"Gentle Dental Care for the Entire Family"
I have known Dr. Hank Willis since he first moved to Bonners Ferry.
Patients know there are other dentists within the region to choose from for their dental care and I am thankful that I have chosen Dr. Willis' dental practice for my dental care. I am like most people, I wanted a dentist that cared not only about my dental health, but me as a patient. I "used" to be a nervous wreck when going to the dentist. Dr. Willis saw that, and his personal approach was above gentle and caring AND he was patient with me while reassuring me through the entire process of my first appointment with him. I have been seeing Dr. Willis for four years now. He is still patient, caring and professional. His staff is highly trained and courteous. They continually receive training with new standards, treatments, and technology. We are blessed to have a dentist in this community who is on the cutting edge and state of the art technology in his practice. I feel appreciated as a patient. If there was a "Hall of Fame" for dentists in Idaho, YOU would be number one! Seriously....I am no longer a nervous wreck, but a very pleased patient that gets great care and service. If you want that as a patient, go see Dr. Willis!
That may seem like a funny thing to say about a dentist, but Dr. Willis is truly an excellent dentist. Now, for the first time in my life, I no longer dread going to the dentist.
Dr. Willis is extremely competent, gentle, and almost pain free. His entire staff is highly competent and cheerful.
I had been saving up my dental work for years, because I had just not found a dentist that I felt was truly excellent. Now, in the past 3 years, I have had Dr. Willis do all of my long overdue dental work incuding a root canal, two crowns, multiple fillings, and a cleaning.
Even if you live many miles away as I do, Dr. Willis is worth the trip.


Jimmy was one of my best friends in high school and since we had all of our art classes together we often spent a lot of time after school working on art with one another. Some nights we both couldn't sleep so we would stay up playing Mario Kart before heading out and grabbing some Taco Bell. Sometimes we would just drive around listening to music. One night in particular we decided to take an alternate route and drive around exploring. We happened upon Barry Road (for those who don't know where that is – it's a long winding road out in the country) and drove on. As we followed the twists of the road, something came out of the trees on the right hand side of the road and smashed into the car. It appeared to be a white figure, although it had no face, and it was running. I screamed. He stopped the car. We looked around but there was no person or animal around and his car was in perfect condition – not a scratch or a dent on it. I asked him if he thought it was a ghost and he said he thought it was. Scared, perplexed, and a little bit entertained we turned around and tried to recreate it that night and many nights to come but we never had the same experience. This story has always been very significant to me because it was one of the few times I had someone with me to vouch for my experiences.
I kinda held off in publishig this one because it so strongly hit home, and reminded me of some similar experiences that I've had. I thought it would be kind of fun to publish mine and this at the same time, but I decided not to because I'm lazy and have no clue when I'll write mine. Didn't wanna leave ya hangin' any longer.
While in high school, to extend our curfew my buddy Ash would tell his parents that he was gonna stay at my house and I said the same thing to mine. We did this countless times. We would stay out all night instead of going home. We'd go driving in the most goofy places… We saw some stuff, man! Lotsa crazy stories, both spooky, and not.
Anyway, good story. Brought back memories.
Vic, call me!!!!! This is reeeeeally near to my house. You know I love the spooky! If you want to check it out, CALL MEEEE!!!!
Bill, seriously, you have a friend named Ash? Jealous.
What is this? R u kidding me?!!


text:
title: Hingamispäeva pärastlõuna
date: 21/07/2018
---

### Selle nädala õppeaine
Ap 6. ptk; Ap 7:48; Hb 5:11–14; Mi 6:1–16, Ap 7. ptk; Ap 8:4–25.

>
Juhtsalm:

> "Aga Jumala sõna levis ning jüngrite arv Jeruusalemmas kasvas väga suureks, ka suur hulk preestreid võttis usu omaks" (Ap 6:7).

Paljud nelipühal usku tulnud inimesed olid hellenistlikud juudid, see tähendab kreeka-rooma keskkonnast pärit juudid, kes elasid nüüd Jeruusalemmas (Ap 2:5, 9–11). Ehkki nad olid juudid, erinesid nad Juudamaa juutidest – Ap 6:1 nimetatud "heebrealastest" – mitmes mõttes. Nähtavaim erinevus oli see, et tavaliselt ei osanud nad aramea keelt, keelt, mida kõneldi Juudamaal.

Oli veel rida teisi erinevusi, nii kultuurilisi kui ka usulisi. Olles sündinud välisriikides, ei olnud nad juurdunud Juudamaa juutlikus traditsioonis või vähemasti polnud nende juured nii sügaval kui Juudamaa juutide omad. Arvatavasti ei olnud nad nii palju seotud templitseremooniatega ja Moosese seaduse nende külgedega, mis käisid ainult Iisraeli riigi kohta.

Olles viibinud enamjao oma elust kreeka-rooma keskkonnas ja elanud külg külje kõrval paganatega, olid nad varmamad aru saama kristliku usu kõiki kaasavast olemusest. Tegelikult kasutas Jumal just paljusid hellenistlikke usklikke selleks, et täita korraldus viia tunnistus tervele ilmamaale.

_Selle nädala õppetükki õppides valmistud hingamispäevaks, 28. juuliks._

redpajama_set_name: RedPajamaGithub


Organizations today face a broadening and increasingly sophisticated set of threats to the security of their sensitive data. These include privileged user abuse and advanced persistent threat (APT) attacks. As organizations grow increasingly reliant on virtual servers, their physical and virtual server security challenges grow in complexity and scale. Sensitive data grows increasingly mobile. It is copied across many virtual machines. It is spread across the storage infrastructure and an increasingly diverse mix of physical platforms and operating systems. In this kind of environment, organizations need the persistent, robust, flexible, and comprehensive safeguards Thales eSecurity provides to ensure sensitive data remains secure and compliant.
When an organization runs sensitive data in fluid virtual machines, that data needs to be secured through robust, persistent, and auditable controls. Without virtual server security, the organization exposes itself to data breaches, penalties for non-compliance and a sullied reputation.
Vormetric Transparent Encryption from Thales eSecurity delivers the virtual server security capabilities you need to safeguard your sensitive assets. Whether you're running VMware, Microsoft Hyper-V, KVM (Kernel-based Virtual Machine), or any other standard virtualization platform, Thales eSecurity can help you address your critical virtual machine security compliance requirements, with unparalleled efficiency and low cost of ownership.
Thales eSecurity enables you to encrypt data at the file system or volume level within virtual machines (VMs) and then use fine-grained, centrally managed policies to control access to protected data. As a result, you can enforce security policies and track access, no matter where data is copied or moved.
Vormetric Transparent Encryption from Thales eSecurity provides fine-grained, policy-based controls that restrict access to encrypted data. Privileged users—whether cloud, virtualization, or storage administrators—can manage systems, without gaining access to encrypted data, unless they have expressly been granted permissions to do so.
For fast rollouts and integration within virtualized machines, both web and command-line level APIs provide access to the Vormetric Data Security environment for policy management, deployment, and monitoring.
By delivering capabilities for segregating security management domains, Thales eSecurity helps organizations address internal security policies and compliance mandates. Further, these security capabilities enable service providers to establish strong boundaries between customer environments, while still leveraging centralized visibility and control of security policies and activities.
Vormetric Transparent Encryption from Thales eSecurity offers support for implementation in a range of Linux and Windows operating systems. All policy and encryption key administration is done through the Vormetric Data Security Manager, which can be deployed as a physical appliance or virtualized service, and either on or off premise, according to your objectives and environments.
Vormetric Data Security Platform products from Thales eSecurity make it efficient to manage data-at-rest security across your entire organization.
Vormetric Application Encryption from Thales eSecurity provides encryption for Enterprise, database, big data and PaaS applications.


text:
Q: Automate adding new column and field names to all csv files in directories This question does not link with any question and does not have any answer to the problem
I need help to add new column in index (0) and field names to all csv files in directories and sub-directories.
I have csv files like this
apple   chocolate    smoothies
orange  dark cocoa   ice-coffee
grapes  brownies     hot-choco
banana  oreo         ice-cream

I want my output csv files to be like this
Key         A       C            S
Record_01   apple   chocolate    smoothies
Record_02   orange  dark cocoa   ice-coffee
Record_03   grapes  brownies     hot-choco
Record_04   banana  oreo         ice-cream

Firstly, I am trying to write the same Record column to all csv files in folder and nested folders.
Secondly, I am trying to write the same Field name to all csvfiles in folder and nested folders
I have tried this code
key = []
for row, Record in enumerate(key):
    print('Record_{}'.format(row, Record))
    row.append(row[0])
    key.appen(row)

for roots, dirs, files in os.walk('path'):
    for csvfiles in files:
        if os.path.splitext(csvfiles)[0] == '.csv':
            with open(f'{csvfiles}', 'r') as csvinput:
                with open(f'{csvfiles}', 'w') as csvoutput:
                    writer = csv.writer(csvoutput, delimiter=',')
                    writer.writerow[key]

I have basically no idea how to add Record_{} by automatically to all csv files.
please help me. Thank you for any answers.

A: You have several options:

*

*Read the CSV data into a Pandas DataFrame specifying no columns. Dummy columns names will be generated. Rename the columns and save the DataFrame back to CSV.

*Open a new file, write the header, read the original file and write it to the new file. Close both files. Delete the original file and rename the new file to the old file name.

redpajama_set_name: RedPajamaStackExchange


The Resort at Cape St Francis is a unique and exclusive paradise on the South African southern coast which draws holidaymakers from all over the world to enjoy the glorious beaches, excellent water-sport facilities, leisurely lifestyle and the magical beauty of the unspoilt surroundings. Cape St Francis with its historic lighthouse, built in 1878, is a tranquil eco-destination with mild temperatures and the longest sunshine hours on the coast. Cape St Francis Resort offers accommodation from luxury beach villas to upmarket beach cottages and affordable apartments.
The Resort is the ideal base from which to explore all that the area has to offer from golf, canal and river cruises, surfing, whale-watching and the penguin rescue facility. Facilities include the award-winning Joe Fish Restaurant and Bar and Hibiscus health and beauty studio.
"We had a lovely time and will definitely return in future. Would recommend this accommodation to anyone!"
"A great time was had by all . I would like to suggest that the kitchen utensils be upgrade and few extra salad bowls mixing bowls etc be added. Be a self catering place it lacks a few of these items . As a added luxury heating the pool would not do the wave warrior any harm !!"
On the N2, 90 km west of Port Elizabeth take the Humansdorp off-ramp. Follow the R330 for 23 km.
Enter Cape St Francis, at first stop sign, turn left and continue to the Cape St Francis Resort reception office.


100 grams of Ground Veal contain 18.58 grams of protein, 13.06 grams of fat, no carbohydrates, and no fiber.
You have 197 calories from 100 grams of Ground Veal, the 10% of your total daily needs. It contains 13.06 grams of fat and 49 mg of Cholesterol.
Some minerals can be present in Ground Veal, such as Potassium (198 mg), Phosphorus (197 mg) or Sodium (103 mg) but no Manganese or Fluoride.
It contains some important vitamins: Vitamin B-3 (5.52 mg), Vitamin B-9 (3 mg) or Vitamin B-12 (2.28 mg).
100 grams of Ground Veal have the 10% of your total daily calorie needs: 197 calories.
An average adults needs 65 grams of total fat per day. 100 grams of Ground Veal have 13.06 gr of total fat, the 20% of your total daily needs.
An average adults needs 2,400 mg.of sodium per day. 100 grams of Ground Veal have 103 mg of salt, the 7% of your total daily needs.
100 grams of Ground Veal contain 66.16 grams of water, the 66% of the total weight.


text:
require "replicator/version"

require 'active_support/core_ext'

module Replicator
  autoload :Consumer, 'replicator/consumer'
  autoload :Producer, 'replicator/producer'
  autoload :Events, 'replicator/events'
  autoload :Packet, 'replicator/packet'

  mattr_accessor :consumers
  mattr_accessor :producers
  mattr_accessor :packet # Extesions could provide specific packet class

  self.consumers, self.producers = {}, []
  self.packet = Packet
end

require 'replicator/config'


Washington, April 11- Iran will likely use upcoming world talks to play for more time as it forges ahead with its nuclear program, a top analyst said Wednesday.
"My suspicion is that not a lot of good is going to come out of these negotiations," Vice President of the American Foreign Policy Council Ilan Berman said on a conference call sponsored by The Israel Project.
Iranian negotiators are slated to hold talks in Istanbul on April 14 with representatives of the five permanent members of the United Nations Security Council and Germany (P5+1) even as President Mahmoud Ahmadinejad mocked the rounds of international sanctions that have crippled Iran's economy. Western diplomats have said the punishing measures may be the main reason that Iran has decided to rejoin negotiations.
The U.N. imposed four rounds of sanctions following years of Iranian refusal to cooperate with the International Atomic Energy Agency, which concluded last year that some Iranian nuclear activities could only be used for weapons development. Despite the IAEA evidence, Tehran denies it is trying to create nuclear weapons.
One reason for the current pessimism, Berman said, is that with U.S. forces out of Iraq and pulling out of Afghanistan, the Iranians see a diminished U.S. threat in the region and would feel less need to compromise on their program. For their part, the Obama administration is unsure of the specific goal with Iran as it seeks to leave the region.
Berman also said that western diplomats have already shifted the goalposts, and the debate is now about what percentage of enriched uranium is acceptable for Iran to have rather than about possession of such nuclear materials in the first place.
Ultimately, coaxing Iran away from its nuclear efforts to peaceful purposes may prove impossible for international diplomats, Berman said.


Brisket is a cut of meat from the breast or lower chest of beef. The beef brisket is one of the nine beef prime cuts. Brisket can be cooked many ways. Basting of the meat is often done during the cooking process. Popular methods include rubbing with a spice rub or marinating the meat, then cooking slowly over indirect heat from charcoal or wood. This is a form of smoking the meat. Barbeque is also a good form of cooking this meat.
Please note that we cut the brisket into two to three pieces to easily ship it, we can not ship whole brisket.
This brisket is delicious. Cooked in the slow cooker, pressure cooker, or Instant Pot, the meat is incredibly tender and flavorful.
Meat is good quality. Packaging is great. I have ordered a second time However, no one has answered questions I asked for my initial order and a second time for my second order just placed 2 days ago. I would give this place a 5 stars if the customer service would pick up.
This is an excellent product--I recommend slow-cooking or a crock-pot for 7+ hours. The fat will render into whatever sauce you have made and yield a delicious, tender dinner.
Loved the brisket! I wish I had gotten more. Excellent quality for the price. I pray Allah (SWT) adds Barakah to your business. Ameen. You're doing something great for the umrah. I wish more people knew about your products.


Quartz White Pearl exterior and Beige interior, SE trim. FUEL EFFICIENT 38 MPG Hwy/29 MPG City! OPTION GROUP 01, Back-Up Camera, Bluetooth, iPod/MP3 Input, "The seats are comfortable and offer a nice balance between soft and firm to help you out on longer rides." -Edmunds.com. Warranty 10 yrs/100k Miles - Drivetrain Warranty; SEE MORE!
OPTION GROUP 01. MP3 Player, Keyless Entry, Remote Trunk Release, Steering Wheel Controls.


Have you been impacted by the government shutdown?
For members with a consumer loan, impacted by the government shutdown, deferral programs are available. Please contact Frank DeAngelis at 617.889.7701 to discuss your specific situation.
The shutdown may impact a member's ability to make scheduled mortgage payments. To assist borrowers who are unable to make their monthly mortgage loan payment as a result of the shutdown, please contact Frank DeAngelis at 617.889.7701 to discuss terms of the Plan.
Metro will waive early withdrawal penalties during the furlough period on existing certificates and club accounts for those affected by the Government shutdown.
Deposit product fees incurred as a result of the Government shutdown will be reimbursed upon the member's request.
If you're looking for a paycheck replacement or another way to access to funds during the Government shutdown, Metro offers a furloughed loan product.
• If the shutdown persists for more than one pay period, an extension of the loan will be re-evaluated.
Available to members and family members of members. Available until Government shutdown ends. Must be able to prove you are a furloughed employee.
*Cannot be withdrawn during the first 6 days after account opening.
**3.99% Annual Percentage Rate (APR) as of 01/24/19. 12 monthly payments of $85.14 per $1,000.00 borrowed. Interest accrues after the first 90 days. Member must be in good standing with Metro Credit Union in order to qualify. Debt protection is available.


Time to start writing before I run out of time. I can't really decide if doing this in the morning or at night is better. Considering night is somewhere way too close to midnight, probably morning. But morning is spent dreading going to work. I don't know. I don't know what I should be writing today. I need to get back into a rhythm. I probably need to write my novel myself. I keep trying to reclaim my motivation, but that's never going to happen, is it? You can't reclaim moments like that. They come around so rarely, and what? I wasted it. I never want to feel like that. I've spent too long in fandom to believe that it's all been a waste. It can't have been. It's practice, yes. It's my way of working towards something.
I just have to do it. That's the thing, isn't it? I'm having trouble with that part. It's the same trouble I always have, a lot of people have, I think. I need the situation to be perfect before I start. I need to know how I'm writing this thing–Am I doing it in a blog or in a doc or just in simplenote. But I don't really need to know this crap to write. That'll all shake out later. Things will work themselves out and I need to stop worrying. I need to just write. I don't have a plot. It's true. I don't even have character names. I don't know exactly who these people are yet, but stuff like that will work itself out.
Oh, holy crap, how did it get that late so fast? I just barely have 400 words and less than ten minutes. So I need to stop thinking and just start typing. I'm listening to Nina Simone's Sinnerman right now. I love this song so much. It's kind of the theme song to the Pierce Brosnan version of the Thomas Crown Affair. That's not the first place I heard it, but the first time I heard it and noticed it. That's an important distinction to make, especially in this world we live in today. So inundated with culture, but never taking it in. We can't take it all in. It's pretty impossible.
That's why the curator is such an important role today. We need those people to sift through and find the good stuff, no, the great stuff, because we don't have time for anything less. I barely have time to watch the great stuff that gets passed on by the curators I know and love. But sometimes, you need to take a moment and just listen to a song. Really listen to a song, not have it on in the background while you do three other things. The first time I really heard Sinnerman, I was, obviously, watching a movie. I was probably on my computer, too. But the song made me sit up. I almost stopped watching the movie. I just wanted to listen to the song. It's more than 10 minutes long, and they used a considerable chunk of it during the movie. The handclaps are my favourite part, I think. The way the rhythm builds. The way the crowd builds. It's pretty awesome. It's like a whole musical in one song. There are movements, variations, not just one melody, one lyric, one line from beginning to end. It's many songs in one. It's not too many, just enough many. It's Nina Simone, but also that crowd of hand clappers, and the drummer working the cymbals, and the piano player, and everyone else making the whole thing a symphony, not just a song. I can't even imagine what it must have been like to hear this song performed in a tiny coffee and smoke filled club back in the 1950s.


text:
Q: How to add a new entry to route table? I have two computers host1 and host2. From host1 I can ping Vitrual Machines with IP addresses like 192.168.122.0/24 fine (there is a bridge for the Virtual Machines). But I cannot do that from host2.
Given that both host1 and host2 are part of eth0 and can ping each other fine, do you think there is a way to ping the IPs in 192.168.122.0/24 from host2? How would you go about it?
I thought about adding something similar to this into the routing table "route add -host 192.168.122.11 gw 172.8.30.150 eth0" but It doesn't work.
Thank you for any hints/ideas on this!

A: If you are saying that you have two separate computers...one of which is housing VMs.  And you can't ping the far vm inside the other host, your gw needs to be the far host...or you also need a route in the router pointing back at the local host and some SRTs to hairpin traffic.  If you are bridging, sometimes a bit more is required.  To do this, I have previously put eth0 in promisc mode and added it to a bridge group with some tap interfaces.  I then use the tap interfaces inside the vm.  This can get past a lot of routing/switching weirdness and the eth0 interface disregarding traffic destined for the VM in some instances


text:
FDA Issued Orders To 10 Companies To Remove Their E-Cigarette Products
in Health News
The Food and Drug Administration issued warning orders to 10 companies to discard their e-cigarette products from the market. The letters were issued to the companies because they were accused of selling illegal products, disposable, and flavored e-cigarettes that appeal to underage and young people.
The FDA sent warning letters to 10 companies to eradicate the production and distribution of e-cigarette products. Among these 7 companies are accused of illegal sale and distribution of flavored nicotine cigarettes. The other three companies are indicted for illegal marketing strategies for their e-cigarettes.
Director of the FDA, Mitch Zeller, at the Center for Tobacco Products, announced that they suspended several in-person activities like vape shop inspections and retail compliance checks in the COVID pandemic.
The enforcement is continued against the unauthorized marketing and sale of flavored e-cigarette products. He further said that the warning letters are a consequence of violations of tobacco laws and regulations that were monitored online by the company.
Companies are advised to remove their disposable and flavored e-cigarettes and their e-liquid products because they are not authorized as per the law and are appealing to youth. The commissioner of the FDA, Dr. Stephen Hahn said is his statement,
"We are concerned about the popularity of these products among youth and want to make it clear to all tobacco product manufacturers and retailers that, even during the ongoing pandemic, the FDA is keeping a close watch on the marketplace and will hold companies accountable."
Accusations on the E-cigarette Companies
The FDA has explicitly accused the Puff Bar (Cool Clouds Distribution Inc.) Myle Vape Inc., and HQD Tech USA for illegal marketing of flavored e-cigarettes. In August 2016, new regulations were issued by the FDA on the marketing and distribution of tobacco products. Even later than that, the companies were distributing their e-cigarette products without authorization.
The three companies were marketing the tobacco products without approval from the FDA at a modified risk.
Additionally, Eleaf USA, Majestic Vapor LLC, Vape Deal LLC, E Cigarette Empire LLC, Breazy Inc., Ohm City Vapes Inc., and Hina Singh Enterprises have also received the warning letters from the agency.
These seven companies are accused of unauthorized sale and distribution of electronic nicotine delivery systems (ENDS) that are specifically targeted to youth. The packaging and marketing of e-liquid products were imitative of food products (Cherry Coke, Twinkies, Popcorn, or Cinnamon toast crunch cereal) that are popular among youth.
Also Read: COVID-19: People Are Taking Steps To Quit Smoking
Dr. Andrew Stiehm, the director, and pulmonologist at Allina Health United Lung and Sleep Clinic (Minnesota) said,
"With vaping, the first time you try it, you may taste cotton candy or cherry, that is tasty. You don't have the taste aversion from trying it again. So it is an easier habit to begin. It becomes a gateway to bigger behaviors."
The 10 companies have 15 working days to provide their action plan to the FDA to address the accusations and to comply with the laws and regulations of the agency. Moreover, there is a chance that the companies might face an injunction, monetary penalties, or seizure if they do not comply.
The imported e-cigarette products might be subject to admission refusal in the United States. The agency has refused the admission of 74 disposable ENDS products into the United States due to violations.
Tags: CigaretteFood and Drug AdministrationSmokingUnited States
by Surayya Jabin
Winter is coming! Well, this might not be the same Winter from Game of Thrones, but it does need attention. If you want to protect yourself...
Honey beats over-the-counter medications and antibiotics to treat irritating cough, sore throat effectively, and the common cold, a new study claims. The upper respiratory tract infections...
Meditation Relaxation Therapy Helps In The Treatment Of Sleep Paralysis
by Namood-e-Sahar
The research team from the United Kingdom and Italy found in a pilot study that meditation relaxation therapy (MR Therapy) helps in the treatment of sleep...
Russian President Vladimir Putin declared on Tuesday that the country had become the first country in the world to approve the Covid-19 vaccine. Gamaleya Institute, Moscow,...
Researchers from the University of California found evidence of neural brain activity in infants that predicts autism spectrum disorder in 18 months old infants. Elsevier published...
Brazilian Study Finds Hydroxychloroquine Ineffective For COVID-19
Not Every Woman Needs A Pain Reliever During Childbirth
Coronavirus: Mothers Unlikely To Infect Baby During Childbirth
9 Natural Remedies For Dark Circles And Eye Bags
Papaya Fruit: Health Benefits and Nutritional Components
Fundamental Principles of PERMA Model For a Happy Life
COVID-19: What is Happy Hypoxemia?
Turmeric: Health Benefits And Nutritional Components
Ebola (Ebola Virus Disease)
Ebola Virus: History, Types, Symptoms, Prevention, Treatment, And Myths
Hantavirus: Symptoms, Causes, Prevention, Treatment, And Myths
Eyes Health

redpajama_set_name: RedPajamaCommonCrawl


How to open a PDF file from a clickable button, link, or image in Excel. This allows you to seamlessly link PDF files with a spreadsheet.
Click away from the cell by selecting another one and now any time you click the original cell, the PDF will open.
Find the PDF on your computer, select it, and hit OK.
Select an empty cell and, now, any time you click that image, the PDF will open.
I find it nice to include and image and a text link for files like PDFs so that users know what kind of file they will open.
The process of linking to a PDF file from Excel is easy but the user must, of course, have a program that can open a PDF file already on their computer.
The link to the PDF in the attached spreadsheet for this tutorial might not work when you download it due to the fact that it was linked on my computer before being uploaded. In that case, just follow the steps above to relink it on your computer and test it out.
Make sure to download the accompanying spreadsheet and PDF to test this out.


If you love Pasta Salad, this Creamy Avocado Pasta Salad will impress you!! There is a secret ingredient packed into the avocado dressing that is healthy and so sneaky, no one will suspect it's there.
This avocado pasta salad recipe is so vibrant, healthy and loaded with sweet corn, peas, cherry tomatoes and creamy avocados! The veggies cook right in the same pot with the pasta so this couldn't be easier!
This recipe couldn't be easier. It comes together quickly and is one of the most tempting looking salads I've come across. It's creamy, vibrant, fresh and so satisfying!
The idea for this came from this oldie but goodie recipe. This avocado pasta salad is ultra smooth and flavorful. Everything about it is good, and also good for you!!
This Pasta Salad has simple ingredients (it is loaded with avocados!) and it comes together quickly. The spinach is the surprise ingredient. It blends into the sauce so well and you have a healthier pasta salad! This pasta salad is loaded with good-for-you ingredients.
1. Cook pasta according to package instructions – I cooked 12 oz pasta in 4 Qts water with 1 Tbsp salt and boiled 10 minutes. About 4 minutes before your pasta is done cooking, add frozen peas and frozen corn and cook 4 minutes or until veggies are crisp tender. Rinse under running cold water to chill the pasta. Drain well and transfer to a large mixing bowl.
2. In a food processor or blender, combine half of your spinach, 2 quartered avocados, 4 Tbsp lemon juice, 1 sliced garlic clove, 1 tsp salt and 1/4 tsp black pepper. Pulse/puree until blended then add remaining spinach and pulse until creamy and smooth.
3. Add avocado dressing to the pasta bowl along with sliced tomatoes and 1 remaining sliced avocado. Toss everything together until evenly coated. Season to taste with more salt and black pepper.
If you're feeling adventurous, you can garnish with chives and crumbled bacon (totally optional, but nice).
P.S. This pasta salad tastes best fresh but leftovers are still good the next day and make for a perfect work or school lunch!
Creamy Shrimp Pasta – better than Olive Garden's!
The secret ingredient in this creamy avocado pasta salad will shock you! This avocado pasta salad is easy, creamy, vibrant, fresh and so satisfying!
Cook pasta according to package instructions - I cooked 12 oz pasta in 4 Qts water with 1 Tbsp salt and boiled 10 min. About 4 min before pasta is done cooking, add frozen peas and corn and cook 4 min until veggies are crisp tender. Rinse under running cold water to chill pasta. Drain well and transfer to large mixing bowl.
In a food processor or blender, combine half of the spinach, 2 quartered avocados, 4 Tbsp lemon juice, 1 sliced garlic clove, 1 tsp salt and 1/4 tsp black pepper. Pulse until blended then add remaining spinach and blend until creamy and smooth.
Add avocado dressing to pasta bowl along with sliced tomatoes and 1 remaining sliced avocado. Toss salad together until evenly coated. Season to taste with more salt and black pepper.
As a busy Mom, I love shortcuts so I always keep a stash of peas and corn in the freezer and often mix them with fresh ingredients in my cooking for a healthy and easy boost of vegetables. They are perfect for pasta salads!
Q: Do you have any go-to cooking shortcuts, like combining frozen veggies with fresh ingredients? Let me know in a comment below!!
My husband and I get together with my best friend and her husband once a week. This recipe looks DELIGHTFUL! I am bringing this as a side tomorrow night! I did pre-make just the pasta with corn and peas. I think that will be fine chilling in the fridge overnight, don't you?
Natasha, do you think the avocado will brown if it is stays overnight. I am thinking of making this salad for my son for school lunch.


text:
require 'redis'
require 'json'

module Bunraku
  module Server
    class Application < Sinatra::Base

      configure do
        set :static, true
        set :root, File.dirname(__FILE__)
        set :public_folder, 'public'
        set :redis_server, 'localhost'
        set :redis_port, 6379
        $redis = Redis.new(:host => settings.redis_server, :port => settings.redis_port)
      end

      helpers do
        def cycle
          %w{even odd}[@_cycle = ((@_cycle || -1) + 1) % 2]
        end

        CYCLE = %w{even odd}
        def cycle_fully_sick
          CYCLE[@_cycle = ((@_cycle || -1) + 1) % 2]
        end
      end

      def get_title(path)
        path.gsub(/^\//, '').capitalize
      end

      def load_nodes(ids)
        if ids.empty?
          []
        else
          nodes = []
          ids.each { |id|
            node = $redis.hgetall("node-#{id}")
            nodes << node
          }
          nodes
        end
      end

      def load_node(id)
        if id.nil?
          return "No node found"
        else
          @detail = $redis.hgetall("node-#{id}")
          @metrics = $redis.hgetall("node-#{id}:metrics")
        end
      end

      def sort_nodes(nodes)
        if nodes.empty?
          []
        else
          nodes.sort_by { |node| node["time"] }.reverse!
        end
      end

      def select_node(nodes,query,value)
        nodes.select { |node| node[query] == value }
      end

      get '/' do
        @title = nil
        nodes = load_nodes($redis.smembers("all-nodes").last(100))
        @sorted = sort_nodes(nodes)
        erb :index
      end

      get '/failed' do
        @title = get_title(request.path_info)
        nodes = load_nodes($redis.smembers("all-nodes"))
        nodes = select_node(nodes,'status','failed')
        @sorted = sort_nodes(nodes)
        redirect '/' if @sorted.empty?
        erb :index
      end

      get '/unchanged' do
        @title = get_title(request.path_info)
        nodes = load_nodes($redis.smembers("all-nodes"))
        nodes = select_node(nodes,'status','unchanged')
        @sorted = sort_nodes(nodes)
        redirect '/' if @sorted.empty?
        erb :index
      end

      get '/changed' do
        @title = get_title(request.path_info)
        nodes = load_nodes($redis.smembers("all-nodes"))
        nodes = select_node(nodes,'status','changed')
        @sorted = sort_nodes(nodes)
        redirect '/' if @sorted.empty?
        erb :index
      end

      get '/node/:node' do |node|
        @node = params[:node]
        nodes = load_nodes($redis.smembers("all-nodes"))
        nodes = select_node(nodes,'node',node)
        @sorted = sort_nodes(nodes)
        erb :node
      end

      get '/node/detail/:id' do |id|
        id = params[:id]
        load_node(id)
        erb :detail
      end

      post '/new/?' do
        id = $redis.incr(:node_counter)

        node = JSON.parse(params[:data])

        node.each { |type,value|
          if type == 'metrics'
            value.each { |n,d|
              $redis.hset("node-#{id}:#{type}", n, d)
            }
          else
            $redis.hset("node-#{id}", type, value)
          end
        }
        $redis.hset("node-#{id}", "id", id)
        $redis.sadd("all-nodes", id)
      end
    end
  end
end


On February 7th, our NEXT TERMINOLOGY ONLINE COURSE - advanced level - will start. The course will end in May 2018, with oral and written exams.
Since we did not raise fees since many years, all online courses will be a bit less cheap this year.


Perfectly Situated Two Family Home on 10,000 feet of land and a quick walk to Riverside Station Green Line, Express Bus, Williams Elementary School, Auburndale Square and just off the Pike and Rt 128 entrances for commuters. The home would appeal to end-user owner occupants desiring rental income in a lovely neighborhood. The property has appeal for builders & developers with plenty of land that stretches deep off the street. Interior is all freshened up and each unit contains 6 rooms and 2 generous bedrooms. High ceilings and large replacement windows bring in plenty of light to fill the space. Kitchen and bath could use updating, but are functional as is, and has been rented for market rates in recent past.


In many places, public transportation is inadequate to get us where we need to be, making car ownership necessary. In the United States at least, car ownership can also be a status symbol. Some families even maintain multiple vehicles for different purposes.
But owning and maintaining a vehicle can be costly. According to AAA, in 2017 the average cost of owning and maintaining a new car was about $8500 per year. The survey done by AAA included interest on a car loan, maintenance, repairs, fuel cost, and depreciation.
Sure, you could decrease this ownership cost by buying a used car without incurring a loan. But don't overlook the cost to insure your vehicle, as well as any property tax and registration fees you will have to pay.
The point is this: cars are expensive to own, so some strategy is in order. Here are some things to do to reduce the cost of car ownership.
Buy used: This is one you have heard before. Exercise caution when buying used, however. Knowing the fair value for the type of car you wish to buy is important, but so is making sure the car was maintained properly. A used car is not much of a deal if it turns out to be a repair nightmare.
Buy new and own for a long time: Buying used isn't for everyone. Some people don't have the knowledge of cars to be able to spot issues when buying used. Buying a new car with a warranty provides peace of mind during the early years of car ownership. The key to making a new car purchase work out financially is to maintain the car diligently and drive that sucker forever.
Compare the total cost of ownership carefully: Some cars are more expensive to drive when it comes to insurance, fuel, and the cost of repairs. The more specialized the type of car, the more it is likely to cost when it comes to repair it. New cars cost far more to insure than cars with a few years on them. And some car models require premium fuel, which will drive up the cost of ownership.
Carefully consider your needs before choosing a car: Do you need a four-wheel drive car to get around in winter? Do you need a spacious backseat for your large family? How safe is the vehicle you are considering? This sounds like an obvious thing to do, but I will admit that I made a dreadful car choice once. Having an electric car was very appealing to me. But I neglected to test drive it up our steep, rugged driveway. The short story is that I only owned that car for a year. I was sad to let it go, and I was mad at myself for being so short-sighted when car shopping.
Keep your car longer and keep it maintained: The longer you can drive one car, the better your average annual cost of driving will be. The caveat is that you keep it maintained. A car that is constantly in need of repair is unreliable, but it also drives up the cost of ownership.
After your car is paid off, keep making payments: Once you have paid off your car, continue making the car payment to a savings account for future car expenses. If you didn't have a car payment in the first place, figure out how much you can afford to direct to a car fund. Preparing in advance will help you reduce costs in the future.
Reduce borrowing costs: If you have no choice but to take out a loan for your car, try to take the shortest duration loan you can manage. Stretching the loan out for more years increases the amount of interest you will pay. Once you have a loan, pay extra each month to reduce the principal as fast as you can.
Buy within your budget: Don't buy more car than you need simply because you can afford the payments. Ideally, you should take a loan for no more than four years for a vehicle. Agreeing to stretch a loan into five or six years gives you the false sense of being able to afford a more expensive car. The average time someone owns a car is just under seven years (source). Therefore, if you could finance for four or fewer years, that gives you a few years to accumulate principal for your next vehicle.
Do you even need a car? Does your family need to have more than one? This is also where careful consideration and diligent research comes in. Depending on where you live and your ability to utilize public transportation, you might be a renegade and just say no to car ownership.
What are your strategies for reducing the cost of car ownership? Share below! Or if you want to start a discussion with some like-minded friends, join the free SimpleMoney Community on Facebook to share your thoughts!


There is nothing quite like Autumn crafts. After the hustle and bustle of summertime fun outside, turning inside a little bit more as the weather cools is very welcomed by this Mama.
And turning inside goes hand in hand with crafting around these parts.
Who am I kidding? A lot of things go hand in hand with crafting around these parts. Pretty much anything, really. I really love crafting with my kids.
We were outside the other day teaching our little Delilah to fly. Our sweet little duckling is all grown up! The boys have been busy getting her settled as an 'outdoor duck' and decided she should likely be able to fly. I really thought it was too soon myself, but thought they would have fun trying all the same.
Well wouldn't you know, as those boys of mine ran down the hill flapping their arms, that sweet little duck took off right after them!
When those boys of mine were busy flying, I was busy cleaning up some branches on that very hill so those little peanuts of mine didn't trip. And once a big pile of branches has been collected, crafting is sure to follow.
We made cherry blossom trees a few Springs ago and I remember just loving them. I thought we could use the same process but use orange, red, and yellow tissue paper for Autumn leaves.
The craft was simple, just as crafting with kids should be. I laid out a few branches on the table, cut up some squares of red, orange, and yellow tissue paper, and grabbed a bottle of white glue and a paint brush.
I thinned the glue down just a touch with water and popped it into a shallow dish.
My little ones painted a twig from their branch and then wrapped a square of the tissue paper around.
And then Ben went to play Lego, but Sam continued again and again and again.
Once the branches were covered in the tissue paper we popped them in a Mason Jar and they look lovely! Such a nice Fall decoration.
I hope you are in the Autumn crafting spirit friends, as I have a bunch of ideas heading your way.
Happy Tuesday! I hope your week is off to a wonderful start. Thank you so much for being here.


text:
On The Psychology Of The Conspiracy Denier
A fascinating take on what makes the conspiracy denier tick…
A closer look at the class that mocks.
By Tim Foyle
Why is it that otherwise perfectly intelligent, thoughtful and rationally minded people baulk at the suggestion that sociopaths are conspiring to manipulate and deceive them? And why will they defend this ill-founded position with such vehemence?
History catalogues the machinations of liars, thieves, bullies and narcissists and their devastating effects. In modern times too, evidence of corruption and extraordinary deceptions abound.
We know, without question, that politicians lie and hide their connections and that corporations routinely display utter contempt for moral norms – that corruption surrounds us.
We know that revolving doors between the corporate and political spheres, the lobbying system, corrupt regulators, the media and judiciary mean that wrongdoing is practically never brought to any semblance of genuine justice.
We know that the press makes noise about these matters occasionally but never pursues them with true vigour.
We know that in the intelligence services and law enforcement wrongdoing on a breathtaking scale is commonplace and that, again, justice is never forthcoming.
We know that governments repeatedly ignore or trample on the rights of the people, and actively abuse and mistreat the people. None of this is controversial.
So exactly what is it that conspiracy deniers refuse to acknowledge with such fervour, righteousness and condescension? Why, against all the evidence, do they sneeringly and contemptuously defend the crumbling illusion that 'the great and good' are up there somewhere, have everything in hand, have only our best interests at heart, and are scrupulous, wise and sincere? That the press serves the people and truth rather than the crooks? That injustice after injustice result from mistakes and oversights, and never from that dread word: conspiracy?
What reasonable person would continue to inhabit such a fantasy world?
The point of disagreement here is only on the matter of scale. Someone who is genuinely curious about the plans of powerful sociopaths won't limit the scope of their curiosity to, for example, one corporation, or one nation. Why would they? Such a person assumes that the same patterns on display locally are likely to be found all the way up the power food chain. But the conspiracy denier insists this is preposterous.
It is painfully obvious that the pyramidical societal and legal structures that humanity has allowed to develop are exactly the kind of dominance hierarchies that undoubtedly favour the sociopath. A humane being operating with a normal and healthy cooperative mindset has little inclination to take part in the combat necessary to climb a corporate or political ladder.
So what do conspiracy deniers imagine the 70 million or more sociopaths in the world do all day, born into a 'game', in which all the wealth and power are at the top of the pyramid, while the most effective attributes for 'winning' are ruthlessness and amorality? Have they never played Monopoly?
Sociopaths do not choose their worldview consciously, and are simply unable to comprehend why normal people would put themselves at such an incredible disadvantage by limiting themselves with conscientiousness and empathy, which are as beyond the understanding of the sociopath as a world without them are to the humane being.
All the sociopath need do to win in the game is lie publicly whilst conspiring privately. What could be simpler? In 2021, to continue to imagine that the world we inhabit is not largely driven by this dynamic amounts to reckless naiveté bordering on insanity. Where does such an inadvertently destructive impulse originate?
The infant child places an innate trust in those it finds itself with – a trust which is, for the most part, essentially justified. The infant could not survive otherwise.
In a sane and healthy society, this deep instinct would evolve as the psyche developed. As self-awareness, the cognitive and reasoning abilities and scepticism evolved in the individual, this innate trust impulse would continue to be understood as a central need of the psyche. Shared belief systems would exist to consciously evolve and develop this childish impulse in order to place this faith somewhere consciously – in values and beliefs of lasting meaning and worth to the society, the individual, or, ideally, both.
Reverence and respect for tradition, natural forces, ancestors, for reason, truth, beauty, liberty, the innate value of life, or the initiating spirit of all things, might all be considered valid resting places in which to consciously place our trust and faith – as well as those derived from more formalised belief systems.
Regardless of the path taken to evolve and develop a personal faith, it is the bringing of one's own consciousness and cognition to this innate impulse that is relevant here. I believe this is a profound responsibility – to develop and cultivate a mature faith – which many are, understandably, unaware of.
What occurs when there is a childish need within us which has never evolved beyond its original survival function of trusting those in our environment who are, simply, the most powerful; the most present and active? When we have never truly explored our own psyches, and deeply interrogated what we truly believe and why? When our motivation for trusting anything or anyone goes unchallenged? When philosophy is left to the philosophers?
I suggest the answer is simple, and that the evidence of this phenomenon and the havoc it is wreaking is all around us: the innate impulse to trust the mother never evolves, never encounters and engages with its counterbalance of reason (or mature faith), and remains forever on its 'default' infant setting.
While the immature psyche no longer depends on parents for its well-being, the powerful and motivating core tenet I have described remains intact: unchallenged, unconsidered and undeveloped. And, in a world in which stability and security are distant memories, these survival instincts, rather than being well-honed, considered, relevant, discerning and up to date, remain, quite literally, those of a baby. Trust is placed in the biggest, loudest, most present and undeniable force around, because instinct decrees that survival depends on it.
And, in this great 'world nursery', the most omnipresent force is the network of institutions which consistently project an unearned image of power, calm, expertise, concern and stability.
In my view, this is how conspiracy deniers are able to cling to and aggressively defend the utterly illogical fantasy that somehow – above a certain undefined level of the societal hierarchy – corruption, deceit, malevolence and narcissism mysteriously evaporate. That, contrary to the maxim, the more power a person has, the more integrity they will inevitably exhibit. These poor deluded souls essentially believe that where personal experience and prior knowledge cannot fill in the gaps in their worldview – in short, where there is a barred door – mummy and daddy are behind it, working out how best to ensure that their little precious will be comfortable, happy and safe forever.
This is the core, comforting illusion at the root of the conspiracy denier's mindset, the decrepit foundation upon which they build a towering castle of justification from which to pompously jeer at and mock those who see otherwise.
This explains why it is that the conspiracy denier will attack any suggestion that the caregiving archetype is no longer present – that sociopaths are behind the barred door, who hold us all in utter contempt or disregard us completely. The conspiracy denier will attack any such suggestion as viciously as if their survival depended on it – which, in a way, within the makeup of their unconscious and precarious psyche, it does.
Their sense of well-being, of security, of comfort, even of a future at all, is completely (and completely unconsciously) invested in this fantasy. The infant has never matured, and, because they are not conscious of this, other than as a deep attachment to their personal security, they will fiercely attack any threat to this unconscious and central aspect of their worldview.
The tediously common refrain from the conspiracy denier is, 'there couldn't be a conspiracy that big'.
The simple retort to such a self-professed expert on conspiracies is obvious: how big?
The biggest 'medical' corporations in the world can go for decades treating the settling of court cases as mere business expenses, for crimes ranging from the suppressing of adverse test events to multiple murders resulting from undeclared testing to colossal environmental crimes.
Governments perform the vilest and most unthinkable 'experiments' (crimes) on their own people without consequence.
Politicians habitually lie to our faces, without consequence.
And on and on. At what point, exactly, does a conspiracy become so big that 'they' just couldn't get away with it, and why? I suggest it's at the point where the cognitive ability of the conspiracy denier falters, and their unconscious survival instinct kicks in. The point at which the intellect becomes overwhelmed with the scope of events and the instinct is to settle back into the familiar comforting faith known and cultivated since the first moment one's lips found the nipple. The faith that someone else is dealing with it – that where the world becomes unknown to us, a powerful and benevolent human authority exists in which we have only to place our faith unconditionally in order to guarantee eternal emotional security.
This dangerous delusion may be the central factor placing humanity's physical security and future in the hands of sociopaths.
To anyone in the habit of dismissing people who are questioning, investigative and sceptical as tin foil hat wearing, paranoid, science-denying Trump supporters, the question is: what do you believe in? Where have you placed your faith and why? How is it that while no one trusts governments, you appear to trust nascent global governance organisations without question? How is this rational?
If you are placing faith in such organisations, consider that in the modern global age, these organisations, as extraordinarily well presented as they are, are simply grander manifestations of the local versions we know we can't trust. They are not our parents and demonstrate no loyalty to humane values. There is no reason to place any faith whatsoever in any of them.
If you haven't consciously developed a faith or questioned why you believe as you do to some depth, such a position might seem misanthropic, but in truth, it is the opposite. These organisations have not earned your trust with anything other than PR money and glossy lies. True power remains, as ever, with the people.
There is a reason why Buddhists strongly advise the placing of one's faith in the Dharma, or the natural law of life, rather than in persons, and that similar refrains are common in other belief systems.
Power corrupts. And, in the world today, misplaced and unfounded trust could well be one of the greatest sources of power there is.
Massive criminal conspiracies exist. The evidence is overwhelming. The scope of those currently underway is unknown, but there is no reason to imagine, in the new global age, that the sociopathic quest for power or the possession of the resources required to move towards it is diminishing. Certainly not while dissent is mocked and censored into silence by gatekeepers, 'useful idiots', and conspiracy deniers, who are, in fact, directly colluding with the sociopathic agenda through their unrelenting attack on those who would shine a light on wrongdoing.
It is every humane being's urgent responsibility to expose sociopathic agendas wherever they exist – never to attack those who seek to do so.
Now, more than ever, it is time to put away childish things, and childish impulses, and to stand up as adults to protect the future of the actual children who have no choice but to trust us with their lives.
This essay has focussed on what I consider to be the deepest psychological driver of conspiracy denial.
There are certainly others, such as the desire to be accepted; the avoidance of knowledge of, and engagement with, the internal and external shadow; the preservation of a positive and righteous self-image: a generalised version of the 'flying monkey' phenomenon, in which a self-interested and vicious class protect themselves by coalescing around the bully; the subtle unconscious adoption of the sociopathic worldview (e.g. 'humanity is the virus'); outrage addiction/superiority complex/status games; a stunted or unambitious intellect that finds validation through maintaining the status quo; the dissociative protective mechanism of imagining that crimes and horrors committed repeatedly within our lifetime are somehow not happening now, not 'here'; and plain old fashioned laziness and cowardice.
My suggestion is that, to some degree, all of these build on the foundation of the primary cause I've outlined here.
commentLeave a Comment on On The Psychology Of The Conspiracy Denier
Legit Impeachment With Merit!
Learning to Trust and Obey


text:
New office building to host growing Boise private equity firm
Rendering of new office building at Ten Mile Crossing, which will house Capital Eleven. Via Capital Eleven
Another new office building will sprout up at Ten Mile Crossing.
The latest office building will go up three stories, with about 73,000 square feet of office space on the large mixed-use campus at the corner of I-84 and Ten Mile.
The latest building will be anchored by Treasure Valley private equity firm Capital Eleven, which has investments in or launched a number of Boise-area startups, including FlipRide, Capital Auto, Fitted, Tovuti, Crave Delivery, and Pivot by KA.
Capital Eleven already has its offices at Ten Mile Crossing but will expand to take up one whole floor of the new building.
[Meridian will allow taller buildings, new office and retail at Ten Mile & Franklin]
"As one of the first tenants at Ten Mile Crossing, we are excited to be expanding into a new building at the center of the Treasure Valley," Capital Eleven co-founder Travis Hawkes said. "It is easily accessible for our employees, investors, and partners and we are thrilled to have room to grow to continue to create jobs and invest in great companies."
City of Meridian permit filings show the building is valued at $9.5 million – before tenant improvements are factored in.
Ten Mile Crossing has roughly 1.2 million square feet of office space, retail, residential and medical facilities, according to a news release. The project is developed by Brighton Corp., SCS and BVA Development.
Project: Ten Mile Crossing, Ten Mile Creek
ICYMI: 'Shake, apply and go:' Boise startup gets patent on at-home makeup system that's like Keurig for beauty
Boise-based systems integrator sold to national firm
KAMUT brand wins Trailmix pitch competition


Climbing from the bathtub onto a wet floor can lead to a fall accident regardless of what how old you are, but such falls could be existence threatening for many seniors. A stroll-in tub may be the solution preferred by many. Water tight doorways allow quick access and reduce tripping dangers. Walk-in bath are available in all sizes and shapes and also the greater finish models feature the same features like a "regular" luxury tub, including whirlpool and water jet features. Some walk-in bath also provide an digitally powered bath seat which makes getting interior and exterior the bathtub even simpler. See a bathroom renovation specialist to find out what options won't meet your needs however your budget too.
A transfer bench is really a removable option to a stroll in bathtub. A transfer bench is affixed to 1 side of the tub and also the user then sits around the seat and only slides or swivels in to the tub. It's clearly essential that a transfer bench are designed for the load of anybody who definitely are utilizing it, so shop carefully. Most traditional transfer benches are only able to securely support an individual who weighs 250 pounds or under, therefore if the individual you're searching for is a touch heavier, ask to determine what are named as bariatric models.
Grab bars are useful for anybody, not only individuals who're seniors. For optimum safety grab bars ought to be set up in multiple locations round the bathroom, not only to the bathtub or shower. Installing grab bars doesn't have to diminish the feel of a properly designed bathroom, because these days they are available in designer colors and finishes and can also be purchased to complement an expensive new faucet.
If you're undertaking your bathroom remodeling project in the home where an seniors or infirm person lives, now may well be a great time to think about widening the doorway just a little. Even when all your family members aren't presently utilizing a motorized wheel chair to assist them to circumvent, that might easily happen somewhere lower the road. Addressing the priority now helps you to save getting to cover another bathroom remodel later on. For simpler bathroom access, consider replacing your traditional bathroom door having a sliding version.


text:
Allison-Antrim Museum
Allison-Antrim MuseumAllison-Antrim MuseumAllison-Antrim Museum
Museum House
Museum Barn
Smith Paintings
Iron Collar
Buffalo Hide and Rifle
Governor Signatures
Carl's Drug Store
Ebbert Spring Park
Enoch Brown Park
Soldiers' Stories
Gordon Crook's Research
Upper Level - Bay 1 - Rotating Exhibits
Upper Level - Bay 2 & 3 - Meeting/Event Space
Bay 1 is the South exhibit area in the barn. It is used for rotating exhibits. The above is student art exhibit.
Bay 2 & 3 is used for public events and meetings. There are about 60 chairs and a number of 8' tables and about 1/2 dozen card tables available for use. A public address system and a large pull-down projector screen is available.
Upper Level - Bay 4 - Civil War Exhibit
Bay 4 is the North exhibit area in the barn. It displays local Civil War artifacts from the Greencastle-Antrim area. There is also a very rare 'Slave Collar' that was made and used locally.
Upper Level - Kitchen
Lower Level - Conservation
A fully functional kitchen is available for events and meetings. It is especially useful to caterers. The kitchen is ADA compliant. It is located in what was the barn 'Granary.
The other granary houses the ADA compliant restrooms.
This room is reserved for the future conservation room where incoming donations are processed. It is also climate controlled and located in lower level. Presently, it holds the archaeological collection from Ebbert Springs.
Lower Level - Storage
There are north and south storage rooms in the lower level of the barn. They house the museum collections from small artifacts to larger farm equipment.
Barnes Library
The Barnes Library is named after A. Isabelle Barnes, a native of Greencastle-Antrim. She was a charter member of the Greencastle-Antrim Civil War Roundtable, established in the mid 1970s. For almost four decades, Isabelle collected books – books on the local history of Greencastle-Antrim, books on the history of Franklin County, books on the history of Pennsylvania, and books on American History. When she moved to Maine Isabelle said, "I have no use for these local history books in Maine, so I am giving my collection to Allison-Antrim Museum." With one magnanimous gesture, Allison-Antrim Museum's library grew tenfold. For this reason, the board of directors decided to name the library "The Barnes Library." Allison-Antrim Museum is sincerely grateful and indebted to Isabelle for her generosity.
The Barnes Library is located in the upper level of the barn.
Copyright © 2020 Allison-Antrim Museum Inc - All Rights Reserved.
365 South Ridge Avenue, Greencastle, PA


text:
package com.google.cloud.shell.v1.samples;

// [START shell_v1_generated_cloudshellserviceclient_removepublickey_async]
import com.google.api.core.ApiFuture;
import com.google.cloud.shell.v1.CloudShellServiceClient;
import com.google.cloud.shell.v1.RemovePublicKeyRequest;
import com.google.longrunning.Operation;

public class AsyncRemovePublicKey {

  public static void main(String[] args) throws Exception {
    asyncRemovePublicKey();
  }

  public static void asyncRemovePublicKey() throws Exception {
    // This snippet has been automatically generated and should be regarded as a code template only.
    // It will require modifications to work:
    // - It may require correct/in-range values for request initialization.
    // - It may require specifying regional endpoints when creating the service client as shown in
    // https://cloud.google.com/java/docs/setup#configure_endpoints_for_the_client_library
    try (CloudShellServiceClient cloudShellServiceClient = CloudShellServiceClient.create()) {
      RemovePublicKeyRequest request =
          RemovePublicKeyRequest.newBuilder()
              .setEnvironment("environment-85904877")
              .setKey("key106079")
              .build();
      ApiFuture<Operation> future =
          cloudShellServiceClient.removePublicKeyCallable().futureCall(request);
      // Do something.
      Operation response = future.get();
    }
  }
}
// [END shell_v1_generated_cloudshellserviceclient_removepublickey_async]


text:
Q: Memory leak (?) while shared element transition I have RecycleView with StaggeredLayoutManager which shows images. these images are displayed via glide:
Glide.with(itemView.getContext()).load(post.image.url)
                .bitmapTransform(new RoundedCornersTransformation(itemView.getContext(), 10, 0, RoundedCornersTransformation.CornerType.TOP))
                .diskCacheStrategy(DiskCacheStrategy.SOURCE)
                .into(image);

I have another activity, where is viewpager with details if these images. I want to make shared element transition between this image, and bigger image displayed in viewpager. However, when I do shared element transition, I can see that memory usage grows while I go to ViewPager's activity and back, and after some transitions OutOfMemory error is thrown. When I run activity without transition, memory is ok. What might cause such a leak?


text:
Coverage round-up: Deeyah Khan's film JIHAD makes the headlines
The latest Fuuse documentary by Deeyah Khan premiered on ITV's Exposure series last night. In the build up to the broadcast, the outspoken filmmaker and human rights activist, Deeyah has been featured in a number of high profile interviews. We've highlighted a few of these articles and videos below...
The Independent - Deeyah Khan interview: The award-winning filmmaker on chronicling British jihadism
The Guardian - What Muslim extremists and rightwing racists have in common
Huffington Post - 'Jihad': My Journey Into the Roots of Islamic Radicalism in Britain
The Observer - 'Recruiter' of UK jihadis: I regret opening the way to Isis
Daily Mail - JIHAD in the Daily Mail Online
Video - Channel 4 News
Video - Good Morning Britain


I spent some time as the General Secretary of Middlesex Schools' Football Association (MSFA) whilst working for the Middlesex County FA (which administers club football in the region). One of the first tasks I undertook was to revamp the MSFA website which was, at that time, exceptionally basic and rarely updated. Becoming the General Secretary of MSFA gave me much greater access to information and enabled me to develop the site into a source of news and information.
I continued to manage the website after relocating to Liverpool and it continues to evolve as MSFA changes the way it administers its competitions. The site is now being used to collect affiliation and cup entry data and new tools will be added to it during 2017-18.
Brent Schools' FA runs representative football teams representing the London Borough of Brent alongside other activities intended to help the young people engaged with the Association to fulfil their potential.
In addition to having designed and constructed the site, I remain actively involved in its day-to-day management, uploading new content when it is provided to me.


Given the thousands of press releases coming out of CES, we almost missed the news that your TiVo DVR/Set-top box will soon integrate in your smart home setup, almost regardless of which platform you've decided to delve into.
Over the next few months, TiVo will have an IFTTT channel with multiple applets to let you automate your TV watching based on other happenings around your house, such as turn on ESPN when you get home, pause when the doorbell rings, and so on. It will also have an Amazon Alexa skill as well as a Google Assistant app. The video below shows a few demos of features, but to be honest, I was left a little confused by the Google Home part of it.


Four day Classical Greece tour to explore the glory of ancient Greece with and visit five UNESCO'S world heritage sites.
After breakfast, pick up from hotel for your four day Classical Greece tour to explore the glory of ancient Greece with and visit five UNESCO'S world heritage sites. We depart from Athens for the Corinth canal (short stop). After our short stop at Corinth, we drive to Epidaurus and visit the Ancient theatre, world famous for its outstanding acoustics. We then proceed to Nauplia (short stop) and through the plain of Argos, we arrive to Mycenae, where we visit its archaeological site and the Tomb of Agamemnon. In the afternoon (optional stop for souvenirs), we depart for Olympia, the birth place of the Olympic Games. Overnight stay in Olympia.
We start our day with a visit to the Olympia archaeological site with the Temple of Zeus, the Temple of Hera, the altar of the Olympic flame, the Stadium and the archaeological Museum. After Olympia, we depart via Patras (short stop) to Rion, cross the Corinthian bay by the new bridge to Antirion. Pass by the city of Nafpaktos and from there on, along the coastal road to Delphi, known in ancient times as the navel of the world. In the afternoon, we visit the picturesque mountain village of Arachova. Overnight stay in Delphi.
In the morning, we visit the archaeological site and the Museum of Delphi. After our navigation through the center of earth and universe for Ancient Greeks, Delphi, we depart for Kalambaka, by crossing the Thessalian plain in Central Greece through the cities of Lamia (short stop) and Trikala. Kalambaka is a small town built at the foot of the majestic Meteora grey rocks, just below the Monasteries.
Today, we depart in the morning for nearby Meteora on a tour of the cliff-top monasteries by visiting two different Byzantine Monasteries, built on top of rock towers and offering fantastic views over the plain. Meteora is included on the UNESCO world heritage monument list. The monasteries are perched on top of towers of sandstone, which provided the monks with totally inaccessible retreats for centuries. Meteora is considered one of the most beautiful areas of Greek mainland and its cluster of monasteries is one of the country's most unique attractions. We return to Athens passing by the heroic monument of Leonidas, the King of Sparta in Thermopylae (short stop for photos). The tour returns to Athens late in the afternoon.


text:
Ро́нчна () — село в Польше в сельской гмине Лишки Краковского повята Малопольского воеводства.

География
Село располагается в 4 км от административного центра гмины села Лишки и в 14 км от административного центра воеводства города Краков. Село связано с Краковом автобусными маршрутами № 239, 259, доходящими до краковского района Сальватор.

В 1975—1998 годах село входило в состав Краковского воеводства.

Население
По состоянию на 2013 год в селе проживало 1810 человек.

Данные переписи 2013 года:

Достопримечательности

 Католическая церковь, построенная в 1984 году.

Примечания

Ссылки
 Główny Urząd Statystyczny — TERYT
 Rozporządzenie Ministra Administracji i Cyfryzacji z dnia 13 grudnia 2012 r. w sprawie wykazu urzędowych nazw miejscowości i ich części. «Dziennik Ustaw». Nr 29, 2013-02-13. Ministerstwo Administracji i Cyfryzacji

Населённые пункты Краковского повета

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
Let $q$ be a power of a prime $p$ and let $k$ be an algebraic closure of the finite field $\mathbb{F}_{q}$. We denote by a column vector $\bm{x}=\lt(x_0,x_1,x_2,x_3)$ a point in the $k$-projective space $\mathbb{P}^3$. For a nonzero $4$-by-$4$ matrix $A$ with entries in $k$, a $k$-Hermitian surface of degree $q+1$ is
defined by
\begin{equation*}
X_A:=\{\bm{x}\in\mathbb{P}^3\ |\ \lt\bm{x}A\bm{x}^{(q)}=0\},
\end{equation*}
where $\bm{x}^{(q)}=\lt({x_0}^q,{x_1}^q,{x_2}^q,{x_3}^q)$. The map $x\mapsto x^q$ is an involutive automorphism on $\mathbb{F}_{q^2}$, which is an analog of the complex conjugate $z\mapsto\bar{z}$ on the complex numbers. If $\lt A=A^{(q)}$ then $X_A$ is defined over $\mathbb{F}_{q^2}$ and is simply called a Hermitian surface. Especially, when $A$ is equal to the identity matrix $I$, the associated surface $X_I$ is the Fermat surface of degree $q+1$. One can easily show that $X_A$ is smooth if and only if $A$ is invertible.

Lines on a Hermitian surface, more generally, linear spaces on a Hermitian hypersurface have been extensively studied in various papers, for example \cite{Se}, \cite{BC}, \cite{Sh}, \cite{KK} and \cite{SSL}. In contrast, there are few results known about rational curves of degree $>1$ on a Hermitian surface.

Let $A$ be an invertible $4$-by-$4$ matrix with entries in $k$ and
let $R_d$ be the set of nonplanar rational curves of degree $d$ contained in $X_A$. In \cite{O} we studied on $R_d$ for $d\le q+1$. The result may be summarized as follows:\\
The set $R_d$ is empty for $d<q+1$. In addition, no conics are also contained in $X_A$. On the other hand, when $d=q+1$ the set $R_d$ is not empty and all curves in $R_d$ are projectively isomorphic over $k$ to an explicitly given curve, and furthermore the number of curves in $R_d$ is exactly $q^4(q^3+1)(q^2-1)$.

In this paper, we extend this result to the case $d\ge q+2$ under the assumption that the curves in $R_d$ can be parametrized by polynomials with at most $4$ terms. We refer to the curves satisfying such condition as ``tetranomial curves'' in this paper. By definition, for a nonplanar tetranomial curve of degree $d$ with the locus $\lt(x_0,x_1,x_2,x_3)$, there are an invertible $4$-by-$4$ matrix $F$ with entries in $k$ and integers $i,j$ with $1\le i<j\le d-1$ such that
\begin{equation*}\label{intro}
\lt(x_0,x_1,x_2,x_3)
=F\,\,\lt(s^d,s^{d-i}t^i,s^{d-j}t^j,t^d),
\end{equation*}
where $s,t$ are homogeneous coordinates of $\mathbb{P}^1$.
We denote by $C(d,i,j)$ a curve for when $F=I$ in the above expression.
Our purpose is to prove the following theorem:

\begin{thm}\label{mainthm}
Let $T_d$ be the set of nonplanar tetranomial curves of degree $d$ on a smooth $k$-Hermitian surface. If $T_d$ is non-empty then $T_d$ is equal to one of the following three cases:
\begin{itemize}
\item[C1)]$d=q+1$,
\begin{equation*}
|T_d|=
\begin{cases}
\infty&{\rm for}\ q=2\\q^4(q^3+1)(q^2-1)&{\rm for}\ q\ge3,\\
\end{cases}
\end{equation*}
all elements of $T_d$ are projectively isomorphic over $k$ to $C(q+1, 1, q)$.
\item[C2)]$d=(q+1)q$ for $q$ even,
\begin{equation*}
|T_d|=
\begin{cases}
\infty&{\rm for}\ q=2\\q^6(q^4-1)(q^2-1)&{\rm for}\ q\ge3,\\
\end{cases}
\end{equation*}
all elements of $T_d$ are projectively isomorphic over $k$ to\\ $C((q+1)q,q+1,q^2+1)$.
\item[C3)]$d=(q+1)q/2$ for $q$ odd,
\begin{equation*}
|T_d|=
\begin{cases}
2q^6(q^4-1)(q^2-1)&{\rm for}\ q\equiv1\ {\rm mod}\ 4\\
4q^6(q^4-1)(q^2-1)&{\rm for}\ q\equiv3\ {\rm mod}\ 4,
\end{cases}
\end{equation*}
all elements of $T_d$ are projectively isomorphic over $k$ to\\ $C((q+1)q/2, (q+1)/2, (q^2+1)/2)$.
\end{itemize}
\end{thm}
Here C1 has already proven for $q\ge3$ in \cite{O}. When $q=2$, C1 states that a smooth $k$-Hermitian cubic surface contains an infinite number of nonplanar rational curves of degree $3$, however, which seems to contradict a result of \cite{O}. Unfortunately, there is actually an oversight in the case $q=2$ of \cite{O}. We recall that the Fermat surface is exceptionally rational when its degree is $3$. Indeed when $q=2$ non-empty $T_d$ splits into infinitely many ${\rm Aut}(X_A)$-orbits, where ${\rm Aut}(X_A)$ is the projective automorphism group of a smooth $k$-Hermitian surface $X_A$ (see Remark \ref{rem1}).
Further we will see that all curves of C2, C3 are singular, although those of C1 are smooth (see Remark \ref{rem2}).
\begin{ex}
{\bf The case ${\bf q=2}$}: There are infinitely many nonplanar rational curves of degree $3$ and infinitely many nonplanar tetranomial curves of degree $6$ on a smooth $k$-Hermitian cubic surface, and their curves are projectively isomorphic over $k$ to $C(3,1,2)$ and $C(6,3,5)$, respectively.\\
{\bf The case ${\bf q=3}$}: There are exactly $18144$ nonplanar rational curves of degree $4$ and $1866240$ nonplanar tetranomial curves of degree $6$ on a smooth $k$-Hermitian quartic surface, and their curves are projectively isomorphic over $k$ to $C(4,1,3)$ and $C(6,2,5)$, respectively.\\
{\bf The case ${\bf q=4}$}: There are exactly $249600$ nonplanar rational curves of degree $5$ and $15667200$ nonplanar tetranomial curves of degree $20$ on a smooth $k$-Hermitian quintic surface, and their curves are projectively isomorphic over $k$ to $C(5,1,4)$ and $C(20,5,17)$, respectively.\\
{\bf The case ${\bf q=5}$}: There are exactly $1890000$ nonplanar rational curves of degree $6$ and $468000000$ nonplanar tetranomial curves of degree $15$ on a smooth $k$-Hermitian sextic surface, and their curves  are projectively isomorphic over $k$ to $C(6,1,5)$ and $C(15,3,13)$, respectively.
\end{ex}
In the next section, we give a necessary condition for which $T_d\neq\emptyset$, and determine the curves in $T_d$ up to $k$-projective isomorphisms. In Section $3$, we associate the set $T_d$ modulo ${\rm Aut}(X_A)$ to a certain set consisting of matrices of rank $4$ by introducing some notations, and show that there is a one-to-one correspondence between them. Moreover we deduce that the number of  ${\rm Aut}(X_A)$-orbits is exactly $1$ if $q\ge3$ otherwise an infinite number. In Section $4$, by computing the stabilizer of a curve in $T_d$ for $q\ge3$, we complete the proof of Theorem \ref{mainthm}.

\section{A condition for which $T_d\neq\emptyset$}
Let $s,t$ be homogeneous coordinates of $\mathbb{P}^1$. We write the vector
\begin{equation*}
\lt(s^{d},s^{d-i}t^{i},s^{d-j}t^{j},t^{d})\in\mathbb{P}^3\ \ {\rm with}\ 1\le i<j\le d-1
\end{equation*}
as $\bm{v}(d,i,j)$ shortly. Let $B$ and $F$ be $4$-by-$4$ invertible matrices with entries in $k$. We
denote by $f_B(d,i,j)$ and $C_F(d,i,j)$ a homogeneous $(q+1)d$-form $\lt\bm{v}(d,i,j)\,B\,\bm{v}(d,i,j)^{(q)}$ and a tetranomial curve having a parametrization given by $F\bm{v}(d,i,j)$, respectively. It is obvious that $C_F(d,i,j)$ is contained in a smooth $k$-Hermitian surface $X_A$ if and only if $f_{\lt FAF^{(q)}}(d,i,j)$ is identically $0$ for all $s,t$. For a positive integer $n$ and the $4$-by-$4$ anti-diagonal matrix $J$ whose entries are entirely $1$, we identify $4$-tuples $(dn,in,jn,B)$ and $(d,d-j,d-i,JBJ)$ with $(d,i,j,B)$ because both $C(dn,in,jn)$ and $C_J(d,d-j,d-i)$ become $C(d,i,j)$ by the changes of coordinates $(s,t)\mapsto(s^{1/n},t^{1/n})$ and $(s,t)\mapsto(t,s)$, respectively. Then we will prove the following lemma:
\begin{lem}\label{mat}
If $f_B(d,i,j)$ is identically $0$ for all $s,t$ then $(d,i,j,B)$ is equal to one of the following three cases:\\
\begin{itemize}
\item[I.] $d=q+1$, $i=1$, $j=q$.
\begin{equation*}
B=
\begin{pmatrix}
0&a_{11}&a_{12}&a_{13}\\
0&a_{21}&a_{22}&a_{23}\\
-a_{11}&-a_{12}&-a_{13}&0\\
-a_{21}&-a_{22}&-a_{23}&0\\
\end{pmatrix},
\end{equation*}
where the first two rows are $k$-linearly independent,
$(a_{11},a_{21})$, $(a_{13},a_{23})\neq(0,0)$, and if $q\ge3$ then $a_{12}=a_{22}=0$.\\
\item[II.] $d=(q+1)q$, $i=q+1$, $j=q^2+1$ for $q$ even.\\
\begin{equation*}
B=
\begin{pmatrix}
0&b_{1}&0&b_{2}\\
0&0&0&b_{3}\\
0&0&-b_{3}&0\\
-b_{1}&-b_{2}&0&0\\
\end{pmatrix},
\end{equation*}
where $b_1,b_3\neq0$, and if $q\ge3$ then $b_2=0$.
\item[III.] $d=(q+1)q/2$, $i=(q+1)/2$, $j=(q^2+1)/2$ for $q$ odd.\\
The matrix $B$ is the same with the case {\rm II}.
\end{itemize}
\end{lem}
\begin{proof}
The exponents of $t$ appearing in $f_B(d,i,j)$ are formed in the following matrix:
\begin{equation*}
E=
\begin{pmatrix}
0&qi&qj&qd\\
i&(q+1)i&i+qj&i+qd\\
j&qi+j&(q+1)j&j+qd\\
d&qi+d&qj+d&(q+1)d
\end{pmatrix}.
\end{equation*}
Here denoting the entries of $B$ by $B_{lm}$ and those of $E$ by $E_{lm}$, $B_{lm}$ indicates the coefficient of the monomial containing $t^{E_{lm}}$. Then by assumption, there are some linear expressions $F_{lml'm'}:=E_{lm}-E_{l'm'}$ with respect to $d,i,j$ such that $F_{lml'm'}=0$, and then $B_{lm}+B_{i'm'}=0$. Hence if we choose three $F_{lml'm'}$'s to be linearly independent over rational numbers then $(d,i,j)=(0,0,0)$, but this is a contradiction. On the other hand, if one chooses two linearly independent $F_{lml'm'}$'s then $d,j$ can be represented by only $i,q$.
In fact we can choose two appropriate $F_{lml'm'}$'s as follows: First of all, note that one can exclude $E_{11},E_{21},E_{34},E_{44}$ since these are not equal to any other entries of $E$, and furthermore $E_{31}$ and $E_{24}$ can be equal to only $E_{12}$ or $E_{22}$ and $E_{33}$ or $E_{43}$, respectively. When one puts $F_{3112}=0$ and $F_{2443}=0$ as $E$ to be rank $4$, one has $d=(q+1)i$, $j=qi$, $F_{4122}=0$, $F_{1433}=0$, and furthermore if $q=2$ then $F_{3213}=0$, $F_{4223}=0$. Then $B$ and $f_B(d,i,j)$ become a form of the case I under the identification $f_B((q+1)i,i,qi)=f_B(q+1,1,q)$. Similarly, putting $F_{4112}=0$ and $F_{3324}=0$, one has $d=qi$, $(q+1)j=(q^2+1)i$, and furthermore if $q=2$ then $F_{4214}=0$, and thus $B$ is of the case II. Noting that $\gcd(q^2+1,q+1)=1$ or $2$ by whether $q$ is even or odd, respectively, one has $d=(q+1)qn$, $i=(q+1)n$, $j=(q^2+1)n$ for the former and $d=(q+1)qn/2$, $i=(q+1)n/2$, $j=(q^2+1)n/2$ for the latter, where $n$ is a positive integer. By the identification, each of them becomes a form of the case II or III. One can easily verify that all possible choices of $(d,i,j,B)$ up to the identification become the case I, II or III, and thus the lemma has been proven.
\end{proof}
We quote the following important proposition (cf. e.g. \cite{Be} and \cite[Proposition 2.5.]{Sh} for a proof):
\begin{prop}\label{prop1}
For each $A\in{\rm GL}_n(k)$, there is an element $B\in{\rm GL}_n(k)$ such that $A={}^\mathrm t\hspace{-0.5mm} BB^{(q)}$. If $\lt A=A^{(q)}$ then $B$ is defined over $\mathbb{F}_{q^2}$.
\end{prop}
By Lemma \ref{mat} and Proposition \ref{prop1}, if $T_d\neq\emptyset$, all curves in $T_d$ are projectively isomorphic over $k$ to $C(q+1,1,q)$, $C(q(q+1),q+1,q^2+1)$ or $C(q(q+1)/2,(q+1)/2,(q^2+1)/2)$.
\begin{rem}\label{rem2}
Every curve $C\in T_{q+1}$ is smooth. In fact, $C$ is projectively isomorphic over $k$ to the curve $C(q+1,1,q)$ whose defining equations are presented as
\begin{equation*}
{x_1}^q={x_0}^{q-1}x_2,\ \ \ {x_2}^{q}={x_1}{x_3}^{q-1},\ \ \ x_1x_2=x_0x_3
\end{equation*}
and whose Jacobian matrix always has rank $2$. Hence $C(q+1,1,q)$ is smooth, and thus so is $C$.
On the other hand, all curves in $T_{q(q+1)}$, $T_{q(q+1)/2}$ are singular, because every $C'\in T_{q(q+1)}$ is projectively isomorphic over $k$ to the curve $C((q+1)q,q+1,q^2+1)$ whose defining equations are presented as
\begin{equation*}
{x_1}^q={x_0}^{q-1}x_3,\ \ \ {x_2}^{q+1}={x_1}{x_3}^q,\ \ \ (x_1x_2)^{q(q+1)/2}={x_0}^{(q+2)(q-1)/2}{x_3}^{(q^2+q+2)/2}
\end{equation*}
and whose Jacobian matrix has rank $1$ at $\lt(0,0,0,1)$. Hence $C((q+1)q,q+1,q^2+1)$ has a singular point, and thus so does $C'$. The case of $T_{q(q+1)/2}$ is the same too.
\end{rem}

\section{The action of ${\rm Aut}(X_A)$ on $T_d$}
We denote the set of the matrices $B$ of I, II, III of Lemma \ref{mat} by $\mathcal{B}_1$, $\mathcal{B}_2$, $\mathcal{B}_3$, respectively.
Further we denote by $Q'_1$ the set of $(q+2)$-by-$(q+2)$ matrices whose entries are $0$ except for the $(i,j)$-entries with $i,j\in\{1,2,q+1,q+2\}$ which form a submatrix belonging to $\mathcal{B}_1$, and for $q$ even, denote by $Q'_2$ the set of $((q+1)q+1)$-by-$((q+1)q+1)$ matrices whose entries are $0$ except for the $(i,j)$-entries with $i,j\in\{1,q+2,q^2+2,(q+1)q+1\}$ which form a submatrix belonging to $\mathcal{B}_2$, and for $q$ odd, denote by $Q'_3$ the set of $((q+1)q/2+1)$-by-$((q+1)q/2+1)$ matrices whose entries are $0$ except for the $(i,j)$-entries with $i,j\in\{1,(q+1)/2+1,(q^2+1)/2+1,(q+1)q/2+1\}$ which form a submatrix belonging to $\mathcal{B}_3$.

For each $M\in Q'_i$ we denote by $M^*$ the $4$-by-$4$ submatrix of $M$ belonging to $\mathcal{B}_i$ for $i=1,2,3$. We also use the same symbol for the inverse correspondence, namely $M^{**}=M$. Then $M\in Q'_i$ if and only if $M^*\in\mathcal{B}_i$.

For a positive integer $d$, we denote by ${\rm Im}(\varphi_d)$ the image of the group homomorphism
$\varphi_d:{\rm GL}_2(k)\ni g\mapsto\varphi_d(g)\in{\rm GL}_{d+1}(k)$ defined by
\begin{equation*}
(u^d,u^{d-1}v,\dots,v^d)=\varphi_d(g)\,\lt(s^d,s^{d-1}t,\dots,t^d)\ {\rm for} \ \lt(u,v)=g\,\lt(s,t).
\end{equation*}

We denote the set of $m$-by-$n$ matrices with entries in $k$ by $\mathcal{M}_{m,n}$, particularly by $\mathcal{M}_m$ if $m=n$. Putting $d_1=q+1$, $d_2=(q+1)q$, $d_3=(q+1)q/2$, we write $\varphi_1$, $\varphi_2$, $\varphi_3$ instead of $\varphi_{d_1}$, $\varphi_{d_2}$, $\varphi_{d_3}$, respectively.

For $i=1,2,3,$ the sets $Q_i$ are defined by
\begin{eqnarray*}
Q_i&:=&\left\{M\in\mathcal{M}_{d_i+1}\ \left|\ \lt\varphi_i(g)M\varphi_i(g)^{(q)}\in Q'_i\ {\rm for}\ {\rm some}\ g\in{\rm GL}_2(k)\right\}\right..
\end{eqnarray*}
By definition $Q'_i\subset Q_i$, and ${\rm Im}(\varphi_i)$ acts on $Q_i$ as $M\mapsto\lt\varphi_i(g)M\varphi_i(g)^{(q)}$. We denote by $\sim$ the equivalence relation induced by this action, and denote by $M^{\varphi_i}$ the equivalence class of $M\in Q_i$.

Further for $i=1,2,3$ we define the sets $S_i$ by
\begin{eqnarray*}
S_i&:=&\left\{F\in\mathcal{M}_{4,d_i+1}\ \left|\ \lt FAF^{(q)}\in Q_i\right\}\right..
\end{eqnarray*}
By definition, ${\rm Im}(\varphi_i)$ acts on $S_i$ as $F\mapsto F\varphi_i(g)$, and then notice that the set $T_{d_i}$ is in one-to-one correspondence with $k^{\times}\backslash S_i/{{\rm Im}(\varphi_i)}$, where $k^{\times}=k\setminus\{0\}$ (cf. \cite[Section 3]{O}).

We consider the action of the group ${\rm Aut}(X_A)$ on $T_{d_i}$, where ${\rm Aut}(X_A)$ is represented by
\begin{equation*}
\{P\in{\rm GL}_4(k)\ |\ {}^\mathrm t PAP^{(q)}=\lambda A,\ \lambda\in k^{\times}\}/k^{\times}I.
\end{equation*}
Then ${\rm Aut}(X_A)$ acts on $T_{d_i}$ by acting on $k^{\times}\backslash S_i/{{\rm Im}(\varphi_i)}$ by the multiplication from the left.

\begin{lem}\label{bico}
For $i=1,2,3$, there is the following bijection:
\begin{equation*}
\begin{array}{ccc}
{\rm Aut}(X_A)k^{\times}\backslash S_i/{{\rm Im}(\varphi_i)}&\longrightarrow& k^{\times}\backslash Q_i/\sim\\
\rotatebox{90}{$\in$}&&\rotatebox{90}{$\in$}\\
{\rm Aut}(X_A)k^{\times}F\,{{\rm Im}(\varphi_i)}&\longmapsto&k^{\times}({}^\mathrm t\hspace{-0.5mm}{F}A{F}^{(q)})^{\varphi_i}.
\end{array}
\end{equation*}
\end{lem}
\begin{proof}
It is obvious by definition that the map is well-defined.\\
{\bf Surjectivity}: For every $k^{\times}M^{\varphi_i}\in k^{\times}\backslash Q_i/\sim$, we can choose $M\in Q_i$ such that $M\in Q'_i$, if necessary, by acting an appropriate $\varphi_i(g)$ on $M$. Then by definition $M^*\in\mathcal{B}_i\subset{\rm GL}_4(k)$, and also $A\in{\rm GL}_4(k)$. Hence by Proposition \ref{prop1}, there are $G,H\in{\rm GL}_2(k)$ such that $M^*=\lt GG^{(q)}$ and $A=\lt HH^{(q)}$, and putting $F^*=H^{-1}G$, one has $\lt F^*A{F^*}^{(q)}=M^*$.
Then we may take $F\in S_i$ such that $\lt FAF^{(q)}=M$ by placing $4$ columns of $F^*$ into the appropriate columns of $F$. \\
{\bf Injectivity}: For $F_1,F_2\in S_i$ such that $k^{\times}(\lt F_1A{F_1}^{(q)})^{\varphi_i}=k^{\times}(\lt F_2A{F_2}^{(q)})^{\varphi_i}$ with $\lt F_1A{F_1}^{(q)}, \lt F_2A{F_2}^{(q)}\in Q'_i$, we have
\begin{equation*}
\lambda\,{\lt\varphi_i(g)^*}\lt {F_1}^{*}A{{F_1}^{*}}^{(q)}{\varphi_i(g)^*}^{(q)}=\lt {F_2}^{*}A{{F_2}^{*}}^{(q)}\ \ {\rm for}\ \lambda\in k^{\times},\ g\in{\rm GL}_2(k),
\end{equation*}
where ${F_1}^*,{F_2}^*,\varphi_i(g)^*\in{\rm GL}_4(k)$ such that
\begin{equation*}
{\lt\varphi_i(g)^*}\lt {F_1}^{*}A{{F_1}^{*}}^{(q)}{\varphi_i(g)^*}^{(q)}=\left({\lt\varphi_i(g)}\lt {F_1}A{{F_1}}^{(q)}{\varphi_i(g)}^{(q)}\right)^*
\end{equation*}
and
$\lt {F_2}^{*}A{{F_2}^{*}}^{(q)}=(\lt F_2A{F_2}^{(q)})^*$. Then $k^{\times}{{F_1}^{*}}\varphi_i(g){{F_2}^*}^{-1}\in{\rm Aut}(X_A)$, and thus $k^{\times}{{F_1}^{*}}\varphi_i(g)=k^{\times}P{F_2}^*$ for some $k^{\times}P\in{\rm Aut}(X_A)$. Therefore
\begin{equation*}
{\rm Aut}(X_A)k^{\times}F_1\,{{\rm Im}(\varphi_i)}={\rm Aut}(X_A)k^{\times}F_2\,{{\rm Im}(\varphi_i)}.
\end{equation*}
This means that the map is injective.
\end{proof}
\begin{lem}\label{lem3}
\begin{equation*}
|k^{\times}\backslash Q_i/\sim|=
\begin{cases}
\infty&\ {\rm for}\ q=2,\ i=1,2,\\
1&\ {\rm for}\ q\ge3,\ i=1,2,3,
\end{cases}
\end{equation*}
and furthermore
when $q\ge3$ one can take a representative $M_i$ of $k^{\times}\backslash Q_i/\sim$ such that\\
\begin{equation*}
{M_1}^*=
\begin{pmatrix}
0&1&0&0\\
0&0&0&1\\
-1&0&0&0\\
0&0&-1&0\\
\end{pmatrix},
\ \ \ \
{M_2}^*={M_3}^*=
\begin{pmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&-1&0\\
-1&0&0&0\\
\end{pmatrix}.
\end{equation*}
\end{lem}
\begin{proof}
For every $k^{\times}M^{\varphi_i}\in k^{\times}\backslash Q_i/\sim$, acting an appropriate $\varphi_i(g)$ on $M$ if necessary, one may assume that $M\in {Q'_i}$, and then $M^*\in\mathcal{B}_i$. We refer to the entries of $M^*$ as $a_{ij}(M^*)$ or $b_i(M^*)$ by following the description of the matrices $B$ of Lemma \ref{mat}.\\
{\bf The case ${\bf q\ge3}$:}
In this case, $(a_{12}(M^*),a_{22}(M^*))=(0,0)$ and $b_2(M^*)=0$. When $i=1$ has been already proven in \cite{O}.
When $i=2,3$, there is a matrix
\begin{equation*}
g=
\begin{pmatrix}
\lambda&0\\
0&\mu\\
\end{pmatrix}
\in {\rm GL}_2(k)
\end{equation*}
such that $\lt\varphi_i(g)M{\varphi_i(g)}^{(q)}={M_i}$. In fact, this implies that
\begin{equation*}
b_1(M^*)\lambda^{d_iq}\mu^{d_i}=1\ \ {\rm and}\ \ b_3(M^*)\lambda^{(d_i-j)(q+1)}\mu^{j(q+1)}=1,
\end{equation*}
where $j=q^2+1$ or $(q^2+1)/2$ by whether $i=2$ or $3$, respectively, and solutions $\lambda,\mu$ of above equations can be found in $k$ since $k$ is algebraically closed. This completes the proof for $i=2,3$.\\
{\bf The case ${\bf q=2}$:} When $i=1$, we consider the following matrices:
\begin{equation*}
M^*(x)=
\begin{pmatrix}
0&x&1&0\\
0&1&0&1\\
x&1&0&0\\
1&0&1&0
\end{pmatrix}
\in\mathcal{B}_1\ \ {\rm for}\ x\in k.
\end{equation*}
Then one has by a direct calculation that if $k^{\times}M(x)^{\varphi_1}=k^{\times}M(x')^{\varphi_1}$ then $x=x'$. Therefore $M(x)$ and $M(x')$ with $x\neq x'$ belong to distinct classes of $k^{\times}\backslash Q_1/\sim$. Since there is an infinite number of choices of $x,x'$, we conclude that $|k^{\times}\backslash Q_1/\sim|=\infty$.
When $i=2$, for all $M\in Q'_2$ we may assume that $b_1(M^*)=b_3(M^*)=1$ by the same argument as $q\ge3$. Then a direct computation shows that if $k^{\times}M^{\varphi_2}=k^{\times}N^{\varphi_2}$ for $M,N\in{Q'_2}$
then $b_{2}(M^*)=b_{2}(N^*)\xi$, where $\xi$ is a cube root of unity. Hence $M$ and $N$ in $Q'_2$ such that $b_{2}(M^*)\neq b_{2}(N^*)\xi$ belong to distinct classes of $k^{\times}\backslash Q_2/\sim$, and thus
$|k^{\times}\backslash Q_2/\sim|=\infty$.
\end{proof}
\begin{rem}\label{rem1}
As we have seen in the above proof,
when $q=2$, for $i=1,2$ the set $k^{\times}\backslash Q_i/\sim$, and thus the set $T_{d_i}$ modulo ${\rm Aut}(X_A)$, is one-dimensional over $k$. That is, $k^{\times}\backslash Q_2/\sim$ is in one-to-one correspondence with $k/\langle\omega\rangle$, where $\langle\omega\rangle$ denotes the cyclic group generated by a primitive cube root $\omega$ of unity. Moreover denoting a matrix $B$ of the case I of Lemma \ref{mat} by
\begin{equation*}
\begin{pmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}
\end{pmatrix},
\end{equation*}
a complete set of representatives of $k^{\times}\backslash Q_1/\sim$ is presented as follows:
\begin{equation*}
\begin{pmatrix}
1&0&0\\
0&0&1
\end{pmatrix},
\ \
\begin{pmatrix}
0&1&1\\
1&0&0
\end{pmatrix},
\ \
\begin{pmatrix}
1&1&0\\
0&0&1
\end{pmatrix},
\ \
\begin{pmatrix}
\lambda&1&0\\
1&0&1
\end{pmatrix}
\ \ {\rm for\ all}\ \lambda\in k.
\end{equation*}
By Lemmas \ref{bico} and \ref{lem3} one sees immediately that $|T_{d_1}|,|T_{d_2}|=\infty$ for $q=2$, and when $q\ge3$, ${\rm Aut}(X_A)$ acts transitively on $T_{d_1},T_{d_2},T_{d_3}$. Here the proof of the case $q=2$ of Theorem \ref{mainthm} has been completed.
\end{rem}

\section{the stabilizer of a curve}
It remains to determine the number of curves in $T_{d_i}$ of the case $q\ge3$ and $i=2,3$. In this case one has that
\begin{equation*}
|T_{d_i}|=|{\rm Aut}(X_A)|/|{\rm Stab}(C_i)|,
\end{equation*}
where ${\rm Stab}(C_i)=k^{\times}\backslash\varGamma(C_i)$ is the subgroup of ${\rm Aut}(X_A)$ stabilizing the curve $C_i$ corresponding to $k^{\times}F_i{{\rm Im}(\varphi_i)}$ such that $\lt F_iA{F_i}^{(q)}=M_i$ for $M_i$ of Lemma \ref{lem3}. We have already known that
${\rm Aut}(X_A)\simeq{\rm PGU}_4(\mathbb{F}_{q^2})$ cf. \cite{O}, and thus
\begin{equation*}
|{\rm Aut}(X_A)|=q^6(q^4-1)(q^3+1)(q^2-1).
\end{equation*}
\begin{lem}\label{lemstb}
The group ${\rm Stab}(C_2)$ is a cyclic group of order $q^3+1$, and ${\rm Stab}(C_3)$ is a cyclic group of order $(q^3+1)/2$ or $(q^3+1)/4$ by whether $q$ modulo $4$ is $1$ or $3$, respectively.
\end{lem}
\begin{proof}
For $i=2,3$ and $\gamma\in\varGamma(C_i)$, there are $\lambda',\lambda''\in k^{\times}$ and $g\in{\rm GL}_2(k)$ such that $\lt\gamma A\gamma^{(q)}=\lambda'A$ and $\gamma=\lambda''{{F_i}^*} \varphi_i(g)^*{{F_i}^*}^{-1}$ with $\lt {F_i}^*A{{F_i}^*}^{(q)}={M_i}^*$.
Hence one has
\begin{equation}\label{eq1}
\lt\varphi_i(g)^*{M_i}^*{\varphi_i(g)^*}^{(q)}=\lambda {M_i}^*\ \ {\rm for\ some}\ \lambda\in k^{\times}.
\end{equation}
We will find a matrix $g$ satisfying this equation. For $i=2$, the matrix $\varphi_i(g)^*$ is of the following form:
\begin{equation*}
\begin{pmatrix}
a^{q^2+q}&0&0&b^{q^2+q}\\
a^{q^2-1}c^{q+1}&\alpha&\beta&b^{q^2-1}d^{q+1}\\
a^{q-1}c^{q^2+1}&\gamma&\delta&b^{q-1}d^{q^2+1}\\
c^{q^2+q}&0&0&d^{q^2+q}\\
\end{pmatrix}
\ \
{\rm for}
\
g=
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix},
\end{equation*}
where
\begin{equation*}
\begin{array}{lll}
\alpha=a^{q^2-q-2}\det(g)^{q+1},&\beta=a^{q-2}b^{q^2-q}d^{q}\det(g),\\
\gamma=0,&\delta=a^{q-2}d^{q^2}\det(g).
\end{array}
\end{equation*}
In the equation \eqref{eq1}, multiplying $g$ by an appropriate element of $k^{\times}$ we may assume that $\lambda=1$. Then by a straightforward computation, it turns out that $g=(a,b;c,d)$ satisfies \eqref{eq1} if and only if $b=c=0$, $a=\xi^{nq^2}$ and $d=\xi^n$, where $n$ is an integer such that $1\le n\le(q^3+1)(q+1)$ and $\xi$ is a primitive $(q^3+1)(q+1)$-th root of unity. Further one has that $\xi^{nq^2}=\xi^n$ if and only if $n$ is a multiple of $q^3+1$. Hence there is an isomorphism
\begin{equation*}
\mathbb{Z}/(q^3+1)\mathbb{Z}\ni\overline{n}\longmapsto k^{\times}{{F_2}^*} \varphi_2((\xi^{\overline{n}q^2},0;0,\xi^{\overline{n}}))^*{{F_2}^*}^{-1}\in{\rm Stab}(C_2).
\end{equation*}
Similarly, for $i=3$ one has that $g=(a,b;c,d)$ satisfies \eqref{eq1} if and only if $b=c=0$, $a=\xi^{nq^2}$ and $d=\xi^n$, where $n$ is an integer such that $1\le n\le(q^3+1)(q+1)/2$ and $\xi$ is a primitive $(q^3+1)(q+1)/2$-th root of unity. Further one has that $\xi^{nq^2}=\xi^n$ if and only if $n$ is a multiple of $(q^3+1)/2$ or $(q^3+1)/4$ by whether $q$ modulo $4$ is $1$ or $3$, respectively. Hence there is an isomorphism from $\mathbb{Z}/2^{-1}(q^3+1)\mathbb{Z}$ or $\mathbb{Z}/4^{-1}(q^3+1)\mathbb{Z}$ to ${\rm Stab}(C_3)$ as described above.
\end{proof}
By Lemma \ref{lemstb} it follows immediately that $|T_{d_2}|=q^6(q^4-1)(q^2-1)$ and $|T_{d_3}|=2q^6(q^4-1)(q^2-1)$ or $4q^6(q^4-1)(q^2-1)$. Thus the proof of Theorem \ref{mainthm} has been completed.

redpajama_set_name: RedPajamaArXiv