url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.siranah.de/html/sail045d.htm | 1,695,595,504,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00203.warc.gz | 74,471,591 | 2,357 | ### Great-Circle Sailing
The following form calculates the location of the destination (L1) given a location of departure (L0), a travelled Distance (D) and Bearing (Z0) for the great-circle track to be sailed. Since great-circle sailing implies a track of non-constant course, the Bearing Z0 is the initial course at the departure.
The picture on the left below will show the situation and explain the abbreviations and terms used.
Location (L0) Distance and Bearing Latitude N/S dd-mm.m Distance (D) Nm Longitude E/W ddd-mm.m Bearing (Z0) ° Destination (L1) Latitude ° Longitude ° D: Great-Circle Distance from L0 to L1 Z0, Z1: Azimuth angles under which the great-circle track intersects with the local Meridian in L0 and L1. The Agles Z0 and Z1 are measured clockwise from the northern branch of the local Meridian (true North) in L0 and L1 respectively and towards the (shortest) great-cirlce segment connecting L0 and L1. This way the resulting Azimuth angles represent true bearings. These true bearings are the initial courses to be sailed to travel from L0 to L1 (B0) or reverse (B1). The great-circle track gives the shortest path between L0 and L1, but the course to be sailed is not constant. The Destination problem can also be solved for a loxodrome track (track of constant course). | 313 | 1,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.868646 |
https://slashdot.org/~TheSexican | 1,493,547,251,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124478.77/warc/CC-MAIN-20170423031204-00469-ip-10-145-167-34.ec2.internal.warc.gz | 828,357,772 | 29,123 | typodupeerror
DEAL: For \$25 - Add A Second Phone Number To Your Smartphone for life! Use promo code SLASHDOT25. Also, Slashdot's Facebook page has a chat bot now. Message it for stories and more. Check out the new SourceForge HTML5 Internet speed test! ×
## Submission + - Texas Instruments doesn't know Avogadro's Number?
lpq writes: Was watching Monk last night and some kid wrote Avogadro's number (or something that looked like it) on a blackboard. They wrote 6.0221415 * 10^23. Today I was trying to remember the extra digits (only carried around 3 digits of accuracy in brain...not a constant I need alot or alot of precision of, usually.
I remembered my Texas Instruments Solar calculator had it as a programmed in constant. All sources I've seen have the exponent as 23, but the number part varies a bit, source to source. The farthest "off" from the others is the one on my TI-36X calculator. Some values I've found:
(all scaled by 10^23, lowest to highest)
6.022 131 67 Texas Instruments (built-in to calculator):
6.022 141 5 Monk TV show (USA Network)
6.022 141 79 NIST (website)
6.022 141 79 Wikipedia Same as NIST
6.022 257 Lapeer County, MI ("http://chem.lapeer.org/Chem1Docs/MolExercise.htm l")
So why all the different answers? Has the value of Av's Num been fluctuating lately like the price of gold or the stock market? Are the other answers "older, accepted values?"
Google is "close" — if they had left it at 7 digits (6.033142) I would have thought it a rounding of the NIST value, but throwing in that 7th decimal place shoots that idea out of the water.
Seems like Lapeer County is most off from everyone else (varying by .00035), being on the high side, but a _CALCULATOR_ company?
TI getting their programmed in constants, "WRONG"?!? What faith should I have in any of their other constants or their formulae and calculations? They seem to be low by ".000 010 12", or to put things in obfuscating perspective:
1.012 x 10^18 or about 1 quintillion?
Is it common for calculator manufacturers to have such different values? Haven't checked other constants — maybe they are better, but it seems "concerning" (if I needed more precise constants, my calculator could be very misleading). It reminded me of the Pentium math bug where it returned the wrong answer on some calculations. We just accept these things as "right" or "correct".
Is quality control dropping? (Calculator made in China). Is it a "foreign-made" quality issue?
Thanks for any insights on why things are so odd....
## Submission + - Why No High-MPG Diesels For The U.S.?
gbulmash writes: "While looking for a high-MPG minivan, wagon, or SUV, I've been finding that the pickings in the U.S. are pretty slim, but that there are plenty of fuel-efficient diesel models in Europe that get even better mileage than some of the larger hybrids for sale in the U.S. With the U.S. having so many people driving so many miles, it seems ridiculous that even Ford is offering highly fuel efficient diesels in Europe that they don't/won't offer here. Is there an actual plausible reason why these models aren't being brought to American markets aside from "marketing objectives"?"
## Submission + - There is no HD DVD, Only DVD in BOTH Formats (betanews.com)
cybrthng writes: "In a single statement Collins made in the now famous "We're not at war with Bluray" article there was one nugget everyone failed to grasp and I quote: "Foremost is compatibility. All new movie titles from Universal and soon Warner will be combination (or twin-format) discs — HD DVD on one side and standard DVD on the other. This means that HD DVD discs will also play on older DVD players, which is crucial for portability.". No Distinction between HD-DVD or DVD means a single SKU with both movies and complete market domination over night. (Well, after re-tooling of fabs). Not only that but its one heck of a segway to allow people to adopt HD-DVD players at their own pace without risk of obsolescence."
## Submission + - Leave the country, get a \$1000+ iPhone bill?
elistan writes: US owners of the iPhone are subject to the standard ATT international rates. For example, in Canada they pay \$0.59 per minute and \$0.0195 per KB. The problem is that the iPhone is chatty. VERY chatty. Not only is there the normal data downloaded while browsing the "real Internet," it likes to do things like make multi-MB transmissions in the middle of the night while its owner is asleep. (Examine one of those multipage iPhone bills if you get a chance, look for large data sessions at odd hours.) 5000 KB would end up costing \$97.50 USD, for example. Traveling to another country could get very expensive very quickly, with the iPhone owner not realizing what's happening. Posts at MacRumors have started of people realizing just that issue. How does \$3000 sound? Can these users only blame themselves for not checking the iPhone usage more closely? Was it reasonable for them to assume that limiting their usage would involve only a reasonable fee? In either case, as it stands the iPhone is unsuitable as an internationally roaming phone. Are ATT's usage rates being outdated by Apple's new technology? Or is it Apple's responsibility to make the phone workable within ATT's fee structure?
## Submission + - Boson explains high-temperature superconductivity
kgb1001001 writes: Just saw this (http://www.engr.uiuc.edu/news/?xId=071508320770) in a newsletter the University of Illinois at Urbana-Champaign. Seems like the key to understanding high-temperature superconductivity might be a new elementary particle. The article discusses how they think that the new boson is an emergent phenomena — maybe you can't find everything by breaking things apart in a particle accelerator.
## Submission + - Multiformat Listening Test at 64kbps1
prospective_user writes: "Do you think you have good ears? Think again.
The community at Hydrogenaudio has prepared a Public Listening Test for comparison of the most popular audio codecs (AAC, Vorbis, and Microsoft's WMA included) in a battle to see how they stand at compressing audio at 64kbps.
Many of the participants right now have expressed their surprise at being unable to determine which is the original and which is the compressed version of 18 samples covering a vast amount of musical styles.
The results of this test (and other that are conducted at Hydrogenaudio) will be used by the developers of the codecs to further improve the "transparency" and let this kind of test be even harder.
Everyone is invited to participate and show how good your listening is!"
## Submission + - Do Not Call Registry gets wake-up call (networkworld.com) 2
coondoggie writes: "If you signed up for the federal or your state's Do Not Call Registry a few years ago, you might want to thing about refreshing it. Pennsylvanians this week got a wake up call, so to speak from the state's Attorney General Tom Corbett who kicked off a public awareness campaign designed to remind people what many have forgotten or never knew — that the 2002 law set registrations to expire after five years. That is of course unless you want to start hearing from those telemarketers as you sit down to dinner. Corbett said about 2 million people signed up in the immediate aftermath of the law taking effect and those who do not act by Sept. 15 will have their numbers dropped from the registry on Nov. 1. The Pennsylvania action is a reminder that the National Do Not Call Registry has a five year life span as well. The Federal Trade Commission is set to being a nation campaign in Spring 2008 to remind all US citizens to refresh their federal Do Not Call Registry standing. http://www.networkworld.com/community/node/18066"
## Submission + - Cooling Your House With Solar Heat
An anonymous reader writes: The German Fraunhofer research institute has created the spin-off company SorTech, which plans to produce air conditioning systems that are run by solar heat. This mind-boggling feat is achieved by a thermo-chemical process called sorption. The technology could help to satisfy the increasing energy consumption used for air conditioning. It seems to be a perfect application for using solar energy: Good efficiency is possible by avoiding a conversion from heat into electricity and back to cooling energy. It also does not need a long term energy storage system, because the energy needed for cooling spikes exactly at the time, when most solar energy is available.
## Submission + - Next-Gen Car Batteries Promise Longer Life (wired.com)
hzero writes: Firefly has replaced the lead plates found inside conventional batteries with a lead-impregnated foam made from carbon graphite ?- one of the few materials that can withstand the highly corrosive sulfuric acid inside batteries. The foam increases the surface area of lead inside the battery, delivering more power and slashing the recharge time, says Firefly CEO Ed Williams.
Equally important, Firefly's approach eliminates the crystals that can build up inside lead-acid batteries. Over time, those crystals reduce the amount of electricity a battery can hold, one of the major reasons electric and hybrid automakers have favored lithium-ion or nickel batteries, even though lead acid is less expensive.
"Our batteries will come back to their full capacity for years," says Williams
## Submission + - There are just 12 types of TV ads (slate.com)
prostoalex writes: "After watching hours of television for one year without skipping commercials, Seth Stevenson claims there are only 12 types of TV ads: "This slide show presents some recent ads exemplifying each of Gunn's 12 basic categories. With a little practice, you, too, will be ticking off the master formats during commercial breaks.""
## Submission + - Fire fighting beetle developed in Germany
Lisandro writes: "German researchers are developing a robotic fire-fighting "beetle". The robot, called OLE, will monitor large areas of forrest, discover fire sources and immediately report and fight them. Built after the pattern of the pill millipede, OLE can roll up into a ball when danger threatens, retracting its six legs."
## Submission + - MySpace finds 29,000 sex offenders among users
StonyandCher writes: MySpace has identified more than 29,000 registered sex offenders among those registered to use its site — more than four times what the company said in May it had found from an investigation, according to North Carolina Attorney General Roy Cooper.
"[The 29,000] includes just the predators who signed up using their real names and not the ones who failed to register or used fake names," Cooper said in the statement. Cooper is one of eight state attorneys general who asked MySpace in May to turn over the names of users who are registered sex offenders.
In May, MySpace reluctantly revealed it had uncovered 7000 sex offenders.
## Submission + - Wisconsin Orders Gas Station to Raise Gas Prices
hahafaha writes: "Raj Bhandari, a gas station owner in Wisconsin, offered a 2 cents/gallon discount for gas to seniors, and 3 cents/gallon to those that supported youth sports. However, the Wisconsin Department of Agriculture is threatening to penalize him for each discounted gallon, with the fine at a judge's discretion. According to the Department, he is violating Wisconsin's Unfair Sales Act, which requires stations to sell gas for about 9.2 percent more than the wholesale price."
## Submission + - How do you DMCA a tattoo off of a person?
dmn writes: Shannon Larratt (of the BMEzine fame) recently posted a challenge at modblog for someone to tattoo the recently published HD DVD key on them and test what happens if they go public. From the first post:
"Any site that lists the AACS key/crack is getting legal takedown notices if someone gets it tattooed, would a lawsuit forcing them to never show it to anyone be launched? Would they try and force a removal?".
Well someone did take the challenge. From the second post:
"So how do you DMCA a tattoo off of a person? Am I allowed to tell people that its a 09-f9-11-02-9d-74-e3-5b-d8-41-56-c5-63-56-88-c0 tattoo? Is a magazine permitted to print photos of this tattoo? Can the tattoo artist or shop (Good Faith Tattoos, Boston) be sued? Can the person wearing the tattoo be sued? Can a corporation force a medical procedure (a tattoo removal)? Can they force him to always wear a shirt? Well, Rich from The New Freedom has decided to be the bait!".
# Slashdot Top Deals
We don't know who it was that discovered water, but we're pretty sure that it wasn't a fish. -- Marshall McLuhan
Working... | 2,814 | 12,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-17 | latest | en | 0.931037 |
https://creativenumerology.com/tag/letting-go/ | 1,713,858,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818468.34/warc/CC-MAIN-20240423064231-20240423094231-00646.warc.gz | 160,541,739 | 21,929 | # WEEKLY FORECAST – February 26, 2023
TO CALCULATE YOUR NUMBER FOR 2023, simply add 7 to your month and day of birth. (Do NOT include your year of birth in this calculation).
Example birthday: April 26:
Month: 4
Day: 2+6 = 8
Year: 2+0+2+3 = 7
4 + 87 =19 (Keep adding until you get a single number).
1+9 = 10
1+0 = 1
In this example, the Year Number for 2023 is 1.
If you are in doubt about your Year Number, use the CONTACT FORM on this website to send me your month and day of birth. (Year of birth is unnecessary in this calculation).
Use your new Yearly Number to read your Monthly and Weekly forecasts throughout 2023. (Your Personal Year runs from calendar year to calendar year – not from birthday to birthday).
SCROLL DOWN TO YOUR PERSONAL NUMBER
WEEK 9 RUNS FROM SUNDAY, FEBRUARY 26 TO SATURDAY, MARCH 4.
NOTE 1: The energy of WEEK 9 involves the way we respond to whatever we are experiencing, and because February 2023 is a 9 MONTH in this 7 YEAR, the emphasis on realism and truth is stronger than ever…
2+0+2+3=7
February = 2
7+2=9.
The emotional vibrations of 9 shake our minds open and expand our awareness. The alternative is to use denial of reality to render our minds so narrow that vital information simply cannot enter. And now, all the emotions we denied in the process of denying reality are coming to the surface. Reality is sinking in, literally.
On Wednesday, WEEK 9 brings us into the 1 MONTH of March, the number of change and innovation, and the beginning of something new… or at least the potential for something new to emerge out of something old. This is a time to LET GO of what was – and work with what is instead.
2+0+2+3=7
March = 3
7+3 =10, and 1+0 = 1.
No matter what your situation happens to be, this is a time to dispense with denial and accept reality fully. This may be a matter of realizing and being grateful for what you’ve already got. Or it could mean facing what you’re up against and finding ways to deal with that. Whatever the case, this is a time to LET GO of any misconceptions that you are clinging to, so that progress is actually possible.
Meanwhile, how people respond and react to what happens this week is a major factor, especially when it comes to the re-emergence of old hostilities, resentments, and hatreds. Forgiveness is not easy, but it is vital to your own wellbeing. Forgiveness does not mean letting down your guard if someone cannot be trusted to treat you fairly. But it does mean accepting what happened, releasing yourself from the shock of that experience, reclaiming your life as your own, and moving on.
If you haven’t already read OUR 19/1 KARMIC RUT which I re-posted last week, WEEK 9 in this 1 MONTH would be a good time to do so. It explains a lot.
NOTE 2: you can gain additional insight this week by reading the 1, 4, 7, and 9 Forecasts as well as your own.
SCROLL DOWN TO YOUR PERSONAL NUMBER
PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you so much. DONATE HERE
WEEK BEGINNING SUNDAY, FEBRUARY 26, 2023
WEEKLY FORECAST / 1 YEAR
Surface appearances can be deceiving, so stay alert and try to picture your actions from someone else’s perspective. Take some time for yourself. Recognize the fact that you could not control a recent situation, and yet here you are fighting off the weight of the misplaced guilt it has produced. Be kind to yourself. You have a right to be who and where you are.
WEEKLY FORECAST / 2 YEAR
Teamwork and cooperation are needed. Only by acting in the best interests of all concerned, can you comfortably travel this slow, emotional, and down-to-Earth vibration. Sometimes, taking a back seat can place you in a more powerful position, but you will only be able to take advantage of this through a patient, considerate, practical, and humble frame of mind.
WEEKLY FORECAST / 3 YEAR
Be friendly to your own ideas, recognize their long-term potential, and trust in your abilities and intelligence. If you think you are lacking good ideas, it’s because you are shutting them down before you have figured out how to make them happen. Your Will is made of your feelings, and where there’s a Will, there’s a way. Be patient. A piece of the puzzle is not in place yet.
PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you so much. DONATE HERE
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with. From \$95.00
CREATIVE NUMEROLOGY YEAR BOOK – Your Journey Through The Cycles Of Time
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your own CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these extraordinary, unpredictable, evolutionary, and often frightening times, each book contains inspiring daily, weekly, and monthly forecasts for your specific yearly cycle. Your Year Book will be used and appreciated every single day for the entire year ahead..
This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
If you don’t know which Year Book to order, use the contact form on this website to send me your MONTH AND DAY OF BIRTH, or the birthdays of those you wish to buy for, and I’ll get back to you as quickly as possible.
Buy the classic numerology book: LIFE CYCLES: your emotional journey to freedom and happiness HERE (Available in PAPERBACK or KINDLE)
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter le livre de numérologie classique : CYCLES DE LA VIE : votre voyage émotionnel vers la liberté et le bonheur – ICI
All of my books are available in most countries that have Amazon. Just search CHRISTINE DELOREY. Here are some links:
INDIA
AUSTRALIA
U.K.
JAPAN
WEEKLY FORECAST / 4 YEAR
Barriers arise when different ideals, beliefs, or cultures refuse to accept each other. One such barrier can now be removed so that peace has a chance to develop. The first step is to genuinely desire a solution, which will be impossible if you are blinded by old judgments. Full acceptance of your current reality, including its long-term potential, for better or worse, will bring much insight.
WEEKLY FORECAST / 5 YEAR
Being able to love yourself is essential if you are to live freely and safely. This form of self-respect enables you to make moment-by-moment decisions which serve the best interests of everyone involved. Now, however, you must rescue this power by recognizing a certain mistake which you are in danger of making or repeating. Be determined to break an old pattern of behavior.
WEEKLY FORECAST / 6 YEAR
There is power in numbers! Two heads are better than one! Cooperation means working together! You can achieve more as a team than you can alone! Peace is a two-way street! There are two sides to every story… Let these little clichés help you to realize how peaceful life can be when everyone’s needs are being met, and everyone’s story is being heard.
PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you so much. DONATE HERE
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with. From \$95.00
CREATIVE NUMEROLOGY YEAR BOOK – Your Journey Through The Cycles Of Time
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your own CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these extraordinary, unpredictable, evolutionary, and often frightening times, each book contains inspiring daily, weekly, and monthly forecasts for your specific yearly cycle. Your Year Book will be used and appreciated every single day for the entire year ahead.
This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
KINDLE \$9.99
If you don’t know which Year Book to order, use the contact form on this website to send me your MONTH AND DAY OF BIRTH, or the birthdays of those you wish to buy for, and I’ll get back to you as quickly as possible.
Buy the classic numerology book: LIFE CYCLES: your emotional journey to freedom and happiness HERE (Available in PAPERBACK or KINDLE)
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter le livre de numérologie classique : CYCLES DE LA VIE : votre voyage émotionnel vers la liberté et le bonheur – ICI
All of my books are available in most countries that have Amazon. Just search CHRISTINE DELOREY. Here are some links:
INDIA
AUSTRALIA
U.K.
JAPAN
WEEKLY FORECAST / 7 YEAR
Things cannot stay as they are, and this is a chance to recognize the fact that reality is constantly changing, and that previous expectations or beliefs are no longer realistic. You can now benefit from a recent change of mind or circumstance. Issues from the past are affecting everything in the present and cannot be avoided. Ongoing adjustment to your plans can help you feel freer and more optimistic than you thought possible.
WEEKLY FORECAST / 8 YEAR
Guilt is telling you that your ego must be kept in check when, really, all you are doing is trying to take back some of your own power. But guilt does not want you to feel powerful – it wants you to feel guilty! As you wait for deeper understanding, treat others as you would like to be treated, and remember that if you hold on to the past and try to move forward at the same time, all you can achieve is standstill.
WEEKLY FORECAST / 9 YEAR
This cycle cannot be rushed because it is made of present time. But you can look back into the past because present time is a reflection of how the past has unfolded. Slow right down! Whatever you are feeling is an indication of what you want – or don’t want. Your understanding of cause-and-effect will grow as you find connections between what is happening now, how it came to be – and what you can turn it into.
PLEASE SUPPORT THIS WORK by making a donation – large or small. This will help to offset the costs of this ad-free site and keep it available to everyone. Thank you so much. DONATE HERE
WHO ARE YOU? WHY ARE YOU HERE? WHERE ARE YOU GOING? This profile will help you to understand your purpose for being here at this precarious evolutionary time in the human journey, what you have to work on, and what you have to work with. From \$95.00
CREATIVE NUMEROLOGY YEAR BOOK – Your Journey Through The Cycles Of Time
You can get a whole year of Monthly/Weekly/Daily Forecasts (with all the calculations made for you) in your own CREATIVE NUMEROLOGY YEAR BOOK. Written specifically to help you steer your way through these extraordinary, unpredictable, evolutionary, and often frightening times, each book contains inspiring daily, weekly, and monthly forecasts for your specific yearly cycle. Your Year Book will be used and appreciated every single day for the entire year ahead..
This is a 9-book collectable set. Buy one book – or more – or buy all 9 books and get one free. (Free Book does not apply to Kindle).
KINDLE \$9.99
If you don’t know which Year Book to order, use the contact form on this website to send me your MONTH AND DAY OF BIRTH, or the birthdays of those you wish to buy for, and I’ll get back to you as quickly as possible.
Buy the classic numerology book: LIFE CYCLES: your emotional journey to freedom and happiness HERE (Available in PAPERBACK or KINDLE)
LIFE CYCLES – FRENCH VERSION
LES CYCLES DE LA VIE – Broché
Acheter le livre de numérologie classique : CYCLES DE LA VIE : votre voyage émotionnel vers la liberté et le bonheur – ICI
All of my books are available in most countries that have Amazon. Just search CHRISTINE DELOREY. Here are some links:
INDIA
AUSTRALIA
U.K.
JAPAN
A POEM FOR THESE TIMES – LET THERE BE PEACE
Join me on
MASTODON | 3,001 | 12,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.906004 |
https://betterthisworld.com/uncategorized/the-contents-of-a-cash-basis-balance-sheet/ | 1,713,020,855,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00222.warc.gz | 124,629,379 | 41,657 | Connect with us
Uncategorized
# The contents of a cash basis balance sheet
Published
on
This blog post is a summary of a Cash basis accounting blog post. It describes briefly the contents of a Cash basis balance sheet.
Cash basis accounting is a useful accounting method for companies that are constantly changing their inventory into cash. The process of moving inventory into cash and then into the book of accounts is the basis of a cash basis balance sheet.
Home » Bookkeeping » Rate Of Return Calculator
Jul 13, 2020
The expected return for an investment portfolio is the weighted average of the expected return of each of its components. Components are weighted by the percentage of the portfolio’s total value that each accounts for. Examining the weighted average of portfolio assets can also help investors assess the diversification of their investment portfolio.
Basics of Probability Distribution
Based on the expected return formula an investor can decide whether to invest in an asset based on the given probable returns. Rate of Return – An asset’s rate of return measures how much money you, as an investor, would have made or lost had you invested in this asset over a specific period of time.
What Is Expected Return?
So the cash flow that you receive, this is DIV1 plus P1, then you subtract P sub 0, this is what you’ve invested, and then you divide that by P sub 0. So basically, if this stock does not pay a dividend, then your only expected cash inflow is, if P1 is greater than P sub 0, you decide to sell and then you get some money. If you’re forced to sell for whatever reason and there’s no dividend, P1 is smaller than P sub 0, then you’ll lose money. If, for example you opt not to sell at all at this point, but you receive all of the dividend, then this is your total yield.
And also you can, right after receipt of this dividend, you can sell the stock at the price P1. So what is your expected return on this kind of investment? Well, it’s pretty clear, your expected return is equal to.
Imagine that an investor is considering buying Microsoft stock. She starts by researching the risk-free rate and finds that the one year U.S. Next, she learns that the stock trades on the Nasdaq, which investors believe will generate annual returns of 12 percent. Finally, she estimates that Microsoft’s stock has a beta of 1.2, meaning that on average, when the Nasdaq gains 1 percent, Microsoft’s stock gains 1.2 percent.
In other words, it is a percentage by which the value of investments is expected to exceed its initial value after a specific period of time. The expected rate of return can be calculated either as a weighted average of all possible outcomes or using historical data of investment performance. Now we are shifting focus and are moving toward stocks. Well, clearly the stock market is huge and it’s extremely important to come up with certain evaluation approaches for stocks.
But for now, we just study stocks with respect to their cash flows. Now, in order to proceed, let’s start with the most simplistic example. Imagine the one period project, so you bought a share of stock at P sub 0 here, at point one.
And the total of 13%, which is a very nice return but clearly this is just an example. Now, what we did so far seems to have nothing to do with PV. We just introduced the most simplistic approaches to expected returns and we analyzed stock cash flows.
It is computed as the expected return divided by the amount invested. It is important to understand the concept of a portfolio’s expected return as it is used by investors to anticipate the profit or loss on an investment.
Real World Example of Expected Return
Hence the expected return calculation is based on historical data and hence may not be reliable in forecasting future returns. It can be looked at as a measure of various probabilities and the likelihood of getting a positive return on one’s investment and the value of that return.
How Investment Risk Is Quantified
Investors sometimes speak of a required rate of return, which is the minimum expected rate of return for a particular investment decision to make sense.
An investor might decide that, given a particular investment’s riskiness and other factors, a minimum required rate of return is 5 percent, 10 percent or something higher.
The expected return doesn’t just apply to a single security or asset. It can also be expanded to analyze a portfolio containing many investments. If the expected return for each investment is known, the portfolio’s overall expected return is a weighted average of the expected returns of its components. This Stock Investment Calculator will calculate the expected rate of return given a stock’s current dividend, the current price per share, and the expected growth rate.
But it doesn’t require a lot of mental advancement to move from this formula to the one that will look very much like the PV formula for the stocks. And from that, we will proceed and come very close to the general stock price and formula. Calculating expected return is not limited to calculations for a single investment.
But expected rate of return is an inherently uncertain figure. As an investor you calculate it by assuming that the asset’s growth and yield in the past will continue unabated into the future. If your stock returned dividends in the past year, it will continue to pay those dividends in future years.
Expected Return Template
Investors sometimes speak of a required rate of return, which is the minimum expected rate of return for a particular investment decision to make sense. This can be based upon the relative rate of return of other, safer investments. An investor might decide that, given a particular investment’s riskiness and other factors, a minimum required rate of return is 5 percent, 10 percent or something higher. Using those expected rates and other information such as relative risk and liquidity of investments, investors can make decisions about where to put their money.
Now, the right to vote is very important, but we ignore it for now. Because we will talk about that in the fourth course of this specialization when we talk about [INAUDIBLE].
For example, say you assume a \$1,000 investment in a stock over a one-year period after which you sold it. Between dividends and the sale of the stock you would have made \$150. The rate of return on this stock would have been 0.15 percent. A portfolio is a grouping of several investments, so its expected percentage return is a weighted average of all expected rate of returns of its components according to their proportion.
Since ERR is based on assumptions that rarely hold true, most investors use ERR to compare the potential returns of one stock investment with another. After all, the growth rate figure used in the ERR formula does account for the actual historical growth of a company’s earnings per share. The expected return can be looked in the short term as a random variable which can take different values based on some distinct probabilities. This random variable has values within a certain range and can only take values within that particular range.
If it grew 10 percent in the past year, it will grow by at least another 10 percent this year. The expected rate of return is a percentage return expected to be earned by an investor during a set period of time, for example, year, quarter, or month.
Plugging the information into the CAPM formula tells the investor that she should expected an annual return of 13.9 percent. The expected rate of return is the amount you expect to lose or gain on an investment over a time period, and this lacks certainty due to market changes, interest rates and other factors. In contrast, the rate of return is how much you actually end up gaining or losing on that investment.
Well, we can get a very simple example for it if P sub 0 is \$100 and P1 is \$105. And then if dividend at 0.1 is expected to be \$8 per share, then clearly we can rewrite this formula and see that that expected return will consist of two parts. Dividend yield will be 8%, Plus capital gain will be 5%.
Also, an investor can use the expected return formula for ranking the asset and eventually make the investment as per the ranking and include them in the portfolio. In short, the higher the expected return, the better is the asset. As was mentioned above, the expected rate of return of a portfolio is the weighted average of the expected percentage return on each security according to their weight. The concept of expected return is part of the overall process of evaluating a potential investment.
If our portfolio of investments has diversified away as much risk as is possible given the costs of diversifying, our portfolio will be attractive to investors. If our bowl does not diversify away enough risk, it will not lie on the Security Market Line for those who we are trying to recruit into buying our portfolio. For a portfolio, you will calculate expected return based on the expected rates of return of each individual asset.
The rate of return expected on an asset or a portfolio. The expected rate of return on a single asset is equal to the sum of each possible rate of return multiplied by the respective probability of earning on each return.
We know that these are public securities, that they are traded, that they are risky. And one thing with respect to what we’ve been learning about NPV is that cash flows offered to stock holders are not certain at all. You know that, basically, stockholders, Have two fundamental rights, one is the right to vote and the other is the right to claim, Dividends. But that means that investors they make their choice about which we talked about in the first week. And they just decide to keep this cash in the company because they believe that for them it’s better to reinvest than to pocket this money and then to look for some other ways of investing.{“@context”:”https://schema.org”,”@type”:”FAQPage”,”mainEntity”:[{“@type”:”Question”,”name”:”How do you do a cash basis on a balance sheet?”,”acceptedAnswer”:{“@type”:”Answer”,”text”:” A cash basis balance sheet is one in which the assets are listed on the left and liabilities are listed on the right.”}},{“@type”:”Question”,”name”:”What accounts are in cash basis?”,”acceptedAnswer”:{“@type”:”Answer”,”text”:” Accounts that are in cash basis are those that use the accrual method of accounting.”}},{“@type”:”Question”,”name”:”What is basis on a balance sheet?”,”acceptedAnswer”:{“@type”:”Answer”,”text”:” The basis is the cost of an asset.”}}]}
#### How do you do a cash basis on a balance sheet?
A cash basis balance sheet is one in which the assets are listed on the left and liabilities are listed on the right.
#### What accounts are in cash basis?
Accounts that are in cash basis are those that use the accrual method of accounting.
#### What is basis on a balance sheet?
The basis is the cost of an asset. | 2,282 | 10,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.940871 |
https://estiv2019.com/qa/question-what-is-roi-strategy.html | 1,627,138,537,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00256.warc.gz | 257,896,407 | 8,215 | # Question: What Is ROI Strategy?
## What is ROI method?
Return on Investment (ROI) is a performance measure used to evaluate the efficiency of an investment or compare the efficiency of a number of different investments.
To calculate ROI, the benefit (or return) of an investment is divided by the cost of the investment.
The result is expressed as a percentage or a ratio..
## What is the average ROI?
The current average annual return from 1923 (the year of the S&P’s inception) through 2016 is 12.25%.
## What is ROI and how is it calculated?
ROI is calculated by subtracting the initial value of the investment from the final value of the investment (which equals the net return), then dividing this new number (the net return) by the cost of the investment, and, finally, multiplying it by 100.
## How do you calculate ROI for a project?
Return on investment is typically calculated by taking the actual or estimated income from a project and subtracting the actual or estimated costs. That number is the total profit that a project has generated, or is expected to generate. That number is then divided by the costs.
## What is ROI example?
Return on investment (ROI) is the ratio of a profit or loss made in a fiscal year expressed in terms of an investment. … For example, if you invested \$100 in a share of stock and its value rises to \$110 by the end of the fiscal year, the return on the investment is a healthy 10%, assuming no dividends were paid.
## How do you read ROI results?
Analysts usually present the ROI ratio as a percentage. When the metric calculates as ROI = 0.24, for instance, the analyst probably reports ROI = 24.0%. A positive result such as ROI = 24.0% means that returns exceed costs. Analysts, therefore, consider the investment a net gain.
## What are the three benefits of ROI?
ROI has the following advantages:Better Measure of Profitability: … Achieving Goal Congruence: … Comparative Analysis: … Performance of Investment Division: … ROI as Indicator of Other Performance Ingredients: … Matching with Accounting Measurements:
## Is 5 a good return on investment?
Safe Investments Historical returns on safe investments tend to fall in the 3% to 5% range but are currently much lower (0.0% to 1.0%) as they primarily depend on interest rates. When interest rates are low, safe investments deliver lower returns.
## What is a good ROI?
GOOD ROI FOR INVESTING. “A really good return on investment for an active investor is 15% annually. It’s aggressive, but it’s achievable if you put in time to look for bargains. ROI, or Return on Investment, measures the efficiency of an investment.
## What is ROI formula in Excel?
Return on investment (ROI) is a calculation that shows how an investment or asset has performed over a certain period. It expresses gain or loss in percentage terms. The formula for calculating ROI is simple: (Current Value – Beginning Value) / Beginning Value = ROI.
## What is a good ROI on a project?
A project is more likely to proceed if its ROI is higher – the higher the better. For example, a 200% ROI over 4 years indicates a return of double the project investment over a 4 year period. Financially, it makes sense to choose projects with the highest ROI first, then those with lower ROI’s.
## How do I do an ROI analysis?
The 4 Steps to Creating an ROI AnalysisStep 1: Discovery. The first thing you’ll want to do is pull your internal team together. … Step 2: Metrics Gathering. During discovery, it’s common to get estimated numbers or ranges for some of the metrics. … Step 3: Collaborative Analysis. … Step 4: Executive Presentation.
## What is a good ROI for a startup?
Invest in startups, and you’ll average 27% annual return on your investments! Well, maybe it’s not quite that easy; however, according to Robert Wiltbank, PhD, 27% returns actually are the average for startup investments in the United States.
## What is a good ROI percentage?
12 percentMost people would agree that, over time, an average annual return of 5 to 12 percent on your passive investment dollars is good, and anything higher than 12 percent is excellent. | 909 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.938483 |
http://www.ck12.org/analysis/Synthetic-Division-of-Polynomials/lesson/Synthetic-Division-of-Polynomials-MAT-ALY/ | 1,490,245,449,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186774.43/warc/CC-MAIN-20170322212946-00264-ip-10-233-31-227.ec2.internal.warc.gz | 463,061,774 | 36,195 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Due to system maintenance, CK-12 will be unavailable on Thursday, 3/9/2017 from 7:00p.m to 7:30p.m. PT.
Synthetic Division of Polynomials
Concise method of dividing polynomials when the divisor is x minus a constant.
Estimated21 minsto complete
%
Progress
Practice Synthetic Division of Polynomials
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated21 minsto complete
%
Synthetic Division of Polynomials
If you have completed the lesson on oblique asymptotes, you probably know that the problem:
Find the oblique asymptote of
could be an excellent reason to groan and grumble due to the long division that would be required.
Isn't there a better way?
Synthetic Division of Polynomials
Synthetic division is an alternative to long division from the previous concept. It can also be used to divide a polynomial by a possible factor, . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.
Examples
Example 1
Earlier, you were given a question about finding a better way to divide polynomials.
If we want to use synthetic division, notice that the factor is not in the form
. Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If , then , which can also be expressed as or . Therefore, we need to use synthetic division twice because there are two complex roots.
To start, put up in the left-hand corner box.
When we perform the synthetic division, we get a remainder of .
Next, we divide results from the last synthetic division with the other complex root. Put up in the left-hand corner box.
As a result, .
Example 2
Divide by .
Using synthetic division, the setup is as follows:
Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is . Notice that when we synthetically divide by , the “leftover” polynomial is one degree less than the original. We could also write .
Example 3
Determine if 4 is a solution to .
Using synthetic division, we have:
The remainder is 304, so 4 is not a solution. Notice if we substitute in , also written , we would have . This leads us to the Remainder Theorem.
Remainder Theorem: If , then is also the remainder when dividing by .
This means that if you substitute in or divide by , what comes out of is the same. is the remainder, but it is also the corresponding value. Therefore, the point would be on the graph of .
Example 4
Determine if is a factor of .
If you use synthetic division, the factor is not in the form . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this example, we need zeros for the term and the term.
This means that is a zero and its corresponding binomial, , is a factor.
Example 5
Is 6 a solution for ? If so, find the real-number zeros (solutions) of the resulting polynomial.
Put a zero placeholder for the term. Divide by 6.
The resulting polynomial is . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.
The solutions to this polynomial are 6, and .
Example 6
Divide by . Write the resulting polynomial with the remainder (if there is one).
Using synthetic division, divide by .
Divide the results from the last step by .
Review
Divide using synthetic division:
Use synthetic substitution to evaluate the polynomial function for the given value:
1. for
2. for
3. for
4. The area of a rectangle is and the length is . What is the width?
5. A group of geologists have taken a collection of samples of a substance from a proposed mining site and must identify the substance. Each sample is roughly cylindrical. The volume of each sample as a function of cylinder height (in centimeters) is . The mass (in grams) of each sample in terms of height can be modeled by . Write an expression that represents the density of the samples. (Hint: )
Divide using synthetic division:
1. divided by
2. divided by
3. divided by
4. divided by
To see the Review answers, open this PDF file and look for section 2.11.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
TermDefinition
Dividend In a division problem, the dividend is the number or expression that is being divided.
divisor In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Polynomial long division Polynomial long division is the standard method of long division, applied to the division of polynomials.
Quotient The quotient is the result after two amounts have been divided.
Remainder A remainder is the value left over if the divisor does not divide evenly into the dividend. | 1,295 | 5,797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 1, "texerror": 0} | 4.625 | 5 | CC-MAIN-2017-13 | latest | en | 0.893092 |
https://testpremier.com/book/3m-graduate-practice-past-questions-for-2022 | 1,685,253,095,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00238.warc.gz | 613,516,718 | 52,613 | Sale!
# 3m Graduate Practice Past Questions for 2023
\$19
-50%
Practice Full-length aptitude test questions for 3m graduate jobs with this updated study pack. Featuring the latest questions of the same style and difficulty as the actual test. Practice now, track your scores, and ace it! What You’ll Get:
• 12 Practice tests adapted from past exams.
• 590 Worked answers step-by-step explanations
• Free assessment day preparation guide (EBOOK)
• Lifetime offline access (PDF), guaranteed success.
• Covered by Testpremier’s Money Back Guarantee* (Limited Time Offer!)
Prep smarter and faster
• Zero risk: 48-hour “No questions asked” money-back guarantee
• Instant PDF download, and free online mock prep for qualifying tests
100% Secured Checkout
### 3m Graduate Practice Past Questions for 2022
Enhance your practice for 3m graduate assessment with extensive practice questions from the 3m Graduate Practice Past Questions for 2022, and featuring all the sections on the actual exams. This book provides insight into what to expect and helps you develop effective study strategies. Kick off your preparation with our 3m Graduate Practice Past Questions for 2022 (all-in-one pack):
• 3m Numerical reasoning test / 3m Assessment
• 3m Verbal reasoning test / 3m Assessment
• 3m Abstract reasoning tests / 3m Assessment
• 3m Logical reasoning tests / 3m Assessment
With step-by-step explanations on every question, and hints on how to solve them faster.
### 3m Graduate Practice pack tests formats; What to expect:
The sections on the assessments may include any of the following test sections, depending on the role that you applied to and also the country you are applying from:
1. Numerical Reasoning
2. Verbal Reasoning
3. Diagramatic
4. Interview
5. Personality Test
### Sample 3m Graduate Practice pack test
#### 3m Graduate Practice pack Numerical Reasoning
Question 1
Approximately what proportion of Beauty & Fragrance are Store-Brand Product?
A. 12.5%
B. 15.8%
C. 16.8%
D. 10.8%
E. 19.7%
Question 2
What proportion of Jewelry & Watches are National Product?
A. 13.41%
B. 18.29%
C. 31.71%
D. 26.83%
E. 9.76%
Question 3
What is the approximate ratio of National Product to Imported products?
A. none
B. 3.73:1
C. 1:3.73
D. 1:2.73
E. 2.73:1
Question 4
What proportion of Clothing & Shoes and Home Products are Imported Products.
A. 50%
B. 51%
C. 54%
D. 53%
E. 52%
1. What proportion of beauty and fragrance are stored-brand products: Beauty and fragrance that are brand = 30 Total stored-brand products = 45 30 60 15 = 190 So, therefore, the proportion of beauty and fragrance that are stored-brand products: 30/190 x 100 = 15.8%.
2. Jewelry and watches that are natural products = 80 Total natural products = 160 150 200 220 80 = 820 So, therefore, the proportion of Jewelry and watches that are natural products are: 80/820 x 100 = 9.76%.
3. Total natural products = 820, total imported products = 2240 The ratio of national products to imported products = 820: 2240 = 820/2240 = 41/114 = 41:112 = 1:2.73 (D)
4. What proportion of clothing and shoe and home products are imported products Clothing and shoes = 680 for imported products Homes products = 345 for imported products, total = 1025 1025/2240 x 100 = 54%.
#### 3m Graduate Practice pack Verbal Reasoning
Sages Store enjoys an international reputation for quality and style. Nowhere is this more important than in the dress and appearance of its staff. The company sets minimum standards of appearance which are demanded of all shop floor staff, although some departments have specific additional requirements. Hair must be clean, tidy and well cut at all times. With very few exceptions, such as “Designer Corner”, which operates a different staff dress code reflecting their particular style, business dress must be worn…Women should wear tailored suits, with a white or cream blouse. Men should wear dark grey trousers together with a white shirt and navy blazer.
Question 1
Women in “Designer Corner” are allowed to wear jewelry.
A. True
B. False
C. Cannot say
Question 2
Business dress must be worn by staff in all departments.
A. True
B. False
C. Cannot say
The early chaos of the home computing industry in the USA, where it developed, probably had a more detrimental effect in Europe than it did in the States. All the innovators in the field were companies that were too small to cope with or understand foreign sales. As a result, all US companies sold exclusively through European distributors, some of which were only interested in making maximum profits in a minimum amount of time. Home computing in Europe got off to a slow start because greedy distributors worked through incompetent suppliers, none of which had any real interest in the long-term future of the technology.
Question 3
Incompetent suppliers were one of the reasons for the slow development of home computing in Europe.
A. True
B. False
C. Cannot say
Question 4
None of the American innovators in the field were able to deal adequately with foreign sales.
A. True
B. False
C. Cannot say
### FAQ
We support local currency payments across the globe. Proudly powered by stripe - The world’s most powerful and secured payment gateway.
A: Testpremier practice packs do not only contain randomly created simulated questions. They infact includes actual tests that appear in the past, which are shared with us by previous candidates of the exams. We guarantee that studying with these materials, you will increase your chance of passing and scoring higher by 99.9% or you get your money back. Please see our terms and conditions for all the details. | 1,354 | 5,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.830519 |
https://www.analyticbridge.datasciencecentral.com/forum/topics/simulating-in-sas?groupUrl=sasandstatisticalprogramming&xg_source=activity | 1,563,411,517,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00388.warc.gz | 572,510,581 | 15,029 | # AnalyticBridge
A Data Science Central Community
# Simulating in SAS
Hi all!!
I would like to know whether anybody has simulated in SAS.. I am trying to simulate a data set with 4 varaibles (3 are categorical) taking into account the correlation among them. My problem is how to consider the correlation...
Many thanks!
Views: 449
### Replies to This Discussion
create crosstab (proc freq) and consider CHISQ.
dont forget to discretize numeric variables (proc rank groups=5; ...) .
here's example from sas doc:
proc freq data=Color order=data;
tables Eyes*(Hair Sex Country) / expected cellchi2 norow nocol chisq;
output out=ChiSqData n nmiss pchi lrchi;
weight Count;
title 'Chi-Square Tests for 3 by 5 Table of Eye and Hair Color and...';
run;
Sorry my answer is a bit late but hope other people find it helpful.
/* This program generates 300 observations of the variables y1, y2, y3, y4 . Beginning with the correlation matrix R and a vector of means m = (m1,m2,m3,m4)' and standard deviations s = (s1,s2,s3,s4)' read in as variables using a CARDS statement.
The means were arbitrarily selected as those in Liu and Gould 2002. The standard deviation vector and the correlation matrix were obtained from the data used by Allison et al (2003).
*/
data MVN_par; /* data for the parameter for the multivariate normal data*/
input r1 r2 r3 r4 means vars ; /* these can be adjusted to cater for the other two scenarios*/
cards;
1 0.986 0.967 0.949 92 14.3 /* These can be extended to an n by n correlation
0.986 1 0.992 0.980 88 14 matrix and n-dimensional means and standard
0.967 0.992 1 0.995 85 14.2 deviations vectors
run;
proc iml;
use MVN_par;
read all var {r1 r2 r3 r4} into R;
read all var {means} into mu;
read all var {vars} into sigma;
p = ncol(R); /* p is the number of variables generated*/
diag_sig = diag(sigma);
DRD = diag_sig * R * diag_sig` ; /* D is the a diagonal matrix whose element are
the standard deviations of each yi */
U = half(DRD);
do i = 1 to 300; /* this can be replaced with k to generate the
dataset with k patients. */
z = rannor(j(p,1,1234)); /* Generating random numbers. Zi i = 1,2,...,p
independent and have N(0,1)distribution.
The var-cov matrix for Z' is the identity matrix.*/
yprime = y`;
yall = yall // yprime;
end;
varnames = {y1 y2 y3 y4}; /* naming the variables*/
create my_MVN from yall (|colname = varnames|);
append from yall;
proc print data = my_MVN; | 729 | 2,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-30 | latest | en | 0.705836 |
https://www.deeconometrist.nl/portfolio-dynamics-101/ | 1,718,530,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00484.warc.gz | 652,864,492 | 42,977 | Select Page
Portfolio dynamics 101
#### June 21, 2022
[supsystic-social-sharing id='1']
#### Written by dominiquearkwook
A portfolio is a collection of financial investments like stocks, bonds, commodities, cash, and cash equivalents, including closed-end funds and exchange-traded funds (ETFs). People generally believe that stocks, bonds, and cash comprise the core of a portfolio. Though this is often the case, it does not need to be the rule. A portfolio may contain a wide range of assets including real estate, art, and private investments. Usually, these types of investments are popular with insurance companies and pension funds, since these kinds of companies want stable returns over a very long period. However, we will not focus on the type of assets in a portfolio, but on the general dynamics in a portfolio consisting of mainly stocks and bonds.
A Portfolio
You may think of an investment portfolio as a pie that has been divided into pieces of varying wedge-shaped sizes, each piece representing a different asset class and/or type of investment. Investors aim to construct a well-diversified portfolio to achieve a risk-return portfolio allocation appropriate for their risk tolerance level. A very famous mathematical framework to achieve such a portfolio is the Markowitz model. This model maximizes the expected return , where is the return of the portfolio, the weight of asset and the return of asset . We maximize this expected return subject to a certain level of variance in the portfolio. So in this model, the variance is used as a proxy for risk. In addition, from this model, we find the well-known efficient frontier depicted in the figure below. A portfolio lying on the efficient frontier represents the combination offering the best possible expected return for a given risk level.
The straight line in the figure is called the capital allocation line and can be compared to the famous CAPM. However, the CAL is calculated as , where depicts a sub-portfolio of risky assets, is a risk-free asset and is a combination of and in one new portfolio.
Although the Markowitz model in modern portfolio theory is of great theoretical importance and gives a lot of intuition in finding the optimal portfolio strategy, it has some major drawbacks. First, the risk, return, and correlation measures used by this model are based on expected values, which means that they are statistical statements about the future. Such measures often cannot capture the true statistical features of the risk and return which often follow highly skewed distributions and can give rise to inflated growth of return. One could use a scenario-based approach to overcome this flaw but this is out of the scope of this article. Another problem is that it relies on the efficient-market hypothesis and uses fluctuations in share price as a substitute for risk. It is well known that the efficient-market hypothesis does not hold in the financial markets in our world.
Dynamics of assets
In order to continue with the portfolio dynamics that will be shown in the next section, a fundamental concept first needs to be explained. It is called stochastic differential equations (SDE) and are of great importance to capture the dynamics of a stock, bond or a portfolio. An SDE is usually dependent on a Wiener process . The only thing that you have to know is that a Wiener process follows a normal distribution with mean 0 and variance dt, where dt is a very small time interval. Then a general SDE used for modelling stocks is often a Geometric Brownian Motion (GBM) and has the following form:
(1)
where is called the drift term and the diffusion term. We can find an explicit solution to this equation, but this is not really interesting. However, SDE’s can help us solve certain problems that model the price dynamics of options really well (like the Black-Scholes model). If you would like to more about this model, read my previous article on differential equations. We will use these SDE’s for our portfolio dynamics.
Portfolio dynamics
Suppose we have two assets, and , that follow a GBM as described before. Now we will combine these two ‘assets’ into a portfolio with value . Then , where is the amount of shares that we have of asset . The amount we invest in each asset changes over time, as share prices fluctuate. Therefore, we have dynamic portfolio strategies. We need to describe how we mathematically represent such strategies. Moreover, it turns out that we should limit ourselves to so-called self-financing portfolios. A self-financing portfolio is defined as:
(2)
Instead of using , we can also define portfolio weights, which was also described in the beginning of the article. The portfolio weight of an asset , can be written as:
(3)
Then the portfolio dynamics can be rewritten as:
(4)
Now, suppose we want to construct a portfolio conisting of 2 stocks following the same GBM. We assume that the two stocks are not correlated and we want to invest 50% in each share, meaning that the portfolio weights will not change (i.e. ). Then, after some simple simplifications, the portfolio dynamics can be written as:
(5)
There is a small problem in this equation, we have two Wiener processes where we should have one. However, this can be solved by noticing that . Then using the fact that we can define we can write
(6)
So we see that there is some diversification. The expected return is the same as the two stocks but the variance is lower. You can play around with different strategies for to find the portfolio dynamics and the associated diversification benefits that you could receive. However, if stocks are correlated with each other, their Wiener processes will also be correlated. This will result in a lower reduction of volatility when you combine the stocks in a portfolio.
Last but not least
This was a short overview of how portfolio dynamics is used to model the value of a portfolio. It is a very powerful tool that is used by many quantitative analysts who are responsible for modelling the value of existing portfolios. Moreover, you can also find the best strategy that fits your risk appetite. For example, you could try to find the optimal strategy that minimizes the volatility of a portfolio or maximizes return. However, it all depends on your level of risk aversion and how much risk you are willing to take. | 1,283 | 6,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.939192 |
https://ph.answers.yahoo.com/question/index?qid=20201022175129AAuwGNA | 1,606,638,019,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197278.54/warc/CC-MAIN-20201129063812-20201129093812-00440.warc.gz | 436,808,745 | 18,513 | # Write an absolute value inequality for the following:?
A solid at 50*C will change to a gas or liquid if the temperature t increases or decreases more than 50*C.
Relevance
• 1 month ago
|T - 50| > 50
---------------------------
When the increase more than 50, the temp is more than 100
|101 - 50| = 51 which is > 50
When the decrease more than -50, the temp is less than 0
|-1 - 50| = 51 which is > 50 | 120 | 411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-50 | latest | en | 0.855059 |
https://www.chegg.com/homework-help/thomas-calculus-mymathlab-sak-ssm-pkg-12th-edition-chapter-16.6-solutions-9780321706676 | 1,550,508,091,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487595.4/warc/CC-MAIN-20190218155520-20190218181520-00629.warc.gz | 789,283,059 | 24,141 | # THOMAS CALCULUS & MYMATHLAB SAK&SSM PKG (12th Edition) View more editions 90% (207 ratings) for Chapter 16.6Solutions for Chapter 16.6
• 8724 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Surface Integrals
In Exercise, integrate the given function over the given surface.
Parabolic cylinder G(x, y, z) = x over the parabolic cylinder
Sample Solution
Chapter: Problem:
• Step 1 of 5
The objective is to integrate the function: over the parabolic cylinder:
Let Then,
• Step 2 of 5
The parameterization equation is,
Then,
So,
Then, compute:
Limits:
• Step 3 of 5
Therefore, the area of the parabolic cylinder is,
To evaluate the indefinite integral: let the substitution be: then,
Now, the indefinite integral will be:
• Step 4 of 5
Thus, the surface integral becomes:
• Step 5 of 5
Therefore, the required value of integration is
Corresponding Textbook
THOMAS CALCULUS & MYMATHLAB SAK&SSM PKG | 12th Edition
9780321706676ISBN-13: 0321706676ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Thomas' Calculus 12th Edition Textbook Solutions | 312 | 1,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-09 | latest | en | 0.618743 |
https://gmatclub.com/forum/if-a-b-0-and-a-b-0-which-of-the-following-must-be-true-315164.html?kudos=1 | 1,585,600,816,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00198.warc.gz | 500,355,881 | 165,463 | GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video
It is currently 30 Mar 2020, 12:40
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# If a + b > 0 and a^b < 0, which of the following must be true?
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 62353
If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 00:08
1
4
00:00
Difficulty:
55% (hard)
Question Stats:
54% (01:26) correct 46% (01:33) wrong based on 122 sessions
### HideShow timer Statistics
Competition Mode Question
If $$a + b > 0$$ and $$a^b < 0$$, which of the following must be true?
I. $$a < 0$$
II. $$b > 0$$
III. $$|b| > |a|$$
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
Are You Up For the Challenge: 700 Level Questions
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 62353
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
28 Jan 2020, 03:41
1
PrankurD wrote:
Bunuel Isn't the question more as "which of the following can be true"
No, it's a must be true question.
If $$a + b > 0$$ and $$a^b < 0$$, which of the following must be true?
I. $$a < 0$$
II. $$b > 0$$
III. $$|b| > |a|$$
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
Given: $$a^b < 0$$. This to be true a MUST be negative because (positive)^(any number) > 0.
Next, since a + b > 0 and a < 0, then b must be positive AND further from 0 than a. So, b < 0 and |b| < |a|.
As you can see all three options must be true.
_________________
##### General Discussion
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3406
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 00:34
1
Quote:
If a+b>0 and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. None of these
$$a^b<0$$
i.e. a<0 because -ve value with any non-even exponent is always Negative
a+b>0
i.e. b > 0 (because a is negative)
Also |b|>|a|
because e.g. a = -2 then b >2 only then a+b will be 0 (as given)
i.e.
I. a<0 (TRUE)
II. b>0 (TRUE)
III. |b|>|a| (TRUE)
_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
My Recent Posted Questions
Q-1-Integer-Divisibility l Q-2-Inequality DS l Q-3-Standard Deviation l Q-4-Inequality
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
Director
Joined: 30 Sep 2017
Posts: 804
GMAT 1: 720 Q49 V40
GPA: 3.8
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 01:11
1
a^b<0 --> means a<0 (i)
a+b>0 and a<0 --> it must be true that b>0 and |b|>|a| (ii) and (iii).
(i), (ii) and (iii) must be true.
Posted from my mobile device
CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 459
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 02:33
1
In this question on inequalities, we can use some basic properties of inequalities and solve this question without too much of difficulty.
Let us try to analyse the data given in the question statement.
1) a+b>0. If the sum of two numbers is more than ZERO, there are not many conclusions we can draw from this situation about the signs of the numbers. The only things we can conclusively say are:
Both the numbers cannot be negative
Both the numbers cannot be ZERO
If a<0, b>0 in such a way that the absolute value of b is more than the absolute value of a.
If b<0, a>0 in such a way that the absolute value of a is more than the absolute value of b.
2) The second piece of information given in the question is $$a^b$$<0. This means that a is definitely negative and b is positive.
From the above, it’s clear that statement I and II are definitely true. Based on this, answer options A and C can be eliminated.
a is negative, b is positive and a+b>0 means that the absolute value of b is definitely more than the absolute value of a i.e. |b| > |a|. Statement III is also definitely true. Answer options B and E can be eliminated
The correct answer option is D.
As you see, there weren’t too many advanced concepts we applied here. We stuck to the basics and we were able to solve it in less than 2 minutes. This is probably why there is so much emphasis on being good with your basics, regardless of the topic.
Hope that helps!
_________________
VP
Joined: 24 Nov 2016
Posts: 1353
Location: United States
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 03:56
1
Quote:
If a+b>0 and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
a+b>0: a,b=++ or a,b=+-,-+
a^b<0: a,b≠0, a,b=-,+
I. true; II. true; III. true: a+b>0, if a,b=-,+, then |b|>|a| or else, a+b<0.
Ans (D)
Senior Manager
Joined: 20 Mar 2018
Posts: 492
Location: Ghana
Concentration: Finance, Real Estate
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 07:42
1
If a+b > 0 and a^b < 0, which of the following must be true?
Constraint: a+b >0 and a^b < 0
Now if a^b <0 ,then a must be negative and b must be odd or cube root
Now if a+b>0, then b must be +ve since a is -ve and b is +ve odd or cube root
where a is not equal to b
I. a < 0 (Must be true)
II. b > 0 (Must be true)
III. |b| > |a|
When a = -2 ,b= 3 —-> a^b <0
and a=-2,b=3 —-> a+b>0
Yes |b| > |a| (Must be true)
Or
|b| > |a| says the distance of b from zero(0) is greater than the distance of a from zero(0)
<——a(-2)——0———b(3)——>
since a+b>0
A. I only
B. I and II only
C. I and III only
D. I, II ,and III
E. None of these
Hit that D
Posted from my mobile device
Director
Joined: 25 Jul 2018
Posts: 647
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 15:49
1
If a+b >0 and $$a^{b}<0$$, which of the following must be true?
--> In order $$a^{b}$$ to be less then zero, --> a must be less than zero (a <0)
We got:
a+b >0 and -a >0
--> b>0
Also, we got:
b>0
-a >0
--> b-a >0
I. a<0
II. b>0
III. |b|>|a|
--> square the both sides--> $$b^{2}-a^{2} >0$$
(b-a)*(b+a) >0
All are TRUE.
Manager
Joined: 26 Dec 2017
Posts: 54
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 21:59
1
Ans: D
I, II and III
as a^b<0..so a must be negative ,so a<0
as a<0 and a+b>0, b must be positive b>0
III is also true
Manager
Joined: 14 Sep 2019
Posts: 220
If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
Updated on: 28 Jan 2020, 00:18
We are given that, a+b>0 and a^b<0,
From a^b<0, we can say that a is negative and b is odd number.
a<0, is ok
b>0, is ok
For being a+b>0, III |b|>|a| is true. Hence: (D)
Originally posted by Jawad001 on 27 Jan 2020, 02:09.
Last edited by Jawad001 on 28 Jan 2020, 00:18, edited 1 time in total.
VP
Joined: 20 Jul 2017
Posts: 1466
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 02:42
a^b < 0 is only possible when
--> a < 0 & b is positive (either positive or negative)
E.g: $$(-2)^3$$, $$(-2)^{-3}$$, . . . .
If $$a + b > 0$$ & if "$$a$$" is negative
--> "$$b$$" must be positive with greater numerical value than that of "$$a$$"
E.g: -2 + 3, -4 + 10 etc
--> |b| > |a| always
--> I & III must be true
Option C
Intern
Joined: 13 Jan 2020
Posts: 18
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 06:32
The given condition (a^b > 0) is only possible when :-
a < 0 & b is positive or negative
E.G. $$(-4)^5, (-4)^{-5},.....$$
.
.
.
If a + b > 0 and a is negative.
Then b is positive and b > a
.
Hence it won't be wrong to say.
$$|b| > |a|$$
.
And therefore I & III must be true.
.
Option C
Posted from my mobile device
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6059
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
Updated on: 28 Jan 2020, 00:52
test the values :
a=-2 and b = 3
we get yes to all options
IMO D; all correct
If a+b>0and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
Originally posted by Archit3110 on 27 Jan 2020, 06:46.
Last edited by Archit3110 on 28 Jan 2020, 00:52, edited 1 time in total.
Director
Joined: 07 Mar 2019
Posts: 905
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
27 Jan 2020, 07:25
If a+b>0 and $$a^b<0$$, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
$$a^b<0$$ where a < 0 and b is odd always. Thus
I. a<0 always True
II. b>0 false when b is even
III. |b|>|a| False when b = - 2 and a = -2 even through $$a^b<0$$
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
GMATPREPSoft1 680(Q48,V35) June 26, 2019
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3406
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
28 Jan 2020, 00:00
Archit3110 wrote:
test the values with
a=5 , b=-3
we get ; no to all answers
a=-2 and b = 3
we get yes to all options
IMO E; none is correct
If a+b>0and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
Archit3110
$$a^b$$ for the chosen values (a=5 and b=-3) is NOT satisfied so the chosen values are incorrect to test the three conditions.
_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
My Recent Posted Questions
Q-1-Integer-Divisibility l Q-2-Inequality DS l Q-3-Standard Deviation l Q-4-Inequality
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 6059
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
28 Jan 2020, 00:51
GMATinsight ; my bad i did an error
(1/a^x = a^-x) ( exponents)
silly error I did..
GMATinsight wrote:
Archit3110 wrote:
test the values with
a=5 , b=-3
we get ; no to all answers
a=-2 and b = 3
we get yes to all options
IMO E; none is correct
If a+b>0and a^b<0, which of the following must be true?
I. a<0
II. b>0
III. |b|>|a|
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. Noe of these
Archit3110
$$a^b$$ for the chosen values (a=5 and b=-3) is NOT satisfied so the chosen values are incorrect to test the three conditions.
GMAT Tutor
Status: Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator
Joined: 26 Jun 2014
Posts: 134
GMAT 1: 740 Q51 V39
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
28 Jan 2020, 01:29
Bunuel wrote:
Competition Mode Question
If $$a + b > 0$$ and $$a^b < 0$$, which of the following must be true?
I. $$a < 0$$
II. $$b > 0$$
III. $$|b| > |a|$$
A. I only
B. I and II only
C. I and III only
D. I, II and III
E. None of these
Are You Up For the Challenge: 700 Level Questions
$$a + b > 0$$ and $$a^b < 0$$
Possibilities for a + b > 0 are:
#1. $$a > 0$$ and $$b > 0$$
=> fails the test for $$a^b < 0$$ - Hence rejected
#2. $$a > 0$$ and $$b < 0$$, but $$|a| > |b|$$ (i.e. $$a$$ is a larger positive and $$b$$ is a smaller negative)
=> fails the test for $$a^b < 0$$ (since any positive number $$a$$ raised to any exponent $$b$$ remains positive) - Hence rejected
#3. $$a < 0$$ and $$b > 0$$, but $$|a| < |b|$$ (i.e. $$b$$ is a larger positive and $$a$$ is a smaller negative)
=> passes the test for $$a^b < 0$$ on condition that $$b$$ is an odd number (since any negative number $$b$$ raised to any odd exponent $$a$$ remains negative) - Hence possible
Working with the statements:
I. $$a < 0$$ - True
II. $$b > 0$$ - True
III. $$|a| < |b|$$ - True
_________________
Sujoy Kumar Datta
Director - CUBIX Educational Institute Pvt. Ltd. (https://www.cubixprep.com)
IIT Kharagpur, TU Dresden Germany
GMAT - Q51 & CAT (MBA @ IIM) 99.98 Overall with 99.99 QA
_________
Feel free to talk to me about GMAT & GRE | Ask me any question on QA (PS / DS)
Let's converse!
Skype: sk_datta
Alt. Email: sujoy.datta@gmail.com
Intern
Joined: 10 Sep 2016
Posts: 1
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
28 Jan 2020, 03:12
Bunuel Isn't the question more as "which of the following can be true"
Intern
Joined: 15 Nov 2018
Posts: 23
If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
29 Jan 2020, 22:49
Bunuel
A quick question.
a^b will be negative when a<0 (as stated in option I) and b>0(as stated in option II). But at the same time, b must also be odd power for a^b to remain negative. If b ends up being even, then the term a^b will become positive even if a is negative.
Based on this reasoning, I rejected option II.
Let me know if my understanding is wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 62353
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink]
### Show Tags
29 Jan 2020, 23:01
Sarkar93 wrote:
Bunuel
A quick question.
a^b will be negative when a<0 (as stated in option I) and b>0(as stated in option II). But at the same time, b must also be odd power for a^b to remain negative. If b ends up being even, then the term a^b will become positive even if a is negative.
Based on this reasoning, I rejected option II.
Let me know if my understanding is wrong.
Does II say anything about even or odd nature of b? It simply says: b > 0.
From a^b < 0, we can only say that a < 0 and b is not even. But b can be odd, not an integer, positive, or negative.
_________________
Re: If a + b > 0 and a^b < 0, which of the following must be true? [#permalink] 29 Jan 2020, 23:01
Display posts from previous: Sort by | 5,394 | 15,528 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.850904 |
https://www.glbasic.com/forum/index.php?PHPSESSID=6s7umdfjauklrdnrtihjepqfk6&topic=4410.0;wap2 | 1,632,107,258,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00040.warc.gz | 823,109,910 | 2,252 | Codesnippets > Math
ROT47 decoder/encoder
(1/1)
Topzombie:
Function
--- Code: (glbasic) ---FUNCTION ROT47\$: TextInput\$
LOCAL TextLen, convertedText\$
LOCAL ROT47NumberThing
TextLen=LEN(TextInput\$)
FOR ABC=0 TO TextLen-1
IF((ASC(MID\$(TextInput\$, ABC, 1))<127) AND (ASC(MID\$(TextInput\$, ABC, 1))>32))
IF(ASC(MID\$(TextInput\$, ABC, 1))+47 > 126)
ROT47NumberThing=(ASC(MID\$(TextInput\$, ABC, 1))+47)-127
convertedText\$=convertedText\$+CHR\$(33+ROT47NumberThing)
ELSE
convertedText\$=convertedText\$+CHR\$(ASC(MID\$(TextInput\$, ABC, 1))+47)
ENDIF
ELSE
convertedText\$=convertedText\$+MID\$(TextInput\$, ABC, 1)
ENDIF
NEXT
RETURN convertedText\$
ENDFUNCTION
--- End code ---
Example:
--- Code: (glbasic) ---GLOBAL String\$, TextFileSize, ASCIIFileSize, ASCII\$, InputStr\$
GLOBAL Output\$
PRINT "Type !input or a Text",10,10
INPUT InputStr\$,10,20
IF InputStr\$="!input"
TextFileSize=GETFILESIZE("input.txt")
OPENFILE(1, "input.txt", 1)
CLOSEFILE 1
Output\$=ROT47\$(String\$)
ELSE
String\$=InputStr\$
Output\$=ROT47\$(String\$)
ENDIF
OPENFILE(1, "output.txt", 0)
WRITELINE 1, Output\$
CLOSEFILE 1
PRINT Output\$,10,10
SHOWSCREEN
KEYWAIT
END
FUNCTION ROT47\$: TextInput\$
LOCAL TextLen, convertedText\$
LOCAL ROT47NumberThing
TextLen=LEN(TextInput\$)
FOR ABC=0 TO TextLen-1
IF((ASC(MID\$(TextInput\$, ABC, 1))<127) AND (ASC(MID\$(TextInput\$, ABC, 1))>32))
IF(ASC(MID\$(TextInput\$, ABC, 1))+47 > 126)
ROT47NumberThing=(ASC(MID\$(TextInput\$, ABC, 1))+47)-127
convertedText\$=convertedText\$+CHR\$(33+ROT47NumberThing)
ELSE
convertedText\$=convertedText\$+CHR\$(ASC(MID\$(TextInput\$, ABC, 1))+47)
ENDIF
ELSE
convertedText\$=convertedText\$+MID\$(TextInput\$, ABC, 1)
ENDIF
NEXT
RETURN convertedText\$
ENDFUNCTION
--- End code ---
bigsofty:
Some sort of simple encryption? :doubt:
Neurox:
--- Quote from: bigsofty on 2010-Apr-11 ---Some sort of simple encryption? :doubt:
--- End quote ---
Yes, ROT47 is a variant of ROT13.
See wikipedia : http://en.wikipedia.org/wiki/Rot13
Bye bye,
Neurox
bigsofty:
ah, I see, handy... very quick and simple, thank you! :good:
spicypixel:
--- Code: (glbasic) ---function str_rot47(\$str){
return strtr(\$str, '!"#\$%&\'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~', 'PQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~!"#\$%&\'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNO');
}
--- End code ---
You could use the above PHP ROT47 function to decrypt a simple name / high score combination for an online table. This will decrypt the encrypted text from the GLBASIC function. You will need to update the GLB function to use NETWEBEND with the URL and add the data to the URL so you can use GET from php to grab the url data sent. :) | 881 | 2,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.126529 |
https://jupyter.brynmawr.edu/services/public/dblank/CS110%20Intro%20to%20Computing/2017-Spring/Lectures/Bouncing%20Ball%20to%20Angry%20Birds.ipynb?download | 1,685,754,425,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00025.warc.gz | 374,567,697 | 8,903 | \n", " Sketch #1:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "// Global variables: \n", "// defined here, used in setup() and draw()\n", "\n", "float x;\n", "float y;\n", "\n", "void setup() {\n", " size(200, 200);\n", " x = width/2;\n", " y = 50;\n", "}\n", "\n", "void drawBall(float x, float y, int w, int h) {\n", " fill(255, 0, 0);\n", " ellipse(x, y, w, h);\n", "}\n", "\n", "void draw() {\n", " drawBall(x, y, 10, 10);\n", "}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Nothing too interesting there. I did use the special variables `width` and `height` that are automatically defined to be the size of the canvas. But other than that, it is just a picture of a red ball sitting quietly in space." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Brownian Motion" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Let's try a slight variation. Do you know the idea of [Brownian Motion](http://en.wikipedia.org/wiki/Brownian_motion) (sometimes called _particle theory_). This is basically just the idea that small objects will randomly move because they are hit will atoms or molecules. \n", "\n", "**Warning**: _I did not say \"Brownie in motion\"... that is a small chocolate treat that is on the move._\n", "\n", "We can simulate Brownian Motion by simply moving the ball a little left, right, or stay in the same place. We use the fact that `random(2) - 1` represents either a -1, 0, or 1." ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "name": "stdout", "output_type": "stream", "text": [ "1.8257991019485953\n", "0.6388284903246306\n", "1.8374085991016196\n", "1.4968469794203065\n", "1.202652726209918\n", "0.27141170253131275\n", "1.3101816043198309\n", "1.178512733650588\n", "1.769124800374796\n", "0.7704968113940065\n" ] } ], "source": [ "%%processing\n", " \n", " \n", "for (int i = 0; i < 10; i++) {\n", " println(random(2));\n", "}" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false, "format": "column" }, "outputs": [ { "data": { "application/javascript": [ "\n", " var component = document.getElementById(\"sketch_10\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"state_10\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"controls_div_10\");\n", " if (component != undefined)\n", " component.remove();\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " // FIXME: Stop all previously running versions (?)\n", " var processingInstance = Processing.getInstanceById(\"canvas_10\");\n", " if (processingInstance != undefined && processingInstance.isRunning())\n", " processingInstance.noLoop();\n", " });\n", "\n", "\n", " var output_area = this;\n", " // find my cell element\n", " var cell_element = output_area.element.parents('.cell');\n", " // which cell is it?\n", " var cell_idx = Jupyter.notebook.get_cell_elements().index(cell_element);\n", " // get the cell object\n", " var cell = Jupyter.notebook.get_cell(cell_idx);\n", "\n", " function jyp_print(cell, newline) {\n", " return function(message) {\n", " cell.get_callbacks().iopub.output({header: {\"msg_type\": \"stream\"},\n", " content: {text: message + newline,\n", " name: \"stdout\"}});\n", " }\n", " }\n", " window.jyp_println = jyp_print(cell, \"\\n\");\n", " window.jyp_print = jyp_print(cell, \"\");\n", "\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " Processing.logger.println = jyp_print(cell, \"\\n\");\n", " Processing.logger.print = jyp_print(cell, \"\");\n", " });\n", "\n", "\n", " " ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\n", "
\n", " Sketch #10:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "// Global variables: \n", "// defined here, used in setup() and draw()\n", "\n", "float x;\n", "float y;\n", "\n", "void setup() {\n", " size(200, 200);\n", " x = width/2;\n", " y = width/2;\n", "}\n", "\n", "void drawBall(float x, float y, int w, int h) {\n", " fill(255, 0, 0);\n", " ellipse(x, y, 10, 10);\n", "}\n", "\n", "void draw() {\n", " drawBall(x, y, 10, 10);\n", " x += random(2) - 1;\n", " y += random(2) - 1;\n", "}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "It looks like the ball is moving randomly about, doesn't it. You can make the effect even stronger by clearing the background before you redraw the ball:" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "data": { "application/javascript": [ "\n", " var component = document.getElementById(\"sketch_11\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"state_11\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"controls_div_11\");\n", " if (component != undefined)\n", " component.remove();\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " // FIXME: Stop all previously running versions (?)\n", " var processingInstance = Processing.getInstanceById(\"canvas_11\");\n", " if (processingInstance != undefined && processingInstance.isRunning())\n", " processingInstance.noLoop();\n", " });\n", "\n", "\n", " var output_area = this;\n", " // find my cell element\n", " var cell_element = output_area.element.parents('.cell');\n", " // which cell is it?\n", " var cell_idx = Jupyter.notebook.get_cell_elements().index(cell_element);\n", " // get the cell object\n", " var cell = Jupyter.notebook.get_cell(cell_idx);\n", "\n", " function jyp_print(cell, newline) {\n", " return function(message) {\n", " cell.get_callbacks().iopub.output({header: {\"msg_type\": \"stream\"},\n", " content: {text: message + newline,\n", " name: \"stdout\"}});\n", " }\n", " }\n", " window.jyp_println = jyp_print(cell, \"\\n\");\n", " window.jyp_print = jyp_print(cell, \"\");\n", "\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " Processing.logger.println = jyp_print(cell, \"\\n\");\n", " Processing.logger.print = jyp_print(cell, \"\");\n", " });\n", "\n", "\n", " " ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\n", "
\n", " Sketch #11:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", " Sketch #18:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "float vx;\n", "float vy;\n", "\n", "float x;\n", "float y;\n", "\n", "float dt; \n", "float t;\n", "\n", "void setup() {\n", " size(200, 200);\n", " x = width/2;\n", " y = 50;\n", " dt = 0.1; \n", " t = 0;\n", " vx = 10.0;\n", " vy = 0.0;\n", "}\n", "\n", "void drawBall(float x, float y, int w, int h) {\n", " fill(255, 0, 0);\n", " ellipse(x, y, 10, 10);\n", "}\n", "\n", "void draw() {\n", " background(0, 255, 0);\n", " float dx = vx * dt; \n", " if (((x + dx) > width/2) || ((x + dx) < 0)) {\n", " vx = vx * -1;\n", " } else {\n", " x = x + dx;\n", " }\n", " \n", " float dy = vy * dt;\n", " if (((y + dy) > height/2) || ((y + dy) < 0)) {\n", " vy = vy * -1;\n", " } else {\n", " y = y + dy;\n", " }\n", "\n", " drawBall(x, y, 10, 10);\n", " \n", " t = t + dt;\n", "}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Gravity" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Now, the final step: let's add gravity. At each timestep, we'll add a gravity component to vy. Over time, that will get bigger and bigger." ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "data": { "application/javascript": [ "\n", " var component = document.getElementById(\"sketch_23\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"state_23\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"controls_div_23\");\n", " if (component != undefined)\n", " component.remove();\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " // FIXME: Stop all previously running versions (?)\n", " var processingInstance = Processing.getInstanceById(\"canvas_23\");\n", " if (processingInstance != undefined && processingInstance.isRunning())\n", " processingInstance.noLoop();\n", " });\n", "\n", "\n", " var output_area = this;\n", " // find my cell element\n", " var cell_element = output_area.element.parents('.cell');\n", " // which cell is it?\n", " var cell_idx = Jupyter.notebook.get_cell_elements().index(cell_element);\n", " // get the cell object\n", " var cell = Jupyter.notebook.get_cell(cell_idx);\n", "\n", " function jyp_print(cell, newline) {\n", " return function(message) {\n", " cell.get_callbacks().iopub.output({header: {\"msg_type\": \"stream\"},\n", " content: {text: message + newline,\n", " name: \"stdout\"}});\n", " }\n", " }\n", " window.jyp_println = jyp_print(cell, \"\\n\");\n", " window.jyp_print = jyp_print(cell, \"\");\n", "\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " Processing.logger.println = jyp_print(cell, \"\\n\");\n", " Processing.logger.print = jyp_print(cell, \"\");\n", " });\n", "\n", "\n", " " ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\n", "
\n", " Sketch #23:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "float g = 9.8; \n", "\n", "float vx;\n", "float vy;\n", "\n", "float x;\n", "float y;\n", "\n", "float dt; \n", "float t;\n", "\n", "void setup() {\n", " size(200, 500);\n", " x = width/2;\n", " y = 50;\n", " dt = 0.1; \n", " t = 0;\n", " vx = 5.0;\n", " vy = 0.0;\n", "}\n", "\n", "void drawBall(float x, float y, int w, int h) {\n", " fill(255, 255, 255);\n", " ellipse(x, y, w, h);\n", "}\n", "\n", "void draw() {\n", " background(0, 0, 128);\n", " // gravity\n", " vy = vy + g * dt;\n", " \n", " float dx = vx * dt; \n", " if (((x + dx) > width) || ((x + dx) < 0)) {\n", " vx = vx * -1;\n", " } else {\n", " x = x + dx;\n", " }\n", " \n", " float dy = vy * dt;\n", " if (((y + dy) > height) || ((y + dy) < 0)) {\n", " vy = vy * -1;\n", " } else {\n", " y = y + dy;\n", " }\n", "\n", " drawBall(x, y, 100, 100);\n", " \n", " t = t + dt;\n", "}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Dampening" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "That is weird... it keeps bouncing! We probably want to lose a little bit of energy each time it \"hits something.\" We merely don't give the full amount when we change directions." ] }, { "cell_type": "code", "execution_count": 27, "metadata": { "collapsed": false }, "outputs": [ { "data": { "application/javascript": [ "\n", " var component = document.getElementById(\"sketch_24\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"state_24\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"controls_div_24\");\n", " if (component != undefined)\n", " component.remove();\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " // FIXME: Stop all previously running versions (?)\n", " var processingInstance = Processing.getInstanceById(\"canvas_24\");\n", " if (processingInstance != undefined && processingInstance.isRunning())\n", " processingInstance.noLoop();\n", " });\n", "\n", "\n", " var output_area = this;\n", " // find my cell element\n", " var cell_element = output_area.element.parents('.cell');\n", " // which cell is it?\n", " var cell_idx = Jupyter.notebook.get_cell_elements().index(cell_element);\n", " // get the cell object\n", " var cell = Jupyter.notebook.get_cell(cell_idx);\n", "\n", " function jyp_print(cell, newline) {\n", " return function(message) {\n", " cell.get_callbacks().iopub.output({header: {\"msg_type\": \"stream\"},\n", " content: {text: message + newline,\n", " name: \"stdout\"}});\n", " }\n", " }\n", " window.jyp_println = jyp_print(cell, \"\\n\");\n", " window.jyp_print = jyp_print(cell, \"\");\n", "\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " Processing.logger.println = jyp_print(cell, \"\\n\");\n", " Processing.logger.print = jyp_print(cell, \"\");\n", " });\n", "\n", "\n", " " ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\n", "
\n", " Sketch #24:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", "
\n", "\n" ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "float g = 9.8; \n", "\n", "float vx;\n", "float vy;\n", "\n", "float x;\n", "float y;\n", "\n", "float dt; \n", "float t;\n", "\n", "void setup() {\n", " size(200, 500);\n", " x = width/2;\n", " y = 50;\n", " dt = 0.1; \n", " t = 0;\n", " vx = 50.0;\n", " vy = 0.0;\n", "}\n", "\n", "void drawBall(float x, float y, int w, int h) {\n", " fill(255, 0, 0);\n", " ellipse(x, y, 10, 10);\n", "}\n", "\n", "void draw() {\n", " // gravity\n", " background(255);\n", " vy = vy + g * dt;\n", " \n", " float dx = vx * dt; \n", " if (((x + dx) > width) || ((x + dx) < 0)) {\n", " vx = vx * -0.8;\n", " } else {\n", " x = x + dx;\n", " }\n", " \n", " float dy = vy * dt;\n", " if (((y + dy) > height) || ((y + dy) < 0)) {\n", " vy = vy * -0.8;\n", " } else {\n", " y = y + dy;\n", " }\n", "\n", " drawBall(x, y, 10, 10);\n", " \n", " t = t + dt;\n", "}" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Interactivity" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Nice! Now let's drop the ball with the mouse:" ] }, { "cell_type": "code", "execution_count": 28, "metadata": { "collapsed": false }, "outputs": [ { "data": { "application/javascript": [ "\n", " var component = document.getElementById(\"sketch_25\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"state_25\");\n", " if (component != undefined)\n", " component.remove();\n", " component = document.getElementById(\"controls_div_25\");\n", " if (component != undefined)\n", " component.remove();\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " // FIXME: Stop all previously running versions (?)\n", " var processingInstance = Processing.getInstanceById(\"canvas_25\");\n", " if (processingInstance != undefined && processingInstance.isRunning())\n", " processingInstance.noLoop();\n", " });\n", "\n", "\n", " var output_area = this;\n", " // find my cell element\n", " var cell_element = output_area.element.parents('.cell');\n", " // which cell is it?\n", " var cell_idx = Jupyter.notebook.get_cell_elements().index(cell_element);\n", " // get the cell object\n", " var cell = Jupyter.notebook.get_cell(cell_idx);\n", "\n", " function jyp_print(cell, newline) {\n", " return function(message) {\n", " cell.get_callbacks().iopub.output({header: {\"msg_type\": \"stream\"},\n", " content: {text: message + newline,\n", " name: \"stdout\"}});\n", " }\n", " }\n", " window.jyp_println = jyp_print(cell, \"\\n\");\n", " window.jyp_print = jyp_print(cell, \"\");\n", "\n", " require([window.location.protocol + \"//calysto.github.io/javascripts/processing/processing.js\"], function() {\n", " Processing.logger.println = jyp_print(cell, \"\\n\");\n", " Processing.logger.print = jyp_print(cell, \"\");\n", " });\n", "\n", "\n", " " ], "text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" }, { "data": { "text/html": [ "\n", "
\n", " Sketch #25:
\n", "
\n", "
\n", "
\n", " \n", " \n", " \n", " \n", " | 5,364 | 16,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.631193 |
https://www.tutorialgateway.org/cpp-program-to-add-two-numbers/ | 1,723,415,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00219.warc.gz | 781,381,702 | 16,560 | # C++ Program to Add Two Numbers
Write a C++ program to add two numbers with multiple examples. The below written code uses an arithmetic addition operator to add num1 and num2.
```#include<iostream>
using namespace std;
int main()
{
int num1 = 10, num2 = 20, sum;
sum = num1 + num2;
cout << "Sum of Two Numbers " << num1 <<" and " << num2 << " = " << sum;
return 0;
}```
``Sum of Two Numbers 10 and 20 = 30``
This C++ code allows users to insert two integer values and then add that numbers.
```#include<iostream>
using namespace std;
int main()
{
int num1, num2, sum;
cout << "Please enter the First Number : ";
cin >> num1;
cout << "Please enter the Second Number : ";
cin >> num2;
sum = num1 + num2;
cout << "Sum of Two Numbers " << num1 <<" and " << num2 << " = " << sum;
return 0;
}```
``````Please enter the First Number : 5
Please enter the Second Number : 220
Sum of Two Numbers 5 and 220 = 225``````
## C++ Program to Add Two Numbers using functions
Here, we created a function that accepts two arguments and returns the addition of those two arguments. Next, we are calling that function inside our main() program.
```// using functions
#include<iostream>
using namespace std;
int add(int x, int y)
{
return x + y;
}
int main()
{
int num1, num2, sum;
cout << "Please enter the First Number : ";
cin >> num1;
cout << "Please enter the Second Number : ";
cin >> num2;
sum = add(num1, num2);
cout << "Sum of Two Numbers " << num1 <<" and " << num2 << " = " << sum;
return 0;
}```
In this C++ Add Two Numbers example using OOPS, we used a separate class with a public method to perform addition.
```#include<iostream>
using namespace std;
public: int add(int x, int y){
return x + y;
}
};
int main()
{
int num1, num2, sum;
cout << "Please enter the First Number : ";
cin >> num1;
cout << "Please enter the Second Number : ";
cin >> num2;
``````Please enter the First Number : 99 | 528 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.550491 |
https://www.mersenneforum.org/showthread.php?s=7d450dce5d8de8d3a81969adb1ee5fa5&t=19725 | 1,610,921,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00221.warc.gz | 899,906,000 | 11,809 | mersenneforum.org "Divides Phi" category
Register FAQ Search Today's Posts Mark Forums Read
2014-09-27, 03:02 #1 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,257 Posts "Divides Phi" category This is an interesting category, and Keller, Broadhurst and others submitted many such primes. The test for it is implemented in the old Y.Gallot's Proth.exe (the newest executable is >10 years old but it runs; its SSE2 code emits an error on modern CPUs, so this can be disabled in 'Options'). Because there is no detailed page for the properties of such numbers, I took a sheet of paper and sat for an hour pondering. Ok, what do we know? 1. Candidates can be primes p = 2q^n+1 with prime q. 2. q ≡ 11 (mod 12) is a well known requirement. (Proof is left to the reader. :wink-wink: ... Seriously though, there probably exists a paper from Keller and probably as early as from 70s) 3. n is odd. 4. $Phi(q^n,2) = {{2^{q^n} -1} \over {2^{q^{n-1}} -1}}$, so we need 2^q^n ≡ 1 (mod p) but 2^q^{n-1} $\ne$ 1 (mod p) UTM database carries some more exotic examples: with k>2 (then q is not restricted to 11 (mod 12) and n is not only odd) or with k*q^n+1 | Phi(q^m,2), where m != n. This category sieves well with sr1sieve (cannot combine multiple bases into one workunit for sr2sieve) and tests with LLR. Final check can be done with the old Proth.exe or with modified LLR (with 2^q $\ne$ 1 (mod p) added test). _____________________ 2 · 10859^87905 + 1 and 2 · 11171^100961+1 are found and are under final check with Proth.exe (this is quite slow)... Both checks finished: both divide Phi(q^n,2). Last fiddled with by Batalov on 2014-09-27 at 17:54 Reason: some fluff removed
2014-09-27, 17:53 #2 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 925710 Posts Keller (with Ingo Buechel) once reported to have searched k=2 for bases 383 (limit 433k), 467 (347k; a prime found in 2006!), 647 (322k), 947 (105k), 67607 (412k !!). These series have no known primes. These efforts could be of some interest to CRUS; I've seen these values still outstanding when I scanned their status yesterday. Last fiddled with by Batalov on 2014-09-27 at 18:29 Reason: fixed a typo (in the quoted message)
2014-09-27, 20:45 #3 paulunderwood Sep 2002 Database er0rr 2×5×353 Posts http://primes.utm.edu/primes/page.php?id=118560 does not appear on http://primes.utm.edu/top20/page.php?id=37 EDIT: it does now. C.C. was away for a day. Last fiddled with by Batalov on 2014-09-29 at 17:28 Reason: (official comment was updated)
2014-09-27, 21:04 #4 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 925710 Posts All of these initial labels get manually promoted by C. "The Prime Mogul " C. Or 'demoted' as the case may be for "Divides xGF(n,a,b)!!!!" that get automatically submitted by the PrimeGrid's engine. It's the weekend. P.S. There's another one!
2014-09-27, 21:15 #5
paulunderwood
Sep 2002
Database er0rr
DCA16 Posts
Quote:
Originally Posted by Batalov P.S. There's another one!
Congrats. Are you going for #1?
2014-09-27, 21:53 #6
Citrix
Jun 2003
1,579 Posts
Quote:
Originally Posted by Batalov Keller (with Ingo Buechel) once reported to have searched k=2 for bases 383 (limit 433k), 467 (347k; a prime found in 2006!), 647 (322k), 947 (105k), 67607 (412k !!). These series have no known primes. These efforts could be of some interest to CRUS; I've seen these values still outstanding when I scanned their status yesterday.
You might be interested in http://www.mersenneforum.org/showthr...t=10354&page=3 (This is an extremely big project to work on:( )
Also there used to be a project to search 2*a^a+1....everything under 40k was tested.
2014-09-28, 01:52 #7
Batalov
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
100100001010012 Posts
Quote:
Originally Posted by paulunderwood Congrats. Are you going for #1?
Certainly.
But this time with a proof of Dividing Phi first, and submitting second. It should show up in the top in about two hours, with overall rank ~271.
2014-09-28, 06:29 #8 paulunderwood Sep 2002 Database er0rr 2·5·353 Posts Congrats again, this time for 681817 digits
2014-10-03, 19:32 #9 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,257 Posts the order-of-2 (mod p) test I have modded LLR to run a naively refactored order-of-2-mod p test instead of Proth.exe (and even though naive, this is a fast solution; as fast as the primality test, 1-2 hr max; this is what I used for the current top1 and top3; Proth.exe would have taken cpu-days). I've scanned all known 155 current and former "Divides Phi" categorized Proth primes and they, of course, all confirmed. After that validity test, I moved on to the untagged prime and composite base q Proth numbers (Proth.exe does not run the order-of-2 test on them), and found half a dozen interesting divisors. The largest of them is Ian's CRUS S695 prime with k=2, and it was large enough to enter the top20. While the initial identification of such numbers is routine (we can check that 2^(q^n) ≡ 1 (mod k*q^n+1)), the determination of exact Phi() that they divide is more cumbersome than with prime q's. One has to divide away all prime factors of q separately and together until 2^(q^n/factors) !≡ 1. For this particular number, 695 = 5 * 139 and while 2^(q^n/139) already !≡ 1, 2^(q^n/5^k) ≡ 1 for 0<=k<=4 and not for k=5. Hence, as reported, "Divides Phi(695^94625/5^4,2)".
2014-10-10, 15:34 #10 paulunderwood Sep 2002 Database er0rr 1101110010102 Posts Congrats yet again, this time for a megaprime: 2 * 59^608685 + 1 Last fiddled with by paulunderwood on 2014-10-10 at 15:34
2014-10-10, 18:25 #11 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100100001010012 Posts Thanks. Of course a prime q=383, 647, 947 or 67607 (no known primes) would have been so much nicer, but after quite a bit of time spent on them nothing yet was found. So, a small fishing expedition was sent out to small q's - even though they have known primes, they are much faster (this is a short version of a long story; 67607, unfortunately is "generic FFT" only, which is very significantly slower).
Similar Threads Thread Thread Starter Forum Replies Last Post NookieN PrimeNet 9 2018-06-18 19:14 Runtime Error PrimeNet 4 2018-01-07 20:20 Buckle Factoring 7 2010-03-29 22:56 jinydu Homework Help 10 2008-08-06 19:17
All times are UTC. The time now is 22:16.
Sun Jan 17 22:16:55 UTC 2021 up 45 days, 18:28, 0 users, load averages: 1.56, 1.50, 1.65 | 2,063 | 6,424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 3, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-04 | latest | en | 0.888508 |
https://artofproblemsolving.com/wiki/index.php?title=2009_USAMO_Problems/Problem_1&oldid=131179 | 1,624,152,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00271.warc.gz | 127,515,362 | 11,635 | # 2009 USAMO Problems/Problem 1
## Problem
Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.
## Solution 1
Let $\omega_3$ be the circumcircle of $PQRS$, $r_i$ to be the radius of $\omega_i$, and $O_i$ to be the center of the circle $\omega_i$, where $i \in \{1,2,3\}$. Note that $SR$ and $PQ$ are the radical axises of $O_1$ , $O_3$ and $O_2$ , $O_3$ respectively. Hence, by power of a point(the power of $O_1$ can be expressed using circle $\omega_2$ and $\omega_3$ and the power of $O_2$ can be expressed using circle $\omega_1$ and $\omega_3$), $$O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2$$ $$O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2$$ Subtracting these two equations yields that $O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2$, so $O_3$ must lie on the radical axis of $\omega_1$ , $\omega_2$.
~AopsUser101
## Solution 2
Define $\omega_i$ and $O_i$ similarly to above. Note that $O_1O_3$ is perpendicular to RS and $O_2 O_3$ is perpendicular to PQ. Thus, the intersection of PQ and RS must be the orthocenter of triangle $O_1O_2O_3$. Define this as point $H$. Extending line $O_3H$ to meet $O_1O_2$, we note that $O_3H$ is perpendicular to $O_1O_2$.
In addition, note that by the radical axis theorem, the intersection of $PQ$ and $RS$ must also lie on the radical axis of $\omega_1$ and $\omega_2$. Because the radical axis of $\omega_1$ and $\omega_2$ is perpendicular to $O_1O_2$ and contains $H$, it must also contain $O_3$, and we are done. | 643 | 1,769 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 61, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-25 | latest | en | 0.749802 |
https://www.physicsforums.com/threads/physics-riddle.868973/ | 1,685,313,610,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00460.warc.gz | 1,052,707,745 | 26,831 | # Physics riddle
• micromass
#### micromass
Staff Emeritus
Homework Helper
http://img-9gag-fun.9cache.com/photo/aX9zqD2_460s.jpg [Broken]
Last edited by a moderator:
jedishrfu and QuantumQuest
B
NOT 2 B
jedishrfu
Without specific data and judging solely from the diagram I intuitively tend to vote for B.
jedishrfu
If the start and stop height differences are the same, wouldn't the answer be C?
atom jana and jedishrfu
If the start and stop height differences are the same, wouldn't the answer be C?
HINT:
NOT 2 B
What was the question again?
If the lines are frictionless then it shouldn't matter
HINT:
Well, when you put it that way...
billy_joule
If the lines are frictionless then it shouldn't matter
Suppose the blue line had a smaller slope at the beginning and end. The ball could take a long time to get across.
And, if the red line had a really large dip, it could take a long time to get across.
I reckon the red will be behind at the top of the first trough and may just about catch up by the end. But, you need specific data.
Brachistochrone is not flat, but everything that goes below is again slower - does it mean trajectories that go "deeper" are slower than brachistochrone, but always faster than the flat/straight one?
Brachistochrone is not flat, but everything that goes below is again slower - does it mean trajectories that go "deeper" are slower than brachistochrone, but always faster than the flat/straight one?
It's easy enough to compare a straight line down a slope with the limiting case of freefall, followed by horizontal motion. If the height of the slope is ##h## and the horizontal distance is ##d##, then the times are:
##t_1^2 = \frac{2(h^2 + d^2)}{gh}## (for the straight line)
The straight line takes longer when:
##h < \frac{3d}{4}##
But,of course, the brachisrochrone is always faster than both.
θOk the answer to this problem was bothering me, so I did some math:
1) Suppose we have two paths with one ball on each path. The both start at height ho at xo and finish at height h1 at x1. h1 is less than ho. xo, and ho are taken to be the 0 datum. Positive h is downwards.
2) Suppose further that we know height profile of the paths h(x) between xo and x1.
How do we calculate which ball will reach x1 first?
My solution:
Neglecting friction and all other interesting physics, we have a relationship between the height and the velocity of the ball, v(x) = √[2*g*h(x)]
We can estimate which ball will get to the finish line first by computing its average velocity in the x direction. Since we know the paths of the ball, we can average over [xo,x1].
Taking the angle θ to be measured between positive x and tangent to h(x), the average can be computed as: {∫√[2*g*h(x)]*cos(θ(x))dx}/(x1-xo) from xo to x1.
In this case we would have to show that the product of those expressions is greater than a case where h(x) is constant (as in case 1 for example).
Now, that integral might be a pain because we have a general function h(x) and a trigonometric function. I thought to myself "How can we relate h(x) to θ?" Ah ha! We know that h(x) = ∫{dh(x)/dx}*dx, and we know that the slope of h(x) can be expressed as the tangent of θ at a point x. So we have an equation in terms of θ, only:
average(v(x)_x) = { ∫√[2*g* (∫tan(θ(x))*dx)] * cos(θ(x))dx }/(x1-xo)
And I hate to quit but this is as far as I got before looking up this integral in mathematica and seeing how many terms are in the solution. The only way I know how to do this by hand is through integration by parts, which would have to be done twice since the angle is parameterized by x.
Anyway, you can see that if h(x) is constant, then the cosine term goes to 1, the tangent term becomes h(x). You need to show that h(x)*cos(θ) on average is greater for the slopey case than h(x) in the flat case.
Pepper Mint
Everyone else seems to be seeing a diagram. Not I.
θOk the answer to this problem was bothering me, so I did some math:
A simpler problem, but relevant to this question would be:
Take two balls with the same initial velocity, to travel a distance ##d## along the track. The first ball goes horizontally. The second ball goes down and up an incline (assume two symmetric straight lines). What's the relationship between the times, the initial velocity and the angle of the incline? What is the optimum angle?
The straight line takes longer when:
##h < \frac{3d}{4}##
Actually now that I think about it, it is trivial - let's say we go down, horizontal, up (sides of a rectangle, or more precisely, sides of a right trapezoid). There always exist a rectangle in which going down takes exactly as long as the horizontal leg in the "flat" case. That means going through the sides of the rectangle will take a bit longer than two times the flat case. That in turn means not every path below the flat one is guaranteed to be faster - or, in other words, B is not guaranteed to be faster than A, and to be sure which is faster we need to know exact shape of B and calculate the time it takes to travel both paths.
Or am I misunderstanding something?
Biker
Actually now that I think about it, it is trivial - let's say we go down, horizontal, up (sides of a rectangle, or more precisely, sides of a right trapezoid). There always exist a rectangle in which going down takes exactly as long as the horizontal leg in the "flat" case. That means going through the sides of the rectangle will take a bit longer than two times the flat case. That in turn means not every path below the flat one is guaranteed to be faster - or, in other words, B is not guaranteed to be faster than A, and to be sure which is faster we need to know exact shape of B and calculate the time it takes to travel both paths.
Or am I misunderstanding something?
Yes, you have to calculate in each case.
The problem here involves an initial velocity, which complicates everything. If the initial velocity is low, then many paths below the horizontal will be faster. The optimum path will be related to the brachistrochrone, but the initial velocity messes things up. But, steeps paths, curves, straight inclines (down and up) will all be better than going along the flat.
As the initial velocity increases, it should still be possible to beat the flat path, but only by going down a short distance. I haven't tried the maths (yet) but the equation involves the relative (proportional) increase in average velocity against the proportional increase in distance. The faster the initial velocity, the smaller the proportional increase you get by going down. If the ball is going fast enough, then nothing will be significantly faster than just going along the flat.
The answer is some paths below the horizontal beat the horizontal path and some do not. The key factors are the initial velocity and the force of gravity (as well as the shape of the path, of course).
Pepper Mint
If friction ignored answer A. On the flat section A continues with a steady speed but during the equivelent section of Bs journey the speed changes continually. Imagine just one part of Bs journey for example the first valley it encounters. Now imagine that, compared to the horizontal distance travelled, the valley was incredibly deep, for example horizontal distance traveled = 20m and depth of valley = 20 000 000m. From these numbers it should be easy to see that in general a journey with variable heights will take a longer time to complete.
http://img-9gag-fun.9cache.com/photo/aX9zqD2_460s.jpg [Broken]
Personally, I would have to find out the time it takes for the blue ball reach N, O, then P and similarly for the red one to reach N, X, O then P before being able to make a comparison.
HINT:
This is unrelated to the problem though.
Is this one of the unsolvable physics problems you once said you did ask on yahoo groups and you sometimes got (incorrect) answers ?
Last edited by a moderator:
From these numbers it should be easy to see that in general a journey with variable heights will take a longer time to complete.
Can take longer, not "will take longer". Actually the fastest trajectory (mentioned earlier, called brachistochrone), is curved with a lowest point somewhere in between.
You don't need to know the initial velocity as long as h(x) is below the datum. Like I said, the straightforward way to figure out which ball will reach the end first is to compute its average velocity. This is essentially computing the average of the product h(x)*cos(theta(x)), and showing it is higher for h(x) is constant.
Hello Borek. Thanks for pointing that out. I guess now that there is not enough information in the question to come up with an answer.
Without further information they won't move at all. (Newton 1)
Infinitum
I would tend to say that since the lower curve doesn't have a huge extra distance to travel, it would get there first because it'll be able to get more momentum.. However, at some point (even in a frictionless environment) if there are too many humps adding too much distance for it to travel, and that'll slow down it's rightward speed too much.
I would tend to say that since the lower curve doesn't have a huge extra distance to travel, it would get there first because it'll be able to get more momentum.. However, at some point (even in a frictionless environment) if there are too many humps adding too much distance for it to travel, and that'll slow down it's rightward speed too much.
I feel like whether the curve counts as "deep" or "shallow" depends not only on its shape, but also on the acceleration involved.
In my example (the one with trapezoid) time required for a vertical leg to finish is $t = \sqrt{\frac{2L} a}$ - so even for a shallow trapezoid there exist an a, L combination that gives a required time of travel.
I'd like to travel a long the curved ones.
If the difference in height is equal on both curves then the difference in potential energy is equal at start and finish line. Since this is the only available energy (assumption) to be transformed in kinetic energy the saldo is zero. So (c) is the correct answer. Taking eventual friction into account it would be (a) for it is shorter, i.e. less friction.
If the difference in height is equal on both curves then the difference in potential energy is equal at start and finish line. Since this is the only available energy (assumption) to be transformed in kinetic energy the saldo is zero. So (c) is the correct answer. Taking eventual friction into account it would be (a) for it is shorter, i.e. less friction.
Then how come brachistochrone exists, if time on all curves is identical?
Then how come brachistochrone exists, if time on all curves is identical?
But it isn't one. The more on kinetic energy gained downwards is lost again upwards. (But I haven't done the math, maybe it doesn't cancel out, however I think it does. I desperately try to remember the analogous experiment in a math museum I once visited, I thought they were equally fast at the finish ...)
But it isn't one. The more on kinetic energy gained downwards is lost again upwards. (But I haven't done the math, maybe it doesn't cancel out, however I think it does.)
The final speed must be the same (without friction) but the ball will take different times for different paths. For example, you can have a path with a time to the lowest point as large as you wish! | 2,689 | 11,376 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.930893 |
https://www.coursehero.com/file/p6uj2/weight-factors-w-1-w-2-etc-You-already-KNOW-this-e-g-your-grade-25-5-15-25-15/ | 1,519,483,814,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815812.83/warc/CC-MAIN-20180224132508-20180224152508-00619.warc.gz | 835,389,988 | 26,477 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
# “weight factors” w 1 w 2 etc you already know
This preview shows pages 9–14. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: “weight factors” w 1 , w 2 etc: You already KNOW this – e. g. your grade: % 25 5 * % 15 % 25 % 15 + × + × = ∑ FINAL LABS GRADE Weights: 25 for Final Exam, 15 for each of 5 labs More precise data points should carry more weight! Idea: weigh the points with the ~ inverse of their error bar 5 10 15 20 25 x 10 20 30 y(x) Weight-adjusted average: How do we average values with different uncertainties? Student A measured resistance 100±1 Ω (x 1 =100 Ω , σ 1 =1 Ω ) Student B measured resistance 105±5 Ω (x 2 =105 Ω , σ 2 =5 Ω ) 2 1 2 2 1 1 w w x w x w x + + = 2 1 1 1 σ = w 2 2 2 1 σ = w N N N i i i w w w x w x w x w w x w x + + + + + + = = ∑ ∑ ... ... 2 1 2 2 1 1 Or in this case calculate for i=1, 2: with “statistical” weights: BOTTOM LINE: More precise measurements get weighed more heavily! 5 10 15 20 25 x 10 20 30 y(x) How good is the agreement between theory and data? χ 2 TEST for FIT (Ch.12) ( ) ( ) ∑ = − = N j j j j x f y 1 2 2 2 σ χ 5 10 15 20 25 x 10 20 30 y(x) χ 2 TEST for FIT (Ch.12) N N y y = ≅ 2 2 σ σ d 2 2 ~ χ χ = d = N - c # of degrees of freedom # of data points # of parameters calculated from data...
View Full Document
{[ snackBarMessage ]}
### Page9 / 17
“weight factors” w 1 w 2 etc You already KNOW this –...
This preview shows document pages 9 - 14. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 630 | 1,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-09 | latest | en | 0.766606 |
https://astronomy.stackexchange.com/questions/19150/is-the-winds-intensity-on-mars-similar-to-earth/40155#40155 | 1,642,994,450,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00711.warc.gz | 165,124,111 | 34,521 | # Is the wind's intensity on Mars similar to Earth?
I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere.
So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity?
• Related questions here: space.stackexchange.com/questions/9301/… and space.stackexchange.com/questions/2621/… The first link has math where the wind-force can be calculated. Nov 24 '16 at 22:29
• ok, so the pressure of the wind it would be 61,25 times lower? nice answer Nov 24 '16 at 22:46
• do you want to post the answer here so I mark it as accepted? Nov 24 '16 at 23:42
• I think your math is right at least, that's what I get too, but as for an answer, I didn't want to post or copy someone else's answer as my own. Nov 25 '16 at 2:38
• @com.prehensible The atmospheric pressure on top of Olympus Mons is 0.0007 x the normal pressure at sea level on Earth, or 0.7 millibar. For comparison, a vacuum pump that you could buy online for 125 USD makes 0.1 millibar, only 7 times better; a pump that costs 50 USD makes 0.2 millibar, or 3.5 times better. Colloquially, I would describe the pressure on top of Olympus Mons as "pretty lousy vacuum". Seems like there's room for a lot of wind speed there before it really becomes threatening. Dec 14 '17 at 0:28
Credit to this question for inspiration, though my calculation methods differ.
The dynamic pressure equation is $$q=0.5\rho v^2$$ where $$q$$ is the pressure, $$\rho$$ is the atmospheric density, and $$v$$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $$q=0.5\rho_e v_e^2$$ and $$q=0.5\rho_m v_m^2$$, set them equal $$q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$$, and solve for $$v_e$$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $$\rho_m=0.020 \space kg/m^3$$ is the atmospheric density for Mars, $$\rho_e=1.225 \space kg/m^3$$ is the atmospheric density on Earth, $$v_m$$ is the wind speed on Mars, and $$v_e$$ is the equivalent wind speed on Earth.
With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get:
v_mars v_earth equivalent
10 1.28
50 6.39
100 12.8
200 25.6
400 51.1
These could be kph, or in fact, any units of velocity. screeenshot
and here's what hat looks like in a plot:
So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth | 789 | 2,789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.941304 |
https://stackoverflow.com/questions/6941904/recursion-schemes-for-dummies | 1,708,545,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00714.warc.gz | 563,439,242 | 45,679 | # Recursion schemes for dummies?
I'm looking for some really simple, easy-to-grasp explanations of recursion schemes and corecursion schemes (catamorphisms, anamorphisms, hylomorphisms etc.) which do not require following lots of links, or opening a category theory textbook. I'm sure I've reinvented many of these schemes unconsciously and "applied" them in my head during the process of coding (I'm sure many of us have), but I have no clue what the (co)recursion schemes I use are called. (OK, I lied. I've just been reading about a few of them, which prompted this question. But before today, I had no clue.)
I think diffusion of these concepts within the programming community has been hindered by the forbidding explanations and examples one tends to come across - for example on Wikipedia, but also elsewhere.
It's also probably been hindered by their names. I think there are some alternative, less mathematical names (something about bananas and barbed wire?) but I have no clue what the cutsier names are for recursion schemes that I use, either.
I think it would help to use examples with datatypes representing simple real-world problems, rather than abstract data types such as binary trees.
• Jeremy Gibbons has several papers that might be the best introduction as they are clear and largely self contained. "Streaming representation changers" (fold and unfold combined), "Fission for Program Comprehension" (paramorphisms and more), "The under appreciated unfold" (anamorphisms). cs.ox.ac.uk/people/publications/date/Jeremy.Gibbons.html Aug 5, 2011 at 7:40
Extremely loosely speaking, a catamorphism is just a slight generalization of `fold`, and an anamorphism is a slight generalization of `unfold`. (And a hylomorphism is just an unfold followed by a fold.). They're presented in a more rigorous form usually, to make the connection to category theory clearer. The denser form lets us distinguish data (the necessarily finite product of an initial algebra) and codata (the possibly infinite product of a final coalgebra). This distinction lets us guarantee that a fold is never called on an infinite list. The other reason for the funny way that catamorphisms and anamorphisms are generally written is that by operating over F-algebras and F-coalgebras (generated from functors) we can write them once and for all, rather than once over a list, once over a binary tree, etc. This in turn helps make clear exactly why they're all the same thing.
But from a pure intuition standpoint, you can think of cata and ana as reducing and producing, and that's about it.
Edit: a bit more
A metamorphism (Gibbons) is like an inside-out hylo -- its a fold followed by an unfold. So you can use it to tear down a stream and build up a new one with a potentially different structure.
Ekmett posted a nice "field guide" to the various schemes in the literature: http://comonad.com/reader/2009/recursion-schemes/
However, while the "intuitive" explanations are straightforward, the linked code is less so, and the blog posts on some of these might be a tad on the complex/forbidding side.
That said, except perhaps for histomorphisms I don't think the rest of the zoo is necessarily something you'd want to think with directly most of the time. If you "get" hylo and meta, you can express nearly anything in terms of them alone. Typically the other morphisms are more restrictive, not less (but therefore give you more properties "for free").
• OK, thanks, but that's just those three - there are others. I hope that someone will add an answer that's about some other recursion schemes. Aug 4, 2011 at 16:14
• Most of the remaining recursion schemes are kind of obscure, except for maybe paramorphisms, which correspond quite nicely to the "induction principles" for types we often see in dependent languages. I haven't quite figured out how all the category theory works out here, but I doubt it would break too horribly :) Aug 4, 2011 at 21:07
• Paramorphism is like a fold but you can peek at the "rest of input". A fold only gives you elementary access during the traversal. Aug 5, 2011 at 7:30
A few references, from the most category-theoretic (but relevant to give a "territory map" that will let you avoid "clicking lots of links") to the simpler & more self-contained:
• As far as the "bananas & barbed wire" vocabulary goes, this comes from the original paper of Meijer, Fokkinga & Patterson (and its sequel by other authors), and it is in sum just as notation-heavy as the less cute alternatives : the "names" (bananas, etc) are just a shortcut to the graphical appearance of the ascii notation of the constructions they are pegged to. For example, catamorphisms (i.e. folds) are represented with `(| _ |)`, and the par-with-parenthesis looks like a "banana", hence the name. This is the paper who is most often called "impenetrable", hence not the first thing I'd look up if I were you.
• The basic reference for those recursion schemes (or more precisely, for a relational approach to those recursion schemes) is Bird & de Moor's Algebra of Programming (the book is unavailable except as a print-on demand, but there are copies available second-hand & it should be in libraries). It contains a more paced & detailed explanation of point-free programming, if still "academic" : the book introduces some category-theoretic vocabulary, though in a self-contained manner. Yet, the exercises (that you wouldn't find in a paper) help.
• Sorting morphisms by Lex Augustjein, uses sorting algorithms on various data structures to explain recursion schemes. It is pretty much "recursion schemes for dummies" by construction:
This presentation gives the opportunity to introduce the various morphisms in a simple way, namely as patterns of recursion that are useful in functional programming, instead of the usual approach via category theory, which tends to be needlessly intimidating for the average programmer.
• Another approach to making a symbols-free presentation is Jeremy Gibbons' chapter Origami Programming in The Fun of Programming, with some overlap with the previous one. Its bibliography gives a tour of the introductions to the topic.
Edit : Jeremy Gibbons just let me know he has added a link to the bibliography of the whole book on the book's webpage after reading this question. Enjoy !
I'm afraid these last two references only give a solid explanation of (cata|ana|hylo|para)morphisms, but my hope is that this would be enough to tear through the algebraic formalism you can find in more notation-heavy publications. I don't know of any strictly non-category-theoretic explanation of (co-)recursion schemes other than those four.
Tim Williams gave a brilliant talk at the London Haskell User Group last night about recursion schemes with a motivating example of each of the ones you mention. Check out the slides:
http://www.timphilipwilliams.com/slides.html
There are references to all the usual suspects (lenses, bananas, barbed wire ala carte etc) at the end of the slides and you could also google "Origami Programming" which is a nice intro that I hadn't come across before.
and the video will be here when it's uploaded: | 1,597 | 7,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-10 | latest | en | 0.951002 |
https://www.jiskha.com/display.cgi?id=1334023720 | 1,503,360,930,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109682.23/warc/CC-MAIN-20170821232346-20170822012346-00062.warc.gz | 922,397,381 | 3,887 | posted by .
How would you solve the problem below using the subsitution method?
12x + 4 = 8y
y = x - 7
Please explain to me and help. If you could show all the steps, that would help a lot. Thank you.
put the y equation in the first equation.
12x+4=8*(x-7)
12x+4=8x-56
subtract 8x from each side, subtract 4 from each side.
Then having solved for x, put that into the original y= equation
Oohhh thank you soo much!!!
## Similar Questions
1. ### Algebra 2 URGENT
Solve the problem using the "z" method. Please Show All Work! u^4+6=-5u^2
2. ### Math
solve the system of linear equations using the addition method. Can you please show me the steps on how you reached the answer. Thank You 2x+20y= -144 12x+5y= 56
This problem is related to Chapter-Sets. Please solve the question using x method. x method means x ∈ A ∩ B Q. If A-B=A then show that A∩B=Ø
(Second time posting this problem)This problem is related to Chapter-Sets. Please solve the question using x method. x method means x ∈ A ∩ B Q. If AUB=Ø, then prove that A=Ø,B=Ø
(Second Time posting this problem)This problem is related to Chapter-Sets. Please solve the question using x method. x method means x ∈ A ∩ B Q. If A-B=A then show that A∩B=Ø
6. ### algebra
solve using the subsitution method 4x+4y=36 4x+4y=12 If the system has no solution or an infinite number of solutions, please help me understand it. | 406 | 1,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-34 | latest | en | 0.886124 |
https://gyankosh.net/msexcel/functions/logical/how-to-use-ifs-function-in-excel/ | 1,627,116,352,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00285.warc.gz | 300,162,418 | 48,229 | # EXCEL FUNCTIONS-IFS
## INTRODUCTION
IFS function comes under the LOGICAL FUNCTIONS category in Excel.
IFS function is very important and is going to be used a lot which makes it one of the very important functions in excel.
THE IFS FUNCTION CHECKS THE DIFFERENT GIVEN CONDITIONS AND RETURNS THE CORRESPONDING VALUE OF THE FIRST TRUE CONDITION.
This function is an extension to the very useful IF FUNCTION.
In this article, we’ll learn about the purpose, formula syntax and examples of the IFS FUNCTION IN EXCEL.
## PURPOSE OF IFS IN EXCEL
IFS FUNCTION is simply an extension to the very useful IF FUNCTION.
IFS Function lets us insert a number of conditions and the associated values if that particular condition is true.
If you have been using the EXCEL for a while, you must have used the NESTED IF FUNCTION. IFS function is a dedicated function for the NESTED IF which is more readable and easy to implement.
For the first time users, let us try to understand the IFS FUNCTION.
Suppose we have different codes for the fruits. Such as
1 – Apple
2 – Banana
3 – Pine Apple
4 – WaterMelon
and so on.
We need to put the condition to check the name of the fruit from the given code using the IFS FUNCTION.
We would put the function as
IFS( ” If code = 1″, Return Apple, ” If code =2 ” Return Banana , If code = 3…. and so on.
We’ll learn the proper formula in the examples below.
Simply stating, we’ll use IFS when we need to put a number of conditions in the same statement or simply speaking it is a MULTIPLE IF.
## PREREQUISITES TO LEARN IFS
THERE ARE A FEW PREREQUISITES WHICH WILL ENABLE YOU TO UNDERSTAND THIS FUNCTION IN A BETTER WAY.
• Basic understanding of how to use a formula or function.
• Basic understanding of rows and columns in Excel.
• The knowledge of IF function will be of great help.
• Of course, Excel software.
Helpful links for the prerequisites mentioned above What Excel does? How to use formula in Excel?
## SYNTAX: IFS FUNCTION
The Syntax for the IFS function is
=IF(TEST CONDITION 1, VALUE IF CONDITION 1 IS TRUE, TEST CONDITION 2, VALUE IF CONDITION 2… )
TEST CONDITION is the test condition which you want to test. The condition must return a TRUE OR FALSE.
VALUE IF CONDITION IS TRUE is the value which will be returned by the function if the corresponding condition is true.
We can enter up to 127 conditions.
If more than two conditions are true, the leftmost condition will be executed and value will be returned.
## EXAMPLE:IFS FUNCTION IN EXCEL
### DATA SAMPLE
For the better understanding of the use of IFS FUNCTION, let us take an example.
We have SPECIAL CODES for a few fruits which are given below.
We want Excel to find out the fruit on the basis of selected code.
We’ll create a simple drop down list and the output cell will show the fruit chosen.
### CREATING A DROPDOWN LIST FOR THE SELECTION OF CODE.
STEPS TO CREATE A DROP DOWN LIST:
*the drop down list creation steps are brief as there is a complete article for the same. [ CLICK HERE TO VISIT]
• Select the cell in which we want to insert the data list.
• Go to DATA TAB and choose DATA VALIDATION>DATA VALIDATION [ UNDER DROP DOWNLIST ].
• Select the Options as
• ALLOW: CHOOSE LIST FROM THE DROP DOWN.
• SOURCE: THE SOURCE OF THE LIST DATA. FOR OUR EXAMPLE, THE CODES ARE AVAILABLE AT A4 TO A8 , HENCE THE RANGE \$A\$4:\$A\$8.
• Click OK.
• The dropdown list will be created as shown in the picture below.
Now , the list is already created. Now we want to put the formula using the function IFS to find out the fruit code immediately after selecting the fruit code.
### SOLUTION:
STEPS TO USE IFS FUNCTION TO CHOOSE THE FRUIT ON THE BASIS OF FRUIT CODE.
• Select the cell where we want the result.
• Put the formula as = IFS (” FIRST CONDITION:, VALUE IF FIRST CONDITION IS TRUE, SECOND CONDITION, VALUE IF SECOND CONDITION IS TRUE , AND SO ON….).
• For our example, the code is present in the cell G3, so the formula become =IFS(G3=1,”APPLE”,G3=2,”BANANA”,G3=3,”PINEAPPLE”,G3=4,”WATERMELON”,G3=5,”PEACH”)
• Click OK.
• We are all set to test our example.
• The following picture shows the formula used.
*EXPLANATION FOLLOWS THE PICTURE BELOW.
STEPS TO USE IFS FUNCTION TO CHOOSE THE FRUIT ON THE BASIS OF FRUIT CODE.
• Select the cell where we want the result.
• Put the formula as = IFS (” FIRST CONDITION:, VALUE IF FIRST CONDITION IS TRUE, SECOND CONDITION, VALUE IF SECOND CONDITION IS TRUE , AND SO ON….).
• For our example, the code is present in the cell G3, so the formula become =IFS(G3=1,”APPLE”,G3=2,”BANANA”,G3=3,”PINEAPPLE”,G3=4,”WATERMELON”,G3=5,”PEACH”)
• Click OK.
• We are all set to test our example.
• The following picture shows the formula used.
*EXPLANATION FOLLOWS THE PICTURE BELOW.
### EXPLANATION:
Let us try to understand the steps which we followed in the discussed example.
The CODE SELECTION will take place in the cell G3 which means that the condition will be applied on the cell G3.We used the formula as =IFS(G3=1,”APPLE”,G3=2,”BANANA”,G3=3,”PINEAPPLE”,G3=4,”WATERMELON”,G3=5,”PEACH”)
The function starts with the FUNCTION NAME IFS.
The first argument is the first condition. We want to know if the code is 1 so we put the argument as G3=1, which will be the instruction to EXCEL to check if G3 is equal to 1 or not, if YES, the corresponding value i.e. second argument will be returned which is APPLE.
If not, then next condition will be tested, which is third argument where we again check if G3=2 or not, if TRUE, fourth argument will be returned as value which is BANANA and so on.
Similarly , all the conditions will be checked i.e. codes from 1 to 5.ONE THING TO BE NOTICED HERE IS THAT WE HAVE USED A LIST, SO THERE ARE NO CHANCES FOR ANY WRONG CODE.The following picture shows the running of the example.
## KNOWLEDGE BYTES
### NESTED IF VS IFS
Nested IF and IFS do the same job but a few differences can be noticed.
• Nested If is used by nesting the IF function one inside the another [ The process known as nesting ]. whereas IFS is the dedicated Function made for the IF Nesting.
• IFS is a new function which is available only in the latest Excel version past 2019 only.
• IFS is more readable whereas Nested IF is somewhat harder to understand.
• Conclusion is that if we have IFS, we should use IFS in place of Nested IF because of its simplicity and readability in case of troubleshooting.
*Although we can have 127 arguments in the IFS function, never use a high number of IF conditions otherwise troubleshooting is going to be tough. | 1,625 | 6,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-31 | longest | en | 0.837599 |
http://ejde.math.txstate.edu/Volumes/2002/18/palamides-tex | 1,484,883,906,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-04/segments/1484560280774.51/warc/CC-MAIN-20170116095120-00446-ip-10-171-10-70.ec2.internal.warc.gz | 89,809,133 | 11,728 | \documentclass[twoside]{article} \usepackage{amsfonts,amsmath} \pagestyle{myheadings} \markboth{\hfil Positive and monotone solutions \hfil EJDE--2002/??} {EJDE--2002/??\hfil Panos K. Palamides \hfil} \begin{document} \title{\vspace{-1in} \parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. ??, pp. 1--16. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Positive and monotone solutions of an m-point boundary-value problem % \thanks{ \emph{Mathematics Subject Classifications:} 34B10, 34B18, 34B15. \hfil\break \indent {\em Key words:} multipoint boundary value problems, positive monotone solution, vector field, \hfill\break\indent sublinear, superlinear, Kneser's property, solution's funel. \hfil\break \indent \copyright 2002 Southwest Texas State University. \hfil\break \indent Submitted January 10, 2002. Published February 18, 2002.} } \date{} \author{Panos K. Palamides} \maketitle \begin{abstract} We study the second-order ordinary differential equation $$y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq 1,$$ subject to the multi-point boundary conditions $$\alpha y(0)\pm \beta y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)\,.$$ We prove the existence of a positive solution (and monotone in some cases) under superlinear and/or sublinear growth rate in $f$. Our approach is based on an analysis of the corresponding vector field on the $(y,y')$ face-plane and on Kneser's property for the solution's funnel. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode@=11 \@addtoreset{equation}{section} \catcode@=12 \section{Introduction} Recently an increasing interest has been observed in investigating the existence of positive solutions of boundary-value problems. This interest comes from situations involving nonlinear elliptic problems in annular regions. Erbe and Tang \cite{ET} noted that, if the boundary-value problem \begin{equation*} -\Delta u=F(|x|,u)\quad\text{in }R<|x|<\hat{R} \end{equation*} with \begin{gather*} u=0\quad\mbox{for }|x| =R,\quad u=0 \quad\mbox{for }|x| =\hat {R}; \quad \mbox{or} \\ u=0 \quad\mbox{for }|x| =R,\quad \frac{\partial u}{\partial|x|} =0 \quad \mbox{for } |x| =\hat{R};\quad\mbox{or} \\ \frac{\partial u}{\partial|x|}=0\quad\mbox{for } |x|=R, \quad u=0 \quad \mbox{for }|x|=\hat{R} \end{gather*} is radially symmetric, then the boundary-value problem can be transformed into the scalar Sturm-Liouville problem \begin{gather} x''(t)=-f(t,x(t)),\quad 0\leq t\leq1, \label{E0} \\ \alpha x(0)-\beta x'(0)=0,\quad \gamma x(1)+\delta x'(1)=0. \label{C0} \end{gather} where $\alpha$, $\beta$, $\gamma$, $\delta$ are positive constants. By a positive solution of (\ref{E0})-(\ref{C0}), we mean a function $x(t)$ which is positive for $00$. \end{enumerate} Also nonlinear boundary constraints have been studied, among others by Thompson \cite{TO} and by the author of this paper and Jackson \cite{JP}. There are common ingredients in these papers: an (assumed) Nagumo-type growth condition on the nonlinearity $f$ or/and the presence of upper and lower solutions. The multi-point boundary-value problem for second-order ordinary differential equations was initiated by Ilin and Moiseev \cite{IM-1,IM-2}. Gupta \cite{GU} studied the three-point boundary-value problems for nonlinear ordinary differential equations. Since then, more general nonlinear multi-point boundary-value problems have been studied by several authors. Most of them used the Leray-Schauder continuation theorem, nonlinear alternatives of Leray-Schauder, coincidence degree theory or a fixed-point theorem on cones. We refer the reader to \cite{CM, FE, GNT,Ma1} for some recent results of nonlinear multipoint boundary-value problems. Let $a_i\geq 0$ for $i=1,\dots ,m-2$ and let $\xi_i$ satisfy $0<\xi _{1}<\xi_{2}<\dots <\xi_{m-2}<1$. Ma \cite{Ma2} applied a fixed-point theorem on cones to prove the existence of a positive solution of \begin{equation*} \begin{gathered} u''+a(t)f(u)=0\\ u(0)=0,\quad u(1)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i) \end{gathered} \end{equation*} under superlinearity or sublinearity assumptions on $f$. He also assumed the following \begin{enumerate} \item[($\Gamma 1$)] $a\in C([0,1],[0,\infty ))$, $f\in C([0,\infty ),[0,\infty ))$, and there exists $t_{0}\in$ $[\xi_{m-2},1]$ such that $a(t_{0})>0$ \item[($\Gamma 2$)] For $i=1,\dots ,m-2$, $a_i\geq 0$ and $\sum_{i=1}^{m-2}a_i\xi_i<1$. \end{enumerate} Recently, Gupta \cite{GU1} obtained existence results for the boundary-value problem \begin{gather*} y''(t)=f(t,y(t),y'(t))+e(t),\quad 0\leq t\leq 1 \\ y(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i), \end{gather*} by using the Leray-Schauder continuation theorem, under smallness assumptions of the form \begin{equation*} |f(t,y,y')| \leq p(t)|y| +q(t)| y'| +r(t)\quad\text{and} \quad C_{1}\| p(t)\| +C_{2}\|q(t)\| \leq1, \end{equation*} with $p(t)$, $q(t)$, $r(t)$ and $e(t)$ in $L^{1}(0,1)$ and $C_{1}$ and $C_{2}$ constants. In this paper, we consider the problem of existence of positive solutions for the m-point boundary-value problem \begin{gather} y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq1, \label{E} \\ \alpha y(0)-\beta y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i). \label{C} \end{gather} We assume $\alpha>0$, $\beta>0$, the function $f$ is continuous, and $$f(t,y,y')\geq0,\quad \text{for all }t\in[0,1],\;y\geq 0\,\; y'\in\mathbb{R}. \label{A1}$$ The presence of the third variable $y'$ in the function $f(t,y,y')$ causes some considerable difficulties, especially, in the case where an approach relies on a fixed point theorem on cones and the growth rate of $f(t,y,y')$ is sublinear or superlinear. We overcome this predicament, by extending below the concept-assumptions (\ref{SL}) and ( \ref{sl}) as follows: Suppose that for any $M>0$, $$\begin{gathered} f_{0,0}:=\lim_{(y,y')\to (0,0)}\max_{0\leq t\leq1} \frac{f(t,y,y')}{y}=0 \\ f_{+\infty}:=\lim_{y\to +\infty}\min_{0\leq t\leq1}\frac {f(t,y,y')}{y}=+\infty, \quad\mbox{for }|y'|\leq M \end{gathered} \label{A2S}$$ i.e. $f$ is \emph{jointly superlinear} at the end point $(0,0)$ and \emph{ uniformly superlinear }at\emph{\ }$+\infty$. Similarly $$\begin{gathered} f_{0}:=\lim_{y\to 0+}\min_{0\leq t\leq1}\frac{f(t,y,y')} {y}=+\infty,\quad\mbox{for } |y'|\leq M.\\ f_{+\infty,+\infty}:=\lim_{(y,y')\to (+\infty,+\infty)} \max_{0\leq t\leq1}\frac{f(t,y,y')}{y}=0, \end{gathered} \label{A2s}$$ i.e. $f$ is \emph{jointly sublinear} at $(+\infty,+\infty)$ and \emph{ uniformly sublinear }at $0$. Furthermore there exist $\bar{l}\in(0,\infty]$, such that for every $\bar{M} >0$ $$\lim_{y'\to -\infty}\max_{0\leq t\leq1}\frac{f(t,y,y')}{ y'}=-\bar{l},\quad\text{for }y\in[0,\bar{M}] \label{A3}$$ i.e. $f(t,y,.)$ is \emph{linear or superlinear }at $-\infty$ \ and for every $\bar{\eta}>0$ $$\lim_{y'\to 0}\min_{0\leq t\leq1}\frac{f(t,y,y')}{y'} =0, \quad\text{for }y\in(0,\bar{\eta}). \label{A4}$$ i.e. $f(t,y,.)$ is \emph{superlinear }at $0$. \begin{remark}\label{R1} \rm Note that the differential equation (\ref{E}) defines a vector field whose properties will be crucial for our study. More specifically, we look at the $(y,y')$ face semi-plane $(y>0)$. From the sign condition on $f$ (see assumption (\ref{A1})), we immediately see that $y''<0$. Thus any trajectory $(y(t),y'(t))$, $t\geq0$, emanating from the semi-line $E_{0}:=\{(y,y'):\alpha y-\beta y'=0,\;y>0\}$ trends'' in a natural way, (when $y'(t)>0$) toward the positive $y$-semi-axis and then (when $y'(t)<0$) trends toward the negative $y'$-semi-axis. Lastly, by setting a certain growth rate on $f$ (say superlinearity) we can control the vector field, so that some trajectory satisfies the given boundary condition $y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)$ at the time $t=1$. These properties will be referred as \emph{The nature of the vector field''} throughout the rest of paper. \end{remark} So the technique presented here is different to that given in the above mentioned papers \cite{GU1, EW,DEH, GNT,ET}, but it is closely related with those in \cite{JP,Ma2}. Actually, we rely on the above ''nature of the vector field'' and on the simple shooting method. Finally, for completeness we refer to the well-known Kneser's theorem (see for example Copel's text-book \cite{Co}). \begin{theorem} \label{Th1} Consider the system $$\;x''=f(t,x,x'),\quad (t,x,x')\in\Omega :=[\alpha,\beta]\times\mathbb{R}^{2n},\label{*}$$ with the function $f$ continuous. Let $\hat{E}_{0}$ be a continuum (compact and connected) set in $\Omega_{0}:=\{(t,x,x')\in \Omega:t=\alpha\}$ and let $\mathcal{X}(\hat{E}_{0})$ be the family of \ all solutions of (\ref{*}) emanating from $\hat{E}_{0}$. If any solution $x\in\mathcal{X}(\hat{E}_{0})$ is defined on the interval $[\alpha,\tau]$, then the set (\emph{cross-section }at the point $\tau$) $\mathcal{X}(\tau;\hat{E}_{0}):=\left\{ (x(\tau),x'(\tau )):x\in\mathcal{X}(\hat{E}_{0})\right\}$ is a continuum in $\mathbb{R}^{2n}$. \end{theorem} Now consider (\ref{E})-(\ref{C}) with the following notation. \begin{gather*} \sigma:=\sum_{i=1}^{m-2}\alpha_i\xi_i<1, \quad \sigma^{\ast}:=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha} \Big\{ \sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\} <1, \\ K_{0}:=\max\Big\{ \frac{2\alpha}{\beta},\;2\big[ \frac{\alpha+\beta}{\beta}- \frac{\sigma}{\xi_{m-2}}\big] \Big\} , \\ \mu_{0}:=\min\Big\{ (1-m^{\ast})\frac{\varepsilon\alpha}{\beta},\;2\big[ \frac{\varepsilon(\alpha+\beta)}{\beta}-1\big] \text{ }\Big\} \end{gather*} where $\beta/(\alpha+\beta)<\varepsilon<1$ and $\sigma^{\ast}0$ such that $$\min_{0\leq t\leq1}f(t,y,y')>-\bar{K}y',\;\;0\leq y\leq H \big( 1+\frac{\alpha}{\beta}\big)\quad\text{and}\quad y'<-H. \label{99}$$ By the superlinearity of $f(t,y,y')$ at $y=+\infty\;$(see condition ( \ref{A2S})), for any $K^{\ast}>K_{0}$ there exists $H^{\ast}>H$ such that $$\min_{0\leq t\leq1}f(t,y,y')>K^{\ast}y,\ y\geq H^{\ast}\quad\text{and } \quad -2H\leq y'\leq\frac{\alpha}{\beta}H. \label{100}$$ Similarly by the superlinearity of $f(t,y,y')$ at $(0,0)$, for any $0<\mu^{\ast}<\mu_{0}$ there is an $\eta^{\ast}>0$ such that 01.\label{MP*} \end{itemize} Furthermore, there exists a positive number $H$ such that $0K>K_{0} there exists H^{\ast }\geq H>0 such that $$\min_{0\leq t\leq 1}f(t,y,y')>Ky,\;\;\ y\geq H\quad \text{and}\quad \frac{\alpha }{\beta }H\geq y'\geq -2H. \label{1}$$ Consider the function \begin{equation*} W(P):=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1), \end{equation*} where y\in \mathcal{X}(P_{1}) is any solution of differential equation ( \ref{E*}) starting at the point P_{1}:=(y_{1},y_{1}')\in E_{0} with y_{1}=H. By the assumption (\ref{A1}) (i.e. the nature of the vector field, see Remark \ref{R1}) it is obvious that y(t)\geq y_{1}=H and y'(t)\leq y_{1}'=\frac{\alpha}{\beta}y_{1}=\frac{\alpha}{ \beta}H,\ for all t in a sufficiently small neighborhood of t=0. Let's suppose that there is t^{\ast}\in(0,1] such that \begin{equation*} y(t)\geq H,\;-2H\leq y'(t)\leq\frac{\alpha}{\beta}H,\;0\leq t\frac{ 2\alpha}{\beta}. Furthermore, by (\ref{2}), H\leq y(t)-\bar{K}y',\quad 0\leq y\leq H \big[ 1+\frac{\alpha}{\beta}\big] \quad\text{and}\quad y'<-H. \label{210} We shall prove that $$\frac{\alpha}{\beta}H\geq y'(t)\geq-\varepsilon^{\ast}H>-2H,\quad 0\leq t\leq1. \label{22}$$ Indeed, since y'(t) is decreasing on \left[ 0,1\right], let's assume that there exist t_{0},t_{1}\in(0,1) such that \begin{equation*} y'(t_{0})=-H\;,\;-\varepsilon^{\ast}H2\big[1+\frac{\alpha}{\beta}-\frac{\sigma}{\xi_{m-2}}\big]. \end{equation*} Similarly by the superlinearity of f(t,y,y') at (0,0), for any \mu>0 there is an \eta>0 such that 0\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{ \beta }{\alpha }\big\{\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1 \big\}. It is now clear that the function W=W(P), P\in[P_{0},P_{1}] is continuous and thus by the Kneser's property (see Theorem \ref{Th1}), (\ref {5}) and (\ref{10}), we get a point P\in[P_{0},P_{1}] (we chose the last one to the left'' of P_{1}) such that W(P)=0. This fact clearly means that there is a solution y\in\mathcal{X}(P) of equation (\ref{E*}), such that \begin{equation*} W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)=0. \end{equation*} It remains to be proved that the so obtaining solution y=y(t) is actually a bounded function. Indeed, by the choice of P, the continuity of y(t) with respect initial values, (\ref{5}) and (\ref{10}), it follows that \begin{equation*} y(t)>0,\quad 0\leq t\leq1, \end{equation*} because if \begin{equation*} y(t)>0,\;0\leq t<1\quad\text{and}\quad y(1)=0, \end{equation*} then W(P)>0. Moreover by the nature of the vector field (see Remark \ref {R1}), there is t_{P}\in\left( 0,1\right) such that the so obtaining solution y\in\mathcal{X}(P) is strictly increasing on \left[ 0,t_{p} \right] , strictly decreasing on \left[ t_{p},1\right] and further is strictly positive on \left[ 0,1\right] . Also it holds y(t)\leq H, 0\leq t\leq1, i.e. 0 \frac{\alpha +\beta }{\beta } and recall that \begin{equation*} \sum_{i=1}^{m-2}\alpha_iy(\xi_i)+\frac{\beta }{\alpha }\Big\{ \sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\}0 such that $$\max_{0\leq t\leq 1}f(t,y,y')<\mu y,\quad y\geq H,\quad \text{and} \quad \frac{\alpha }{\beta }H\geq y'\geq m^{\ast }\frac{\alpha }{ \beta }H. \label{71}$$ Let's consider a point P_{0}:=(y_{0},y_{0}')\in E_{0} with y_{0}=H. We will prove first that for any solution y\in \mathcal{X}(P_{0}) , $$H\leq y(t)\leq \varepsilon_{0}H\quad \text{and}\quad \frac{m^{\ast }\alpha }{\beta }H\leq y'(t)\leq \frac{\alpha }{\beta }H,\quad 0\leq t\leq 1. \label{72}$$ Let us suppose that this is not the case. Then by the assumption (\ref{A1}), there is t^{\ast }\in [ 0,1] such that \begin{gathered} H\leq y(t)\leq\varepsilon_{0}H,\quad \frac{m^{\ast}\alpha}{\beta }H\leq y'(t)\leq\frac{\alpha}{\beta}H,\quad 0\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta }{\alpha}\big\{ \sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\big\}. On the other hand, since f_{0}=+\infty , for any K>\max \{\frac{2(\alpha -\beta )}{\beta },\frac{2\alpha }{\beta }\} there exist \eta \in (0,H) such that \min_{0\leq t\leq 1}f(t,y,y')>Ky,\;0\max \{\frac{2(\alpha -\beta )}{\beta },\frac{ 2\alpha }{\beta }\} we can easily prove that $$\frac{\eta }{2}\leq y(t)\leq \eta ,\quad 0\leq t\leq 1. \label{76}$$ We choose now \varepsilon_{0}^{\ast }\in \left( 1,2\right) and then\ by Assumption (\ref{A4}), there exist \bar{\eta}_{0}\in (0,\eta ) and 0-\eta ,\quad 0\leq t\leq 1. \label{78} Indeed since y'(t) is decreasing on \left[ 0,1\right] and \varepsilon_{0}^{\ast }\in \left( 1,2\right) is arbitrary, let's assume that there exist t_{0},t_{1}\in [ 0,1] such that y^{\prime }(t_{0})=-\bar{\eta}_{0}, \begin{equation*} -2\bar{\eta}_{0}<-\varepsilon_{0}^{\ast }\bar{\eta}_{0}\leq y^{\prime }(t)\leq -\bar{\eta}_{0},\quad t_{0}\leq t\frac{y(\xi^{\ast})}{\xi^{\ast}},$ where clearly $\xi^{\ast}=\xi_{m-2}$ and this contradicts the concavity of the solution $y=y(t)$. Furthermore we must seek the monotone (obviously increasing) solutions of (\ref{E})-(\ref{C}), only for the case $\sum_{i=1}^{m-2}\alpha_i\geq1$, since otherwise we get $0=W(P)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i)-y(1)<\Big[ \sum_{i=1} ^{m-2}\alpha_i-1\Big] y(1)<0.$ The question of existence of such a monotone solution remains open. However we can obtain a strictly decreasing solution for the boundary-value problem $$\begin{gathered} y''(t)=-f(t,y(t),y'(t)),\quad 0\leq t\leq1, \\ \alpha y(0)+\beta y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i). \end{gathered} \label{C*}$$ where $\alpha\geq0$ $\$and $\beta>0$. \end{remark} \begin{remark} \label{R3} \rm Suppose that the concept of jointly sublinearity is modified to $$\begin{gathered} f_{0}:=\lim_{y\to 0+}\min_{0\leq t\leq1}\frac{f(t,y,y')} {y}=+\infty,\quad \mbox{for } |y'|\leq M.\\ f_{\infty,-\infty}:=\lim_{(y,y')\to (+\infty,-\infty)} \max_{0\leq t\leq1}\frac{f(t,y,y')}{y}=0. \end{gathered} \label{A2*}$$ Then, following almost the same line as above (under the obvious modifications) we may prove the next theorem. \end{remark} \begin{theorem} Assume that (\ref{A1}) holds and further $\sigma^{\ast}=\sum_{i=1}^{m-2}\alpha_i\xi_i+\frac{\beta}{\alpha} \Big\{\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}-1\Big\} <1.$ Then the boundary-value problem (\ref{C*}) has a positive strictly decreasing solution provided that: \begin{itemize} \item The function $f$ is superlinear (see (\ref{A2S})) along with (\ref{A3}), or \item The function $f$ is sublinear (see (\ref{A2*})), (\ref{A4}) is true and in addition, $\sum_{i=1}^{m-2}\alpha_i\xi_i\big[ \frac{1}{\xi_{m-2}}-\frac{\alpha }{\beta}\big] >1.$ \end{itemize} Furthermore there exists a positive number $H$ such that $00.$ \end{remark} Finally consider the boundary-value problem $$\begin{gathered} y''+f(t,y,y')=0, y'(0)=0,\quad y(1)=\sum_{i=1}^{m-2}\alpha_iy(\xi_i). \end{gathered} \label{D}$$ Then following almost the same lines as above, we may prove the next theorem. \begin{theorem} Assume that (\ref{A1}) holds and $\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{1}}<1.$ Then the boundary-value problem (\ref{D}) has a positive strictly decreasing solution provided that \begin{itemize} \item The function $f$ is superlinear (see (\ref{A2S})) along with (\ref{A3}), or \item The function $f$ is sublinear (see (\ref{A2*})), (\ref{A4}) holds and in addition $\sum_{i=1}^{m-2}\alpha_i\frac{\xi_i}{\xi_{m-2}}>1.$ \end{itemize} Furthermore there exists a positive number $H$ such that \[ 0 | 6,609 | 17,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 16, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-04 | latest | en | 0.631466 |
http://mathhelpforum.com/algebra/139950-factor-expression-containing-rational-exponents-print.html | 1,519,513,299,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815951.96/warc/CC-MAIN-20180224211727-20180224231727-00160.warc.gz | 218,050,180 | 3,417 | # Factor an expression containing rational exponents
• Apr 18th 2010, 06:13 PM
desiderius1
Factor an expression containing rational exponents
Factor the expression. Express your answer so that only positive exponents occur.
$
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}
$
$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)
$
so far I've done:
$
(24x+12)^{(1/2)}(5x+1)^{(11/7)}+(18x+9)^{(3/2)}(5x+1)^{(4/7)}
$
$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(4x+4)(5x+1)^{(7/7)}(3x+3)
$
Is that right so far and if so, what do I do next?
Thanks.
• Apr 18th 2010, 06:35 PM
skeeter
Quote:
Originally Posted by desiderius1
Factor the expression. Express your answer so that only positive exponents occur.
$
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}
$
$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)
$
so far I've done:
$
\textcolor{red}{(24x+12)^{(1/2)}}(5x+1)^{(11/7)}+\textcolor{red}{(18x+9)^{(3/2)}}(5x+1)^{(4/7)}
$
these two factors are incorrect ... you cannot distribute the constants into factors w/ exponents not equal to 1.
$4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)}
$
common factors are $(6x+3)^{1/2}(5x+1)^{4/7}$ ...
$(6x+3)^{1/2}(5x+1)^{4/7}\left[4(5x+1) + 3(6x+3)\right]$
$(6x+3)^{1/2}(5x+1)^{4/7}\left[(20x+4) + (18x+9)\right]$
$(6x+3)^{1/2}(5x+1)^{4/7}\left[38x + 13\right]
$
...
• Apr 18th 2010, 07:12 PM
desiderius1
Another factor question:
$
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}
$
$
\frac{4(14+x)}{7x^{(1/2)}}
$
Wouldn't $8x^{(-1/2)} = \frac{1}{\sqrt{8x}}$?
I don't see how they got the $7x^{(1/2)}$ on the bottom.. when the exponent isn't negative.
• Apr 18th 2010, 07:22 PM
skeeter
Quote:
Originally Posted by desiderius1
Another factor question:
$
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}
$
$
\frac{4(14+x)}{7x^{(1/2)}}
$
Wouldn't $8x^{(-1/2)} = \frac{1}{\sqrt{8x}}$?
I don't see how they got the $7x^{(1/2)}$ on the bottom.. when the exponent isn't negative.
$
8x^{-1/2}+\frac{4}{7}x^{1/2}
$
$\frac{8}{x^{1/2}} + \frac{4x^{1/2}}{7}$
common denominator is $7x^{1/2}$ ...
• Apr 18th 2010, 07:36 PM
desiderius1
$
8 + 4x^{1/2}
$
How do you get 4(14 + x)?
• Apr 18th 2010, 07:48 PM
desiderius1
oh I see, you multiply 8 and 7 to get 56.
56/14 = 4. | 1,074 | 2,248 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-09 | longest | en | 0.660385 |
https://stats.stackexchange.com/questions/81396/clustering-algorithms-that-operate-on-sparse-data-matricies | 1,638,888,250,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00008.warc.gz | 576,489,204 | 31,677 | # Clustering algorithms that operate on sparse data matricies [closed]
I'm trying to compile a list of clustering algorithms that are:
1. Implemented in R
2. Operate on sparse data matrices (not (dis)similarity matrices), such as those created by the sparseMatrix function.
There are several other questions on CV that discuss this concept, but none of them link to R packages that can operate directly on sparse matrices:
So far, I've found exactly one function in R that can cluster sparse matrices:
# skmeans: spherical kmeans
From the skmeans package. kmeans using cosine distance. Operates on dgTMatrix objects. Provides an interface to a genetic k-means algorithm, pclust, CLUTO, gmeans, and kmndirs.
Example:
library(Matrix)
set.seed(42)
nrow <- 1000
ncol <- 10000
i <- rep(1:nrow, sample(5:100, nrow, replace=TRUE))
nnz <- length(i)
M1 <- sparseMatrix(i = i,
j = sample(ncol, nnz, replace = TRUE),
x = sample(0:1 , nnz, replace = TRUE),
dims = c(nrow, ncol))
M1 <- M1[rowSums(M1) != 0, colSums(M1) != 0]
library(skmeans)
library(cluster)
clust_sk <- skmeans(M1, 10, method='pclust', control=list(verbose=TRUE))
summary(silhouette(clust_sk))
The following algorithms get honerable mentions: they're not quite clustering algorithms, but operate on sparse matrices.
# apriori: association rules mining
From the arules package. Operates on "transactions" objects, which can be coerced from ngCMatrix objects. Can be used to make recommendations.
example:
library(arules)
M1_trans <- as(as(t(M1), 'ngCMatrix'), 'transactions')
rules <- apriori(M1_trans, parameter =
list(supp = 0.01, conf = 0.01, target = "rules"))
summary(rules)
# irlba: sparse SVD
From the irlba package. Does SVD on sparse matrices. Can be used to reduced the dimensionality of sparse matrices prior to clustering with traditional R packages.
example:
library(irlba)
s <- irlba(M1, nu = 0, nv=10)
M1_reduced <- as.matrix(M1 %*% s$v) clust_kmeans <- kmeans(M1, 10) summary(silhouette(clust_kmeans$cluster, dist(M1_reduced)))
# apcluster: Affinity Propagation Clustering
library(apcluster)
sim <- crossprod(M1)
sim <- sim / sqrt(sim)
clust_ap <- apcluster(sim) #Takes a while
What other functions are out there?
• Do you mean sparse as in "lots of zeros" or as in "lots of missing values"? Jan 7 '14 at 14:52
• This question appears to be off-topic according to multiple criteria at stats.stackexchange.com/help/dont-ask: every answer would be equally valid, you expect more answers in addition to those provided, and there is no actual problem to be solved.
– whuber
Jan 7 '14 at 14:55
• I realise this got closed, but I've been tripping over all your questions on this as I browse SO as I had a similar problem ;) I found this library which uses affinity propensity that can work with sparse matrices: bioinf.jku.at/software/apcluster Jul 9 '15 at 11:28
• @MarkeD Thanks a lot! It's really too bad software recommendations are off-topic here, as I've found nowhere else online to ask for them.
– Zach
Jul 9 '15 at 12:56
• once again very useful question is closed :( if you dont know the answer just dont vote for close! Apr 4 '16 at 23:03
I don't use R. It is often very slow and has next to no indexing support. But software recommendations are considered off-topic anyway.
Note that plenty of algorithms don't care how you store your data. If you prefer to have a sparse matrix, that should be your choice, not the algorithms choice.
People that use too much R tend to get stuck in thinking in matrix operations (because that is the only way to write fast code in R). But that is a limited way of thinking. For example k-means: it doesn't care. In particular, it doesn't use pairwise distances at all. It just needs a way to compute the variance contribution; which is equivalent to computing the squared Euclidean distance.
Or DBSCAN. All it needs is a "neighbor" predicate. It can work with arbitrary graphs; it's just that Euclidean distance and the Epsilon threshold is the most common way of computing the neighborhood graph it uses.
P.S. Your question isn't very precise. Do you refer to sparse data matrixes or sparse similarity matrixes?
• sparse data matrixes
– Zach
Jan 6 '14 at 20:35
• Most algorithms can operate on sparse data matrixes. E.g., AGNES, PAM, DBSCAN, OPTICS, CLARA, ... Oct 11 '17 at 21:29
• Not sure why you even answered if you don't even know R. Jul 31 '19 at 16:08
• I know R. Probably even better than the average R user. I know non-standard evaluation in R and I know that most of the modules are written in C, so when you pass a sparse matrix, it is first copied into a sense matrix before passing it to the actual code. And every package uses a different way of doing so... That is not efficient. You don't choose R if you need efficiency or good integration or backwards compatibility or coordinated development. Jul 31 '19 at 17:40 | 1,252 | 4,876 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | latest | en | 0.703427 |
http://www.java-gaming.org/index.php?topic=2516.msg23914 | 1,529,654,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00232.warc.gz | 441,034,375 | 18,805 | Java-Gaming.org Hi !
Featured games (91) games approved by the League of Dukes Games in Showcase (757) Games in Android Showcase (229) games submitted by our members Games in WIP (844) games currently in development
News: Read the Java Gaming Resources, or peek at the official Java tutorials
Pages: [1]
ignore | Print
Basic question, a beginner ! (Read 4678 times) 0 Members and 1 Guest are viewing this topic.
alvaro
Junior Newbie
Java games rock!
« Posted 2003-11-10 06:40:41 »
Hi,
can somebody tell me when this formula can be used, where can I get more information about it?
dir_x = Math.cos(alpha*(Math.PI/180));
dir_y = Math.sin(alpha*(Math.PI/180));
....
ball.pos_x+=dir_x ...
ball.pos_y+=dir_y ...
regards
Alvaro
Orangy Tang
JGO Kernel
Medals: 57
Projects: 11
« Reply #1 - Posted 2003-11-10 06:58:39 »
Looks like an equation to generate a unit circle, what do you actually want it to do?
[ TriangularPixels.com - Play Growth Spurt, Rescue Squad and Snowman Village ] [ Rebirth - game resource library ]
Herkules
Senior Devvie
Friendly fire isn't friendly!
« Reply #2 - Posted 2003-11-10 07:02:00 »
You use it when calculating the x/y translation when you have a angle of direction given in degrees [0,360].
Did I get the question right?
HARDCODE -- DRTS/FlyingGuns/JPilot/JXInput -- skype me: joerg.plewe
Orangy Tang
JGO Kernel
Medals: 57
Projects: 11
« Reply #3 - Posted 2003-11-10 07:12:09 »
Oh yeah, convert an angle to a vector, that makes more sense. I had assumed that the alpha meant it was being interpolated over, instead its actually an input angle.
Guessing the question is still the tricky part though
[ TriangularPixels.com - Play Growth Spurt, Rescue Squad and Snowman Village ] [ Rebirth - game resource library ]
cfmdobbie
Senior Devvie
Medals: 1
Who, me?
« Reply #4 - Posted 2003-11-10 12:48:43 »
Okay, while we're playing some kind of bizzare Jeopardy, try this:
Programmer Tom is writing some code and uses the decimal number 1597463007. What is his method likely intended to do?
Hellomynameis Charlie Dobbie.
kevglass
« JGO Spiffy Duke »
Medals: 319
Projects: 25
Exp: 22 years
Coder, Trainee Pixel Artist, Game Reviewer
« Reply #5 - Posted 2003-11-10 16:19:04 »
Random number generator
Kev
Jeff
JGO Coder
Got any cats?
« Reply #6 - Posted 2003-11-10 17:25:23 »
Thats fundemental trig. Try any trigonometry textbook.
Alpha is your angle of direction in degrees. pi/180 is a conversion from degress to radians, which is what all of the trig functions want.
sin == opposite/hypotenuse == Y part of that angle (assuming a total movement of 1, hypotenuse is actually the total distance moved, opposite is the Y component.) Note that this assumes a
trig style coordinate system where poitive Y is at the top and negative Y at the bottom. if you are working in screen space you'll need to reverse it.
cos == adjacent/hypotenuse == X part of that component.
So your simple answer is. given a ship position x,y and angle of motion alpha, that math calculates the new position if the ship moves 1 unit in alpha direction.
Basic trig. if you can't handle this you will have a LOT of "fun" trying to write game code....
Got a question about Java and game programming? Just new to the Java Game Development Community? Try my FAQ. Its likely you'll learn something!
http://wiki.java.net/bin/view/Games/JeffFAQ
Jeff
JGO Coder
Got any cats?
« Reply #7 - Posted 2003-11-10 17:28:56 »
Btw...
This question has nothing to do with physcis and really belongs in "Newless Clubies" which is our total beginner topic.
Got a question about Java and game programming? Just new to the Java Game Development Community? Try my FAQ. Its likely you'll learn something!
http://wiki.java.net/bin/view/Games/JeffFAQ
alvaro
Junior Newbie
Java games rock!
« Reply #8 - Posted 2003-11-11 10:54:51 »
regards
Jeff
JGO Coder
Got any cats?
« Reply #9 - Posted 2003-11-12 00:43:48 » | 1,123 | 3,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-26 | latest | en | 0.803412 |
https://www.debate.org/debates/The-Big-Bang-may-have-occured/1/ | 1,556,051,309,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613603.65/warc/CC-MAIN-20190423194825-20190423220825-00257.warc.gz | 648,900,229 | 11,054 | All Big Issues
The Instigator
Pro (for)
Winning
6 Points
The Contender
Con (against)
Losing
3 Points
# The Big Bang may have occured
Do you like this debate?NoYes-1
Post Voting Period
The voting period for this debate has ended.
after 2 votes the winner is...
PwningEinstein
Voting Style: Open Point System: 7 Point Started: 12/15/2013 Category: Science Updated: 5 years ago Status: Post Voting Period Viewed: 778 times Debate No: 42411
Debate Rounds (4)
Pro Rules:1) Con will argue in favor of the resolution that the Big Bang did not occur, as its mainpoints may not be inherent from a religious standpoint, or scientific facts. 2) No trolling.3) Con shall type "No arguments shall be made as agreed upon" at round 4.Failure to follow these rules will result in a 7 point forfeiture. Report this Argument Con how is earth orbitting the sun? that doesnt just suddenly happen in a big bangReport this Argument Pro Rebuttals:"how is earth orbitting the sun? that doesnt just suddenly happen in a big bang"The Earth is compelled to orbit the Sun at the same distance away from the Sun because the Sun's gravitational pull is pulling Earth inwards. However, its rotation compensates, in which it pulls itself outward.Report this Argument Con `Newton's second law is not true because an airplane is faster than a car, while an airplane has more inertia and mass!` Report this Argument Pro Rebuttals:"Newton's second law is not true because an airplane is faster than a car, while an airplane has more inertia and mass!"According to Newton's second law of physics[1], the net force has to be in proportion to, and in the same direction as, the object acted upon.In other words, the net force a car is acted upon is weaker than the net force an airplane is acted upon.Also, why did you change the subject?Sources: [1] http://en.wikipedia.org...'s_laws_of_motion Report this Argument Con Also, in an environment without friction you would have this spinning dot going so fast it would then explode. If this happened, then all of the particles and matter being expelled from this "spinning dot" would all have to spin in the same direction as the dot they exploded from. This is a known law of science, which those who believe in Evolution cannot do away with. It is known as the Conservation of angular momentum. This matter which is said to have created the planets would all need to spin in the same direction as the object it came from. So therefore, all of the planets should be spinning in the same direction. However two of them are not. Venus and Uranus spin backwards. Some planets even have moons that not only spin backwards, but travel backward around their planets. Report this Argument Pro Rebuttals:"Also, in an environment without friction you would have this spinning dot going so fast it would then explode. If this happened, then all of the particles and matter being expelled from this "spinning dot" would all have to spin in the same direction as the dot they exploded from. This is a known law of science, which those who believe in Evolution cannot do away with. It is known as the Conservation of angular momentum." People do so differently, but I will assume that this is one of the mainpoints.Yes, indeed. According to the laws of physics (Conservation of angular momentum[1], in particular), especially in an environment without friction, when a piece comes off a spinning object, that piece must spin in the same direction.And you assume that the spinning dot span in this direction:However, some do not. Venus and Uranus spin backwards.However, that is not what happened.Asteroids came flying out after the cataclysmic explosion due to the extreme density. The asteroids span in the same direction.Asteroids go through a process where they fuse, forming stars, etc.However, when so happens, they no longer spin, thus not violating the laws of conservation of angular momentum. If any piece spins, it must come off the formation."Some planets even have moons that not only spin backwards, but travel backward around their planets."The planets, stars, etc. did not come out right after the cataclysmic explosion. They were formed.The Earth, for instance, is compelled to orbit the Sun in the same distance because the Sun's gravitational pull is pulling it towards its center. However, the Earth's rotation compensates. It's rotation pulls it outward.Sources:[1] http://en.wikipedia.org...Report this Argument Con "No arguments shall be made as agreed upon"Report this Argument | 981 | 4,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-18 | latest | en | 0.91494 |
https://www.jiskha.com/questions/14541/why-is-this-problem-37-1-correct-sam-scored-45-points-againd-memphis-in-an-nba-game-he | 1,620,494,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00074.warc.gz | 855,070,300 | 4,807 | # Math
Why is this problem 37,1% correct. Sam scored 45 points againd Memphis in an NBA game. He made 13 of 37 shots what was his shooting percentage? Multiple choice test. 37% 37.4% 37.1% 37.3%
1. 👍
2. 👎
3. 👁
## Similar Questions
1. ### MATH
Write an inquality for the situation. Sam scored at least 26 points. A. K > 26 B. K => 26 C. K =< 26 D. K> 26 Hey guys please help. ASAP
Which sentence uses a verb in the interrogative mood? A) Did you score a touchdown at the football game? B) At last week's football game, you scored two touchdowns! C) I wondered if you scored a touchdown at the football game. D)
3. ### Math help please
1. Each time a touchdown is scored in a football game, 6 points are added to the score of the scoring team. A team already has 12 points. What rule represents the number of points as an arithmetic sequence when n is the number of
4. ### ASVAB Arithmetic
In Tim's last 5 games here's the number of three-point field goals scored per game. 9 6 0 9 3 How many three-point field goals would Tim have to kick in the next game to bring his average (arithmetic mean) to six points per game?
1. ### math
Lisa and Katie are playing a card game, and a total of 900 points has been scored. Lisa scored 150 more points than Katie. If you let l= the number of points that Lisa scored, and k= the number of points that Katie scored, then
2. ### Math
A back-to-back stem and leaf plot showing the points scored by each player on two different basketball teams is shown below. Points scored in a game Team one. | | Team 2 4 3 | 0 | 4 9 9 8 2 0 | 1 | 0 1 7 6 1 | 2 | 3 3 8
3. ### Math
Craig scored 12 points in a game. Marla scored twice as many points as Craig but 5 fewer points than Nelson scored. How many points did Nelson score? a. 2x12+5 b. 2 x 12 -5 c. 1/2 x 12 + 5 d. 2 x (12 + 5) Answer-b Can you please
4. ### Math
In the last basketball game, Kevin scored 2 less than a third of his team's points. Part A: Let n represent the number of points Kevin's team scored. Write an expression for the number of points Kevin scored. Part B: Kevin scored
1. ### Dot Plots
Susan kept track of the number of points she scored in each game during basketball season. How many games did Susan score 10 points or more? Number of Points Scored 2= 2 dots 4= 3 dots 6= 2 dots 8= 4 dots 10= 1 dot 12= 4 dots 14=
2. ### Math
Basketball player scored 12 times during one game. He scored a total of 18 points, two for each two-point Shot and one for each free throw. How many two-point shots did he make? How many free throws?
3. ### algebra
a volleyball team scored 17 points in its first game than in its third game . in the second game, the team scored 23 points. The total numbers of points scored was less than 60. What is the greatest number of points the team could
4. ### mathematics
Kris and Julio played a card game. Together, they scored 36 points in one game. Kris scored 2 times as many points as Julio. How many points did Kris and Julio each score? Write an equation and solve. Explain your work. | 851 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-21 | latest | en | 0.957797 |
https://dsp.stackexchange.com/questions/40303/how-to-look-at-rubidium-frequency-standard-using-the-oscilloscope | 1,621,162,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00105.warc.gz | 244,419,119 | 39,400 | # How to look at Rubidium frequency standard using the oscilloscope?
I have the Rubidium frequency standard: http://www.thinksrs.com/products/PRS10.htm
and the Oscilloscope: https://www.atecorp.com/products/tektronix/tds3034
From the Oscilloscope, I want to look at 1PPS and 10MHz from the Rubidium frequency standard, How to do that?
Thank you very much.
From the Oscilloscope, I want to check 1PPS and 10MHz from the Rubidium frequency standard, How to do that?
Not at all.
Accuracy: 200 ppm
That's relatively bad, even for cheap measurement equipment.
From the rubidium oscillator's datasheet:
Accuracy at shipping: $\pm 5 \cdot 10^{-11}$
which is
$$\frac{2\cdot 10^{-4}}{5\cdot 10^{-11}}\,$$
i.e. 4 Million times more accurate than your oscilloscope promises to be.
I'd very much expect a comparable difference in oscillator phase noise.
In other words: your rubidium clock is so much better than your oscilloscope that you can only measure your oscilloscope by observing the rubidium clock, not the other way around.
Generally, you'll have a hard time finding an oscillator that beats your rubidium clock in accuracy, so that you can asses it. You can compare long-term against a good GPS-disciplined oscillator, or against other, more expensive atomic clocks.
• Thanks for quick response, but I would like to know a little more about rubidium clock, how to observe the rubidium clock? – Nate Duong Apr 17 '17 at 14:48
• what do you mean, "observe"? – Marcus Müller Apr 17 '17 at 14:49
• I want to see microsecond variation, is it vary? how much is it vary? How much differences there from PPS output to rubidium clock? Microsecond in variation I should be able to see if I were change this scope trigger on 1 of these PPS output and then I look at the others. And since system is 300MHz bandwidth and I can see of the other of 3 nanoseconds of variation, so microsecond variation in a few, if I trigger on 1 and watch the other, what does it do? That is my goal to check Rubidium frequency standard using the oscilloscope – Nate Duong Apr 17 '17 at 14:51
• read my answer. You can not use your oscilloscope to verify something that is 4 million times better than your oscilloscope. That's like using a scale designed to weigh elephants to verify the weight of a single dust particle. The whole point of having a frequency standard is to calibrate the rest of your equipment. – Marcus Müller Apr 17 '17 at 14:55
• I repeat myself: You can not check your rubidium clock with your oscilloscope. That'd be like trying to check the diameter of a single hair by putting it next to a cricket field. Seriously. – Marcus Müller Apr 17 '17 at 15:00
but I just want to check this rubidium are still working or not because once we dropped it on the floo
Short Rant:
OK. we have a sixteen messages conversation under my answer after you wrote a question, and only four full days after asking, it occurs to you that oops I dropped my multi-thousand-dollar frequency standard might be a relevant piece of info???
So, this is clearly no signal processing problem. You might have ruined a piece of equipment that someone else might rely on with their PhD or job. Find someone who's actually competent enough to diagnose the device, if in doubt, call the company that produced it. They have calibration services, and will be able to do that properly.
A "possibly broken", "checked only by a so-and-so competent" measurement standard is worse than not having a measurement standard; it might ruin future research. It's mandatory you inform the person in charge of the lab, and clearly mark the device as being of unknown state of functionality. | 887 | 3,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.919185 |
http://oeis.org/A075271 | 1,571,111,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655864.19/warc/CC-MAIN-20191015032537-20191015060037-00007.warc.gz | 140,523,142 | 4,488 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A075271 a(0) = 1 and, for n >= 1, (BM)a(n) = 2*a(n-1), where BM is the BinomialMean transform. 12
1, 3, 17, 211, 5793, 339491, 41326513, 10282961907, 5181436229441, 5258784071302723, 10717167529963833681, 43779339268428732008723, 358114286723184561034838497, 5862685570087914880854259126371, 192026370558313054275618817346778353 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The BinomialMean transform BM is defined by (BM)a(n) = (M^n)a(0) where (M)a(n) is the mean (a(n) + a(n+1))/2, or, alternatively, by (BM)a(n) = (Sum_{k=0..n} binomial(n,k)*a(k))/(2^n). The BinomialMean transform of this sequence is given in A075272. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..50 Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1. FORMULA O.g.f. as a continued fraction: A(x) = 1/(1 + x - 2^2*x/(1 - 2*(2 - 1)^2*x/(1 + x - 2^4*x/(1 - 2*(2^2 - 1)^2*x/(1 + x - 2^6*x/(1 - 2*(2^3 - 1)^2*x/(1 + x - 2^8*x/(1 - 2*(2^4 - 1)^2*x/(1 + x - ... ))))))))). Cf. A075272. - Peter Bala, Nov 10 2017 EXAMPLE Given that a(0)=1 and a(1)=3. Then (BM)a(2) = (1 + 2*3 + a(2))/4 = 2a(1) = 6, hence a(2)=17. MAPLE iBM:= proc(p) proc(n) option remember; add(2^(k)*p(k)*(-1)^(n-k) *binomial(n, k), k=0..n) end end: a:= iBM(aa): aa:= n-> `if`(n=0, 1, 2*a(n-1)): seq(a(n), n=0..16); # Alois P. Heinz, Sep 09 2008 MATHEMATICA iBM[p_] := Module[{proc}, proc[n_] := proc[n] = Sum[2^k*p[k]*(-1)^(n-k) * Binomial[n, k], {k, 0, n}]; proc]; a = iBM[aa]; aa[n_] := If[n == 0, 1, 2*a[n-1]]; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Nov 08 2015, after Alois P. Heinz *) Table[Sum[QFactorial[k, 2] Binomial[n + 1, k]/2, {k, 0, n + 1}], {n, 0, 15}] (* Vladimir Reshetnikov, Oct 16 2016 *) CROSSREFS Cf. A075272. Sequence in context: A210898 A009494 A267659 * A194925 A072350 A181032 Adjacent sequences: A075268 A075269 A075270 * A075272 A075273 A075274 KEYWORD eigen,nonn AUTHOR John W. Layman, Sep 11 2002 EXTENSIONS More terms from Alois P. Heinz, Sep 09 2008 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified October 14 23:59 EDT 2019. Contains 328025 sequences. (Running on oeis4.) | 1,007 | 2,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-43 | latest | en | 0.497798 |
http://piclist.com/techref/microchip/math/radix/index.htm | 1,725,727,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00445.warc.gz | 26,562,020 | 11,871 | Also:
Archive:
• Thank's a lot for such an useful site ! I got several routines there and all of them work perfectly well, saving hours of brain storming for me with probably not such so good results... Denis, from France+
• You are VERY welcome! Please do consider shareing any code you do end up writing
Questions:
• KILLspambrianszczepanik@spam@ at yahoo.ca asks: " im looking for a way to convert an eight digit decimal number to two 16 bit words using only 16 bit architecture" +
• KILLspamszaroletta@spam@ at hotmail.com asks: " I am looking for relocatable code, or a macro which converts a hex number like FE8A into the corresponding ASCII-Code."
Convert the Hex digits to binary (16bits) and then convert the binary to ASCII.
Apply one of the "2 digit Hex to Binary 8 bits" routines to the first two hex digits and then again to the second, thereby producing two 8 bit values that are then passed to any one of the "16bit to ASCII Decimal 5 digits" routines. +
Interested:
See:
• shares this code:
```;some code snippets that someone might find useful...
;This routine will return the number of decimal
;hundreds from an 8-bit binary
;Input: w
;Output: w
;RAM:2
;Cycles: 12-24
GETHNDS movwf t1
clrf w2
gethnds_loop movlw .100
incf w2,f
subwf t1,f
btfsc STATUS,C
goto gethnds_loop
decf w2,w
return
;---
;This routine will return the number of decimal
;tens from an 8-bit binary
;Loop based, so it might take some time...
;It needs getones too
GETTENS movwf t1
clrf w2
gettens_loop movlw .10
incf w2,f
subwf t1,f
btfsc STATUS,C
goto gettens_loop
decf w2,w
goto getones
;---
;This routine will return the number of decimal
;ones from an 8-bit binary
GETONES movwf w2
movlw .10
deltens_loop subwf w2,f
btfsc STATUS,C
goto deltens_loop
return
```
+
+
Code:
• ```Compact and fast 10bit to BCD conversion for converting
;****************************************************
;* 10bit BINARY to BCD *
;****************************************************
;
;An implimentation of the 'Shift and add-3' Algorithm
; for small MicroChip 'PIC' microcomputers
;
;ALGORITHM
;
;1)Shift binary number one bit left (into a BCD 'result'
; (initially empty)
;2)If any 4 bit BCD column is greater than 5[hex] add 3[hex]
;3)Goto 1) {until the LSB binary bit is shifted out}
;
;<------MSB LSB----->
;------------- BCD ------------] [------ binary -----
;
; TENS UNITS
; BCD column BCD column MSB bit bit
; 0 0 0 0 0 0 0 0 1 0 1
;
; <------ / <---/ <---/
;Inalisation
;**************
Processor 16F818
include <p16f818.inc>
cblock 0x020 ;Define variable block starting at \$020
binH ;bin is the 8bit binary value to be converted
bin ;The 2bit binary MSB's to be converted
bcdH ;Thousands (always blank)/ Hundreds nybbles
bcdL ;Tens / Units nybbles
counter
temp
endc
_bin2bcd movlw d'8'
movwf counter
clrf bcdL
clrf bcdH
;Save time by not shifting in first 6 bits (always '0's)
;--------------------------------------------------------
swapf binH,1 ;Chop off first nybble (TEN THOUSANDS)
rlf binH,1 ;Shift out first 2 MSB's(always '0's)
rlf binH,1
;Ssave more ime by no test and add +3' for first TWO shifts
;---------------------------------------------------------------------
rlf binH,F ;Shifting 'binH' left through carry into BCD nybbles
rlf bcdL,F
rlf binH,F
;Iteration loop shifts the 4 'columns' (10TH, THO, HUN, TEN, UNT) 1 bit left
;Tests each coulmn (4bit nybble) if > 5, adds '3'
;Shifts in the next MSB of the binary for conversion on the right.
;(for remaining EIGHT shifts)
Next_bit movfw bcdL
movwf temp
movfw bcdL
btfsc temp,3 ;Test MSB of Units+3 nybble in 'temp
addlw 0x03 ;Add 3 to U nybble if units+3 MSB = 1
btfsc temp,7 ;Test MSB of Tens+3nybble in 'temp'
movwf bcdL
movfw bcdH
mmovfw bcdH
btfsc temp,3 ;Test MSB of Hundreds+3 nybble in temp
btfsc temp,7 ;Test MSB of Thousand+3 nybble in temp
movwf bcdH
rlf bin,F ;Shift in next MSB from bin into T/UNITS
rlf bcdL,F ;Shift next MSB from TENS/UNITS to TH/H
rlf bcdH,F ;Shift up
decfsz counter,F
goto Next_bit
return
end
```
+
file: /Techref/microchip/math/radix/index.htm, 17KB, , updated: 2019/8/26 09:43, local time: 2024/9/7 09:45, TOP NEW HELP FIND: 98.81.24.230:LOG IN
©2024 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions?Please DO link to this page! Digg it! / MAKE! PIC Microcontoller Radix Math Methods
After you find an appropriate page, you are invited to your to this massmind site! (posts will be visible only to you before review) Just type a nice message (short messages are blocked as spam) in the box and press the Post button. (HTML welcomed, but not the <A tag: Instead, use the link box to link to another page. A tutorial is available Members can login to post directly, become page editors, and be credited for their posts.
Attn spammers: All posts are reviewed before being made visible to anyone other than the poster.
Did you find what you needed? "No. I'm looking for: " "No. Take me to the search page." "No. Take me to the top so I can drill down by catagory" "No. I'm willing to pay for help, please refer me to a qualified consultant"
PICList 2024 contributors: o List host: MIT, Site host massmind.org, Top posters @none found - Page Editors: James Newton, David Cary, and YOU! * Roman Black of Black Robotics donates from sales of Linistep stepper controller kits. * Ashley Roll of Digital Nemesis donates from sales of RCL-1 RS232 to TTL converters. * Monthly Subscribers: Gregg Rew. on-going support is MOST appreciated! * Contributors: Richard Seriani, Sr.
. | 1,778 | 5,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.711232 |
https://softuni.bg/forum/answers/details/55883 | 1,642,541,879,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00092.warc.gz | 550,467,571 | 14,635 | anton_fotev 9 Точки
## Runtime error на *Kamino Factory
Колеги, пробвах по три различни начина да реша задачата. И трите пъти ми дава Runtime error на поне половината отговори.
Прилагам и трите варианта на решението.
условието на задачата:
9. *Kamino Factory
The clone factory in Kamino got another order to clone troops. But this time you are tasked to find the best DNA
sequence to use in the production.
You will receive the DNA length and until you receive the command "Clone them!" you will be receiving a DNA
sequences of ones and zeroes, split by "!" (one or several).
You should select the sequence with the longest subsequence of ones. If there are several sequences with same
length of subsequence of ones, print the one with the leftmost starting index, if there are several sequences with
same length and starting index, select the sequence with the greater sum of its elements.
After you receive the last command "Clone them!" you should print the collected information in the following
format:
"Best DNA sample {bestSequenceIndex} with sum: {bestSequenceSum}."
"{DNA sequence, joined by space}"
Input / Constraints
- The first line holds the length of the sequences – integer in range [1…100];
- On the next lines until you receive "Clone them!&" you will be receiving sequences (at least one) of ones and
zeroes, split by "!" (one or several).
Output
The output should be printed on the console and consists of two lines in the following format:
"Best DNA sample {bestSequenceIndex} with sum: {bestSequenceSum}."
"{DNA sequence, joined by space}"
Examples
5
1!0!1!1!0
0!1!1!0!0
Clone them!
Best DNA sample 2 with sum: 2.
0 1 1 0 0
We receive 2 sequences with same length of
subsequence of ones, but the second is printed,
because its subsequence starts at index[1].
4
1!1!0!1
1!0!0!1
1!1!0!0
Clone them!
Best DNA sample 1 with sum: 3.
1 1 0 1
We receive 3 sequences. Both 1 and 3 have same
length of subsequence of ones -> 2, and both start
from index[0], but the first is printed, because its sum
is greater.
Линк към джъджа: https://judge.softuni.bg/Contests/Practice/Index/1206#8
1. Вариант на решението
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int n = int.Parse(Console.ReadLine());
int[] maxSeries = new int[n];
int maxSequense = 0;
int maxIndexPosition = n - 1;
int maxSumOfLine = 0;
int lineMax = 0;
int line = 0;
while (true)
{
string input = Console.ReadLine();
if (input == "Clone them!")
{
break;
}
line++;
int[] series = input
.Split('!')
.Select(int.Parse)
.ToArray();
int sequense = 0;
int indexPosition = n-1;
int sumOfLine = 0;
int currentSequense = 0;
for (int i = n - 1; i >= 0; i--)
{
if (series[i] == 1)
{
sumOfLine++;
currentSequense++;
if (currentSequense >= sequense)
{
sequense = currentSequense;
indexPosition = i;
}
}
else
{
currentSequense = 0;
}
} // end for
bool currentlineBestOfMaxLine = IsCurrentlineBestOfMaxLine
(maxSequense, maxIndexPosition, maxSumOfLine, sequense, indexPosition, sumOfLine);
if (currentlineBestOfMaxLine)
{
maxSequense = sequense;
maxIndexPosition = indexPosition;
maxSumOfLine = sumOfLine;
maxSeries = series;
lineMax = line;
}
} // end while
Console.WriteLine("Best DNA sample {0} with sum: {1}.", lineMax, maxSumOfLine);
Console.WriteLine(string.Join(" ", maxSeries));
Console.WriteLine();
}
public static bool IsCurrentlineBestOfMaxLine
(int maxSequense, int maxIndexPosition, int maxSumOfLine, int sequense, int indexPosition, int sumOfLine)
{
if (maxSequense < sequense)
{
return true;
}
else if (maxSequense == sequense && maxIndexPosition > indexPosition)
{
return true;
}
else if (maxSequense == sequense && maxIndexPosition == indexPosition && maxSumOfLine < sumOfLine)
{
return true;
}
else
{
return false;
}
}
}
2. Втори вариант - с използване на лист, за да избегна презаписването на масиви.
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int sequenceLingth = int.Parse(Console.ReadLine());
string dna = Console.ReadLine();
List<int[]> allDNA = new List<int[]>();
//int line = 0;
while (dna != "Clone them!")
{
/// line++;
int[] series = dna
.Split('!')
.Select(int.Parse)
.ToArray();
if (series.Length == sequenceLingth)
{
}
} // end while
int maxSequense = 0;
int maxIndexPosition = sequenceLingth - 1;
int maxSumOfLine = 0;
int lineMax = 0;
for (int lineDna = 0; lineDna < allDNA.Count; lineDna++)
{
int sequense = 0;
int indexPosition = sequenceLingth-1;
int sumOfLine = 0;
int currentSequense = 0;
for (int i = allDNA[lineDna].Length - 1; i >= 0; i--)
{
if (allDNA[lineDna][i] == 1)
{
sumOfLine++;
currentSequense++;
}
else
{
if (currentSequense >= sequense)
{
sequense = currentSequense;
indexPosition = i;
}
currentSequense = 0;
}
} // end internal for
bool currentlineBestOfMaxLine = IsCurrentlineBestOfMaxLine
(maxSequense, maxIndexPosition, maxSumOfLine, sequense, indexPosition, sumOfLine);
if (currentlineBestOfMaxLine)
{
maxSequense = sequense;
maxIndexPosition = indexPosition;
maxSumOfLine = sumOfLine;
lineMax = lineDna;
}
} // end external for
Console.WriteLine("Best DNA sample {0} with sum: {1}.", lineMax +1, maxSumOfLine);
Console.WriteLine(string.Join(" ", allDNA[lineMax]));
}
public static bool IsCurrentlineBestOfMaxLine
(int maxSequense, int maxIndexPosition, int maxSumOfLine, int sequense, int indexPosition, int sumOfLine)
{
if (maxSequense < sequense)
{
return true;
}
else if (maxSequense == sequense)
{
if (maxIndexPosition > indexPosition)
{
return true;
}
else if (maxIndexPosition == indexPosition)
{
if (maxSumOfLine < sumOfLine)
{
return true;
}
else
{
return false;
}
}
else
{
return false;
}
}
else
{
return false;
}
}
}
3. Трети вариант - без буул и метод, за да облекча пресмятанията на процесора
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
public static void Main()
{
int sequenceLingth = int.Parse(Console.ReadLine());
string dna = Console.ReadLine();
List<int[]> allDNA = new List<int[]>();
//int line = 0;
while (dna != "Clone them!")
{
/// line++;
int[] series = dna
.Split('!')
.Select(int.Parse)
.ToArray();
if (series.Length == sequenceLingth)
{
}
} // end while
int maxSequense = 0;
int maxIndexPosition = sequenceLingth - 1;
int maxSumOfLine = 0;
int lineMax = 0;
for (int lineDna = 0; lineDna < allDNA.Count; lineDna++)
{
int sequense = 0;
int indexPosition = sequenceLingth-1;
int sumOfLine = 0;
int currentSequense = 0;
for (int i = allDNA[lineDna].Length - 1; i >= 0; i--)
{
if (allDNA[lineDna][i] == 1)
{
sumOfLine++;
currentSequense++;
}
else
{
if (currentSequense >= sequense)
{
sequense = currentSequense;
indexPosition = i;
}
currentSequense = 0;
}
} // end internal for
if (maxSequense < sequense)
{
maxSequense = sequense;
maxIndexPosition = indexPosition;
maxSumOfLine = sumOfLine;
lineMax = lineDna;
}
else if (maxSequense == sequense)
{
if (maxIndexPosition > indexPosition)
{
maxSequense = sequense;
maxIndexPosition = indexPosition;
maxSumOfLine = sumOfLine;
lineMax = lineDna;
}
else if (maxIndexPosition == indexPosition)
{
if (maxSumOfLine < sumOfLine)
{
maxSequense = sequense;
maxIndexPosition = indexPosition;
maxSumOfLine = sumOfLine;
lineMax = lineDna;
}
}
}
} // end external for
Console.WriteLine("Best DNA sample {0} with sum: {1}.", lineMax +1, maxSumOfLine);
Console.WriteLine(string.Join(" ", allDNA[lineMax]));
}
}
Тагове:
0 | 2,195 | 7,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-05 | latest | en | 0.883597 |
https://www.proprofs.com/quiz-school/story.php?title=atomic-bomb-quiz | 1,709,552,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00534.warc.gz | 935,564,004 | 98,686 | # The Atomic Bomb Quiz
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Yellow_bottle
Y
Yellow_bottle
Community Contributor
Quizzes Created: 1 | Total Attempts: 330
Questions: 10 | Attempts: 330
Settings
This is a short quiz to test you on whatever you have learnt from our atomic bomb awareness kit. Enjoy!
• 1.
### What was the radius of total destruction (during the Hiroshima bombing)?
• A.
2 miles
• B.
1 mile
• C.
1.5 miles
B. 1 mile
Explanation
The radius of total destruction during the Hiroshima bombing was 1 mile. This means that everything within a 1-mile radius of the bomb's epicenter was completely destroyed. This includes buildings, infrastructure, and anything else in the blast zone. The immense power and impact of the bomb caused widespread devastation and loss of life within this radius.
Rate this question:
• 2.
### How high above the ground did the 'Little Boy' explode?
• A.
1700 feet
• B.
2000 feet
• C.
1900 feet
• D.
1750 feet
C. 1900 feet
Explanation
The 'Little Boy' exploded at a height of 1900 feet above the ground.
Rate this question:
• 3.
### For beta particle emission to occur, the number of protons must be ______ than the number of neutrons.
• A.
Higher
• B.
Lower
• C.
It is not neutrons, but electrons
• D.
None of the above
B. Lower
Explanation
Beta particle emission occurs when a neutron in the nucleus of an atom is converted into a proton and an electron. The electron is then emitted from the nucleus as a beta particle. This process reduces the number of neutrons in the nucleus and increases the number of protons. Therefore, for beta particle emission to occur, the number of protons must be lower than the number of neutrons.
Rate this question:
• 4.
### Which type of radiation has the highest penetrating power?
• A.
Gamma rays
• B.
Alpha particles
• C.
Beta particles
• D.
All the same
A. Gamma rays
Explanation
Gamma rays have the highest penetrating power among the given options. This is because gamma rays are high-energy electromagnetic waves that can easily pass through most materials due to their short wavelength and high frequency. On the other hand, alpha and beta particles are subatomic particles that have mass and charge, making them less penetrating compared to gamma rays.
Rate this question:
• 5.
• A.
3
• B.
2
• C.
1
A. 3
Explanation
Rate this question:
• 6.
### How does hair loss occur?
• A.
The hair falls out in clumps
• B.
The hair disintegrates
• C.
The hair becomes thinner before breaking off
C. The hair becomes thinner before breaking off
Explanation
Hair loss occurs when the hair becomes thinner before breaking off. This means that the individual hair strands gradually become weaker and thinner over time, leading to eventual hair loss. It is a common pattern in various types of hair loss, such as androgenetic alopecia (male or female pattern baldness), where the hair follicles shrink and produce thinner hair strands until they eventually stop producing new hair altogether.
Rate this question:
• 7.
### Can the hair regrow?
• A.
No
• B.
Yes
B. Yes
Explanation
There will be some root hair stem cells that survive in the epithelial layer, which can regenerate the hair matrix.
Rate this question:
• 8.
### How are radiation-induced cataracts formed?
• A.
The cells in the eye are destroyed.
• B.
The retina is damaged.
• C.
The cells in the eye are damaged.
C. The cells in the eye are damaged.
Explanation
Radiation-induced cataracts are formed when the cells in the eye are damaged. Exposure to radiation can cause changes in the structure and function of these cells, leading to the development of cataracts. This damage can disrupt the normal functioning of the lens, causing it to become cloudy and impairing vision.
Rate this question:
• 9.
### Which of the following has been proven to be an effect of radiation exposure?
• A.
Hair falls out in clumps
• B.
The person changes to become a monster as a result of genetic mutation
• C.
The development of cancer.
C. The development of cancer.
Explanation
Exposure to radiation has been proven to cause the development of cancer. Radiation can damage the DNA in cells, leading to mutations that can result in uncontrolled cell growth and the formation of tumors. This is a well-documented effect of radiation exposure and has been observed in numerous studies and cases of individuals exposed to high levels of radiation, such as survivors of atomic bombings or nuclear accidents.
Rate this question:
• 10.
### What is the name of the plane that dropped the atomic bomb?
• A.
B-20 Bomber
• B.
Little Boy
• C.
Enola Gay
C. Enola Gay
Explanation
Enola Gay is the correct answer because it was the name of the plane that dropped the atomic bomb on the Japanese city of Hiroshima on August 6, 1945. The Enola Gay was a Boeing B-29 Superfortress bomber, specifically modified for this mission. It was piloted by Colonel Paul Tibbets and played a significant role in the history of World War II.
Rate this question:
Related Topics | 1,244 | 5,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-10 | latest | en | 0.904281 |
https://estebantorreshighschool.com/faq-about-equations/mean-free-path-equation.html | 1,679,697,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00602.warc.gz | 303,642,245 | 10,929 | ## What do you mean by mean free path and write its formula?
Mean free path, average distance an object will move between collisions. The actual distance a particle, such as a molecule in a gas, will move before a collision, called free path, cannot generally be given because its calculation would require knowledge of the path of every particle in the region.
## What is the mean free path of air?
The reason for this lies in the great number of collisions that a gas particle sustains along its way. The mean free path is the average distance that a particle can travel between two successive collisions with other particles.
## How is mean free path affected by temperature and pressure?
Application of temperature will increase the space between molecules by decreasing the density hence the free main path will increase while application of pressure will decrease the space between molecules thereby increasing the density and again affecting the path.
## What do you think happens to mean free path as pressure increases?
So, as pressure increases number of collisions increase. Hence, mean free path decreases.
## What do you mean by free path?
From Wikipedia, the free encyclopedia. In physics, the mean free path is the average distance travelled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions), which modifies its direction or energy or other particle properties.
## Does mean free path depend on pressure?
The mean free path equation depends upon the temperature and pressure as well as the molecular diameter.
## What increases mean free path?
Factors affecting mean free path Density: As gas density increases, the molecules become closer to each other. Increasing the number of molecules or decreasing the volume causes density to increase. This decreases the mean free path.
## What is mean collision time?
At the individual particle level, the collision time is the mean time required for the direction of motion of an individual type particle to deviate through approximately as a consequence of collisions with particles of type .
## What is meant by ideal gas?
The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules: Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container.
## Why does collision diameter decrease with temperature?
As the temperature is increased the molecules are moving faster, but the average distance between them is not affected. The mean time between collisions decreases, but the mean distance traveled between collisions remains the same. The collision diameter, d.
## Which of the following has longest mean free path?
hydrogen
You might be interested: Linear equation definition
## What is the mean free path of the gas temperature is doubled at constant volume?
if the mean free path of atom is doubled at constant temperature then the pressure of gas will become. Hello!! When mean free path is doubled, pressure becomes half. So the answer would be simply p/2.
## How the mean free path is related to the density and pressure of gas?
Hence, mean free path varies inversely as density of the gas. It can easily proved that the mean free path varies directly as the temperature and inversely as the pressure of the gas.
### Releated
#### Depreciation equation
What are the 3 depreciation methods? There are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production. What do you mean by depreciation? Definition: The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation. How do you […]
#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […] | 816 | 4,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.92564 |
https://trends.edugorilla.com/neet-problems-waves-part-1-2013-to-2017-chapterwise-solutions-by-rohit-dahiya/ | 1,657,154,709,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00454.warc.gz | 621,448,504 | 39,402 | # NEET Problems | Waves (Part 1) | 2013 to 2017 | Chapterwise Solutions by Rohit Dahiya - Videos
29
246
Link to share: https://youtu.be/-vxKOcr0wkI
NEET and AIPMT Solutions – Waves (Part 1) – 2013 to 2017
Q01 – NEET 2013 – 00:20 – 03:11
Q02 – NEET 2013 – 03:12 – 05:36
Q03 – NEET 2013 – 05:37 – 09:48
Q04 – AIPMT 2014 – 09:49 – 13:59
Q05 – AIPMT 2014 – 14:00 – 19:02
Q06 – AIPMT 2014 – 19:03 – 24:15
Q07 – AIPMT 1 2015 – 24:16 – 28:37
Q08 – AIPMT 2 2015 – 28:38 – 32:51
Q09 – AIPMT 2 2015 – 32:52 – 37:17
Q10 – AIPMT 2 2015 – 37:18 – 41:37
Q01 – NEET 2013 – A wave travelling in the +ve x-direction having displacement along y-direction as 1m, wavelength 2? m and frequency of 1/π Hz is represented by :
Q02 – NEET 2013 – If we study the vibration of a pipe open at both ends, then the following statement is not true:
(1) Pressure change will be maximum at both ends
(2) Open end will be antinode
(3) Odd harmonics of the fundamental frequency will be generated
(4) All harmonics of the fundamental frequency will be generated
Q03 – NEET 2013 – A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz, The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency 513 Hz, The unknown frequency is
(1) 260 Hz (2) 254 Hz (3) 246 Hz (4) 240 Hz
Q04 – AIPMT 2014 – The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are – (velocity of sound = 340 ms–1)
(1) 4 (2) 5 (3) 7 (4) 6
Q05 – AIPMT 2014 – If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by:
Q06 – AIPMT 2014 – A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be:
(1) 1332 Hz (2) 1372 Hz (3) 1412 Hz (4) 1454 Hz
Q07 – AIPMT 1 2015 – The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
(1) 120 cm (2) 140 cm (3) 80 cm (4) 100 cm
Q08 – AIPMT 2 2015 – 4.0 g of a gas occupies 22.4 litres at STP. The specific heat capacity of the gas at constant volume is 5.0 JK-1mol-1. If the speed of sound in this gas at STP is 952 m/s, then the heat capacity at constant pressure is
(Take gas constant R = 8.3 JK-1 mol-1)
(1) 8.5 JK-1 mol-1 (2) 8.0 JK-1 mol-1 (3) 7.5 JK-1 mol-1 (4) 7.0 JK-1 mol-1
Q09 – AIPMT 2 2015 – A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequencies for this string is :
(1) 105 Hz (2) 155 Hz (3) 205 Hz (4) 10.5 Hz
Q10 – AIPMT 2 2015 – A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m/s at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m/s) is:
(1) 97 Hz (2) 100 Hz (3) 103 Hz (4) 106 Hz
_________________________________________________________________
https://www.coachengg.com
COACHENGG4U
Follow us on INSTAGRAM
https://www.instagram.com/coachengg
________________________________________________________________
source | 1,207 | 3,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-27 | longest | en | 0.822382 |
https://zbmath.org/?q=an%3A0771.53003 | 1,628,027,099,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00060.warc.gz | 1,140,282,460 | 10,199 | # zbMATH — the first resource for mathematics
On the evolution of curves via a function of curvature. I: The classical case. (English) Zbl 0771.53003
Summary: The problem of curve evolution as a function of its local geometry arises naturally in many physical applications. A special case of this problem is the curve shortening problem which has been extensively studied. Here, we consider the general problem and prove an existence theorem for the classical solution. The main theorem rests on lemmas that bound the evolution of length, curvature, and how far the curve can travel.
##### MSC:
53A04 Curves in Euclidean and related spaces
##### Keywords:
curve shortening problem
Full Text:
##### References:
[1] Angenent, S; Angenent, S, Parabolic equations for curves and surfaces, II, University of wisconsin-Madison technical summary reports, nos. 88-19, University of wisconsin-Madison technical summary reports, nos. 89-24, (1989) [2] Ben-Jacobi, E; Goldenfield, N; Langer, J; Schon, G, Dynamics of interfacial pattern formation, Phys. rev. lett., 51, No. 21, 1930-1932, (1983) [3] Brower, R.C; Kessler, D.A; Koplik, J; Levine, H, Geometrical models of interface evolution, Phys. rev. A, 29, 1335-1342, (1984) [4] Gage, M; Hamilton, R.S, The heat equation shrinking plane curves, J. differential geom., 23, 69-96, (1986) · Zbl 0621.53001 [5] Gage, M, On an area-preserving evolution equation for plane curves, Contemp. math., 51, 51-62, (1986) [6] Grayson, M.A, The heat equation shrinks embedded plane curves to round points, J. differential geom., 26, 285-314, (1987) · Zbl 0667.53001 [7] Kimia, B.B; Tannenbaum, A; Zucker, S.W, Toward a computational theory of shape: an overview, (), 402-407 [8] Kimia, B.B, Conservations laws and a theory of shape, () [9] Langer, J.S, Instabilities and pattern formation in crystal growth, Rev. modern phys., 52, No. 1, 1-28, (1980) [10] Lax, P.D, Shock waves and entropy, (), 603-634 [11] Lax, P.D, Hyperbolic systems of conservation laws and the mathematical theory of shock waves, () · Zbl 0108.28203 [12] Osher, S; Sethian, J, Fronts propagating with curvature dependent speed: algorithms based on Hamilton-Jacobi formulations, J. comput. phys., 79, 12-49, (1988) · Zbl 0659.65132 [13] Sethian, J.A, Curvature and the evolution of fronts, Comm. math. phys., 101, 487-499, (1985) · Zbl 0619.76087 [14] Sivashinsky, G.I, Nonlinear analysis of hydrodynamic instability in laminar flames. I. derivation of basic equations, Acta astronautica, 4, 1177-1206, (1977) · Zbl 0427.76047
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 827 | 2,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.794267 |
https://www.chemicalforums.com/index.php?topic=73203.0 | 1,660,129,774,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00185.warc.gz | 619,021,550 | 8,502 | August 10, 2022, 07:09:34 AM
Forum Rules: Read This Before Posting
### Topic: Calculating enthalpy change for two solution mixture (Read 4585 times)
0 Members and 1 Guest are viewing this topic.
#### KYR_Singularity
• Very New Member
• Posts: 2
• Mole Snacks: +0/-0
##### Calculating enthalpy change for two solution mixture
« on: February 11, 2014, 03:13:53 PM »
25.0 cm3 of 2.00 mol dm-3 HCl(aq) was mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH(aq). The temperature increased from 22.5°C to 34.5°C. Find the enthalpy change of reaction for the following equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l).
how do you work out the energy change in this reaction ?
please show all the work out , thank you
#### zsinger
• Full Member
• Posts: 374
• Mole Snacks: +18/-60
• Gender:
##### Re: Calculating enthalpy change for two solution mixture
« Reply #1 on: February 11, 2014, 06:15:26 PM »
See: Hess's Law. Theres your hint….now lets see some work
Zack
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"
#### KYR_Singularity
• Very New Member
• Posts: 2
• Mole Snacks: +0/-0
##### Re: Calculating enthalpy change for two solution mixture
« Reply #2 on: February 12, 2014, 05:22:10 PM »
HCL + NaOH -> NaCl + H2O
(25+25) x 4.18 x 12 = 2508 J
25 Volume reacted
25 x2 / 1000 = 0.05 mol
2508/0.05=50.16KJ/mol
^H = -50.16KJ/mol
is this right ? thanks for the tip anyway
#### Borek
• Mr. pH
• Deity Member
• Posts: 27142
• Mole Snacks: +1762/-405
• Gender:
• I am known to be occasionally wrong.
##### Re: Calculating enthalpy change for two solution mixture
« Reply #3 on: February 12, 2014, 05:41:18 PM »
Doesn't look bad, although you may want to take water produced to the total mass.
Some will consider it nitpicking, but it is almost a 2% difference.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info
#### zsinger
• Full Member
• Posts: 374
• Mole Snacks: +18/-60
• Gender:
##### Re: Calculating enthalpy change for two solution mixture
« Reply #4 on: February 12, 2014, 07:23:38 PM »
Borek is correct. Otherwise, you got it!
-Zack
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"
#### Benzene
• Regular Member
• Posts: 49
• Mole Snacks: +3/-0
##### Re: Calculating enthalpy change for two solution mixture
« Reply #5 on: February 12, 2014, 08:18:19 PM »
In Hess's Law, it states that the sum δH products-δH reactants = δH reaction
Is the δH of the products and reactants calculated from the δH of formation? or combustion?
#### zsinger
• Full Member
• Posts: 374
• Mole Snacks: +18/-60
• Gender: | 866 | 2,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.772849 |
https://baroody.org/GeometryHonors/Class%20Notes/Chapter%202/TestTopics/quizTopics.html | 1,580,121,641,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00086.warc.gz | 336,873,819 | 4,285 | you are here > Class Notes - Chapter 2 - Quiz Topics
Printable version
Quiz Topics
Here's what you need to know for the upcoming quiz...this covers material from Section 2.1 through Section 2.4!!
Definitions
• Everything defined in Chapter 1
• Perpendicular lines (rays, segments)
• Complementary angles
• Supplementary angles
Theorems
• If two angles are right angles, then they are congruent (Right Angle Theorem)
• If two angle are straight angles, then they are congruent (Straight Angle Theorem)
• If a conditional statement is true, then its contrapositive is also true (you don't need to be able to prove this!!)
• If angles are supplementary to the same angle, then they are congruent
• If angles are supplementary to congruent angles, then they are congruent
• If angles are complementary to the same angle, then they are congruent
• If angles are complementary to congruent angles, then they are congruent
Constructions
• Chapter 1 constructions
• Perpendicular to a line from a point not on the line
• Perpendicular to a line from a point on the line
Types of Problems
• Simultaneous equations
• Solving for angle/complement/supplement problems
• Theorem proofs (make sure you can prove Theorems 1-7 not including Theorem 3)
• Proofs using all theorems and definitions
• Constructions | 308 | 1,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-05 | latest | en | 0.868787 |
https://www.jiskha.com/display.cgi?id=1250732594 | 1,516,132,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00109.warc.gz | 934,494,113 | 4,279 | # math
posted by .
Brian O'Reilly earns twice as much each week as a tutor than he does pumping gas. His total weekly wages are \$150 more than that of his younger sister. She earns one quarter as much as Brian does as a tutor. How much does Brian earn as a tutor? Use a Chart.
• math -
let Brian's wage pumping gas be \$x
then his wage tutoring is \$2x
his sister's wage is 2x/4 = x/2
"His total weekly wages are \$150 more than that of his younger sister"
If found that a lot of students have difficulty with that kind of a statement.
They don't know whether to add/subtact the \$150.
I used to have my students translate the above into something like...
His wages are greater than the sister's wages by 150
---> x + 2x > x/2 by 150
so to make that statement into an equation we would add the 150 to the smaller side or ...
x+2x = x/2 + 150
3x = x/2+150
6x = x+300
5x = 300
x = 60
so Brian earns 2x as a tutor or \$120
check:
Brian pumping gas : 60
Brian tutoring : 120
total of his wages = 180
sister : half of his tutoring or 30
does Brian make 150 more than the sister ? YES!
• math -
thank you so much
• math -
Mona Yahuso earns threee times as much as an actuary as she does as a writer. Her total income is \$40,000 more then that of their brother. He earns half as much as Mona dose as an actuary. What is Mona's salary as a actuary?
## Similar Questions
1. ### math
Brian O'Reilly earns twice as much each week as a tutor than he does pumping gas. His total weekly wages are \$150 more than that of his younger sister. She earns one quarter as much as Brian does as a tutor. How much does Brian earn …
2. ### Math
Brian O'Reily earns twice as much each week as a tutor than he does pumping gas. His total weekly wages are \$150 more than that of his younger sister. She earns one quarter as much as Brian does as a tutor. How much does Brian earn …
3. ### Math
For this problem, I need help creating a chart for this: Brian O'Reily earns twice as much each week as a tutor than he does pumping gas. His total weekly wages are \$150 more than that of his younger sister. She earns one quarter as …
4. ### math
Universal Exporting has three warehouse employees: John Abner earns \$422 per week, Anne Clark earns \$510 per week, and Todd Corbin earns \$695 per week. The company’s SUTA tax rate is 5.4%, and the FUTA rate is 6.2% minus the SUTA. …
5. ### math
Universal Exporting has three warehouse employees: John Abner earns \$422 per week, Anne Clark earns \$510 per week, and Todd Corbin earns \$695 per week. The company’s SUTA tax rate is 5.4%, and the FUTA rate is 6.2% minus the SUTA. …
6. ### math
Universal Exporting has three warehouse employees: John Abner earns \$422 per week, Anne Clark earns \$510 per week, and Todd Corbin earns \$695 per week. The company’s SUTA tax rate is 5.4%, and the FUTA rate is 6.2% minus the SUTA. …
7. ### math
Universal Exporting has three warehouse employees: John Abner earns \$422 per week, Anne Clark earns \$510 per week, and Todd Corbin earns \$695 per week. The company’s SUTA tax rate is 5.4%, and the FUTA rate is 6.2% minus the SUTA. …
8. ### math
three sisters run a business. they charge \$260 for each job. If the oldest sister earns 50% more than the middle sister and if the youngest sister earns 50% less than the oldest sister, how much does each sister earn per job?
9. ### accounting
Amazon Appliance Company has three installers. Larry earns \$355 per week, Curly earns \$460 per week, and Moe earns \$585 per week. The company's SUTA rate is 5.4%, and the FUTA rate is 6.2% minus the SUTA. As usual, these taxes are …
10. ### Math
Brian is baking cookies for Zachary's birthday party if an Bryan uses 1 1/4 cups of flour for every 10 cookies if Brian needs to make 50 cookies and he have 4 cups of flour how much more flower Brian need to finish baking
More Similar Questions | 1,043 | 3,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-05 | latest | en | 0.988417 |
https://magedkamel.com/18b-cb-value-bracing-at-the-midpoint/ | 1,726,862,983,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00758.warc.gz | 344,788,344 | 49,544 | # Engineering Oasis
" Civil Engineering data"
# 18b-Cb value-bracing at the midpoint of a beam-uniform load
Last Updated on July 29, 2024 by Maged kamel
## Cb value-bracing at the midpoint of a beam-uniform load.
In this post, we will estimate Cb value-bracing at the midpoint of a beam-uniform load. the case where we have three braces one at the midpoint and two at the supports.
In the previous post, number 17, we estimated the CB value or the coefficient for a moment for a simply supported beam with two braces at the supports under uniform load.
The next slide image shows the different cases for a simply supported beam with different CB values based on the bracing locations in the span.
### Cb value-bracing at the midpoint of a beam-uniform load-simple beam case.
We can see that we have a brace in the mid-span of a simple beam under uniform load W. The midpoint is point C.
To estimate CB value-bracing at the midpoint of a beam-uniform load, we divide part Ac into four quarters and calculate the value of moments at the different points.
#### What is the maximum value for a moment?
The moment value at point A’ is estimated as equal to ((7/(16*8))*wL^2, where L is the span and w is the uniform load per kip or meter based on the used units.
The moment value at point B’ is estimated as equal to (3/32)*wL^2, and the moment value at point C’ is estimated as equal to ((15/(16*8))*wL^2. The moment at point c is equal to (1/8)*wL^2. Exploring the moment values for segment AC we find that the maximum value is WL^2/8.
Now, we will apply the equation to get Cb value-bracing at the midpoint of a beam-uniform load. The maximum moment is w*l2/8, which will be multiplied by 12.50 for the numerator. For the denominator, we will sum 2.5*Max +3Ma’, 4Mb’, and 3Mc’.
The Cb value for a simply supported beam will be found to be equal to 1.3, which matches the Cb value for the second case in Table 3-1.
The Cb value for segment Bc will be equal to 1.3 because of the symmetry.
If you wish to compare the value of Cb based on the equation to the old value, we will find that the old value of Cb will be equal to 1.75 since m1/M2 will be equal to zero.
The next slide details how to find the old Cb value.
### Cb value-bracing at the midpoint of a beam-uniform load-fixed end beam case.
The second case is a fixed-end beam with three braces, two at the support and one at the midpoint.
We will find the Cb value-bracing at the midpoint of a beam under uniform load. As we know from our structural analysis study for a fixed-end beam under uniform load, the fixed end moment of such a beam will be equal to W*L2/12, where w is the uniform load, and L is the span of the mean.
The middle moment can be found by superposing the positive WL2/8 and the fixed end moments of W*L2/12.
To estimate CB value-bracing at the midpoint of a beam-uniform load, we divide part AC into four quarters and assess the value of moments at the different points. The next slide image includes the steps to find M’a and M’b moment values.
The next step is to estimate the moment for c’ in the third quarter for segment Ac. The value of the moment at c’ will be equal to (13/384)*w*L2. The value of mc is w*L2/24.
I have listed all the values for moments at points A, a’, b’, c’, and point c. Considering the maximum value of the estimated moment value, which will be found to be equal to Wl2/12, we consider the absolute value of moments.
#### What is the maximum value for a moment?
The denominator values are prepared using (2.5M max+3*Ma’+4*Mb’+3Mc’) as a multiplier of w*L2. The final value is (7/16)*w*L2.
Apply the equation for Cb to get the coefficient of moment for the fixed end beam with three supports, two at the supports and one at the middle point. The value is 2.38, which matches the value shown. for segemnts AC and CB.
Please refer to table 8.2.1.1 Computation of Cb from A Beginner Guide to Structural Steel Manual 15th Edition.
This is the complete list of all posts related to Cb:
1-Introduction to Cb-Bending coefficient part-1 for steel.-post 17.
2- Cb-The coefficient of bending part 2 for steel beams-post 18.
3-Cb-The coefficient of bending part-3 for steel beams-post 18a -Previous post
4-Cb value-bracing at the midpoint of a beam-uniform load-Post 18b-This post.
5-Cb value bracing at third points of a beam-U load-Post-18C-Next post. | 1,132 | 4,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-38 | latest | en | 0.888501 |
https://www.geeksforgeeks.org/expert/62/ | 1,618,180,137,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00057.warc.gz | 891,685,024 | 29,115 | All Expert Articles
Problem – Assembly level program in 8085 which converts a binary number into ASCII number. Example – Assumptions – Binary number which have to convert… Read More
Problem – Write a program in 8086 microprocessor to find out the largest among 8-bit n numbers, where size “n” is stored at memory address… Read More
Message dialogs provide information to the user. Message dialogs are created with the JOptionPane.showMessageDialog() method. We call the static showMessageDialog() method of the JOptionPane class… Read More
Below is the example of Date toString() method. Example: <script> // Here a date has been assigned // while creating Date object var dateobj = … Read More
Download links for previous years ISRO original Papers and official Keys: Original Paper Official Keys ISRO 2018 ISRO 2018 Key ISRO 2017 Dec ISRO 2017… Read More
Redundant links are used to provide back up path when one link goes down but Redundant link can sometime cause switching loops. The main purpose… Read More
The fabs() function returns the absolute value of the argument. Mathematically |a|. If a is value given in the argument. Syntax: double fabs(double a); float… Read More
What is the probability that a randomly selected bit string of length 10 is a palindrome? (A) 1/64 (B) 1/32 (C) 1/8 (D) 1/4 Answer:… Read More
oct() function is one of the built-in methods in Python3. The oct() method takes an integer and returns it’s octal representation in a string format.… Read More
In QuickSort, ideal situation is when median is always chosen as pivot as this results in minimum time. In this article, Merge Sort Tree is… Read More
If one uses straight two-way merge sort algorithm to sort the following elements in ascending order: 20, 47, 15, 8, 9, 4, 40, 30, 12,… Read More
Functions are one of the building blocks of any programming language and JavaScript has taken the Functions to a whole new level. Functions are said… Read More
sar : System Activity Report It can be used to monitor Linux system’s resources like CPU usage, Memory utilization, I/O devices consumption, Network monitoring, Disk… Read More
In Javascript(ES6), there are four ways to test equality which are listed below: Using ‘==’ operator Using ‘===’ operator SameValueZero: used mainly in sets, maps… Read More
Introduction to Stream, Java Intstream, Java Longstream, Java Doublestream anyMatch() noneMatch() mapToLong() findAny() forEachOrdered() forEach() allMatch() filter() findFirst() flatMapToInt() mapToInt() map() peek() counting() Iterator() Generate()… Read More
Reason: To Ensure Backward Compatibility What is Backward Compatibility? Backward compatibility is a term used to describe software or hardware that is compatible with previous… Read More
The levenshtein() function is an inbuilt function in PHP. The levenshtein() function is used to calculate the levenshtein distance between two strings. The Levenshtein distance… Read More
In our previous post on fibonacci series, we have seen many approaches to generate fibonacci numbers. In this approach, we shall be generating fibonacci numbers… Read More
Given a number num containing n digits. The problem is to find the next greater number using the same set of digits in num on… Read More
A lexical analyzer uses the following patterns to recognize three tokens T1, T2, and T3 over the alphabet {a,b,c}. T1: a?(b∣c)*a T2: b?(a∣c)*b T3: c?(b∣a)*c… Read More | 814 | 3,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-17 | latest | en | 0.789406 |
https://fr.mathworks.com/matlabcentral/profile/authors/2950957?detail=cody | 1,718,848,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00575.warc.gz | 231,920,720 | 24,667 | # Guillaume
### University of Brighton
Last seen: 2 mois il y a Actif depuis 2012
Followers: 2 Following: 0
All
#### Feeds
Afficher par
A résolu
Cell joiner
You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string fr...
presque 5 ans il y a
A résolu
Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
presque 6 ans il y a
A résolu
How to subtract?
*± ± ± ± ± ± ± ± ± ± ±* * Imagine you need to subtract one...
plus de 6 ans il y a
A résolu
Pair Primes
Let's define pair primes as follow; * *For 2 digits numbers:* 11 and 17 are pair primes because both of them are 2 digits pri...
plus de 6 ans il y a
A résolu
5th Time's a Charm
Write a function that will return the input value. However, your function must fail the first four times, only functioning prope...
plus de 6 ans il y a
A résolu
"Cody" * 5 == "CodyCodyCodyCodyCody"
*Alice*: What? *"Cody" * 5 == "CodyCodyCodyCodyCody"*? You've gotta be kidding me! *Bob*: No, I am serious! Python supports...
plus de 6 ans il y a
A résolu
Number of Even Elements in Fibonacci Sequence
Find how many even Fibonacci numbers are available in the first d numbers. Consider the following first 14 numbers 1 1 2...
plus de 6 ans il y a
A résolu
The 5th Root
Write a function to find the 5th root of a number. It sounds easy, but the typical functions are not allowed (see the test su...
plus de 6 ans il y a
A résolu
Relation between functions "dec2bin" & "dec2binvec"
Here it's an easy problem we try to find the relation between the two functions "dec2bin" & "dec2binvec", so here you must write...
environ 7 ans il y a
A résolu
Replace Negative(-) by 0 and positive by 1
In a given Matrix Replace all element having *Negative sign with 0* and *Positive elements with 1* .
environ 7 ans il y a
A résolu
A different counting method
Given an array (x) of integers, the "counting" array (y) is showing the number of identical consecutive integers in x in front o...
environ 7 ans il y a
A résolu
Evaluating continued fractions
Given row vector c=[c0 c1 c2 c3 ...] evaluate the continued fraction x=c0+1/(c1+1/(c2+1/(c3+...))) If c is a ...
environ 7 ans il y a
A résolu
Isothermal Expansion
Given the initial pressure and volume of an ideal gas, calculate the new volume, given the new pressure. Hint: <https://en.wi...
environ 8 ans il y a
A résolu
Sum two real numbers
It seems easy, but... You cannot use +, -, plus, diff, cumsum, *, prod, times, etc.
environ 8 ans il y a
A résolu
Does this dress make me look fat
For the input string "Does xyz make me look fat" output the string "No, xyz does not make you look fat"
environ 8 ans il y a
A résolu
Fifteen Parity Check
The Matlab function fifteen initializes the 4x4 array with randperm(16), which produces 50% unsolvable puzzles. A <https://en.wi...
environ 8 ans il y a
A résolu
MATLAB Prison: Summing light bulbs
On one wall in the MATLAB prison there is a row of n numbered light bulbs. Each bulb is controlled by a switch. Every morning, n...
environ 8 ans il y a
A résolu
Number of divisors of a given number
Given a Number n, return the number of his divisors without listing them example: n=14 ; Divisors={1,7,2,14} ; y=4 n=...
environ 8 ans il y a
A résolu
Full combinations
Given n input vectors x1, x2, …, xn, generate a p*n matrix y whose rows contain all element-wise combinations of the vectors x1,...
environ 8 ans il y a
A résolu
Accessing values in a cell
You are given c, which is a 1xN cell array, and a and b, which are each two 1xM vectors. Your job is to give the bth value in t...
environ 8 ans il y a
A résolu
How many ways to write a number
Given two positive numbers n and k, where n>=k. In how many ways can we write n as sum of k positive numbers. Same numbers but d...
environ 8 ans il y a
A résolu
GJam March 2016 IOW: Polynesiaglot Small
This Challenge is derived from <http://code.google.com/codejam/contest/8274486/dashboard#s=p2 GJam March 2016 Annual I/O for Pol...
plus de 8 ans il y a
A résolu
the "power matrix" of two vetcors
Given two row vectors x,y of lengths m and n (resp.), create an m x n matrix whose i,j entry is x(i)^y(j).
plus de 8 ans il y a
A résolu
Raise a polynomial to a power
In Matlab, polynomials are represented by a vector of coefficients. For example, the polynomial p=a + b*x + c*x^2 is represente...
plus de 8 ans il y a
A résolu
Is my wife really right?
For every input, output the string 'yes' once. Example: [yes1, yes2] = YesSheIs('Am I right?', 'Do you love me?') yes1 = '...
plus de 8 ans il y a
A résolu
GJam March 2016 IOW: Cody's Jams
This Challenge is derived from <http://code.google.com/codejam/contest/8274486/dashboard GJam March 2016 Annual I/O for Women Co...
plus de 8 ans il y a
A résolu
Create formatted value string Cell array of a vector
This Challenge is to create a cell array that contains strings of vector components in a given format. c=vector2formattedstri...
plus de 8 ans il y a
A résolu
Find the gcm of n given values
Create a function that given n integer values greater than zero, finds the two numbers with the greatest common divisor and retu...
plus de 8 ans il y a
A résolu
Convert integer to base26 using letters
Write a function that converts a decimal integer to base26 using the letters of the english alphabet, i.e. 0->'a', 1->'b', 2->'c...
plus de 8 ans il y a
A résolu
Volume of a cube
Find the volume of a cube.
plus de 8 ans il y a | 1,645 | 5,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-26 | latest | en | 0.562521 |
https://softmath.com/algebra-software/subtracting-exponents/what-is-a-lineal-metre.html | 1,685,984,421,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00383.warc.gz | 589,305,408 | 9,633 | what is a lineal metre
Related topics:
math 100 chapter 1 review | free online saxon pre-algebra answers | 6th grade math print out worksheets | c-rod fractions | free equation worksheets dealing with the substitution method | simplifying functional expressions | saxon algebra 2 answers and work | Free Worksheet On Adding Like Terms | properties of "square roots" activities | glencoe mcgraw hill algebra 1 answer sheets | mix fraction to decimal
Author Message
Fuvanec
Registered: 02.12.2001
From:
Posted: Saturday 18th of Jul 14:01 Hi math wizards! I am about one week through the semester, and getting a bit worried about my course work. I just don’t seem to understand the stuff I am learning, especially things to do with what is a lineal metre. Could somebody out there please teach me with relations, adding fractions and like denominators. I can’t afford to look for a tutor, but if anyone knows about other ways of learning topics like parallel lines or multiplying fractions painlessly , please help me out Much appreciated
Back to top
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Sunday 19th of Jul 07:35 I don’t think I know of any website where you can get your solutions of what is a lineal metre checked within hours. There however are a couple of websites which do offer help , but one has to wait for at least 24 hours before expecting any reply .What I know for sure is that, this software called Algebrator, that I used during my college career was really good and I was quite happy with it. It almost gives the type of results you need.
Back to top
TC
Registered: 25.09.2001
From: Kµlt °ƒ Ø, working on my time machine
Posted: Monday 20th of Jul 11:12 I have tried out various software. I would boldly say that Algebrator has assisted me to come to grips with my difficulties on converting fractions, trigonometry and simplifying fractions. All I did was to merely key in the problem. The answer appeared almost instantly showing all the steps to the result. It was quite straightforward to follow. I have relied on this for my algebra classes to figure out Intermediate algebra and Pre Algebra. I would highly recommend you to try out Algebrator.
Back to top
lalddobcat
Registered: 14.09.2004
From: New Zealand
Posted: Wednesday 22nd of Jul 07:16 Oh really? Incredible . You mean it’s that effortless? I must without doubt try it. Please tell me where I can access this program?
Back to top
Matdhejs
Registered: 08.12.2001
From: The Netherlands
Posted: Wednesday 22nd of Jul 12:36 I remember having problems with geometry, linear inequalities and multiplying matrices. Algebrator is a truly great piece of algebra software. I have used it through several algebra classes - Basic Math, College Algebra and Remedial Algebra. I would simply type in the problem from a workbook and by clicking on Solve, step by step solution would appear. The program is highly recommended.
Back to top
DVH
Registered: 20.12.2001
From:
Posted: Wednesday 22nd of Jul 17:13 https://softmath.com/news.html. There you go. Hopefully you will not have to leave math.
Back to top | 739 | 3,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-23 | longest | en | 0.954821 |
https://beanaroundtheworld.wordpress.com/2011/10/07/epidemiology-odds-ratio-or/ | 1,685,278,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00159.warc.gz | 169,028,426 | 28,246 | # Epidemiology – Odds Ratio (OR)
Definition
The Odds Ratio is a measure of association which compares the odds of disease of those exposed to the odds of disease those unexposed.
Formulae
• OR = (odds of disease in exposed) / (odds of disease in the non-exposed)
Example
I often think food poisoning is a good scenario to consider when interpretting ORs: Imagine a group of 20 friends went out to the pub – the next day a 7 were ill. They suspect that it may have been something they ate, maybe the fish casserole… here are the numbers:
Cases (ill) Controls (not ill) Total Exposed (ate fish) 5 3 8 Unexposed (didn’t eat fish) 2 10 12 7 13 20
• Odds of exposure in cases = a/c = 5/2 = 2.5
• Odds of exposure in controls = b/d = 3/10 = 0.3
• Odds Ratio = (a/c) / (b/d) = 2.5/0.3 = 8.33
Interpretation: What does this mean?
• OR of 1 would suggests that there is no difference between the groups; i.e. there would be no association between the suggested exposure (fish) and the outcome (being ill)
• OR of > 1 suggests that the odds of exposure are positively associated with the adverse outcome compared to the odds of not being exposed
• OR of < 1 suggests that the odds of exposure are negatively associated with the adverse outcomes compared to the odds of not being exposed. Potentially, there could be a protective effect
In the example above, we can conclude that those who ate the fish casserole (exposure) were 8.3 times more likely (OR = 8.3) to be ill (outcome), compared to those who did not eat the fish casserole. Of course this is an entirely ficticious example, and I have nothing against fish
• Appropriate to analyse associations between groups from case-control and prevalent (or cross-sectional) data.
• For rare diseases (or diseases with long latency periods) the OR can be an approximate measure to the RR (relative risk)
• Doesn’t require denominator (i.e. total number in population) unlike measuring risk
• Good method to estimate the strength of an association between exposures and outcomes
• Association does not infer causation! *epidemiology golden rule*
### 8 responses to “Epidemiology – Odds Ratio (OR)”
1. Good post. I learn something new and challenging on blogs I stumbleupon
articles from other writers and practice something from their websites.
2. Christie Kanka
Great analogy, really helped me to understand odds ration in a simple straightforward manner. Thank you! Christie
3. Taz Rundle
Thanks for this definition, my text book (Jekel’s Epidemiology, Biostatistics,Preventive Medicine, and Public Health) was not getting meaning across. You did.
4. Thanks for all your help! Using this for my research!
5. Olli Raitakari
Very nice example, thanks. But think that your conclusion was not accurate: “.. we can conclude that those who ate the fish casserole (exposure) were 8.3 times more likely..”. If the effect estimate had been risk ratio, this would be true but in this example, because the outcome was so common(about 30%), the odds ratio value is much greater than the actual risk. But anyhow, very nice definition. I will use this in my teachings.
6. Peter
Under your table of odds ratio you have (a/c) / (b/d) = (a*d / b*c)
It is more commonly written as (a/b) / (c/d) = (a*d / b*c) but they produce the same results. Nice article.
7. Thank you very helpful. S. Lindani | 841 | 3,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.948873 |
https://brilliant.org/problems/when-all-language-meet-together/ | 1,476,999,278,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717954.1/warc/CC-MAIN-20161020183837-00373-ip-10-171-6-4.ec2.internal.warc.gz | 847,378,409 | 10,055 | # When all language meet together
Algebra Level 5
There are 100 countries participating in an olympiad. Take any natural $$n$$ such that each of 100 country is willing to communicate in exactly $$n$$ languages. What is the minimum possible value of $$n$$ if at least one language is common to 20 countries and no language is common to all 100 countries.
Give your answer as $$\dfrac{n}{4}$$
× | 97 | 396 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-44 | latest | en | 0.913354 |
http://sciencedocbox.com/Physics/80316162-Number-theory-and-divisibility.html | 1,603,967,066,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00054.warc.gz | 92,331,652 | 24,210 | # Number Theory and Divisibility
Size: px
Start display at page:
Transcription
1
2 Number Theory and Divisibility Recall the Natural Numbers: N = {1, 2, 3, 4, 5, 6, } Any Natural Number can be expressed as the product of two or more Natural Numbers: 2 x 12 = 24 3 x 8 = 24 6 x 4 = 24 Factors of 24 Factors of 24 Factors of 24 Factors of 24: Natural Numbers that multiply to give us 24
3 Number Theory and Divisibility Factors of 24: 1, 2, 3, 4, 6, 8, 12, and 24 1 x 24 = 24 2 x 12 = 24 3 x 8 = 24 4 x 6 = 24 6 x 4 = 24 8 x 3 = x 2 = x 1 = 24
4 Divisibility a is divisible by b, if the operation of dividing a by b leaves a remainder of 0 b a b is a divisor of a b divides a The following are Equivalent Statements: 24 is divisible by 8 8 is a divisor of 24 8 divides
5 Ex. Divisors of 24 Divisibility 1 24 because 24 1 = 24 with no remainder 2 24 because 24 2 = 12 with no remainder 3 24 because 24 3 = 8 with no remainder 4 24 because 24 4 = 6 with no remainder 6 24 because 24 6 = 4 with no remainder 8 24 because 24 8 = 3 with no remainder because = 2 with no remainder because = 1 with no remainder 5 24 because 24 5 gives us a nonzero remainder 7 24 because 24 5 gives us a nonzero remainder
6 Divisibility True or False? 4 16 True since 16 4 = False since we get a remainder 7 14 True since 14 7 = True since 15 3 = True since = 104
7 Divisibility Rules
8 Divisibility True or False? Use the Rules of Divisibility 8 48,324 False, last 3 digits 324 not divisible by ,324 True, divisible by 2 and ,324 False, 4 does divide 48,324 since 4 24
9 Prime Factorization Prime Number: a Natural Number greater than 1 that has only itself and 1 as factors. Ex. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, Composite Number: a Natural Number greater than 1 that is divisible by a number other than 1 and itself. Ex. 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, Note: 1 is neither prime nor composite by definition
10 Prime Factorization Every Composite Number can be expressed as the product of prime numbers. Expressing a number in this form is called Prime Factorization. Ex. 45 = 3 x 3 x 5 18 = 2 x 3 x 3 42 = 7 x 3 x 2
11 Fundamental Theorem of Arithmetic Every composite number can be expressed as a product of prime numbers in one and only one way. (Order does not matter) This means Prime Factorization is unique to every composite number.
12 Factor Tree A method to find the prime factorization of a composite number Step 1) Select two numbers (other than 1) that are factors of your number Step 2) Repeat this process for the factors that are composite numbers until you only have primes left
13 Factor Tree Ex. Find the prime factorization of = 2 2 x 5 2 x 7 Ex. Find the prime factorization of 120
14 Greatest Common Divisor gcd(a,b) = the largest number that is a divisor (factor) of a and b. If gcd(a,b) = 1, then a and b are Relatively Prime Finding GCD using Prime Factorization 1) Write prime factorization of each number 2) Select each prime factor with the smallest exponent that is common to each prime factorizations 3) The GCD is the product of the numbers in part 2
15 Greatest Common Divisor Find the GCD of 16 and = 2 2 x 3 16 = 2 4 gcd(12,16) = 2 2 = 4
16 Greatest Common Divisor Find the GCD of 40 and 24
17 Greatest Common Divisor For an intramural league, you need to divide 192 men and 288 women into all-male and all-female teams so that each team has the same number of people. What is the largest number of people that can be placed on a team?
18 Greatest Common Divisor 192 men divided into teams: number of men per team is a divisor of women divided into teams: number of women per team is a divisor of 288 The number of people per team must be the same for both men and women. We are looking for the largest number for this occur, i.e. the largest number that divides both 192 and 288 without a remainder
19 Greatest Common Divisor We want to find the GCD of 192 and Therefore, the greatest number of people that can be placed into teams is 2 5 x 3 = 96
20 Least Common Multiple lcm(a,b) = smallest Natural Number that is divisible by a and b Find LCM by making a list of multiples of each number Ex. Find lcm(15,20) Multiples of 15: {15, 30, 45, 60, 75, 90, 105, 120, } Multiples of 20: {20, 40, 60, 80, 100, 120, 140, 160, } 60 and 120 are common multiples. The least common multiple is 60.
21 Least Common Multiple Method 2: Finding LCM using Prime Factorization 1) Write the prime factorization of each number 2) Select every prime raised to the greatest power that it occurs 3) The product of numbers in Step 2 is the LCM
22 Least Common Multiple Find the LCM of 16 and 12 Recall that: 12 = 2 2 x 3 and 16 = 2 4 Prime 2 has highest exponent 4 Prime 3 has highest exponent 1 Therefore lcm(16,12) = 2 4 x 3 = 48
23 Least Common Multiple A movie theater runs two documentary films continuously. One documentary runs for 40 minutes and a second documentary runs for 60 minutes. Both movies begin at 3PM. When will the movies begin again at the same time?
24 Least Common Multiple The documentaries repeat and we want to find when they both start at the same time. Common multiples of 40 and 60 will give us the minutes after 3PM when both movies start at the same time. We want to find the LCM, which is the first time they start together again. lcm(40,60) = minutes = 2 hours 2 hours from 3PM is 5PM
25 Practice Problems Page 256 Divisible? #1, 7, 11, 17, 21 Prime Factorization: #25, 27, 31, 35, 39, 41 GCD: #45, 49, 51, 53, 55 LCM: #57, 59, 61, 65, 67 Applications: #91, 98
### N= {1,2,3,4,5,6,7,8,9,10,11,...}
1.1: Integers and Order of Operations 1. Define the integers 2. Graph integers on a number line. 3. Using inequality symbols < and > 4. Find the absolute value of an integer 5. Perform operations with
### Discrete Structures Lecture Primes and Greatest Common Divisor
DEFINITION 1 EXAMPLE 1.1 EXAMPLE 1.2 An integer p greater than 1 is called prime if the only positive factors of p are 1 and p. A positive integer that is greater than 1 and is not prime is called composite.
### Divisibility, Factors, and Multiples
Divisibility, Factors, and Multiples An Integer is said to have divisibility with another non-zero Integer if it can divide into the number and have a remainder of zero. Remember: Zero divided by any number
### Ch 4.2 Divisibility Properties
Ch 4.2 Divisibility Properties - Prime numbers and composite numbers - Procedure for determining whether or not a positive integer is a prime - GCF: procedure for finding gcf (Euclidean Algorithm) - Definition:
### 2 Elementary number theory
2 Elementary number theory 2.1 Introduction Elementary number theory is concerned with properties of the integers. Hence we shall be interested in the following sets: The set if integers {... 2, 1,0,1,2,3,...},
### The set of integers will be denoted by Z = {, -3, -2, -1, 0, 1, 2, 3, 4, }
Integers and Division 1 The Integers and Division This area of discrete mathematics belongs to the area of Number Theory. Some applications of the concepts in this section include generating pseudorandom
### 5.1. Primes, Composites, and Tests for Divisibility
CHAPTER 5 Number Theory 5.1. Primes, Composites, and Tests for Divisibility Definition. A counting number with exactly two di erent factors is called a prime number or a prime. A counting number with more
### Arithmetic, Algebra, Number Theory
Arithmetic, Algebra, Number Theory Peter Simon 21 April 2004 Types of Numbers Natural Numbers The counting numbers: 1, 2, 3,... Prime Number A natural number with exactly two factors: itself and 1. Examples:
### Slide 1 / 69. Slide 2 / 69. Slide 3 / 69. Whole Numbers. Table of Contents. Prime and Composite Numbers
Slide 1 / 69 Whole Numbers Table of Contents Slide 2 / 69 Prime and Composite Numbers Prime Factorization Common Factors Greatest Common Factor Relatively Prime Least Common Multiple Slide 3 / 69 Prime
### SEVENTH EDITION and EXPANDED SEVENTH EDITION
SEVENTH EDITION and EXPANDED SEVENTH EDITION Slide 5-1 Chapter 5 Number Theory and the Real Number System 5.1 Number Theory Number Theory The study of numbers and their properties. The numbers we use to
### Introduction to Number Theory
Introduction to Number Theory Number theory is about integers and their properties. We will start with the basic principles of divisibility, greatest common divisors, least common multiples, and modular
### CISC-102 Fall 2017 Week 6
Week 6 page 1! of! 15 CISC-102 Fall 2017 Week 6 We will see two different, yet similar, proofs that there are infinitely many prime numbers. One proof would surely suffice. However, seeing two different
### a the relation arb is defined if and only if = 2 k, k
DISCRETE MATHEMATICS Past Paper Questions in Number Theory 1. Prove that 3k + 2 and 5k + 3, k are relatively prime. (Total 6 marks) 2. (a) Given that the integers m and n are such that 3 (m 2 + n 2 ),
### 8 Primes and Modular Arithmetic
8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.
### Prime Factorization and GCF. In my own words
Warm- up Problem What is a prime number? A PRIME number is an INTEGER greater than 1 with EXACTLY 2 positive factors, 1 and the number ITSELF. Examples of prime numbers: 2, 3, 5, 7 What is a composite
### The numbers 1, 2, 3, are called the counting numbers or natural numbers. The study of the properties of counting numbers is called number theory.
6.1 Number Theory Number Theory The numbers 1, 2, 3, are called the counting numbers or natural numbers. The study of the properties of counting numbers is called number theory. 2010 Pearson Education,
### REVIEW Chapter 1 The Real Number System
REVIEW Chapter The Real Number System In class work: Complete all statements. Solve all exercises. (Section.4) A set is a collection of objects (elements). The Set of Natural Numbers N N = {,,, 4, 5, }
Summer Packet th into 6 th grade Name Addition Find the sum of the two numbers in each problem. Show all work.. 62 2. 20. 726 + + 2 + 26 + 6 6 Decimal Addition: Remember to line up the decimals before
### Proofs. Methods of Proof Divisibility Floor and Ceiling Contradiction & Contrapositive Euclidean Algorithm. Reading (Epp s textbook)
Proofs Methods of Proof Divisibility Floor and Ceiling Contradiction & Contrapositive Euclidean Algorithm Reading (Epp s textbook) 4.3 4.8 1 Divisibility The notation d n is read d divides n. Symbolically,
### LP03 Chapter 5. A prime number is a natural number greater that 1 that has only itself and 1 as factors. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
LP03 Chapter 5 Prime Numbers A prime number is a natural number greater that 1 that has only itself and 1 as factors. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, Question 1 Find the prime factorization of 120.
### Quantitative Aptitude
WWW.UPSCMANTRA.COM Quantitative Aptitude Concept 1 1. Number System 2. HCF and LCM 2011 Prelims Paper II NUMBER SYSTEM 2 NUMBER SYSTEM In Hindu Arabic System, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7,
### Number Theory. Number Theory. 6.1 Number Theory
6.1 Number Theory Number Theory The numbers 1, 2, 3, are called the counting numbers or natural numbers. The study of the properties of counting numbers is called number theory. 2 2010 Pearson Education,
### Math 110 FOUNDATIONS OF THE REAL NUMBER SYSTEM FOR ELEMENTARY AND MIDDLE SCHOOL TEACHERS
4-1Divisibility Divisibility Divisibility Rules Divisibility An integer is if it has a remainder of 0 when divided by 2; it is otherwise. We say that 3 divides 18, written, because the remainder is 0 when
### MTH 310, Section 001 Abstract Algebra I and Number Theory. Sample Midterm 1
MTH 310, Section 001 Abstract Algebra I and Number Theory Sample Midterm 1 Instructions: You have 50 minutes to complete the exam. There are five problems, worth a total of fifty points. You may not use
### Number Sense. Basic Ideas, Shortcuts and Problems #1-20 from the Sequence Chart
UIL Number Sense Contest Basic Ideas, Shortcuts and Problems #1-20 from the Sequence Chart Larry White UIL State Number Sense Contest Director texasmath@centex.net http://www.uiltexas.org/academics/number-sense
### Number Theory Basics Z = {..., 2, 1, 0, 1, 2,...} For, b Z, we say that divides b if z = b for some. Notation: b Fact: for all, b, c Z:
Number Theory Basics Z = {..., 2, 1, 0, 1, 2,...} For, b Z, we say that divides b if z = b for some z Z Notation: b Fact: for all, b, c Z:, 1, and 0 0 = 0 b and b c = c b and c = (b + c) b and b = ±b 1
### The Euclidean Algorithm and Multiplicative Inverses
1 The Euclidean Algorithm and Multiplicative Inverses Lecture notes for Access 2009 The Euclidean Algorithm is a set of instructions for finding the greatest common divisor of any two positive integers.
### Finding Prime Factors
Section 3.2 PRE-ACTIVITY PREPARATION Finding Prime Factors Note: While this section on fi nding prime factors does not include fraction notation, it does address an intermediate and necessary concept to
### Homework #2 solutions Due: June 15, 2012
All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is
### NOTES ON SIMPLE NUMBER THEORY
NOTES ON SIMPLE NUMBER THEORY DAMIEN PITMAN 1. Definitions & Theorems Definition: We say d divides m iff d is positive integer and m is an integer and there is an integer q such that m = dq. In this case,
### Section 3-4: Least Common Multiple and Greatest Common Factor
Section -: Fraction Terminology Identify the following as proper fractions, improper fractions, or mixed numbers:, proper fraction;,, improper fractions;, mixed number. Write the following in decimal notation:,,.
### Math.3336: Discrete Mathematics. Primes and Greatest Common Divisors
Math.3336: Discrete Mathematics Primes and Greatest Common Divisors Instructor: Dr. Blerina Xhabli Department of Mathematics, University of Houston https://www.math.uh.edu/ blerina Email: blerina@math.uh.edu
### Discrete Math. Instructor: Mike Picollelli. Day 10
Day 10 Fibonacci Redux. Last time, we saw that F n = 1 5 (( 1 + ) n ( 5 2 1 ) n ) 5. 2 What Makes The Fibonacci Numbers So Special? The Fibonacci numbers are a particular type of recurrence relation, a
### Discrete Mathematics GCD, LCM, RSA Algorithm
Discrete Mathematics GCD, LCM, RSA Algorithm Abdul Hameed http://informationtechnology.pk/pucit abdul.hameed@pucit.edu.pk Lecture 16 Greatest Common Divisor 2 Greatest common divisor The greatest common
### HCF & LCM Solved Sums
HCF & LCM Solved Sums 1. The least common multiple of 24, 36, and 40 is A ) 340 b ) 360 c) 230 d) 400 2 24 36 40 => 2 X 2 X 2 X 3 X 1 X 3 X 5 2 12 18 20 => 360 2 6 9 10 3 3 9 5 1 3 5 Ans : 360 2. The LCM
### Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions
Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions Warm-up Problems 1. What is a prime number? Give an example of an even prime number and an odd prime number. A prime number
### 18 Divisibility. and 0 r < d. Lemma Let n,d Z with d 0. If n = qd+r = q d+r with 0 r,r < d, then q = q and r = r.
118 18. DIVISIBILITY 18 Divisibility Chapter V Theory of the Integers One of the oldest surviving mathematical texts is Euclid s Elements, a collection of 13 books. This book, dating back to several hundred
### Section 1.3 Review of Complex Numbers
1 Section 1. Review of Complex Numbers Objective 1: Imaginary and Complex Numbers In Science and Engineering, such quantities like the 5 occur all the time. So, we need to develop a number system that
### Chapter 3: The Euclidean Algorithm and Diophantine. Math 138 Burger California State University, Fresno
Chapter 3: The Euclidean Algorithm and Diophantine Equations Math 138 Burger California State University, Fresno Greatest Common Divisor d is the greatest common divisor of integers a and b if d is the
### Quiz 1, Mon CS 2050, Intro Discrete Math for Computer Science
Quiz 1, Mon 09-6-11 CS 050, Intro Discrete Math for Computer Science This quiz has 10 pages (including this cover page) and 5 Problems: Problems 1,, 3 and 4 are mandatory ( pages each.) Problem 5 is optional,
### KNOWLEDGE OF NUMBER SENSE, CONCEPTS, AND OPERATIONS
DOMAIN I. COMPETENCY 1.0 MATHEMATICS KNOWLEDGE OF NUMBER SENSE, CONCEPTS, AND OPERATIONS Skill 1.1 Compare the relative value of real numbers (e.g., integers, fractions, decimals, percents, irrational
### 1. (16 points) Circle T if the corresponding statement is True or F if it is False.
Name Solution Key Show All Work!!! Page 1 1. (16 points) Circle T if the corresponding statement is True or F if it is False. T F The sequence {1, 1, 1, 1, 1, 1...} is an example of an Alternating sequence.
### 4 Powers of an Element; Cyclic Groups
4 Powers of an Element; Cyclic Groups Notation When considering an abstract group (G, ), we will often simplify notation as follows x y will be expressed as xy (x y) z will be expressed as xyz x (y z)
### Exam 2 Review Chapters 4-5
Math 365 Lecture Notes S. Nite 8/18/2012 Page 1 of 9 Integers and Number Theory Exam 2 Review Chapters 4-5 Divisibility Theorem 4-1 If d a, n I, then d (a n) Theorem 4-2 If d a, and d b, then d (a+b).
### Chapter 3: Section 3.1: Factors & Multiples of Whole Numbers
Chapter 3: Section 3.1: Factors & Multiples of Whole Numbers Prime Factor: a prime number that is a factor of a number. The first 15 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
### Applied Cryptography and Computer Security CSE 664 Spring 2017
Applied Cryptography and Computer Security Lecture 11: Introduction to Number Theory Department of Computer Science and Engineering University at Buffalo 1 Lecture Outline What we ve covered so far: symmetric
### Sect Complex Numbers
161 Sect 10.8 - Complex Numbers Concept #1 Imaginary Numbers In the beginning of this chapter, we saw that the was undefined in the real numbers since there is no real number whose square is equal to a
### 1 Paid Copy Don t Share With Anyone
HCF & LCM Solved Sums 1. The least common multiple of 24, 36, and 40 is A ) 340 b ) 360 c) 230 d) 400 2 24 36 40 => 2 X 2 X 2 X 3 X 1 X 3 X 5 2 12 18 20 => 360 2 6 9 10 3 3 9 5 1 3 5 Ans : 360 2. The LCM
### Exercises Exercises. 2. Determine whether each of these integers is prime. a) 21. b) 29. c) 71. d) 97. e) 111. f) 143. a) 19. b) 27. c) 93.
Exercises Exercises 1. Determine whether each of these integers is prime. a) 21 b) 29 c) 71 d) 97 e) 111 f) 143 2. Determine whether each of these integers is prime. a) 19 b) 27 c) 93 d) 101 e) 107 f)
### CISC-102 Winter 2016 Lecture 11 Greatest Common Divisor
CISC-102 Winter 2016 Lecture 11 Greatest Common Divisor Consider any two integers, a,b, at least one non-zero. If we list the positive divisors in numeric order from smallest to largest, we would get two
### Masters Tuition Center
1 REAL NUMBERS Exercise 1.1 Q.1. Use Euclid s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Solution. (i) In 135 and 225, 225 is larger integer. Using Euclid
### Chapter 2 (Part 3): The Fundamentals: Algorithms, the Integers & Matrices. Integers & Algorithms (2.5)
CSE 54 Discrete Mathematics & Chapter 2 (Part 3): The Fundamentals: Algorithms, the Integers & Matrices Integers & Algorithms (Section 2.5) by Kenneth H. Rosen, Discrete Mathematics & its Applications,
### Integers and Division
Integers and Division Notations Z: set of integers N : set of natural numbers R: set of real numbers Z + : set of positive integers Some elements of number theory are needed in: Data structures, Random
### Mathematics of Cryptography
Modulo arithmetic Fermat's Little Theorem If p is prime and 0 < a < p, then a p 1 = 1 mod p Ex: 3 (5 1) = 81 = 1 mod 5 36 (29 1) = 37711171281396032013366321198900157303750656 = 1 mod 29 (see http://gauss.ececs.uc.edu/courses/c472/java/fermat/fermat.html)
### Math League SCASD. Meet #2. Number Theory. Self-study Packet
Math League SCASD Meet #2 Number Theory Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements
### Basic elements of number theory
Cryptography Basic elements of number theory Marius Zimand 1 Divisibility, prime numbers By default all the variables, such as a, b, k, etc., denote integer numbers. Divisibility a 0 divides b if b = a
### Basic elements of number theory
Cryptography Basic elements of number theory Marius Zimand By default all the variables, such as a, b, k, etc., denote integer numbers. Divisibility a 0 divides b if b = a k for some integer k. Notation
### CHAPTER 3. Number Theory
CHAPTER 3 Number Theory 1. Factors or not According to Carl Friedrich Gauss (1777-1855) mathematics is the queen of sciences and number theory is the queen of mathematics, where queen stands for elevated
### cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska
cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska LECTURE 12 CHAPTER 4 NUMBER THEORY PART1: Divisibility PART 2: Primes PART 1: DIVISIBILITY Basic Definitions Definition Given m,n Z, we say
### Summer Math Packet for Students Entering 6th Grade. Please have your student complete this packet and return it to school on Tuesday, September 4.
Summer Math Packet for Students Entering 6th Grade Please have your student complete this packet and return it to school on Tuesday, September. Work on your packet gradually. Complete one to two pages
### Direct Proof Divisibility
Direct Proof Divisibility Lecture 15 Section 4.3 Robb T. Koether Hampden-Sydney College Fri, Feb 8, 2013 Robb T. Koether (Hampden-Sydney College) Direct Proof Divisibility Fri, Feb 8, 2013 1 / 20 1 Divisibility
### This is a recursive algorithm. The procedure is guaranteed to terminate, since the second argument decreases each time.
8 Modular Arithmetic We introduce an operator mod. Let d be a positive integer. For c a nonnegative integer, the value c mod d is the remainder when c is divided by d. For example, c mod d = 0 if and only
### Direct Proof Divisibility
Direct Proof Divisibility Lecture 15 Section 4.3 Robb T. Koether Hampden-Sydney College Fri, Feb 7, 2014 Robb T. Koether (Hampden-Sydney College) Direct Proof Divisibility Fri, Feb 7, 2014 1 / 23 1 Divisibility
### NUMBER THEORY AND CODES. Álvaro Pelayo WUSTL
NUMBER THEORY AND CODES Álvaro Pelayo WUSTL Talk Goal To develop codes of the sort can tell the world how to put messages in code (public key cryptography) only you can decode them Structure of Talk Part
### Introduction to Information Security
Introduction to Information Security Lecture 5: Number Theory 007. 6. Prof. Byoungcheon Lee sultan (at) joongbu. ac. kr Information and Communications University Contents 1. Number Theory Divisibility
### August 15, M1 1.4 Common Factors_Multiples Compacted.notebook. Warm Up MI 36. Jun 20 10:53 AM
Warm Up MI 36 8 14 18 Jun 20 10:53 AM 1 Assignment Jun 20 12:36 PM 2 Practice 7 13 A = bh 7 x 13 91 7 7 A = ½bh ½(7 x 7) ½(49) 24.5 Jun 20 12:36 PM 3 Practice 6 4 8 A=½bh 4 6x8 24 A=bh 4x8 32 4 5 8 8 A=bh
### Arithmetic. Integers: Any positive or negative whole number including zero
Arithmetic Integers: Any positive or negative whole number including zero Rules of integer calculations: Adding Same signs add and keep sign Different signs subtract absolute values and keep the sign of
### COT 3100 Applications of Discrete Structures Dr. Michael P. Frank
University of Florida Dept. of Computer & Information Science & Engineering COT 3100 Applications of Discrete Structures Dr. Michael P. Frank Slides for a Course Based on the Text Discrete Mathematics
### and LCM (a, b, c) LCM ( a, b) LCM ( b, c) LCM ( a, c)
CHAPTER 1 Points to Remember : REAL NUMBERS 1. Euclid s division lemma : Given positive integers a and b, there exists whole numbers q and r satisfying a = bq + r, 0 r < b.. Euclid s division algorithm
### COMP239: Mathematics for Computer Science II. Prof. Chadi Assi EV7.635
COMP239: Mathematics for Computer Science II Prof. Chadi Assi assi@ciise.concordia.ca EV7.635 The Euclidean Algorithm The Euclidean Algorithm Finding the GCD of two numbers using prime factorization is
### Intermediate Math Circles February 14, 2018 Contest Prep: Number Theory
Intermediate Math Circles February 14, 2018 Contest Prep: Number Theory Part 1: Prime Factorization A prime number is an integer greater than 1 whose only positive divisors are 1 and itself. An integer
### Section 4. Quantitative Aptitude
Section 4 Quantitative Aptitude You will get 35 questions from Quantitative Aptitude in the SBI Clerical 2016 Prelims examination and 50 questions in the Mains examination. One new feature of the 2016
### not to be republished NCERT REAL NUMBERS CHAPTER 1 (A) Main Concepts and Results
REAL NUMBERS CHAPTER 1 (A) Main Concepts and Results Euclid s Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 r < b. Euclid s Division
Q 1 Find the square root of 729. 6. Squares and Square Roots Q 2 Fill in the blank using the given pattern. 7 2 = 49 67 2 = 4489 667 2 = 444889 6667 2 = Q 3 Without adding find the sum of 1 + 3 + 5 + 7
### { independent variable some property or restriction about independent variable } where the vertical line is read such that.
Page 1 of 5 Introduction to Review Materials One key to Algebra success is identifying the type of work necessary to answer a specific question. First you need to identify whether you are dealing with
### Park Forest Math Team. Meet #2. Number Theory. Self-study Packet
Park Forest Math Team Meet #2 Number Theory Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements
### 3 The fundamentals: Algorithms, the integers, and matrices
3 The fundamentals: Algorithms, the integers, and matrices 3.4 The integers and division This section introduces the basics of number theory number theory is the part of mathematics involving integers
### GRE Quantitative Reasoning Practice Questions
GRE Quantitative Reasoning Practice Questions y O x 7. The figure above shows the graph of the function f in the xy-plane. What is the value of f (f( ))? A B C 0 D E Explanation Note that to find f (f(
### Number Theory Proof Portfolio
Number Theory Proof Portfolio Jordan Rock May 12, 2015 This portfolio is a collection of Number Theory proofs and problems done by Jordan Rock in the Spring of 2014. The problems are organized first by
### MTH310 EXAM 2 REVIEW
MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not
### MATHEMATICS IN EVERYDAY LIFE 8
MATHEMATICS IN EVERYDAY LIFE Chapter : Square and Square Roots ANSWER KEYS EXERCISE.. We know that the natural numbers ending with the digits,, or are not perfect squares. (i) ends with digit. ends with
### Number Tree LCM HCF Divisibility Rules Power cycle Remainder Theorem Remainder of powers a n b n Last and Second last digit Power of Exponents Euler s
Vedic Numbers Number Tree LCM HCF Divisibility Rules Power cycle Remainder Theorem Remainder of powers a n b n Last and Second last digit Power of Exponents Euler s Theorem Fermet s Theory Wilson Theorem
### Algebra Summer Review Packet
Name: Algebra Summer Review Packet About Algebra 1: Algebra 1 teaches students to think, reason, and communicate mathematically. Students use variables to determine solutions to real world problems. Skills
### Primes and Modular Arithmetic! CSCI 2824, Fall 2014!!
Primes and Modular Arithmetic! CSCI 2824, Fall 2014!!! Scheme version of the algorithm! for finding the GCD (define (gcd a b)! (if!(= b 0)!!!!a!!!!(gcd b (remainder a b))))!! gcd (812, 17) = gcd(17, 13)
### EDULABZ INTERNATIONAL NUMBER SYSTEM
NUMBER SYSTEM 1. Find the product of the place value of 8 and the face value of 7 in the number 7801. Ans. Place value of 8 in 7801 = 800, Face value of 7 in 7801 = 7 Required product = 800 7 = 00. How
### FACTORS AND MULTIPLES
FACTORS AND MULTIPLES.(A) Find the prime factors of : (i) (ii) (iii) Ans. (i) (ii) (B.) If P n means prime - factors of n, find : (i) P (ii) P (iii) P (iv) P Ans. (i) F =,,, P (Prime factor of ) = and.
### Unit 1. Number Theory
Unit 1 Number Theory 1-1 Divisibility Rules Divisible by: Rule 2 The number is even (it ends in 0, 2, 4, 6 or 8) 3 The sum of its digits is divisible by 3 (eg 741: 7 + 4 + 1 = 12) 4 The last two digits
### Mathematics for Computer Science Exercises for Week 10
Mathematics for Computer Science Exercises for Week 10 Silvio Capobianco Last update: 7 November 2018 Problems from Section 9.1 Problem 9.1. Prove that a linear combination of linear combinations of integers
### Executive Assessment. Executive Assessment Math Review. Section 1.0, Arithmetic, includes the following topics:
Executive Assessment Math Review Although the following provides a review of some of the mathematical concepts of arithmetic and algebra, it is not intended to be a textbook. You should use this chapter
### Intermediate Math Circles Number Theory II Problems and Solutions
WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Intermediate Math Circles Number Theory II Problems and Solutions 1. The difference between the gcd and lcm of the numbers 10,
### Maths Book Part 1. By Abhishek Jain
Maths Book Part 1 By Abhishek Jain Topics: 1. Number System 2. HCF and LCM 3. Ratio & proportion 4. Average 5. Percentage 6. Profit & loss 7. Time, Speed & Distance 8. Time & Work Number System Understanding
### Chapter 2. Divisibility. 2.1 Common Divisors
Chapter 2 Divisibility 2.1 Common Divisors Definition 2.1.1. Let a and b be integers. A common divisor of a and b is any integer that divides both a and b. Suppose that a and b are not both zero. By Proposition
### Solutions to Assignment 1
Solutions to Assignment 1 Question 1. [Exercises 1.1, # 6] Use the division algorithm to prove that every odd integer is either of the form 4k + 1 or of the form 4k + 3 for some integer k. For each positive
### Algebra Introduction to Polynomials
Introduction to Polynomials What is a Polynomial? A polynomial is an expression that can be written as a term or a sum of terms, each of which is the product of a scalar (the coefficient) and a series
### Prime Factorizaon. Algebra with Career Applicaons. Find GCF using Prime Factorizaon. Greatest Common Factor. Unit7Vol1.notebook.
Prime Factorizaon Algebra with Career Applicaons Blitzer 5.1 Prime Factorizaon, GCF, LCM A prime number is a number that is divisible by 1 and itself. Examples of primes are 2, 3, 5, 7, 11, 13, 17, 19,
### Midterm 1. Your Exam Room: Name of Person Sitting on Your Left: Name of Person Sitting on Your Right: Name of Person Sitting in Front of You:
CS70 Discrete Mathematics and Probability Theory, Fall 2018 Midterm 1 8:00-10:00pm, 24 September Your First Name: SIGN Your Name: Your Last Name: Your Exam Room: Name of Person Sitting on Your Left: Name
### Know the Well-ordering principle: Any set of positive integers which has at least one element contains a smallest element.
The first exam will be on Monday, June 8, 202. The syllabus will be sections. and.2 in Lax, and the number theory handout found on the class web site, plus the handout on the method of successive squaring
### Chapter 5. Number Theory. 5.1 Base b representations
Chapter 5 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization, | 8,800 | 31,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-45 | latest | en | 0.907298 |
https://transum.org/Maths/Exercise/Ratio/Recipe.asp?Level=5 | 1,716,757,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00577.warc.gz | 506,836,453 | 12,783 | # Recipe Ratios
## Tom Yam Gung.
Spicy Lemon Shrimp Soup (Serves 2)
### Ingredients
80g Shrimp
60g mushrooms
1 stalk of lemon grass (sliced)
4 or 5 lime leaves
Bunch of Thai parsley
1 tbsp Tom Yam Chili paste (Can buy in a jar)
1½ tbsp Sugar
5 tbsp Milk
3 cups of pork stock
1 tbsp Lemon Juice
1½ tbsp Fish Sauce
2 tbsp Tamarind Juice
What is the ratio of the weights of shrimps to mushrooms in its simplest form? What is the ratio of the quantity of lemon juice to fish sauce in its simplest form? What is the ratio of the number of tablespoons of sugar to tamarind juice in its simplest form? How many grams of mushrooms will be required if the recipe has been scaled up to use up 200g of shrimp? How many tablespoons of sugar will be required if the recipe has been scaled up to use up 18 cups of pork stock?
Check
## Method
Bring the stock to the boil and add mushrooms, tamarind juice, fish sauce, sugar, lemon grass, the leaves, shrimp, chili paste
After about 3 minutes add the milk and turn off the heat
Add 1 tbsp lemon juice and serve straight away. There should be a unique sweet and spicy flavor. Garnish with Thai Parsley.
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 21 October 'Starter of the Day' page by Mr Trainor And His P7 Class(All Girls), Mercy Primary School, Belfast:
"My Primary 7 class in Mercy Primary school, Belfast, look forward to your mental maths starters every morning. The variety of material is interesting and exciting and always engages the teacher and pupils. Keep them coming please."
Comment recorded on the 9 April 'Starter of the Day' page by Jan, South Canterbury:
"Thank you for sharing such a great resource. I was about to try and get together a bank of starters but time is always required elsewhere, so thank you."
Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.
The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.
Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.
#### Nine Digits
Arrange the given digits one to nine to make three numbers such that two of them add up to the third. This is a great puzzle for practicing standard pen and paper methods of three digit number addition and subtraction.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
Subscribe
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.
It may be worth remembering that if Transum.org should go offline for whatever reason, there is a mirror site at Transum.info that contains most of the resources that are available here on Transum.org.
When planning to use technology in your lesson always have a plan B!
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
Scan the QR code below to visit the online version of this activity.
https://www.Transum.org/go/?Num=759
## Description of Levels
Close
Level 1 - Cow Pad Gung (Fried Rice with Shrimp) Increase quantities by integer multiplication
Level 2 - Massaman Gai (Chicken and Potato Curry) Increase quantities by multiplication
Level 3 - Gai Gorae (Spicy Grilled Chicken) Decrease quantities
Level 4 - Gwittyo Tom Yam Gung (Spicy Lemon Shrimp Soup with Noodles) Mixed questions
Level 5 - Tom Yam Gung (Spicy Lemon Shrimp Soup) Hardfer mixed questions
Level 6 - Geng Jute Jang Ngok (Glass Noodle Soup With Pork) Puzzling questions
More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
The recipes used in this exercise are from Hot Sak's.
## Example
This video is from Jonathan Robinson
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
Close | 1,488 | 6,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.93085 |
https://www.tutorialspoint.com/abs-function-for-complex-number-in-cplusplus | 1,679,735,227,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00178.warc.gz | 1,156,146,751 | 9,381 | # abs() function for complex number in c++ ?
The abs function in C++ is used to find the absolute value of a complex number. The absolute value of a complex number (also known as modulus) is the distance of that number from the origin in the complex plane. This can be found using the formula −
For complex number a+bi:
mod|a+bi| = √(a2+b2)
The abs() function returns the result of the above calculation in C++. It is defined in the complex library that is needed to be included.
## PROGRAM TO SHOW USE OF abs() FUNCTION FOR COMPLEX NUMBERS IN C++
#include <iostream>
#include <complex>
using namespace std;
int main () {
float a= 13.0 , b = 5.0;
complex<double> complexnumber (a, b);
cout << "The absolute value of " << a<<"+"<<b<<"i" << " is: ";
cout << abs(complexnumber) << endl;
return 0;
}
## Output
The absolute value of 13+5i is: 13.9284 | 229 | 853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.695115 |
https://koorio.com/a29b632c0b1c59eef9c7b403.html | 1,532,229,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00206.warc.gz | 691,494,359 | 7,478 | koorio.com
# 2014高考最后信息卷-广西数学文
\$% " &'(
+,/! " # \$0!"#\$%&'
!!
)*+,
!"#\$% # &'( )%# *&'( +,\$ -\$ # ."/0 # # % & "\$ ! " \$1 2(3 .456789 "% !% & :;<=>?@ABCDEFGHIJKLM .4N 2 / OC2P2Q2(RS% #TU(&V2PW 7! '% ( XBY2(R SZ[(I\2P]^_< % # O 7 8 9 "% &:;<=>?@ABQ2(RS`
-./012345/678 9 : 0 # ;<=>?/ (\2(abGN2 78
@ # AB0 ! C ) " D
E AB0 FG0C # IH0 & D C) 9 I H 0 J . 5 K L A M N OPEMQ ! H0 " D! RS0TUV5 ! 1234 "* 5 "&\$ 67 # ! # #\$' \$* # #+! #+\$ !! # # %" ,%# #%! (%# ''#%\$ -% # ''#'\$ # # #
! 869# #+# %" :;3< !! #/# ,!# #+#%#%# !!!!
.% # #(\$ #
-! # #+#%#%" !!! >? %* '! # :@>?A #(\$ # 槡
(! # "%#%# # .! # #(# = +#%#%" #
# # ,! #% !!! %* ! ! # -! %* # # #( !!!! ! # 槡
(! %*
# # "%#% ! # 槡
# # .! "%#% %* ! ! #
# 12BC & BC' 5 & <' DE: \$! #'# \$# # FG8HIBC ,! FIBC -! HI8FGBC (! J8FGK8HIBC .!
0 126L?M :NL **# 5?M :O ' &! ( ( ( " P:QA ) ' " *! ) ! 1 ,! +# &
! !
(! ! & &
-! +# 1 &
.! +) "
% ! ! (" 12RST# :UBVWTX Q Y , \RS )! +(" # /% +) #/&*" Z [ ! + ! *# ( ( +
5aRST:bcA T:]^_AY , :Y`
! ! # % ,% + *# & \$ ! ! # % (% + *# \$ & ! ! # % -% + *# ' ) ! ! # % .% + *# ) '
* \$ ) ! " # \$ !"#\$%&' !"# ! ( ! # ) ! +, ! -.
\$ ( \$ ( defg ( 12 * # 5 ( 0! ( \$( ' ! / * # '( # 2 hi:jkl67 2 ,! ) & mP9 # 2! #/ 2 (! + ) & !\$ # ! no9p:q?P< # 槡 (2P (!
' ( (% # !
# ) -! ) &
# ) .! + ) &
(1P (# -! .! "P rstA ( :ubv " {\$ 5|}~ -+" 1! \$ , - wxiT " , yz ( -* (( \$ , :?A
' ( ,% )
(0P ,!
'' 槡 -% ( # !
!' 槡 .% ( # !
\$ r>? %*3 :??f?d? # "! 4 5 ## (" ??x?:>?;?9A # ,! 4 5 #/ \$ %*3 ) # (! 4 5 #+ \$ %*3 ) # -! 4 5 ! #/ \$ %*3 ' # .! 4 5 ! #+ %*3
???( d?\$:??? ? ? ?( 5d?\$:? )
\$ ' ????|?? * U?? * ???|?????? ( ? ' ?D?? ! ?????? ( ??H? # #! ?D??? ( 58?:??b?? ,! # 2 ? !! (! ! # ? !! -! ! 0 ? !! .! ' &?
! ??T ,) ?^_A .( ¤_ . \?# ?§¨:¨T/ ?, ?7" * _0 A # !! ) #( \$ U_ ( % *槡 * * * ) * ) ) ) ! ! ! 8?¨T/ ?:??_ ( \# 5# 0 " 0 \$ ! 0 "/0 \$ # /# # *\$( # :?l??<
# ,! )+'槡 '( )/'槡 '\$ !! # -! )+'槡 '( )/'槡 ', !!
+ (! )+'槡 '( )/'槡 ', + .! )+'槡 '( )/'槡 '\$
@ # WAB0 ! C 1 " D
X YZ0 FG0C \$ H0 IH0 & D C! "D!
# " 0 ?# :no9p ( :°?< +# 5±? (* !!!!! ! # '! #+ (\$ # &( 1 ) # ( \$ ( 12_ " 1 ) ?¨T #/! # \$! %+!*" ? 5 ! /\$ :#?lA !!!! !
#\$!( , ±? #( # &! #/ % ?? B C+ %'\$(
? ? > ? 3*' 5 a ? ? > ? 3* #/% : # ? l A &(
-+! #/ 2\$"( %/ ' #/ % :#?lA !!!!! ! ? ? (· "* ? "- . d e " \ "- *!( # )! \$* ,* - A ? 4 ? : ? _( \$ ,( ( 5\$ " ,*) " 6 " \$*!槡 '( \$ ,*'( , U_?:?e??< !!!!! ! /\$
! * \$ )\$ ! " # \$ !"#\$%&' " !"# ! ( ! ) # ! +, ! -. %
[ \/0 FG0C ) H0 C0 \/]4.^_`a bacdefghi ! "D!
# +, 9 : 0 # > ? / 7 8 ????G # # 0! " G\$ " " " " " " " " "
126L?M & :O) PQA 5 6??M & p( ( ( '\5 ! ( # 0 ! + + ( + 5 ) ! )' )( '* &+ #* )' #* !( ! '* # ?( \$ ) Q+ )# ·2 · 6) * ? 6) ! ( + 2 2 2 \$ )* )/# )( #/ !/ . / )(
# +, 9 : 0 # > ? / 7 8 ????G # # 2! ! G\$ " " " " " " " " "
( ( 12 0" fg * # \ " \$ , p( +#( '\$ * # 7 8 3"( 3 4 5"\$ *#! 槡 # ?i "\$ # ?i "( \ (*槡 ? 0" \$( , ?x:sG?A(( +( 2( '( \$ , :e?:#?l ! \$
# ????G # # 1! ! G\$ ?/±?p?*????? # ?p?? ) ? ( ?? \$ ? ( ??I?p?? \$ ? ? ? ? ? "?( × 7 ? ? ? I( ?p????Q????8???? ???0?? ? ? ? : ? ? ? ? ( ?? ! # ???: \$ ? ? ? p à ? ? ? ? á ( \ \$ ? ? ? p ? ? ! ? ? ? ?* !????: \$ ?? # ???: \$ ???p???8?7 ! ?:?? ! \$
! * \$ )\$ ! " # \$ !"#\$%&' " !"# ! ( ! ' )# ! +, ! -. %
# +, 9 : 0 # > ? / 7 8 ????G # ! "! ! G\$ " " " " " " " " "
12|}? " _- ? \$ ,+"#\$#, ""# . d e " \$ ,( " \$*&( " ,*\$( \$ ,*'( ""# *\$( # p(
" \$ ?! # ? - <" ?? ) \$ :p_ ( " , , -\$ # 1 de \$ # - #?( # è\$ ?mei \$+, * -+\$# :jkl ! \$ " \$ &
# ????G # ! #! ! G\$ ! \$ \$ \$ 12>? 7# :[T/ ¤_ # #\$ * ( # / + #/ 2 :???_ # #( # "( 2+# ! 7# # ? ( :l \$ ! # ( è+*! > ? .# x?é# 869 2(" ? ( #\$ *#+ #\$ /2 +2, # +2( 2, \$ 7# #( !2+ # ?2 :#?l ! .# # +.# # # #' 2 êDE ( #\$ !\$ '
# +, 9 : 0 # > ? / 7 8 ????G # ! !! ! G\$ " " " " " " " " "
' ( 槡 \$ ( 12?Y , :??^_ .## ì¤_ "# xí?A??? ! +槡 '( " #( \$ ! # ??Y , :bc \$ & :¨T ? ? Y 7 * U _ ( # ( ¤_ -# T ? 89 p _ A : ( _ \$# è "( \$ / , 8 9 +#( "\$ \$ ' ?¨T/ :bc ! / \$? ( .:
! * \$ )\$ ! " # \$ !"#\$%&' " !"# ! ( ! \$ )# ! +, ! -. %
### 2018年全国统一招生考试最新高考信息卷数学(文)试卷含答案
2018年全国统一招生考试最新高考信息卷数学(文)试卷含答案 - 绝密 ★ 启用前 2018 年最新高考信息卷 文科数学 注意 事项: 1、 本试卷分第Ⅰ卷(选择题) ...
### 2018年全国高考最新信息卷文科数学(五)(解析版附后)
2018年全国高考最新信息卷文科数学(五)(解析版附后) - 2018 年全国高考最新信息卷 文科数学 (五) (解析版附后) 第Ⅰ卷一、选择题:本大题共 12 小题,每...
### 2018届全国高考信息卷(八)(文)数学试卷
2018届全国高考信息卷(八)(文)数学试卷 - 2018 届全国高考信息卷(八) (文科)数学试卷 本试题卷共 10 页,23 题(含选考题) 。全卷满分 150 分。考试用时...
### 2018届全国高考考前信息卷(二)数学试卷(文科)
2018届全国高考考前信息卷(二)数学试卷(文科) - 2018 届全国高考考前信息卷(二) 数学(文科) 本试题卷共 10 页,23 题(含选考题) 。全卷满分 150 分。考试...
### 2018届全国高考考前信息卷(六)数学试卷文科
2018届全国高考考前信息卷(六)数学试卷文科 - 2018 届全国高考考前信息卷(六) 数学文科试卷 本试题卷共 10 页,23 题(含选考题) 。全卷满分 150 分。考试用... | 3,020 | 4,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-30 | latest | en | 0.147471 |
https://www.jiskha.com/questions/1811250/write-an-expression-to-describe-a-rule-for-the-sequence-then-find-the-100th-term-in-the | 1,623,855,652,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00504.warc.gz | 751,929,316 | 6,343 | # Math
Write an expression to describe a rule for the sequence. Then find the 100th term in the sequence.
4, 9, 14, 19, 24, 29, . . .
I'm pretty confused h e l p
I think it's B(5n-1;499) but I'm unsure please check my answer for me!!
1. 👍
2. 👎
3. 👁
1. looks like an arithmetic sequence where
a =4 and d = 5
term(n) = a + (n-1)d = 4 + (n-1)(5) = 4 + 5n - 5 = 5n -1
term(100) = 5(100) - 1
= ....
You are correct
1. 👍
2. 👎
👤
Reiny
2. Thank you so much! That's really helpful since I wasn't sure ^^
1. 👍
2. 👎
2. -2,592 -15,552 -93,312
3. 63
4. The sequence is geometric because it decreases by a factor of 1/18
5. K= 28; j= 4
6. 5n - 1; 499
7. 2,100
8. Is complicated but D
9. 4 kilometers
10. y=9+19x
11. k= -6j
12. Mr. Davis was looking for a parking space
13. At 20 minutes, she took a 10 minute break
14. Is another complicated one but D again
15. \$1,732.50; \$2,887.50
16. \$2665.61
17. p= M/9n
18. J= PM/300
19. 45 years
Sorry if you don't believe me but these are 100 % the answers except for 8 and 14 cause I don't know what it will look like for you also the last 2 are not here cause I want you to do them like I did myself
1. 👍
2. 👎
4. goofyproduct 100 percent right
1. 👍
2. 👎
5. don't bother wasting your time and putting down answers, the test changes the answers for each of us... so ofc, were arent cheating :|
g'day
-OGW πΎ
1. 👍
2. 👎
6. I luva ur naymmmm gurrllll πΎ
OGWOWWWW
1. 👍
2. 👎
## Similar Questions
1. ### algebra
Could someone check my work; 1. Generate the first five terms in the sequence using the explicit formula. yn = β5n β 5 β30, β25, β20, β15, β10 30, 25, 20, 15, 10 β10, β15, β20, β25, β30(my answer) 10, 15,
Question 1 Write the first four terms of the sequence whose general term is given. an = 3n - 1 Answer 2, 3, 4, 5 2, 5, 8, 11 -2, -5, -8, -11 4, 7, 10, 13 3 points Question 2 Write the first four terms of the sequence whose general
4. ### math
24. Four friends attempted to write the explicit expression for the nth term of the sequence 2, 5, 10, 17, β¦. 24. Four friends attempted to write the explicit expression for the nth term of the sequence 2, 5, 10, 17, β¦. If n
1. ### algebra
Describe the pattern in each sequence and determine the next term of the sequence. 11, 17, 23, 29, β¦ The pattern in each sequence is that after each sequence, the number is added by 6. The next term of the sequence would be 35.
2. ### Math Check My Work Please
Find the value of the variables in the table. x n 4 6 10 12 y 1 9 m 21 25 m = 13, n = 0 m = 11, n = 0* m = 15, n = 3 m = 13, n = 3 6. Write an expression to describe a rule for the sequence. Then find the 100th term in the
3. ### math sequence
the first 5 terms of a linear sequence are given below: 8,6,4,2,0... What is the 100th term in the sequence? A. -192 B. -190 C. -108 D. -90 I could figure this out in like 5 minutes, but I know there's a pattern I can't remember
4. ### patterns
the first five terms of a sequence are shown below: 8,6,4,2,0 ... What is the 100th term in the sequence? A. -192 B. -190 C. -108 D. -90 I know how to figure this out but it would take a few minutes. There UST be some sort of
1. ### Math
4. Which explains why the sequence 81, 3, 1/9,... is arithmetic or geometric? A)The sequence is arithmetic because it decreases by a factor of 1/27 B)The sequence is geometric because it decreases by a factor of 1/27 C)The
2. ### math
Write an expression to describe a rule for the sequence. Then find the 100th term in the sequence. 3,10,17,24,31,38... A. n+7;707 B. 7n;700 C. 7n-4;696++ D. 6n-4;596
3. ### pre-calculus urgent
The first term of an arithmetic sequence is 6, and the tenth term is 2. (a) Find the common difference d. d = Find the 100th term of the sequence. a100 = (b) Find the partial sum of the first ten terms. S10 = how do i do this,
4. ### Urgent math
how do I do this can any body show me how? An arithmetic sequence begins 4, 9, 14, 19, 24, . . . . (a) Find the common difference d for this sequence. d = (b) Find a formula for the nth term an of the sequence. an = (c) Find the | 1,460 | 4,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-25 | latest | en | 0.87261 |
https://www.coursehero.com/file/20831/Calc10-4/ | 1,540,200,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515029.82/warc/CC-MAIN-20181022092330-20181022113830-00522.warc.gz | 915,283,139 | 190,049 | # Calc10_4 - 10.4 Projectile Motion Fort Pulaski GA Photo by...
This preview shows pages 1–10. Sign up to view the full content.
10.4 Projectile Motion Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2002 Fort Pulaski, GA
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
One early use of calculus was to study projectile motion . In this section we assume ideal projectile motion: Constant force of gravity in a downward direction Flat surface No air resistance (usually)
We assume that the projectile is launched from the origin at time t =0 with initial velocity v o . o v Let o o v = v ( 29 ( 29 then cos sin o o o v v α α = + v i j The initial position is: r 0 0 0 o = + = i j α
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
o v α Newton’s second law of motion: Vertical acceleration f ma = 2 2 f d r m dt =
o v α Newton’s second law of motion: The force of gravity is: Force is in the downward direction f ma = f mg = - j 2 2 f d r m dt =
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
o v α Newton’s second law of motion: The force of gravity is: f ma = f mg = - j 2 2 f d r m dt = mg = - j 2 2 d r m dt
o v α Newton’s second law of motion: The force of gravity is: f ma = f mg = - j 2 2 f d r m dt = mg = - j 2 2 d r m dt
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2 2 d r g dt = - j Initial conditions: r r v when o o dr t o dt = = = 2 o 1 r v r 2 o gt t = - + + j 2 1 r 2 gt = - j 0 + ( 29 ( 29 cos sin o o v t v t α α + + i j o v dr gt dt = - + j
( 29 ( 29 2 1 r cos sin 0 2 o o gt v t v t α α = - + + + j i j ( 29 ( 29 2 1 r cos sin 2 o o v t v t gt α α = + - i j Vector equation for ideal projectile motion:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 810 | 3,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-43 | latest | en | 0.824769 |
https://www.futurestarr.com/blog/other/6-out-of-30-as-a-percentage-or | 1,660,765,648,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00037.warc.gz | 695,652,315 | 26,161 | FutureStarr
6 Out of 30 As a Percentage OR
## 6 Out of 30 As a Percentage
via GIPHY
We recently hired a developer for our front-end team, which currently is only four people. But we are getting close to launching our next major product, so we need him to start building basic prototypes. What's a reasonable conversion factor we should use to determine how many developers to hire in the meantime?
### Percentage
via GIPHY
This percentage calculator is a tool that lets you do a simple calculation: what percent of X is Y? The tool is pretty straightforward. All you need to do is fill in two fields, and the third one will be calculated for you automatically. This method will allow you to answer the question of how to find a percentage of two numbers. Furthermore, our percentage calculator also allows you to perform calculations in the opposite way, i.e., how to find a percentage of a number. Try entering various values into the different fields and see how quick and easy-to-use this handy tool is. Is only knowing how to get a percentage of a number is not enough for you? If you are looking for more extensive calculations, hit the advanced mode button under the calculato
So what is percentage good for? As we wrote earlier, a percentage is a way to express a ratio. Say you are taking a graded exam. If we told you that you got 123 points, it really would not tell you anything. 123 out of what? Now, if we told you that you got 82%, this figure is more understandable information. Even if we told you, you got 123 out of 150; it's harder to feel how well you did. A week earlier, there was another exam, and you scored 195 of 250, or 78%. While it's hard to compare 128 of 150 to 195 of 250, it's easy to tell that 82% score is better than 78%. Isn't the percent sign helpful? After all, it's the percentage that counts! (Source: www.omnicalculator.com)
### Number
via GIPHY
As your maths skills develop, you can begin to see other ways of arriving at the same answer. The laptop example above is quite straightforward and with practise, you can use your mental maths skills to think about this problem in a different way to make it easier. In this case, you are trying to find 20%, so instead of finding 1% and then multiplying it by 20, you can find 10% and then simply double it. We know that 10% is the same as 1/10th and we can divide a number by 10 by moving the decimal place one place to left (removing a zero from 500). Therefore 10% of £500 is £50 and 20% is £100.
Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%-- this literally means 30 per 100. So you could view this as 30/100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. Now we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30/100-- and you could view this times 6/1-- this is equal to 180/100. And let's see. We can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9/5, which is the same thing as 1 and 4/5. And then if we wanted to write this as a decimal, 4/5 is 0.8. And if you want to verify that, you could verify that 5 goes into 4-- and there's going to be a decimal. So let's throw some decimals in there. It goes into 4 zero times. So we don't have to worry about that. It goes into 40 eight times. 8 times 5 is 40. Subtract. You have no remainder, and you just have 0's left here. So 4/5 is 0.8. You've got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would've gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6-- let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal. So I'm only going to have one number to the right of the decimal here. So I just count one number. It's going to be 1.8. So either way you think about it or calculate it, 30% of 6 is 1.8. (Source: www.khanacademy.org)
## Related Articles
• #### C All Star Futures Game 2015"
August 17, 2022 | Muhammad Zain
• #### Future Stars Camp Bryn Mawr
August 17, 2022 | laiba idris
• #### AA How To Do Fractions On A Ti 83 Plus
August 17, 2022 | Fahad Nawaz
• #### Where Can I Buy A Scientific Calculators
August 17, 2022 | Fahad Nawaz
• #### Entry Level Nurse Resume Sample.
August 17, 2022 | Ayaz Hussain
• #### How To Write A Prisoner A Letter
August 17, 2022 | Javeria Ijaz
• #### donny osmond net worth.
August 17, 2022 | Ayaz Hussain
• #### Future Stars Gymnastics Competition 2019 OR
August 17, 2022 | Ayaz Hussain
• #### Sample 2 Page Resume Format OR
August 17, 2022 | Ayaz Hussain
• #### Bank Teller Resume Sample ORR
August 17, 2022 | Sami Ullah
• #### Premiere Date Announced for Lifetime TV Show
August 17, 2022 | Mr. JA
• #### Good Front End Developer Resume OR
August 17, 2022 | Danish Ali
• #### Mercedes S Class 2016 For Sale
August 17, 2022 | Ayaz Hussain
• #### Printing Calculator Onlines 2022-2025
August 17, 2022 | Fahad Nawaz
• #### how many cheez its are in a box in 2022
August 17, 2022 | Abid Ali | 1,518 | 5,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-33 | latest | en | 0.95745 |
https://socratic.org/questions/how-do-you-solve-abs-8-6c-1 | 1,624,223,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00079.warc.gz | 475,596,424 | 5,813 | # How do you solve abs(8 - 6c ) = 1?
##### 1 Answer
Feb 19, 2017
$\frac{7}{6} \le c \le \frac{3}{2}$
#### Explanation:
The given equation is equivalent to:
$- 1 \le 8 - 6 c \le 1$
Let's subtract 8 and get:
$- 9 \le - 6 c \le - 7$
then change signs and inequality:
$7 \le 6 c \le 9$
and divide all terms by 6:
$\frac{7}{6} \le c \le {\cancel{9}}^{3} / {\cancel{6}}^{2}$
$\frac{7}{6} \le c \le \frac{3}{2}$ | 178 | 417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-25 | latest | en | 0.642324 |
https://www.scribd.com/document/236700124/g10m-measurement | 1,566,093,930,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00300.warc.gz | 965,343,153 | 60,763 | You are on page 1of 20
# MEASUREMENT
## Recap: Measurement Concepts
Prisms
Cylinders
Scale Factors
Spheres
Pyramids
Cones
Building a 3-D Space Recap: Perimeter and Area
1
RECAP: MEASUREMENT
CONCEPTS
Building a 3-D Space
Recap: Perimeter and Area
2
PRISMS
Total Surface Area
TSA = (L x B) + (L x B) + (L x H) + (L x H) + (B x H) + (B x H)
= 2LB + 2LH + 2BH
Sum the areas of the panels of the rectangle to get the
total surface area
Volume
= (L x B x H)
3
Example
(a) Calculate the surface area and volume of the
rectangular prism if it is closed on all sides. l = 8cm;
b = 3cm ; h = 4cm
Surface Area = 2LB + 2LH + 2BH
Surface Area = 2(8) (3) + 2(8) (4) + 2(3) (4)
Surface Area = 48+64+24
Surface Area = 136 cm
(b) Calculate the surface area if the prism is open on top.
Surface Area = LB + 2LH + 2BH (TOP MISSING)
Surface Area = (8) (3) + 2(8) (4) + 2(3) (4)
Surface Area = 112 cm
Volume &
Surface Area of
a Right Prism
Challenge! Diagonal Length of a Rectangular Prism
4
EXERCISE
Calculate the surface area and volume of a rectangular
prism in each of the following cases.
(a) the prism is closed and the dimensions are:
Length =13 cm. Breadth = 6 cm and Height = 9 cm
(b) the prism is closed and the dimensions are:
Length = 12 m, Breadth = 5 m and Height = 7 m
(c) the prism is open on top and the dimensions are:
Length = 13 cm, Breadth = 6 cm and Height = 9 cm
(d) the prism is open on top and the dimensions are:
Length =12 m, Breadth = 5 m and Height = 1 m
Volume of Triangular Prisms
5
CYLINDERS
Total Surface Area
The Radius of a Circle Circumference of a Circle
rh r TSA 2 2
2
A Cylinder & its Net
Proof: Curved Surface of a Cylinder = Rectangle
6
Volume
h r V
2
## Volume & Surface Area of a
Cylinder Example
Calculate the Surface Area & Volume of Cylinders
7
Example
Given a cylinder with r = 4cm and h = 6m:
(a) Calculate the surface area of the cylinder
assuming a closed cylinder
(b) Calculate the surface area of the cylinder
assuming an open-topped cylinder
(c) Calculate the volume of the cylinder
2
2
) 80 (
48 32
2 2
cm TSA
TSA
rh r TSA
3
2
2
) 96 (
) 6 ( ) 4 (
cm V
V
h r V
2
2
2
) 64 (
48 16
) 6 )( 4 ( 2 ) 4 (
2
cm TSA
TSA
TSA
rh r TSA
8
EXERCISE
Calculate the surface area and volume of a cylinder in
each of the following cases:
(a) the cylinder is closed and the dimensions are:
radius = 5 cm and height = 3 cm.
(b) the cylinder is closed and the diameter = 12 cm
and the height = 7 cm.
(c) the cylinder is open and the dimensions are:
radius = 15 cm and height = 8 mm.
(d) the cylinder is open and the diameter = 4 cm
and the height = cm.
9
It sometimes happens that the dimensions of the prism
are changed.
For example, the length, breadth and height might be
doubled.
The surface area and volume will then obviously be
different from the original prism.
The number which is multiplied by each dimension is
called a scale factor.
Enlarge the triangle by a scale factor of 3
SCALE FACTORS
10
Example
If the length is x units and the
breadth is y units, then the
area of the rectangle is:
A = xy
If we now double the length
length and breadth by a scale
factor of 2), then the area of the
rectangle will change to:
A = (2x)(2y)
A = (2)(2)xy
A = 4xy
Therefore the new area is 4
times the original area.
Finding the New
Enlarged Area
11
Note:
In general if the length and breath of a
rectangle is multiplied by a scale factor
of k units then the area of the rectangle
will have been multiplied by k squared.
The new area is k squared times the
original area.
Scale Factors and Area
Fractional Scale Factor Enlargements
12
Example
Calculate the surface area and
volume of the rectangular prism
if the length, breadth and height
are multiplied by a scale factor
of 5 units.
Surface Area =2(5L)(5B) + 2(5L)(5H) + 2(5B)(5H)
Surface Area = 2(5 x 8)(5 x 3) + 2(5 x 8)(5 x 4) +
2(5 x 3)(5 x 4)
Surface Area = 3400 cm
Volume = (5L)(5B)(5H)
Volume = (5 x 8)(5 x 3)(5 x 4)
Volume = 12000 cm
Therefore the new surface area is 5 x 5 times
greater than the original surface area.
13
Note:
If the length, breadth and height are multiplied by a scale
factor of k units, then:
Surface Area = 2(kL)(kB) + 2(kL)(kH) + 2(kB)(kH)
Surface Area = k [2LB + 2LH + 2BH]
Volume = (k L) (k B) (k H)
Volume = k [LBH]
In general if the length, breadth and height of a
Rectangular prism are multiplied by a scale factor of k
units then the
surface area will be multiplied by k squared
volume will be multiplied by k cubed.
3
Identify the Scale Factor of Enlargement for each Circle
14
Consider the cylinder with the given dimensions:
r = 3cm ; h = 6cm.
(a) Calculate the surface area if the radius is doubled.
(b) Calculate the surface area if the radius is multiplied
by 4cm.
(c) Calculate the surface area if the height is trebled.
(d) Calculate the volume if the radius is doubled.
(e) Calculate the volume if the radius is halved.
EXERCISE
15
Some Fun with Shapes: An Impossible Fork!
Area, Circumference & Volume Problems
Challenge!
16
SPHERES
Surface area = 4 r
2
Volume =
3
4
3
r
Calculate the Surface Area & Volume of Spheres
Volume of a Sphere Example
17
PYRAMIDS
Slant edges
Slant height
Perpendicular height
18
Surface Area =area of base
+ (perimeter of base) x slant height
Volume = 1/3 area of base
x perp height
Calculate the SA & Volume of
a Square Pyramid
19
CONES
TSA = area of base + area of curved surface
= r
2
+ r s
(s is slant height)
Volume =
1
/
3
( area of base ) x ht
=
1
/
3
r
2
h
Calculating the Height of a Cone
Recap: Identify the Shapes
Calculate the Volume of the Cone
20 | 1,744 | 5,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-35 | latest | en | 0.832622 |
https://electronics.stackexchange.com/questions/662573/solving-for-current-in-rlc-circuit-with-laplace-and-steady-state-solution | 1,726,275,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00465.warc.gz | 206,206,609 | 44,796 | Solving for current in RLC circuit with Laplace and steady-state solution
The following parallel RLC circuit is example 8.8 in Fundamentals of Electric Circuits by Charles Alexander.
In the example in this text, the current through the inductor i(t) is found to be:
$$i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978})$$
This solution is found by directly solving the second order differential equation. I want to solve this same circuit using Laplace transforms. Following the methods in the textbook, I have performed a Laplace transform on this circuit:
simulate this circuit – Schematic created using CircuitLab
Please note that even though 4 A current supply is always I, I have converted it to from 4 A to 4/s; I hope that is correct. Also, the initial current through the inductor is 4 A, leading to another 4/s current supply in the opposite direction.
In this circuit, the current through the inductor is:
$$I(s) = \frac{0.75 s}{s(s^2 + 12.5 s + 6.25)}$$
The inverse Laplace transform of I(s) is:
$$i(t) = 0.0655(e^{-0.5218t} - e^{-11.978})$$
In this second solution, the "4" is missing. I have a two questions:
Should Laplace only give the transient solution?
Even though the current supply on the left does not change at t=0, why must it change to 4/s in the Laplace domain?
• What is 30.u(-t) ? Commented Apr 14, 2023 at 13:33
• @Antonio51 Sorry, I should add that to my question. u(t) is the step function, so u(-t) indicates that the voltage supply shuts off at t=0. Commented Apr 15, 2023 at 1:11
I wish you would have disclosed your process. It would have been appreciated.
Grounding the bottom node, find this KCL for the switched node:
\begin{align*} \frac{v_t}{R_1}+\frac{v_t}{R_2}+C_1\frac{\text{d}\,v_t}{\text{d}t}+\frac1{L_1}\int v_t\:\text{d}t&= 0\:\text{A} \\\\ \frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t&= 0\:\text{A} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\mathscr{L}\left\{\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t\right\}+\mathscr{L}\left\{\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\mathscr{L}\left\{\frac{\text{d}}{\text{d}t}v_t\right\}+\frac1{L_1\,C_1}\mathscr{L}\left\{\vphantom{\frac{\text{d}}{\text{d}t}}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \left\{\vphantom{\frac{\text{d}}{\text{d}t}}s^2V_s-sv_{_0}-v_{_0}^{'}\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\left\{\vphantom{\frac{\text{d}}{\text{d}t}}sV_s-v_{_0}\right\}+\frac1{L_1\,C_1}V_s&= \mathscr{L}\left\{0\:\text{A}\right\} \end{align*}
Or,
\begin{align*} \left[s^2+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)s+\frac1{L_1\,C_1}\right]V_s&= \left[s+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\right]v_{_0}+v_{_0}^{'} \end{align*}
Since $$\v_{_0}=15\:\text{V}\$$ and the current supplied by $$\C_1\$$ at $$\t=0\$$ must be $$\-1.5\:\text{A}\$$, it's clear that $$\\frac{\text{d}}{\text{d}t}v_{t=0}=\frac{-1.5\:\text{A}}{8\:\text{mF}}=-187.5\:\frac{\text{V}}{\text{s}}\$$. The above then becomes:
\begin{align*} \left[s^2+12.5s+6.25\right]V_s&= 15s \end{align*}
Now for your mistake. I suspect that you elected to believe that $$\V_s=s L_1 I_s\$$ and found that:
\begin{align*} s\left[s^2+12.5s+6.25\right]I_s&= 0.75s \end{align*}
However, that's not correct.
Instead $$\V_s=\mathscr{L}\left\{v_t\right\}=L_1 \mathscr{L}\left\{i_t^{'}\right\}=L_1\left[sI_s-i_{_0}\right]\$$.
So, as $$\L_1=20\:\text{H}\$$:
\begin{align*}\require{cancel} \left[s^2+12.5s+6.25\right]L_1\left[sI_s-i_{_0}\right]&= 15s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s-i_{_0}\right]&= 0.75s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s\right]&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ s\left[s^2+12.5s+6.25\right]I_s&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ I_s&=\frac{0.75\cancel{s}}{\cancel{s}\left[s^2+12.5s+6.25\right]}+\frac{\cancel{\left[s^2+12.5s+6.25 \right]}4}{s\cancel{\left[s^2+12.5s+6.25\right]}} \\\\ &=\frac{0.75}{s^2+12.5s+6.25}+\frac4{s} \end{align*}
So, your Laplace equation should be $$\I_s=\frac4{s}+\frac{0.75}{s^2+12.5\, s+6.25}\$$.
If you had solved this in the time domain, you'd have been using annihilation and finding a need for hyperbolic sine (hyperbolic because the system is over-damped.)
$$i_t=4.0 + 0.130930734141595\cdot\exp\left(-6.25\,t\right)\cdot\sinh\left(5.7282196186948\,t\right)$$
The way you wrote it is also correct as the hyperbolic sine has a built-in $$\\frac12\$$ factor and expanding it would change 0.130930734141595 to 0.0654653670707975 (and provide your difference pair of exponentials.)
• This is it exactly, this was my mistake. Commented Apr 13, 2023 at 18:48
• @EElmo Thanks for letting me know. And I thought as much. The reason I bothered writing (for such a short question that really didn't expose all of your thinking) is that I could almost see exactly where you made a mistake. You gave me enough for that much. But just barely enough. And it didn't dawn on me until a few hours after, doing other housework here. A little more writing wouldn't have hurt and it might have sped my ability to spot it. Just some kind advice for next time. Also, I'm impressed. You've pretty much mastered this kind of work! I don't believe many here can say as much. Commented Apr 13, 2023 at 21:15
• @EElmo Just FYI. I really like the 9th edition of Fundamentals of Differential Equations by Nagle, Saff & Snider. The pacing is excellent and very well-managed, in my opinion, for self-learners as well as for those taking classes. (You don't need it, though.) Beyond that, almost anything Gilbert Strang writes is pretty darned decent (if a little bit unexpectedly folksy -- which is something I actually like about him but some others don't, so much.) He provides insights and how to gather them up that many here at this site should acquire (directed graph math patterns and Kalman, e.g.) Commented Apr 13, 2023 at 21:23
• I will buy this book. Are older editions ok? Commented Apr 15, 2023 at 11:33
• @EElmo I only have the 9th edition. I can't speak to earlier versions. In the 9th edition, the section that covers the above is "7.5 Solving Initial Value Problems" in chapter 7 which is on Laplace Transforms. Commented Apr 15, 2023 at 20:09
It seems that it should be this ...
Final current into L1 = 4 + 30/20 = 5.5 A ?
Or this ?
• Antonio51 - Hi, (a) Your comment to the downvoter was sarcastic & unfriendly so it broke the Code of Conduct & has therefore been deleted. Please read that important rule. Don't break any part of it again. (b) The FAQ article about being downvoted explains what type of comment response is allowed, if you are sure you want to comment after a downvote. (c) If I had to guess a reason for the downvote, it's because the answer did not address the specific question about using Laplace transforms. An answer should answer that question. TY Commented Apr 14, 2023 at 13:53 | 2,674 | 7,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-38 | latest | en | 0.868741 |
http://android-apps.com/apps/hexagon-campaign/ | 1,506,411,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695113.88/warc/CC-MAIN-20170926070351-20170926090351-00480.warc.gz | 18,790,722 | 11,130 | # Hexagon Campaign
Price \$1.00 Brain & Puzzle 1.0.5 rgw_ch 0 views 0 votes 0 0 On
The Hexagon Election Campaign GameSpend money to win elections: The game world is made up of hexagonal areas. An area can be empty or belong to either one of the RED or the BLUE party. Each party can spend money (hexabucks) on hexagons to influence them. The computer announces how many hexabucks can be spent in the next round. The player chooses on which hexagon he or she wants to spend them. (a) On an empty hexagon, the value of this hexagon will be the sum of hexabucks spent. The campaign will influence each of the up to six neighboring hexagons: (a1) if empty, it will be converted to the color of the player, with a value of 1(a2) if of the same color as the player, it will increase its value by 1(a3) if of the opposite color, will decrease its value by 1. If this would result in a zero value, it will be converted to the player's color instead. (b) On a Hexagon of the player's color, this hexagon's value will be increased by the amount spent. This will not influence neighboring hexagons.(c) On a Hexagon of the opposite color, this hexagon's value will be decreased by the amount spent. If the resultis negative, the hexagon is converted to the player's color. In this case, the move will influence neighboring hexagon as if it was empty. The game is over when: (a) No empty hexagons are left. (b) 60 Rounds are played.(whatever comes first)\n\nHave fun!
4.7(54)
0(0)
5(7) | 389 | 1,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-39 | latest | en | 0.923366 |
https://www2.mdpi.com/2504-3110/1/1/5 | 1,709,652,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948235171.95/warc/CC-MAIN-20240305124045-20240305154045-00672.warc.gz | 1,049,821,466 | 77,291 | Next Article in Journal
Exact Discretization of an Economic Accelerator and Multiplier with Memory
Next Article in Special Issue
Stokes’ First Problem for Viscoelastic Fluids with a Fractional Maxwell Model
Previous Article in Journal
A Fractional Complex Permittivity Model of Media with Dielectric Relaxation
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article
# Dynamics and Stability Results for Hilfer Fractional Type Thermistor Problem
1
Department of Mathematics, Sri Ramakrishna Mission Vidyalaya College of Arts and Science, Coimbatore 641020, India
2
Department of Mathematics, Embry-Riddle Aeronautical University, Daytona Beach, FL 32114, USA
*
Author to whom correspondence should be addressed.
Fractal Fract. 2017, 1(1), 5; https://doi.org/10.3390/fractalfract1010005
Submission received: 22 August 2017 / Revised: 5 September 2017 / Accepted: 6 September 2017 / Published: 9 September 2017
(This article belongs to the Special Issue The Craft of Fractional Modelling in Science and Engineering)
## Abstract
:
In this paper, we study the dynamics and stability of thermistor problem for Hilfer fractional type. Classical fixed point theorems are utilized in deriving the results.
MSC:
26A33; 26E70; 35B09; 45M20
## 1. Introduction
Fractional differential equations (FDEs) occur in many engineering systems and scientific disciplines such as the mathematical modelling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, etc. FDEs also provide as an efficient tool for explanations of hereditary properties of different resources and processes. As a result, the meaning of the FDEs has been of great importance and attention, and one can refer to Kilbas [1], Podlubny [2] and the papers [3,4,5,6,7,8,9]. Recently, the Hilfer fractional derivative [10] for FDEs has become a very active area of research. R. Hilfer initiated the Hilfer fractional derivative. This is used to interpolate both the Riemann–Liouville and the Caputo fractional derivative for the theory and applications of the Hilfer fractional derivative (see, e.g., [6,10,11,12,13,14,15,16] and references cited therein). Analogously, we prefer the Hilfer derivative operator that interpolates both the Riemann–Liouville and the Caputo derivative.
English scientist Michael Faraday first discovered the concept of thermistors in 1833 while reporting on the semiconductor behavior of silver sulfide. From his research work, he noticed that the silver sulfides resistance decreased as the temperature increased. This later leads to the commercial production of thermistors in the 1930s when Samuel Ruben invented the first commercial thermistor. Ever since, technology has improved; this made it possible to improve manufacturing processes along with the availability of advanced quality material.
A thermistor is a thermally sensitive resistor that displays a precise and predictable change in resistance proportional to small changes in body temperature. How much its resistance will change is dependent upon its unique composition. Thermistors are part of a larger group of passive components. Unlike their active component counterparts, passive devices are incapable of providing power gain, or amplification to a circuit. Thermistors can be found everywhere in airplanes, air conditioners, in cars, computers, medical equipment, hair dryers, portable heaters, incubators, electrical outlets, refrigerators, digital thermostats, ovens, stove tops and in all kinds of appliances. Ice sensors and aircraft wings, if ice builds up on the wings, the thermistor senses this temperature drop and a heater will be activated to remove the ice. Flight tests need to be completed on a particular date, hence there may not be enough time to create a flight test technique on that date. However, it is possible to take a number of recommendations on the needs of any future flight plan to examine the nature of thermistor thermometer at high subsonic and supersonic speeds. In general, the unusual behaviour of the thermistor thermometer is caused by the possibility of vortices and an aerodynamic disturbance generating non-uniform flow, happening in the chamber with sensing element. The thermistors are small, which makes them very delicate to such effects [17,18].
A thermistor is a temperature dependent resistor and comes in two varieties, negative temperature coefficient (NTC) and positive temperature coefficient (PTC), although NTCs are most commonly used. With NTC, the resistance variation is inverse to the temperature change i.e.,: as temperature goes up, resistance goes down. NTC Thermistors are nonlinear, and their resistance decreases as temperature increases. A phenomenon called self-heating may affect the resistance of an NTC thermistor. When current flows through the NTC thermistor, it absorbs the heat causing its own temperature to rise. In [19], Khan et al. investigated the coupled p-Laplacian fractional differential equations with nonlinear boundary conditions. Wenjing Song and Wenjie Gao studied the existence of solutions for a nonlocal initial value problem to a p-Laplacian thermistor problems on time scales in [20]. Later, Moulay Rchid Sidi Ammi and Delfim F. M. Torres developed and applied a numerical method for the time-fractional nonlocal thermistor problem in [21]. They investigated the existence and uniqueness of a positive solution to generalized nonlocal thermistor problems with fractional-order derivatives in [22]. Recently, Moulay Rchid Sidi Ammi and Delfim F. M. Torres [23] discussed the existence and uniqueness results for a fractional Riemann–Liouville nonlocal thermistor problem on arbitrary time scales. Interested readers can refer to recent papers [22,23,24,25,26] treating a nonlocal thermistor problem.
Motivated by the aforementioned papers, we study the existence, uniqueness and Ulam–Hyers stability types of solutions for Hilfer type thermistor problem of the form
$D 0 + α , β u ( t ) = λ f ( u ( t ) ) ∫ 0 T f ( u ( x ) ) d x 2 , t ∈ J : = [ 0 , T ] , I 0 + 1 − γ u ( 0 ) = u 0 , γ = α + β − α β ,$
where $D 0 + α , β$ is the Hilfer fractional derivative of order $α$ and type $β$, $0 < α < 1$, $0 ≤ β ≤ 1$ and let $J = [ 0 , T ]$, X be a Banach space, $f : J × X → X$ is a given continuous function. The operator $I 0 + 1 − γ$ denotes the left-sided Riemann–Liouville fractional integral of order $1 − γ$. Choosing $λ$ such that $0 < λ < L T α + 1 − γ ( C 1 T ) 2 Γ ( α + 1 ) + 2 C 2 2 L T α + 3 − γ ( C 1 T ) 2 Γ ( α + 1 ) − 1$ is discussed in Section 4.
It is seen that (1) is equivalent to the following nonlinear integral equation
$u ( t ) = u 0 Γ ( γ ) t γ − 1 + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s .$
The stability of the functional equations were first introduced in a discourse conveyed in 1940 at the University of Wisconsin. The issue made by Ulam is as per the following: Under what conditions does there exist an additive mapping near an approximately additive mapping? [5,27,28,29]. The first reply to the topic of Ulam was given by Hyers in 1941 on account of Banach spaces. Ever since, this type of stability was known as the Ulam–Hyers stability. Rassias [29] gave a generalization of the Hyers theorem for linear mappings. Many mathematicians later extended the issue of Ulam in different ways. Recently, Ulam’s problem was generalized for the stability of differential equations. A comprehensive interest was given to the study of the Ulam and Ulam–Hyers–Rassias stability of all kinds of functional equations [5,8,9,30]. An exhaustive interest was given to the investigation of the Ulam and Ulam–Hyers–Rassias stability of all kinds of functional Equation (1).
The paper is organized as follows. In Section 2, we introduce some definitions, notations, and lemmas that are used throughout the paper. In Section 3, we will prove existence and uniqueness results concerning problem (1). Section 4 is devoted to the Ulam–Hyers stabilities of problem (1).
## 2. Basic Concepts and Results
In this section, we introduce notations, definitions, and preliminary facts that are used throughout this paper. For more details on Hilfer fractional derivative, interested readers can refer to [6,10,12,13,15,31].
Definition 1.
Let $C [ J , X ]$ denote the Banach space of all continuous functions from $[ 0 , T ]$ into X with the norm
$u C : = sup u ( t ) : t ∈ J .$
We denote $L 1 R +$, the space of Lebesgue integrable functions on J.
By $C γ [ J , X ]$ and $C γ 1 [ J , X ]$, we denote the weighted spaces of continuous functions defined by
$C γ [ J , X ] : = f ( t ) : J → X | t γ f ( t ) ∈ C [ J , X ] ,$
with the norm
$f C γ = t γ f ( t ) C ,$
and
$f C γ n = ∑ k = 0 n − 1 f k C + f ( n ) C γ , n ∈ N .$
Moreover, $C γ 0 [ J , X ] : = C γ [ J , X ]$.
Now, we give some results and properties of fractional calculus.
Definition 2
([1,16]). The left-sided mixed Riemann–Liouville integral of order $α > 0$ of a function $h ∈ L 1 R +$ is defined by
where $Γ ( · )$ is the (Euler’s) Gamma function defined by
$Γ ( ξ ) = ∫ 0 ∞ t ξ − 1 e − t d t ; ξ > 0 .$
Notice that for all $α , α 1 , α 2 > 0$ and each $h ∈ C [ J , X ]$, we have $I 0 + α h ∈ C [ J , X ]$, and
Definition 3
([1,16]). The Riemann–Liouville fractional derivative of order $α ∈ ( 0 , 1 ]$ of a function $h ∈ L 1 R +$ is defined by
Let $α ∈ ( 0 , 1 ]$, $γ ∈ [ 0 , 1 )$ and $h ∈ C 1 − γ [ J , X ]$. Then, the following expression leads to the left inverse operator as follows:
Moreover, if $I 0 + 1 − α h ∈ C 1 − γ 1 [ J , X ]$, then the following composition
Definition 4
([1,16]). The Caputo fractional derivative of order $α ∈ ( 0 , 1 ]$ of a function $h ∈ L 1 R +$ is defined by
In [10], Hilfer studied applications of a generalized fractional operator having the Riemann–Liouville and the Caputo derivatives as specific cases (see also [6,32]).
Definition 5 (Hilfer derivative).
Let $0 < α < 1$, $0 ≤ β ≤ 1$, $h ∈ L 1 R +$, $I 0 + ( 1 − α ) ( 1 − β ) ∈ C γ 1 [ J , X ]$. The Hilfer fractional derivative of order α and type β of h is defined as
Properties. Let $0 < α < 1$, $0 ≤ β ≤ 1$, $γ = α + β − α β$, and $h ∈ L 1 R +$.
• The operator $( D 0 + α , β h ) ( t )$ can be written as
Moreover, the parameter $γ$ satisfies
$0 < γ ≤ 1 , γ ≥ α , γ > β , 1 − γ < 1 − β ( 1 − α ) .$
• The generalization (3) for $β = 0$ coincides with the Riemann–Liouville derivative and for $β = 1$ with the Caputo derivative
$D 0 + α , 0 = D 0 + α , a n d D 0 + α , 1 = c D 0 + α .$
• If $D 0 + β ( 1 − α ) h$ exists and in $L 1 R +$, then
Furthermore, if $h ∈ C γ [ J , X ]$ and $I 0 + 1 − β ( 1 − α ) h ∈ C γ 1 [ J , X ]$, then
• If $D 0 + γ h$ exists and in $L 1 R +$, then
In order to solve our problem, the following spaces are presented
$C 1 − γ α , β [ J , X ] = f ∈ C 1 − γ [ J , X ] , D 0 + α , β f ∈ C 1 − γ [ J , X ] ,$
and
$C 1 − γ γ [ J , X ] = f ∈ C 1 − γ [ J , X ] , D 0 + γ f ∈ C 1 − γ [ J , X ] .$
It is obvious that
$C 1 − γ γ [ J , X ] ⊂ C 1 − γ α , β [ J , X ] .$
Corollary 1
([31]). Let $h ∈ C 1 − γ [ J , X ]$. Then, the linear problem
$D 0 + α , β x ( t ) = h ( t ) , t ∈ J = [ 0 , T ] , I 0 + 1 − γ x ( 0 ) = x 0 , γ = α + β − α β ,$
has a unique solution $x ∈ L 1 R +$ given by
$x ( t ) = x 0 Γ ( γ ) t γ − 1 + 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 h ( s ) d s .$
From the above corollary, we conclude the following lemma.
Lemma 1.
Let $f : J × X → X$ be a function such that $f ∈ C 1 − γ [ J , X ]$. Then, problem (1) is equivalent to the problem of the solutions of the integral Equation (2).
Theorem 1
(Schauder fixed point theorem [31,33]). Let B be closed, convex and nonempty subset of a Banach space E. Let $P : B → B$ be a continuous mapping such that $P ( B )$ is a relatively compact subset of E. Then, P has at least one fixed point in B.
Now, we study the Ulam stability, and we adopt the definitions in [4,30,34] of the Ulam–Hyers stability, generalized Ulam–Hyers stability, Ulam–Hyers–Rassias stability and generalized Ulam–Hyers–Rassias stability.
Consider the following Hilfer type termistor problem
$D 0 + α , β u ( t ) = λ f ( u ( t ) ) ∫ 0 T f ( u ( x ) ) d x 2 , t ∈ J : = [ 0 , T ] ,$
and the following fractional inequalities:
$D 0 + α , β z ( t ) − λ f ( z ( t ) ) ∫ 0 T f ( z ( x ) ) d x 2 ≤ ϵ , t ∈ J ,$
$D 0 + α , β z ( t ) − λ f ( z ( t ) ) ∫ 0 T f ( z ( x ) ) d x 2 ≤ ϵ φ ( t ) , t ∈ J ,$
$D 0 + α , β z ( t ) − λ f ( z ( t ) ) ∫ 0 T f ( z ( x ) ) d x 2 ≤ φ ( t ) , t ∈ J .$
Definition 6.
Equation (4) is Ulam–Hyers stable if there exists a real number $C f > 0$ such that, for each $ϵ > 0$ and for each solution $z ∈ C 1 − γ γ [ J , X ]$ of Inequality (5), there exists a solution $u ∈ C 1 − γ γ [ J , X ]$ of Equation (4) with
$z ( t ) − u ( t ) ≤ C f ϵ , t ∈ J .$
Definition 7.
Equation (4) is generalized Ulam–Hyers stable if there exists $ψ f ∈ C ( [ 0 , ∞ ) , [ 0 , ∞ ) ) , ψ f ( 0 ) = 0$ such that, for each solution $z ∈ C 1 − γ γ [ J , X ]$ of Inequality (5), there exists a solution $u ∈ C 1 − γ γ [ J , X ]$ of Equation (4) with
$z ( t ) − u ( t ) ≤ ψ f ϵ , t ∈ J .$
Definition 8.
Equation (4) is Ulam–Hyers–Rassias stable with respect to $φ ∈ C 1 − γ [ J , X ]$ if there exists a real number $C f > 0$ such that, for each $ϵ > 0$ and for each solution $z ∈ C 1 − γ γ [ J , X ]$ of Inequality (6), there exists a solution $u ∈ C 1 − γ γ [ J , X ]$ of Equation (4) with
$z ( t ) − u ( t ) ≤ C f ϵ φ ( t ) , t ∈ J .$
Definition 9.
Equation (4) is generalized Ulam–Hyers–Rassias stable with respect to $φ ∈ C 1 − γ [ J , X ]$ if there exists a real number $C f , φ > 0$ such that, for each solution $z ∈ C 1 − γ γ [ J , X ]$ of Inequality (7), there exists a solution $u ∈ C 1 − γ γ [ J , X ]$ of Equation (4) with
$z ( t ) − u ( t ) ≤ C f , φ φ ( t ) , t ∈ J .$
Remark 1.
A function $z ∈ C 1 − γ γ [ J , X ]$ is a solution of Inequality (5) if and only if there exist a function $g ∈ C 1 − γ γ [ J , X ]$ (which depends on solution z) such that
1.
$g ( t ) ≤ ϵ , ∀ t ∈ J .$
2.
$D 0 + α , β z ( t ) = λ f ( z ( t ) ) ∫ 0 T f ( z ( x ) ) d x 2 + g ( t ) , t ∈ J$.
Remark 2.
It is clear that:
1.
Definition 6Definition 7.
2.
Definition 8Definition 9.
3.
Definition 8 for $φ ( t ) = 1$Definition 6.
Lemma 2
([3]). Let $v : [ 0 , T ] → [ 0 , ∞ )$ be a real function and $w ( · )$ is a nonnegative, locally integrable function on $[ 0 , T ]$ and there are constants $a > 0$ and $0 < α < 1$ such that
$v ( t ) ≤ w ( t ) + a ∫ 0 t v ( s ) ( t − s ) α d s .$
Then, there exists a constant $K = K ( α )$ such that
$v ( t ) ≤ w ( t ) + K a ∫ 0 t w ( s ) ( t − s ) α d s ,$
for every $t ∈ [ 0 , T ]$.
## 3. Existence Results
The following existence result for Hilfer type thermistor problem (1) is based on Schauder’s fixed point theorem. Let us consider the following assumptions:
Assumption 1.
Function $f : J × X → X$ of problem (1) is Lipschitz continuous with Lipschitz constant L such that $c 1 ≤ f ( u ) ≤ c 2$, with $c 1$ and $c 2$ two positive constants.
Assumption 2.
There exists an increasing function $φ ∈ C 1 − γ [ J , X ]$ and there exists $λ φ > 0$ such that, for any $t ∈ J ,$
$I 0 + α φ ( t ) ≤ λ φ φ ( t ) .$
Our main result may be presented as the following theorem.
Theorem 2 (existence).
Under the above Assumption 1, problem (1) has at least one solution $u ∈ X$ for all $λ > 0$.
Proof.
Consider the operator $P : C 1 − γ [ J , X ] → C 1 − γ [ J , X ]$ is defined by
$( P u ) ( t ) = u 0 Γ ( γ ) t γ − 1 + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s .$
Clearly, the fixed points of P are solutions to (1). The proof will be given in several steps.
Step 1: The operator P is continuous. Let $u n$ be a sequence such that $u n → u$ in $C 1 − γ [ J , X ]$. Then, for each $t ∈ J$,
$t 1 − γ ( P u n ) ( t ) − ( P u ) ( t ) ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u n ( s ) ) ∫ 0 T f ( u n ( x ) ) d x 2 − f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( u n ( x ) ) d x 2 f ( u n ( s ) ) − f ( u ( s ) ) + f ( u ( s ) ) 1 ∫ 0 T f ( u n ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( u n ( x ) ) d x 2 f ( u n ( s ) ) − f ( u ( s ) ) d s + λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( u n ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ I 1 + I 2 ,$
where
$I 1 = λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( u n ( x ) ) d x 2 f ( u n ( s ) ) − f ( u ( s ) ) d s , I 2 = λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( u n ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s .$
We estimate $I 1$ and $I 2$ terms separately. By Assumption 1, we have
$I 1 ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( u n ( x ) ) d x 2 f ( u n ( s ) ) − f ( u ( s ) ) d s ≤ λ t 1 − γ ( c 1 T ) 2 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u n ( s ) ) − f ( u ( s ) ) d s ≤ L λ t 1 − γ ( c 1 T ) 2 Γ ( α ) u n − u C 1 − γ ∫ 0 t ( t − s ) α − 1 d s ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) u n − u C 1 − γ .$
Then,
$I 1 ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) u n − u C 1 − γ ,$
$I 2 = λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( u n ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u n ( x ) ) d x 2 − ∫ 0 T f ( u ( x ) ) d x 2 ∫ 0 T f ( u n ( x ) ) d x 2 ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ t 1 − γ c 2 ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ∫ 0 T f ( u n ( x ) ) d x 2 − ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ t 1 − γ c 2 ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ∫ 0 T ( f ( u n ( x ) ) − f ( u ( x ) ) ) d x ∫ 0 T ( f ( u n ( x ) ) + f ( u ( x ) ) ) d x d s ≤ 2 λ c 2 2 T t 1 − γ ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ∫ 0 T f ( u n ( x ) ) − f ( u ( x ) ) d x d s ≤ 2 λ c 2 2 T L t 1 − γ ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ∫ 0 T u n ( x ) − u ( x ) d x d s ≤ 2 λ c 2 2 L T 2 t 1 − γ ( c 1 T ) 4 Γ ( α ) u n − u C 1 − γ ∫ 0 t ( t − s ) α − 1 d s ≤ 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 4 Γ ( α + 1 ) u n − u C 1 − γ .$
It follows that
$I 2 ≤ 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 4 Γ ( α + 1 ) u n − u C 1 − γ .$
To substitute (10) and (11) into (9), we have
$t 1 − γ ( P u n ) ( t ) − ( P u ) ( t ) ≤ I 1 + I 2 ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) + 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 4 Γ ( α + 1 ) u n − u C 1 − γ .$
Then,
$P u n − P u C 1 − γ ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) + 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 4 Γ ( α + 1 ) u n − u C 1 − γ .$
Here, independently of $λ$, the right-hand side of the above inequality converges to zero as $u n → u$. Therefore, $P u n → P u$. This proves the continuity of P.
Step 2: The operator P maps bounded sets into bounded sets in $C 1 − γ [ J , X ]$.
Indeed, it is enough to show that, for $r > 0$, there exists a positive constant l such that $u ∈ B r u ∈ C 1 − γ [ J , X ] : u ≤ r$, we have $( P u ) C 1 − γ ≤ l$. Set $M = sup B r f ( c 1 T ) 2$:
$t 1 − γ ( P u ) ( t ) ≤ u 0 Γ ( γ ) + λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ u 0 Γ ( γ ) + λ M t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 d s ≤ u 0 Γ ( γ ) + λ M T 1 − γ + α Γ ( α ) .$
Thus,
$P u C 1 − γ ≤ u 0 Γ ( γ ) + λ M T 1 − γ + α Γ ( α ) : ≤ l .$
Step 3:P maps bounded sets into equicontinuous set of $C 1 − γ [ J , X ]$.
Let $t 1 , t 2 ∈ J , t 1 < t 2$, $B r$ be a bounded set of $C 1 − γ [ J , X ]$ and $u ∈ B r$. Then,
$t 2 1 − γ ( P u ) ( t 2 ) − t 1 1 − γ ( P u ) ( t 1 ) ≤ λ Γ ( α ) t 2 1 − γ ∫ 0 t 2 ( t 2 − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s − t 1 1 − γ ∫ 0 t 1 ( t 1 − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ t 2 1 − γ Γ ( α ) ∫ t 1 t 2 ( t 2 − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s + λ Γ ( α ) ∫ 0 t 2 t 2 1 − γ ( t 2 − s ) α − 1 − t 1 1 − γ ( t 1 − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ λ c 2 t 2 1 − γ ( c 1 T ) 2 Γ ( α ) ∫ t 1 t 2 ( t 2 − s ) α − 1 d s + λ c 2 ( c 1 T ) 2 Γ ( α ) ∫ 0 t 2 t 2 1 − γ ( t 2 − s ) α − 1 − t 1 1 − γ ( t 1 − s ) α − 1 d s ≤ λ c 2 t 2 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) ( t 2 − t 1 ) 1 − α + λ c 2 ( c 1 T ) 2 Γ ( α ) ∫ 0 t 2 t 2 1 − γ ( t 2 − s ) α − 1 − t 1 1 − γ ( t 1 − s ) α − 1 d s .$
Beacause the right-hand side of the above inequality does not depend on u and tends to zero when $t 2 → t 1$, we conclude that $P ( B r ¯ )$ is relatively compact. Hence, B is compact by the Arzela–Ascoli theorem. Consequently, since P is continuous, it follows by Theorem 1 that problem (1) has a solution. The proof is completed. ☐
## 4. The Ulam–Hyers–Rassias Stability
In this section, we investigate generalized Ulam–Hyers–Rassias stability for problem (1). The stability results are based on the Banach contraction principle.
Lemma 3 (Uniqueness).
Assume that the Assumption 1 is hold. If
$L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) + 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 2 Γ ( α + 1 ) < 1 ,$
then problem (1) has a unique solution.
Proof.
Consider the operator $P : C 1 − γ [ J , X ] → C 1 − γ [ J , X ]$:
$( P u ) ( t ) = u 0 Γ ( γ ) t γ − 1 + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s .$
It is clear that the fixed points of P are solutions of problem (1).
Letting $u , v ∈ C 1 − γ [ J , X ]$ and $t ∈ J$, then we have
$t 1 − γ ( P v ) ( t ) − ( P u ) ( t ) ≤ λ t 1 − γ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( v ( s ) ) ∫ 0 T f ( v ( x ) ) d x 2 − f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) + 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 2 Γ ( α + 1 ) v − u C 1 − γ .$
Then,
$P v − P u C 1 − γ ≤ L λ T α + 1 − γ ( c 1 T ) 2 Γ ( α + 1 ) + 2 λ c 2 2 L T α + 3 − γ ( c 1 T ) 2 Γ ( α + 1 ) v − u C 1 − γ .$
Choosing $λ$ such that $0 < λ < L T α + 1 − γ ( C 1 T ) 2 Γ ( α + 1 ) + 2 C 2 2 L T α + 3 − γ ( C 1 T ) 2 Γ ( α + 1 ) − 1$, the map $P : C 1 − γ [ J , X ] → C 1 − γ [ J , X ]$ is a contraction. From (12), it follows that P has a unique fixed point, which is a solution of problem (1). ☐
Theorem 3.
In Assumption 1 and (12), problem (1) is Ulam–Hyers stable.
Proof.
Let $ϵ > 0$ and let $z ∈ C 1 − γ γ [ J , X ]$ be a function that satisfies Inequality (5) and let $u ∈ C 1 − γ γ [ J , X ]$ be the unique solution of the following Hilfer type thermistor problem
$D 0 + α , β u ( t ) = λ f ( u ( t ) ) ∫ 0 T f ( u ( x ) ) d x 2 , t ∈ J : = [ 0 , T ] , I 0 + 1 − γ u ( t ) = I 0 + 1 − γ z ( t ) = u 0 ,$
where $α ∈ ( 0 , 1 )$, $β ∈ [ 0 , 1 ]$. From Lemma 1, we have
$u ( t ) = u 0 Γ ( γ ) t γ − 1 + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s .$
By integration of (5), we obtain
$z ( t ) − u 0 Γ ( γ ) t γ − 1 − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 d s ≤ ϵ T α Γ ( α + 1 ) ,$
for all $t ∈ J$. From the above, it follows:
$z ( t ) − u ( t ) ≤ z ( t ) − u 0 Γ ( γ ) t γ − 1 − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 d s + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 − f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ ϵ T α Γ ( α + 1 ) + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( z ( x ) ) d x 2 f ( z ( s ) ) − f ( u ( s ) ) d s + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( z ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s .$
For computational convenience, we set
$K 1 = λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( z ( x ) ) d x 2 f ( z ( s ) ) − f ( u ( s ) ) d s , K 2 = λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( z ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s .$
We estimate $K 1$, $K 2$ terms separately. By Assumption 1, we have
$K 1 ≤ λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( z ( x ) ) d x 2 f ( z ( s ) ) − f ( u ( s ) ) d s ≤ λ ( c 1 T ) 2 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) − f ( u ( s ) ) d s ≤ λ L ( c 1 T ) 2 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s ,$
$K 2 ≤ λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 − ∫ 0 T f ( u ( x ) ) d x 2 ∫ 0 T f ( z ( x ) ) d x 2 ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ 2 λ c 2 2 T L ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ∫ 0 T z ( x ) − u ( x ) d x d s ≤ 2 λ c 2 2 T L ( c 1 T ) 4 Γ ( α ) z − u C 1 − γ ∫ 0 t ( t − s ) α − 1 d s ≤ 2 λ c 2 2 T L ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s .$
To substitute (16) and (17) into (15), we get
$z ( t ) − u ( t ) ≤ ϵ T α Γ ( α + 1 ) + λ L ( c 1 T ) 2 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s + 2 λ c 2 2 T 2 L ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s ≤ ϵ T α Γ ( α + 1 ) + λ L ( c 1 T ) 2 + 2 λ c 2 2 T 2 L ( c 1 T ) 4 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s ,$
and, to apply Lemma 2, we have
$z ( t ) − u ( t ) ≤ T α Γ ( α + 1 ) 1 + ν T α Γ ( α + 1 ) λ L ( c 1 T ) 2 + 2 λ c 2 2 T 2 L ( c 1 T ) 4 ϵ : = C f ϵ ,$
where $ν = ν ( α )$ is a constant, which completes the proof of the theorem. Moreover, if we set $ψ ( ϵ ) = C f ϵ$; $ψ ( 0 ) = 0$, then problem (1) is generalized Ulam–Hyers stable. ☐
Theorem 4.
In Assumptions 1, 2 and (12), problem (1) is Ulam–Hyers–Rassias stable.
Proof.
Let $z ∈ C 1 − γ γ [ J , X ]$ be solution of Inequality (6) and let $z ∈ C 1 − γ γ [ J , X ]$ be the unique solution of the following Hilfer type thermistor problem
$D 0 + α , β u ( t ) = λ f ( u ( t ) ) ∫ 0 T f ( u ( x ) ) d x 2 , t ∈ J : = [ 0 , T ] , I 0 + 1 − γ u ( t ) = I 0 + 1 − γ z ( t ) = u 0 ,$
where $α ∈ ( 0 , 1 )$, $β ∈ [ 0 , 1 ]$. From Lemma 1, we have
$u ( t ) = u 0 Γ ( γ ) t γ − 1 + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s .$
By integration of (6) and Assumption 2, we obtain
$z ( t ) − u 0 Γ ( γ ) t γ − 1 − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 d s ≤ ϵ λ φ φ ( t ) ,$
for all $t ∈ J$. From the above, it follows:
$z ( t ) − u ( t ) ≤ z ( t ) − u 0 Γ ( γ ) t γ − 1 − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 d s + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( z ( s ) ) ∫ 0 T f ( z ( x ) ) d x 2 − f ( u ( s ) ) ∫ 0 T f ( u ( x ) ) d x 2 d s ≤ ϵ λ φ φ ( t ) + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 ∫ 0 T f ( z ( x ) ) d x 2 f ( z ( s ) ) − f ( u ( s ) ) d s + λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( u ( s ) ) 1 ∫ 0 T f ( z ( x ) ) d x 2 − 1 ∫ 0 T f ( u ( x ) ) d x 2 d s .$
To substitute (16) and (17) into (19), we get
$z ( t ) − u ( t ) ≤ ϵ λ φ φ ( t ) + λ L ( c 1 T ) 2 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s + 2 λ c 2 2 T 2 L ( c 1 T ) 4 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s ≤ ϵ λ φ φ ( t ) + λ L ( c 1 T ) 2 + 2 λ c 2 2 T 2 L ( c 1 T ) 4 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 z ( s ) − u ( s ) d s ,$
and, to apply Lemma 2, we have
$z ( t ) − u ( t ) ≤ 1 + ν 1 λ φ λ L ( c 1 T ) 2 + 2 λ c 2 2 T 2 L ( c 1 T ) 4 λ φ ϵ φ ( t ) = C f ϵ φ ( t ) ,$
where $ν 1 = ν 1 ( α )$ is a constant. It completes the proof of Theorem 4. ☐
## Acknowledgments
The authors are grateful to anonymous referees for several comments and suggestions.
## Author Contributions
All of the authors contributed to the conception and development of this manuscript.
## Conflicts of Interest
The authors declare no conflict of interest.
## References
1. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of Fractional Differential Equations; Elsevier: Amsterdam, The Netherlands, 2006; Volume 204. [Google Scholar]
2. Podlubny, I. Fractional Differential Equations; Academy Press: Cambridge, MA, USA, 1999; Volume 198. [Google Scholar]
3. Benchohra, M.; Henderson, J.; Ntouyas, S.K.; Ouahab, A. Existence results for fractional order functional differential equations with infinite delay. J. Math. Anal. Appl. 2008, 338, 1340–1350. [Google Scholar] [CrossRef]
4. Benchohra, M.; Bouriah, S. Existence and stability results for nonlinear boundary value problem for implicit differential equations of fractional order. Moroc. J. Pure Appl. Anal. 2005, 1, 22–37. [Google Scholar] [CrossRef]
5. Ibrahim, R.W. Generalized Ulam–Hyers stability for fractional differential equations. Int. J. Math. 2012, 23, 1250056. [Google Scholar] [CrossRef]
6. Hilfer, R.; Luchko, Y.; Tomovski, Z. Operational method for the solution of fractional differential equations with generalized Riemann-Lioville fractional derivative. Fract. Calc. Appl. Anal. 2009, 12, 289–318. [Google Scholar]
7. Vivek, D.; Kanagarajan, K.; Sivasundaram, S. Dynamics and stability of pantograph equations via Hilfer fractional derivative. Nonlinear Stud. 2016, 23, 685–698. [Google Scholar]
8. Wang, J.; Lv, L.; Zhou, Y. Ulam stability and data dependence for fractional differential equations with Caputo derivative. Electron. J. Qual. Theory Differ. Equ. 2011, 63, 1–10. [Google Scholar] [CrossRef]
9. Wang, J.; Zhou, Y. New concepts and results in stability of fractional differential equations. Commun. Nonlinear Sci. Numer. Simul. 2012, 17, 2530–2538. [Google Scholar] [CrossRef]
10. Hilfer, R. (Ed.) Fractional time evolution. In Application of Fractional Calculus in Physics; World Scientific: Singapore, 1999; pp. 87–130. [Google Scholar]
11. Abbas, S.; Benchohra, M.; Lagreg, J.E.; Alsaedi, A.; Zhou, Y. Existence and Ulam stability for fractional differential equations of Hilfer-Hadamard type. Adv. Differ. Equ. 2017. [Google Scholar] [CrossRef]
12. Furati, K.M.; Kassim, M.D.; Tatar, N.E. Existence and uniqueness for a problem involving Hilfer fractional derivative. Comput. Math. Appl. 2012, 64, 1616–1626. [Google Scholar] [CrossRef]
13. Furati, K.M.; Kassim, M.D.; Tatar, N.E. Non-existence of global solutions for a differential equation involving Hilfer fractional derivative. Electron. J. Differ. Equ. 2013, 2013, 235. [Google Scholar]
14. Gu, H.; Trujillo, J.J. Existence of mild solution for evolution equation with Hilfer fractional derivative. Appl. Math. Comput. 2014, 257, 344–354. [Google Scholar] [CrossRef]
15. Wang, J.R.; Yuruo Zhang, Y. Nonlocal initial value problems for differential equations with Hilfer fractional derivative. Appl. Math. Comput. 2015, 266, 850–859. [Google Scholar] [CrossRef]
16. Samko, S.G.; Kilbas, A.A.; Marichev, O.I. Fractional Integrals and Derivatives, Theory and Applications; Gordon and Breach: Amsterdam, The Netherlands, 1987. [Google Scholar]
17. Kwok, K. Complete Guide to Semiconductor Devices; Mc Graw-Hill: New York, NY, USA, 1995. [Google Scholar]
18. Maclen, E.D. Thermistors; Electrochemical Publication: Glasgow, UK, 1979. [Google Scholar]
19. Khan, A.; Li, Y.; Shah, K.; Khan, T.S. On Coupled p-Laplacian Fractional Differential Equations with Nonlinear Boundary Conditions. Complexity 2017. [Google Scholar] [CrossRef]
20. Song, W.; Gao, W. Existence of solutions for nonlocal p-Laplacian thermistor problems on time scales. Bound. Value Probl. 2013. [Google Scholar] [CrossRef]
21. Sidi Ammi, M.R.; Torres, D.F.M. Galerkin spectral method for the fractional nonlocal thermistor problem. Comput. Math. Appl. 2017, 73, 1077–1086. [Google Scholar] [CrossRef]
22. Sidi Ammi, M.R.; Torres, D.F.M. Existence and uniqueness of a positive solution to generalized nonlocal thermistor problems with fractional-order derivatives. Differ. Equ. Appl. 2012, 4, 267–276. [Google Scholar]
23. Ammi, M.R.S.; Torres, D.F.M. Existence and uniqueness results for a fractional Riemann-liouville nonlocal thermistor problem on arbitrary time scales. J. King Saud Univ. Sci. 2017, in press. [Google Scholar]
24. Sidi Ammi, M.R.; Torres, D.F.M. Numerical analysis of a nonlocal parabolic problem resulting from thermistor problem. Math. Comput. Simul. 2008, 77, 291–300. [Google Scholar] [CrossRef]
25. Sidi Ammi, M.R.; Torres, D.F.M. Optimal control of nonlocal thermistor equations. Int. J. Control 2012, 85, 1789–1801. [Google Scholar] [CrossRef]
26. Liang, Y.; Chen, W.; Akpa, B.S.; Neuberger, T.; Webb, A.G.; Magin, R.L. Using spectral and cumulative spectral entropy to classify anomalous diffusion in SephadexTM gels. Comput. Math. Appl. 2017, 73, 765–774. [Google Scholar] [CrossRef]
27. Andras, S.; Kolumban, J.J. On the Ulam–Hyers stability of first order differential systems with nonlocal initial conditions. Nonlinear Anal. Theory Methods Appl. 2013, 82, 1–11. [Google Scholar] [CrossRef]
28. Jung, S.M. Hyers-Ulam stability of linear differential equations of first order. Appl. Math. Lett. 2004, 17, 1135–1140. [Google Scholar] [CrossRef]
29. Muniyappan, P.; Rajan, S. Hyers-Ulam-Rassias stability of fractional differential equation. Int. J. Pure Appl. Math. 2015, 102, 631–642. [Google Scholar] [CrossRef]
30. Rus, I.A. Ulam stabilities of ordinary differential equations in a Banach space. Carpathian J. Math. 2010, 26, 103–107. [Google Scholar]
31. Abbas, S.; Benchohra, M.; Sivasundaram, S. Dynamics and Ulam stability for Hilfer type fractional differential equations. Nonlinear Stud. 2016, 4, 627–637. [Google Scholar]
32. Kamocki, R.; Obcznnski, C. On fractional Cauchy-type problems containing Hilfer derivative. Electron. J. Qual. Theory Differ. Equ. 2016, 50, 1–12. [Google Scholar] [CrossRef]
33. Granas, A.; Dugundji, J. Fixed Point Theory; Springer: New York, NY, USA, 2003. [Google Scholar]
34. Vivek, D.; Kanagarajan, K.; Sivasundaram, S. Theory and analysis of nonlinear neutral pantograph equations via Hilfer fractional derivative. Nonlinear Stud. 2017, 24, 699–712. [Google Scholar]
## Share and Cite
MDPI and ACS Style
Vivek, D.; Kanagarajan, K.; Sivasundaram, S. Dynamics and Stability Results for Hilfer Fractional Type Thermistor Problem. Fractal Fract. 2017, 1, 5. https://doi.org/10.3390/fractalfract1010005
AMA Style
Vivek D, Kanagarajan K, Sivasundaram S. Dynamics and Stability Results for Hilfer Fractional Type Thermistor Problem. Fractal and Fractional. 2017; 1(1):5. https://doi.org/10.3390/fractalfract1010005
Chicago/Turabian Style
Vivek, D., K. Kanagarajan, and Seenith Sivasundaram. 2017. "Dynamics and Stability Results for Hilfer Fractional Type Thermistor Problem" Fractal and Fractional 1, no. 1: 5. https://doi.org/10.3390/fractalfract1010005 | 12,681 | 33,773 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 193, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.876789 |
https://www.lotterypost.com/thread/154143/2 | 1,481,420,330,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543614.1/warc/CC-MAIN-20161202170903-00366-ip-10-31-129-80.ec2.internal.warc.gz | 958,633,050 | 18,911 | Welcome Guest
You last visited December 10, 2016, 7:11 pm
All times shown are
Eastern Time (GMT-5:00)
# Straight Into The Heart Of "Absolute Versatility": Part II
Topic closed. 48 replies. Last post 10 years ago by JAP69.
Page 2 of 4
missouri
United States
Member #34763
March 7, 2006
756 Posts
Offline
Posted: April 10, 2007, 11:04 pm - IP Logged
Could I ask what numbers each letter represents?
that's what I was asking too! I'm really trying to get this but with the letters instead of numbers I'm just not getting it.
missouri
United States
Member #34763
March 7, 2006
756 Posts
Offline
Posted: April 10, 2007, 11:11 pm - IP Logged
I grouped them from the very beginning in this way:
A (012)
B (3456)
C (789)
There are now only three numbers to worry anything about.
Thank you!
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 10, 2007, 11:15 pm - IP Logged
that's what I was asking too! I'm really trying to get this but with the letters instead of numbers I'm just not getting it.
Now, anytime that you see the digit "3" in the Pick 3 or 4 it is now every bit as much a 4, 5 or a 6; a 4 is every bit as much as a 3, 5,or a 6: a 5 is every bit as much as a 3, 4 or a 6; a 6 is every bit as much as a 3, 4 or a 5.
Among those 4 digits (3, 4, 5 and 6) there is only one to worry about now.
The single letter A represents either a 0, 1 or a 2 occurring.
The single letter B represents either a 3, 4, 5 or 6 occurring.
The single letter C represents either a 7, 8, or 9 occurring.
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 10, 2007, 11:16 pm - IP Logged
Thank you!
You are very welcome.
South Carolina
United States
Member #6
November 4, 2001
8790 Posts
Online
Posted: April 10, 2007, 11:25 pm - IP Logged
I grouped them from the very beginning in this way:
A (012)
B (3456)
C (789)
There are now only three numbers to worry anything about.
Thanks for posting your letter to number conversion.
I do work with numbers in groups. But I grouped mine differently for my own reasons.
For a three number group I use this. I use a number to represent a group.
1____1-3-6-8
2____024
5____579
I also do a four number group keeping the even odd and high low numbers in their own group.
MAGA
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 10, 2007, 11:41 pm - IP Logged
Thanks for posting your letter to number conversion.
I do work with numbers in groups. But I grouped mine differently for my own reasons.
For a three number group I use this. I use a number to represent a group.
1____1-3-6-8
2____024
5____579
I also do a four number group keeping the even odd and high low numbers in their own group.
Jap69, that is impressive just to look at.
I don't get the 1 2 and 5 references, but to the right are Pick 3 and 4 type numbers I'm assuming.
I don't mean to be nosey though.
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 10, 2007, 11:45 pm - IP Logged
If I can, tomorrow I'll try to redo that fiasco with the formatting problems for the sake of clarifying what I had intended.
See ya everyone.
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 11, 2007, 12:03 pm - IP Logged
A B A
C B B
A B A
A A C
A C B
B C B
B B C
C B B
********
A B A
C B B
A B A
A A C
A C B
B C B
B B C
C B B
Two examples of different progressions occurring within the same span of numbers:
In the first example (at the very top) you will notice a progression occurring (illustrated in red in the extreme left column) and another progression occurring (illustrated in blue in the extreme right column). Both of those are to be perceived from the bottom going in an upward direction. Please notice that in the (red) progression the C is the "first occurrence" of a C and in the (blue) progression the A is the first occurrence of an A. Can you see that both progressions are "balanced"?
(red: remember that it is meant to be perceived from the bottom going up)
ABA
CBB
BBC
(blue: remember to perceive it from the bottom going up and from the extreme right column)
CBB
ABA
BBC
(Because that one depends on being percieved from the extreme right column, technically you can invert it like this so the human eye can perceive it in the more natural direction of left to right.)
Like this:
BBC
ABA
CBB
See how both the red and blue progressions are completed and balanced?
***
In the second example (beneath the top example) please note the occurrence of two progressions illustrated in green. (They are to be perceived moving in a downward direction):
ABA
AAC
ACB
***
ABA
BBC
CBB
Because both the middle and extreme right columns can be said to "dominate" for those two progressions, please view the progressions from either of those two columns.
Please note that both are "completed" and "balanced".
I want you to notice the presence of versatility within both examples.
Can you see how a number can be two or more things at the same moment because it can be involved in two or more progressions at the same moment?
If you have any difficulties understanding these examples please tell me and I will try to help you with it (in as much as I am able to).
Still problems with formatting correctly: all B's in the middle column of top example are supposed to be "black" and not red.
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 11, 2007, 12:12 pm - IP Logged
Just disregard the last post.
It's not only the B's with the incorrect colors -- there are some other incorrect colors as well throughout.
You won't understand the idea(s) correctly that I am trying to show you with all those wrong colors.
If anybody has any ideas about how to correct for the incorrect colors let me know.
Florida
United States
Member #46570
September 14, 2006
558 Posts
Offline
Posted: April 11, 2007, 12:23 pm - IP Logged
A B A
C B B
A B A
A A C
A C B
B C B
B B C
C B B
********
A B A
C B B
A B A
A A C
A C B
B C B
B B C
C B B
Two examples of different progressions occurring within the same span of numbers:
In the first example (at the very top) you will notice a progression occurring (illustrated in red in the extreme left column) and another progression occurring (illustrated in blue in the extreme right column). Both of those are to be perceived from the bottom going in an upward direction. Please notice that in the (red) progression the C is the "first occurrence" of a C and in the (blue) progression the A is the first occurrence of an A. Can you see that both progressions are "balanced"?
(red: remember that it is meant to be perceived from the bottom going up)
ABA
CBB
BBC
(blue: remember to perceive it from the bottom going up and from the extreme right column)
CBB
ABA
BBC
(Because that one depends on being percieved from the extreme right column, technically you can invert it like this so the human eye can perceive it in the more natural direction of left to right.)
Like this:
BBC
ABA
CBB
See how both the red and blue progressions are completed and balanced?
***
In the second example (beneath the top example) please note the occurrence of two progressions illustrated in green. (They are to be perceived moving in a downward direction):
ABA
AAC
ACB
***
ABA
BBC
CBB
Because both the middle and extreme right columns can be said to "dominate" for those two progressions, please view the progressions from either of those two columns.
Please note that both are "completed" and "balanced".
I want you to notice the presence of versatility within both examples.
Can you see how a number can be two or more things at the same moment because it can be involved in two or more progressions at the same moment?
If you have any difficulties understanding these examples please tell me and I will try to help you with it (in as much as I am able to).
Still problems with formatting correctly: all B's in the middle column of top example are supposed to be "black" and not red.
Hi all...
Maybe you can clarify this for me ...
A=0,1,2
B=3,4,5,6
C=7,8,9
Last nights Pick 3 # in FL was 658
6 = B
5 = B
8 = C
Does this mean that ABA or CBB number combos may come out ?
For some reason, I see 892 and/or 594 coming out for tonight..oh well.
The 9 is a predominant # for tonights FL pick3... just feels right. Perhaps I should wheel 89254 and use those 10 combos...
Getting off-topic ,sorry ... Will review this more and see if I can put this puzzle together if possible.
Omniscient
United States
Member #48618
January 3, 2007
87 Posts
Offline
Posted: April 11, 2007, 3:49 pm - IP Logged
Hi all...
Maybe you can clarify this for me ...
A=0,1,2
B=3,4,5,6
C=7,8,9
Last nights Pick 3 # in FL was 658
6 = B
5 = B
8 = C
Does this mean that ABA or CBB number combos may come out ?
For some reason, I see 892 and/or 594 coming out for tonight..oh well.
The 9 is a predominant # for tonights FL pick3... just feels right. Perhaps I should wheel 89254 and use those 10 combos...
Getting off-topic ,sorry ... Will review this more and see if I can put this puzzle together if possible.
Omniscient
Hello Omniscient.
I wish it was as simple as seeing this occur in any last two drawings:
BBC
ABA
and then expecting that the CBB will appear the very next drawing to complete and balance it.
It doesn't work that way. AV is following it's own particular logic. All the progressions are very specific.
It is only occassionally that you will find the three that are needed to balance and complete the progression occurring in three consecutive drawings.
If I could just get those screwy colors to come up correctly!
I was hoping to eventually reach a point in which anyone who wanted to help me could work with me on some actual Pick 3 type lottery results. It's possible that any number of people could offer suggestions on the fly that just might prove to be solutions to some of the problems I'm having with the phenomenon. I mean, it might just be a "fresh pair of eyes" that will see some of these solutions.
Regards.
Michigan
United States
Member #34209
March 1, 2006
265 Posts
Offline
Posted: April 11, 2007, 5:48 pm - IP Logged
I sent you a PM Grumple.
missouri
United States
Member #34763
March 7, 2006
756 Posts
Offline
Posted: April 11, 2007, 5:54 pm - IP Logged
Can you give us an example using numbers. I don't know about anyone else but it would sure help me. I get what you're saying about A=012 ect. but that doesn't make A only one number to worry about cause there are still three numbers that equal A. And how do you know when A is the first appearance cause 01and 2 come out all the time so how far do you go back to find A?
South Carolina
United States
Member #6
November 4, 2001
8790 Posts
Online
Posted: April 11, 2007, 6:26 pm - IP Logged
Jap69, that is impressive just to look at.
I don't get the 1 2 and 5 references, but to the right are Pick 3 and 4 type numbers I'm assuming.
I don't mean to be nosey though.
"I don't get the 1 2 and 5 references, but to the right are Pick 3 and 4 type numbers I'm assuming."
the 1 2 and 5 are the numbers I use to identify each group.
I track my draw results in software and I need to use numbers for the groups.
All numbers to the right are the lottery numbers.
______________________________________________
I do not know what you are doing but this is the way I do it.
If you are coloring the A'S. Hi-lite the first A go select your color from the chart. Color your first A , When you select a color that color will remain present on the color selection in front of you.
Then proceed to the next A which you want to be the same color, Hi lite the A and go to color selection and just click on The color in front of you.
There is no need to repull the color selections when doing the same color.
Then when you want to do another color just pull the color chart and select your color. That color will remain in front of you until you want another color.
MAGA
South Carolina
United States
Member #6
November 4, 2001
8790 Posts
Online
Posted: April 11, 2007, 6:40 pm - IP Logged
Hello Omniscient.
I wish it was as simple as seeing this occur in any last two drawings:
BBC
ABA
and then expecting that the CBB will appear the very next drawing to complete and balance it.
It doesn't work that way. AV is following it's own particular logic. All the progressions are very specific.
It is only occassionally that you will find the three that are needed to balance and complete the progression occurring in three consecutive drawings.
If I could just get those screwy colors to come up correctly!
I was hoping to eventually reach a point in which anyone who wanted to help me could work with me on some actual Pick 3 type lottery results. It's possible that any number of people could offer suggestions on the fly that just might prove to be solutions to some of the problems I'm having with the phenomenon. I mean, it might just be a "fresh pair of eyes" that will see some of these solutions.
Regards.
"I was hoping to eventually reach a point in which anyone who wanted to help me could work with me on some actual Pick 3 type lottery results. It's possible that any number of people could offer suggestions on the fly that just might prove to be solutions to some of the problems I'm having with the phenomenon. I mean, it might just be a "fresh pair of eyes" that will see some of these solutions."
I do not know how much you have ciphered out with your number groups. I could post some things I have figured out.
MAGA
Page 2 of 4 | 3,617 | 13,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-50 | longest | en | 0.933859 |
https://brilliant.org/discussions/thread/infinite-prime-theorem-euclids-theorem/ | 1,618,812,229,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00138.warc.gz | 253,050,500 | 13,514 | # 'Infinite Prime Theorem' – Euclid's Theorem
## Problem:
Prove that there is an infinite number of primes.
## Solution:
The problem was originally called Euclid’s Theorem, named after Euclid, who proved that there were an infinite amount of primes in his book Elements (Book: IX, Proposition: 20) using contradiction. The following is an adaptation of his proof:
Assuming that there is a finite list of primes, let $m$ denote the product of all such primes and be one more than it.
$m=2\times 3\times 5\times 7\times...\times p+1$
It is then shown that $m$ must either be a prime or have prime factors larger than $p$.
Case $m$ is prime:
If $m$ is prime, and is the product of all primes in a finite list plus 1, then it must be larger than the largest prime $p$, meaning that there is a larger prime bigger than the one defined as the largest, and therefore, there cannot be a finite number of prime numbers.
Case $m$ is composite:
If $m$ is composite, it follows that it cannot have a prime factor smaller than $p$:
$m$ cannot be divided by 2 as it is $2(3\times 5\times 7\times...\times p)+1$, or one more than a multiple of 2.
$m$ cannot be divided by 3 as it is $3(2\times 5\times 7\times...\times p)+1$, or one more than a multiple of 3.
$m$ cannot be divided by 5 as it is $5(2\times 3\times 7\times...\times p)+1$, or one more than a multiple of 5.
It follows that $m$ cannot be divided by the assumed largest prime $p$ (in a finite list to comply with contradiction assumption that there is a finite number of primes) as it is $p(2\times 3\times 7\times...)+1$, or one more than a multiple of $p$.
Note by Just C
1 week, 4 days ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
## Comments
Sort by:
Top Newest
Nice! I think you meant to say "let $m$ denote the product of all such primes". :)
- 1 week, 4 days ago
Log in to reply
Yeah, that was a mistake. Thanks for pointing it out! :)
- 1 week, 4 days ago
Log in to reply
No problem!
- 1 week, 4 days ago
Log in to reply
×
Problem Loading...
Note Loading...
Set Loading... | 1,058 | 3,817 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-17 | longest | en | 0.936682 |
https://mail.haskell.org/pipermail/haskell-cafe/2008-July/045409.html | 1,656,619,864,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103877410.46/warc/CC-MAIN-20220630183616-20220630213616-00164.warc.gz | 427,818,501 | 2,764 | # [Haskell-cafe] Trouble with non-exhaustive patterns
C.M.Brown cmb21 at kent.ac.uk
Mon Jul 21 07:46:16 EDT 2008
```Hi Fernando,
I hope you don't mind, but I've moved this over to the Haskell-beginners
mailing list, where I think this kind of question will be more
appropriate.
In Haskell, it helps to think of functions in terms of an input and an
output, that is, what is the thing that is going into the function; and
what is the thing that is coming out of the function?
In your function, final, the input is clearly a list of something (we can
denote this by the Haskell type [a] which means a list of some type,
a). Its return type is clearly an element of the list, so that must be the
something type (the 'a' from the [a]). This gives the Haskell
type:
final :: [a] -> a
(final takes a list of 'a' and gives back a single 'a'. The 'a' is a
type variable, and it is used to denote that anything can be put in its
place, so we can give final a list of integers, characters, whatever).
Now, let's take a look at your definition of final. If we take a closer
look, in fact only two equations satisfy this type:
final [a] = a
final (_:t) = final t
The other takes a list and returns a list. The equation,
final [] = []
takes an empty list and returns an empty list (its type is therefore
[a] -> [a]).
This is why you got an error, as Haskell doesn't know how what
to do with the conflicting equation. What we need is the final element of
the list. How do we do that?
Let's think of the simple cases first. The final element of a list
containing a single element is just that element,
final [a] = a
But what about if the list contains more elements? Or is an empty list?
The empty list may be confusing, as an empty list contains no elements, so
in effect, we can't return anything. We can, however, return an error
message.
fun [] = error "empty List"
And the final element of any list, must be the final element of its tail:
final (_:t) = final t
this gives us:
final :: [a] -> a
final [] = error "Empty List"
final [a] = a
final (_:t) = final t
I hope that gives some insight.
Kind regards,
Chris.
On Mon, 21 Jul 2008, Fernando Rodriguez wrote:
>
> Hi,
>
> I defiend the following function to get the last element of a list:
>
> final [a] = a
> final (_:t) = final t
>
> and it works as expected. Since I didn't want to have a non exhaustive pattern,
> I added the following case:
>
> final [] = [] - I consider that the end of an empty list is the empty list
> final [a] = a
> final (_:t) = final t
>
> Suddenly, the function stoped working with a rather cryptic (for a newbie
> at least) error message:
>
> *Temp> final [4,5]
>
> <interactive>:1:9:
> No instance for (Num [a])
> arising from the literal `5' at <interactive>:1:9
> Possible fix: add an instance declaration for (Num [a])
> In the expr*Temp> ession: 5
> In the first argument of `final', namely `[4, 5]'
> In the expression: final [4, 5]
>
> What have I done so wrong?
> | 826 | 2,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.92639 |
https://justaaa.com/statistics-and-probability/88307-1-use-the-given-values-of-n-and-p-to-find-the | 1,721,403,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00082.warc.gz | 284,923,320 | 10,595 | Question
# 1. Use the given values of n and p to find the minimum usual value muμminus−2sigmaσ...
1. Use the given values of n and p to find the minimum usual value
muμminus−2sigmaσ
and the maximum usual value
muμplus+2sigmaσ.
Round to the nearest hundredth unless otherwise noted.
nequals=10141014;
pequals=0.860.86
x P(x) Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. 0 0.0350.035 1 0.1510.151 2 0.3140.314 3 0.3140.314 4 0.1510.151 5 0.0350.035
Does the table show a probability distribution? Select all that apply.
A.
Yes, the table shows a probability distribution.
B.
No, the random variable x is categorical instead of numerical.
C.
No, the random variable x's number values are not associated with probabilities.
D.
No, the sum of all the probabilities is not equal to 1.
E.
No, not every probability is between 0 and 1 inclusive.
Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
muμequals=nothing
child(ren) (Round to one decimal place as needed.)
B.
The table does not show a probability distribution.
Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
sigmaσequals=nothing
child(ren) (Round to one decimal place as needed.)
B.
The table does not show a probability distribution.
A. Yes, the table shows a probability distribution.
x P(X=x) xP(x) x2P(x) 0 0.035 0.00000 0.00000 1 0.151 0.15100 0.15100 2 0.314 0.62800 1.25600 3 0.314 0.94200 2.82600 4 0.151 0.60400 2.41600 5 0.035 0.17500 0.87500 total 2.5000 7.5240 E(x) =μ= ΣxP(x) = 2.5000 E(x2) = Σx2P(x) = 7.5240 Var(x)=σ2 = E(x2)-(E(x))2= 1.274000 std deviation= σ= √σ2 = 1.12872
from above:
mean μ=2.5
σ= 1.11
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 691 | 2,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.703902 |
https://www.aqua-calc.com/one-to-all/surface-density/preset/microgram-per-square-millimeter/1 | 1,695,828,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00679.warc.gz | 713,138,622 | 9,948 | # Convert [µg/mm²] to other units of surface density
## micrograms/millimeter² [µg/mm²] surface density conversions
1 µg/mm² = 1 × 10-18 microgram per square picometer µg/mm² to µg/pm² 1 µg/mm² = 1 × 10-14 microgram per square angstrom µg/mm² to µg/Ų 1 µg/mm² = 1 × 10-12 microgram per square nanometer µg/mm² to µg/nm² 1 µg/mm² = 1 × 10-6 microgram per square micron µg/mm² to µg/µ² 1 µg/mm² = 1 × 10-6 microgram per square micrometer µg/mm² to µg/µm² 1 µg/mm² = 100 micrograms per square centimeter µg/mm² to µg/cm² 1 µg/mm² = 1 000 000 micrograms per square meter µg/mm² to µg/m² 1 µg/mm² = 1 000 micrograms per hectare µg/mm² to µg/ha 1 µg/mm² = 645.16 micrograms per square inch µg/mm² to µg/inch² 1 µg/mm² = 92 903.04 micrograms per square foot µg/mm² to µg/ft² 1 µg/mm² = 836 127.36 micrograms per square yard µg/mm² to µg/yd² 1 µg/mm² = 1 × 10-21 milligram per square picometer µg/mm² to mg/pm² 1 µg/mm² = 1 × 10-17 milligram per square angstrom µg/mm² to mg/Ų 1 µg/mm² = 1 × 10-15 milligram per square nanometer µg/mm² to mg/nm² 1 µg/mm² = 1 × 10-9 milligram per square micron µg/mm² to mg/µ² 1 µg/mm² = 1 × 10-9 milligram per square micrometer µg/mm² to mg/µm² 1 µg/mm² = 0.001 milligram per square millimeter µg/mm² to mg/mm² 1 µg/mm² = 0.1 milligram per square centimeter µg/mm² to mg/cm² 1 µg/mm² = 1 000 milligrams per square meter µg/mm² to mg/m² 1 µg/mm² = 0.65 milligram per square inch µg/mm² to mg/in² 1 µg/mm² = 92.9 milligrams per square foot µg/mm² to mg/ft² 1 µg/mm² = 836.13 milligrams per square yard µg/mm² to mg/yd² 1 µg/mm² = 1 × 10-24 gram per square picometer µg/mm² to g/pm² 1 µg/mm² = 1 × 10-20 gram per square angstrom µg/mm² to g/Ų 1 µg/mm² = 1 × 10-18 gram per square nanometer µg/mm² to g/nm² 1 µg/mm² = 1 × 10-12 gram per square micron µg/mm² to g/µ² 1 µg/mm² = 1 × 10-12 gram per square micrometer µg/mm² to g/µm² 1 µg/mm² = 1 × 10-6 gram per square millimeter µg/mm² to g/mm² 1 µg/mm² = 0.0001 gram per square centimeter µg/mm² to g/cm² 1 µg/mm² = 1 gram per square meter µg/mm² to g/m² 1 µg/mm² = 0.001 gram per square inch µg/mm² to g/in² 1 µg/mm² = 0.09 gram per square foot µg/mm² to g/ft² 1 µg/mm² = 0.84 gram per square yard µg/mm² to g/yd² 1 µg/mm² = 1 × 10-27 kilogram per square picometer µg/mm² to kg/pm² 1 µg/mm² = 1 × 10-23 kilogram per square angstrom µg/mm² to kg/Ų 1 µg/mm² = 1 × 10-21 kilogram per square nanometer µg/mm² to kg/nm² 1 µg/mm² = 1 × 10-15 kilogram per square micron µg/mm² to kg/µ² 1 µg/mm² = 1 × 10-15 kilogram per square micrometer µg/mm² to kg/µm² 1 µg/mm² = 1 × 10-9 kilogram per square millimeter µg/mm² to kg/mm² 1 µg/mm² = 1 × 10-7 kilogram per square centimeter µg/mm² to kg/cm² 1 µg/mm² = 0.001 kilogram per square meter µg/mm² to kg/m² 1 µg/mm² = 10 kilograms per hectare µg/mm² to kg/ha 1 µg/mm² = 6.45 × 10-7 kilogram per square inch µg/mm² to kg/in² 1 µg/mm² = 9.29 × 10-5 kilogram per square foot µg/mm² to kg/ft² 1 µg/mm² = 0.001 kilogram per square yard µg/mm² to kg/yd² 1 µg/mm² = 0.1 centner per hectare µg/mm² to centner/ha 1 µg/mm² = 1 × 10-30 tonne per square picometer µg/mm² to t/pm² 1 µg/mm² = 1 × 10-26 tonne per square angstrom µg/mm² to t/Ų 1 µg/mm² = 1 × 10-24 tonne per square nanometer µg/mm² to t/nm² 1 µg/mm² = 1 × 10-18 tonne per square micron µg/mm² to t/µ² 1 µg/mm² = 1 × 10-18 tonne per square micrometer µg/mm² to t/µm² 1 µg/mm² = 1 × 10-12 tonne per square millimeter µg/mm² to t/mm² 1 µg/mm² = 1 × 10-10 tonne per square centimeter µg/mm² to t/cm² 1 µg/mm² = 1 × 10-6 tonne per square meter µg/mm² to t/m² 1 µg/mm² = 6.45 × 10-10 tonne per square inch µg/mm² to t/in² 1 µg/mm² = 9.29 × 10-8 tonne per square foot µg/mm² to t/ft² 1 µg/mm² = 8.36 × 10-7 tonne per square yard µg/mm² to t/yd² 1 µg/mm² = 3.53 × 10-26 ounce per square picometer µg/mm² to oz/pm² 1 µg/mm² = 3.53 × 10-22 ounce per square angstrom µg/mm² to oz/Ų 1 µg/mm² = 3.53 × 10-20 ounce per square nanometer µg/mm² to oz/nm² 1 µg/mm² = 3.53 × 10-14 ounce per square micron µg/mm² to oz/µ² 1 µg/mm² = 3.53 × 10-14 ounce per square micrometer µg/mm² to oz/µm² 1 µg/mm² = 3.53 × 10-8 ounce per square millimeter µg/mm² to oz/mm² 1 µg/mm² = 3.53 × 10-6 ounce per square centimeter µg/mm² to oz/cm² 1 µg/mm² = 0.04 ounce per square meter µg/mm² to oz/m² 1 µg/mm² = 2.28 × 10-5 ounce per square inch µg/mm² to oz/in² 1 µg/mm² = 0.003 ounce per square foot µg/mm² to oz/ft² 1 µg/mm² = 0.03 ounce per square yard µg/mm² to oz/yd² 1 µg/mm² = 2.2 × 10-27 pound per square picometer µg/mm² to lb/pm² 1 µg/mm² = 2.2 × 10-23 pound per square angstrom µg/mm² to lb/Ų 1 µg/mm² = 2.2 × 10-21 pound per square nanometer µg/mm² to lb/nm² 1 µg/mm² = 2.2 × 10-15 pound per square micron µg/mm² to lb/µ² 1 µg/mm² = 2.2 × 10-15 pound per square micrometer µg/mm² to lb/µm² 1 µg/mm² = 2.2 × 10-9 pound per square millimeter µg/mm² to lb/mm² 1 µg/mm² = 2.2 × 10-7 pound per square centimeter µg/mm² to lb/cm² 1 µg/mm² = 0.002 pound per square meter µg/mm² to lb/m² 1 µg/mm² = 1.42 × 10-6 pound per square inch µg/mm² to lb/in² 1 µg/mm² = 0.0002 pound per square foot µg/mm² to lb/ft² 1 µg/mm² = 0.002 pound per square yard µg/mm² to lb/yd² 1 µg/mm² = 1.54 × 10-23 grain per square picometer µg/mm² to gr/pm² 1 µg/mm² = 1.54 × 10-19 grain per square angstrom µg/mm² to gr/Ų 1 µg/mm² = 1.54 × 10-17 grain per square nanometer µg/mm² to gr/nm² 1 µg/mm² = 1.54 × 10-11 grain per square micron µg/mm² to gr/µ² 1 µg/mm² = 1.54 × 10-11 grain per square micrometer µg/mm² to gr/µm² 1 µg/mm² = 1.54 × 10-5 grain per square millimeter µg/mm² to gr/mm² 1 µg/mm² = 0.002 grain per square centimeter µg/mm² to gr/cm² 1 µg/mm² = 15.43 grains per square meter µg/mm² to gr/m² 1 µg/mm² = 15.43 grains per hectare µg/mm² to gr/ha 1 µg/mm² = 0.01 grain per square inch µg/mm² to gr/inch² 1 µg/mm² = 1.43 grains per square foot µg/mm² to gr/ft² 1 µg/mm² = 12.9 grains per square yard µg/mm² to gr/yd² 1 µg/mm² = 1.1 × 10-30 short ton per square picometer µg/mm² to short tn/pm² 1 µg/mm² = 1.1 × 10-26 short ton per square angstrom µg/mm² to short tn/Ų 1 µg/mm² = 1.1 × 10-24 short ton per square nanometer µg/mm² to short tn/nm² 1 µg/mm² = 1.1 × 10-18 short ton per square micron µg/mm² to short tn/µ² 1 µg/mm² = 1.1 × 10-18 short ton per square micrometer µg/mm² to short tn/µm² 1 µg/mm² = 1.1 × 10-12 short ton per square millimeter µg/mm² to short tn/mm² 1 µg/mm² = 1.1 × 10-10 short ton per square centimeter µg/mm² to short tn/cm² 1 µg/mm² = 1.1 × 10-6 short ton per square meter µg/mm² to short tn/m² 1 µg/mm² = 7.11 × 10-10 short ton per square inch µg/mm² to short tn/in² 1 µg/mm² = 1.02 × 10-7 short ton per square foot µg/mm² to short tn/ft² 1 µg/mm² = 9.22 × 10-7 short ton per square yard µg/mm² to short tn/yd² 1 µg/mm² = 9.84 × 10-31 long ton per square picometer µg/mm² to long tn/pm² 1 µg/mm² = 9.84 × 10-27 long ton per square angstrom µg/mm² to long tn/Ų 1 µg/mm² = 9.84 × 10-25 long ton per square nanometer µg/mm² to long tn/nm² 1 µg/mm² = 9.84 × 10-19 long ton per square micron µg/mm² to long tn/µ² 1 µg/mm² = 9.84 × 10-19 long ton per square micrometer µg/mm² to long tn/µm² 1 µg/mm² = 9.84 × 10-13 long ton per square millimeter µg/mm² to long tn/mm² 1 µg/mm² = 9.84 × 10-11 long ton per square centimeter µg/mm² to long tn/cm² 1 µg/mm² = 9.84 × 10-7 long ton per square meter µg/mm² to long tn/m² 1 µg/mm² = 6.35 × 10-10 long ton per square inch µg/mm² to long tn/in² 1 µg/mm² = 9.14 × 10-8 long ton per square foot µg/mm² to long tn/ft² 1 µg/mm² = 8.23 × 10-7 long ton per square yard µg/mm² to long tn/yd² 1 µg/mm² = 1.57 × 10-28 stone per square picometer µg/mm² to st/pm² 1 µg/mm² = 1.57 × 10-24 stone per square angstrom µg/mm² to st/Ų 1 µg/mm² = 1.57 × 10-22 stone per square nanometer µg/mm² to st/nm² 1 µg/mm² = 1.57 × 10-16 stone per square micron µg/mm² to st/µ² 1 µg/mm² = 1.57 × 10-16 stone per square micrometer µg/mm² to st/µm² 1 µg/mm² = 1.57 × 10-10 stone per square millimeter µg/mm² to st/mm² 1 µg/mm² = 1.57 × 10-8 stone per square centimeter µg/mm² to st/cm² 1 µg/mm² = 0.0002 stone per square meter µg/mm² to st/m² 1 µg/mm² = 0.07 stone per hectare µg/mm² to st/ha 1 µg/mm² = 1.02 × 10-7 stone per square inch µg/mm² to st/inch² 1 µg/mm² = 1.46 × 10-5 stone per square foot µg/mm² to st/ft² 1 µg/mm² = 0.0001 stone per square yard µg/mm² to st/yd² 1 µg/mm² = 3.22 × 10-26 troy ounce per square picometer µg/mm² to oz t/pm² 1 µg/mm² = 3.22 × 10-22 troy ounce per square angstrom µg/mm² to oz t/Ų 1 µg/mm² = 3.22 × 10-20 troy ounce per square nanometer µg/mm² to oz t/nm² 1 µg/mm² = 3.22 × 10-14 troy ounce per square micron µg/mm² to oz t/µ² 1 µg/mm² = 3.22 × 10-14 troy ounce per square micrometer µg/mm² to oz t/µm² 1 µg/mm² = 3.22 × 10-8 troy ounce per square millimeter µg/mm² to oz t/mm² 1 µg/mm² = 3.22 × 10-6 troy ounce per square centimeter µg/mm² to oz t/cm² 1 µg/mm² = 0.03 troy ounce per square meter µg/mm² to oz t/m² 1 µg/mm² = 0.03 troy ounce per hectare µg/mm² to oz t/ha 1 µg/mm² = 2.07 × 10-5 troy ounce per square inch µg/mm² to oz t/inch² 1 µg/mm² = 0.003 troy ounce per square foot µg/mm² to oz t/ft² 1 µg/mm² = 0.03 troy ounce per square yard µg/mm² to oz t/yd² 1 µg/mm² = 2.68 × 10-27 troy pound per square picometer µg/mm² to troy/pm² 1 µg/mm² = 2.68 × 10-23 troy pound per square angstrom µg/mm² to troy/Ų 1 µg/mm² = 2.68 × 10-21 troy pound per square nanometer µg/mm² to troy/nm² 1 µg/mm² = 2.68 × 10-15 troy pound per square micron µg/mm² to troy/µ² 1 µg/mm² = 2.68 × 10-15 troy pound per square micrometer µg/mm² to troy/µm² 1 µg/mm² = 2.68 × 10-9 troy pound per square millimeter µg/mm² to troy/mm² 1 µg/mm² = 2.68 × 10-7 troy pound per square centimeter µg/mm² to troy/cm² 1 µg/mm² = 0.003 troy pound per square meter µg/mm² to troy/m² 1 µg/mm² = 0.003 troy pound per hectare µg/mm² to troy/ha 1 µg/mm² = 1.73 × 10-6 troy pound per square inch µg/mm² to troy/inch² 1 µg/mm² = 0.0002 troy pound per square foot µg/mm² to troy/ft² 1 µg/mm² = 0.002 troy pound per square yard µg/mm² to troy/yd² 1 µg/mm² = 6.43 × 10-25 pennyweight per square picometer µg/mm² to dwt/pm² 1 µg/mm² = 6.43 × 10-21 pennyweight per square angstrom µg/mm² to dwt/Ų 1 µg/mm² = 6.43 × 10-19 pennyweight per square nanometer µg/mm² to dwt/nm² 1 µg/mm² = 6.43 × 10-13 pennyweight per square micron µg/mm² to dwt/µ² 1 µg/mm² = 6.43 × 10-13 pennyweight per square micrometer µg/mm² to dwt/µm² 1 µg/mm² = 6.43 × 10-7 pennyweight per square millimeter µg/mm² to dwt/mm² 1 µg/mm² = 6.43 × 10-5 pennyweight per square centimeter µg/mm² to dwt/cm² 1 µg/mm² = 0.64 pennyweight per square meter µg/mm² to dwt/m² 1 µg/mm² = 0.64 pennyweight per hectare µg/mm² to dwt/ha 1 µg/mm² = 0.0004 pennyweight per square inch µg/mm² to dwt/inch² 1 µg/mm² = 0.06 pennyweight per square foot µg/mm² to dwt/ft² 1 µg/mm² = 0.54 pennyweight per square yard µg/mm² to dwt/yd²
#### Foods, Nutrients and Calories
BANANA BREAD SPROUTED BITES, UPC: 851489003330 contain(s) 188 calories per 100 grams (≈3.53 ounces) [ price ]
11 foods that contain Inositol. List of these foods starting with the highest contents of Inositol and the lowest contents of Inositol
#### Gravels, Substances and Oils
Sand, Fine weighs 1 999 kg/m³ (124.79349 lb/ft³) with specific gravity of 1.999 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Magnesium diiodide octahydrate [MgI2 ⋅ 8H2O] weighs 2 098 kg/m³ (130.97386 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-402B, liquid (R402B) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
A pound per foot-second (lb/ft/s) is a non-SI (non-System International) measurement unit of dynamic viscosity.
The frequency is defined as an interval of time, during which a physical system, e.g. electrical current or a wave, performs a full oscillation and returns to its original momentary state, in both sign (direction) and in value, is called the oscillation period of this physical system.
g/m³ to long tn/metric c conversion table, g/m³ to long tn/metric c unit converter or convert between all units of density measurement.
#### Calculators
Weight to Volume conversions for sands, gravels and substrates | 5,160 | 12,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.288539 |
https://www.reference.com/web?q=2%20Variable%20Equations&qo=pagination&o=600605&l=dir&sga=1&qsrc=998&page=2 | 1,638,408,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00621.warc.gz | 1,038,962,862 | 30,145 | Related Search
Web Results
Improve your math knowledge with free questions in "Find a value using two-variable equations" and thousands of other math skills.
For example, to solve the equation 3x + 2 = 23 , you'd solve for (isolate) the x variable. Why? Because it's the only variable in there!
math.stackexchange.com/questions/391546/why-do-we-need-2-equations-to-solve-2-variables-3-equations-to-solve-3-variable
Assume that we have n linear equation in n variables. We can write these as: a1,1x1+⋯+a1,nxn=c1a2,1x1+⋯+a2,nxn=c2⋮an,1x1+⋯+an,nxn=cn.
artofproblemsolving.com/wiki/index.php/System_of_equations
1 Solve 2 variable equations in less than 5 seconds!!! 2 Solving Linear Systems. 2.1 Gaussian Elimination. 2.1.1 Problem; 2.1.2 Solution. 2.2 Substitution.
artofproblemsolving.com/wiki/index.php/Equation
1 Solve 2 variable equations in less than 5 seconds!!! 2 Word Problem AMC 8 Algebra Video; 3 Linear equations; 4 Linear equations; 5 Quadratic equations ...
www.webmath.com/solver2.html
This page will show you how to solve two equations with two unknowns. There are many ways of doing this, ... Equation #2: = Type the variables to solve for:.
betterlesson.com/lesson/456769/more-with-solving-two-variable-equations
Students will formalize their understanding of two variable equations by examining their graphs. Plan your 60-minute lesson in Math or two variables with ...
pressbooks.bccampus.ca/algebraintermediate/chapter/graph-linear-equations-in-two-variables
To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- ...
www.geeksforgeeks.org/linear-equation-in-two-variables
Mar 22, 2021 ... Therefore, Linear Equation in two variables can be written in the general form of, ax + by + c = 0, where a, b, c are the constants and x, ... | 504 | 1,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.840473 |
https://www.flightpedia.org/convert/7-seconds-to-decade.html | 1,643,273,609,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00128.warc.gz | 803,226,025 | 4,560 | # Convert 7.0 Seconds to Decades
• Q: How many Seconds in a Decade?
• Q: How do you convert 7 Second (s) to Decade (decade)?
7 Second is equal to 2.2e-08 Decade. Formula to convert 7 s to decade is 7 / 315360000
• Q: How many Seconds in 7 Decades? | 87 | 251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.800935 |
http://forums.wolfram.com/mathgroup/archive/2011/Jan/msg00414.html | 1,723,631,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00462.warc.gz | 10,835,833 | 8,448 | Re: Parallelize & Functions That Remember Values They Have Found
• To: mathgroup at smc.vnet.net
• Subject: [mg115516] Re: Parallelize & Functions That Remember Values They Have Found
• From: thomas <thomas.muench at gmail.com>
• Date: Thu, 13 Jan 2011 03:27:22 -0500 (EST)
```Dear Guido,
I have faced a similar problem recently. As a way around this, I collected the definitions known to the remote kernels in the following way:
f[n_] := f[n] = Prime[n]
DistributeDefinitions[f];
ParallelEvaluate[f[n], {n, 500000}];(*now all f's are known remotely*)
DownValues[f]=Flatten[ParallelEvaluate[DownValues[f]]];(*now all f's are known centrally*)
result = Table[f[n], {n, 500000}];
This collection of data can take quite some time and eat up the advantages you gain by parallelization. So it is only worth doing this if your real code gains enough speed by parallel evaluation. The best is to experiment with that!
Even though it works, it seems quite cumbersome to me. I feel that there should be a better way.
thomas
On Wednesday, January 12, 2011 10:08:44 AM UTC+1, Guido Walter Pettinari wrote:
> Dear group,
>
> I am starting to discover the magic behind Parallelize and
> ParallelTable, but I still have got many problems. The latest one
> occurred when I tried to parallelize a function that is supposed to
> store his values, i.e. those defined as f[x_] := f[x] = .....
>
> You can reproduce my problem by running the following snippet twice:
>
> f[n_] := f[n] = Prime[n]
> DistributeDefinitions[f];
> result = ParallelTable[f[n], {n, 500000}] // AbsoluteTiming;
> elapsed = result[[1]]
>
> On my machine, the first execution takes 2 seconds. Since I defined f
> as f[x_]:=f[x], I expect the second execution to take much less than
> that, but it actually takes around 1.8s. The third one takes
> something less than that (say 1.4s), and so on. After many
> executions, the execution time stabilizes to 0.6 seconds.
>
> Incidentally, 0.6 seconds is the time that a normal Table takes (on
> the second execution) to run the same code:
>
> Exit[]
> f[n_] := f[n] = Prime[n]
> result = Table[f[n], {n, 500000}] // AbsoluteTiming;
> elapsed = result[[1]]
>
> It looks like my 4 kernels are storing the downvalues of f[x]
> separately, so that each of them stores only a (random) quarter of the
> f-values every time the code is run. When all of them have all of the
> 500.000 f-values, which happens after many executions, the execution
> time finally reaches 0.6s.
>
> Is there a way to make all the f-values stored by the 4 kernels
> available? Maybe a function that "collapses" all the information
> gathered by the kernels into the main kernel, i.e. a
> DeDistributeDefinitions function? Or maybe a way to access the memory
> of all 4 kernels? I tried to SetSharedFunction on f[x], but it just
> made the calculation extremely long.
>
> I will be grateful for any suggestion.
>
> Thank you for your attention,
>
> Guido W. Pettinari
```
• Prev by Date: Re: How to change the directory for the docs?
• Next by Date: Re: Having some trouble with plot and solve
• Previous by thread: Parallelize & Functions That Remember Values They Have Found
• Next by thread: Re: Parallelize & Functions That Remember Values They Have Found | 866 | 3,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.912017 |
http://www.wyzant.com/resources/answers/22387/word_problems_and_turning_them_into_algebraic_equation_to_slove | 1,409,169,882,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829754.11/warc/CC-MAIN-20140820021349-00285-ip-10-180-136-8.ec2.internal.warc.gz | 701,049,308 | 11,399 | Search 75,453 tutors
0 0
# word problems and turning them into algebraic equation to slove
Yet another farm problem to set up and solve.....
Tina looks out into the barnyard and sees pigs, chickens and ducks. She counts 38 heads and 86 feet. Knowing that that the sum of the number of pigs and chickens is 2 more than the number of ducks. (all animals are normal-each has 1 head, etc) set up the using algebra equations and solve.
Let p = the number of pigs, c = number of chickens, and d = number of ducks.
p+c = d+2 ......(1)
P+c+d = 38 ......(2), balance by the total number of heads
4p+2c+2d = 86, balance by the total number of legs
/2, 2p+c+d = 43 ......(3)
(3)-(2): p = 5
(2)-(1): d = 36-d => d = 18
c = 38-p-d = 38-18-5 = 15
Answer: There are 5 pigs, 15 chickens and 18 ducks.
Hi Nancy;
Tina looks out into the barnyard and sees pigs, chickens and ducks. She counts 38 heads and 86 feet. Knowing that that the sum of the number of pigs and chickens is 2 more than the number of ducks. (all animals are normal-each has 1 head, etc) set up the using algebra equations and solve.
FIRST EQUATION...P+C+D=38
4P+2C+2D=86
SECOND EQUATION...2P+C+D=43
P+C=D+2
THIRD EQUATION...P+C-D=2
------------
SECOND EQUATION...2P+C+D=43
FIRST EQUATION...P+C+D=38
Let's subtract the first equation from the second...
P=5
SECOND EQUATION...2(5)+C+D=43
SECOND EQUATION...10+C+D=43
THIRD EQUATION...P+C-D=2
THIRD EQUATION...5+C-D=2
Let's add the second and third equations...
15+2C+0=45
Let's subtract 15 from both sides..
2C=30
Let's divide both sides by 2...
C=15
FIRST EQUATION...P+C+D=38
5+15+D=38
20+D=38
D=18
SECOND EQUATION...2P+C+D=43
2(5)+15+18=43
10+15+18=43
25+18=43
43=43
THIRD EQUATION..P+C-D=2
5+15-18=2
20-18=2
2=2 | 616 | 1,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2014-35 | longest | en | 0.879357 |
http://support.microsoft.com/kb/103493 | 1,408,653,176,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500821289.49/warc/CC-MAIN-20140820021341-00136-ip-10-180-136-8.ec2.internal.warc.gz | 185,618,647 | 23,103 | # Algorithm Used for QUARTILE() Function
Article translations
Close
Expand all | Collapse all
## SUMMARY
In Microsoft Excel, the QUARTILE() function returns a specified quartile in an array of numeric values. QUARTILE() accepts 2 arguments: Array and Quart. Array is the range of values for which you want to find the quartile value. Quart indicates the value you want to return, where:
``` 0 Minimum value (Same as MIN())
1 1st quartile - 25th percentile
2 2nd quartile - 50th percentile (Same as MEDIAN())
3 3rd quartile - 75th percentile
4 4th quartile - 100th percentile (Same as MAX())```
NOTE: In Microsoft Excel versions 5.0 and later, you can use the Function Wizard to insert the QUARTILE() function, by clicking Function on the Insert menu. The Function Wizard gives you information about the function, as well as required and optional arguments.
Following is the algorithm used to calculate QUARTILE():
1. Find the kth smallest member in the array of values, where:
``` k=(quart/4)*(n-1))+1
```
If k is not an integer, truncate it but store the fractional portion (f) for use in step 3.
``` quart = value between 0 and 4 depending on which quartile
you want to find.
n = number of values in the array
```
2. Find the smallest data point in the array of values that is greater than the kth smallest, the (k+1)th smallest member.
3. Interpolate between the kth smallest and the (k+1)th smallest values:
``` Output = a[k]+(f*(a[k+1]-a[k]))
a[k] = the kth smallest<BR/>
a[k+1] = the k+1th smallest
```
### Example
To find the 3rd quartile in the array of values, 0,2,3,5,6,8,9, follow these steps:
1. Find k and f:
``` k=TRUNC((3/4*(7-1))+1)=5<BR/>
f=(3/4*(7-1))-TRUNC(3/4*(7-1))=.5
```
2. The 5th (kth) smallest value is 6, and the (5+1)th smallest value is 8.
3. Interpolate:
``` 6+(.5*(8-6))=7
```
## REFERENCES
"Function Reference," version 4.0, pages 342-343
## Properties
Article ID: 103493 - Last Review: August 15, 2003 - Revision: 1.1
##### APPLIES TO
• Microsoft Excel 97 Standard Edition
• Microsoft Excel 98 for Macintosh
##### Keywords:
KB103493
Retired KB Content Disclaimer | 639 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2014-35 | latest | en | 0.783614 |
https://documen.tv/question/find-the-slope-of-the-line-containing-the-given-pair-of-points-if-the-slope-is-undefined-state-s-24083795-51/ | 1,652,883,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00364.warc.gz | 262,597,658 | 15,860 | ## Find the slope of the line containing the given pair of points. If the slope is undefined, state so. (-1,-7) and (-6,1)
Question
Find the slope of the line containing the given pair of points. If the slope is undefined, state so.
(-1,-7) and (-6,1)
in progress 0
5 months 2021-09-03T22:33:59+00:00 1 Answers 8 views 0
m =
Step-by-step explanation:
m = slope
m =
m =
m =
m = | 122 | 387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-21 | latest | en | 0.839999 |
https://wolfcrow.com/notes-by-dr-optoglass-airy-disk-and-pixel-density-of-the-human-eye/ | 1,579,894,806,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00519.warc.gz | 738,305,767 | 18,256 | Categories
# Notes by Dr. Optoglass: Airy Disk and Pixel Density of the Human Eye
Topics Covered:
• The Airy Disk
• Diffraction
• Pixel size of the Eye
• Pixel density or resolution of the Eye
Everybody thinks they know what art should be. But very few of them have the sense that is necessary to experience painting, that is the sense of sight, that sees colors and forms as living reality in the picture – Otto Dix
What shape are pixels? On displays they are rectangular. On camera sensors they are probably amoebic! We know our photoreceptor cells are shaped like rods and cones. Printers recognize droplets as circular. Which is it?
No matter what, light that passes through a circular aperture (a lens usually) has the characteristics of a circle. Hence a lot of optic testing relies on the mathematics that govern this model.
The Airy disk is named after George Airy. This is what it looks like:
The 3D model tells us how the intensity of light is distributed. The width of the main circle is the smallest possible dot that can be produced by any given lens at a certain aperture. The rings are diffraction patterns.
Diffraction is light bending around corners:
They look similar, don’t they? So why is diffraction important?
• You can’t escape from diffraction.
• Every lens needs a slit (aperture), so no lens can escape diffraction.
• Even a perfect lens (which doesn’t exist) is affected by diffraction.
• Any camera system that relies on a lens (even a perfect one) is affected by diffraction.
• Every camera system has a sensor that is of a certain resolution.
• The lens used must either be equal to the sensor or greater than the sensor in resolution. The lens shouldn’t be the weak link in the chain.
• A system that has a lens that is better than the sensor and all other optical elements in the chain is said to be diffraction limited.
• It is diffraction limited because the lens (and all other optical elements, like filters, etc) is so much better than the sensor in resolution that at this point only diffraction can spoil the image.
Therefore, a camera system in which resolution is not limited by imperfections in the lens but only by diffraction is said to be diffraction limited.
If you have read what Professor Sampler has to say about this, you’ll remember that it’s only when things get smaller that quantum mechanics comes into play. The effect of diffraction is almost invisible if the slit (aperture) is large. That’s why we don’t observe its effects in the real world. But when the aperture gets really small (try it by almost squeezing your eyes shut – everything blurs) the effects of diffraction are not negligible.
Big question: Is our eye diffraction limited?
To answer this question we’ll need a formula to calculate the resolution of the human eye. Just like for everything else in science and engineering, we have not one, but two formulas to choose from:
The Rayleigh Criterion where ? is the angular resolution ? is the wavelength of light in meters and D is the diameter of the lens’ aperture in meters
The Dawes’ Limit R = 11.6/D where D is the diameter of the lens’ aperture in centimeters R is the angular resolution in arc seconds
The size of the human pupil can vary from 3mm to 9mm.
A typical human eye will respond to wavelengths from about 390 to 750 nm, with maximum sensitivity at around 555 nm.
What does that give us?
• According to Rayleigh about 0.2 arc min to 1 arc min
• According to Dawes about 0.2 arc min to 0.6 arc min
What both agree on fundamentally is that the eye cannot resolve beyond 0.2 arc minutes due to diffraction. As we have seen, in studies the eye does not resolve beyond 0.4 arc minutes anyway, and it would be a rare individual who can better 0.4 arc minute.
To answer the question: No, the eye is not a diffraction limited system. Our sensor (retina) is better than our lens.
Pixel Density
So how good is our ‘sensor’? We have already seen that most of the cones are found in the fovea, with the foveola having the highest cone density.
What is the size of one disk that 0.2 arc minutes would subtend on the foveola, assuming the diameter of the eye is 22.22 mm? It is about 1.3 ?m (microns). A 0.4 arc minute vision would subtend 2.6 ?m (microns).
What is the area of the fovea? It’s about 3.14mm2. The area of our airy disk is ? x (1.3 x 10-3)2 = 5.3 x 10-6 mm2. How many disks can fit our fovea? 600,000.
If we try the same calculation with 2.6 ?m (microns), we get the total cones in the fovea to be about 150,000. Which one is right?
The fovea subtends about 2 degrees of human vision, and if each disk subtends 0.2 arc minutes, how many disks can fit into 2o? About 1 million. That should tell us that about 2 ?m (microns) is a good average. Remember, in The Human Eye Part II, we learnt that a cone can vary in size, between 0.5 to 4.0 ?m. What’s the average? Yes, 2 ?m.
This is also confirmed by tests on the cone density in the fovea, which at maximum is about 350,000. This gives us a ‘pixel’ value of about 1.7 or 2 ?m.
If the diameter of the fovea is 1mm, then the ‘pixels’ per mm = 500. You could translate that into 500 lines per mm or 250 lp/mm or about 12,700 ppi.
Modern sensors have a pixel size (pitch) of 4 ?m. This gives us a theoretical maximum of about 250 lines per mm (125 lp/mm) (52.5 Megapixels) for a 35mm sensor (8750 x 6000).
Therefore, here are the results:
• Density at its absolute sharpest (0.5?m) = about 50,000 ppi or 2,000 lpm or 1,000 lp/mm
• Density at 1?m = 25,400 ppi or 1,000 lpm or 500 lp/mm
• Density at 2?m = 12,700 ppi or 500 lpm or 250 lp/mm
• Density at 4?m = 6,350 ppi or 250 lpm or 125 lp/mm
Modern sensors are getting there!
Takeaways:
• A point light through a lens is described in terms of an airy disk pattern.
• A camera system in which resolution is not limited by imperfections in the lens but only by diffraction is said to be diffraction limited.
• The pixel size of the eye can be said to be about 2?m (microns).
• The pixel density (or resolution in these terms) of the eye can be said to be about 12,700 ppi or 500 lpm or 250 lp/mm.
Links for further study: | 1,577 | 6,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-05 | latest | en | 0.943916 |
https://mathworld.wolfram.com/KroneckerDelta.html | 1,679,799,865,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00177.warc.gz | 432,765,867 | 5,301 | TOPICS
# Kronecker Delta
The simplest interpretation of the Kronecker delta is as the discrete version of the delta function defined by
(1)
The Kronecker delta is implemented in the Wolfram Language as KroneckerDelta[i, j], as well as in a generalized form KroneckerDelta[i, j, ...] that returns 1 iff all arguments are equal and 0 otherwise.
It has the contour integral representation
(2)
where is a contour corresponding to the unit circle and and are integers.
In three-space, the Kronecker delta satisfies the identities
(3) (4) (5) (6)
where Einstein summation is implicitly assumed, , 2, 3, and is the permutation symbol.
Technically, the Kronecker delta is a tensor defined by the relationship
(7)
Since, by definition, the coordinates and are independent for ,
(8)
so
(9)
and is really a mixed second-rank tensor. It satisfies
(10) (11) (12) (13) (14)
Delta Function, Permutation Symbol, Permutation Tensor
## Related Wolfram sites
http://functions.wolfram.com/IntegerFunctions/KroneckerDelta/
## Explore with Wolfram|Alpha
More things to try:
## Cite this as:
Weisstein, Eric W. "Kronecker Delta." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/KroneckerDelta.html | 310 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-14 | latest | en | 0.77542 |
https://cuberootgeek.com/index.html | 1,610,906,926,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00696.warc.gz | 291,149,840 | 4,406 | # Alliance
### Simplify Square Root
You can do this using our simplify square root calculator seen above or manually using two methods: Factoring and long division…
### Multiplying Square Roots
To multiply and calculate square roots, just multiply the numbers inside the radical signs and then apply factoring…
### Negative Square Roots
When a number involves a negative square root, it is called an imaginary number…
## Finding Square Roots
You can find square roots by forming groups of similar numbers or by long division. The group method is pretty straightforward but works only for perfect squares. | 125 | 618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-04 | latest | en | 0.821306 |
https://forums.wolfram.com/mathgroup/archive/1993/Jan/msg00061.html | 1,657,028,879,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00043.warc.gz | 308,164,546 | 7,522 | How to Animate Points???
• To: mathgroup at yoda.physics.unc.edu
• Subject: How to Animate Points???
• From: Susan Rempe <rempe at carson.u.washington.edu>
• Date: Mon, 25 Jan 1993 13:28:42 -0800 (PST)
```Hi Mathgroup,
I have the x,y,z coordinates for a point in space at an
initial position (req) and a change in that position (coordlist).
I want to watch that point move using the equations
x(t)=x + delta x *Sin[w*t]
y(t)=y + delta y *Sin[w*t]
z(t)=z + delta z *Sin[w*t]
I tried by first identifying just one point,
anipts=
Point[
{req[[1,1]]+coordlist[1][[1,1]] Sin[t],
req[[1,2]]+coordlist[1][[1,2]] Sin[t],
req[[1,3]]+coordlist[1][[1,3]] Sin[t]
}]
then making a bunch of graphs
Table[Show[Graphics3D[anipts],DisplayFunction->Identity],{t,0,5}]
and then trying to animate the result using
Animate[Table[Show[Graphics3D[anipts]]],{t,0,5}]
and
ShowAnimation[Table[Show[Graphics3D[anipts]]],{t,0,5}]
among other things.
Nothing works. Can anyone offer a suggestion??
Susan Rempe
```
• Prev by Date: "Chaos and Fractals" programs
• Next by Date: Re: Exporting Mathematica output to word processing program
• Previous by thread: "Chaos and Fractals" programs
• Next by thread: EllipticTheta[3,real,Exp[I real] ] | 403 | 1,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.728689 |
https://gmatclub.com/forum/according-to-a-survey-of-graduating-medical-students-51868.html?fl=similar | 1,487,901,773,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171271.48/warc/CC-MAIN-20170219104611-00497-ip-10-171-10-108.ec2.internal.warc.gz | 727,328,291 | 58,809 | According to a survey of graduating medical students : GMAT Sentence Correction (SC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 23 Feb 2017, 18:02
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# According to a survey of graduating medical students
Author Message
TAGS:
### Hide Tags
Manager
Joined: 27 Jul 2007
Posts: 115
Followers: 1
Kudos [?]: 8 [0], given: 0
### Show Tags
06 Sep 2007, 05:36
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
According to a survey of graduating medical students conducted by the Association of American Medical Colleges, minority graduates are nearly four times more likely than are other graduates in planning to practice in socioeconomically deprived areas.
(A) minority graduates are nearly four times more likely than are other graduates in planning to practice
(B) minority graduates are nearly four times more likely than other graduates who plan on practicing
(C) minority graduates are nearly four times as likely as other graduates to plan on practicing
(D) it is nearly four times more likely that minority graduates rather than other graduates will plan to practice
(E) it is nearly four times as likely for minority graduates than other graduates to plan to practice
If you have any questions
New!
Manager
Joined: 27 Jul 2007
Posts: 115
Followers: 1
Kudos [?]: 8 [0], given: 0
### Show Tags
06 Sep 2007, 05:46
Lets say there's another option choice:
F) minority graduates are nearly four times more likely than are other graduates to plan on practicing
Manager
Joined: 20 Jun 2007
Posts: 157
Followers: 1
Kudos [?]: 19 [0], given: 0
### Show Tags
06 Sep 2007, 05:46
I don't think you need the as likely as, more likely than distintion here - 4 sentences are wrong for other reasons:
A - "than are other graduates in planning to practice" is awkward
B - "than other graduates who plan on practicing " changes meanings
D - "that minority graduates rather than other graduates will plan to practice " awkward
E - "as likely for minority graduates than " as .. than
VP
Joined: 10 Jun 2007
Posts: 1459
Followers: 7
Kudos [?]: 263 [0], given: 0
Re: SC- more likely than vs as likely as [#permalink]
### Show Tags
06 Sep 2007, 05:46
farend wrote:
According to a survey of graduating medical students conducted by the Association of American Medical Colleges, minority graduates are nearly four times more likely than are other graduates in planning to practice in socioeconomically deprived areas.
(A) minority graduates are nearly four times more likely than are other graduates in planning to practice
(B) minority graduates are nearly four times more likely than other graduates who plan on practicing
(C) minority graduates are nearly four times as likely as other graduates to plan on practicing
(D) it is nearly four times more likely that minority graduates rather than other graduates will plan to practice
(E) it is nearly four times as likely for minority graduates than other graduates to plan to practice
C for me.
In A, "are" should be there
In B, "who" makes the sentence incomplete
In D, when you use "four times", I believe you have to use as. For example, She is twice as likely as...not She's twice more
E is missing "as"
Manager
Joined: 27 Jul 2007
Posts: 115
Followers: 1
Kudos [?]: 8 [0], given: 0
### Show Tags
06 Sep 2007, 06:07
You are right. C is the OA. What to choose if F) were an option ?
VP
Joined: 10 Jun 2007
Posts: 1459
Followers: 7
Kudos [?]: 263 [0], given: 0
### Show Tags
06 Sep 2007, 06:16
I still wouldn't pick it. I don't think the use of "four times more...than" is correct here. The use of "four times more" means 4x and it should be use in sentence that has an actual countable number, not comparing the difference. For example,
He kicked the door four times more.
Just my opinion. Feel free to discuss.
Senior Manager
Joined: 04 Jun 2007
Posts: 373
Followers: 1
Kudos [?]: 64 [0], given: 0
### Show Tags
06 Sep 2007, 10:56
correct idiom is 'as likely as' so we need 'four times as likely as'- clear C
06 Sep 2007, 10:56
Similar topics Replies Last post
Similar
Topics:
According to a survey of graduating medical students 4 01 Jul 2008, 02:42
1 According to a survey of graduating medical students 7 22 Jun 2008, 16:32
6 According to a survey of graduating medical students 5 18 Jun 2008, 17:14
According to a survey of graduating medical students 6 30 Oct 2007, 06:12
According to a survey of graduating medical students 13 22 May 2007, 18:46
Display posts from previous: Sort by | 1,336 | 5,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-09 | longest | en | 0.937463 |
https://www.meritnation.com/ask-answer/question/a-diwali-rocket-is-ejecting-0-05-kg-of-gases-per-second-at-a/work-and-energy/3603780 | 1,591,045,282,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00591.warc.gz | 807,049,668 | 9,204 | # A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m s -1 . The accelerating force on the rocket is( a ) 20 dyne ( b ) 20 newton( c ) 20 kg wt ( d ) sufficient data not given
We have,
dm/dt = 0.05 kg/s
v = 400 m/s
We know,
P = mv
=> dP/dt = d(mv)/dt = v(dm/dt) + m(dv/dt)
=> F = v(dm/dt) + 0
=> F = (400)(0.05)
=> F = 20 N
• 4
What are you looking for? | 157 | 390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-24 | latest | en | 0.730122 |
https://tagvault.org/blog/8am-to-3pm-how-many-hours-guide/ | 1,709,205,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00028.warc.gz | 543,700,261 | 25,975 | # 8am to 3pm – How Many Hours? (Guide)
Do you ever wonder how many hours there are between 8am and 3pm? Whether you need to calculate the time duration for scheduling or planning purposes, understanding how to determine the number of hours between these two points is essential. In this guide, we will walk you through the process of calculating the hours from 8am to 3pm, providing you with a clear understanding of the correct method to use.
### Key Takeaways:
• The duration between 8am and 3pm is 7 hours.
• Converting times to a 24-hour format makes time calculations easier and more accurate.
• Precision in time calculations requires considering both hours and minutes.
• A calculator can simplify the process of calculating time differences.
• For non-sequential times, a different approach is required to determine the duration accurately.
## Understanding Time Calculation
When it comes to calculating hours, it’s essential to have a clear understanding of how time is measured and how to convert between different formats. Converting times to a 24-hour format can simplify the process and help avoid confusion between AM and PM.
In a 24-hour clock format, each hour is represented by a number from 0 to 23. For example, 8AM is equivalent to 08:00, while 3PM is equivalent to 15:00. By converting the times to this format, we can easily calculate the duration between them.
Let’s take the example of calculating the hours between 8AM and 3PM using the 24-hour format. Subtracting 8:00 from 15:00 gives us a total of 7 hours. By converting the times to a 24-hour format, we eliminate any ambiguity and can accurately determine the time difference.
### The Importance of Precision in Time Calculations
When dealing with time calculations, it’s important to consider both hours and minutes to obtain an accurate result. Simple subtraction of hours would not provide the precise duration between two given times. By including minutes in the calculation, we can achieve a more precise and reliable result.
For example, if we were to calculate the time difference between 8:30AM and 3:45PM, we would need to consider the additional 45 minutes. The duration would then be 7 hours and 15 minutes. By accounting for the minutes along with the hours, we ensure accuracy in our time calculations.
### Converting Times to a 24-Hour Format
Converting times from a 12-hour clock (AM/PM) to a 24-hour clock format can be done by adding or subtracting 12 from the given time, depending on whether it is AM or PM. This conversion helps ensure accurate calculations and eliminates any confusion that may arise due to the AM/PM distinction.
For example, to convert 3PM to a 24-hour format, we add 12 hours, resulting in 15:00. Similarly, 8AM remains the same in the 24-hour format as it is already represented as 08:00. Converting times to a 24-hour format is a crucial step in accurately calculating the time difference.
Understanding time calculation and converting times to a 24-hour format are fundamental skills when it comes to calculating hours. These skills enable us to accurately determine the duration between two given times, ensuring precise scheduling, planning, and coordination.
## Importance of Precision in Time Calculations
When it comes to time calculations, precision is key. It is not enough to simply calculate the number of hours between two given times; one must also consider the minutes to obtain an accurate result. Let’s explore why precision is important in time calculations and how to achieve it.
When we calculate time differences, we often focus solely on the hours. However, neglecting the minutes can lead to inaccurate results. For example, if we calculate the duration between 8:15 AM and 3:30 PM by subtracting the hours alone, we would arrive at a result of 7 hours. However, by considering the minutes as well, we can determine that the actual duration is 7 hours and 15 minutes.
To ensure precision in time calculations, it is crucial to account for both hours and minutes. By doing so, we can provide more accurate information and avoid any confusion or discrepancies that may arise from neglecting the minutes. Whether it’s for scheduling, coordinating activities, or planning events, precision in time calculations is essential.
To summarize, precision in time calculations requires us to consider both the hours and minutes. By accurately accounting for the minutes, we can obtain precise and reliable results. So, the next time you need to calculate the duration between two given times, remember to pay attention to the minutes as well, ensuring the accuracy of your calculations.
## Converting Times to a 24-Hour Format
Converting times from a 12-hour clock (AM/PM) to a 24-hour clock format is a crucial step in accurately calculating time differences. This conversion helps to eliminate any confusion between AM and PM and ensures precise calculations. For example, 3PM in a 12-hour clock is equivalent to 15:00 in a 24-hour clock.
The 24-hour clock follows a simple and consistent format, with the hours ranging from 00:00 to 23:59. The format uses two digits for hours and two digits for minutes, separated by a colon. When converting from a 12-hour clock, we can simply add 12 to the hour if it’s in the PM time frame. For instance, 8AM becomes 08:00 in the 24-hour format, while 8PM becomes 20:00.
Converting to a 24-hour clock format is especially useful when calculating time differences that span over several hours or even days. By using the 24-hour format, we can easily subtract the starting time from the ending time to obtain the precise duration. This method ensures accuracy in our calculations and eliminates any ambiguity regarding time representation.
### Example:
“I need to calculate the time difference between 10:30AM and 11:45PM for an event. By converting these times to a 24-hour format, we have 10:30 and 23:45, respectively. Subtracting the starting time from the ending time, we find that the duration is 13 hours and 15 minutes.”
Converting times to a 24-hour format simplifies time calculations and ensures accurate results. It is an essential step in determining the duration between two given times, providing a standardized and consistent representation of time. By following the correct format, we can confidently calculate time differences and effectively manage our schedules and plans.
## Factors to Consider in Time Calculations
When performing time calculations, there are several factors that need to be taken into consideration to ensure accurate results. Two of these factors are daylight saving time and time zones. Let’s explore each of these factors in more detail:
### Daylight Saving Time
Daylight saving time is a practice in many countries where the clocks are adjusted forward by one hour during the warmer months to provide more daylight in the evenings. This means that during daylight saving time, the time difference between two given points may vary depending on whether the time falls within the daylight saving time period or not. It is essential to consider whether daylight saving time is in effect when calculating time differences to ensure accuracy.
### Time Zones
Time zones are regions of the world that have the same standard time. There are 24 time zones in total, and each zone is approximately 15 degrees of longitude apart. When calculating time differences across different time zones, it is crucial to account for the difference in standard time. This is especially important when dealing with international time calculations or when determining the duration between two points in different regions.
By considering factors such as daylight saving time and time zones, time calculations can be made with greater accuracy. It is essential to keep these factors in mind when performing calculations for scheduling, planning, or coordinating activities that involve different locations and time zones.
Factor Description
Daylight Saving Time The practice of adjusting clocks forward by one hour during warmer months.
Time Zones Regions of the world with the same standard time, separated by approximately 15 degrees of longitude.
## Using a Calculator for Time Difference Calculation
When it comes to calculating time differences accurately, using a calculator can be a great tool. With the help of a calculator, you can easily subtract the starting time from the ending time to obtain the desired result in hours. By inputting the times in the correct format and utilizing the subtraction function, a calculator ensures that the calculation is precise and error-free.
Using a calculator for time difference calculation also saves you from the hassle of manual calculations, which can be prone to mistakes. The calculator provides a quick and efficient way to obtain accurate results, especially when dealing with complex time intervals or multiple time calculations.
Whether you’re trying to determine the duration between two events, calculate work hours, or schedule appointments, using a calculator simplifies the process and gives you confidence in the accuracy of your calculations. It’s a handy tool that can be easily accessed on your smartphone, computer, or even a dedicated time calculation device.
Benefits of using a calculator for time difference calculation:
Accuracy: A calculator ensures precise calculations, eliminating the possibility of manual errors.
Efficiency: It provides a quick and convenient way to calculate time differences, saving you time and effort.
Reliability: Using a calculator gives you confidence in the accuracy of your calculations, especially for complex time intervals.
Accessibility: Calculators are widely available on various devices, making it easy to perform time difference calculations wherever you are.
## Applicability to Different Time Frames
Calculating hours for any time frame is not limited to specific start and end times. The method discussed earlier can be applied to determine the duration between any given starting and ending times. Whether it’s calculating the hours between 9AM and 5PM for a regular workday or determining the time difference between 2PM and 10PM for an evening event, the same approach can be used.
However, it is important to note that this method is applicable for sequential times only. In other words, it assumes that the starting time occurs before the ending time. If you’re dealing with non-sequential times, such as calculating the duration between 10PM and 5AM the next day, a different approach is required.
### Finding the Duration for Non-Sequential Times
When it comes to calculating the duration between non-sequential times, it’s necessary to account for the change in date. For example, if you want to calculate the hours between 10PM and 5AM the next day, you’ll need to consider that the ending time goes into the next day.
To calculate the duration for non-sequential times, you can break it down into two separate calculations. First, calculate the remaining hours from the starting time until midnight. Then, calculate the hours from midnight to the ending time on the following day. Add these two durations together to get the total number of hours between the non-sequential times.
By following these methods, you can accurately calculate the hours for any time frame, whether the times are sequential or non-sequential. This knowledge can be valuable for various purposes, such as scheduling shifts, planning events, or determining the duration of activities.
## Other Methods for Time Difference Calculation
While manual calculation can be effective for determining the time difference between two given points, there are alternative methods available that can provide accurate results with ease. Online time calculators and software programs are valuable tools that can simplify the process and save time. These resources offer a user-friendly interface and allow for quick and efficient time difference calculations.
Online time calculators are accessible through various websites and can be used directly in a web browser. They typically require inputting the starting and ending times and provide an instant result, displaying the duration in hours, minutes, and seconds. These calculators eliminate the need for manual calculations and reduce the chance of errors.
Software programs designed specifically for time calculations are another option worth considering. These programs offer additional features and functionalities, such as the ability to calculate time differences for multiple time zones, account for daylight saving time, and handle non-sequential times. They can be particularly useful for complex time calculations or for professionals who regularly deal with time-related tasks.
### Benefits of Alternative Methods
• Convenience: Online time calculators and software programs provide a hassle-free way to calculate time differences, eliminating the need for manual calculations.
• Accuracy: These tools offer precise results, reducing the likelihood of errors that may occur when performing calculations manually.
• Efficiency: With the help of online calculators or software programs, time difference calculations can be completed quickly, saving valuable time.
• Additional Features: Software programs often come with advanced features that cater to specific time-related tasks, allowing for a more comprehensive approach to time calculations.
By utilizing alternative methods such as online time calculators or software programs, individuals can simplify the process of calculating time differences and ensure accurate results. These tools offer convenience, accuracy, and efficiency, making them invaluable resources in time-related tasks and projects.
Method Pros Cons
Manual Calculation – No external dependencies
– Basic method
– Potential for errors
– Time-consuming
Online Time Calculators – Convenient and accessible
– Instant results
– Limited customization
– Dependence on internet connection
– Customization options
– Requires installation
– Learning curve
## Consideration of Minutes in Addition to Hours
When calculating time durations, it is essential to consider not only the hours but also the minutes to ensure accuracy. By including the minutes in our calculations, we can obtain a more precise measurement of time. This is particularly important when the duration between two given times is less than an hour.
To calculate minutes along with hours, we need to evaluate the difference between the starting and ending minutes. Let’s consider an example: if the starting time is 8:45 AM and the ending time is 10:30 AM, we subtract the minutes. This gives us a duration of 1 hour and 45 minutes. It is worth noting that when calculating minutes, it is crucial to account for both the starting and ending times.
Accuracy in time calculations is vital for various scenarios, such as scheduling appointments, determining travel times, or tracking work hours. Ignoring the minutes can lead to incorrect estimations and potential inconveniences. By taking into account both hours and minutes, we can achieve greater precision and ensure that our calculations reflect the true duration of time.
Starting Time Ending Time Duration
8:45 AM 10:30 AM 1 hour and 45 minutes
9:15 AM 10:05 AM 50 minutes
1:30 PM 3:45 PM 2 hours and 15 minutes
By considering minutes in addition to hours, we can accurately calculate the duration between two given times and ensure our time calculations are precise.
## Duration Calculation for Non-Sequential Times
Calculating the duration between non-sequential times requires a different approach compared to sequential times. When the starting and ending times are not in chronological order, we need to consider both the date and time to accurately determine the time difference.
For example, if we want to calculate the duration between 10 PM on April 5th and 6 AM on April 6th, we can’t simply subtract the starting time from the ending time. Instead, we need to calculate the time difference for each day separately and then add them together.
When dealing with non-sequential times, it is important to consider both the hours and minutes as well as the date. Taking into account the date ensures that we accurately calculate the duration between the two given times without overlooking any part of the time span.
### Example:
To further illustrate this approach, let’s consider the following scenario:
Starting Time: 9 PM on July 10th
Ending Time: 6 AM on July 11th
We can divide the calculation into two parts:
1. Calculation for July 10th: Subtract 9 PM (21:00) from 11:59 PM (23:59) to calculate the duration for July 10th. In this case, the duration would be 2 hours and 59 minutes.
2. Calculation for July 11th: Subtract 12:00 AM (00:00) from 6 AM (06:00) to calculate the duration for July 11th. This results in a duration of 6 hours.
By adding the durations for both days together, we find that the total duration between 9 PM on July 10th and 6 AM on July 11th is 8 hours and 59 minutes.
### Table:
Date Starting Time Ending Time Duration
July 10th 9 PM (21:00) 11:59 PM (23:59) 2 hours 59 minutes
July 11th 12:00 AM (00:00) 6 AM (06:00) 6 hours
Total Duration 8 hours 59 minutes
As shown in the example and table above, accurately calculating the duration between non-sequential times involves considering each day separately and taking into account both the time and date. This approach ensures precise measurement of the time difference and provides an accurate result.
## Conclusion
In conclusion, the duration between 8AM and 3PM is 7 hours. By understanding the process of calculating time differences and converting times to a 24-hour format, you can accurately determine the duration between any two given times. It is important to consider both hours and minutes to obtain precise results.
When calculating time differences, factors such as daylight saving time and time zones may need to be taken into account. However, for the specific calculation of hours between 8AM and 3PM, these factors are not relevant.
Alternative methods, such as using online time calculators or software programs, can provide accurate results for time difference calculations. Additionally, it is crucial to note that the method discussed is applicable for sequential times only, not for non-sequential times.
To ensure accuracy in time calculations, always consider the minutes along with the hours. This will provide a more precise duration. Whether you’re scheduling, planning, or coordinating, understanding how to calculate time differences is essential. Follow the correct method and confidently determine the duration between two given times.
## FAQ
### 8am to 3pm is how many hours?
There are 7 hours from 8am to 3pm.
### How do you calculate hours from 8am to 3pm?
To calculate the hours, subtract the starting time (8am) from the ending time (3pm).
### Why is it important to convert times to a 24-hour format?
Converting times to a 24-hour format helps in accurately calculating time differences and avoiding confusion between AM and PM.
### Do I need to consider minutes when calculating time differences?
Yes, considering both hours and minutes ensures accuracy in time calculations and provides a more precise duration.
### How do I convert times from a 12-hour clock to a 24-hour clock?
To convert, simply change AM and PM times to their equivalent in a 24-hour format. For example, 3PM is equivalent to 15:00.
### Do factors like daylight saving time and time zones affect calculating hours between 8am and 3pm?
No, for this specific calculation, these factors are not relevant. They may need to be considered in other time difference calculations.
### Can I use a calculator to calculate the hours between 8am and 3pm?
Yes, using a calculator to subtract the starting time from the ending time ensures accurate results and simplifies the calculation process.
### Is the method mentioned applicable for calculating hours between any starting and ending times?
Yes, the method can be applied to calculate hours between any given sequential starting and ending times.
### Are there alternative methods for calculating time differences?
Yes, online time calculators and software programs can provide accurate results for time difference calculations.
### Should I consider minutes along with hours when calculating time differences?
Yes, considering minutes along with hours ensures accuracy in time calculations and provides a more precise duration.
### Can the method discussed be used to calculate the duration between non-sequential times?
No, for non-sequential times, a different approach is required to accurately determine the time difference. | 4,130 | 20,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.931632 |
https://resources.quizalize.com/view/quiz/review-for-vocabulary-quiz21-e26ca4a8-8684-4319-bcab-85768be8a0b7 | 1,695,454,857,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00086.warc.gz | 519,932,514 | 14,658 | Review for vocabulary quiz2/1
Quiz by Linasette
Feel free to use or edit a copy
includes Teacher and Student dashboards
### Measure skillsfrom any curriculum
Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill.
• edit the questions
• save a copy for later
• start a class game
• view complete results in the Gradebook and Mastery Dashboards
• automatically assign follow-up activities based on students’ scores
• assign as homework
• share a link with colleagues
• print as a bubble sheet
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
10 questions
• Q1
A letter or symbol used to represent a number
Variable
point of intersection
Parallel Lines
Intersecting lines
30s
• Q2
two lines that will never intersect and have no ordered pairs in commons; no solution
intersecting lines
point of intersection
parallel lines
collinear
30s
• Q3
two lines that have one point in common: one solution
intersecting lines
variable
parallel lines
collinear lines
30s
• Q4
the spot, or point in common; one solution
collinear
substitution
intersecting lines
point of intersection
30s
• Q5
the graph of two equations results in the same line: infinitely many solutions
parallel
intersecting lines
collinear
substitution
30s
• Q6
to replace one element of mathematical equaiton or expression with another
substitution
system of linear equations
intersecting lines
collinear lines
30s
• Q7
a set or collections of equations that are solved simultaneously
intersecting lines
collinear lines
system of linear equations
substitution
30s
• Q8
to take a term away
substitution
elimination
variable
point of intersection
30s
• Q9
also known as elimination method; the goal with this method of solving systems of equations is to eliminate one set of variables from your problem
collinear lines
systems of linear equations
substitution method
30s
• Q10
the goal with this method of solving systems of equations is to solve one equation for one variable, then substitute that solution in the other equation and solve
substitution method
elimination
parallel lines | 503 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.896939 |
http://cboard.cprogramming.com/cplusplus-programming/33760-random-number-between-0-variable-value.html | 1,475,311,768,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662698.85/warc/CC-MAIN-20160924173742-00281-ip-10-143-35-109.ec2.internal.warc.gz | 40,841,200 | 13,040 | # Thread: random number between 0 and variable value
1. ## random number between 0 and variable value
Hey guys,
I'm writing an RPG, and i need to generate a random number between 0 and a certain values content.
eg.
Code:
```x = 10;
random = (rand()x); //to generate a number between 0 and 10```
That doesn't seem to work, but it gives you an idea of what i want to do, i've searched, but a lot of stuff comes up when you enter random numbers!
Thanks.
-HM
2. Code:
`rand() % x; //???`
Oh and make sure to seed before you do that.
3. thanks for the speedy reply, i knew it would be as simle as that :-/
btw, i always seed.
4. Originally posted by HybridM
thanks for the speedy reply, i knew it would be as simle as that :-/
btw, i always seed.
Just reminding you since you didn't seem to include that in your demostration.
5. What is seeding? Why seed?
6. Code:
`srand(time(NULL));`
makes the numbers more random...
7. Code:
```#include <time.h>
#include <stdlib.h>
...
srand(time(NULL));
x=(rand()%(N+1)); // Generate random # between 0 and N
x=(rand()%(end-start+1)+start); // Generate random # between start and end
...```
Hope this helps...
** edit..fixed typo
8. ## Accually...
Code:
`x=(rand()%N);`
That would generate a random number between 0 and N-1.
9. But I thought % <- was a modulus... how does this have to do with a random number?
Other then that... what is the difference betwem rand() and srand() along with what does this line do: srand(time(NULL));
10. Originally posted by Munkey01
Code:
`x=(rand()%N);`
That would generate a random number between 0 and N-1.
oops...yep...you are right...damn typos *edited to fix*
srand() seeds the random # generator with a value. If you do not srand() before you generate numbers then the same 'random' numbers might show up each time your program runs.
EX:
c:\> text.exe
r1=4
r2=10
r3=1
c:\> text.exe
r1=4
r2=10
r3=1
I hope this explains it...the numbers are 'random' but only within a single execution (if it is not seeded via srand()) because it uses that same value as the seed value each time. Seeding it with something out of time() is the best way we can seed it with a different seed each time.
11. Originally posted by biz
Code:
```#include <time.h>
#include <stdlib.h>
...
srand(time(NULL));
x=(rand()%(N+1)); // Generate random # between 0 and N
x=(rand()%(end-start+1)+start); // Generate random # between start and end
...```
Hope this helps...
** edit..fixed typo
Sorry, that didnt help much. What does srand() and time(NULL) do? It would be great if i just knew what those functions do.
12. srand() seeds rand
time(NULL) seeds ran with time
sorry that's the best i can do, someone else can explain better
13. rand() produces a series of numbers based on a starting number (it uses the previously generated number to generate the next number). You can change this starting number using srand(). If the starting number is always the same you'll get the same series of numbers each time. If you seed it with a different number (which time() gives you) you'll get a different series each occurance (assuming you're not seeding rand() at exactly the same time). | 832 | 3,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-40 | latest | en | 0.913357 |
https://www.coursehero.com/file/6283854/chapter7/ | 1,512,965,118,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00355.warc.gz | 711,912,605 | 127,958 | chapter7 - Approximations to Probability Distributions...
This preview shows pages 1–5. Sign up to view the full content.
Approximations to Probability Distributions: Limit Theorems
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Sequences of Random Variables Interested in behavior of functions of random variables such as means, variances, proportions For large samples, exact distributions can be difficult/impossible to obtain Limit Theorems can be used to obtain properties of estimators as the sample sizes tend to infinity Convergence in Probability – Limit of an estimator Convergence in Distribution – Limit of a CDF Central Limit Theorem – Large Sample Distribution of the Sample Mean of a Random Sample
Convergence in Probability The sequence of random variables, X 1 ,…,X n , is said to converge in probability to the constant c , if for every ε >0, Weak Law of Large Numbers (WLLN): Let X 1 ,…,X n be iid random variables with E(X i )= μ and V(X i )= σ 2 < . Then the sample mean converges in probability to μ : 1 ) | (| lim = - ε c X P n n ( 29 ( 29 n X X X P X P n i i n n n n n = = = - = - 1 where 1 lim or 0 lim μ
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
( 29 ( 29 ( 29 μ ε σ Prob 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 lim | | lim 1 | | 1 : Let 1 | | 1 ) | (| 1 1 ) | (| ) 1 ( 1 1 ) ( : Inequality s Chebyshev' 2200 = - = = = - = = = = = - - - - - + - = = = = n n X n n X X n X X n X X X X X X X X X n X n X k n k n k n k k n k k X P k k X P k k X P k k k X k P n n X V X E
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
Page1 / 15
chapter7 - Approximations to Probability Distributions...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 564 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-51 | latest | en | 0.837265 |
https://www.instructables.com/id/The-5-3D-Printer/ | 1,571,654,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00525.warc.gz | 948,029,898 | 23,484 | # The \$5 3D Printer
1,898
8
2
You're interested in 3D printing but you don't have a printer of your own because it's too expensive. You're in luck! This Instructable shows you how to build one for a measly \$5 (or less if you already have some parts).
The results might not be as good as a \$2,000.00 printer, but along the way you'll learn how 3D printers work (hint: they are not magic), gain an enhanced appreciation for what these machines do and maybe even decide to build a better one!
Here's what you'll need:
• A piece of graph paper
• A hot glue gun
• Hot glue sticks (several)
I'd also recommend printing out a hardcopy of the code provided in step 5, it's a lot easier to work with on paper.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Prepare the Build Platform
What is a build platform?
The \$5 3D printer uses a flat, mostly stationary table to construct the printed model. In some designs, the table is moved by motors and can be heated to prevent defects from forming in the model due to uncontrolled cooling.
For your build platform you'll need some graph paper with numbered lines. You can do this by hand or you can download and print a PDF of a pre-numbered graph paper from here:
Once you have the numbered graph paper, you'll want to affix it to something so it won't move around by accident. Taping it to the table works, but using something like a piece of cardboard will allow you to move it when you want to without having it wander around on its own.
Make sure you setup the build platform in a comfortable work area.
## Step 2: Select a Model
What is a model?
In this context, a model is an electronic file that contains instructions that describe a 3-dimensional object to print. Models can be designed in a number of ways - including 3D scanning. 3D scanning captures physical objects in three dimensions like a camera captures physical objects in two dimensional files.
The model is "sliced" into a file containing a language a 3D printer understands. This language is called Gcode, and we'll be looking at it more closely later. This Gcode is interpreted by the printer's control electronics and firmware, and translated into movement.
In the \$5 3D printer, you are the printer's electronics and firmware (no Arduino required)!
You'll be reading the model's Gcode directly, so I recommend starting with the simple model included in this Instructable. Of course if you're ambitious, you can select something more complex or create a design of your own. The key is that you have a Gcode file to work with.
As mentioned earlier, printing a hardcopy of the Gcode allows you to mark it up and having your computer go to sleep in the middle of the printing process is frustrating.
## Step 3: Select a Hot End and Extruder
What is a hot end? What is an extruder?
The printer's hot end is a nozzle which heats the material being printed (typically plastic, in our case glue) and deposits the material on the build platform. The extruder is what pushes the material into the hot end.
In the \$5 3D printer, the hot end is a hot glue gun, and the trigger is the extruder!
Any glue gun will work, but I recommend using an inexpensive low-temp gun. The smaller the nozzle the better, but it's really not that important. Also long glue sticks will make things easier (reloading frequently while printing can make things more difficult).
## Step 4: Prepare the Axis
What is an axis?
On a 3D printer, the axis refers to the mechanism that positions the hot end relative to the build platform. The \$5 printer uses an axis in which the hot end moves in all three axis (X, Y and Z) using the Cartesian coordinate system. Other printers move the hot end in the X and Z axis and move the build platform in the Y axis, or some combination of each.
In the \$5 3D printer, your arm is the axis!
Position yourself close to the print bed (the graph paper) in a position where you can comfortably reach the entire area of the bed with the nozzle of the glue gun pointed down. Make sure you can squeeze the trigger of the gun in any position of the bed, and and also while lifting the gun various heights as well.
Also make sure the cord (if your glue gun has one) isn't going to drag across the surface of the bed as you work.
## Step 5: Print!
Fire up the hot glue gun and get it to temperature. Once it's ready, begin "processing" the Gcode by reading the first instruction and then carrying out that action using the hot glue gun and the grid of the graph paper. It may be easier to mark the start and end points for each line with a pen before laying down the glue.
Here's how to read a line of Gcode:
`G1 X57.81 Y13.13 E262.76105`
The Gcode starts with G0 or G1. There is a difference, but for the \$5 printer hey both mean go.
Next is X, Y and Z. These tell you where to go on the graph paper. This is kind of like a combination of Battleship and connect-the-dots.
Finally there is E. E tells the printer to extrude. Using the \$5 printer, that means pulling the trigger of the glue gun to squeeze out some glue as you move from one point to the next. If there's no E, that means move without laying down any material.
There are many other Gcode instructions available, you can learn more about them from Wikipedia.
To execute this line of Gcode, you would start with the X axis and find the 57th line (you can round these numbers, we are not as precise as robots) along the long side of your graph paper. Then, you would find the 13th line on the short side (the Y axis). Where these two lines intersect is where you'll position the nozzle of your hot end (the hot glue gun). Since there is an E on this sample line of Gcode, we would extrude material on the bed as we move from the current location to the destination the code specifies.
Now that you know how to read and execute Gcode, you can start printing the example model by executing each line below one at a time. If the command includes an E value, be sure to extrude glue as you move from the previous to the next location.
I find it's helpful to print off the Gcode so you can cross-off or check each line as you go so you don't loose your place.
```G0 F6000 X6.13 Y32.47
G1 F900 X38.47 Y32.47 E240.02457
G1 X57.81 Y13.13 E262.76105
G1 X59.93 Y15.25 E265.25577
G1 X43.95 Y31.23 E284.04527
G1 X44.46 Y32.47 E285.15692
G1 X59.13 Y32.47 E297.35341
G1 X59.13 Y45.47 E308.16296
G1 X55.36 Y45.47 E311.29939
G1 X57.56 Y44.40 E313.33332
G1 X58.16 Y41.77 E315.57717
G1 X56.48 Y39.66 E317.82088
G1 X53.78 Y39.66 E320.06594
G1 X52.46 Y41.32 E321.83098
G1 X52.10 Y41.77 E322.30757
G1 X52.57 Y43.85 E324.08486
G1 X52.70 Y44.41 E324.55722
G1 X54.90 Y45.47 E326.58743
G1 X44.46 Y45.47 E335.26833
G1 X43.95 Y46.70 E336.37998
G1 X59.93 Y62.68 E355.16948
G1 X57.81 Y64.80 E357.66361
G1 X38.47 Y45.47 E380.39950
G1 X6.13 Y45.47 E407.29281
G1 X6.13 Y32.47 E418.10236
G0 F6000 X44.11 Y38.96
G1 F900 X44.51 Y40.98 E419.81559
G1 X44.57 Y41.27 E420.06281
G1 X45.71 Y42.97 E421.76431
G1 X45.88 Y43.22 E422.01245
G1 X47.58 Y44.36 E423.71810
G1 X47.83 Y44.53 E423.96440
G1 X49.84 Y44.93 E425.67094
G1 X51.13 Y45.20 E426.76747
G1 X51.13 Y32.75 E437.11802
G1 X47.83 Y33.40 E439.91812
G1 X46.12 Y34.54 E441.62492
G1 X45.87 Y34.71 E441.87307
G1 X44.73 Y36.41 E443.57756
G1 X44.57 Y36.66 E443.82386
G1 X44.17 Y38.66 E445.52356
G1 X44.11 Y38.96 E445.77404
G0 F6000 X52.57 Y36.85
G1 F900 X52.70 Y37.41 E446.24640
G1 X55.13 Y38.58 E448.48710
G1 X57.56 Y37.40 E450.73152
G1 X58.16 Y34.77 E452.97537
G1 X56.48 Y32.66 E455.21908
G1 X53.78 Y32.66 E457.46414
G1 X52.46 Y34.32 E459.22919
G1 X52.10 Y34.77 E459.70577
G1 X52.57 Y36.85 E461.48307
G0 F6000 X46.27 Y33.54
```
This completes the outline of the first layer as shown in the third photo. This outline of the layer is sometimes called the wall. Once the wall is complete, material is laid down to fill in the space inside the wall. This material is called infill.
Infill usually takes the form of diagonal lines within the wall of the layer (you can see an example of this in the fourth photo). Infill can take other forms as well, but we'll stick with diagonal lines to keep things simple. The number of lines can also vary, more lines make the part stronger, whereas fewer lines make it lighter, uses less material and may also print faster.
As you can imagine, it takes a lot of lines of Gcode to describe the infill, and as such I've left them out of this example. Instead, I leave it to you to draw the horizontal lines within the wall as shown in the fourth photo. Repeat this process until the layer is completely filled, being sure to leave the mouth and eyes open as seen in the fifth photo.
Once the infill is complete, a line of Gcode typically instructs the printer to move the Z axis up and start the next layer. In our example print, every layer is identical and since we cannot be as precise as a robot, we start on top of the last layer, laying down an identical pattern. Repeat this process 10 times to complete the model.
As you can see in the last photo, I ran out red and had to change colors part-way through the print, a technique used by 3D printers to print multicolored or multi-material models.
Once you've laid-down 10 layers, the model is complete!
## Step 6: Results
Even if you're not impressed with the output of the \$5 3D Printer, you should now have an understanding of how 3D printers turn digital models into physical objects (as well as a new level of appreciation for the robots that print these models).
You've also learned about the major components of a deposition-type of 3D printer (build platform, controller/firmware, axis, extruder & hot end), and got a taste of the native language these printers speak (Gcode). There are other types of 3D printers that share some of these components, and many machines that understand the Gcode. You can apply what you've learned here to further understand how these other machines work and what areas of 3D technologies you'd like to learn more about.
This Instructable is based on an exercise included in a book I'm writing on 3D printing. If you enjoyed this project and would like to learn more about 3D printing, please visit https://leanpub.com/3dprintingandreprap and sign-up to be notified when the book is published.
Participated in the
3D Printing Contest 2016
## Recommendations
• ### Mold Making & Casting Class
22,755 Enrolled
## 2 Discussions
Good pont-maker. Not the real meat and potatoes thing, like I bought stuff for, but good idea btw!
Nice explanation
I see the point of the instructable, not as an actual printer more of an expalnation. | 2,950 | 10,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.954668 |
http://slidegur.com/doc/40735/projections-and-coordinates | 1,579,910,189,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00123.warc.gz | 150,693,352 | 17,509 | ### Projections and Coordinates
```Grid Systems
Transverse Mercator Projection
Universal Transverse Mercator
South Dakota
Zone 16
• Based on Transverse Mercator (cylindrical) projection
• World divided into 60 zones 6 degrees wide
• Distortion is minimal within each zone
• Maps of different areas use best zone
Copyright
© 2009for
by Maribeth
• Best
maps covering small3-3area in one zone
H. Price
UTM Zones
UTM Pole to Pole
Halfway to the Pole
Using UTM
Using UTM
Using UTM
Three Kinds of North
• True North – Along Meridians
• Magnetic North – What a compass detects
– Important in field
– Not important in most GIS
• Grid North – Along N-S grid lines
– Minor importance for compass work
– Be aware there’s a difference!
– Military uses exclusively
Cautions About UTM
• Military maps use two letter codes for each
100-km square, but maps will have
information to enable conventional UTM
• With military grid references, number of digits
indicates level of precision
• Older maps will sometimes have obsolete
grids
• Datum, datum, datum!
Military Grid System
•
•
•
•
•
•
45N 89 W = E 342,369, N 4,984,896 Zone 16T
Digraph = CQ
1 km accuracy = 16T CQ 42 84
100 m accuracy = 16T CQ 423 848
10 m accuracy = 16T CQ 4236 8489
1 m accuracy = 16T CQ 42369 84896
• Ticks on USGS topo map = 342,000, 4,984,000
Where Zones Meet
Why a Grid?
Latitude /Longitude
Grid System
North varies from place to
place on the map
Grid north is always the same
direction
Angular units differ in scale
between N-S and E-W
Grid scale the same in all
directions
E-W angular units vary in scale
with latitude
Grid squares are always the
same size and shape
Hexadecimal Scale
Decimal Scale
State Plane System
• States divided into
one or more
zones identified
by a unique FIPS
number
• Uses several types
of projections
• E-W zones
generally Conic,
N-S zones
generally UTM
Copyright © 2009 by Maribeth
H. Price
3-15
Projections for large scale maps
• Local, city, county maps, smaller states
– Projection systems virtually eliminate distortion
– Choose appropriate UTM or State Plane zone
Copyright © 2009 by Maribeth
H. Price
3-16
Wisconsin
State
Plane
Zones
Wisconsin
Grid
Systems
Projections for smaller scales
• Distortion is inevitable, so
purpose drives the choice
– Equidistant maps when
distances are important
– Equal area maps when areas
are important
– Conformal or compromise
projections for general
purpose maps
Copyright © 2009 by Maribeth
H. Price
Coordinate system names generally
indicate the locale and purpose it is
optimized for. Use for clues to choice.
3-19
Metes and Bounds System
Public Land Survey System
• AKA Congressional System
• Established 1785
• Does not apply to:
– 13 Original Colonies
– Derivative States (VT, KY, TN, ME, WV)
– Texas (Former independent country)
– Hawaii (Uses Kingdom of Hawaii system)
• Land Division
– 6 x 6 mile townships, 36 sections, quarter sections
USA Public Land Surveys
NonCongressional
Grids
Before
Greenwich
Arbitrary Geography
More Arbitrary
Geography
Grid vs. No Grid
Gridded Landscape
Where Wisconsin’s Grid Starts
Where California’s Grid Starts
Section Numbering
Wisconsin
Townships
Township Labels
Section
Descriptions
Limitations of
the
Congressional
Land Survey
System
Congressional Land Survey System
• Not an accurate grid!
• Locations within subdivisions may be
imprecise
• Data points used in GIS may be tied to system
• Authoritative surveys are forever
– French strips in LA, MO, WI
– Spanish and Mexican land grants
French Long Lot System
French Long Lot System
California: PLSS and Land Grants
California: PLSS and Land Grants
Oh, Canada
• Eastern Canada: Metes and Bounds, Long Lot
• Northern Ontario: 6 and 10-mile townships
• Western Canada: Dominion Land Survey
–
–
–
–
–
–
–
Modeled on PLSS
6-mile townships
Townships numbered N-S with Arabic numerals
Ranges numbered E-W with Arabic or Roman numerals
Road allowances between sections
Sections zigzag from 1 in SE to 36 in NE (Opposite US)
¼-1/4 sections numbered from 1 in SE to 16 in NE
Dangers of Cheap Work
Missed It By That Much
``` | 1,111 | 4,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-05 | latest | en | 0.769848 |
http://www.absoluteastronomy.com/topics/Gaussian_function | 1,719,191,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00584.warc.gz | 31,842,746 | 4,007 | Gaussian function
Overview
In mathematics
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...
, a Gaussian function (named after Johann Carl Friedrich Gauss
Carl Friedrich Gauss
Johann Carl Friedrich Gauss was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.Sometimes referred to as the Princeps mathematicorum...
) is a function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...
of the form:
for some real
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
constants a, b, c > 0, and e ≈ 2.718281828 (Euler's number
E (mathematical constant)
The mathematical constant ' is the unique real number such that the value of the derivative of the function at the point is equal to 1. The function so defined is called the exponential function, and its inverse is the natural logarithm, or logarithm to base...
).
The graph
Graph of a function
In mathematics, the graph of a function f is the collection of all ordered pairs . In particular, if x is a real number, graph means the graphical representation of this collection, in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. Graphing on a Cartesian plane is...
of a Gaussian is a characteristic symmetric "bell curve" shape that quickly falls off towards plus/minus infinity. | 437 | 2,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.89243 |
https://casino-best.top/craps/how-to-play-craps-place-bets.php | 1,601,066,864,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228707.44/warc/CC-MAIN-20200925182046-20200925212046-00388.warc.gz | 310,238,753 | 11,499 | 4692
How to play craps place bets
Sep 21, · The player may bet on any of these numbers, and if it is rolled before a seven, the bet wins. Place bets are just like odds bets, except no pass line bet is required, and don't pay as much odds. To be specific, place bets on the 6 and 8 pay 7 to 6, the 5 and 9 pay 7 to 5, and the 4 and 10 pay 9 to 5. I'm very proud of my play for free craps. Normally, Place bets are considered off on the come-out, but you want them working because the shooter is rolling lots of numbers. You tell the dealer, “I want all my Place bets working on the come-out.” The dealer puts an ON button on one of your Place bets to indicate that they’re all .
Craps Place Bets
Your Place bet chips remain on the table. Since everyone roots for a 7 to show on the come-out for Pass Line bets, the casino automatically turns off all the Place bets for the come-out roll. SevenOut says:. The shooter rolls a When you win a place bet the dealer will push you your winnings but the original bet stays where it is and keeps winning if the number keeps rolling.
• All casinos are built on the principle of the most tangled labyrinths. In the gaming halls there are no direct passages. Because the longer a person walks along the aisles, the more likely that he will play in the meeting slot machines and gambling.
• The annual profit from the gaming industry in the US is 18 billion dollars.
Y You can make a Place bet on any of the point numbers, which are the 4, 5, 6, 8, 9, and As with most other bets on the table, the Place bet plays against the 7. Therefore, once you make a Place bet, the only numbers that mean anything in terms of winning or losing are the number you bet on and the 7.
After making a Place bet, the outcome of each roll is either: 1 the Place number appears and you win; 2 a 7 appears and you lose; or 3 any other number appears, which means nothing in terms of winning or losing. After the come-out roll, the button is no longer needed so the dealer removes it from your chips and your Place bet remains on and working until it wins or loses, or until you turn it off again. The house advantage for Place bets is relatively low, which makes it one of the better bets on the table we discuss good versus bad bets in another article.
If you need to review our other article about basic craps math, now is a good time to do it. Number Pairings Place Odds Payoff Odds True Odds 4 and 10 or 5 and 9 or or 6 and 8 or or Because odds are like fractions, I adjusted all of the odds except the odds on the 4 and 10 to equivalent odds so you can easily compare them to the true odds.
This gives you a better idea of how small the house advantage is for these bets. What this tells us is that the Place bet on the 6 or 8 is one of the smartest bets on the craps table i. The bet multiple depends on the Place odds. Answer the following three questions. Your wife is nagging you to go to the lounge for drinks and dancing, so you decide to blow it on a Place 4 bet. If you make the maximum Place bet amount in question 2 and you win, how much do you win?
View Details
The pass line bet with odds is the best craps bet you can make because it has the lowest house edge. However, players can also wager on the numbers 4, 5, 6, 8, 9, or 10 at any time with a Place Bet. Place bets are very valuable when a shooter is rolling many numbers, as you'll get paid each time the number repeats. To make a place bet you put your chips in front of you on the layout and tell the inside dealer which numbers you want to cover.
The dealer will then move your chips to the numbered box you want to wager on. Place bets are not self-service bets. The dealer keeps track of which bet belongs to which player by placing them strategically inside the box. After you place a number it must roll before a seven appears in order to win. When the shooter sevens-out, you lose all of your place bets. When you win a place bet the dealer will push you your winnings but the original bet stays where it is and keeps winning if the number keeps rolling.
Your original bet will stay on the place number until the shooter sevens out or until you ask the dealer to take your bet down. You can take down your place bets anytime you want to. If the shooter makes their point the place bet will be turned-off for the come-out roll. Want it to work win or lose on the come-out?
You've got to tell the dealer before the shooter releases the dice! The house edge on the 6 and 8 is only 1. The house edge on the 5 and 9 is 4-percent. These numbers require bets in multiples of five dollars. The house edge on the 4 and 10 is 6. A Buy Bet is very similar to a place bet, but with slightly better odds for the player in the 4 and Since the minimum chip value at the casino is a dollar you will be paying at least that much. This is because you are being paid 2 to 1 instead of 9 to 5.
Some casinos only charge you the commission on buy bets when you win. Ask the dealer at the table if this is their policy. If you have a choice of casinos to play in, always choose the one that only charges a commission on winning buy bets.
Tips to win big in craps
Most dealers probably prefer that you tip as-you-go, becauseyou might bust out and not have any money left at the end of your session to tipwith, but this is not to your advantage, of course. The dealers will notice your early bets for them and make sure you get goodservice, and by delaying future tipping to the end of the session, you can avoida situation where you tipped heavily during a hot streak and then hit a bad streak,with the dealers winding up with more money than you.
If you take or lay odds in their behalf you will be appreciated by the dealers evenmore. Dealers cringe when players make bad bets for them, because such bets waste a goodchunk of the tip. Such wagers discouragethe dealers as much as the players. You willbe surprised at how quickly the dealers snap to this because you have just toldthem that you and they are going to parlay this wager if it hits.
Video
Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,456 | 6,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.941553 |
http://forum.alchemyforums.com/archive/index.php/t-1616.html?s=3571bf1b2c387713b4b2e6e7adc989a6 | 1,537,610,284,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158279.11/warc/CC-MAIN-20180922084059-20180922104459-00167.warc.gz | 89,353,343 | 8,077 | PDA
View Full Version : Hyperdimentions
Ghislain
06-04-2010, 12:07 AM
Hi
In an earlier post by Aleilius he mentioned Hyperdimensional physics.
here
What is Hyperdimensionality or Hyperdimensional physics?
Is it a science and if so are there theories and if so what are they?
I have been trying to find an explanation for this in layman’s terms, but unfortunately unsuccessfully.
I once saw a YouTube clip about 2 dimensions (http://www.youtube.com/watch?v=BWyTxCsIXE4&feature=related)...is this an extrapolation of that idea and if so is it all theory or are there some facts?
Thanks
Ghislain
Edit: In the clip the circle came into the three dimentional world and saw a whole new perspective.
What do you think we might see if we moved into a fourth?
Aleilius
06-04-2010, 01:39 AM
Hey, I believe RCH explains hyperdimensional physics very well. Although he kind of paints a picture of his own version, but for the most part, it explains what one needs to know to gain a general understanding of this subject.
Take a look at the link here: http://www.enterprisemission.com/hyper1.html.
This interview is interesting:
http://www.dennishollingsworth.us/archives/image/SaturnNASA_AP_468x397.jpg
Now ask yourself, what is really happening here? Why, and how is that hexagon/hexagram manifesting? This cannot be explained very easily without the use of hyperdimensional physics, and the geometrical nature of the building blocks of matter.
Next we can get into nuclear & quantum geometry:
http://blazelabs.com/magicnumbers.pdf
http://blazelabs.com/f-p-develop.asp
http://www.quantum-geometry.com/background.asp
vega33
06-04-2010, 02:29 AM
Now ask yourself, what is really happening here? Why, and how is that hexagon/hexagram manifesting? This cannot be explained very easily without the use of hyperdimensional physics, and the geometrical nature of the building blocks of matter.
Lots of things form hexagonal shapes, I don't think theres any need to invoke extra dimensions to explain it, at least in the physical sense. Think about snowflakes and the hexagonal fractals they exhibit merely as a result of the coherence of their hydrogen bonding matrix.
Aleilius
06-04-2010, 02:48 AM
Lots of things form hexagonal shapes, I don't think theres any need to invoke extra dimensions to explain it, at least in the physical sense. Think about snowflakes and the hexagonal fractals they exhibit merely as a result of the coherence of their hydrogen bonding matrix.
I cannot prove it to you, but it is my personal belief that this is the case. This hexagonal hyperdimensional signature isn't limited to the "coherence of their hydrogen bonding matrix." This can be seen throughout nature & the universe. Actually, I would say that the hexagonal formation of the coherent hydrogen bonding matrix is the direct result of hyperdimensional physics.
Did you read all the links I posted? Did you listen to the interviews I posted?
I also recommend listening to this interview (there are 8 parts): http://www.youtube.com/watch?v=kzid6jVGuCo&feature=related (http://www.youtube.com/watch?v=kzid6jVGuCo&feature=related)
Why do beehives utilize the hexagon? Ever thought about it? Sure, there could be a mundane answer, but there could also be an occult answer.
Source: http://www.halexandria.org/dward118.htm
Nevertheless, <http://www.enterprisemission.com/hyper1.html> is worth reviewing in detail (including its some five or six detailed, elaborate webpages). Hoagland notes, among many other things, that the anomalous energy being radiated by the giant planets of Jupiter, Saturn, Uranus, and Neptune can be explained by Hyperdimensional Physics. In essence, these planets’ energy output is “over unity”, i.e. they are giving off more energy than is being absorbed from the Sun energy impinging upon them. Furthermore, when Uranus and Neptune are “normalized” (i.e. their different distances from the Sun are taken into account), these two planets are roughly equal in their output. Hoagland then explains that all of this can be accounted for if we assume:
In one of the interviews it was mentioned that at the very center of Saturn energy is manifesting from another dimension.
http://www.gogeometry.com/geometry/hexagon_nature_perfect_shape.htm
http://threesixty360.wordpress.com/2008/04/10/hexagons-in-nature-the-giants-causeway/
Consider benzene, and how it is often the foundation of organic molecules.
Also consider the "star" formation that forms in the antimony regulus process. Why is it forming? What is the significance? Sure, purity is part of the answer, but it cannot merely be easily explained to any satisfactory degree.
"The term 'starred' was here employed by Newton in its most literal sense. For if the antimony has been properly purified as in this instance, it forms long and slender crystals. During cooling the crystals in turn form triangular branches around a central point, taking on the aspect of a silver star." - http://www.themystica.com/mystica/articles/n/newton.html
"... cause when life is awakened you get a different level of antimony, from here on it crystallizes according to this level. It's the hexagonal (form) and it's the star you can see in the Regulus." - http://www.alchemywebsite.com/pon-02.html
Aleilius
06-04-2010, 04:41 AM
This is one of the supposed nuclear structures for gold:
Doesn't the supposed nuclear structure for gold remind you of something? Perhaps this:
http://www.vitriolum.net/images/aleilius/hyperdimensionalhexagram.jpg
vega33
06-04-2010, 05:48 AM
I cannot prove it to you, but it is my personal belief that this is the case. This hexagonal hyperdimensional signature isn't limited to the "coherence of their hydrogen bonding matrix." This can be seen throughout nature & the universe. Actually, I would say that the hexagonal formation of the coherent hydrogen bonding matrix is the direct result of hyperdimensional physics.
Did you read all the links I posted? Did you listen to the interviews I posted?
I used to be a Hoagland fan back in the day, heard him on Art Bell a lot, went to his website, saw the Mars mission stuff when it first came out. Its entertaining stuff and easy to go along with... whenever something is presented with a conspiracy angle it tends to absolve one from critical thinking (note I'm not saying you haven't thought his work through here, only that this tends to be the usual way people approach his work).
The thing is though, multiple dimensions beyond 3 or 4 are a mathematical abstraction. You can for instance distinguish out to in as a dimension, or a continuum between spirit and matter as a dimension, any continuum that represents a quality or quantity/location can be mapped onto a dimension. However, for the purposes of normal discussion, most people tend to look at dimension as a spatial characteristic, and to map a non spatial continuum as if it were a spatial continuum often merely creates confusion.
In one of the interviews it was mentioned that at the very center of Saturn energy is manifesting from another dimension.
OK, how do you define that dimension? What is the spatial movement or alteration of an object in space time you believe is happening to create the hexagon?
Also consider the "star" formation that forms in the antimony regulus process. Why is it forming? What is the significance? Sure, purity is part of the answer, but it cannot merely be easily explained to any satisfactory degree.
"The term 'starred' was here employed by Newton in its most literal sense. For if the antimony has been properly purified as in this instance, it forms long and slender crystals. During cooling the crystals in turn form triangular branches around a central point, taking on the aspect of a silver star." - http://www.themystica.com/mystica/articles/n/newton.html
"... cause when life is awakened you get a different level of antimony, from here on it crystallizes according to this level. It's the hexagonal (form) and it's the star you can see in the Regulus." - http://www.alchemywebsite.com/pon-02.html
Sure, there is sacred geometrical significance to such chemical events. In my experience though, when higher dimensions are used in mathematics, its merely a way of getting out of answering the question of what exactly is going on. You can create a mathematical abstraction in 10 dimensions to explain reality, but the real question is what do these 10 continuums represent, and how do they interrelate? In mathematics, imaginary numbers, quaternions etc, all have their own methods of multiplication and addition, etc, ways they transform one another. But if an n-dimensional model is to have any usefulness to us, we must frame it in terms of our reality of 3 dimensions of space and one dimension of time, correlated with whatever the other continuums such as energy represent. Using a mathematical abstraction often merely confuses what is being described. In facr if I was to posit a conspiracy in science that has evolved in the past century or so, it would be the use of complex mathematics to mask the elegantly simple nature of the universe from the common man, and to place the power of manipulating society in the hands of the obscure few who can speak this priestly language..
Aleilius
06-04-2010, 05:59 AM
Hi vega33, you bring up too many good points for me to mention. I'm afraid that I do not know the answers. These are the great questions that eludes man, or at the very least: I am still thinking. :)
Ghislain
06-04-2010, 02:36 PM
The clips sound like something created for a shopping channel selling RCH’s CD’s, presentations and
web page.
If there is information out there and it is believed to be repressed I would need to know why would
anyone want to suppress it? Also how would they manage this in a world so full of free
communication? One would think that decent provable information would be leaked...each of us have
their own hidden agendas. There has to be one person in every group that believes the public
is better served with full information. Or am I a little naive? :(
The only reason I could imagine NASSA would have for not publishing photographic material is that
other countries would get this info for free without the expense of the space missions. What other
reason might there be? That seems a fair reason too...perhaps sell the info then release it.
manager to close the window. I tried the link numerous times, but no joy.
The link to ‘ Earthlink.net’ was information overload and as I am not a chemist or nuclear physicist I
could not understand the relevance of what was being proffered. I tried to read the last few
paragraphs hoping it would be a conclusion to previous text and it mentioned "static nuclear
theory" which when searched for on the net brought about just one result...Earthlink.net
I have a problem when information only comes from one source. Anyone with a fair knowledge of
an obscure science can write a convincing article that can be completely free from any reality. Many
witch doctors and puffers used similar practices.
In the article about the hexagon on Saturn’s North Pole RCH states that the hexagon phenomenon is
not present in the South Pole. Below are links to possible explanations.
Explanation of events at Saturn’s South Pole. (http://www.holoscience.com/news.php?article=4egjus1n&pf=YES)
Article ends with, “The primary energy source in the universe has been overlooked.”
Explanation of events at Saturn’s North Pole. (http://news.sciencemag.org/sciencenow/2010/04/saturns-strange-hexagon-recreate.html)
Article ends with,“Although the lab experiment does not explain what force is driving this particular jet stream”
Both articles are devoid of a hard fact explanation to the source of these phenomena, but though there is no explanation
of the source driving these particular phenomena it takes a large leap of faith to take it into another dimension when there
is no explanation of what this other dimension even consists of.
most people tend to look at dimension as a spatial characteristic
Could a fourth dimension be a direction that connects other separate three dimensional worlds?
How would the ‘circle’, in the first post YouTube link, explain the third dimension without experiencing it?
Did radiation not detectable by human senses alone only come into existence when it was physically proved?
Is Hyperdymensional Physics just a makeshift explanation of later to be found mathematical facts
within the third dimensional world?
Why do beehives utilize the hexagon?
It may be to do with it being an optimal shape for strength and efficiency.
As difficult as they were to follow, thanks for the links Aleilius. Is food for thought
Ghislain
Edit: I have read and re-read many articles on Quantum Physics and still have just above zero understanding of it.
If you understand this Aleilius I have to take my hat off to you.
vega33
06-04-2010, 08:50 PM
Explanation of events at Saturn’s South Pole. (http://www.holoscience.com/news.php?article=4egjus1n&pf=YES)
Article ends with, “The primary energy source in the universe has been overlooked.”
This would link in with my personal pet theory. Like Leedskalnin who disbelieved in the existence of the electron, and Ighina who thought that scientists have an incorrect picture of the atom, both of them had a similar reasoning for their rejection: observation of such negatively entropic systems by techniques of bombardment perturbs them, alters their shape temporarily. You don't observe a fast moving object by speeding it up, why then would you bombard an atom with light or magnetic fields in an attempt to analyze it? Leedskalnin in fact claimed that the electron was an abnormal condition of the atom. Ighina tried to observe the atom by slowing down its level of vibration, by surrounding it by canals of other atoms which absorbed different frequencies of light. Light bombarding a body will tend to excite it.
The idea can be generalized into something along the lines of Heisenberg's uncertainty theory... observation often demands interaction in the case of things so small, and what you will see is the interaction of the object with your machine, not the object in its normal state. Similar case with other negatively entropic systems where there is tight communication between apparent parts. We can go so far with invasive tools to analyze a person's condition - barium meals, x-rays, MRIs etc... but what we end up with is like a photographic negative, a mirror image, and one that doesn't take into account those aspects that can't be measured well without disrupting them (such as biophotons, communication between cells/organs). This doesn't mean the unmeasurable aspects aren't there, but it does mean there are things about life that may be measurable only by extremely subtle methods, or through inference of conclusions from experiment. Perhaps these "unmeasurable aspects" are what alchemy tries to discover.
The purpose of this long statement is to try to explain that where sciences like astronomy/cosmology come up with ideas like "dark matter" to satisfy their mass equations, there may be hidden forces or hidden variables at work, things not taken into account, inconsistencies in the equations. We often assume modern physics to be infallible because it allows us to run our refrigerator, our tv and our computer without problems, but being able to model something mathematically and predict results doesn't mean direct experience or understanding of the thing-in-itself. So how does one measure things like orgone energy? How does one measure whats going on at Saturn's poles? Well, you look for what you missed in your lab model. Ask what is present in space that isn't present in the lab. The answer may be as simple as taking the planet's mass/rotation into account, or the ionic winds from the sun, or some other energy.source that its capturing.
I believe RCH pointed this out when he looked at the inconsistencies with Jupiter seemingly putting out more energy than it got in. Unfortunately, his answer was energy coming from other dimensions, not that our science had overlooked something or might not be entirely correct. My own answer would involve negatively entropic, formative forces, which occur at the poles because vortices, rotation tends to create or channel form. I'd suspect that answering the question of what causes such bodies to rotate would answer what causes the hexagon. | 3,617 | 16,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-39 | longest | en | 0.917447 |
https://fr.mathworks.com/matlabcentral/cody/problems/1036-cell-counting-how-many-draws/solutions/1743843 | 1,569,176,517,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00286.warc.gz | 464,287,702 | 15,521 | Cody
# Problem 1036. Cell Counting: How Many Draws?
Solution 1743843
Submitted on 6 Mar 2019 by malues
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
games = {'D','D','A','H','D','H'}; draws = 3; assert(isequal(how_many_draws(games),draws))
2 Pass
games = {'D','D'}; draws = 2; assert(isequal(how_many_draws(games),draws))
3 Pass
games = {'H','H','A'}; draws = 0; assert(isequal(how_many_draws(games),draws))
4 Pass
games = {'D','H','H','A','D','H','H','A','D','H','H','A','D','D'}; draws = 5; assert(isequal(how_many_draws(games),draws)) | 232 | 671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.776888 |
http://situspokerkita.com/the-odds-of-winning-a-lottery-3/ | 1,721,742,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00745.warc.gz | 27,529,989 | 10,471 | Posted on
The Odds of Winning a Lottery
Lottery is a type of gambling in which people buy tickets and hope to win a prize. The prizes can range from cash to goods or services. The winners are chosen through a random selection process. The odds of winning are very low, but people still buy tickets. In the United States, state-run lotteries raise more than \$80 billion a year. Many of the proceeds are distributed to education, public welfare, and other programs. Some states allocate the profits to specific projects, while others reinvest the money into future lottery draws.
The term lottery is derived from the Dutch word lot meaning “fate” or “fateful.” The earliest known lotteries date back to the Chinese Han Dynasty between 205 and 187 BC. Modern lotteries are often run by governments or private organizations and are a popular form of taxation. Historically, they have raised funds for public buildings and works of art as well as military campaigns.
In colonial America, lotteries were a common source of funding for private and public ventures, including roads, canals, bridges, schools, churches, colleges, and other public institutions. Lotteries also played a significant role in financing the American Revolutionary War, and the founding of Princeton and Columbia Universities were financed by lotteries.
While playing the lottery is not a wise or rational decision from a purely financial point of view, it can be fun for some individuals. The entertainment value or other non-monetary benefits can offset the disutility of a monetary loss, and therefore making the purchase of a lottery ticket may be an acceptable risk.
The odds of winning the lottery are determined by a combination of probability and demand for the tickets. Some of the demand comes from the desire to be a part of history by being one of the few who have won the jackpot, but a majority of it is simply due to chance. The likelihood of winning is calculated using the probability formula p(n, k) / (n – k). The k in this formula is the number of balls drawn. A higher k value means the probability of drawing a certain ball is lower.
A key to the success of any lottery is the ability to balance the odds and frequency of winning with the cost of organization and promotion. The prizes must be large enough to attract players, but not so large that they prevent the lottery from generating revenue and profits. In addition, a percentage must be deducted to cover expenses and pay the profits to the state or sponsor.
Organizing the lottery involves selecting and licensing retailers, training employees to use lottery terminals, promoting the lottery, and distributing promotional materials. In addition to these duties, the state must also ensure that the lottery is conducted fairly and impartially. It must also select and distribute high-tier prizes, pay winners, and administer and enforce state lottery law.
Lottery winnings can be received in either a lump sum or an annuity payment. The choice of which option is best depends on your financial goals and the applicable rules of each state. In general, a lump sum is better for short-term spending needs, while an annuity offers a steady stream of income over time. | 639 | 3,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.97332 |
http://www.learningelectronics.net/vol_6/chpt_6/7.html | 1,521,655,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647681.81/warc/CC-MAIN-20180321180325-20180321200325-00541.warc.gz | 418,828,053 | 4,197 | 555 audio oscillator
PARTS AND MATERIALS
• Two 6 volt batteries
• One capacitor, 0.1 µF, non-polarized (Radio Shack catalog # 272-135)
• One 555 timer IC (Radio Shack catalog # 276-1723)
• Two light-emitting diodes (Radio Shack catalog # 276-026 or equivalent)
• One 1 MΩ resistor
• One 100 kΩ resistor
• Two 510 Ω resistors
• Audio detector with headphones
• Oscilloscope (recommended, but not necessary)
A oscilloscope would be useful in analyzing the waveforms produced by this circuit, but it is not essential. An audio detector is a very useful piece of test equipment for this experiment, especially if you don't have an oscilloscope.
CROSS-REFERENCES
Lessons In Electric Circuits, Volume 4, chapter 10: "Multivibrators"
LEARNING OBJECTIVES
• How to use the 555 timer as an astable multivibrator
• Working knowledge of duty cycle
SCHEMATIC DIAGRAM
ILLUSTRATION
INSTRUCTIONS
The "555" integrated circuit is a general-purpose timer useful for a variety of functions. In this experiment, we explore its use as an astable multivibrator, or oscillator. Connected to a capacitor and two resistors as shown, it will oscillate freely, driving the LEDs on and off with a square-wave output voltage.
This circuit works on the principle of alternately charging and discharging a capacitor. The 555 begins to discharge the capacitor by grounding the Disch terminal when the voltage detected by the Thresh terminal exceeds 2/3 the power supply voltage (Vcc). It stops discharging the capacitor when the voltage detected by the Trig terminal falls below 1/3 the power supply voltage. Thus, when both Thresh and Trig terminals are connected to the capacitor's positive terminal, the capacitor voltage will cycle between 1/3 and 2/3 power supply voltage in a "sawtooth" pattern.
During the charging cycle, the capacitor receives charging current through the series combination of the 1 MΩ and 100 kΩ resistors. As soon as the Disch terminal on the 555 timer goes to ground potential (a transistor inside the 555 connected between that terminal and ground turns on), the capacitor's discharging current only has to go through the 100 kΩ resistor. The result is an RC time constant that is much longer for charging than for discharging, resulting in a charging time greatly exceeding the discharging time.
The 555's Out terminal produces a square-wave voltage signal that is "high" (nearly Vcc) when the capacitor is charging, and "low" (nearly 0 volts) when the capacitor is discharging. This alternating high/low voltage signal drives the two LEDs in opposite modes: when one is on, the other will be off. Because the capacitor's charging and discharging times are unequal, the "high" and "low" times of the output's square-wave waveform will be unequal as well. This can be seen in the relative brightness of the two LEDs: one will be much brighter than the other, because it is on for a longer period of time during each cycle.
The equality or inequality between "high" and "low" times of a square wave is expressed as that wave's duty cycle. A square wave with a 50% duty cycle is perfectly symmetrical: its "high" time is precisely equal to its "low" time. A square wave that is "high" 10% of the time and "low" 90% of the time is said to have a 10% duty cycle. In this circuit, the output waveform has a "high" time exceeding the "low" time, resulting in a duty cycle greater than 50%.
Use the audio detector (or an oscilloscope) to investigate the different voltage waveforms produced by this circuit. Try different resistor values and/or capacitor values to see what effects they have on output frequency or charge/discharge times. | 848 | 3,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-13 | longest | en | 0.847336 |
https://www.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=1848 | 1,627,271,324,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00689.warc.gz | 786,063,720 | 13,365 | ### 1: Number and Algebra
#### 1.1: Number and place value
1.1.2: Recognise, represent and order numbers to at least tens of thousands
1.1.3: Apply place value to partition, rearrange and regroup numbers to at least tens of thousands to assist calculations and solve problems
1.1.4: Investigate number sequences involving multiples of 3, 4, 6, 7, 8, and 9
1.1.5: Recall multiplication facts up to 10 × 10 and related division facts
#### 1.2: Fractions and decimals
1.2.1: Investigate equivalent fractions used in contexts
1.2.2: Count by quarters halves and thirds, including with mixed numerals. Locate and represent these fractions on a number line
1.2.3: Recognise that the place value system can be extended to tenths and hundredths. Make connections between fractions and decimal notation
#### 1.4: Patterns and algebra
1.4.1: Explore and describe number patterns resulting from performing multiplication
1.4.2: Solve word problems by using number sentences involving multiplication or division where there is no remainder
1.4.3: Use equivalent number sentences involving addition and subtraction to find unknown quantities
### 2: Measurement and Geometry
#### 2.1: Using units of measurement
2.1.2: Compare objects using familiar metric units of area and volume
#### 2.2: Shape
2.2.2: Compare and describe two dimensional shapes that result from combining and splitting common shapes, with and without the use of digital technologies
### 3: Statistics and Probability
#### 3.2: Data representation and interpretation
3.2.2: Construct suitable data displays, with and without the use of digital technologies, from given or collected data. Include tables, column graphs and picture graphs where one picture can represent many data values
3.2.3: Evaluate the effectiveness of different displays in illustrating data features including variability
Correlation last revised: 9/16/2020
* Copyright 2010 National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. | 493 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.848289 |
https://www.ownmainerealstate.com/how-do-i-calculate-my-monthly-mortgage-payment/ | 1,675,946,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00435.warc.gz | 946,202,083 | 53,284 | # How Do I Calculate My Monthly Mortgage Payment
Do I Need To Get Preapproved For A Mortgage E4 Annual Income Navy Section 1 Termite clearance california termite Inspection Services For Escrow in Southern California – In the event that the purchase contract states that the seller will provide a termite report, most lenders will require a copy of the termite report and require section 1 (active infestation or infection) clearance prior to the closing of escrow.Military Times – Military Times Pay Center. From basic, to flight and COLA – we’ve got you covered. Below is everything you need to determine how much you’ll make in the military. current general pay calculator. Please select your paygrade, years in service, location and if you have dependents. This tool will show you the total of your BASIC, BAS, BAH and.Getting preapproved for a mortgage before you go home shopping isn’t required, but it is a good idea, especially in a seller’s market, where competition among buyers is intense.
. of your rent income on paying down principal on your mortgage – instead of holding interest-only debt or no debt at all – then you’re building wealth with every monthly debt payment. To calculate.
150 000 Salary How Much House How much should I make to get a \$450,000 house? – Quora – The third item will determine how much loan you can qualify for. Lenders use a number called "debt to income ratio" (DTI) to approve So if you want house worth \$450,000, your salary or income must be I am assuming that your annual salary is \$150,000. It is just an assumption to clear some things.
Stretching out payments over more years (up to 30) will generally result in lower monthly payments. The longer you take to pay off your mortgage, the higher the. not included in these figures.) You.
The monthly payment is \$599.55. Calculate the following values so that you can plug them into the payment formula: n = 360 (30 years times 12 monthly payments per year) i = .005 (6 percent annually expressed as .06, divided by 12 monthly payments per year (For more details, see how to convert percentages to decimal format)
Home Loan With No Credit February 7, 2017 – Some borrowers come to the FHA loan process with a long credit history, while other borrowers are just getting started. That leads to a very important question. Is it possible for a borrower be turned down for an FHA loan because of a lack of credit history? typical fha loan applicant questions in this area go something like.
Use this mortgage calculator to calculate your monthly mortgage payments quickly and easily. Enter your home location and the desired home price in.
Free mortgage payoff calculator to evaluate options and schedules to pay off a mortgage earlier, such as extra monthly payments, a one-time extra payment, a bi-weekly payment, or simply paying back the mortgage altogether. Also gain some understanding of the pros and cons of paying off a mortgage earlier, or explore many other calculators covering math, fitness, health, and more.
What can I do to lower my monthly mortgage payments?-Frank The leap into homeownership is. considered excellent and will make you the most attractive borrower. Calculate: How much home can your.
Transunion Maximum Credit Score Free Online Mortgage Calculator With Extra Payments Extra Payment Calculator | Is It the Right Thing to Do? – Extra payment calculator with payment schedule calculates interest savings due to one lump sum or multiple extra payments.. You’ll have a \$92,929 loan balance on the debt free date when no extra payments are made.. you can use this mortgage calculator and it will consider the various items.Points On Home Loan A mortgage point equals 1 percent of your total loan amount – for example, on a \$100,000 loan, one point would be \$1,000. Mortgage points are essentially a form of prepaid interest you can choose to pay up front in exchange for a lower interest rate and monthly payments (a practice known as "buying down" your interest rate).
You could calculate the payment using a quick online calculator, but if you want to see how all of the variables work together, you can do it by hand using the mortgage monthly payment formula.
Use SmartAsset's free mortgage loan calculator to find out your monthly payments.. Enter your details below to estimate your monthly mortgage payment with.
Loan To Income Ratio Mortgage king county conforming loan limits King County Conforming Loan Limit | Texasclerks – Washington Conforming, FHA & VA Loan Limits by County – FHFA sets conforming loan limits on a regional basis, by county within Washington. There are two types of conventional loans, conforming and non-conforming. Conforming loans are equal to or less than the published conforming loan limits.Your debt-to-income ratio (DTI) is a representation of your cash flow. It shows. A mortgage loan may be the largest loan you'll ever obtain. You should do as.
To calculate how much interest you’ll pay on a mortgage each month, you can use the monthly interest rate. Generally, you’ll find this by dividing your annual interest rate by 12. Then, multiply this by the amount of principal outstanding on the loan.
Everyone’s financial situation is different, and it may suit you to pay off your mortgage. But do keep in mind that they are not always unlimited. Sometimes home loans with an extra repayment. | 1,108 | 5,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-06 | latest | en | 0.922714 |
https://stats.stackexchange.com/questions/140330/multiple-linear-regression-does-bic-drop-vaguely-collinear-variables | 1,579,306,745,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591431.4/warc/CC-MAIN-20200117234621-20200118022621-00491.warc.gz | 682,064,364 | 30,613 | # Multiple linear regression: does BIC drop (vaguely) collinear variables?
Say I have the following multiple linear regression:
Y ~ X1 + X2 + X3 + X4
All X variables are independent, but X1 and X2 look kind of linearly related when plotted against one another (e.g., R-squared ~0.5, if I fit a simple linear regression to X1 ~ X2). Will BIC or AIC throw out one of the variables because this? How much collinearity is too much? Is there a formal procedure for identifying and dealing with such collinearity?
A2 & A3: There is no unique answer to this question. Most of the time we will look at Variance inflation factor i.e. $VIF_j=\dfrac{1}{1-R^2_{j}}$, where $R^2_{j}$ is the R-squared of the model with $X_j$ as the response over all other remaining independent variables. As a rule of thumb, if the $VIF>10$ then there is a serious multicollinearity problem. Although, in some text books they consider cut-off value of 5. | 249 | 930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.887849 |
https://b-ok.org/book/511310/a7c26b | 1,558,691,447,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00111.warc.gz | 393,156,210 | 42,194 | Main Elementary Number Theory with Applications
# Elementary Number Theory with Applications
Year: 2007
Edition: 2
Language: english
Pages: 801
ISBN 13: 978-0-12-372487-8
File: PDF, 8.10 MB
You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them.
1
### Instructor's Manual (0471726478) for Advanced Engineering Mathematics
Year: 2006
Language: english
File: PDF, 2.31 MB
2
### Basic Algebra [modern]
Year: 2007
Language: english
File: PDF, 5.86 MB
``` List of Symbols
Symbol
Z
x∈S
x∈
/S
Z+
N
W
a<b
a>b
a≤b
a≥b
min{x, y}
max{x, y}
|x|
x
x
i=m
m
m
ai =
ai = ai
i=k
i=k
k
ai
i∈I
aij
P
i=m
ai =
i=k
n!
n
r
tn
sn
pn
hn
Tn
Sn
Pn
Hn
a div b
a mod b
m
i=k
ai =
m
ai
Meaning
set of integers
x belongs to set S
x does not belong to set S
set of positive integers
set of positive integers
set of whole numbers
a is less than b
a is greater than b
a < b or a = b
a > b or a = b
the minimum of x and y
the maximum of x and y
the absolute value of x
the floor of the real number x
the ceiling of the real number x
ak + ak+1 + · · · + am
Page
(3)
(3)
(3)
(3)
(3)
(4)
(4)
(4)
(5)
(5)
(5)
(5)
(5)
(6)
(6)
(9)
the sum of the values of ai as i runs over the various values in I
(11)
the sum of the values of aij , where i and j satisfy properties P
(11)
ak ak+1 · · · am
(13)
n factorial
(13)
binomial coefficient
(33)
triangular number
square number
pentagonal number
hexagonal number
tetrahedral number
square pyramidal number
pentagonal pyramidal number
hexagonal pyramidal number
the quotient when a is divided by b
the remainder when a is divided by b
(40)
(44)
(46)
(48)
(49)
(50)
(51)
(51)
(71)
(71)
k
Symbol
a|b
ab
|A|
A∪B
A∩B
A
N = (ak ak−1 . . . a1 a0 )b
Rn
π(x)
Fn
Ln
|A|
fn
(a, b)
(a1 , a2 , . . . , an )
pa n
[a, b]
[a1 , a2 , . . . , an ]
a ≡ b (mod m)
a ≡ b (mod m)
[r]
a−1
ρ(n)
In
n#
ϕ(n)
τ (n)
σ (n)
Mp
μ(n)
λ(n)
ordm a
ψ(d)
indα a
(a/p)
(a/m)
(a/n)
Meaning
a is a factor of b
a is not a factor of b
the number of elements in set A
the union of sets A and B
the intersection of sets A and B
the complement of set A
base-b representation of N
repunit with n ones
the number of primes ≤ x
the nth Fibonacci number
the nth Lucas number
the determinant of matrix A
the nth Fermat number
the greatest common factor of a and b
the greatest common factor of a1 , a2 , . . . , and an
pa exactly divides n
the least common multiple of a and b
the least common multiple of a1 , a2 , . . . , and an
a is congruent to b modulo m
a is not congruent to b modulo m
the congruence class represented by r
an inverse of a modulo m
the digital root of n
the identity matrix of order n
the product of primes ≤ n
Euler’s phi function
the number of positive factors of n
the sum of the positive factors of n
Mersenne number 2p − 1
Möbius function
Liouville function
the order of a modulo m
the number of incongruent residues of order d modulo p
the index of a to the base α
Legendre symbol
Jacobi symbol
Kronecker symbol
Page
(74)
(74)
(76)
(76)
(76)
(76)
(83)
(96)
(110)
(129)
(136)
(138)
(139)
(155)
(162)
(183)
(184)
(187)
(212)
(212)
(216)
(234)
(291)
(316)
(325)
(342)
(365)
(366)
(381)
(398)
(405)
(456)
(470)
(483)
(501)
(527)
(549)
Elementary Number Theory with
Applications
Second Edition
Elementary Number Theory with
Applications
Second Edition
Thomas Koshy
AMSTERDAM • BOSTON • HEIDELBERG • LONDON
NEW YORK • OXFORD • PARIS • SAN DIEGO
SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO
Academic Press is an imprint of Elsevier
Academic Press is an imprint of Elsevier
30 Corporate Drive, Suite 400, Burlington, MA 01803, USA
525 B Street, Suite 1900, San Diego, California 92101-4495, USA
84 Theobald’s Road, London WC1X 8RR, UK
∞
This book is printed on acid-free paper.
No part of this publication may be reproduced or transmitted in any form or by any means, electronic
or mechanical, including photocopy, recording, or any information storage and retrieval system, without
permission in writing from the publisher.
Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford,
UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333, E-mail: permissions@elsevier.com. You may
also complete your request on-line via the Elsevier homepage (http://elsevier.com), by selecting “Support
& Contact” then “Copyright and Permission” and then “Obtaining Permissions.”
Koshy, Thomas.
Elementary number theory with applications / Thomas Koshy. – 2nd ed.
p. cm.
Includes bibliographical references and index.
ISBN 978-0-12-372487-8 (alk. paper)
1. Number theory. I. Title.
QA241.K67 2007
512.7–dc22
2007010165
British Library Cataloguing-in-Publication Data
A catalogue record for this book is available from the British Library.
ISBN: 978-0-12-372487-8
For information on all Academic Press publications
Printed in the United States of America
07 08 09 10
9 8 7 6 5 4 3 2 1
Dedicated to
my sister, Aleyamma Zachariah, and my brother,
M. K. Tharian; and to the memory of
Professor Edwin Weiss, Professor Donald W. Blackett,
and Vice Chancellor A. V. Varughese
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
A Word to the Student . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xiii
xxi
Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Fundamental Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
3
9
15
26
32
39
49
52
57
60
62
65
66
Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
2.1 The Division Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
80
89
98
103
128
139
143
146
148
151
153
The Summation and Product Notations . . . . . . . . . . . . . . . . . . . . .
Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polygonal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Pyramidal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Catalan Numbers .
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2.2
Base-b Representations (optional) . . . . . . . . . . . . . . . . . . . . . . . . .
2.3
Operations in Nondecimal Bases (optional) . . . . . . . . . . . . . . . . . . .
2.4
2.5
2.6
2.7
Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Prime and Composite Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . .
Fibonacci and Lucas Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . .
Fermat Numbers . .
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
vii
viii
Contents
3
Greatest Common Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155
3.1
3.2
3.3
3.4
3.5
Greatest Common Divisor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155
The Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
166
The Fundamental Theorem of Arithmetic . . . . . . . . . . . . . . . . . . . .
173
Least Common Multiple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
188
205
207
209
210
210
Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
211
4.1 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 The Pollard Rho Factoring Method . . . . . . . . . . . . . . . . . . . . . . . . .
211
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
230
238
240
241
243
244
245
Congruence Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
247
5.1
5.2
5.3
5.4
5.5
5.6
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Divisibility Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
247
Modular Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
253
Check Digits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
259
The p-Queens Puzzle (optional) . . . . . . . . . . . . . . . . . . . . . . . . . .
273
Round-Robin Tournaments (optional) . . . . . . . . . . . . . . . . . . . . . . .
277
.
.
.
.
.
.
.
.
.
.
.
.
282
288
289
291
291
292
Systems of Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . .
295
6.1 The Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
295
The Perpetual Calendar (optional)
Chapter Summary . . . . . . . . . . . . . . .
Review Exercises . . . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . . . .
Computer Exercises . . . . . . . . . . . . . .
Enrichment Readings . . . . . . . . . . . . .
6
.
.
.
.
.
.
.
.
.
.
.
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
5
.
.
.
.
.
.
184
.
.
.
.
.
.
Linear Diophantine Equations
Chapter Summary . . . . . . . . . . . .
Review Exercises . . . . . . . . . . . .
Supplementary Exercises . . . . . . .
Computer Exercises . . . . . . . . . . .
Enrichment Readings . . . . . . . . . .
6.2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
General Linear Systems (optional) . . . . . . . . . . . . . . . . . . . . . . . . .
303
ix
Contents
6.3
2 × 2 Linear Systems (optional)
Chapter Summary . . . . . . . . . . . . . .
Review Exercises . . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . . .
Computer Exercises . . . . . . . . . . . . .
Enrichment Readings . . . . . . . . . . . .
7
.
.
.
.
.
.
307
313
314
316
318
318
Three Classical Milestones . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
321
7.1
7.2
7.3
7.4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
321
326
337
341
348
350
351
352
353
Multiplicative Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
355
Fermat’s Little Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Pseudoprimes (optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1
8.2
8.3
8.4
8.5
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Euler’s Phi Function Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
355
365
373
381
398
406
408
409
411
412
Cryptology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
413
9.1
9.2
9.3
9.4
9.5
416
425
430
434
443
448
450
451
The Tau and Sigma Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Perfect Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Mersenne Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Möbius Function (optional)
Chapter Summary . . . . . . . . . . . . .
Review Exercises . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . .
Computer Exercises . . . . . . . . . . . .
Enrichment Readings . . . . . . . . . . .
9
.
.
.
.
.
.
Wilson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Euler’s Theorem . .
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
8
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Affine Ciphers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Hill Ciphers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exponentiation Ciphers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The RSA Cryptosystem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Knapsack Ciphers .
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
x
Contents
10
Computer Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Enrichment Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
452
453
Primitive Roots and Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . .
455
10.1
10.2
10.3
10.4
10.5
The Order of a Positive Integer . . . . . . . . . . . . . . . . . . . . . . . . . .
455
Primality Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
464
Primitive Roots for Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
467
Composites with Primitive Roots (optional) . . . . . . . . . . . . . . . . . .
474
The Algebra of Indices .
Chapter Summary . . . . . . . . .
Review Exercises . . . . . . . . .
Supplementary Exercises . . . .
Computer Exercises . . . . . . . .
Enrichment Readings . . . . . . .
11
.
.
.
.
.
.
482
489
491
492
493
493
Quadratic Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
495
11.1
11.2
11.3
11.4
11.5
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
495
The Legendre Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
501
Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
515
The Jacobi Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
527
.
.
.
.
.
.
535
543
546
548
549
550
Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
551
12.1 Finite Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Infinite Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
552
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
565
575
576
578
578
578
Miscellaneous Nonlinear Diophantine Equations . . . . . . . . . .
579
13.1 Pythagorean Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
579
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
13
.
.
.
.
.
.
Quadratic Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quadratic Congruences with Composite Moduli (optional)
Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
Computer Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Enrichment Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
xi
Contents
13.2 Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.3 Sums of Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.4 Pell’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
590
602
613
621
623
626
628
628
A
Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
631
A.1 Proof Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.2 Web Sites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
631
638
T
Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
641
T.1
T.2
T.3
T.4
Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
642
649
652
653
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
657
Solutions to Odd-Numbered Exercises . . . . . . . . . . . . . . . . . . .
665
Chapter 1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
665
677
689
696
702
707
711
718
728
731
737
746
748
Chapter Summary . . . . .
Review Exercises . . . . .
Supplementary Exercises
Computer Exercises . . . .
R
S
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Factor Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Values of Some Arithmetic Functions . . . . . . . . . . . . . . . . . . . . . . .
Least Primitive Roots r Modulo Primes p . . . . . . . . . . . . . . . . . . . . .
Chapter 2 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 3 Greatest Common Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 4 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5 Congruence Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 6 Systems of Linear Congruences . . . . . . . . . . . . . . . . . . . . . . .
Chapter 7 Three Classical Milestones . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 8 Multiplicative Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 9 Cryptology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 10 Primitive Roots and Indices . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 11 Quadratic Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 12 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 13 Miscellaneous Nonlinear Diophantine Equations . . . . . . . . . . . .
Credits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
757
761
Preface
Man has the faculty of becoming completely absorbed in one subject,
no matter how trivial and no subject is so trivial that it will not assume
infinite proportions if one’s entire attention is devoted to it.
— TOLSTOY, War and Peace
F
or over two thousand years, number theory has fascinated and inspired both
amateurs and mathematicians alike. A sound and fundamental body of knowledge, it has been developed by the untiring pursuits of mathematicians all
over the world. Today, number theorists continue to develop some of the most sophisticated mathematical tools ever devised and advance the frontiers of knowledge.
Many number theorists, including the eminent nineteenth-century English number theorist Godfrey H. Hardy, once believed that number theory, although beautiful,
had no practical relevance. However, the advent of modern technology has brought
a new dimension to the power of number theory: constant practical use. Once considered the purest of pure mathematics, it is used increasingly in the rapid development of technology in a number of areas, such as art, coding theory, cryptology, and
computer science. The various fascinating applications have confirmed that human
ingenuity and creativity are boundless, although many years of hard work may be
needed to produce more meaningful and delightful applications.
The Pursuit of a Dream
This book is the fruit of years of dreams and the author’s fascination for the subject,
its beauty, elegance, and historical development; the opportunities it provides for
both experimentation and exploration; and, of course, its marvelous applications.
This new edition, building on the strengths of its predecessor, incorporates a
number of constructive suggestions made by students, reviewers, and well-wishers. It
is logically conceived, self-contained, well-organized, nonintimidating, and written
with students and amateurs in mind. In clear, readable language, this book offers an
overview of the historical development of the field, including major figures, as well
xiii
xiv
Preface
as step-by-step development of the basic concepts and properties, leading to the more
Audience and Prerequisites
The book is designed for an undergraduate course in number theory for students
majoring in mathematics and/or computer science at the sophomore/junior level and
for students minoring in mathematics. No formal prerequisites are required to study
the material or to enjoy its beauty except a strong background in college algebra.
The main prerequisite is mathematical maturity: lots of patience, logical thinking,
and the ability for symbolic manipulation. This book should enable students and
number theory enthusiasts to enjoy the material with great ease.
Coverage
The text includes a detailed discussion of the traditional topics in an undergraduate number theory course, emphasizing problem-solving techniques, applications,
pattern recognition, conjecturing, recursion, proof techniques, and numeric computations. It also covers figurate numbers and their geometric representations, Catalan
numbers, Fibonacci and Lucas numbers, Fermat numbers, an up-to-date discussion
of the various classes of prime numbers, and factoring techniques. Starred () optional sections and optional puzzles can be omitted without losing continuity of development.
Included in this edition are new sections on Catalan numbers and the Pollard
rho factoring method, a subsection on the Pollard p − 1 factoring method, and a
short chapter on continued fractions. The section on linear diophantine equations
now appears in Chapter 3 to provide full prominence to congruences.
A number of well-known conjectures have been added to challenge the more ambitious students. Identified by the conjecture symbol ? in the margin, they should
provide wonderful opportunities for group discussion, experimentation, and exploration.
Examples and Exercises
Each section contains a wealth of carefully prepared and well-graded examples and
exercises to enhance student skills. Examples are developed in detail for easy understanding. Many exercise sets contain thought-provoking true/false problems, numeric problems to develop computational skills, and proofs to master facts and the
various proof techniques. Extensive chapter-end review exercise sets provide comprehensive reviews, while chapter-end supplementary exercises provide challenging
opportunities for the curious-minded to pursue.
Preface
xv
Starred () exercises are, in general, difficult, and doubly starred () ones are
more difficult. Both can be omitted without losing overall understanding of the concepts under discussion. Exercises identified with a c in the margin require a knowledge of elementary calculus; they can be omitted by students with no calculus background.
Historical information, including biographical sketches of about 50 mathematicians,
is woven throughout the text to enhance a historical perspective on the development of number theory. This historical dimension provides a meaningful context for
prospective and in-service high school and middle school teachers in mathematics.
An index of the biographies, keyed to pages in the text, can be found inside the back
cover.
Applications
This book has several unique features. They include the numerous relevant and
thought-provoking applications spread throughout, establishing a strong and meaningful bridge with geometry and computer science. These applications increase student interest and understanding and generate student interaction. In addition, the
book shows how modular systems can be used to create beautiful designs, linking number theory with both geometry and art. The book also deals with barcodes,
zip codes, International Serial Book Numbers, European Article Numbers, vehicle
identification numbers, and German bank notes, emphasizing the closeness of number theory to our everyday life. Furthermore, it features Friday-the-thirteenth, the
p-queens puzzle, round-robin tournaments, a perpetual calendar, the Pollard rho factoring method, and the Pollard p − 1 factoring method.
Flexibility
The order and selection of topics offer maximum flexibility for instructors to select
chapters and sections that are appropriate for student needs and course lengths. For
example, Chapter 1 can be omitted or assigned as optional reading, as can the optional sections 6.2, 6.3, 7.3, 8.5, 10.4, and 11.5, without jeopardizing the core of
development. Sections 2.2, 2.3, and 5.4–5.6 also can be omitted if necessary.
Foundations
All proof methods are explained and illustrated in detail in the Appendix. They provide a strong foundation in problem-solving techniques, algorithmic approach, and
proof techniques.
xvi
Preface
Proofs
Most concepts, definitions, and theorems are illustrated through thoughtfully selected
examples. Most of the theorems are proven, with the exception of some simple ones
left as routine exercises. The proofs shed additional light on the understanding of the
topic and enable students to develop their problem-solving skills. The various proof
techniques are illustrated throughout the text.
Proofs Without Words
Several geometric proofs of formulas are presented without explanation. This unique
feature should generate class discussion and provide opportunities for further exploration.
Pattern Recognition
An important problem-solving technique used by mathematicians is pattern recognition. Throughout the book, there are ample opportunities for experimentation and exploration: collecting data, arranging them systematically, recognizing patterns, making conjectures, and then establishing or disproving these conjectures.
Recursion
By drawing on well-selected examples, the text explains in detail this powerful strategy, which is used heavily in both mathematics and computer science. Many examples are provided to ensure that students are comfortable with this powerful problemsolving technique.
Numeric Puzzles
Several fascinating, optional number-theoretic puzzles are presented for discussion
and digression. It would be a good exercise to justify each. These puzzles are useful
for prospective and in-service high school and middle school teachers in mathematics.
Algorithms
A number of algorithms are given as a problem-solving technique in a straightforward fashion. They can easily be translated into computer programs in a language
of your choice. These algorithms are good candidates for class discussion and are
boxed in for easy identification.
Preface
xvii
Computer Assignments
Relevant and thought-provoking computer assignments are provided at the end of
each chapter. They provide hands-on experience with concepts and enhance the opportunity for computational exploration and experimentation. A computer algebra
system, such as Maple or Mathematica, or a language of your choice can be used.
Chapter Summary
At the end of each chapter, you will find a summary that is keyed to pages in the text.
This provides a quick review and easy reference. Summaries contain the various
definitions, symbols, and properties.
Each chapter ends with a carefully prepared list of readings from various sources for
further exploration of the topics and for additional enrichment.
Relevant annotated web sites are listed in the Appendix. For instance, up-to-date
information on the discovery of Mersenne primes and twin primes is available on
the Internet. This enables both amateurs and professionals to access the most recent
discoveries and research.
Special Symbols
The square denotes the end of a proof and an example. The conjecture symbol ?
indicates an unresolved problem.
Index of Symbols
Inside the front cover, you will find, for quick reference, a list of symbols and the
page numbers on which they first occur.
Odd-Numbered Solutions
The solutions to all odd-numbered exercises are given at the end of the text.
Solutions Manual for Students
The Student’s Solutions Manual contains detailed solutions to all even-numbered exercises. It also contains valuable tips for studying mathematics, as well as for preparing and taking examinations.
xviii
Preface
Instructor’s Manual
The Instructor’s Manual contains detailed solutions to all even-numbered exercises,
sample tests for each chapter, and the keys for each test. It also contains two sample
final examinations and their keys.
Highlights of this Edition
They include:
•
•
•
•
•
•
•
•
•
•
•
Catalan numbers (Sections 1.8, 2.5, and 8.4)
Linear diophantine equations with Fibonacci coefficients (Section 3.5)
Pollard rho factoring method (Section 4.3)
Vehicle identification numbers (Section 5.3)
German bank notes (Section 5.3)
Factors of 2n + 1 (Section 7.2)
Pollard p − 1 factoring method (Section 7.2)
Pascal’s binary triangle and even perfect numbers (Section 8.4)
Continued fractions (Chapter 12)
Well-known conjectures
Expanded exercise sets
Acknowledgments
I am grateful to a number of people for their cooperation, support, encouragement,
and thoughtful comments during the writing and revising of this book. They all have
played a significant role in improving its quality.
To begin with, I am indebted to the following reviewers for their boundless enthusiasm and constructive suggestions:
Steven M. Bairos
Peter Brooksbank
Roger Cooke
Joyce Cutler
Daniel Drucker
Maureen Femick
Burton Fein
Justin Wyss-Gallifent
Napolean Gauthier
Richard H. Hudson
Robert Jajcay
Roger W. Leezer
Data Translation, Inc.
Bucknell University
University of Vermont
Framingham State College
Wayne State University
Minnesota State University at Mankato
Oregon State University
University of Maryland
The Royal Military College of Canada
University of South Carolina
Indiana State University
California State University at Sacramento
Preface
I. E. Leonard
Don Redmond
Dan Reich
Helen Salzberg
Seung H. Son
David Stone
M. N. S. Swamy
Fernando Rodriguez Villegas
Betsey Whitman
Raymond E. Whitney
xix
University of Alberta
Southern Illinois University
Temple University
Rhode Island College
Georgia Southern University
Concordia University
University of Texas at Austin
Framingham State College
Lock Haven University
Thanks also to Roger Cooke of the University of Vermont, Daniel Drucker of Wayne
State University, Maureen Fenrick of Minnesota State University at Mankato, and
Kevin Jackson-Mead for combing through the entire manuscript for accuracy; to
Daniel Drucker of Wayne State University and Dan Reich of Temple University for
class-testing the material; to the students Prasanth Kalakota of Indiana State University and Elvis Gonzalez of Temple University for their comments; to Thomas E.
Moore of Bridgewater State College and Don Redmond of Southern Illinois University for preparing the solutions to all odd-numbered exercises; to Ward Heilman
of Bridgewater State College and Roger Leezer of California State University at
Sacramento for preparing the solutions to all even-numbered exercises; to Margarite
Roumas for her superb editorial assistance; and to Madelyn Good and Ellen Keane at
the Framingham State College Library, who tracked down a number of articles and
books. My sincere appreciation also goes to Senior Editors Barbara Holland, who
initiated the original project, Pamela Chester, and Thomas Singer; Production Editor
Christie Jozwiak, Project Manager Jamey Stegmaier, Copyeditor Rachel Henriquez,
and Editorial Assistant Karen Frost at Harcourt/Academic Press for their cooperation, promptness, support, encouragement, and confidence in the project.
Finally, I must confess that any errors that may yet remain are my own responsibility. However, I would appreciate hearing about any inadvertent errors, alternate
solutions, or, better yet, exercises you have enjoyed.
Thomas Koshy
tkoshy@frc.mass.edu
A Word to the Student
Mathematics is music for the mind;
music is mathematics for the soul.
— ANONYMOUS
The Language of Mathematics
To learn a language, you have to know its alphabet, grammar, and syntax, and you
have to develop a decent vocabulary. Likewise, mathematics is a language with its
own symbols, rules, terms, definitions, and theorems. To be successful in mathematics, you must know them and be able to apply them; you must develop a working vocabulary, use it as often as you can, and speak and write in the language of
math.
This book was written with you in mind, to create an introduction to number
theory that is easy to understand. Each chapter is divided into short sections of approximately the same length.
Problem-Solving Techniques
Throughout, the book emphasizes problem-solving techniques such as doing experiments, collecting data, organizing them in an orderly fashion, recognizing patterns, and making conjectures. It also emphasizes recursion, an extremely powerful
problem-solving strategy used heavily in both mathematics and computer science.
Although you may need some practice to get used to recursion, once you know how
to approach problems recursively, you will appreciate its power and beauty. So do
not be turned off, even if you have to struggle a bit with it initially.
The book stresses proof techniques as well. Theorems are the bones of mathematics. So, for your convenience, the various proof methods are explained and illustrated in the Appendix. It is strongly recommended that you master them; do the
worked-out examples, and then do the exercises. Keep reviewing the techniques as
often as needed.
Many of the exercises use the theorems and the techniques employed in their
proofs. Try to develop your own proofs. This will test your logical thinking and
xxi
xxii
A Word to the Student
analytical skills. In order to fully enjoy this beautiful and elegant subject, you must
feel at home with the various proof methods.
Getting Involved
Basketball players such as Michael Jordan and Larry Bird did not become superstars by reading about basketball or watching others play. Besides knowing the rules
and the objects needed to play, they needed countless hours of practice, hard work,
and determination to achieve their goal. Likewise, you cannot learn mathematics by
do it yourself every day, just as skill is acquired in a sport. You can learn mathematics in small, progressive steps only, building on skills you already have developed.
Suggestions for Learning
Here are a few suggestions you should find useful in your pursuit:
• Read a few sections before each class. You might not fully understand the material, but you will certainly follow it far better when your professor discusses
it in class. Besides, you will be able to ask more questions in class and answer
more questions.
• Always go to class well prepared. Be prepared to answer and ask questions.
• Whenever you study from the book, make sure you have a pencil and enough
scrap paper next to you for writing the definitions, theorems, and proofs and
for doing the exercises.
• Study the material taught in class on the same day. Do not just read it as if you
were reading a novel or a newspaper. Write down the definitions, theorems,
and properties in your own words without looking in your notes or the book.
Rewrite the examples, proofs, and exercises done in class, all in your own
words. If you cannot do them on your own, study them again and try again;
continue until you succeed.
• Always study the relevant section in the text and do the examples there, then
do the exercises at the end of the section. Since the exercises are graded in
order of difficulty, do them in order. Do not skip steps or write over previous
steps; this way you will be able to progress logically, locate your errors, and
correct your mistakes. If you cannot solve a problem because it involves a
new term, formula, or some property, then re-study the relevant portion of
the section and try again. Do not assume that you will be able to do every
problem the first time you try it. Remember, practice is the best shortcut to
success.
A Word to the Student
xxiii
Solutions Manual
The Student’s Solutions Manual contains additional tips for studying mathematics,
preparing for an examination in mathematics, and taking an examination in mathematics. It also contains detailed solutions to all even-numbered exercises.
A Final Word
Mathematics, especially number theory, is no more difficult than any other subject.
If you have the willingness, patience, and time to sit down and do the work, then you
will find number theory worth studying and this book worth studying from; you will
find that number theory can be fun, and fun can be number theory. Remember that
learning mathematics is a step-by-step matter. Do your work regularly and systematically; review earlier chapters every week, since things must be fresh in your mind
to apply them and to build on them. In this way, you will enjoy the subject and feel
confident to explore more. I look forward to hearing from you with your comments
and suggestions. In the meantime, enjoy the book.
Thomas Koshy
1
Fundamentals
Tell me and I will forget.
Show me and I will remember.
Involve me and I will understand.
— CONFUCIUS
T
he outstanding German mathematician Karl Friedrich Gauss (1777–1855)
once said, “Mathematics is the queen of the sciences and arithmetic the queen
of mathematics.” “Arithmetic,” in the sense Gauss uses it, is number theory,
which, along with geometry, is one of the two oldest branches of mathematics. Number theory, as a fundamental body of knowledge, has played a pivotal role in the
development of mathematics. And as we will see in the chapters ahead, the study of
number theory is elegant, beautiful, and delightful.
A remarkable feature of number theory is that many of its results are within the
reach of amateurs. These results can be studied, understood, and appreciated without much mathematical sophistication. Number theory provides a fertile ground for
both professionals and amateurs. We can also find throughout number theory many
fascinating conjectures whose proofs have eluded some of the most brilliant mathematicians. We find a great number of unsolved problems as well as many intriguing
results.
Another interesting characteristic of number theory is that although many of its
results can be stated in simple and elegant terms, their proofs are sometimes long
and complicated.
Generally speaking, we can define “number theory” as the study of the properties
of numbers, where by “numbers” we mean integers and, more specifically, positive
integers.
Studying number theory is a rewarding experience for several reasons. First, it
has historic significance. Second, integers, more specifically, positive integers, are
1
2
CHAPTER 1 Fundamentals
Pythagoras (ca. 572–ca. 500 B.C.), a Greek philosopher and mathematician, was born on the Aegean island of Samos. After extensive travel and studies, he
returned home around 529 B.C. only to find that Samos
was under tyranny, so he migrated to the Greek port
of Crontona, now in southern Italy. There he founded
the famous Pythagorean school among the aristocrats of the city. Besides being an academy for philosophy, mathematics, and natural science, the school
The Island of Samos
became the center of a closely knit brotherhood sharing arcane rites and observances. The brotherhood
A Greek Stamp
ascribed all its discoveries to the master.
A philosopher, Pythagoras taught that number was the essence of everything, and
Honoring
he associated numbers with mystical powers. He also believed in the transmigration of the
Pythagoras
soul, an idea he might have borrowed from the Hindus.
Suspicions arose about the brotherhood, leading to the murder of most of its members. The school was
destroyed in a political uprising. It is not known whether Pythagoras escaped death or was killed.
the building blocks of the real number system, so they merit special recognition.
Third, the subject yields great beauty and offers both fun and excitement. Finally,
the many unsolved problems that have been daunting mathematicians for centuries
provide unlimited opportunities to expand the frontiers of mathematical knowledge.
Goldbach’s conjecture (Section 2.5) and the existence of odd perfect numbers (Section 8.3) are two cases in point. Modern high-speed computers have become a powerful tool in proving or disproving such conjectures.
Although number theory was originally studied for its own sake, today it has
intriguing applications to such diverse fields as computer science and cryptography
(the art of creating and breaking codes).
The foundations for number theory as a discipline were laid out by the Greek
mathematician Pythagoras and his disciples (known as the Pythagoreans). The
Pythagorean brotherhood believed that “everything is number” and that the central
explanation of the universe lies in number. They also believed some numbers have
mystical powers. The Pythagoreans have been credited with the invention of amicable numbers, perfect numbers, figurate numbers, and Pythagorean triples. They
classified integers into odd and even integers, and into primes and composites.
Another Greek mathematician, Euclid (ca. 330–275 B . C .), also made significant
contributions to number theory. We will find many of his results in the chapters to
follow.
We begin our study of number theory with a few fundamental properties of integers.
1.1 Fundamental Properties
3
Little is known about Euclid’s life. He was on the faculty at the University of Alexandria and founded the Alexandrian School of Mathematics. When the Egyptian ruler
King Ptolemy I asked Euclid, the father of geometry, if there were an easier way to
learn geometry than by studying The Elements, he replied, “There is no royal road
to geometry.”
1.1
Fundamental Properties
The German mathematician Hermann Minkowski (1864–1909) once remarked, “Integral numbers are the fountainhead of all mathematics.” We will come to appreciate
how important his statement is. In fact, number theory is concerned solely with integers. The set of integers is denoted by the letter Z:†
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}
Whenever it is convenient, we write “x ∈ S” to mean “x belongs
√ to the set S”;
/ Z.
“x ∈
/ S” means “x does not belong to S.” For example, 3 ∈ Z, but 3 ∈
We can represent integers geometrically on the number line, as in Figure 1.1.
Figure 1.1
The integers 1, 2, 3, . . . are positive integers. They are also called natural numbers or counting numbers; they lie to the right of the origin on the number line. We
denote the set of positive integers by Z+ or N:
Z+ = N = {1, 2, 3, . . .}
† The letter Z comes from the German word Zahlen for numbers.
4
CHAPTER 1 Fundamentals
Leopold Kronecker (1823–1891) was born in 1823 into a well-to-do family in Liegnitz,
Prussia (now Poland). After being tutored privately at home during his early years and
then attending a preparatory school, he went on to the local gymnasium, where he
excelled in Greek, Latin, Hebrew, mathematics, and philosophy. There he was fortunate to have the brilliant German mathematician Ernst Eduard Kummer (1810–1893)
as his teacher. Recognizing Kronecker’s mathematical talents, Kummer encouraged
him to pursue independent scientific work. Kummer later became his professor at the
universities of Breslau and Berlin.
In 1841, Kronecker entered the University of Berlin and also spent time at the
University of Breslau. He attended lectures by Dirichlet, Jacobi, Steiner, and Kummer.
Four years later he received his Ph.D. in mathematics.
Kronecker’s academic life was interrupted for the next 10 years when he ran his uncle’s business. Nonetheless, he managed to correspond regularly with Kummer. After becoming a member of the Berlin Academy of
Sciences in 1861, Kronecker began his academic career at the University of Berlin, where he taught unpaid until
1883; he became a salaried professor when Kummer retired.
In 1891, his wife died in a fatal mountain climbing accident, and Kronecker, devastated by the loss, succumbed to bronchitis and died four months later.
Kronecker was a great lover of the arts, literature, and music, and also made profound contributions to number theory, the theory of equations, elliptic functions, algebra, and the theory of determinants. The vertical bar
notation for determinants is his creation.
The German mathematician Leopold Kronecker wrote, “God created the natural
numbers and all else is the work of man.” The set of positive integers, together with 0,
forms the set of whole numbers W:
W = {0, 1, 2, 3, . . .}
Negative integers, namely, . . . , −3, −2, −1, lie to the left of the origin. Notice
that 0 is neither positive nor negative.
We can employ positive integers to compare integers, as the following definition
shows.
The Order Relation
Let a and b be any two integers. Then a is less than b, denoted by a < b, if there
exists a positive integer x such that a + x = b, that is, if b − a is a positive integer.
When a < b, we also say that b is greater than a, and we write b > a.†
† The symbols < and > were introduced in 1631 by the English mathematician Thomas Harriet
(1560–1621).
1.1 Fundamental Properties
5
If a is not less than b, we write a ≮ b; similarly, a ≯ b indicates a is not greater
than b.
It follows from this definition that an integer a is positive if and only if a > 0.
Given any two integers a and b, there are three possibilities: either a < b, a = b,
or a > b. This is the law of trichotomy. Geometrically, this means if a and b are any
two points on the number line, then either point a lies to the left of point b, the two
points are the same, or point a lies to the right of point b.
We can combine the less than and equality relations to define the less than or
equal to relation. If a < b or a = b, we write a ≤ b.† Similarly, a ≥ b means either
a > b or a = b. Notice that a < b if and only if a ≥ b.
We will find the next result useful in Section 3.4. Its proof is fairly simple and is
an application of the law of trichotomy.
THEOREM ‡ 1.1
Let min{x, y} denote the minimum of the integers x and y, and max{x, y} their maximum. Then min{x, y} + max{x, y} = x + y.§
PROOF (by cases)
case 1
Let x ≤ y. Then min{x, y} = x and max{x, y} = y, so min{x, y}+max{x, y} =
x + y.
Let x > y. Then min{x, y} = y and max{x, y} = x, so min{x, y}+max{x, y} =
y + x = x + y.
case 2
The law of trichotomy helps us to define the absolute value of an integer.
Absolute Value
The absolute value of a real number x, denoted by |x|, is defined by
x if x ≥ 0
|x| =
−x otherwise
For example, |5| = 5, |−3| = −(−3) = 3, |π| = π , and |0| = 0.
Geometrically, the absolute value of a number indicates its distance from the
origin on the number line.
Although we are interested only in properties of integers, we often need to
deal with rational and real numbers also. Floor and ceiling functions are two such
number-theoretic functions. They have nice applications to discrete mathematics and
computer science.
† The symbols ≤ and ≥ were introduced in 1734 by the French mathematician P. Bouguer.
‡ A theorem is a (major) result that can be proven from axioms or previously known results.
§ Theorem 1.1 is true even if x and y are real numbers.
6
CHAPTER 1 Fundamentals
Floor and Ceiling Functions
The floor of a real number x, denoted by x, is the greatest integer ≤ x. The ceiling
of x, denoted by x , is the least integer ≥ x.† The floor of x rounds down x, whereas
the ceiling of x rounds up. Accordingly, if x ∈
/ Z, the floor of x is the nearest integer to
the left of x on the number line, and the ceiling of x is the nearest integer to the right
of x, as Figure 1.2 shows. The floor function f (x) = x and the ceiling function
g(x) = x are also known as the greatest integer function and the least integer
function, respectively.
Figure 1.2
For example, π = 3, log10 3 = 0, −3.5 = −4, −2.7 = −3, π = 4,
log10 3 = 1, −3.5 = −3, and −2.7 = −2.
The floor function comes in handy when real numbers are to be truncated or
rounded off to a desired number of decimal places. For example, the real number π =
3.1415926535 . . . truncated to three decimal places is given by 1000π/1000 =
3141/1000 = 3.141; on the other hand, π rounded to three decimal places is
1000π + 0.5/1000 = 3.142.
There is yet another simple application of the floor function. Suppose we divide
the unit interval [0, 1) into 50 subintervals of equal length 0.02 and then seek to
determine the subinterval that contains the number 0.4567. Since 0.4567/0.02 +
1 = 23, it lies in the 23rd subinterval. More generally, let 0 ≤ x < 1. Then x lies in
the subinterval x/0.02 + 1 = 50x + 1.
The following example presents an application of the ceiling function to everyday life.
EXAMPLE 1.1
(The post-office function) In 2006, the postage rate in the United States for a firstclass letter of weight x, not more than one ounce, was 39¢; the rate for each additional
ounce or a fraction thereof up to 11 ounces was an additional 24¢. Thus, the postage
p(x) for a first-class letter can be defined as p(x) = 0.39 + 0.24 x − 1 , 0 < x ≤ 11.
For instance, the postage for a letter weighing 7.8 ounces is p(7.8) = 0.39 +
0.24 7.8 − 1 = \$2.07.
† These two notations and the names, floor and ceiling, were introduced by Kenneth E. Iverson in the
early 1960s. Both notations are variations of the original greatest integer notation [x].
1.1 Fundamental Properties
7
Some properties of the floor and ceiling functions are listed in the next theorem.
We shall prove one of them; the others can be proved as routine exercises.
THEOREM 1.2
Let x be any real number and n any integer. Then
1.
2.
3.
4.
n−1
n
=
if n is odd.
5.
2
2
n
n+1
6.
=
if n is odd.
2
2
n = n = n
x = x + 1 (x ∈
/ Z)
x + n = x + n
x+n = x +n
PROOF
Every real number x can be written as x = k + x , where k = x and 0 ≤ x < 1. See
Figure 1.3. Then
Figure 1.3
x + n = k + n + x = (k + n) + x
x + n = k + n,
since 0 ≤ x < 1
= x + n
E X E R C I S E S
1. The English mathematician Augustus DeMorgan,
who lived in the 19th century, once remarked that he
was x years old in the year x2 . When was he born?
Evaluate each, where x is a real number.
x
(x = 0)
2. f (x) =
|x|
3. g(x) = x + −x
4. h(x) = x + −x
Determine whether:
5. −−x = x
6. − −x = x
7. There are four integers between 100 and 1000 that are
each equal to the sum of the cubes of its digits. Three
of them are 153, 371, and 407. Find the fourth number. (Source unknown.)
1.1
8. An n-digit positive integer N is a Kaprekar number
if the sum of the number formed by the last n digits
in N 2 , and the number formed by the first n (or n − 1)
digits in N 2 equals N. For example, 297 is a Kaprekar
number since 2972 = 88209 and 88 + 209 = 297.
There are five Kaprekar numbers < 100. Find them.
9. Find the flaw in the following “proof”:
Let a and b be real numbers such that a = b. Then
ab = b2
a2 − ab = a2 − b2
Factoring, a(a − b) = (a + b)(a − b). Canceling
a − b from both sides, a = a + b. Since a = b,
this yields a = 2a. Canceling a from both sides,
we get 1 = 2.
8
CHAPTER 1 Fundamentals
D. R. Kaprekar (1905–1986) was born in Dahanu, India, near Bombay. After losing his mother at the age of
eight, he built a close relationship with his astrologer-father, who passed on his knowledge to his son. He attended Ferguson College in Pune, and then graduated from the University of Bombay in 1929. He was awarded
the Wrangler R. P. Paranjpe prize in 1927 in recognition of his mathematical contributions. A prolific writer in
recreational number theory, he worked as a schoolteacher in Devlali, India, from 1930 until his retirement in
1962.
Kaprekar is best known for his 1946 discovery of the Kaprekar constant 6174. It took him about three
years to discover the number: Take a four-digit number a, not all digits being the same; let a denote the number
obtained by rearranging its digits in nondecreasing order and a denote the number obtained by rearranging its
digits in nonincreasing order. Repeat these steps with b = a − a and its successors. Within a maximum of eight
steps, this process will terminate in 6174. It is the only integer with this property.
10. Express 635,318,657 as the sum of two fourth powers
in two different ways. (It is the smallest number with
this property.)
11. The integer 1105 can be expressed as the sum of two
squares in four different ways. Find them.
12. There is exactly one integer between 2 and 2 × 1014
that is a perfect square, a cube, and a fifth power. Find
it. (A. J. Friedland, 1970)
13. The five-digit number 2xy89 is the square of an integer. Find the two-digit number xy. (Source: Mathematics Teacher)
14. How many perfect squares can be displayed on a 15digit calculator?
15. The number sequence 2, 3, 5, 6, 7, 10, 11, . . . consists
of positive integers that are neither squares nor cubes.
Find the 500th term of this sequence. (Source: Mathematics Teacher)
Prove each, where a, b, and n are any integers, and x is a
real number.
16. |ab| = |a| · |b|
17. |a
+b| ≤ |a| + |b|
n−1
n
=
if n is odd.
18.
2
2
n+1
n
=
if n is odd.
19.
2
2
2
n
n2 − 1
20.
=
if n is odd.
4
4
2
2
n +3
n
=
21.
if n is odd.
4
4
n
n
+
=n
22.
2
2
23. x = x + 1 (x ∈
/ Z)
24. x = −−x
25. x + n = x + n
26. x + x + 1/2 = 2x
27. x/n = x/n
The distance from x to y on the number line, denoted by
d(x, y), is defined by d(x, y) = |y − x|. Prove each, where
x, y, and z are any integers.
28. d(x, y) ≥ 0
29. d(0, x) = |x|
30. d(x, y) = 0 if and only if x = y
31. d(x, y) = d(y, x)
32. d(x, y) ≤ d(x, z) + d(z, y)
33. Let max{x, y} denote the maximum of x and y, and
min{x, y} their minimum, where x and y are any integers. Prove that max{x, y} − min{x, y} = |x − y|.
34. A round-robin tournament has n teams, and each team
plays at most once in a round. Determine the minimum number of rounds f (n) needed to complete the
1.2 The Summation and Product Notations
9
Joseph Louis Lagrange (1736–1813), who ranks with Leonhard Euler as one of the
greatest mathematicians of the 18th century, was the eldest of eleven children in a
wealthy family in Turin, Italy. His father, an influential cabinet official, became bankrupt due to unsuccessful financial speculations, which forced Lagrange to pursue a
profession.
As a young man studying the classics at the College of Turin, his interest in mathematics was kindled by an essay by astronomer Edmund Halley on the superiority of
the analytical methods of calculus over geometry in the solution of optical problems.
In 1754 he began corresponding with several outstanding mathematicians in Europe.
The following year, Lagrange was appointed professor of mathematics at the Royal
Artillery School in Turin. Three years later, he helped to found a society that later
became the Turin Academy of Sciences. While at Turin, Lagrange developed revolutionary results in the calculus of variations, mechanics, sound, and probability, winning the prestigious Grand Prix
of the Paris Academy of Sciences in 1764 and 1766.
In 1766, when Euler left the Berlin Academy of Sciences, Frederick the Great wrote to Lagrange that “the
greatest king in Europe” would like to have “the greatest mathematician of Europe” at his court. Accepting the
invitation, Lagrange moved to Berlin to head the Academy and remained there for 20 years. When Frederick died
in 1786, Lagrange moved to Paris at the invitation of Louis XVI. Lagrange was appointed professor at the École
Normale and then at the École Polytechnique, where he taught until 1799.
Lagrange made significant contributions to analysis, analytical mechanics, calculus, probability, and number
theory, as well as helping to set up the French metric system.
1.2
The Summation and Product Notations
We will find both the summation and the product notations very useful throughout
the remainder of this book. First, we turn to the summation notation.
The Summation Notation
Sums, such as ak + ak+1 + · · · + am , can be written in a compact form using the
summation symbol
(the Greek uppercase letter sigma), which denotes the word
sum. The summation notation was introduced in 1772 by the French mathematician
Joseph Louis Lagrange.
A typical term in the sum above can be denoted by ai , so the above sum is the
i=m
sum of the numbers ai as i runs from k to m and is denoted by
ai . Thus
i=k
i=m
i=k
ai = ak + ak+1 + · · · + am
10
CHAPTER 1 Fundamentals
The variable i is the summation index. The values k and m are the lower and upper
limits of the index i. The “i =” above the
is usually omitted:
i=m
ai =
i=k
m
ai
i=k
For example,
2
i(i − 1) = (−1)(−1 − 1) + 0(0 − 1) + 1(1 − 1) + 2(2 − 1) = 4
i=−1
The index i is a dummy variable; we can use any variable as the index without
affecting the value of the sum, so
m
ai =
i=
EXAMPLE 1.2
Evaluate
3
m
aj =
j=
m
ak
k=
j2 .
j=−2
SOLUTION
3
j2 = (−2)2 +(−1)2 +02 +12 +22 +32 = 19
j=−2
The following results are extremely useful in evaluating finite sums. They can
be proven using mathematical induction, presented in Section 1.3.
THEOREM 1.3
Let n be any positive integer and c any real number, and a1 , a2 , . . . , an and b1 ,
b2 , . . . , bn any two number sequences. Then
n
i=1
c = nc
(1.1)
1.2 The Summation and Product Notations
n
n
(cai ) = c
ai
i=1
11
(1.2)
i=1
n
n
n
(ai + bi ) =
ai +
bi
i=1
i=1
(1.3)
i=1
(These results can be extended to any lower limit k ∈ Z.)
The following example illustrates this theorem.
EXAMPLE 1.3
2
Evaluate
[(5j)3 − 2j].
j=−1
SOLUTION
2
[(5j)3 − 2j] =
j=−1
2
(5j)3 − 2
j=−1
= 125
2
j
j=−1
2
j=−1
j
3
−2
2
j
j=−1
= 125[(−1)3 + 03 + 13 + 23 ] − 2(−1 + 0 + 1 + 2)
= 996
Indexed Summation
The summation notation can be extended to sequences with index sets I as their
ai denotes the sum of the values of ai as i runs over the
domains. For instance,
i∈I
various values in I.
As an example, let I = {0, 1, 3, 5}. Then (2i + 1) represents the sum of the
values of 2i + 1 with i ∈ I, so
i∈I
(2i + 1) = (2 · 0 + 1) + (2 · 1 + 1) + (2 · 3 + 1) + (2 · 5 + 1) = 22
i∈I
Often we need to evaluate sums of the form
aij , where the subscripts i and j
P
satisfy certain properties P. (Such summations are used in Chapter 8.)
12
CHAPTER 1 Fundamentals
For example, let I = {1, 2, 3, 4}. Then
(2i + 3j) denotes the sum of the
values of 2i + 3j, where 1 ≤ i < j ≤ 4. This can be abbreviated as (2i + 3j) pro1≤i<j≤4
i<j
vided the index set is obvious from the context. To find this sum, we must consider
every possible pair (i, j), where i, j ∈ I and i < j. Thus,
(2i + 3j) = (2 · 1 + 3 · 2) + (2 · 1 + 3 · 3) + (2 · 1 + 3 · 4) + (2 · 2 + 3 · 3)
i<j
+ (2 · 2 + 3 · 4) + (2 · 3 + 3 · 4)
= 80
EXAMPLE 1.4
Evaluate
d, where d|6 means d is a factor of 6.
d≥1
d|6
SOLUTION
d = sum of positive integers d, where d is a factor of 6
d≥1
d|6
= sum of positive factors of 6
= 1 + 2 + 3 + 6 = 12
Multiple summations arise often in mathematics. They are evaluated in a
aij is evaluated as
right-to-left fashion. For example, the double summation
i j
aij , as demonstrated below.
i
EXAMPLE 1.5
j
Evaluate
2
1
(2i + 3j).
i=−1 j=0
SOLUTION
1
1
2
2
(2i + 3j) =
(2i + 3j)
i=−1 j=0
i=−1 j=0
=
1
(2i + 3 · 0) + (2i + 3 · 1) + (2i + 3 · 2)
i=−1
1.2 The Summation and Product Notations
=
1
13
(6i + 9)
i=−1
= 6 · (−1) + 9 + (6 · 0 + 9) + (6 · 1 + 9)
= 27
We now turn to the product notation.
The Product Notation
i=m
is used to denote sums, the product ak ak+1 · · · am is denoted by
ai . The
i=k
product symbol
is the Greek capital letter pi. As in the case of the summation
notation, the “i =” above the product symbol is often dropped:
Just as
i=m
i=k
ai =
m
ai = ak ak+1 · · · am
i=k
Again, i is just a dummy variable.
The following three examples illustrate this notation.
The factorial function, which often arises in number theory, can be defined
using the product symbol, as the following example shows.
EXAMPLE 1.6
The factorial function f (n) = n! (read n factorial) is defined by n! = n(n − 1) · · · 2 · 1,
n
where 0! = 1. Using the product notation, f (n) = n! =
k.
k=1
EXAMPLE 1.7
Evaluate
5
(i2 − 3).
i=2
SOLUTION
5
(i2 − 3) = (22 − 3)(32 − 3)(42 − 3)(52 − 3)
i=2
= 1 · 6 · 13 · 22 = 1716
14
CHAPTER 1 Fundamentals
Just as we can have indexed summation, we can also have indexed multiplication, as the following example shows.
EXAMPLE 1.8
Evaluate
(i + j), where I = {2, 3, 5, 7}.
i,j∈I
i<j
SOLUTION
Given product = product of all numbers i + j, where i, j ∈ {2, 3, 5, 7} and i < j
= (2 + 3)(2 + 5)(2 + 7)(3 + 5)(3 + 7)(5 + 7)
= 5 · 7 · 9 · 8 · 10 · 12 = 302,400
The following exercises provide ample practice in both notations.
E X E R C I S E S
Evaluate each sum.
6
1.
i
3.
5.
7.
9.
i=1
4
(j − 1)
j=0
4
4.
(3n − 2)
n=0
4
6.
3k
k=−2
3
2.
8.
(3k)2
10.
k=−1
4
1.2
17. Sums of the form S =
(3 + k)
k=0
4
i=−1
2
3
j(j − 2)
j=−2
3
3(k2 )
k=−2
5
(3 − 2k)k
k=1
n
(ai − ai−1 ) are called
i=m+1
telescoping sums. Show that S = an − am .
1
18. Using Exercise 17 and the identity
=
i(i + 1)
n
1
1
1
−
, derive a formula for
.
i
i+1
i(i
+
1)
i=1
19. Using Exercise 17 and the identity (i + 1)2 − i2 =
n
i.
2i + 1, derive a formula for
i=1
20. Using Exercise 17 and the identity (i + 1)3 − i3 =
n
i2 .
3i2 + 3i + 1, derive a formula for the sum
i=1
Rewrite each sum using the summation notation.
11. 1 + 3 + 5 + · · · + 23
12. 31 + 32 + · · · + 310
13. 1 · 2 + 2 · 3 + · · · + 11 · 12
14. 1(1 + 2) + 2(2 + 2) + · · · + 5(5 + 2)
Determine whether each is true.
n
n
i=
(n + m − i)
15.
16.
i=m
n
i=m
i=m
n
xi =
i=m
xn+m−i
21. Using the ideas in Exercises 19 and 20, derive a forn
mula for
i3 .
i=1
Evaluate each.
5
6
(2i + 3j)
22.
24.
26.
i=1 j=1
5
6
i=1 j=1
3
(i2 − j + 1)
(i + 1)
i=0
23.
25.
3
i
(i + 3)
i=1 j=1
6
5
(i2 − j + 1)
j=1 i=1
5
27.
(j2 + 1)
j=3
15
1.3 Mathematical Induction
28.
50
43.
(−1)k
k=0
37.
39.
Evaluate each, where lg x = log2 x.
1023
45.
lg(1 + 1/n)
46.
47.
48.
i,j∈I
i|j
40.
4
42.
n=1
n
(1 + 1/n)
lg(1 + 1/n)
k · k! (Hint: Use Exercise 17.)
999
k!.
k=1
(3j − 3j−1 )
j=1
50. Find the hundreds digit in the sum
999
k · k!.
k=1
(Hint: Use Exercise 48.)
n
∞
51. Compute 10000 + 2 .
2n+1
n=0
(Hint: x + 1/2 = 2x − x; Source: Mathematics
Teacher, 1993.)
aij
j=1 i=1
1.3
n=1
1024
49. Find the tens digit in the sum
Expand each.
3
2
aij
41.
i=1 j=1
2
3
n=1
1023
k=1
i,j∈I
i≤j
(2i + 3j )
(ai + aj + ak )
1≤i<j<k≤3
j
38.
i
i,j∈I
i<j
44.
p≤10
(i + 2j)
(ai + aj )
1≤i<j≤3
Evaluate each, where p ∈ {2, 3, 5, 7, 11, 13} and
I = {1, 2, 3, 5}.
3
k!
30.
p
29.
k=0
p≤10
p
32.
(3i − 1)
31.
i∈I
p≤10
12
33.
d
34.
d
d≥1
d≥1
d|12
d|12
1
36.
1
35.
d≥1
d|18
Mathematical Induction
The principle of mathematical induction† (PMI) is a powerful proof technique that
we will use often in later chapters.
Many interesting results in mathematics hold true for all positive integers. For
example, the following statements are true for every positive integer n and all real
numbers x, y, and xi :
• (x · y)n = xn · yn
• log(x1 · · · xn ) =
n
log xi
i=1
† The term mathematical induction was coined by Augustus DeMorgan (1806–1871), although the
Venetian scientist Francesco Maurocylus (1491–1575) applied it much earlier, in proofs in a book
he wrote in 1575.
16
CHAPTER 1 Fundamentals
n
n(n + 1)
2
i=1
n−1
i rn − 1
•
(r = 1)
r =
r−1
i=0
•
i=
How do we prove that these results hold for every positive integer n? Obviously,
it is impossible to substitute each positive integer for n and verify that the formula
holds. The principle of induction can establish the validity of such formulas.
Before we plunge into induction, we need the well-ordering principle, which we
accept as an axiom. (An axiom is a statement that is accepted as true; it is consistent
with known facts; often it is a self-evident statement.)
The Well-Ordering Principle
Every nonempty set of positive integers has a least element.
For example, the set {17, 23, 5, 18, 13} has a least element, namely, 5. The elements of the set can be ordered as 5, 13, 17, 18, and 23.
By virtue of the well-ordering principle, the set of positive integers is well ordered. You may notice that the set of negative integers is not well ordered.
The following example is a simple application of the well-ordering principle.
EXAMPLE 1.9
Prove that there is no positive integer between 0 and 1.
Suppose there is a positive integer a between 0 and 1. Let S = {n ∈ Z+ | 0 < n < 1}.
Since 0 < a < 1, a ∈ S, so S is nonempty. Therefore, by the well-ordering principle,
S has a least element , where 0 < < 1. Then 0 < 2 < , so 2 ∈ S. But 2 < ,
which contradicts our assumption that is a least element of S. Thus, there are no
positive integers between 0 and 1.
The well-ordering principle can be extended to whole numbers also, as the following example shows.
EXAMPLE 1.10
Prove that every nonempty set of nonnegative integers has a least element.
PROOF (by cases)
Let S be a set of nonnegative integers.
Suppose 0 ∈ S. Since 0 is less than every positive integer, 0 is less than
every nonzero element in S, so 0 is a least element in S.
case 1
1.3 Mathematical Induction
17
case 2
Suppose 0 ∈
/ S. Then S contains only positive integers. So, by the wellordering principle, S contains a least element.
Thus, in both cases, S contains a least element.
Weak Version of Induction
The following theorem is the cornerstone of the principle of induction.
THEOREM 1.4
Let S be a set of positive integers satisfying the following properties:
1. 1 ∈ S.
2. If k is an arbitrary positive integer in S, then k + 1 ∈ S.
Then S = N.
Suppose S = N. Let S = {n ∈ N | n ∈
/ S}. Since S = ∅, by the well-ordering principle, S contains a least element . Then > 1 by condition (1). Since is the
/ S . Therefore, − 1 ∈ S. Consequently, by condition (2),
least element in S , − 1 ∈
( − 1) + 1 = ∈ S. This contradiction establishes the theorem.
This result can be generalized, as the following theorem shows. We leave its
proof as an exercise.
THEOREM 1.5
Let n0 be a fixed integer. Let S be a set of integers satisfying the following conditions:
• n0 ∈ S.
• If k is an arbitrary integer ≥ n0 such that k ∈ S, then k + 1 ∈ S.
Then S contains all integers n ≥ n0 .
Before we formalize the principle of induction, let’s look at a trivial example.
Consider an infinite number of identical dominoes arranged in a row at varying distances from each other, as in Figure 1.4(a). Suppose we knock down the first domino.
What happens to the rest of the dominoes? Do they all fall? Not necessarily. See Figures 1.4(b) and 1.4(c).
So let us assume the following: The dominoes are placed in such a way that the
distance between two adjacent dominoes is less than the length of a domino; the first
domino falls; and if the kth domino falls, then the (k + 1)st domino also falls. Then
they all would fall. See Figure 1.4(d).
This illustration can be expressed symbolically. Let P(n) denote the statement
that the nth domino falls. Assume the following statements are true:
18
CHAPTER 1 Fundamentals
Figure 1.4
• P(1).
• P(k) implies P(k + 1) for an arbitrary positive integer k.
Then P(n) is true for every positive integer n; that is, every domino would fall. This
is the essence of the following weak version of the principle.
THEOREM 1.6
(The Principle of Mathematical Induction)
the following conditions, where n ∈ Z:
Let P(n) be a statement satisfying
1. P(n0 ) is true for some integer n0 .
2. If P(k) is true for an arbitrary integer k ≥ n0 , then P(k + 1) is also true.
Then P(n) is true for every integer n ≥ n0 .
PROOF
Let S denote the set of integers ≥ n0 for which P(n) is true. Since P(n0 ) is true,
n0 ∈ S. By condition (2), whenever k ∈ S, k + 1 ∈ S, so, by Theorem 1.5, S contains
all integers ≥ n0 . Consequently, P(n) is true for every integer n ≥ n0 .
Condition (1) in Theorem 1.6 assumes the proposition P(n) is true when n = n0 .
Look at condition (2): If P(n) is true for an arbitrary integer k ≥ n0 , it is also true for
n = k + 1. Then, by repeated application of condition (2), it follows that P(n0 + 1),
P(n0 + 2), . . . hold true. In other words, P(n) holds for every n ≥ n0 .
Theorem 1.6 can be established directly from the well-ordering principle. See
Exercise 44.
19
1.3 Mathematical Induction
Proving a result by induction involves two key steps:
• basis step Verify that P(n0 ) is true.
• induction step Assume P(k) is true for an arbitrary integer k ≥ n0
(inductive hypothesis).
Then verify that P(k + 1) is also true.
A word of caution: A question frequently asked is, “Isn’t this circular reasoning?
Aren’t we assuming what we are asked to prove?” In fact, no. The confusion stems
from misinterpreting step 2 for the conclusion. The induction step involves showing
that P(k) implies P(k + 1); that is, if P(k) is true, then so is P(k + 1). The conclusion
is “P(n) is true for every n ≥ n0 .” So be careful.
Interestingly, there were television commercials for Crest toothpaste based on
induction involving toothpastes and penguins.
Some examples will show how useful this important proof technique is.
EXAMPLE 1.11
Prove that
1 + 2 + 3 + ··· + n =
n(n + 1)
2
for every positive integer n.
PROOF (by induction)
Let P(n) be the statement that
n
i = [n(n + 1)]/2.
i=1
To verify that P(1) is true (note: Here n0 = 1):
1
When n = 1, RHS = [1(1 + 1)]/2 = 1 =
i = LHS.† Thus, P(1) is true.
basis step
i=1
† LHS and RHS are abbreviations of left-hand side and right-hand side, respectively.
(1.4)
20
CHAPTER 1 Fundamentals
induction step Let k be an arbitrary positive integer. We would like to show that
P(k) implies P(k + 1). Assume P(k) is true; that is,
k
i=
i=1
k(k + 1)
2
← inductive hypothesis
To show that P(k) implies P(k + 1), that is,
k+1
i=1
the LHS of this equation:
LHS =
k+1
i=1
i=
k
i + (k + 1)
xi =
i=1
k(k + 1)
+ (k + 1),
2
(k + 1)(k + 2)
=
2
= RHS
=
Note:
i=1
k+1
k
xi + xk+1 .
i=1
by the inductive hypothesis
So, if P(k) is true, then P(k + 1) is also true.
Thus, by induction, P(n) is true for every integer n ≥ 1; that is, the formula holds
for every positive integer.
Figure 1.5 demonstrates formula (1.4) without words.
Figure 1.5
Often we arrive at a formula by studying patterns, then making a conjecture, and
then establishing the formula by induction, as the following example shows.
EXAMPLE 1.12
Conjecture a formula for the sum of the first n odd positive integers and then use
induction to establish the conjecture.
1.3 Mathematical Induction
21
SOLUTION
First, we study the first five such sums, and then look for a pattern, to predict a
formula for the sum of the first n odd positive integers.
The first five such sums are
1 = 12
1 + 3 = 22
1 + 3 + 5 = 32
1 + 3 + 5 + 7 = 42
1 + 3 + 5 + 7 + 9 = 52
There is a clear pattern here, so we conjecture that the sum of the first n odd positive
integers is n2 ; that is,
n
(2i − 1) = n2
(1.5)
i=1
We shall now prove it by the principle of induction.
PROOF
When n = 1,
n
(2i − 1) =
i=1
1
(2i − 1) = 1 = 12 , so the result holds when n = 1.
i=1
Now, assume the formula holds when n = k:
holds when n = k + 1, consider the sum
k+1
k
(2i − 1) = k2 . To show that it
i=1
(2i − 1). We have
i=1
k+1
k
(2i − 1) =
(2i − 1) + [2(k + 1) − 1]
i=1
i=1
2
= k + (2k + 1)
by the inductive hypothesis
= (k + 1)
2
Consequently, if the formula holds when n = k, it is also true when n = k + 1.
Thus, by induction, the formula holds for every positive integer n.
Figure 1.6 provides a visual illustration of formula (1.5).
Figure 1.6
22
CHAPTER 1 Fundamentals
Returning to induction, we find that both the basis and the induction steps are essential in the induction proof, as the following two examples demonstrate.
EXAMPLE 1.13
Consider the “formula” 1 + 3 + 5 + · · · + (2n − 1) = (n − 2)2 . Clearly it is true when
n = 1. But it is not true when n = 2.
Conclusion? That the truth of the basis step does not ensure that the statement
1 + 3 + 5 + · · · + (2n − 1) = (n − 2)2 is true for every n.
The following example shows that the validity of the induction step is necessary,
but not sufficient, to guarantee that P(n) is true for all desired integers.
EXAMPLE 1.14
Consider the “formula” P(n): 1 + 3 + 5 + · · · + (2n − 1) = n2 + 1. Suppose P(k) is
k
true: (2i − 1) = k2 + 1. Then
i=1
k+1
i=1
(2i − 1) =
k
(2i − 1) + [2(k + 1) − 1]
i=1
= (k2 + 1) + (2k + 1)
= (k + 1)2 + 1
So if P(k) is true, P(k + 1) is true. Nevertheless, the formula does not hold for any
positive integer n. Try P(1).
An interesting digression: Using induction, we “prove” in the following example
that every person is of the same sex.
EXAMPLE 1.15
“Prove” that every person in a set of n people is of the same sex.
PROOF
Let P(n): Everyone in a set of n people is of the same sex. Clearly, P(1) is true. Let
k be a positive integer such that P(k) is true; that is, everyone in a set of k people is
of the same sex.
To show that P(k + 1) is true, consider a set A = {a1 , a2 , . . . , ak+1 } of k + 1
people. Partition A into two overlapping sets, B = {a1 , a2 , . . . , ak } and C =
1.3 Mathematical Induction
23
{a2 , . . . , ak+1 }, as in Figure 1.7. Since B and C contain k elements, by the inductive hypothesis, everyone in B is of the same sex and everyone in C is of the same
sex. Since B and C overlap, everyone in B ∪ C † must be of the same sex; that is,
everyone in A is of the same sex.
Figure 1.7
Therefore, by induction, P(n) is true for every positive integer n.
Note: Clearly the assertion that everyone is of the same sex is false. Can you find the
flaw in the “proof?” See Exercise 35.
Strong Version of Induction
We now present the stronger version of induction.
Sometimes the truth of P(k) might not be enough to establish that of P(k + 1).
In other words, the truth of P(k + 1) may require more than that of P(k). In such
cases, we assume a stronger inductive hypothesis that P(n0 ), P(n0 + 1), . . . , P(k) are
all true; then verify that P(k + 1) is also true. This strong version, which can be
proven using the weak version (see Exercise 43), is stated as follows.
THEOREM 1.7
(The Second Principle of Mathematical Induction) Let P(n) be a statement satisfying the following conditions, where n ∈ Z:
1. P(n0 ) is true for some integer n0 .
† B ∪ C denotes the union of the sets B and C; it contains the elements in B together with those in C.
24
CHAPTER 1 Fundamentals
2. If k is an arbitrary integer ≥ n0 such that P(n0 ), P(n0 + 1), . . ., and P(k) are
true, then P(k + 1) is also true.
Then P(n) is true for every integer n ≥ n0 .
PROOF
Let S = {n ∈ Z | P(n) is true}. Since P(n0 ) is true by condition (1), n0 ∈ S.
Now, assume P(n0 ), P(n0 + 1), . . . , P(k) are true for an arbitrary integer k. Then
n0 , n0 + 1, . . . , k belong to S. So, by condition (2), k + 1 also belongs to S. Therefore,
by Theorem 1.5, S contains all integers n ≥ n0 . In other words, P(n) is true for every
integer n ≥ n0 .
The following example illustrates this proof technique.
EXAMPLE 1.16
Prove that any postage of n (≥ 2) cents can be made with two- and three-cent stamps.
PROOF (by strong induction)
Let P(n) denote the statement that any postage of n cents can be made with two- and
three-cent stamps.
basis step (Notice that here n0 = 2.) Since a postage of two cents can be made
with one two-cent stamp, P(2) is true. Likewise, P(3) is also true.
induction step Assume P(2), P(3), P(4), . . . , P(k) are true; that is, any postage of
two through k cents can be made with two- and three-cent stamps.
To show that P(k + 1) is true, consider a postage of k + 1 cents. Since k + 1 =
(k − 1) + 2, a postage of k + 1 cents can be formed with two- and three-cent stamps if
a postage of k − 1 cents can be made with two- and three-cent stamps. Since P(k − 1)
is true by the inductive hypothesis, this implies P(k + 1) is also true.
Thus, by the strong version of induction, P(n) is true for every n ≥ 2; that is, any
postage of n (≥ 2) cents can be made with two- and three-cent stamps.
The following exercises and subsequent chapters offer ample practice in both
versions of induction.
E X E R C I S E S
Determine whether each set is well ordered. If it is not,
explain why.
1. Set of negative integers.
2. Set of integers.
1.3
3. {n ∈ N | n ≥ 5}
4. {n ∈ Z | n ≥ −3}
Prove each.
5. Let a ∈ Z. There are no integers between a and a + 1.
25
1.3 Mathematical Induction
6. Let n0 ∈ Z, S a nonempty subset of the set T = {n ∈
Z | n ≥ n0 }, and ∗ be a least element of the set
T ∗ = {n − n0 + 1 | n ∈ S}. Then n0 + − 1 is a least
element of S.
7. (Archimedean property) Let a and b be any positive integers. Then there is a positive integer n such
that na ≥ b.
(Hint: Use the well-ordering principle and contradiction.)
8. Every nonempty set of negative integers has a largest
element.
9. Every nonempty set of integers ≤ a fixed integer n0
has a largest element.
(Twelve Days of Christmas) Suppose you sent your
love 1 gift on the first day of Christmas, 1 + 2 gifts on the
second day, 1 + 2 + 3 gifts on the third day, and so on.
10. How many gifts did you send on the 12th day of
Christmas?
of Christmas?
12. Prove that 1 + 2 + · · · + n = [n(n + 1)]/2 by considering the sum in the reverse order.† (Do not use
mathematical induction.)
Using mathematical induction, prove each for every integer n ≥ 1.
13.
n
(2i − 1) = n2
i=1
n
n(n + 1)(2n + 1)
14.
i2 =
6
i=1
† An interesting personal anecdote is told about Gauss. When
by his teacher to compute the sum of the first 100 positive integers. Supposedly, the teacher did so to get some time to grade
papers. To the teacher’s dismay, Gauss found the answer in a
few moments by pairing the numbers from both ends:
The sum of each pair is 101 and there are 50 pairs. So the total
sum is 50 · 101 = 5050.
n
n(n + 1) 2
i3 =
2
i=1
n − 1)
n
a(r
,
16.
ari−1 =
r−1
i=1
15.
Evaluate each sum.
30
(3k2 − 1)
17.
19.
k=1
n
i/2
r = 1
18.
20.
i=1
50
(k3 + 2)
k=1
n
i/2
i=1
Find the value of x resulting from executing each algorithm fragment, where
variable ← expression
means the value of expression is assigned to variable.
21. x ← 0
for i = 1 to n do
x ← x + (2i − 1)
22. x ← 0
for i = 1 to n do
x ← x + i(i + 1)
23. x ← 0
for i = 1 to n do
for j = 1 to i do
x←x+1
Evaluate each.
24.
26.
n
i
i=1 j=1
n
i
i=1 j=1
n
22i
28.
i=1
n
n
30.
i=1 j=1
i
j2
ij
25.
27.
n
i
i=1 j=1
n
i
i=1 j=1
n
29.
i2
i=1
n
n
31.
j
(2j − 1)
2i+j
i=1 j=1
32. A magic square of order n is a square arrangement
of the positive integers 1 through n2 such that the sum
of the integers along each row, column, and diagonal
is a constant k, called the magic constant. Figure 1.8
shows two magic squares, one of order 3 and the other
of order 4. Prove that the magic constant of a magic
n(n2 + 1)
.
square of order n is
2
26
CHAPTER 1 Fundamentals
Figure 1.8
According to legend, King Shirham of India was so
pleased by the invention of chess that he offered
to give Sissa Ben Dahir, its inventor, anything he
wished. Dahir’s request was a seemingly modest one:
one grain of wheat on the first square of a chessboard,
two on the second, four on the third, and so on. The
king was delighted with this simple request but soon
realized he could never fulfill it. The last square alone
would take 263 = 9,223,372,036,854,775,808 grains
of wheat. Find the following for an n × n chessboard.
33. The number of grains on the last square.
34. The total number of grains on the chessboard.
35. Find the flaw in the “proof” in Example 1.15.
Find the number of times the assignment statement
x ← x + 1 is executed by each loop.
36. for i = 1 to n do
for j = 1 to i do
x←x+1
37. for i = 1 to n do
for j = 1 to i do
for k = 1 to i do
x←x+1
1.4
38. for i = 1 to n do
for j = 1 to i do
for k = 1 to j do
x←x+1
39. for i = 1 to n do
for j = 1 to i do
for k = 1 to i do
for l = 1 to i do
x←x+1
40. Let an denote the number of times the statement
x ← x + 1 is executed in the following loop:
for i = 1 to n do
for j = 1 to i/2 do
x←x+1
⎧
⎪
n2
⎪
⎪
if n is even
⎨
4
Show that an =
⎪
⎪
n2 − 1
⎪
⎩
otherwise.
4
Evaluate each.
41.
1024
lg n
n=1
42.
1024
lg n
n=1
43. Prove the strong version of induction, using the
weak version.
44. Prove the weak version of induction, using the
well-ordering principle.
45. Let S denote the sum of the elements in the nth
n
set of the sequence of sets of squares {1}, {4, 9},
{16, 25, 36}, . . .. Find a formula for Sn . (J. M. Howell, 1989)
Recursion
Recursion is one of the most elegant problem-solving techniques. It is so powerful
a tool that most programming languages support it.
We begin with the well-known handshake problem:
There are n guests at a party. Each person shakes hands with everybody else exactly once. How many
1.4 Recursion
27
If we decide to solve a problem such as this, the solution may not be obvious.
However, it is possible that the problem could be defined in terms of a simpler version of itself. Such a definition is an inductive definition. Consequently, the given
problem can be solved provided the simpler version can be solved. This idea is pictorially represented in Figure 1.9.
Figure 1.9
Recursive Definition of a Function
Let a ∈ W and X = {a, a + 1, a + 2, . . .}. An inductive definition of a function f
with domain X consists of three parts:
• Basis step A few initial values f (a), f (a + 1), . . . , f (a + k − 1) are specified.
Equations that specify such initial values are initial conditions.
• Recursive step A formula to compute f (n) from the k preceding functional
values f (n−1), f (n−2), . . . , f (n−k) is made. Such a formula is a recurrence
relation (or recursive formula).
• Terminal step Only values thus obtained are valid functional values. (For
convenience, we drop this clause from the recursive definition.)
In a recursive definition of f , f (n) may be defined using the values f (k), where
k = n, so not all recursively defined functions can be defined inductively; see Exercises 25–31.
Thus, the recursive definition of f consists of a finite number of initial conditions
and a recurrence relation.
Recursion can be employed to find the minimum and maximum of three
or more real numbers. For instance, min{w, x, y, z} = min{w, {min{x, min{y, z}}}};
max{w, x, y, z} can be evaluated similarly. For example,
min{23, 5, −6, 47, 31} = min{23, min{5, min{−6, min{47, 31}}}} = −6
and
max{23, 5, −6, 47, 31} = max{23, max{5, max{−6, max{47, 31}}}} = 47
The next three examples illustrate the recursive definition.
28
CHAPTER 1 Fundamentals
EXAMPLE 1.17
Define recursively the factorial function f .
SOLUTION
Recall that the factorial function f is defined by f (n) = n!, where f (0) = 1. Since
n! = n(n − 1)!, it can be defined recursively as follows:
f (0) = 1
f (n) = n · f (n − 1),
n≥1
← initial condition
← recurrence relation
Suppose we would like to compute f (3) recursively. We must continue to apply
the recurrence relation until the initial condition is reached, as shown below:
Since f (0) = 1, 1 is substituted for f (0) in equation (1.8) and f (1) is computed:
f (1) = 1 · f (0) = 1 · 1 = 1. This value is substituted for f (1) in equation (1.7) and
f (2) is computed: f (2) = 2 · f (1) = 2 · 1 = 2. This value is now returned to equation (1.6) to compute f (3): f (3) = 3 · f (2) = 3 · 2 = 6, as expected.
EXAMPLE 1.18
(The handshake problem) There are n guests at a party. Each person shakes hands
with everybody else exactly once. Define recursively the number of handshakes h(n)
SOLUTION
Clearly, h(1) = 0, so let n ≥ 2. Let x be one of the guests. The number of handshakes
made by the remaining n − 1 guests among themselves, by definition, is h(n − 1).
Now person x shakes hands with each of these n − 1 guests, yielding n − 1 handshakes. So the total number of handshakes made equals h(n − 1) + (n − 1), where
n ≥ 2.
Thus, h(n) can be defined recursively as follows:
h(1) = 0
h(n) = h(n − 1) + (n − 1),
← initial condition
n≥2
← recurrence relation
1.4 Recursion
EXAMPLE 1.19
29
(Tower of Brahma† ) According to a legend, at the beginning of creation, God
stacked 64 golden disks on one of three diamond pegs on a brass platform in the
temple of Brahma at Benares,‡ India (see Figure 1.10). The priests on duty were
asked to move the disks from peg X to peg Z, using Y as an auxiliary peg, under the
following conditions:
Figure 1.10
• Only one disk can be moved at a time.
• No disk can be placed on the top of a smaller disk.
The priests were told the world would end when the job was completed.
Suppose there are n disks on peg X. Let bn denote the number of moves needed
to move them from peg X to peg Z, using peg Y as an intermediary. Define bn recursively.
SOLUTION
Clearly b1 = 1. Assume n ≥ 2. Consider the top n − 1 disks at peg X. By definition,
it takes bn−1 moves to transfer them from X to Y using Z as an auxiliary. That leaves
the largest disk at peg X; it takes one move to transfer it from X to Z. See Figure 1.11.
Figure 1.11
† A puzzle based on the Tower of Brahma was marketed by the French mathematician François-
Edouard-Anatole Lucas in 1883 under the name Tower of Hanoi.
‡ Benares is now known as Varanasi.
30
CHAPTER 1 Fundamentals
Now the n − 1 disks at Y can be moved from Y to Z using X as an intermediary
in bn−1 moves, so the total number of moves needed is bn−1 + 1 + bn−1 = 2bn−1 + 1.
Thus bn can be defined recursively as follows:
bn =
if n = 1
1
2bn−1 + 1 if n ≥ 2
← initial condition
← recurrence relation
For example,
b4 = 2b3 + 1
= 2[2b2 + 1] + 1
= 4[2b1 + 1] + 2 + 1
= 4b2 + 2 + 1
= 8b1 + 4 + 2 + 1 = 8(1) + 4 + 2 + 1 = 15,
so it takes 15 moves to transfer 4 disks from X to Z.
Notice that the recursive definition of a function f does not provide us with
an explicit formula for f (n) but establishes a systematic procedure for finding it.
The iterative method of finding a formula for f (n) involves two steps: 1) apply the
recurrence formula iteratively and look for a pattern to predict an explicit formula;
2) use induction to prove that the formula does indeed hold for every possible value
of the integer n.
The following example illustrates this method.
EXAMPLE 1.20
Solve the recurrence relation in Example 1.18.
SOLUTION
Using iteration, we have:
h(n) = h(n − 1) + (n − 1)
= h(n − 2) + (n − 2) + (n − 1)
= h(n − 3) + (n − 3) + (n − 2) + (n − 1)
..
.
= h(1) + 1 + 2 + 3 + · · · + (n − 2) + (n − 1)
= 0 + 1 + 2 + 3 + · · · + (n − 1)
n(n − 1)
, by Example 1.11
=
2
(We can verify this using induction.)
31
1.4 Recursion
E X E R C I S E S
In Exercises 1–6, compute the first four terms of the sequence defined recursively.
1. a1 = 1
an = an−1 + 3, n ≥ 2
2. a0 = 1
an = an−1 + n, n ≥ 1
3. a1 = 1
n
an−1 , n ≥ 2
an =
n−1
4. a1 = 1, a2 = 2
an = an−1 + an−2 , n ≥ 3
5. a1 = 1, a2 = 1, a3 = 2
an = an−1 + an−2 + an−3 , n ≥ 4
6. a1 = 1, a2 = 2, a3 = 3
an = an−1 + an−2 + an−3 , n ≥ 4
Define recursively each number sequence.
(Hint: Look for a pattern and define the nth term
an recursively.)
7. 1, 4, 7, 10, 13, . . .
8. 3, 8, 13, 18, 23, . . .
9. 0, 3, 9, 21, 45, . . .
10. 1, 2, 5, 26, 677, . . .
An arithmetic sequence is a number sequence in which
every term except the first is obtained by adding a fixed
number, called the common difference, to the preceding
term. For example, 1, 3, 5, 7, . . . is an arithmetic sequence
with common difference 2. Let an denote the nth term of
the arithmetic sequence with first term a and common difference d.
11. Define an recursively.
12. Find an explicit formula for an .
13. Let Sn denote the sum of the first n terms of the sequence. Prove that
Sn =
n
2a + (n − 1)d
2
A geometric sequence is a number sequence in which
every term, except the first, is obtained by multiplying the
previous term by a constant, called the common ratio.
For example, 2, 6, 18, 54, . . . is a geometric sequence with
common ratio 3. Let an denote the nth term of the geometric sequence with first term a and common ratio r.
14. Define an recursively.
1.4
15. Find an explicit formula for an .
16. Let Sn denote the sum of the first n terms of the sequence. Prove that Sn = [a(rn − 1)]/(r − 1), where
r = 1. Do not use induction.
Use the following triangular array of positive integers to
1
2
4
7
3
5
8
6
9
10
..
.
17. Let an denote the first term in row n, where n ≥ 1.
Define an recursively.
18. Find an explicit formula for an .
19. Find the sum of the numbers in row n.
20. Which row contains the number 2076?
Let an denote the number of times the assignment statement x ← x + 1 is executed by each nested for loop. Define an recursively.
21. for i = 1 to n do
for j = 1 to i do
x←x+1
22. for i = 1 to n do
for j = 1 to i do
for k = 1 to i do
x←x+1
23. Using Example 1.19, predict an explicit formula
for bn .
24. Using induction, prove the explicit formula for bn in
Exercise 23.
The 91-function f , invented by John McCarthy, is defined
recursively on W as follows:
x − 10
if x > 100
f (x) =
f (f (x + 11)) if 0 ≤ x ≤ 100
Compute each:
25. f (99)
26. f (98)
27. f ( f (99))
32
CHAPTER 1 Fundamentals
John McCarthy (1927– ), one of the fathers of artificial intelligence (AI), was born
in Boston. He graduated in mathematics from Caltech and received his Ph.D. from
Princeton in 1951. After teaching at Princeton, Stanford, Dartmouth, and MIT, he returned to Stanford as full professor. While at Princeton, he was named a Proctor
Fellow and later Higgins Research Instructor in mathematics. At Stanford, he headed
its Artificial Intelligence Laboratory.
McCarthy coined the term artificial intelligence while at Dartmouth. He developed
LISP (LISt Programming), one of the most widely used programming languages in Al.
In addition, he helped develop ALGOL 58 and ALGOL 60. In 1971, he received the
prestigious Alan M. Turing Award for his outstanding contributions to data processing.
28. f ( f (91))
29. Show that f (99) = 91.
30. Prove that f (x) = 91 for 90 ≤ x ≤ 100.
31. Prove that f (x) = 91 for 0 ≤ x < 90.
A function of theoretical importance in the study of algorithms is Ackermann’s function, named after the German mathematician and logician Wilhelm Ackermann
(1896–1962). It is defined recursively as follows, where
m, n ∈ W:
⎧
⎪
if m = 0
⎪
⎨n + 1
A(m, n) = A(m − 1, 1)
if n = 0
⎪
⎪
⎩ A(m − 1, A(m, n − 1)) otherwise
Compute each.
32. A(0, 7)
33. A(1, 1)
35. A(2, 2)
Prove each for every integer n ≥ 0.
36. A(1, n) = n + 2
37. A(2, n) = 2n + 3
38. Predict a formula for A(3, n).
39. Prove the formula in Exercise 38 for every integer
n ≥ 0.
40. Let {u } be a number sequence with u = 4 and
n
0
un = f (un−1 ), where f is a function defined by the
following table and n ≥ 1. Compute u9999 . (Source:
Mathematics Teacher, 2004)
x
f (x)
1.5
34. A(5, 0)
1
4
2
1
3
3
4
5
5
2
The Binomial Theorem
Binomials are sums of two terms, and they occur often in mathematics. This section
shows how to expand positive integral powers of binomials in a systematic way. The
coefficients in binomial expansions have several interesting properties.
Let us begin with a discussion of binomial coefficients.
1.5 The Binomial Theorem
Binomial Coefficients
coefficient†
33
n
is defined by
r
Let n and r be nonnegative integers. The binomial
n
n!
=
if r ≤ n, and is 0 otherwise; it is also denoted by C(n, r) and nCr.
r
r!(n − r)!
For example,
5
5!
=
3!(5 − 3)!
3
5·4·3·2·1
=
= 10
3·2·1·2·1
It follows from the definition that
n
n
=1=
.
0
n
Thereare many
instances when we need to compute the binomial coefficients
n
n
and
. Since
r
n−r
n
n!
=
(n − r)![n − (n − r)]!
n−r
n
n!
n!
=
=
=
(n − r)!r! r!(n − r)!
r
thereis no
both;
need
to evaluate
this significantly reduces our workload. For exam25
25
25
ple,
=
=
= 53,130.
20
25 − 20
5
The following theorem shows an important recurrence relation satisfied by binomial coefficients. It is called Pascal’s identity, after the outstanding French mathematician and philosopher Blaise Pascal.
† The term binomial coefficient was introduced by the German algebraist Michel Stifel (1486–1567).
In his best-known work, Arithmetica Integra (1544), Stifel gives the binomial coefficients for n ≤ 17.
The bilevel parentheses notation for binomial coefficient was introduced by the German mathematician and physicist Baron Andreas von Ettinghausen (1796–1878). Von Ettinghausen, born in
Heidelberg, attended the University of Vienna in Austria. For two years he worked as an assistant
in mathematics and physics at the University. In 1821 he became professor of mathematics, and in
1835, professor of physics and director of the Physics Institute. Thirteen years later, he became the
director of the Mathematical Studies and Engineering Academy in Vienna.
A pioneer in mathematical physics, von Ettinghausen worked in analysis, algebra, differential
geometry, mechanics, optics, and electromagnetism.
34
CHAPTER 1 Fundamentals
Blaise Pascal (1623–1662) was born in Clermont-Ferrand, France. Although he
showed astounding mathematical ability at an early age, he was encouraged by his
father to pursue other subjects, such as ancient languages. His father even refused to
teach him any sciences and relented only when he found that Pascal by age 12 had
discovered many theorems in elementary geometry. At 14, Blaise attended weekly
meetings of a group of French mathematicians which later became the French Academy. At 16, he developed important results in conic sections and wrote a book on
them.
Observing that his father would spend countless hours auditing government accounts, and feeling that intelligent people should not waste their time doing mundane
things, Pascal, at the age of 19, invented the first mechanical calculating machine.
THEOREM 1.8
n
(Pascal’s Identity) Let n and r be positive integers, where r ≤ n. Then
=
r
n−1
n−1
+
.
r−1
r
PROOF
We shall simplify the RHS and show that it is equal to the LHS:
n−1
n−1
(n − 1)!
(n − 1)!
+
=
+
r−1
r
(r − 1)!(n − r)! r!(n − r − 1)!
(n − r)(n − 1)!
r(n − 1)!
+
=
r(r − 1)!(n − r)! r!(n − r)(n − r − 1)!
(n − r)(n − 1)!
r(n − 1)!
+
=
r!(n − r)!
r!(n − r)!
n!
(n − 1)![r + (n − r)] (n − 1)!n
=
=
=
r!(n − r)!
r!(n − r)! r!(n − r)!
n
=
r
Pascal’s Triangle
n
, where 0 ≤ r ≤ n, can be arranged in the form
r
of a triangle, called Pascal’s triangle,† as in Figures 1.12 and 1.13.
The various binomial coefficients
† Although Pascal’s triangle is named after Pascal, it actually appeared as early as 1303 in a work by
the Chinese mathematician Chu Shi-Kie.
1.5 The Binomial Theorem
Figure 1.12
Figure 1.13
Figure 1.14
Figure 1.14 shows the Chinese and Japanese versions of Pascal’s triangle.
35
36
CHAPTER 1 Fundamentals
Pascal’s triangle has many intriguing properties:
• Every row begins with and ends in 1.
• Pascal’s triangle is symmetric about a vertical line through the middle. This
is so by Theorem 1.8.
• Any interior number in each row is the sum of the numbers immediately to its
left and to its right in the preceding row; see Figure 1.13. This is so by virtue
of Pascal’s identity.
• The sum of the numbers in any row is a power of 2. Corollary 1.1 will verify
this.
• The nth row can be used to determine 11n . For example, 113 = 1331 and
114 = 14,641. To compute higher powers of 11, you should be careful since
some of the numbers involve two or more digits. For instance, to compute 115
list row 5:
From right to left, list the single-digit numbers. When we come to a two-digit
number, write the ones digit and carry the tens digit to the number on the left.
Add the carry to the number to its left. Continue this process to the left. The
resulting number, 161,051, is 115 .
• Form a regular hexagon with vertices on three adjacent rows (see Figure 1.15).
Find the products of numbers at alternate vertices. The two products are
equal. For example, 10 · 15 · 4 = 6 · 20 · 5. Surprised? Supplementary Exercise 10 confirms this property, known as Hoggatt–Hansell identity, named
after V. E. Hoggatt, Jr., and W. Hansell, who discovered it in 1971; so the
product of the six numbers is a square.
Figure 1.15
37
1.5 The Binomial Theorem
The following theorem shows how the binomial coefficients can be used to find
the binomial expansion of (x + y)n .
THEOREM 1.9
(The Binomial Theorem)†
tive integer. Then
(x + y)n
Let
xand y be any real numbers, and n any nonnegan
n n−r r
x y.
=
r=0 r
PROOF (by weak induction)
When n = 0, LHS
= (x + y)0
0
r 0−r r
x y = x0 y0 = 1, so LHS =
= 1 and RHS =
r=0 0
RHS.
Assume P(k) is true for some k ≥ 0:
k
k k−r r
k
x y
(x + y) =
r
r=0
Then
(1.10)
(x + y)k+1
= (x + y)k (x + y)
k
k k−r r
x y (x + y),
=
r
by equation (1.10)
r=0
k
k k−r r+1
x y
r
r
r=0
r=0
k
k−1
k k+1 k k+1−r r
k k+1
k k−r r+1
=
x
+
y +
+
y
x
x y
0
k
r
r
=
k
k
xk+1−r yr +
r=1
r=0
k
k
k
k + 1 k+1 k k+1−r r
k + 1 k+1
x
xk+1−r yr +
=
x
+
y+
y
r
r−1
0
k+1
r=1
r=1
k
k
k + 1 k+1 k
k + 1 k+1
+
xk+1−r yr +
=
x
+
y
r
r−1
0
k+1
r=1
k
k + 1 k+1 k + 1 k+1−r r
k + 1 k+1
x
=
x
+
y+
x ,
r
0
k+1
by Theorem 1.8
r=1
k+1
k + 1 k+1−r r
x
=
y
r
r=0
Thus, by induction, the formula is true for every integer n ≥ 0.
† The binomial theorem for n = 2 can be found in Euclid’s work (ca. 300 B . C .).
38
CHAPTER 1 Fundamentals
It follows from the binomial theorem that the binomial coefficients in the expansion of (x + y)n are the various numbers in row n of Pascal’s triangle.
The binomial theorem can be used to establish several interesting identities involving binomial coefficients, as the following corollary shows.
n
n
COROLLARY † 1.1
r=0
r
= 2n
That is, the sum of the binomial coefficients is 2n .
This follows by letting x = 1 = y in the binomial theorem.
The following exercises provide opportunities to explore additional relationships.
E X E R C I S E S
(Twelve Days of Christmas) Suppose that ``` | 32,747 | 102,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-22 | latest | en | 0.698284 |
http://programming4.us/desktop/386.aspx | 1,721,377,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514900.59/warc/CC-MAIN-20240719074314-20240719104314-00125.warc.gz | 24,612,073 | 10,630 | DESKTOP
# Algorithms for Compiler Design: EXAMPLES for Bottom-up Parsing
7/24/2010 7:48:36 PM
##### 5.7 EXAMPLES EXAMPLE 1 Construct an SLR(1) parsing table for the following grammar: First, augment the given grammar by adding a production S1 → S to the grammar. Therefore, the augmented grammar is: Next, we obtain the canonical collection of sets of LR(0) items, as follows: The transition diagram of this DFA is shown in Figure 1. The FOLLOW sets of the various nonterminals are FOLLOW(S1) = {\$}. Therefore: Using S1 → S, we get FOLLOW(S) = FOLLOW(S1) = {\$} Using S → xAy, we get FOLLOW(A) = {y} Using S → xBy, we get FOLLOW(B) = {y} Using S → xAz, we get FOLLOW(A) = {z} Therefore, FOLLOW(A) = {y, z}. Using A → qS, we get FOLLOW(S) = FOLLOW(A) = {y, z}. Therefore, FOLLOW(S) = {y, z, \$}. Let the productions of the grammar be numbered as follows: The SLR parsing table for the productions above is shown in Table 1. Table 1: SLR(1) Parsing Table Action Table GOTO Table � x Y Z q \$ S A B I0 S2 R3/R4 ��� 1 �� I1 �� Accept ����� I2 ��� S5 �� 3 4 I3 � S6 S7 ����� I4 � S8 ������ I5 S2 R5/R6 R5 �� 9 �� I6 � R1 R1 � R1 ��� I7 � R3 R3 � R3 ��� I8 � R2 R2 � R2 ��� I9 � R4 R4 ����� EXAMPLE 2 Construct an SLR(1) parsing table for the following grammar: First, augment the given grammar by adding the production S1 → S to the grammar. The augmented grammar is: Next, we obtain the canonical collection of sets of LR(0) items, as follows: The transition diagram of the DFA is shown in Figure 2. The FOLLOW sets of the various nonterminals are FOLLOW(S1) = {\$}. Therefore: Using S1 → S, we get FOLLOW(S) = FOLLOW(S1) = {\$} Using S → 0S0, we get FOLLOW(S) = { 0 } Using S → 1S1, we get FOLLOW(S) = {1} So, FOLLOW(S) = {0, 1, \$}. Let the productions be numbered as follows: The SLR parsing table for the production set above is shown in Table 2. Table 2: SLR Parsing Table for Example 1 Action Table GOTO Table � 0 1 \$ S I0 S2 S3 � 1 I1 ��� accept I2 S2 S3 � 4 I3 S6 S3 � 5 I4 S7 ��� I5 � S8 �� I6 S2/R3 S3/R3 R3 4 I7 R1 R1 � R1 I8 R2 R2 � R2 EXAMPLE 3 Consider the following grammar, and construct the LR(1) parsing table. The augmented grammar is: The canonical collection of sets of LR(1) items is: The parsing table for the production above is shown in Table 3. Table 3: Parsing Table for Example 2 Action Table GOTO Table � A B \$ S I0 S2 S3 R3 1 I1 �� Accept � I2 S5 S6/R3 � 4 I3 S8/R3 S9 � 7 I4 � S10 �� I5 S5 S6/R3 � 11 I6 S8/R3 S9 � 12 I7 S13 ��� I8 S5 S6/R3 � 14 I9 S8/R3 S9 � 15 I10 S2 S3 R3 16 I11 � S17 �� I12 S18 ��� I13 S2 S3 R3 19 I14 � S20 �� I15 � S21 �� I16 �� R1 � I17 S5 S6/R3 � 22 I18 S5 S6/R3 � 23 I19 �� R2 � I20 S8/R3 S9 � 24 I21 S8/R3 S9 � 25 I22 � R1 �� I23 � R2 �� I24 R1 ��� I25 R2 ��� The productions for the grammar are numbered as shown below: EXAMPLE 4 Construct an LALR(1) parsing table for the following grammar: The augmented grammar is: The canonical collection of sets of LR(1) items is: There no sets of LR(1) items in the canonical collection that have identical LR(0)-part items and that differ only in their lookaheads. So, the LALR(1) parsing table for the above grammar is as shown in Table 4. Table 4: LALR(1) Parsing Table for Example 3 Action Table GOTO Table � a b c d \$ S A I0 � S3 � S4 � 1 2 I1 ���� Accept �� I2 S5 ������ I3 ��� S7 � 1 � I4 R5 � S8 ���� I5 ���� R1 �� I6 S10 � S9 ���� I7 �� R5 ���� I8 ���� R3 �� I9 ���� R2 �� I10 ���� R4 �� The productions of the grammar are numbered as shown below: S → Aa S → bAc S → dc S → bda A → d EXAMPLE 5 Construct an LALR(1) parsing table for the following grammar: The augmented grammar is: The canonical collection of sets of LR(1) items is: Since no sets of LR(1) items in the canonical collection have identical LR(0)-part items and differ only in their lookaheads, the LALR(1) parsing table for the above grammar is as shown in Table 5. Table 5: LALR(1) Parsing Table for Example 4 Action Table GOTO Table � a b c d \$ S A B I0 S4 S5 � S6 � 1 2 3 I1 ���� Accept ��� I2 S7 ������� I3 �� S8 ����� I4 ��� S10 �� 9 � I5 ��� S12 ��� 11 I6 R5 � R6 ����� I7 ���� R1 ��� I8 ���� R3 ��� I9 �� S13 ����� I10 �� R5 ����� I11 S14 ������� I12 R6 ������� I13 ���� R2 ��� I14 ���� R4 ��� The productions of the grammar are numbered as shown below: S → Aa S → aAc S → Bc S → bBa A → d B → d EXAMPLE 6 Construct the nonempty sets of LR(1) items for the following grammar: The collection of nonempty sets of LR(1) items is shown in Figure 3. var sc_project=11388663; var sc_invisible=1; var sc_security="7db37af3"; var scJsHost = (("https:" == document.location.protocol) ? "https://secure." : "http://www."); document.write("<sc"+"ript type='text/javascript' src='" + scJsHost+ "statcounter.com/counter/counter.js'></"+"script>");
Other
Top 10
REVIEW
- First look: Apple Watch
- 3 Tips for Maintaining Your Cell Phone Battery (part 1)
- 3 Tips for Maintaining Your Cell Phone Battery (part 2) | 1,672 | 4,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.726895 |
http://www.dositey.com/math/ls20/107/MoneyOverview.htm | 1,469,528,484,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00085-ip-10-185-27-174.ec2.internal.warc.gz | 406,314,824 | 3,562 | Money Math is a mini-course that helps students learn counting, adding, and making sums up to \$1.00 with pennies, nickels, dimes, and quarters.
This is a self-guided course consisting of seven units. Each unit starts with an introductory lesson, and then gradually progresses to more challenging problems involving money mathematics.
The first unit Pennies is offered for FREE. Try the Free DEMO Unit 1
What is in Money Math?
Money Math consists of 7 units.
Units Unit 1 Pennies Unit 2 Nickels Unit 3 Pennies and Nickels - A review unit where the lesson and exercises involve problems with combination of pennies and nickels. Unit 4 Dimes Unit 5 Pennies, Nickels, and Dimes - A review unit where the lesson and exercises involve problems with combination of pennies, nickels, and dimes. Unit 6 Quarters Unit 7 Pennies, Nickels, Dimes, and Quarters A review unit where the lesson and exercises involve problems with combination of pennies, nickels, dimes, and quarters.
Learning Sessions
Each unit consists of up to six learning sessions (LSs). One LS consists of an interactive program and several follow-up printable worksheets. The average time a student spends on an LS is between 10 and 30 minutes.
LS 1 - Meet this Coin
Introductory lesson where students learn to recognize and count the coins.
LS 2 - Count the Money
An interactive exercise where students count the coins on the screen and write the total sum.
LS 3 - Make the Money Amounts
An interactive exercise where students have to make the assigned amount of money by providing the right number and combination of coins.
LS 4 - Test
Take online test. The test scores are saved in a progress report.
Some units have two additional LSs featuring exercises with higher amounts of money.
The first unit Pennies is offered for FREE. Try the Free DEMO Unit 1
Home | 416 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-30 | longest | en | 0.918175 |
https://kuieck.edublogs.org/tag/apps/ | 1,603,414,150,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00026.warc.gz | 404,204,643 | 12,410 | Geometric Shapes
Our second graders are in the middle of their study of geometry including measuring in centimeters and inches as well as building and identifying 2D and 3D shapes ( triangle, square, rectangle, quadrilateral, pentagon, hexagon, cube, pyramid, rectangular prism, cone and cylinder). Last week we explored these shapes in a fun way…. with marshmallows!!!
Here are a few apps that would provide practice for your child in this area (click on name for link to app):
GEOBOARD APP
GEOMETRY 4 KIDS APP
MONTESSORI GEOMETRY APP
New Math Unit :)!
This week our second graders will be revisiting addition and subtraction with and without regrouping, but this time we will be solving equations that have three digits. Here is the parent letter that will be coming today :).
Continue to work on addition and subtraction fact fluency at home through the links on our blog, tenmarks, xtramath and various apps. Here is a screenshot of the apps that we use at school to increase our fact fluency. If you have an ipad, ipod, or iphone, these apps would be a GREAT option to download on your device. Many of the apps are free and some are only \$0.99.
Well… I am VERY excited to be back with my kiddos tomorrow!!! We had a nasty chest infection in our house last week and we are FINALLY making our way back to healthy :).
We have a BUSY schedule schedule this week, with lots of learning adventures and fun activities.
Writer’s Workshop: We are a little behind from last week, so we will be wrapping up our study of reviews on Tuesday with a celebration… YEAH!!! We will then be refreshing our memories in the area of small moment writing ( personal narrative stories) in preparation for our end of the year writing “assessment”. It is truly amazing how much each of the children has grown as a writer.
Daily 5: This week we will be looking at various non fiction texts and using a few of the comprehension strategies that we have learned to answer questions about the text. I will also be starting end of the year reading assessments ( HARD TO BELIEVE!!!!!). Through informal conferences that I have had with children, I am thrilled with their growth and can’t wait to marvel at their reading!
Math: Our topic of study will be triple digit addition and subtraction problems. This will require quick recall of both + and – facts. Any extra practice that you could provide at home would be helpful. Here are a few apps that are available for the ipod touch and iphone that are fun for the kids and GOOD for them :):
* SPEEDO MATH ( FREE)
Science: We will spend a few days this week reviewing how water moves through various landforms and bodies of water. Our assessment will be on Thursday. Look for a study guide to come home on Wednesday. Our next unit of study will be PLANTS… YEAH!!!!
Word Study: After this week, we only have TWO new word lists…. I know, try not to shed a tear :).
Here are our words for this week ( Unit 25): should, give, big, air, home
Also, we will be participating in the TEAM SPENCER FUN RUN on Friday May 4. Please remember to wear orange and bring any donations that you have for the Meyer family to school by Friday…… WOOHOOO!!!
Here’s to a great week!!!
Mrs. Kuieck 🙂
Hi Families,
As we are learning about two digit subtraction, it is very important to build the fluency of our basic subtraction facts. You can help your child at home by doing flashcards, playing subtraction games, online games and apps available on an ipod touch, ipad or iphone.
Here are a few websites for subtraction practice:
SUBTRACTION GAMES
MORE SUBTRACTION GAMES
At school we are also using a few subtraction apps on the ipods. Here are the apps that we are using and they are FREE :)!
Math Tutor 2 Lite | 878 | 3,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-45 | latest | en | 0.940567 |
http://mathhelpforum.com/trigonometry/125103-new-trigonometry-print.html | 1,527,301,798,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00443.warc.gz | 182,736,417 | 4,875 | # New to trigonometry
Printable View
• Jan 23rd 2010, 07:20 PM
davidman
New to trigonometry
Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.
There are two values that fulfil $\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
they are $\displaystyle \theta=\:\:,$
second part of the question:
for $\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,
if $\displaystyle \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then
$\displaystyle \theta_3-\theta_1=\theta_4-\theta_2=$
and $\displaystyle \theta_2-\theta_1=\theta_4-\theta_3=$
thanks for any help.
• Jan 23rd 2010, 07:32 PM
Prove It
Quote:
Originally Posted by davidman
Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.
There are two values that fulfil $\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
they are $\displaystyle \theta=\:\:,$
second part of the question:
for $\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,
if $\displaystyle \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then
$\displaystyle \theta_3-\theta_1=\theta_4-\theta_2=$
and $\displaystyle \theta_2-\theta_1=\theta_4-\theta_3=$
thanks for any help.
First, draw yourself an equilateral triangle of side length $\displaystyle 2$ units.
Then draw a straight line perpendicular to the base and going through the "top" angle.
You will now have two $\displaystyle 30^\circ, 60^\circ, 90^\circ$ triangles.
Their dimensions are $\displaystyle 1, \sqrt{3}, 2$ (you can check this using Pythagoras).
Remember that $\displaystyle \sin{\theta} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}$.
Which angle (out of $\displaystyle 30^\circ$ and $\displaystyle 60^\circ$) has the Opposite side with length $\displaystyle \sqrt{3}$ units and the Hypotenuse as $\displaystyle 2$ units?
• Jan 23rd 2010, 07:34 PM
VonNemo19
Quote:
Originally Posted by davidman
Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.
There are two values that fulfil $\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
they are $\displaystyle \theta=\:\:,$
second part of the question:
for $\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,
if $\displaystyle \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then
$\displaystyle \theta_3-\theta_1=\theta_4-\theta_2=$
and $\displaystyle \theta_2-\theta_1=\theta_4-\theta_3=$
thanks for any help.
I think that you mean to say that $\displaystyle \sin\theta=\frac{\sqrt{3}}{2}$ has two solutions for $\displaystyle 0\leq\theta\leq2\pi$ ? And you wish to find the solutions?
Well, $\displaystyle \sqrt3/2$ is given by evaluating $\displaystyle \sin\theta$ at what is commonly referred to as one of the "special angles". Unfortunately, I can't draw too well on this forum, but here's something for you to look at:
Math Concepts Explained: Special Angles in Trigonometry
• Jan 23rd 2010, 07:35 PM
bigwave
Q1,q2
Quote:
Originally Posted by davidman
Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.
There are two values that fulfil $\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
they are $\displaystyle \theta=\:\:,$
second part of the question:
for $\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$ there are 4 values,
if $\displaystyle \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the ta_4$ then
$\displaystyle \theta_3-\theta_1=\theta_4-\theta_2=$
and $\displaystyle \theta_2-\theta_1=\theta_4-\theta_3=$
thanks for any help.
$\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
$\displaystyle \frac{\pi}{3},\frac{2\pi}{3}$
these are quadrants 1 and 2 respectfully
• Jan 23rd 2010, 08:45 PM
davidman
Quote:
Originally Posted by Prove It
Which angle (out of $\displaystyle 30^\circ$ and $\displaystyle 60^\circ$) has the Opposite side with length $\displaystyle \sqrt{3}$ units and the Hypotenuse as $\displaystyle 2$ units?
$\displaystyle sin60=\frac{\sqrt{3}}{2}\:\:sin30=\frac{1}{2}\:\:s in90=1$
Thanks!
Quote:
Originally Posted by VonNemo19
I think that you mean to say that $\displaystyle \sin\theta=\frac{\sqrt{3}}{2}$ has two solutions for $\displaystyle 0\leq\theta\leq2\pi$ ? And you wish to find the solutions?
Exactly that, yes.
Quote:
Originally Posted by bigwave
$\displaystyle sin\theta=\frac{\sqrt{3}}{2}$
$\displaystyle \frac{\pi}{3},\frac{2\pi}{3}$
these are quadrants 1 and 2 respectfully
this brings me to my next question... how do you get two answers when I only get one from evaluating triangles with "special angles"? And how to go from degrees to radian... ?
• Jan 23rd 2010, 09:00 PM
bigwave
you may be looking at a triangle in just one quadrant
see the attached very familiar demonstration of "special angles"
$\displaystyle (\cos\theta,\sin\theta)$
• Jan 23rd 2010, 09:05 PM
Prove It
OK, now that you have found the "focus angle", we now go to a unit circle approach.
A unit circle is a circle of radius 1.
Have a look at the picture below:
http://upload.wikimedia.org/wikipedi...angles.svg.png
For any angle made with the positive $\displaystyle x$ axis (call it $\displaystyle \theta$), the vertical distance is $\displaystyle \sin{\theta}$ and the horizontal distance is $\displaystyle \cos{\theta}$.
If your angle is $\displaystyle 60^\circ = \frac{\pi}{3}^C$, can you see another angle which has the same vertical distance?
• Jan 23rd 2010, 09:17 PM
davidman
Ok, I see how there would be two solutions. Thinking in degrees, $\displaystyle 180-\theta=\theta_2$
and I googled converting radians/degrees and I must admit that was pretty straightforward...
however, about the second half of the original question,
$\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$
considering the first part of the question, $\displaystyle 2\theta=60\:\:, \theta=30$ makes sense the way I see it. But how are there four solutions and not two?
• Jan 23rd 2010, 09:23 PM
bigwave
radians x $\displaystyle \left(\frac{180^o}{\pi}\right) =$ degrees
degrees x $\displaystyle \left(\frac{\pi}{180^o}\right)=$ radians
http://upload.wikimedia.org/wikipedi...ersion.svg.png
http://en.wikipedia.org/wiki/File:De...Conversion.svg
• Jan 23rd 2010, 09:40 PM
Prove It
Quote:
Originally Posted by davidman
Ok, I see how there would be two solutions. Thinking in degrees, $\displaystyle 180-\theta=\theta_2$
and I googled converting radians/degrees and I must admit that was pretty straightforward...
however, about the second half of the original question,
$\displaystyle sin2\theta=\frac{\sqrt{3}}{2}$
considering the first part of the question, $\displaystyle 2\theta=60\:\:, \theta=30$ makes sense the way I see it. But how are there four solutions and not two?
OK, for any given domain, the number of solutions depends on the period of the function.
Can you see that if you go around the circle (i.e. if you travel $\displaystyle 360^\circ$) you get to the same angle?
If you travel another $\displaystyle 360^\circ$, you get back to the same angle again.
And so on, and so on...
So if you wanted to show all solutions to
$\displaystyle \sin{x} = \frac{\sqrt{3}}{2}$
then $\displaystyle x = \left\{ 60^\circ, 120^\circ \right \} + 360^\circ n$, where $\displaystyle n$ is some integer representing the number of times you have gone around the unit circle.
Now if $\displaystyle x = 2\theta$ then
$\displaystyle 2\theta = \left \{ 60^\circ, 120^\circ \right\} + 360^\circ n$.
Now, to solve for $\displaystyle \theta$ we have to divide both sides of the equation by $\displaystyle 2$. This includes the $\displaystyle 360^\circ$...
So $\displaystyle \theta = \left \{ 30^\circ, 60^\circ \right\} + 180^\circ n$.
So letting $\displaystyle n = 0$ we have
$\displaystyle \theta = \left \{ 30^\circ, 60^\circ \right \}$.
If we let $\displaystyle n = 1$ we have
$\displaystyle \theta = \left \{ 210^\circ, 240^\circ \right \}$.
If we let $\displaystyle n = 2$ we have
$\displaystyle \theta = \left \{ 390^\circ, 420^\circ \right \}$. These solutions are out of the original $\displaystyle 360^\circ$.
So the four solutions are
$\displaystyle \theta = \left \{ 30^\circ, 60^\circ, 210^\circ, 240^\circ \right \}$.
• Jan 23rd 2010, 09:46 PM
davidman
Quote:
Originally Posted by Prove It
Awesome explanation.
Ooooooooohhhhhhhhhh, I get it now! (Nod)
$\displaystyle \theta_1=\frac{\pi}{6}\:,\:\theta_2=\frac{\pi}{3}\ :,\:\theta_3=\frac{7}{6}\pi\:,\:\theta_4=\frac{12} {9}\pi$ | 2,609 | 8,644 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.761405 |
http://lvbet.com/casino-blog/blackjack/blackjack-odds-and-probability/ | 1,638,365,881,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00286.warc.gz | 47,938,624 | 19,529 | # Blackjack Odds and Probability
The card game blackjack is a gambling game of chance, typically played online or at land casinos. The game can appear deceptively simplistic to new players, but those who are experienced in it know the many intricacies and patterns of blackjack. Professional blackjack players are able to use the blackjack odds of winning to their advantage and utilise some complex mathematics to gain increased odds against the house. Read on to answer the question, what are the odds of winning blackjack?
The card game blackjack is a gambling game of chance, typically played online or at land casinos. The game can appear deceptively simplistic to new players, but those who are experienced in it know the many intricacies and patterns of blackjack. Professional blackjack players are able to use the blackjack odds of winning to their advantage and utilise some complex mathematics to gain increased odds against the house.
## The goal of Blackjack
The goal of Blackjack is simply to reach 21 or the sum closest to 21 with your cards. You are given two cards to start out with and you can either keep those two cards as they are and “stand”, or ask for further cards and “hit”. If you hit and the card you receive launches the sum total of the value of your cards over 21, then you “bust” and lose the game. If you hit 21 with your first two cards, you have achieved blackjack and win the hand – unless the dealer also has blackjack! So, how do you increase your odds in blackjack? If you’re keen on basic blackjack strategy and you want to learn how to win at blackjack, read our blog!
### Increasing the chances of winning
The complexities of blackjack become apparent when you try and develop specific winning strategies. Many players will simply take the game at face value as a game of chance, and let the outcome be the outcome. Professional players, however, will always use the odds and mathematics to their advantage, in an effort to increase their chances of winning the game over the house and reaping the rewards. Understanding some basic strategy blackjack to win, and how to get a better advantage over the house, can help you to make the most of your game.
### Odds for various scenarios
Becoming a Blackjack master requires not only understanding the mathematical implications of blackjack odds for multiple scenarios that you may face when playing the game, but also understanding what steps to take once you understand those odds. There’s not always one correct basic strategy and even the most prepared players take losses. Still, understanding the odds and some complex strategies will give you an increased advantage over the house that will make Blackjack a lot more fun and a lot more lucrative for the right and savvy player.
Any player who wishes to master blackjack must first understand the inherent advantage the house has over the player. If the house didn’t have the advantage, there would be no reason for them to hold the game. This is simply because there would be no profit in it for them and they would lose all their money. However, that doesn’t mean you can’t win. By having the edge, the house will win in the long run and win over the majority of games. But that doesn’t mean that you can’t be in that minority of games where the house loses. To give yourself the best chance, you need to be prepared, play blackjack smart and know when to stop playing.
### The house egde in Blackjack
As we’ve already noted, a casino or online website wouldn’t be hosting a game of blackjack if it wasn’t readily apparent to them that they were going to make more money from it than they lose. This is an obvious fact. A casino isn’t a charity, it is a business and a form of entertainment.
While understanding and mastering odds and strategies will give you an increased advantage, players still have to realise the inherent advantage of the house at all times. Unless you are a mathematics savant who is able to learn the shady art of counting cards, you are likely never going to be able to totally dominate the casino. Still, by understanding these blackjack odds and strategies you will be able to increase your own chances of winning.
The house’s advantage comes from the fact that, in blackjack, the player is the one who is going to have the make the first decision in every scenario. That means that the player is the one who has to decide first whether they will hit and take another card or let their hand stand. If the player chooses to hit, that means that the player stands a chance of busting out. Thus, the player has a strong likelihood of losing the game before the house is even forced to make a single decision regarding their own hand.
While the house has the inherent advantage over the player, there are actually some ways in which the player has advantages over the house and even more ways that the player can both increase and utilise these advantages to maybe get a slight upper hand. Let’s examine these in the next section.
### The player’s advantages in blackjack
By understanding the small and not always apparent advantages the player may have we can begin to see how we may turn the tables and attempt to get the upper hand, or at least increase our odds.
While the player has to make decisions earlier than the dealer, the player is freer in the choices they can make regarding their hand. No matter what the value of their hand is, the player gets to decide whether they want to hit or stand. Comparatively, the house is obliged to continually hit until they reach a value of 17. This means that the dealer is slightly more likely to go bust as they have less control over when they can stop hitting. This means that maximum patience and level-headedness on the player’s part is key. Still, there is always a risk that the player will stand and the house will get 21.
• The player can “double-down” on their hand – The player has the unique option to “double-down”, which means that they can double the amount they’ll win if they feel that they have the upper hand against the house with their cards. This opens up the potential for bigger wins if the player knows the right time to double down.
• The player has the opportunity to split pairs– The player has the unique ability over the house to split pairs. This means that the player has the ability to play two hands at once and potentially win on both. The dealer has no such ability.
• The player often has the option to surrender – Lastly, depending on house rules, players will often have the ability to surrender their hand and get back half of the money that has been wagered. While this means that you are going to be sacrificing half of your stake, it also means that you aren’t going to be losing your entire stake.
### How to increase the chances of success
Now that we understand the basic advantages and disadvantages both the player and the house have over each other, as far as their position and abilities, let’s take a look at how understanding, and being well-versed in, blackjack odds can increase our advantage. Certainly, increasing the chances of success in the average game of blackjack is all about the player becoming acquainted with, and understanding, the mathematical complexities of the odds of winning.
• Knowing the odds can make all the difference – regardless of what cards you have been dealt, knowing the odds of winning is incredibly important for making your next move and increasing your chances of winning. Experienced professionals should always be aware of what odds they have to win and what odds they have to bust. Of course, this is going to require a good deal of memorisation and an understanding of some basic strategy.
• Avoid busting – as should be obvious to most players from the get-go, especially if they’ve spent some time with the game, the higher the value of your hand, the higher your chances of busting are if you choose to hit another card. The same goes for the dealer. If the dealer has 16, they are going to be required to hit and the chances of them busting are greater than the chances of them winning. This is an ideal situation for the player but it’s not always the case. If the player has 16, they are not required to hit. The player will have to really think about whether or not they want to hit, stand, or possibly surrender in this situation because if they hit they have a very high chance of busting.
• Knowing if the dealer has a strong likelihood of busting – let us consider that you have a low-value hand, say 13, and the dealer has an upcard of 6. You may feel that it is a good idea to hit, as 13 is nowhere near to 21; but that might not always be the case. If the dealer has a hand value of 6, that means that there is a strong likelihood that the dealer is going to be forced to hit twice and, therefore, the dealer is more likely to bust. If you let your hand of 13 stand in this situation then there’s a very strong chance that you will win the game based on the dealer busting from potentially having to hit twice.
• Focus on the dealer’s hand as much as your own – of course, in the above situation, if the dealer has an Ace in the hole, you’ve lost; but the odds of this happening are stacked against the dealer. For this reason, it is incredibly important to focus on the dealer’s hand as much as your own. Inexperienced players with a hand value of 13 may only pay attention to their own cards, see that they have a minimal chance of going bust if they hit and a strong chance of getting their hand value that much closer to 21. The odds of them going bust are far higher than the odds of the dealer having an Ace in the hole.
• Knowing when to stand – of course, in the above situation, if the dealer has an Ace in the hole, you’ve lost; but the odds of this happening are stacked against the dealer. For this reason, it is incredibly important to focus on the dealer’s hand as much as your own. Inexperienced players with a hand value of 13 may only pay attention to their own cards, see that they have a minimal chance of going bust if they hit and a strong chance of getting their hand value that much closer to 21. The odds of them going bust are far higher than the odds of the dealer having an Ace in the hole.
Just by understanding these basic concepts you will be well on your way to taking some odds back from the house. You still aren’t guaranteed a win and, in fact, the odds are still more often than not ever so slightly in the houses favour, but your blackjack odds can always be increased if you play your cards just right. | 2,186 | 10,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-49 | latest | en | 0.970361 |
https://community.deeplearning.ai/t/random-stochastic-environment-reinforcement-learning/309467 | 1,708,542,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00303.warc.gz | 190,316,448 | 6,694 | # Random (Stochastic) Environment Reinforcement Learning
Hi Mentor,
In Reinforcement learning random stochastic environment optional video lecture, how can be the sequence of different rewards can be mapped to bellman equation ? ie how the max(s’,a’) mapped to the second term in the random sequence of different rewards ?
Example: From state 4, below are the random sequence of different rewards
If it’s so on like, here is our Bellman equation, here E[MaxQ(s’,a’)] how its mapped to the second term in the above random sequence of different rewards
Hello @Anbu,
As prof. andrew said in the lecture video that In the stochastic problem, there would be a sequence of different rewards instead of a single sequence of rewards. Therefore, we are interested in maximizing the expected (average) return across all possible sequences of different rewards because it is random.
Given our limited information about the agent’s next step from state 4, so we just take the average.
This is how the Bellman equation is modified, if you take an action ‘a’ in state ‘s’, the next state ‘s’’ would be random, so you would expect the average of future rewards, denoted as E[MaxQ(s’, a’)]. So the total reward return from state ‘s’ is the sum of immediate reward of state ‘s’ and discount factor gamma what you expect to get average of future returns."
Best,
Mujassim
MaxQ(s’,a’) is basically nothing but best possible return from the next state s’. if stochastics random problem means, assume current state the rover is in state 4, if the next step random means it goes to the state 2, state 4 since random. Then MaxQ(s’,a’) indicates we need to see the best possible return from state 2 and state 4, then do average of the results E[MaxQ(s’,a’)]. This is my understanding of MaxQ(s’,a’)
But MaxQ(s’,a’) does not make sense right because for stochastic problem, we will tends to 1000 different sequence of rewards and we are going to do average of the returns from 1000 sequence of rewards. If its so what is the role play of max here ? Because max is a sign of optimal return from the next state s’ right sir but we are not doing the seek for optimal return (highest possible return)
If you give some example would be helpful
Hey @Anbu,
In this statement, you are missing the next action, i.e., `a'`, and this is in fact what the `max` operation deals with. From a particular next state `s'`, in this environment there are 2 possible actions, left and right. So the `max` operation will choose the return corresponding to the maximising information.
Rest, there are a lot of details regarding the basics of RL which have been overlooked in this week, due to multiple reasons such as time constraints, interpretability, etc. A simple example would be, this week only deals with optimal conditions, i.e., optimal state and action values, optimal policies, etc; but in fact, the optimality conditions are obtained only after the agent goes through non-optimal conditions, which are represented by non-optimal policies and the sort. I hope this resolves your query.
Cheers,
Elemento | 692 | 3,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.881744 |
https://kmatematikatolyesi.com/2018/11/26/real-mathematics-strange-worlds-5/ | 1,679,947,767,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00237.warc.gz | 394,741,521 | 38,713 | # Real Mathematics – Strange Worlds #5
WHERE AM I?
Let’s assume that you have an object shaped like a sphere. Take a pen and draw a starting point on the object. Then, select a direction and move in that direction from your point. When your tour is completed, you’ll be ending up at your starting point in your original direction.
This is an obvious fact which would never change no matter which starting point or direction you choose.
Orientability: Mathematical definition of orientability is really complicated, even for a mathematics graduate. Which is why I will use an example to define it: Imagine that you are travelling to east from Hamburg with a zeppelin. If you continue travelling, and if you are lucky enough to survive, you would eventually end up at Hamburg while your zeppelin faces east. (Please leave you flat-earthers) Actually, it doesn’t matter which direction or city you choose, result will always be the same. If an object or a surface has this specialty, then we say that it has orientability.
It is absurd to think that you finish travelling as a mirror image of your starting state. If you move to the east from a point, you won’t accidently end up at the same point with your direction flipped to west. In other words, you won’t finish taking a tour on a sphere as your mirror image.
Two types of mirror images. Obviously I am talking about the one on the right.
Is it possible to have an object that doesn’t have orientability? Is it possible to go east and when you arrive at your starting point you realize that you are standing upside down?
YES!
Möbius Strip
Cut a rectangle shaped paper. This paper has two faces.
There is no interchanging from one face to another. It means, if you start moving on one face, no matter where or in which direction you started, you won’t end up in the other face. We can call these faces “roads”. So a paper has two different roads.
A paper can have two faces (roads). You can’t go from I to II.
Now glue the ends of this paper together. You will have a cylinder. A cylinder has two faces like a flat paper. It means that a cylinder has two different roads. It is not possible to use one road and end up in the other road. This is why a cylinder has orientability.
Use a transparent paper and make a cylinder. Mark a starting point on your cylinder and move it around your finger until you are at your starting point. You would end up in the same conditions.
Let’s go back: Take your flat rectangle paper and while bringing their ends together twist one end 180 degrees. Mathematicians call this shape a Möbius strip.
While constructing, that 180 degree twist on one end is the only difference between a cylinder and a Möbius strip. But, this difference has absolutely mesmerizing results. First thing to notice is that there is only one face in a Möbius strip. That means a Möbius strip has only one road.
Let’s use a transparent paper again and construct a Möbius strip. Select a starting point and move the strip around your finger. When you end up at your starting point, you’ll see that “up” became “down”, “down” became “up”. In other words, you ended up as your mirror image on a Möbius strip.
Up-right became up-left as if there is a mirror between them.
Then, there is no orientability for Möbius strips. Also, if you make a second tour on it, you would end up at the starting point as your original state.
A little bit of history
It is surprising to learn that Möbius strip was first discovered around 160 years ago. This simple but mysterious shape took its name from the German mathematician August Möbius. Although, Johann Listing, who is another German mathematician, was the first person who published about Möbius strip.
August Möbius (left) and Johann Listing (right).
That is because August Möbius and Johann Listing discovered about Möbius strip independently almost at the same time. After looking through their personal notes we understood that August Möbius discovered the strip 2 months before Listing did. Even though Listing was the first person to use the word “topology”, 2 months gave August Möbius a better mortality.
One wonders…
1. Construct two Möbius strips. While twisting them, twist one of them to the right and the other one to the left 180 degrees. Do you see any difference?
2. Construct a cylinder and cut it into its middle in the direction parallel to its longest edge. You will end up with two cylinders which are little replicas of the original cylinder.
Try the same thing on a Möbius strip. What is the result? Why did you have that result?
M. Serkan Kalaycıoğlu | 1,010 | 4,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-14 | latest | en | 0.942349 |
http://physicsmathforums.com/archive/index.php/t-1083.html | 1,369,069,374,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699113041/warc/CC-MAIN-20130516101153-00086-ip-10-60-113-184.ec2.internal.warc.gz | 208,490,832 | 2,680 | PDA
View Full Version : Design of vapor-liquid separation drums
mbeychok
08-11-2006, 04:06 PM
A vapor-liquid separator drum is a vertical vessel into which a liquid and vapor mixture (or a flashing liquid) is fed and wherein the liquid is separated by gravity, falls to the bottom of the vessel, and is withdrawn. The vapor travels upward at a design velocity which minimizes the entrainment of any liquid droplets in the vapor as it exits the top of the vessel.
The size a vapor-liquid separator drum (or knock-out pot, or flash drum, or compressor suction drum) should be dictated by the anticipated flow rate of vapor and liquid from the drum. The following sizing methodology is based on the assumption that those flow rates are known.
Use a vertical pressure vessel with a length-to-diameter ratio of about 3 to 4, and size the vessel to provide about 5 minutes of liquid inventory between the normal liquid level and the bottom of the vessel (with the normal liquid level being at about the vessel's half-full level).
Calculate the vessel diameter by the Souders-Brown equation to determine the maximum allowable vapor velocity:
V = (k) [ (ρL - ρV) / ρV ]^0.5
where:
V = maximum allowable vapor velocity, m/sec
ρL = liquid density, kg/m³
ρV = vapor density, kg/m³
k = 0.107 m/s (when the drum includes a de-entraining mesh pad)
Then A, the cross-sectional area of the drum, in m²:
= (vapor flow rate, in m³/s) / (vapor velocity V, in m/s)
and D, the drum diameter, in m
= ( 4 A / 3.1416 )^ 0.5
The GPSA Engineering Data Book recommends the following k values for vertical drums with horizontal mesh pads (at the denoted operating pressures):
0 barg: 0.107 m/s
7 barg: 0.107 m/s
21 barg: 0.101 m/s
42 barg: 0.092 m/s
63 barg: 0.083 m/s
105 barg: 0.065 m/s
GPSA Notes:
1. K = 0.107 at 7 barg; subtract 0.003 for every 7 bar above 7 barg
2. For glycol or amine solutions, multiply above K values by 0.6 – 0.8.
3. Typically use one-half of the above K values for approximate sizing of vertical separators without mesh pads.
4. For compressor suction scrubbers and expander inlet separators, multiply K by 0.7 – 0.8The drum should have a vapor outlet at the top, liquid outlet at the bottom, and feed inlet at somewhat above the half-full level. At the vapor outlet, provide a de-entraining mesh pad within the drum such that the vapor must pass through that mesh before it can leave the drum. Depending upon how much liquid flow you expect, the liquid outlet line should probably have a level control valve.
As for the mechanical design of the drum (i.e., materials of construction, wall thickness, corrosion allowance, etc.), use the same methodology as for any pressure vessel.
Milt Beychok
(Visit me at www.air-dispersion.com)
creastiy@sina.com
05-20-2008, 05:04 AM | 729 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.835875 |
https://www.examrace.com/ISS/ISS-Paper-1-and-2-MCQs/Discrete-Distributions-Part-1.html | 1,653,240,156,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00714.warc.gz | 869,829,471 | 6,883 | # Statistics MCQs – Discrete Distributions Part 1
Get top class preparation for ISS right from your home: get questions, notes, tests, video lectures and more- for all subjects of ISS.
1. A new car salesperson knows that he sells cars to one in every twenty customers who enter the showroom. What is the probability that he will sell a new car to exactly two of the next three customers?
a. 0.007
b. 0.021
c. 0.003
d. 0.010
e. 0.001
2. A new car salesperson knows that he sells cars to one in every twenty customers who enter the showroom. What is the probability that he will sell a new car to exactly two of the next five customers?
a. 0.007
b. 0.021
c. 0.003
d. 0.010
e. 0.001
3. A new car salesperson knows that he sells cars to one in every thirty customers who enter the showroom. What is the probability that he will sell a new car to exactly two of the next three customers?
a. 0.007
b. 0.021
c. 0.003
d. 0.010
e. 0.001
4. A new car salesperson knows that he sells cars to one in every thirty customers who enter the showroom. What is the probability that he will sell a new car to exactly two of the next five customers?
a. 0.007
b. 0.021
c. 0.003
d. 0.010
e. 0.001
5. A new car salesperson knows that he sells cars to one in every twenty customers who enter the showroom. What is the probability that he will sell a new car to exactly three of the next five customers?
a. 0.007
b. 0.021
c. 0.003
d. 0.010
e. 0.001
6. Approximately 84 % of persons living in Cape Town who are aged 70 to 84 live in elderly care facilities. If four persons are randomly selected from this population, what is the probability that exactly two of the four live in elderly care facilities?
a. 0.108
b. 0.244
c. 0.007
d. 0.319
e. 0.379
7. Approximately 72 % of persons living in Cape Town who are aged 70 to 84 live in elderly care facilities. If four persons are randomly selected from this population, what is the probability that exactly two of the four live in elderly care facilities?
a. 0.108
b. 0.244
c. 0.007
d. 0.319
e. 0.379
8. Approximately 84 % of persons living in Cape Town who are aged 70 to 84 live in elderly care facilities. If six persons are randomly selected from this population, what is the probability that exactly two of the six live in elderly care facilities?
a. 0.108
b. 0.244
c. 0.007
d. 0.319
e. 0.379
9. Approximately 64 % of persons living in Cape Town who are aged 70 to 84 live in elderly care facilities. If four persons are randomly selected from this population, what is the probability that exactly two of the four live in elderly care facilities?
a. 0.108
b. 0.244
c. 0.007
d. 0.319
e. 0.379
10. Approximately 84 % of persons living in Cape Town who are aged 70 to 84 live in elderly care facilities. If four persons are randomly selected from this population, what is the probability that exactly three of the four live in elderly care facilities?
a. 0.108
b. 0.244
c. 0.007
d. 0.319
e. 0.379
11. The listed occupations of stockholders of a national computer company included 9 % who were housewives. If six of these stockholders are randomly selected, what is the probability that none are housewives?
a. 0.568
b. 0.011
c. 0.083
d. 0.282
e. 0.073
12. The listed occupations of stockholders of a national computer company included 9 % who were housewives. If six of these stockholders are randomly selected, what is the probability that exactly three are housewives?
a. 0.568
b. 0.011
c. 0.083
d. 0.282
e. 0.073
13. The listed occupations of stockholders of a national computer company included 9 % who were housewives. If six of these stockholders are randomly selected, what is the probability that exactly two are housewives?
a. 0.568
b. 0.011
c. 0.083
d. 0.282
e. 0.073
14. The listed occupations of stockholders of a national computer company included 19 % who were housewives. If six of these stockholders are randomly selected, what is the probability that none are housewives?
a. 0.568
b. 0.011
c. 0.083
d. 0.282
e. 0.073
15. The listed occupations of stockholders of a national computer company included 19 % who were housewives. If six of these stockholders are randomly selected, what is the probability that exactly three are housewives?
a. 0.568
b. 0.011
c. 0.083
d. 0.282
e. 0.073
16. A large manufacturing company that produces CD players believes that 1 out of every 20 CD players is defective. To ensure quality control, a random sample of 120 CD players were selected and tested. A large quality control investigation would be launched if more than 10 out of the 120 CD players selected are defective. What is the probability that exactly ten out of the 120 CD players are defective?
a. 0.040
b. 0.105
c. 0.163
d. 0.107
e. 0.063
17. A large manufacturing company that produces CD players believes that 1 out of every 20 CD players is defective. To ensure quality control, a random sample of 120 CD players were selected and tested. A large quality control investigation would be launched if more than 10 out of the 120 CD players selected are defective. What is the probability that exactly 8 out of the 120 CD players are defective?
a. 0.040
b. 0.105
c. 0.163
d. 0.107
e. 0.063
18. A large manufacturing company that produces CD players believes that 1 out of every 20 CD players is defective. To ensure quality control, a random sample of 120 CD players were selected and tested. A large quality control investigation would be launched if more than 10 out of the 120 CD players selected are defective. What is the probability that exactly five out of the 120 CD players are defective?
a. 0.040
b. 0.105
c. 0.163
d. 0.107
e. 0.063
19. A large manufacturing company that produces CD players believes that 1 out of every 10 CD players is defective. To ensure quality control, a random sample of 120 CD players were selected and tested. A large quality control investigation would be launched if more than 10 out of the 120 CD players selected are defective. What is the probability that exactly ten out of the 120 CD players are defective?
a. 0.040
b. 0.105
c. 0.163
d. 0.107
e. 0.063
20. A large manufacturing company that produces CD players believes that 1 out of every 10 CD players is defective. To ensure quality control, a random sample of 120 CD players were selected and tested. A large quality control investigation would be launched if more than 10 out of the 120 CD players selected are defective. What is the probability that exactly 8 out of the 120 CD players are defective?
a. 0.040
b. 0.105
c. 0.163
d. 0.107
e. 0.063 | 1,848 | 6,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-21 | latest | en | 0.923085 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.