ALI JULIA Brazilian Ombre Straight Hair Weft 3 Bundles 100% Real Human Hair Weave Extensions 95-100g/pc (16 18 20 T1B/4/27) - Hair weight&length 95-100g/bundle,16inch~26inchmultiple choices. Delivery time3~7 working days. Hair colort1b/4/27ombre. Hair material100% human hair,grade 6a brazilian straight human hair. Hair advantage no shedding,no tangles,lasting silky look and feel softwith a factory history of more than 10 years, we have strict quality control and skilled workers.
Yaweida Hair 8~28 inches Brazilian Virgin Human Hair Extension Silky Straight, Pack of Three, 100g/Bundle, 7A Natural Color Weft(10 10 10) - Shippingit usually takes about 3-7 working days by fedex or dhl to your destination. Yaweida is our registered brand factory price, if you need large quantity, please contact us and we will offer you best prices with best quality. Unprocessed brazilian virgin hair,true 100% human hair with no artificial or non-human hair mixed in holding good texture for long time and no shedding, lasting silky shining and soft. The brazilian straight human hair is shine,soft,thick and no tangle and no sheddingwith high technic and our hair texture last long timebrazilian hair is one of the most popular types of hair extensions in the market and is also naturally compatible with many hair types. About 100g/bundlefrom 10 inches to 18 inches,3 bundles are enough to make a full head,while from 20 inches to 26 inches, it is better to take 4 bundles to make a full headif you like thicker or longer hair,it's best to wear one more bundleas the longer the length is,the thinner the hair is.
Free Part Lace Closure with Baby Hair Brazilian Straight Human Hair Closure Natural Color 4×4 Inch - 3 medium brown swiss lace blends well with most popular brazilian straight closure. 2 free/middle/3 part closure with baby hair, you can part it as your desire, natural hairline is more natural-looking. 5 excellent service, if you have any problem,just feel free to contact me,i will do my best to help you. 1 nice straight lace closure human hair, soft and clean to wear, 130% density, don't bleached knots. 4 not suggest to dye the closure human hair since it may affect the lace life.
Eayon Hair® 6A Virgin Hair Human Lace Front Wig Brazilian Remy Human Straight Hair Lace Wigs with Baby Hair For African Americans 130% Density Naturl Color 20inch - 2high hair density130% full and thick feeling,for higher density (150% and 180% ) ,should be customized has 3 combs and straps(default), wig cap size can be ajusted. 5this brazilian lace front wig has natural frontal and hair lines around perimeter is natural baby hair making wigs look natural. 4this human lace wigs hair is natural color(close to brown,if you like the natural black,pls contact us). 3cap size(circumference) middle size(225″ in stock),small size(2175″ shoulb be customized),large size (2275″ should be customized ) lace color medium brown (in stock) , dark brown and light brown should be customized . 1this is 100% unprocessed virgin human hair lace front wig,baby hair around perimeter to give a natural hairlinehand tied 134 free part lace front.
ATOZWIG Newness Brazilian Straight Human Hair Weave Bundles 7A Grade Remy Hair Weaving #8 Color Hot Beauty Cheap Soft Thick End Hair 2pcs/lot - Curl, crimp, natural wave, up-do, or straight, it will adapt to any style day in and day out. Our continuing commitment to our clients is to provide the finest grade human hair that has the versatility to keep up with today's changing fashion and styles. We guarantee that every single strand is 100% human and no substitutes are used to cut costs and lower prices. We carry only 100% raw virgin human hair collected from around the world. With proper care and maintenance, and the use of the correct hair care products it is not uncommon for our products to last up to two full years. Our real true virgin human hair is completely natural and no chemical processes are used. Our hair is specifically designed to allow our clients the freedom of choice. We have our own team of hair collectors that painstakingly hand pick only the highest quality raw materials. This allows the extensions to blend in with your own natural hair, especially when your own hair starts to grow out after a few months. This is how we are able to provide a product with the strictest quality standards for an unbeatable price. All of our hair originates from the single person to ensure that the entire bundle is the same in texture, feel, and look. "Atozwig- ""7a 100% virgin human hair"" atozwig features [atozwig brand] atozwig [atozwig material] 100% unprocessed brazilian virgin hair straight [atozwig weight] 50g/bundle,3-5 pieces cover full head [atozwig color] natural color can be dyed as shown on picture [atozwig style]picture shows [atozwig length] 8 inch to 28 inch ". It is then shipped to our own factory where it is assembled and packaged. The cuticles are the body's natural protection and are essential in maintaining the integrity of the hair structure.
Dreambeauty Mixed Length 3Bundles 300g Virgin Brazilian Straight Human Hair Extension Unprocessed for Black Women 22″+24″+26″ - 35oz per bundle, 100g/pc, need 3 or 4 bundles to make a full wig. 100% unprocessed top quality human hair, brazilian virgin hair, none synthetic hair mixed, 100% guarantee. Straight texture, can restyle to any other texture. High quality, no shedding,no tangle 10years experience in human hair. Natural color, can be dyed to other color you need if need other color please leave message.
VRBest Hair Virgin Brazilian Straight Human Hair Extensions 3 Bundles Unprocessed Brazilian Virgin Hair Weave Bundles Natural Black Color (12 14 16 inches) - Hair material 100% unprocessed human hair,brazilian virgin human hair. Hair weight&length 95-105g/bundle, 3 bundles/order,multiple choices,8inch-26inchfor the length,stretching the hair to be straight and then measure. Vrbest hair — trustworthy hair brand we have not authorized any other sellers to sell our products on amazon only vrbest hair can guarantee top quality, please double check before you orderdon't order the fake one. Hair color natural black color, can be dyed and bleached. Hair qualityno shedding,no tangles,can be straightened,curled, bleached and styled as your own hair.
Lin Hair Unprocessed Virgin Brazilian Straight Human Hair Extensions Brazilian Natural Hair Weave Weft Bundles #1b 4pcs 12 14 16 18inch - 3, 5 day expedited shipping by fedex. 95g, 100g/ bundle, natural color (#1b), 12inch, 26inch in stock. 100% unprocessed virgin brazilian human hair is guaranteed. Virgin & raw human hair from healthy donors,no lice,full cuticle,healty ends. Can be bleached or dyed, no chemical or steam process minimum shed, no tangle.
Cool2day 100% Virgin Brazilian Straight Human Hair Weave Extensions,Natural Color,100G,Grade 6A(18″) - 100% virgin brazilian straight human hair extensions feature item type hair extensions hair extension type straight hair grade 6a length8″,10″, 12″, 14″, 16″, 18″, 20″, 22″, 24″, 26″,28″,30″ colornatural color material100% virgin brazilian human hair net weight 100g approx per bundle package including 1 x 100% straight human hair.
Beautyplus Hair Charming girls' Brazilian Straight Human Hair Extensions with Closure (16 18 20+12 Inch Closure) 100% 7a Grade Unprocessed Virgin Hair 3 Bundles - 4hair weight and hair color100g/pc (35oz/pc) including package is 110-130g/pc,natural black hair. 2hair grade7a grade unprocessed virgin hair. 5 shipment and delivery time by fedex, please pay attention to the logistics information on amazon items returning or changing is acceptable within 7 days and all the payment will be returned within 35 working days after we receive the hair with original. 1hair material and hair type100% unprocessed human hair, peruvian、brazilian、malaysian virgin hair. 3hair feature100% real human hair soft ,smooth, gloosy, full cuticle ,double strong hair weft natural hair, no shed, no tangle, no smell ,no knots, no lice with thick bottom, no short hair can be curled, straightened hair durability at least 8 months100% top hand selected human hair without synthetic and animal hairlarge stock and prompt delivery.
Guangxun Hair Wholesale Straight Virgin Hair Weave 4 Bundles 7A Brazilian 100% Unprocessed Remy Human Hair Weft Extensions 95-100g/pc Natural Black (14 16 18 20) - 1hair material100% unprocessed 7a brazilian virgin human hair,no synthetic or animal hair mixed. 2hair colornatural color-off black,can be dyed&bleached,mixed length 10-30inch in stock. 3hair qualityvery clean and healthy,no shedding,no tangle,no lice,no bad smell. 5about shippingfree shipping,3-5 working day by dhl/fedex,15 days no reason to return. 4hair bundles4 bundles/lot,95-105grams per bundle,wholesale price,good customer service.
Allove Hair 8A Brazilian Silky Straight Hair Bundles 100% Virgin Human Hair Wave 3bundles Unprocessed Human Hair Extensions Natural Black Color 300g in Total(14 16 18) - Satisfaction guaranteed, we pride ourselves in having the best brazilian virgin straight hair for each size as such, each order comes with a 30-day money back guarantee for satisfaction. Inspires natural beauty, unprocessed virgin hiar with strong elasticity, 8a brazilian straight human hair made of good quality virgin hair to achieve the glossy looking each piece with strictly quality inspection, can be used over 1 years with proper care, top quality product with an excellent reputation. Scroll down for further special offers &product promotions, your best choice of virgin straight hair comes with a premium quality- 8a brazilian unprocessed virgin straight human hair with natural black color, available in 8-28 inch, pack of 3, 100% brazilian virgin remy human hair, can be dyed and bleached, 100 grams per bundles. Fast&free shipping, sold by allove hair will ship from china by dhl, 3-5 business days arrive sold by fba, only takes 2 days arrive. Thick & silky, 8a unprocessed virgin brazilian straight human hair bundles, one-directional cuticles, all hair cut from one donor no chemical processed, no dry, no follicle, double machine weft soft, thick, smooth and luster have strong weft and healthy ends.
Colorful Queen Brazilian Virgin Hair Straight Remy Human Hair 3 Bundles Weaves 7A Unprocessed Hair Extensions Natural Color 12 14 16Inch - Hair colornatural color,can be dyed alwayshair bundles weight 95-100g/bundlebrazilian straight hair 44 lace closure, density 120%-130%,about 25-40g swiss lace, medium brown,strong and invisible. Hair package brazilian hair straight 3 bundles£¬total 3 pieces/lot. Hair material100% unprocessed human hair,grade 7a brazilian straight virgin human hair bundles. Hair quality no shedding, no tangles, no lices, can be straightened, curled, bleached and styled as your own hair (for the length,stretching the hair to be straight and then measure). Shipment orders will be shipped within 24 hours except weekend and the delivery time for free standard shipping is about 2 to 5 business days.
[email protected] 8A Virgin Brazilian Straight Human Hair Bundles, 4pcs 100% Real remy Brazilian Hair,Unprocessed Brazilian Straight Hair - Our continuing commitment to our clients is to provide the finest grade human hair that has the versatility to keep up with today's changing fashion and styles. With proper care and maintenance, and the use of the correct hair care products it is not uncommon for our products to last up to two full years. Our hair is specifically designed to allow our clients the freedom of choice. The cuticles are the body's natural protection and are essential in maintaining the integrity of the hair structure. This is how we are able to provide a product with the strictest quality standards for an unbeatable price. Curl, crimp, natural wave, up-do, or straight, it will adapt to any style day in and day out. It is then shipped to our own factory where it is assembled and packaged. Our real true virgin human hair is completely natural and no chemical processes are used. All of our hair originates from the single person to ensure that the entire bundle is the same in texture, feel, and look. We have our own team of hair collectors that painstakingly hand pick only the highest quality raw materials. We carry only 100% raw virgin human hair collected from around the world. This allows the extensions to blend in with your own natural hair, especially when your own hair starts to grow out after a few months. We guarantee that every single strand is 100% human and no substitutes are used to cut costs and lower prices. "Atozwig- ""7a 100% virgin human hair"" atozwig features [atozwig brand] atozwig [atozwig material] 100% unprocessed brazilian virgin hair straight [atozwig weight] 50g/bundle,3-5 pieces cover full head [atozwig color] natural color can be dyed as shown on picture [atozwig style]picture shows [atozwig length] 8 inch to 28 inch ".
Haoyuan 10~28 inches 100% Unprocessed Virgin Brazilian Straight Human Hair Weave Extension 3 Bundles Total 300g Natural Black Color (22 24 26 28, Natural Black 4 Bundles) - Delivery time 3~5working days. Human hair color natural color, can be dyed always. Grade aaaaaa (6a), natural color, 95-100g (335-352oz) per bundle, measure the length of hair after it is straightly stretched. 100% unprocessed virgin hair cut from one donor directly, full cuticle aligned, no chemical process strong and well double weft constructed, no shedding, no tangle. Brazilian hair body wave can be dyed, bleached, curled, straightened natural color (#1b) soft 10-28inch in stock.
Unice Hair 3 Way Part 1pc 4×4 Lace Closure with Virgin Brazilian Straight Human Hair 3 Bundles Weaves Natural Color (12 14 16+10inch closure) - Unice hair- the united states trustworthy brandon amazon,we have not authorized any other sellers to sell our productsunice hair allowed only. Natural color, can be straightened, curled, bleached, dyed and styled as your own hair. Straight hair extension about 95-100g/pc lace closure 44,30-35g free part for the length stretching the hair to be straight and then measure. 100% human virgin hair, double weft, soft, healthy, no smell, tangle-free, no shedding. Delivery time 3-5 working days.
ALi Queen 1 Bundles of 20″ Brazilian Straight Virgin Human Weave Hair 100% Unprocessed Human Hair Extensions - We accept customise orders, make it as you like ali queen has its own factory, the quality and service is the soul of the long existing of our enterprise please feel freely to purchase we also wholesale hair, if you want, please contact us. Hair color and longevity natural black color, can be washed, dyed, permed, heat styled real human hair has a longer lifespan than any other type of hair it will last longer under the correct condition of caring. Delivery fulfilled by amazon(2-5 business days) fulfilled by seller expedited (2-5 business days), standard (5-15 business days) if there is any problem with the delivery, please contact us. Hair condition 100% brazilian human hair extension, no shedding, no tangles, no spilt ends. Hair length & weight 20 inches length & 100g (±5g) each bundle very soft, healthy and thickif you want to full fill one head, 3pcs will be ok if the hair over 18″, 4pcs should be better.
Shengji Hair 360 Brazilian Straight Human Hair 22×4 Lace Front Closure for Black Women (14 inch) - No tangle, no shedding, no smell, soft. Free shipping by dhl, fedex and ups which only need 3-5 days, you can choose waht you want. 130% density top lace frontal closure with baby hair 14 inch. All customers order from us and if not satisfied with our product, you can apply for return, as long as keep it in original condition (including extra lace was uncutted), we will accept the return and process full refund and return shipping cost for customer. Straight 100% virgin human hair 360 lace front closure bleached knots.
Beata Hair 130% Density Brazilian Straight Human Hair Lace Frontal Closure 13×4 Free Part Bleached Knots Baby Hair 14inch - After ran fingers though the full human hair, very seldom shedding exsist. Nice lace frontal lightly bleached with baby hair around easy to bleach knots better as you want after bleaching the knots, please use some oil to take good care of it sometimes bleaching knots will affect the soft and silky. If you have a question about the brazilian frontal, we are happy to help within 6 hours. Free part 13×4 inch ear to ear lace frontal enough to coverage full head and doesn't need any plucking. Natural look hair line and middle brown lace matches well with young black women.
Uniwigs Hair Mixed Length 3bundles 300g Virgin Brazilian Straight Human Hair Extension (22inches 24inches 26inches) Natural Color Can be Dyed - Feature can be dyed, highlighted, bleached, curled, straightened and styled customized color is accepted. Material 100% brazilian virgin remy human hair. Note uniwigs brand is the real owner for this item any other sellers are not allowed to sell our items. Package 300g 3 bundles. Hair texture straight.
Irina Hair Products 100% Unprocessed Virgin Brazilian Straight Human Hair Weave Extensions 3Bundle/lot Natural Color 95-100g/pc (10 10 10inch) - Hair stylestraight hair bundles 3pcs/lot or 1pcs free part lace closure. Grand opening sale, best price, fulfilled by amazontry us and you won't regret. Material 100% unprocessed human hair, grade 7a brazilian virgin hair. Hair colour1b natural black. Irina hair, the united states trustworthy brand we have not authorized any other sellers to sell our products on amazon.


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sonrisa salvaje
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sonrisa salvaje's Activity
Terry replied to a thread Jeff Beck Dead at 78 in House of Music
I would have liked to have seen that. Lately, I've been going backward and relistening to that 50's rockabilly stuff.
Terry replied to a thread New Roth Show episode in Main VH/DLR Discussion
Like, not that the guy didn't have his own problems and aspects of his personality that were less than ingratiating at times - just like anybody else...
Terry replied to a thread David Crosby Dead at 81 in House of Music
I recall reading years back that he had more than a few health problems. I haven't even thought about him in musical terms for over thirty years....
Terry replied to a thread DLR on Craig Ferguson's Late, Late Show in Main VH/DLR Discussion
Sam left the band in 1996. He did, what, the one tour and the three new tracks in 2003/2004. That whole experience was pretty fucked up and...
I don't even have anything negative to say about people who like Sammy Hagar anymore. It's not my job or business to tell people they are wrong for...
Didn't (and still don't) much care for Van Halen's Jump, much less a bluegrass version of it. Thought the 2003 Diamond Dave release was bunk.
Thing of it is, who cares anymore one way or the other? Someone thinking Roth is shit and preferring Hagar doesn't faze me in the least. It's not...
Terry replied to a thread Lisa Marie Presley Dead at 54 in Max's Non VH/DLR Related Stuff
Yep, well, those Elvis fans who worshipped him like a god needed to make their pilgrimage to the shrine at Graceland: can't properly pay their...
Apparently, Elvis had an underage sweet tooth. But Elvis gets - from his fans, anyway - a perpetual pass because he's Elvis. His fans won't...
Well, Elvis had the advantage of being largely able to control the display of his image from 1973 to 1977, so large swaths of the public weren't...
I thought the 1968 special was alright, mama. At least with that special Elvis was trying to recapture some of the glimmer of what made him...
She had one single I sort of remember...was it the late 1990s, or the early 2000s (or the mid 2000s)? At any rate, it was a kind of rock/pop with a...
Terry replied to a thread Robbie Bachman of BTO...dead in House of Music
Every day, in every way...they even worked overtime...woo!
Terry replied to a thread David Lee Roth - Full Show in Las Vegas 01/09/2020 in Main VH/DLR Discussion
Nope, they don't grow on trees, that's for sure.
Elvis was already in the waning years of his career when I was growing up. Still famous, but his best years in showbusiness to my ears - his first 2...
Well, that was the thing, in that Lomenzo, Young, Luzier...you had distinctive personalities onstage backing Roth. Same went with the EEAS lineup. ...
I dunno...is the drummer of BTO really up there with Wille Nelson and Keith Richards? I mean, in terms of notable rock star deaths, mmmmm...not so...
I think I'm right there with Zah and Sesh...the Roth backing band circa 2020 was competent but little to nothing beyond that. Nondescript backing...
Yeah, after I saw her on tv kissing that pedo in the 1990s, I had no more use for Lisa Marie: a batshit loon like her father, minus the talent.
I had always meant to see him live. He'd play once in a blue moon over the last 20 years in my area. Never did get around to checking him out. ...
Terry replied to a thread Upcoming U2 album will feature 40 new versions of old U2 songs in House of Music
Not even gonna bother listening to the clip: as you say, just leave well enough alone.
Which they might be a bit late for, mostly due to Stanley's declining voice. I mean, honestly, the show I saw last year...yeah, ideally Criss and...
And why not? Long as there is a buck to be made, long as people are showing up, long as they can still do it, fuck it. Attendance isn't...
Terry replied to a thread Van Halen related things you'd like to see in 2023? in Main VH/DLR Discussion
The fabled 5150/WB 'vault' material... Yeah, if whatever exists ends up being made available, I'll have an interest...although probably not if it...
That'd be cool to see.
Well, you get older...significantly older than, say, your mid-teens - which is how old I was when Roth was at his cultural zenith in the mid-1980s -...
vandeleur - 07-09-2014 07:46 AM View Conversation
Happy birthday mate , this is turning into an annual event
PETE'S BROTHER - 07-09-2013 12:22 PM View Conversation
All the best sonrisa
vandeleur - 07-09-2012 02:19 PM View Conversation
happy birthday to ya
About sonrisa salvaje
Black Sand
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Weekend Box Office Report: 'The Croods' and 'Olympus Has Fallen' Rise to the Top
You can't imagine two films more different than 'The Croods' and 'Olympus Has Fallen,' but right now the two of them sit on top of the box office chart, both temporarily linked by the fact that a bunch of people apparently wanted to see them this weekend. It's the first time in forever that two films have opened to over $30 million at the same time...but it wasn't golden for all of the new releases.
Film Weekend Per Screen
1 The Croods $44,700,000 $11,048 $44,700,000
2 Olympus Has Fallen $30,500,000 $9,845 $30,500,000
3 Oz the Great and Powerful $22,031,000 (-46.6%) $5,790 $177,559,000
4 The Call $8,700,000 (-49.2) $3,470 $30,904,000
5 Admission $6,446,000
$2,984 $6,446,000
6 Spring Breakers $5,000,000 (+1,801.1) $4,529 $5,407,000
7 The Incredible Burt Wonderstone $4,275,000 (-58.0) $1,353 $17,365,000
8 Jack the Giant Slayer $2,965,000 (-53.1) $1,158 $59,052,000
9 Identity Thief $2,545,000 (-42.4) $1,175 $127,727,000
10 Snitch $1,930,000 (-44.9) $1,068 $40,343,000
Let's start with the film at the top of the pile. 'The Croods' has a bizarre concept that put off one adult for every child it intrigued, but never underestimate an animated movie after a lengthy dry spell of family entertainment. Sure, 'Oz the Great and Powerful' helped satiate kids and their ever-annoyed parents, but animation is what really brings the young ones in. With a $44.7 million opening, 'The Croods' has a long way to go if it wants to reach the heights of your typical Pixar or DreamWorks film, but it's a nice start. $150 million is the minimum number it'll need to look like a success and if it plays through April, it should be able to get there.
While all of the kids went to see 'The Croods,' their older brothers flocked to 'Olympus Has Fallen' to watch Gerard Butler play "Die Hard in the White House." Since it's modestly budgeted by Big Action Movie standards (around $70 million), it's most likely going to break even, but audiences need to actually like the movie if it stands a chance of being a hit. The more interesting story here is what this means for the upcoming 'White House Down,' which is literally an identical concept with a bigger budget. Will audiences have White House Siege Movie fatigue once summer rolls around? That's such a bizarrely specific question that it can only exist in the film industry.
Meanwhile, last week's champions took relatively graceful falls to third and fourth place. 'Oz the Great and Powerful' took a large but expected drop, taking in $22 million for a $177 million total. $200 million is a certainty but $250 million remains a possibility if April doesn't bring too many surprises. Right below it, 'The Call' took a sizable drop, but it's made $30 million and it was made for pennies, so it's going to be okay. It's not a smash, but it's a nice jolt into the arm for Halle Berry's career. Does she have a future as a classy horror/thriller heroine? It's possible.
The weekend's third major release, 'Admission,' flopped into the fray in spot five with a $6.4 million opening. The film didn't cost too much, but it's evidence that while Paul Rudd and Tina Fey are truly some of the best and most talented comic personas working today, they simply can't open a movie on their names alone. And that's sad.
In sixth place, 'Spring Breakers' rode last week's wildly successful limited release to a $5 million weekend, paying for its minuscule budget, baffling thousands of teenagers and making director Harmony Korine laugh uproariously as he plotted his next practical joke of a film. 'Spring Breakers' has officially divided both critics and audiences, making it easily the weirdest and most fun movie to talk about of the year so far. It's hard to imagine it making much more money, but bless it for actually opening wide.
As for the rest of the top ten? Meh.
'The Incredible Burt Wonderstone' continued to be dead on arrival, taking a disastrous drop for a $4.2 million weekend. 'Jack the Giant Slayer' has made more, but a $60 million gross simply isn't enough when your movie cost upwards of $200 million. On the very bottom, 'Identity Thief' is ready to leave the top ten as a box office smash and 'Snitch' will exit as a minor success.
Next week, 'G.I. Joe: Retaliation' opens after a lengthy delay. It's probably going to make a lot of money.
Filed Under: Admission, Olympus Has Fallen, The Croods
Categories: Box Office, Movie News, ScreenCrush


I don't really post many hair product reviews as I tend to stick to the same handful but lately I've noticed my hair's not been playing ball. My regular shampoo is one of a creamy consistency and I'm told that by using that can cause product build-up along with everyday dirt and grease. Clear shampoos tend to wipe away some of that which is why I alternate between my regular shampoo and a clear one which I use a few times a month. I used to use a Marks and Spencer Mint and Rosemary clear shampoo but they've stopped stocking it so I had to find a replacement. I've used this Pantene Pro V Clarifying shampoo before about a year ago and it worked then so I thought I'd put my faith in something I know worked.
Now Pantene aren't your friendliest purse purchases at nearly £5 a bottle. There was an offer on at Boots for two bottles for £6 last week but unfortunately my Boots stocks every other Pantene product under the sun apart from this one so I had to get it from Superdrug instead. Luckily I used my student discount and points so I paid £3 in the end.
"Our purifying shampoo provides just what your hair needs: it invigorates hair from root to tip by removing grease, everyday build-up and styling residues to leave it beautifully clean and healthy looking."
The word 'clarifying' can leave some people running as they imagine that it's going to strip your hair. While I can't promise that it doesn't (as I don't use it everyday), I don't feel like my hair is being stripped. I am a little apprehensive as I do colour my hair but I haven't had a full head of colour applied since July 2011 so it would take a lot to strip the colour I have on now!
The shampoo itself is a colourless clear substance. I find I can feel it cleansing my hair straight after application and my hair feels lighter after use. My hair's much more easier to manage and feels a lot cleaner.
I'm not a Pantene fan normally, but this product is a god send. It does what it says on the tin which for a beauty product isn't always the case - so I'm happy!
I like Pantene but have not tried this one yet, I will give it a go!
I have a giveaway on my blog at the mo which includes a MAC lippy if you want to check it out?


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Customer service and a timely delivery of your order is most important to us. All of our employees at Countryside continually strive to meet and exceed our customers expectations at all times. Our staff will make every effort to get you the right part at the right time every time you order with us. Customer loyalty is what continues to grow our business and we want to be your first choice for OEM replacement parts.


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# -----------------------------------------
# In this section, you can install any system dependencies required
# to run your App.  For instance, you could place an apt-get update or
# install line here, a git checkout to download code, or run any other
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# RUN apt-get update

RUN sudo apt-get install nano \
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	&& sudo apt-get update \
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# -----------------------------------------

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WORKDIR /kb/module

RUN make all

ENTRYPOINT [ "./scripts/entrypoint.sh" ]

CMD [ ]

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text:
\section*{Introduction}
The purpose of this short article is to introduce some concepts showing an intriguing relationship between Perelman's energy-functional, the stability of Ricci-flat spaces, and the ADM-mass from general relativity. These ideas will be exploited further in our subsequent papers \cite{H2} and \cite{HSiep}.\\

To start with, in his famous paper \cite{P}, Perelman introduced the energy-functional
\begin{equation}
\mathcal{F}(g,f)=\int_M\left(R+\abs{\nabla f}^2\right)e^{-f}dV,
\end{equation}
for a metric $g$ and a function $f$ on a closed manifold $M$. If $g$ evolves by Hamilton's Ricci flow and $e^{-f}$ by the adjoint heat equation, we have the fundamental monotonicity formula
\begin{equation}
\partial_t\mathcal{F}=2\int_M\abs{\Rc+\nabla^2f}^2e^{-f}dV\geq 0,
\end{equation}
with equality precisely on steady gradient Ricci solitons. Ricci solitons (for a recent survey see \cite{C}) are the fixed points of the Ricci flow up to diffeomorphism and scaling, and play a crucial role in understanding the formation of singularities.\\

In fact, all steady solitons on a \emph{closed} manifold are Ricci-flat, but there exist nontrivial gradient steadies on \emph{complete} manifolds. A nice example is the rotationally symmetric steady gradient Ricci soliton on $\mathbb{R}^n (n\geq 3)$ discovered by Bryant \cite{Bry}. The Bryant soliton has positive curvature and looks like a paraboloid, where the spheres of geodesic radius $r$ have diameter of order $\sqrt{r}$. The scalar curvature decays like $\tfrac{1}{r}$ and the potential $f$ behaves asymptotically like $-r$. From these asymptotics it is immediate that $\mathcal{F}=\infty$ on the Bryant soliton.\\

Nevertheless, we manage to make sense of the $\mathcal{F}$-functional in the situation where $(M,g_s,f_s)$ is a non-compact steady gradient Ricci soliton (e.g. the Bryant soliton) and $g$ and $f$ are a nearby metric and function on $M$. The idea is to consider the relative energy $\mathcal{F}^{(g_s,f_s)}(g,f):=\mathcal{F}(g,f)-\mathcal{F}(g_s,f_s)$, where $\infty-\infty$ gives a finite quantity when suitably interpreted. Actually, we find it technically more convenient to give a slightly different definition, and we call the resulting functional the \emph{renormalized energy} $\mathcal{F}^{(g_s,f_s)}$ (Definition \ref{renormalizedf}).\\

It turns out that $(g_s,f_s)$ is a critical point of this functional if the possibly infinite total measure $\int_Me^{-f}dV$ is kept fixed (Theorem \ref{firstvariation}). The second variation at $(g_s,f_s)$ is of fundamental importance (Theorem \ref{2ndvarthm}). It motivates us to introduce the \emph{stability inequality} for steady gradient Ricci solitons,
\begin{equation}\label{stabineq}
\int_M\left[-\tfrac{1}{2}\abs{\nabla h}^2+\Rm(h,h)\right]e^{-f_s}dV\leq 0
\end{equation}
for all $h\in\ker\div_{f_s}$ with compact support. Here $\Rm(h,h)=R_{ijkl}h_{ik}h_{jl}$ and we restrict to variations $h$ satisfying $\div(e^{-f_s}h)=0$. Applications of inequality (\ref{stabineq}) appear in \cite{HSiep}. In particular, we will show that many Ricci-flat cones in small dimensions are unstable, and discuss a conjecture of Tom Ilmanen relating the stability of Ricci-flat cones, the existence of positive scalar curvature deformations, and the non-uniqueness of Ricci flow with conical initial data (see also \cite{H} for ancient Ricci flows coming out of unstable closed Ricci-flat manifolds, and \cite{SS} for Ricci flows coming out of positively curved cones).\\

For asymptotically flat manifolds $(M,g_{ij})$ with nonnegative scalar curvature, we also define a geometric invariant $\lambda_{AF}(g)$ (Definition \ref{renormalizedlambda}), motivated by Perelman's $\lambda$-functional for closed manifolds. This invariant is closely related with the positive mass theorem in general relativity. Recall that the ADM-mass \cite{ADM,Ba} is defined as
\begin{equation}\label{admmass}
m_{ADM}(g):=\lim_{r\to\infty}\int_{S_r}\left(\partial_jg_{ij}-\partial_ig_{jj}\right)dA^{i}.
\end{equation}
Physical arguments suggest that the mass is always nonnegative. Schoen and Yau proved that this is indeed the case in dimension $n\leq 7$ using minimal surface techniques \cite{SY}. Huisken and Ilmanen found a different proof for $n=3$ based on the inverse mean curvature flow \cite{HI}. Witten discovered a proof for spin manifolds of arbitrary dimension using Dirac spinors \cite{W}. Recently, Lohkamp announced a proof of the positive mass theorem valid in all dimensions and without spin assumption based on singular minimal surface techniques \cite{L}.\\

For the sake of exposition, the following discussion contains some facts well known to experts. Indeed, it is well known that conformal deformations can decrease the mass and that deformations in direction of the Ricci curvature can be used to prove the rigidity statement in the positive mass theorem. Our main motivation is to explain the relationship with the Ricci flow and Perelman's $\lambda$-functional.\\

This said, it is an interesting problem to find quantitative lower bounds for the ADM-mass in terms of other geometric quantities. In fact, the Penrose inequality gives such a quantitative lower bound in terms of the area of the outermost minimal surface, at least for $n\leq 7$ \cite{HI,Bra,BL}. Our geometric invariant $\lambda_{AF}(g)$ also gives a quantitative lower bound for the ADM-mass, i.e.
\begin{equation}\label{lmineq}
m_{ADM}(g)\geq \lambda_{AF}(g),
\end{equation}
at least for spin-manifolds or in dimension $n\leq 7$. For spin-manifolds this directly follows from Witten's formula (Theorem \ref{lambdamassineq}), and in fact the positive mass theorem combined with a conformal transformation implies a slightly stronger lower bound (Theorem \ref{posmassineq}). The discrepancy comes from the difference between the operator $-4\triangle+R$ and the conformal Laplacian $-\tfrac{4(n-1)}{n-2}\triangle+R$ and disappears in the limit $n\to\infty$. Our motivation for stating the inequality in the weaker form (\ref{lmineq}) is that this is better adapted to Ricci flow techniques. As an application, we give a Ricci flow proof of the rigidity statement in the positive mass theorem (Theorem \ref{rigidity}).\\

Guided by these ideas, we can revisit the pressing question if there exists a geometric flow that decreases the ADM-mass. Indeed, we discovered such a flow in dimension three, and this flow conjecturally squeezes out all the mass of an asymptotically flat $3$-manifold with nonnegative scalar curvature. Our flow is based on conformal rescalings and the Ricci flow with surgery and will be discussed in \cite{H2}.\\

This article is organized as follows: Section \ref{variationalstructure} is about $\mathcal{F}^{(g_s,f_s)}$ and the stability inequality. Section \ref{positivemass} is about $\lambda_{AF}$ and the positive mass theorem.\\

\emph{Acknowledgements.} I thank Tom Ilmanen and Richard Schoen for interesting discussions, Simon Brendle, Stanley Deser, Reto M\"{u}ller, Michael Siepmann and Miles Simon for useful comments, and the Swiss National Science Foundation for partial financial support.

\section{The renormalized $\mathcal{F}$-functional}\label{variationalstructure}

Let $(M,g_s,f_s)$ be a \emph{steady gradient Ricci soliton}. This means that $(M,g_s)$ is a smooth, connected, complete Riemannian manifold and $f_s:M\to \mathbb{R}$ is a smooth function such that the following equation holds:
\begin{equation}
\Rc(g_s)+\Hess_{g_s}(f_s)=0.
\end{equation}
Steady gradient Ricci solitons always have nonnegative scalar curvature and correspond to eternal Ricci flows moving only by a diffeomorphism \cite{Z}.

\begin{definition}\label{renormalizedf}
Let $c_s:=R(g_s)+\abs{\nabla f_s}_{g_s}^2\geq 0$ be the \emph{central charge} (or auxilary constant) of the steady soliton. We define the \emph{renormalized energy}
\begin{equation}
\mathcal{F}^{(g_s,f_s)}(g,f):=\int_M \left(R(g)+\abs{\nabla f}_g^2-c_s\right)e^{-f}dV_g.
\end{equation}
\end{definition}
By construction, this is well defined and finite if $(g-g_s,f-f_s)$ has compact support (or decays sufficiently fast), in particular, $\mathcal{F}^{(g_s,f_s)}(g_s,f_s)=0$.\\

The variational structure of compact Ricci-flat metrics has been discussed in \cite{CHI} and \cite{H}. We will now carry out a similar discussion for non-compact gradient steady Ricci solitons using our renormalized functional $\mathcal{F}^{(g_s,f_s)}$. Let $g$ be a metric, $h$ a symmetric 2-tensor, and $f$ and $l$ functions on $M$.

\begin{theorem}\label{firstvariation}
If $(g-g_s,f-f_s)$ and $(h,l)$ have compact support (or decay sufficiently fast), then the first variation of $\mathcal{F}^{(g_s,f_s)}$ at $(g,f)$ is well defined and given by the following formula:
\begin{align}\label{firstvarformula}
&\tfrac{d}{d\varepsilon}|_0 \mathcal{F}^{(g_s,f_s)}(g+\varepsilon h,f+\varepsilon l)\\
&\qquad\qquad=\int_M\left[-\langle h,\Rc+\nabla^2f\rangle+\left(\tfrac{1}{2}\mathrm{tr} h-l\right)\left(2\triangle f-\abs{\nabla f}^2+R-c_s\right)\right]e^{-f}dV.\nonumber
\end{align}
In particular, $(g_s,f_s)$ is a critical point, if the (possibly infinite) total measure $\int_M e^{-f}dV$ is kept fixed in the sense that $\int_M\left(\tfrac{1}{2}\mathrm{tr} h-l\right)e^{-f}dV=0$.
\end{theorem}

\begin{proof}
Equation (\ref{firstvarformula}) follows from a computation as in \cite{P}. Moreover, by the traced soliton equation and the definition of $c_s$ we have
\begin{equation}\label{auxaux}
2\triangle_{g_s}f_s-\abs{\nabla f_s}_{g_s}^2+R_{g_s}-c_s=-2c_s,
\end{equation}
and the last statement follows easily.
\end{proof}

In the next theorem, we will use the notation $\div_f(\cdot)=e^{f}\div(e^{-f}\cdot)$.

\begin{theorem}\label{2ndvarthm}
If $(h,l)$ has compact support (or decays sufficiently fast), then:
\begin{align}
&\tfrac{d^2}{d\varepsilon^2}|_0 \mathcal{F}^{(g_s,f_s)}(g_s+\varepsilon h,f_s+\varepsilon l)\\
&\qquad=\int_M\left[-\tfrac{1}{2}\abs{\nabla h}^2+\Rm(h,h)+\abs{\div_{f_s} h}^2\right]e^{-f_s}dV\nonumber\\
&\qquad+2\int_M\left[\abs{\nabla\left(\tfrac{1}{2}\mathrm{tr} h-l\right)}^2+\left(\tfrac{1}{2}\mathrm{tr} h-l\right)\div_{f_s}\div_{f_s} h-c_s\left(\tfrac{1}{2}\mathrm{tr} h-l\right)^2\right]e^{-f_s}dV.\nonumber
\end{align}
\end{theorem}

\begin{proof}
We write $(g,f)=(g_s,f_s)$ and $(g_\varepsilon,f_\varepsilon)=(g+\varepsilon h,f+\varepsilon l)$. Similar as in \cite[Lemma 2.3]{CZ}, using the soliton equation we obtain
\begin{equation}
\tfrac{d}{d\varepsilon}|_0\left(\Rc_{g_\varepsilon}+\Hess_{g_\varepsilon}f_\varepsilon\right)
=-\tfrac{1}{2}\triangle_fh -\Rm(h,.)-\div_f^\ast\div_f h+\nabla^2(l-\tfrac{1}{2}\mathrm{tr} h),
\end{equation}
where $\triangle_f=\triangle-\nabla f\cdot \nabla$, and $\div_f^\ast$ is the formal $L^2(M,e^{-f}dV)$-adjoint of $\div_f$. Another computation using the soliton equation shows
\begin{align}
&\tfrac{d}{d\varepsilon}|_0\left(2\triangle_{g_\varepsilon} f_\varepsilon-\abs{\nabla f}^2_{g_\varepsilon}+R_{g_\varepsilon}\right)=2\triangle_f(l-\tfrac{1}{2}\mathrm{tr} h)+\div_f\div_f h.
\end{align}
Using this, Theorem \ref{firstvariation}, the soliton equation, and (\ref{auxaux}) we obtain
\begin{align}
&\tfrac{d^2}{d\varepsilon^2}|_0 \mathcal{F}^{(g,f)}(g_\varepsilon,f_\varepsilon )\\
&\qquad=\int_M\left[\tfrac{1}{2}\langle h,\triangle_f h\rangle+\Rm(h,h)+\langle h,\div_f^\ast\div_f h\rangle-\langle h,\nabla^2(l-\tfrac{1}{2}\mathrm{tr} h)\rangle\right]e^{-f}dV\nonumber\\
&\qquad+\int_M\left[\left(\tfrac{1}{2}\mathrm{tr} h-l\right)\left(2\triangle_f\left(l-\tfrac{1}{2}\mathrm{tr} h\right)+\div_f\div_f h\right)-2c_s\left(\tfrac{1}{2}\mathrm{tr} h-l\right)\right]e^{-f}dV,\nonumber
\end{align}
and the claim follows from partial integration.
\end{proof}
Restricting to $h\in\ker\div_{f_s}$ (this corresponds to choosing a slice for the action of the diffeomorphism group) and setting $l=\tfrac{1}{2}\mathrm{tr} h$ (this corresponds to keeping the measure $e^{-f}dV$ fixed), Theorem \ref{2ndvarthm} motivates the stability inequality (\ref{stabineq}).

\begin{remark}
In particular, the \emph{stability inequality for Ricci-flat manifolds or cones} is
\begin{equation}
\int_M 2\Rm(h,h)dV\leq \int_M\abs{\nabla h}^2dV
\end{equation}
for all $h\in\ker\div$ with compact support. As mentioned in the introduction, important consequences of this inequality are discussed in \cite{HSiep}.
\end{remark}

\section{$\lambda_{AF}$ and the positive mass theorem}\label{positivemass}

For perturbations of $(M,g_s,f_s)=(\mathbb{R}^n,\delta,0)$ with nonnegative scalar curvature, or more generally for asymptotically flat manifolds with nonnegative scalar curvature we define a functional $\lambda_{AF}$ as follows:

\begin{definition}\label{renormalizedlambda}

Assume $(M^n,g_{ij})$ is a complete asymptotically flat manifold of order $\tau>\tfrac{n-2}{2}$ with nonnegative scalar curvature. We define
\begin{equation}
\lambda_{AF}(g):=\inf \int_M\left(4\abs{\nabla w}^2+Rw^2\right)dV,
\end{equation}
where the infimum is taken over all $w\in C^\infty(M)$ such that $w=1+O(r^{-\tau})$ at infinity (here the $O$ notation includes the condition that the derivatives also decay appropriately).
\end{definition}

\begin{remark}
We could also define renormalized $\lambda$-functionals for perturbations of non-flat steadies $(M,g_s,f_s)$. However, in that case we find it technically more convenient to work with $\mathcal{F}^{(g_s,f_s)}$ and to fix the measure $e^{-f}dV$.
\end{remark}

\begin{theorem}\label{lambdamassineq}
Assume $(M^n,g_{ij})$ is a complete asymptotically flat spin manifold of order $\tau>\tfrac{n-2}{2}$ with nonnegative scalar curvature. Then
\begin{equation}\label{keyineq}
m_{ADM}(g)\geq \lambda_{AF}(g).
\end{equation}
\end{theorem}

\begin{proof}
By Witten's formula \cite{W,Ba,LP} for the mass of spin manifolds,
\begin{equation}\label{witten}
m_{ADM}(g)=\int_M\left(4\abs{\nabla\psi}^2+R\abs{\psi}^2\right)dV,
\end{equation}
where $\psi$ is a Dirac spinor with asymptotically unit norm. Thus, recalling Definiton \ref{renormalizedlambda}, using $w=\abs{\psi}$ as a test function, and using Kato's inequality, the claim immediately follows.
\end{proof}

\begin{remark}
Note that $\lambda_{AF}$ and $m_{ADM}$ are finite if and only if the scalar curvature is integrable.
\end{remark}

\begin{remark}
The physical interpretation of Witten's formula (\ref{witten}) is that the Dirac spinor with asymptotic boundary conditions is a test field that measures the mass of the gravitational field. Similarly, one of our motivations for Definiton \ref{renormalizedlambda} is to minimize over all test fields $w=e^{-f/2}$ with the asymptotic boundary conditions coming from the trivial potential $f_s=0$. Instead of the Dirac field, we use a Klein-Gordon field. This test field interpretation of the mass also motivates (\ref{lmineq}).
\end{remark}

As mentioned in the introduction the inequality (\ref{lmineq}) is well adapted to Ricci flow techniques, but not sharp in general. From the refined Kato inequality for Dirac spinors, $\abs{\nabla\abs{\psi}}\leq\sqrt{1-1/n}\abs{\nabla\psi}$, one gets a slightly better but still not sharp bound. To find the sharp inequality, let us consider the example of the spatial Schwarzschild metric $g_{ij}=\left(1+cr^{2-n}\right)^{4/(n-2)}\delta_{ij}$ on $M=\mathbb{R}^n\setminus \{0\}$, where $c=\tfrac{m}{4(n-1)\abs{S^{n-1}}}$. Since this manifold has two ends, what we really should do is impose the boundary conditions $w\to 0$ for $r\to 0$ and $w\to 1$ for $r\to\infty$. The rotationally symmetric solution of $\triangle w=0$ with the given asymptotic boundary conditions is $w(r)=\left(1+cr^{2-n}\right)^{-1}$. Now, a straightforward computation shows that
\begin{equation}
4\tfrac{n-1}{n-2}\int_M\abs{\nabla w}^2dV=4(n-1)(n-2)\abs{S^{n-1}}c^2\int_0^\infty(1+cr^{2-n})^{-2}r^{1-n}dr=m
\end{equation}
equals the ADM-mass of the end in consideration. This example motivates the following theorem.

\begin{theorem}\label{posmassineq}
Assume $(M^n,g_{ij})$ is a complete asymptotically flat manifold of order $\tau>\tfrac{n-2}{2}$ with nonnegative scalar curvature. Then there exists a unique positive solution of the equation
\begin{equation}
\left(-\tfrac{4(n-1)}{n-2}\triangle+R\right)w=0,
\end{equation}
with $w\to1$ at infinity. If moreover $M$ is spin or $n\leq 7$, then we have the lower bound
\begin{equation}\label{keyineq}
m_{ADM}(g)\geq \int_M\left(\tfrac{4(n-1)}{(n-2)}\abs{\nabla w}^2+Rw^2\right)dV.
\end{equation}
\end{theorem}

\begin{proof}
The proof is closely related with the article by Zhang-Zhang \cite{ZZ}, so we will be rather sketchy. The first part of the theorem follows from the fact that $R\geq 0$ by assumption. Now consider the conformal metric $\tilde{g}=w^{4/(n-2)}g$. Note that $R(\tilde{g})=0$ and that $(M,\tilde{g})$ is asymptotically flat. We compute
\begin{align}
m_{ADM}(\tilde{g})&=m_{ADM}(g)-\tfrac{4(n-1)}{(n-2)}\lim_{r\to\infty}\int_{S_r}\partial_rwdA\\
&=m_{ADM}(g)- \int_M\left(\tfrac{4(n-1)}{(n-2)}\abs{\nabla w}^2+Rw^2\right)dV,
\end{align}
and the claim follows from the positive mass theorem applied to $\tilde{g}$.
\end{proof}

\begin{remark}
It is interesting to observe that Kato's inequality for Witten's spinor becomes sharp in $L^2$ in the limit $n\to\infty$.\end{remark}

\begin{remark}
It is straightforward to generalize Theorem \ref{posmassineq} to manifolds with multiple ends. The asymptotic boundary value is one at the end in consideration and zero at the other ends. In particular, this improves Bray's lower bound \cite[Thm. 8]{Bra}
\begin{equation}
m_{ADM}(g)\geq \inf\int_M8\abs{\nabla w}^2dV\qquad (n=3),
\end{equation}
where the infimum is taken over all smooth functions with the boundary conditions just mentioned. More importantly, we get a non-zero lower bound even in the one-ended case if the scalar curvature does not vanish identically.
\end{remark}

As an application of Theorem \ref{posmassineq} we give a Ricci flow proof of the following rigidity statement in the positive mass theorem:

\begin{theorem}\label{rigidity}
Assume $(M^n,g_{ij})$ is a complete asymptotically flat manifold of order $n-2$ with nonnegative scalar curvature, and that $M$ is spin or $n\leq 7$. Then $m_{ADM}(g)=0$ implies that $(M,g)$ is isometric to $\mathbb{R}^n$.
\end{theorem}

\begin{proof}
Suppose towards a contradiction that $\Rc\neq0$ somewhere. Consider the Ricci flow  $g(t)$ starting at $g$. The flow exists for a short time and preserves $R\geq 0$, the asymptotic flatness, and $m_{ADM}$ \cite{DM,OW}. In fact, from the evolution equation
\begin{equation}
\partial_tR=\triangle R+2\abs{\Rc}^2
 \end{equation}
 we see that $R$ becomes strictly positive. Thus, together with Theorem \ref{posmassineq} we obtain
 \begin{equation}
 m_{ADM}(g)=m_{ADM}(g(t))>0,
 \end{equation}
 a contradiction. So $(M,g)$ is Ricci-flat, and it is easy to conclude that it is in fact isometric to $\mathbb{R}^n$.
\end{proof}

\begin{remark}
As mentioned in the introduction, the ideas discussed in this section motivate our mass decreasing flow in dimension three \cite{H2}.
\end{remark}

\makeatletter
\def\@listi{%
  \itemsep=0pt
  \parsep=1pt
  \topsep=1pt}
\makeatother
{\fontsize{10}{11}\selectfont

redpajama_set_name: RedPajamaArXiv


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Richburg of Bushland, Texas, met Ponce in a hospital in the offseason. Beckham, who was Ponce's favorite player, met the young man a couple of days before the Giants reported to training camp on Thursday.
Ponce made an impression on both.
Richburg remembers walking into the room on Apr. 8 in the Harrington Cancer Center, meeting the child and recalling his big smile.
Beckham was not available to be interviewed Tuesday after practice, but it was clear the death of Ponce affected him.
Beckham wrote that meeting Ponce taught him more than he could have imagined.
Talking about Ponce last week, Richburg seemed to know that his chances of surviving were not good.
"Obviously you hope and pray that you see some good come out of it and there is a turn for the better, but that's not the way it ended up," Richburg said.
His hope is that the family can find some peace.
Richburg intends to wear his Team Jayro bracelet going forward. He added Beckham might celebrate a touchdown with a special dance.
"Maybe in Dallas in memory of Jayro," Richburg said of the Giants' opening game this season against the Cowboys.


Intermesures - Daily Vessel, retaining tray.
The daily vessel is designed to supply the generator set, motor-driven pump, or fire kit (with fuel, oil, coolant, etc.) during operation thanks to the type of device described below.
The retaining tray is used only in case of a leaking vessel, by means of a CF 30 type device.
CF 3: Direct display electric retransmission gauge.
CF 4: Direct display contactor.
CF 4 RD: Direct display and electrical retransmission contactor.


Aid & International Development Forum (AIDF) are pleased to open nominations for the Global Innovator of the Year Award.
This competition celebrates individuals as well as teams who have introduced game-changing initiatives, created research or inventions that have made an impact on the humanitarian aid and development sectors globally over the past year.
The finalists will be showcased at the AIDF Global Summit and the winner of the award will be announced on 5 September 2018 at the 10th Anniversary AIDF Global Summit in Washington D.C, USA.
Join us at the 10th Anniversary AIDF Global Summit where the award winner will be announced.