_1-c_1}{x_2-c_2}\right|+
\varepsilon_{Re}+\varepsilon_{Ze}\left|\displaystyle\frac{x_1-c_1}{x_2-c_2}\right|
\nonumber\\
&\le& \sqrt{2}\,\,\varepsilon_{Re}\sqrt{2}+
\varepsilon_{Re}+\varepsilon_{Ze}\sqrt{2}
=3\varepsilon_{Re}+\varepsilon_{Ze}\sqrt{2}<\epsilon .
\end{eqnarray}

\medskip
It follows from (\ref{eq27}), (\ref{eq28}) and (\ref{eq29}) that (\ref{eq17}) holds in Case 1.

\bigskip
{\bf Case 2}: $x\stackrel{1}{\le}c$, in which case, a similar computation shows that (\ref{eq17}) holds in Case 2.

\bigskip
This proves that (\ref{eq17}) holds for each $x\in N_{1}^*(c;\,\delta)\cap [a, b]_{1}$.

\hfill\raisebox{1mm}{\framebox[2mm]{}}

\bigskip

redpajama_set_name: RedPajamaArXiv


text:
Jump To062: "Scum and villainy"063: "What a hole"064: Mos Eisley Checkpoint065: "You may pass"066: "This place is a little... rough"067: "I don't like you either"068: "No blasters!"069: "Hai!" We meet Chewbacca070: "Solo's the name" We meet Han071: "A passage to Alderann"072: Han Shot First. Here's the proof!073: "What a piece of junk!"074: "Take 'Em Men!" Docking Bay 94075: "Fire her up Chewie - we're haulin' jets!"076: The Millennium Falcon Blasts Off from Mos Eisley!077: Farewell Tatooine... *sigh*
068: "No blasters!"
by admin on May 31, 2016 at 2:09 pm
Characters: Ben Kenobi, Dr. Evazan, Luke Skywalker, Wuher
Art (?) Notes
"BEN KENOBI
JEDI-NIGHT"
This most juvenile of all of the pages must have been drawn in December 1977 – or very early 1978.
I'm astonished at just how rudimentary the drawing is! You could be forgiven for wondering if they were even drawn beforeseeing the film. Not possible though. I don't think the comic or the novel with its lovely full-colour photos were available in Ireland until early 1978.
Needless to say this is one of my faves.
Ben might originally have been drawn with 'eyes in the back o his head'
Can you tell me reader: which way is that very undersized Obi-Wan supposed to be facing? I'm wondering if he originally had his back to the baddie and craftily activated his lightsaber backwards. Very cool. I'd seen those 'Trinity' spaghetti westerns around that time and he did stuff like that with his pistol. Anyhow, I ask the question partly because of his oddly drawn body and because his head's been redrawn later in blue biro.
Blue biro digitally removed (right). I think my re-do made it look even weirder!
Corrrrr… gruesome!
Yeah – that's what 9 year old boys want isn't it? But you know, this scene was very gruesome for a PG-rated film in the seventies. I don't know how old you are, but I know that that sort of graphic violence is nothing nowadays. Kids can watch virtually anything – unfazed. I believe Lucas did have trouble from the studio over this scene because it might narrow the potential audience. But what was he originally intending to do? If you read the novel you might find some clues.
"…the man found himself lying prone againt the bar, moaning and whimpering as he stared at the stump of an arm.
…the rodent thing had been cleft cleanly in half down the middle, its two halves falling away in opposite directions.
The giant multi-ocular creature still stood staring, dazed, at the old human who stood motionlesss before it (…) the creature fired once, blowing a hole in the door. Then the torso peeled away as neatly as had the body of the rodent, its two cauterised sections falling in opposite directions to lie motionless on the cool stone.
– 1977 Novelisation, by Alan Dean Foster
Meet 'Wuher'
Above: the late Ted Burnett as the grumpy bartender. AndHoward Chaykin's surprisingly faithful rendering of him!
Euckk. Did you know that the actor who played the bartender originally had a London accent? he was one of the many cast members that George would re-dub because of his fears that it was all sounding too British. I won't fool myself that Ted Burnett felt gutted after he saw the final film – I wouldn't be surprised if he never even saw it. For him, in such a minor supporting role he was possibly happy enough with his few quid at the end of the day. Job done, nice and quick. If so, he wouldn't have been the only person involved who didn't expect the film to turn out to be anything special. Carrie Fisher for example.
Kenny Baker actually often recals saying "We thought it was a load of rubbish. What's it all about? It's seems a bit straaaange." But whether Ted recognised its greatness or not – whether he ever gave it much thought until his passing in 2001: this page is for you Teddy.
└ Tags: 1978, action, age 9, Ben Kenobi, blood, Cantina, Dr. Evazan, gore, Jedi, lightsaber, Luke Skywalker, MARKER, MARKER: big black permanent, MARKER: highlight, Mos Eisley, PAPER: MSD (tables), REVISIONS, Tatooine, Version 1, violence, Wuher


It is capable of handling the whole enterprise mobility management. All major platforms including iOS, Android, Windows (Phones and PCs), macOS. Sony, HTC, and Apple's Device Enrollment Program (DE.
flag=T&rep_id=1103 The market for ultra-mobile computing devices is expected to expand at a promising pace in the next few years as the trend of mobility gathers further. UMD market can be segmente.
Microsoft's Windows. built" for Apple's machine. Excel for iPad screenshot. Source: Microsoft. The other product announcement was a new licensing option supporting mobile device management. Microso.
BlackBerry pioneered enterprise mobility with. enterprise platform. Apple (NASDAQ:AAPL) (iOS) and Google (GOOG, GOOGL) (Android) were built for the consumer and adapted, to some degree, for enterpr.
There are various future developments in the field of enterprise mobility field along with the challenges of app management, accompanied by place of Google and Apple in the enterprise. actually mea.
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Citrix released the XenApp 6.5 Mobility Pack as part of its Receiver for Android 3.0 update in December, improving the way Windows applications are delivered and used on mobile devices. Most mobile cl.
–(BUSINESS WIRE)–Today, Citrix released its Enterprise Mobility Cloud. to 36 percent of the region. Apple saw its iOS penetration drop from 56 percent to 43 percent share of enrolled devices in t.
We even see car companies try to move upstream from being makers of cars to being providers of mobility. Traditionally, this is where Apple succeeded. or hardware but by commercial partnerships. Th.
It has agreed to take off Google's (GOOG) hands a storied U.S. brand — Motorola Mobility — picking. discontinued support for Windows XP, prompting IT managers to replace millions of aging Windows b.
Commissioned by Citrix to produce insight into mobility. organization. Apple iOS remains popular, with 54 percent supporting or planning to support the platform, and 48 percent seeing an increase i.
In March 2011, Apple sued Motorola Mobility in US District Court in Wisconsin. fees for wireless and video patents Motorola wants for each sale of an Xbox 360 or Windows PC. That trial begins Nov.
On-demand applications need to rapidly scale with a business, rather than working against it," said Francis Dinha, CEO of OpenVPN Inc. "Business mobility through remote. all OS platforms ranging fr.
It was obvious back then that Android was eating Microsoft's lunch and that it was beginning to nibble at Apple's heels, too. But while Microsoft wanted phone makers to pay for the Windows Phone OS.
The acquisition, however, appears to be a bid to bulk up Android's patent strength, which will benefit from Motorola's deep portfolio of mobility patents. Apple, Google's rival. their bets and perh.
Seventy-seven percent of all new smartphones and tablets activated in the enterprise last quarter were Apple. Q4 2012 Windows reigns, of course, on the enterprise desktop, but it is a marginal, nic.
When you choose a product, a pop-up window lets you select size. or other methods such as Android Pay or Apple Pay. If the.
As much as mobility continues to exist and grow. Various mobile platforms, such as Mac OS X, Windows, iOS and Android, have already made improved system upgrades to enable app development companies.
The upright, elegant exterior shape is highlighted by a huge BMW "kidney" grille and tall side windows. with the Open Mobi.


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