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https://bloggerboy.co.in/2020/06/04/time-management/?shared=email&msg=fail | 1,597,087,539,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737168.97/warc/CC-MAIN-20200810175614-20200810205614-00106.warc.gz | 204,294,670 | 55,732 | You are playing PUBG or using YouTube, Tik Tok or any other social media. Your forehead hurts and your eyes feels heavy as you lift your head towards clock.
And then you realize its 12:00 AM, and you have successfully wasted 4 hours on your smartphone/desktop.
In dismay, you jump on your bed cursing yourself and start thinking about your day. But suddenly your dream breaks and you get up in the morning at 10 AM.
It takes you a moment to remember where you left off and you tilt your head back with a heavy sigh. You aren’t even halfway done with the work you intended to finish.
If this seems familiar to you in any way then you are terrible time manager.
Time management is one of the biggest problems of our generation, we are unable to make use of our time wisely, mostly because of tons of distractions.
So, this article will discuss about some practical tips which will ease your time management skill. But before jumping directly to the tips, look at some example and lessons we should learn.
#### A common example
Suppose tomorrow is your exam and you are still busy on social media and surfing the internet.
What is it?
Its total waste of time and it shows your time is not utilized properly. And in the end result you will not be able to score good marks.
#### Lessons we should know
If you make fun of time today, then time will also make fun of you in future.
So it is better that you use the time wisely.
Billionaire, entrepreneur and most successful person, what they have done, in their life they have given the first priority to their work. And they have utilized most of their time in doing their work.
But at the same time an average person does not make the best of time available. If a person understands the value of time in his/ her life then he/ she will, for sure, become a successful person.
#### Calculations of Time and Money
Suppose you are getting \$86400 daily. Now it depends on you that how you are spending each dollar.
You would be thinking of how you will get \$86400 daily? Let me explain to you.
One day has 24 hours and each hour has 3600 seconds. So in the whole day, each and every person gets 3600×24=86400 seconds. So, as the wise men knows “Time is money” so 1 sec = \$1. According to that, everyone gets \$86400 per day.
It varies from person to person on how much that person makes use of it or wastes the opportunity. That’s why time management is a must in our life to be a successful person.
### Stories of two great personalities
#### 1. Bill Gates
Everybody knows that Bill Gates was a college drop out.
Yes, he was a college drop out but what a lot of people don’t know is that when Bill Gates goes to bed at night, he does not remove his shoes.
He shared a reason for this weird habit. He thinks that “if he removes his shoes now then, he will have to wear them the next morning which will lead to waste of his time“.
#### 2. Mark Zuckerberg and his T-Shirt
Everybody knows about Mark Zuckerberg, the CEO of Facebook, he mostly wears a grey T-shirt only.
Why? Don’t tell me that he doesn’t have enough money to buy other T-shirts.
The reason behind him wearing the same T-shirt is that he is not interested in wasting his time to decide what to wear.
In this way, successful persons give respect to time and manage their time.
Now, lets talk about the steps by which you can develop the skills of time management. You simply have to work on your habits, so I would suggest that before reading further check our article on :
Make a Good Habit or get rid of Bad Habits
### Follow the given steps to Manage Your Time
#### 1. Plan Ahead
15 minutes before going to bed, plan about the next day. The important things which you are going to do on the next day.
So when you will wake up the next morning, you don’t have to waste time on making plans. So, in the morning without wasting time you can start your work according to your plan.
#### 2. Spend your mornings on Most Important Tasks (MITs)
For starters, you usually have the most amount of energy in the morning. So, it’s better to tackle the most important tasks when you are not drained. And with a lot of energy you will be able to complete the task before time and more efficiently.
#### 3. Stop being perfect
When you are perfectionist, nothing will ever be good enough. That means you will keep going back to same task over and over again. How productive do you think your day will be as a result?
#### 4. Say ‘NO’ without hesitation
I know that you don’t want to upset anyone. But first complete your work and then go to help others. Or else you will be going and completing your friend’s work and your work will be pending. So you will not be able to complete your work on the given deadline. So never hesitate to say ‘NO’.
#### 5. Find inspiration
Use inspirational sources like a TED Talks or biography or you can follow the motivational pages on instagram. We have our motivational page on instagram. https://instagram.com/the_strongmindset You can check it out and follow the page if you like it.
#### 6. Create a time audit
Find out where you spend your time and try to evaluate the report. It will help you to spend your time in most important tasks. The report which you will make will show you that how much time you are wasting and how much time you are using it. And by doing so you can divide your time accordingly.
#### 7. Put a time limit on tasks
Put a time limit on every task. Don’t keep on doing whole task for whole day. If you do so then your productivity will decrease. Setting time constraints for completing tasks helps you to be more focused and efficient.
Making the small extra effort to decide on how much time you need to allot for each task can also help you recognize potential problem before they arise. That way you can make plans for dealing with them.
#### 8. Add a done list to your to-do list
Sometimes many unexpected tasks just pop up during the day. So write down them in a separate list next to your to-do list for some extra satisfaction at the end of your day.
#### 9. Block out distractions
Switch off your mobile phone and keep it away from you when you are working on focused tasks. If your phone will be with you then you will not be able to focus and concentrate on your work. So it is necessary to keep your mobile phone and other distractions away from you.
#### 10. Don’t Multitask
Researches have proven that Multitasking is Myth. If you believe that you can do more than one thing at a time and you are utilizing the time then, you are totally wrong.
Whenever you do more than one work at a time you will not be able to concentrate on any of them and it will be more time consuming in completing the work. So it’s better to avoid multitasking.
#### 11. Schedule breaks between tasks
Scientists says, the human brain can only focus for about 90 minutes at a time. Thus, it is better to schedule a break at least every 90 minutes or at every 50 minutes to avoid burnout and maintain high productivity throughout your day.
#### 12. Eliminate bad habits
One of the biggest time wasters we have are our bad habits. Whether it’s Netflix binge – watching, excessively surfing social media, playing games, going out frequently with friends, etc. Those bad habits take away the precious little time that we do have. Use your time wisely by eliminating your bad habits if you are serious about achieving big goals in life.
#### 13. Meditate or exercise every morning
You might think that this won’t help you. But meditating and exercising every single morning gives you a balanced life. It will help you cut out the toxins of your life and you will for sure observe your energy, stamina and mental focus take a drastic shift.
Read our article on ancient exercise “SUN SALUTATION” or “SURYA NAMASKAR“.
Everything about Sun Salutation or Surya Namaskar
#### 14. Organize yourself
Utilize your calendar for a long-term time management. Write down the deadlines for projects, or for tasks that are part of completing overall project. Not only do you need to organize your time but you should also organize your work space.
Think about which days might be the best to dedicate to specific tasks. This will lead to a more efficient working environment.
#### 15. Sleep well
Some people think sacrificing sleep is a good way to hack productivity and wring a couple of extra hours out of the day. This is not the case. Most people need 7 – 8 hours of sleep for their bodies and minds to function optimally. So, listen to your body and don’t underestimate the value of sleep.
Do follow these steps to manage your time. One day, you will, for sure, be able to manage your time and you will also be able to lead a successful life.
Please fill the feedback form to help us to improve our blog. If you think that our blog has some troubles then please mention them in the form, we welcome your suggestions. | 1,957 | 8,917 | {"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} | 2.96875 | 3 | CC-MAIN-2020-34 | latest | en | 0.954092 |
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# 1/10 scale building materials
Happy DIYs are here again
(+3, -4) [vote for, against]
When engaged on DIY projects (which is to say, pretty much permanently), I tend to spend many, many hours in the planning stage, making dozens of sketches, diagrams and calculations so that, in theory, I know exactly how much of everything I'll need, and how the finished article will look. For someone of my limited artistic talents, however, the complexities inherent in 3-dimensions can mean that a construction that I *thought* would work turns out to be geometrically impossible. Also, calculating the number of building blocks required for a wall, for example, is complicated by the mortar, and allowances for half-blocks, and it can be difficult to be sure that a wall will look right.
Stores routinely supply test-pots of paint (at relatively inflated prices) so that customers can check the look of a colour in situ; often the manufacturer gives a discount on the price of a full-size tin on production of a receipt for the test-pot. Perhaps builders' merchants could sell accurate scale-models of commonly-used building items - cement blocks, bricks, timber - so that handymen could make mock-ups of their projects. 1:10 scale seems about right; small enough not to take up too much room, but large enough that a standard house-brick is not too fiddly. Parts could be injection-moulded polystyrene, mortar could be a strip of the correct thickness and width, cut to length as required. Buy a pallet of 72 blocks for £5 (also 1/10 scale), then get a £2 reduction on the real thing.
— angel, Jun 18 2003
Modeling Supplies http://www.plastruct.com/
Enter PS-103 as catalog code. [angel, Oct 05 2004, last modified Oct 21 2004]
Balsa wood sizes http://www.zimsweb....alsawoods/sizes.htm
Just do the math and voilà! [Cr0esus47, Oct 05 2004, last modified Oct 21 2004]
Hate to rain on your parade because this is a very good idea! But unfortunately it's VERY baked! Architects have been building scale models for years and there are plenty of pre-made materials available to them. They are available in pretty much all standard scales. I remember a project i was working on a few years ago at Livermore labs where they fully constructed a 1/4"=1'0" scale model of a new lab building before starting contruction. All major (and even some minor) building systems where constructed so they could find out if there were any conflicts or major problems with the design. Talk about a very expensive way to coordinate between trades!
[waugs] Sorry about the link, it's kinda what i was looking for as an example, but unfortunately i don't have the time here at work to find exactly the right link. But honestly, trust me...it's all out there! (I know....I don't ever trust anyone that says "trust me" either!) I'll try to find a better link when I have more time, probably after i get home tonight.
Oh, the irony! A true DIYer would make his/her own scale material...
— phoenix, Jun 18 2003
...including cutting down hundreds of bonsai trees to make the tiny lumber.
— krelnik, Jun 18 2003
I'd rather see 10x scale popsicle sticks.
— rapid transit, Jun 18 2003
But could you eat the 2Kg popsicle on the end of the stick before it melted?
— Cedar Park, Jun 18 2003
[justadesigner]: I can find nothing in your link to satisfy my need. The simulated 'cement blocks', for example, (shown in detail in my link to the same site) consist of a section of pre-built wall in 1/100 scale, which is fine for making dioramas or model train layouts, but much too small for my purpose. I chose 1/10 scale for a very good reason; it's large enough to enable the individual pieces to be handled satisfactorily, but small enough to be sensible.
[Mr Burns]: Lego blocks are not scale models of actual building blocks.
[mr_imagonna]: Please show me "any model store" which sells 44mm x 21.5mm x 10mm blocks (preferably under the business scheme I cited).
Generally, model-building suppliers sell materials which would enable me to make my own 1/10 scale models, which is rather not the point.
— angel, Jun 19 2003
[edited again]Miniature bricks are widely available enough for you to get many calls of baked. Just google for the term and you'l find lots of people that make 2cm long clay bricks. Thats about 1:10, isn't it?
But, you don't want realistic fired clay bricks, do you? You want brick shaped, objects that you can use to cheaply mock up a design. 1:10 placeholders of all the components that might make it into the finished work, and that is something I haven't seen. 1 plastic 1:10 croissant heading in your direction. Do with it what you will.
— st3f, Jun 19 2003
<edited out the rant> inserting baby rantling here: wouldn't it be easier just to do a crude sketch, like draw a rectangle, write in the dimensions, use a calculator for the brick problem? For the structural questions, I highly reccomend the many books on the subject. I've been able to learn how to build just about anything in wood framed residentiol construction out of books. I even drew a set of plans, got a permit, built a garage with knowledge gleaned from books. And I ain't even that smart, compared to the average halfbaker, so I've come to understand.
— oxen crossing, Jun 19 2003
I might draw the line at *working* appliances, but scaled down baths, sinks, etc would certainly be useful, particularly if you're not sure that they would fit through the doors. A prospective purchaser might be prepared to buy the larger (more expensive) bath if he could be sure that it would go round that corner on the landing.
— angel, Jun 20 2003
Don't forget to feed the tiny constructors in tinytown. 1/30 lb. (each) of meat, salad & beans along with 1/15th a bun for individual construction w-w-workers. BYOB.
— thumbwax, Jun 20 2003
It's the grains of sand 1/10th the size of normal grains of sand that I have problems with. Sand's hard enough to get out of stuff as it is.
— PeterSilly, Jun 20 2003
"... it's large enough to enable the individual pieces to be handled satisfactorily, but small enough not to be sensible ..."
?!
— pjd, Jun 20 2003
What you need is not available. At least I've never seen it.
Instead, [phoenix] has a good idea.
If you are really 3D-challenged, I suggest you work in full 1:1 scale. That way, you'll see if it really fits.
Buy strips, sheets and blocks of expanded styrene sheet. It's cheap, easy to transport and easy to work with. A hotwire cutter, stanley knife, pins and paste is all you need to mock up items really fast.
Structural integrity is not so critical because your mockups will be ultra-light weight and don't have to last a lifetime.
There are some really innovative techniques you can use. Use thick cardboard guides wherever possible. You can trace around real objects in the stores.
You can even make a hand lathe for turning foam (use a fine grain) from a horizontally mounted swivel or a spike.
For really complex forms, chicken wire, spray filler foam and putty will do.
— FloridaManatee, Jun 20 2003
[pjd]: Sorry, please remove the "not" from that sentence.
— angel, Jun 20 2003
Doll's House suppliers could be the way to go here, angel. They do all sorts of surprising things to scale. 'taint cheap though!
— DrBob, Jun 20 2003
Well, I know they do ~1:50 scale furniture, but I still haven't found what I'm looking for. (You, too.)
— angel, Jun 20 2003
I've seen stuff in 1:12 scale for fixtures & fittings type stuff but not Bricks and Mortar (the jam)
— DrBob, Jun 20 2003
Completely baked. Just need to change your scale/building materials. I made a 1/16th replica of my father's cottage using balsa wood, wood panelling (for the walls) popsicle sticks & interestingly black sanpaper. The black sandpaper I cut into little tiny rectangles, about 1 inch by 4 inches and used as roofing material. If you need more building tips, just e-mail me at: croesus47@hotmail.com. Don't ask me, however, how to prevent the nick & cuts from working with the materials ;-)
— Cr0esus47, Jun 20 2003
Lots of people are telling me that I could make my own small-scale building parts, and others are offering hints on *representational* models (dolls' houses and such). My requirement is for miniature parts to enable me (and other DIY weekend handymen, not keen model-makers) to mock up the *functional* design, not the appearance. I am also suggesting a business model under which this might be operated.
— angel, Jun 20 2003
I'm not sure how you'll get the 1/10th scale nails, or if you can get the actual wood (pine, oak, etc) cut small enough. If you want materials that function *exactly* like the full scale project, then your scale of 1:10 would be a tad unreasonable. To put it into perspective, your big 10-penny spikes used for contruction at most are 15cm (and I'm being generous here) That means when reduced in size, you'll have a thin needle the width of your thumb to nail into a board no thicker than your finger. That's if you use REALLY BIG nails. Most construction is done with nails approx 8-10 cm long, with boards 10cm wide. reduce in scale, and the nail is 0.8 to 1.0 cm long, nailed to a board 1 cm wide (Metric makes these scale conversions a snap) So if what you're talking about involves just a quick mock-up, you have to be aware that these quick mock-up materials won't have the same physical properties as the full-scale counterparts. It's a compromise between authenticity and practicality. I doubt you'd be acle to get a set of bricks that behave *exactly* to scale because the density would have to be consistent, and I don't think we can reduce the size of grains of cement powder by 90%.
— Cr0esus47, Jun 20 2003
I don't want them to have the same physical properties as the real thing, I just want them to be the same proportions. I'm not planning on mixing little sand with little cement to mortar little bricks together, I just want to be able to build my mock-up wall or whatever to scale, so I can see how many blocks I'll need to make the wall, how the timbers are going to fit, and whether it's going to be practical to put the garbage can there. The "wood" doesn't need to be real wood (in fact, I said up there --^ that the parts could be polystyrene), and I certainly don't need little nails. Building blocks that are 44 x 21.5 x 10 mm, beams that are 0.2" by 0.4" x 9.6", panels that are 244 x 122 x 1.2 mm. I buy some of these, at twice the sensible price, and get half of it refunded when I buy the real blocks, beams and panels.
— angel, Jun 20 2003
//I'm not planning on mixing little sand with little cement to mortar little bricks together// - aw, go on - you know it makes sense.
— PeterSilly, Jun 20 2003
I'm with you angel. I don't know about the exact marketing scheme, but I like the concept. I'm picturing a full scale model of a house in everyone's garage, ready for tinkering before you start messing with the real house.
— Worldgineer, Jun 20 2003
I'm not against the idea of having these products available, far from it. What I'm saying is that if you go into any *decent* hobby or DIY shop these should already be available. For my 1/16th cabin replica, I took the time to actually measure the cabin (life size) and reduce the scale. Doing the reverse is just as simple. Plan the dimensions, do the math, buy the small parts (you'd be amazed how much a 1/8" X 1/4" stick of balsa wood can *look* like a 2" X 4" in the proper scale) then build your project. My replica, although not entirely complete, could very easily be mistaken at a distance for the real thing. I should get some pictures of it...
— Cr0esus47, Jun 20 2003
1:10 scale probably should be the best scale for this, but 1/8"=1'-0" is probably the going norm for those who have not embraced the logic of the metric system.
I was told in school almost 20 years ago that we would be going metric. What happened?
— Zimmy, Jun 20 2003
Fer chrissake man, use a spreadsheet to CALCULATE how much you need- and if you can't use a spreadsheet, spend the money to get a course on excel or similar..
— RusNash, Jun 20 2003
The idea makes total sense, it's useful, and it's not something I've seen to exist in the fashion you suggest. I don't understand the resistance to it.
— Tabbyclaw, Jul 09 2004
I once built a working model of the engine of the Titanic using baked beans
— xenzag, Sep 30 2005
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https://eduvark.com/syllabus-mathematics-common-proficiency-test-40951.html | 1,701,367,269,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00526.warc.gz | 257,466,632 | 16,712 | 2023 2024 EduVark Syllabus of Mathematics of Common Proficiency Test
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June 19th, 2014, 01:33 PM
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Syllabus of Mathematics of Common Proficiency Test
I want to give the exam of Common Proficiency Test and for that I want to get the syllabus of Mathematics of Common Proficiency Test so can you provide me that?
As you want to get the syllabus of Mathematics of Common Proficiency Test so here is the information of the same for you:
Ratio and proportion, Indices, Logarithms.
Equations
Linear – simultaneous linear equations up to three variables, quadratic and cubic equations in one variable, equations of a straight line, intersection of straight lines, graphical solution to linear equations.
Inequalities
Graphs of inequalities in two variables common region.
Simple and Compound Interest including annuity, Applications
Basic concepts of Permutations and Combinations
Sequence and Series – Arithmetic and geometric progressions
Sets, Functions and Relations
Limits and Continuity, Intuitive Approach
Basic concepts of Differential and Integral Calculus (excluding trigonometric functions)
Statistical description of data
Textual, Tabular & Diagrammatic representation of data.
Frequency Distribution.
Graphical representation of frequency distribution – Histogram, Frequency Polygon, Ogive
Measures of Central Tendency and Dispersion
Arithmetic Mean, Median – Partition Values, Mode, Geometric Mean and Harmonic, Mean, Standard deviation, Quartile deviation
Correlation and Regression
Probability and Expected Value by Mathematical Expectation
Theoretical Distributions
Biomial, Poisson and Normal.
Sampling Theory
Basic Principles of sampling theory , Comparison between sample survey and complete enumeration , Errors in sample survey , Some important terms associated with sampling , Types of sampling , Theory of estimation , Determination of sample size .
Index Numbers
Last edited by Neelurk; June 8th, 2020 at 08:38 AM. | 417 | 2,021 | {"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-2023-50 | latest | en | 0.834724 |
https://cg-tower.com/using-your-senses-to-gather-information-is-called/ | 1,638,059,051,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00211.warc.gz | 247,479,447 | 5,626 | Scientists that all eras use every one of their senses to observe the world roughly them. Ghislain & Marie David de Lossy/Cultura/Getty Images
Let"s breakdown the meaning of cg-tower.com.
You are watching: Using your senses to gather information is called
### Part 1
cg-tower.com is practical. Although cg-tower.com sometimes requires learning native textbooks or professors in lecture halls, that primary activity is discovery. Discovery is one active, manual process, not something excellent by scholars isolated from the human being in cream color towers. That is both a find for information and also a quest to explain how information fits together in meaningful ways. And also it nearly always looks for answers to an extremely practical questions: exactly how does human activity affect worldwide warming? Why room honeybee populaces suddenly declining in north America? What permits birds to move such lengthy distances? how do black color holes form?
### Part 2
cg-tower.com is based upon observation. Researchers use all of their senses to gather information around the world about them. Sometimes they conference this information directly, with no intervening tool or apparatus. Other times they usage a item of equipment, such as a telescope or microscope, come gather info indirectly. Either way, researchers will compose down what castle see, hear and feel. These recorded observations are referred to as data.
### Part 3
Data can reveal the structure of something. This is quantitative data, i beg your pardon describes an item numerically. The adhering to are examples of quantitative data:
The human body temperature of a ruby-throated hummingbird is 105 levels Fahrenheit (40.5 degrees Celsius).The diameter of Jupiter is 88,846 miles (142,984 kilometers).
Notice the quantitative data consist of a number complied with by a unit. The unit is a standardized method to measure a specific dimension or quantity. For example, the foot is a unit the length. For this reason is the meter. In cg-tower.com, the worldwide System (SI) that units, the modern type of the metric system, is the an international standard.
See more: 12 Coffee Chain Drinks You Can I Drink Starbucks While Pregnant Women
### Part 4
Data can also reveal behavior. This is qualitative data, which are written descriptions around an thing or organism. Man James Audubon, the 19th-century naturalist, ornithologist and painter, is renowned for the qualitative monitorings he made around bird behavior, such as this one:
Generally, researchers collect both quantitative and also qualitative data, which contribute equally to the body of knowledge connected with a certain topic. In various other words, quantitative data is not more important or more valuable since it is based on an accurate measurements .
### Try ours Sudoku Puzzles!
let setupModule = function (d) content = d.filter(c => !relatedIds.contains(c.id) );let coreContent = content.filter(c => typeof c.category != "undefined");let articleCore = coreContent.filter(c => <"article", "list", "article-with-html-code", "article-structured">.includes(c.type.toLowerCase()))let continuousIds = contArticleIds.concat(articleCore.map(c => c.id));\$dispatch("populate-pending", item : continuousIds );let baseEvent = event : "raw-event",eventCategory : "page-interaction",eventAction : "loaded";let progEvent = eventLabel : "related-programmatic" ;dataLayer.push(...baseEvent,...progEvent);if (relatedIds.length) let editorialEvent = eventLabel : "related-editorial" ;dataLayer.push(...baseEvent,...editorialEvent);if (typeof promotedRecirc != "undefined" && promotedRecirc.length == 3) promoted = true;content.unshift(promotedRecirc<0>, promotedRecirc<1>, promotedRecirc<2>);let promotedEvent = eventLabel : "related-promoted" ;dataLayer.push(...baseEvent,...promotedEvent);\$nextTick(() => let imgs = document.getElementById("related-hero").querySelectorAll(".lazyload");HSW.utilities.lazyLoadElements(imgs);dataLayer.push(event : "object-load",eventAction : "related-content"););if (typeof relatedContent !== "undefined") setupModule(relatedContent);else HSW.utilities.loadJson(HSW.domainFormatted + "/related/mixed/" + 440972 + "?w=300",(d) => setupModule(d);))()"> | 908 | 4,225 | {"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-2021-49 | longest | en | 0.909017 |
https://www.chegg.com/homework-help/algebra-2-student-edition-1st-edition-chapter-7.4-problem-17pa-solution-9780078279997 | 1,561,403,990,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00275.warc.gz | 691,818,586 | 47,164 | Solutions
Algebra 2, Student Edition
# Algebra 2, Student Edition (1st Edition) Edit edition Problem 17PA from Chapter 7.4
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 4
To use synthetic substitution to find and for each function
By remainder theorem, should be the remainder when you divide the polynomial
by
2 -8 -2 5
6 -6 -24
2 -2 -8 -19
. The remainder is -19
• Chapter , Problem is solved.
Corresponding Textbook
Algebra 2, Student Edition | 1st Edition
9780078279997ISBN-13: 0078279992ISBN: | 163 | 566 | {"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-2019-26 | latest | en | 0.801778 |
http://felix11h.github.io/blog/some_stochastic_processes | 1,518,924,079,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811352.60/warc/CC-MAIN-20180218023321-20180218043321-00724.warc.gz | 142,404,630 | 5,517 | # Expected value and variance of some stochastic processes
As an exercise, I illustrate how expected value and variance of some stochastic processes evolve in time.
## Poisson process
Let $$(N_t)_{t \in \left[0,\infty\right)}$$ be a Poisson process with rate $$\lambda$$. Then $$\operatorname{E}(N_t) = \lambda t$$ and $$\operatorname{Var}(N_t) = \lambda t$$. Below several paths of a Poisson process are shown. These are well contained within $$\operatorname{E}(N_t)$$ plus and minus the three time standard deviation $$3 \sigma$$, where $$\sigma = \sqrt{\lambda t}$$ from above.
The figure was produced by the following Python code
import numpy as np
class Poisson_process(object):
def __init__(self, rate, dt=0.01):
self.N = 0
self.t = 0
self.dt = dt
self.rate = rate
def sample_path(self, T, reset=None):
path = []
if reset != None:
self.N = reset
for t in np.arange(0,T,self.dt):
k = np.random.poisson(self.rate*self.dt)
if k>0:
self.N+=1
path.append(self.N)
return path
rate = 1.
dt = 0.01
pp = Poisson_process(rate, dt=dt)
T = 25
ts = np.arange(0,T,dt)
import matplotlib as mpl
mpl.use('Agg')
import pylab as pl
from matplotlib import rc
rc('text', usetex=True)
pl.rcParams['text.latex.preamble'] = [
r'\usepackage{tgheros}', # helvetica font
r'\usepackage{sansmath}', # math-font matching helvetica
r'\sansmath' # actually tell tex to use it!
r'\usepackage{siunitx}', # micro symbols
r'\sisetup{detect-all}', # force siunitx to use the fonts
]
fig = pl.figure()
fig.set_size_inches(4,3)
pl.plot(ts,ts, 'k', label=r'$\mu$')
pl.plot(ts, ts+3*np.sqrt(rate*ts), 'r', lw = 1., label=r'$\mu \pm 3 \sigma$')
pl.plot(ts, ts-3*np.sqrt(rate*ts), 'r', lw = 1.)
for k in range(12):
path = pp.sample_path(T, reset=0)
pl.plot(ts, path, 'gray')
pl.legend(loc='upper left', frameon=False, prop={'size':12})
pl.xlabel('time')
pl.savefig("pp_sample.png", dpi=300, bbox_inches='tight')
## Wiener process
Let $$(X_t)_{t \in \left[0, \infty\right)}$$ be a Wiener process. It is $$\operatorname{E}(X_t) = 0$$ and $$\operatorname{Var}(X_t) = t$$. Below 15 sampled paths of the Wiener process are shown. It's trace are well contained within $$2 \sigma$$ or $$3 \sigma$$, where $$\sigma = \sqrt{t}$$ is the standard deviation.
The figure was generated by the following Python code
import numpy as np
class Wiener_process(object):
def __init__(self, dt=0.01):
self.X = 0
self.t = 0
self.dt = dt
def sample_path(self, T, reset=None):
path = []
if reset != None:
self.X = reset
for t in np.arange(0,T,self.dt):
x = np.random.normal(0,np.sqrt(self.dt))
self.X+=x
path.append(self.X)
return path
dt = 0.01
wp = Wiener_process(dt=dt)
T = 50
ts = np.arange(0,T,dt)
import matplotlib as mpl
mpl.use('Agg')
import pylab as pl
from matplotlib import rc
rc('text', usetex=True)
pl.rcParams['text.latex.preamble'] = [
r'\usepackage{tgheros}', # helvetica font
r'\usepackage{sansmath}', # math-font matching helvetica
r'\sansmath' # actually tell tex to use it!
r'\usepackage{siunitx}', # micro symbols
r'\sisetup{detect-all}', # force siunitx to use the fonts
]
fig = pl.figure()
fig.set_size_inches(5,3)
for k in range(15):
path = wp.sample_path(T, reset=0)
pl.plot(ts, path, 'gray')
pl.plot(ts, 2*np.sqrt(ts), 'r', linestyle='dashed', lw=1., label=r'$2 \sigma$')
pl.plot(ts, -2*np.sqrt(ts), 'r', linestyle='dashed', lw=1.)
pl.plot(ts, 3*np.sqrt(ts), 'r', lw = 1., label=r'$3 \sigma$')
pl.plot(ts, -3*np.sqrt(ts), 'r', lw = 1.)
pl.ylim(-22.5,22.5)
pl.legend(loc='upper left', frameon=False, prop={'size':12})
pl.xlabel('time')
pl.savefig("wp_sample.png", dpi=300, bbox_inches='tight')
## Ornstein-Uhlenbeck process
Let $$(X_t)_{t \in [0,\infty)}$$ be an Ornstein-Uhlenbeck process, that is a process defined by the stochastic differential equation
\begin{align} dX_t=\theta (\mu_{ou}-X_t) dt + \sigma_{ou} dW_t, \end{align}
with $$X_0 = a$$. The expected of value this process is
\begin{align} \operatorname{E}(X_t) = ae^{-\theta t} + \mu_{ou}(1-e^{-\theta t}) \end{align}
and the variance is
\begin{align} \operatorname{Var}(X_t) = \frac{\sigma_{ou}^2}{2 \theta} (1 - e^{-2\theta t}). \end{align}
import numpy as np
class OU_process(object):
def __init__(self,X_0, mu, theta, sigma, dt=0.01):
self.X = X_0
self.dt = dt
self.mu = mu
self.theta = theta
self.sigma = sigma
self.setup()
def setup(self):
self.emdt = np.exp(-self.theta*self.dt)
self.a = np.sqrt(self.sigma**2/(2*self.theta) *
(1 - np.exp(-2*self.theta*self.dt)))
def step(self):
X_dt = self.X*self.emdt+self.mu*(1 - self.emdt) + \
self.a*np.random.normal()
self.X = X_dt
def path(self, T, reset=None):
if reset != None:
self.X = reset
path = []
for t in np.arange(0,T,self.dt):
path.append(self.X)
self.step()
return path
dt = 0.01
X_0 = -2
mu = 1.
theta = 1.
sigma = 1.
OU = OU_process(X_0, mu, theta, sigma, dt)
import matplotlib as mpl
mpl.use('Agg')
import pylab as pl
from matplotlib import rc
rc('text', usetex=True)
pl.rcParams['text.latex.preamble'] = [
r'\usepackage{tgheros}', # helvetica font
r'\usepackage{sansmath}', # math-font matching helvetica
r'\sansmath' # actually tell tex to use it!
r'\usepackage{siunitx}', # micro symbols
r'\sisetup{detect-all}', # force siunitx to use the fonts
]
T = 10
ts = np.arange(0,T,dt)
mean = np.array([X_0*np.exp(-theta*t) + \
mu*(1-np.exp(-theta*t)) for t in ts])
var = np.array([sigma**2/(2*theta)* \
(1-np.exp(-2*theta*t)) for t in ts])
fig = pl.figure()
fig.set_size_inches(5,3)
for k in range(5):
path = OU.path(T, reset=X_0)
pl.plot(ts, path, 'gray')
pl.plot(ts, mean+2*np.sqrt(var), 'r', linestyle='dashed',
lw=1., label=r'$\mu \pm 2 \sigma$')
pl.plot(ts, mean-2*np.sqrt(var), 'r', linestyle='dashed',
lw=1.)
pl.ylim(-3,3)
pl.legend(loc='lower right', frameon=False, prop={'size':12})
pl.savefig("ou_sample.png", dpi=300, bbox_inches='tight') | 1,895 | 5,887 | {"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": 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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-09 | longest | en | 0.503999 |
https://unitconversion.io/in3-to-ft3-conversion | 1,679,949,534,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00108.warc.gz | 671,291,723 | 9,506 | #### Inches Cubed to Feet Cubed Conversion
This is our conversion tool for converting inches cubed to feet cubed.
To use the tool, simply enter a number in any of the inputs and the converted value will automatically appear in the opposite box.
in3
### =
ft3
##### How to convert Inches Cubed (in3) to Feet Cubed (ft3)
Converting Inches Cubed (in3) to Feet Cubed (ft3) is simple. Why is it simple? Because it only requires one basic operation: multiplication. The same is true for many types of unit conversion (there are some expections, such as temperature). To convert Inches Cubed (in3) to Feet Cubed (ft3), you just need to know that 1in3 is equal to ft3. With that knowledge, you can solve any other similar conversion problem by multiplying the number of Inches Cubed (in3) by . For example, 8in3 multiplied by is equal to ft3.
#### Fast Conversions
1 in3 = ft3 5 in3 = ft3 10 in3 = ft3 15 in3 = ft3 25 in3 = ft3 100 in3 = ft3 1000 in3 = ft3 1 ft3 = in3 5 ft3 = in3 10 ft3 = in3 15 ft3 = in3 25 ft3 = in3 100 ft3 = in3 1000 ft3 = in3
#### Blog
Boost Your Email Open Rates by 50%: The Power of Personalization in Email Marketing The Ultimate Android Performance Testing Guide How to Measure Customer Relationship Management Success The Impact of Digital Transformation on Business: How Technology Is Reshaping Industries Upgrade Your Vaping Experience with These Top 5 CBD Products - Review from AskGrowers | 382 | 1,424 | {"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} | 3.0625 | 3 | CC-MAIN-2023-14 | latest | en | 0.80262 |
https://slideplayer.com/slide/778527/ | 1,532,155,597,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00437.warc.gz | 779,515,634 | 19,966 | # 1 A gender and helping study with a different outcome.
## Presentation on theme: "1 A gender and helping study with a different outcome."— Presentation transcript:
1 A gender and helping study with a different outcome
2 Here is another set of results from the experiment on helping.
3
4 A loglinear model was fitted to the data. Here is a test of its goodness-of-fit
5
6 Question 1 Does this chi-square value measure the goodness-of-fit of a saturated model?
7 Answer No. When a saturated model is applied, chi-square has no degrees of freedom and has a value of zero.
8 Shortly, I shall show you a table of tests of K-way and Higher Order Effects
9 Question 2 Examine the table. Is the opposite-sex dyadic hypothesis supported by these test results?
10
11 Answer No. The opposite-sex dyadic hypothesis predicts a three-way interaction of Participants Sex, Interviewers Sex and Help. The p-value for the three-way interaction (0.514) does not support this expectation.
12 Here is a table of the backward elimination statistics
13
14 Question 3. How many models are described here?
15 Answer This table is difficult to follow. FOUR models are described: 1.Interviewer*Participant*Help – the saturated model. 2.Int*Part, Int*Help, Part*Help. All two-way interactions. 3.Int*Part, Int*Help. Part* Help dropped. 4.Int*Help, Part. Int * Part dropped. Opposite each model, there is a chi-square value with so-many df.
16 Answer … Remember that this chi-square refers to the RESIDUALS associated with the terms that have been LEFT OUT. Opposite the final model Int*Help, Part, is the chi-square value 2.435, with df = 3. This chi- square measures the sizes of the residuals when the terms Int*Help*Part (df = 1), Help*Part (df = 1) and Part*Int (df =1) have been removed from the model. Thats why it has 3 degrees of freedom.
17 Question 4. In the final model, where did Participant come from?
18 Answer The main effect of Participant has really been there all the time; but now it needs to be mentioned explicitly in the generating class, because all the interactions involving it have now been removed from the model.
19 The generating class In the output, we are told that the generating class is Interviewer*Help, Participant.
20 Question 5 Does the final model include a term for the main effect of the Help factor?
21 Answer It must do, according to the hierarchical principle. If there is an interaction term, all lower-order effects among the same factors must also be included in the model. The presence of the Interviewer*Help term implies the presence in the model of the main effects of Interviewer and Help.
22 Question 6 Can you write out an equation for the final loglinear model, expressing the terms verbally, rather than in algebraic symbols? The generating class of the final model is Interviewer*Help, Participant
23 The final loglinear model Theres always a constant. The model contains a main effect of Help. There is an Interviewer × Help interaction. By the hierarchical principle, there must also be main effects of Interviewer and Help. Theres a main effect of Participant.
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Similar presentations | 719 | 3,212 | {"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} | 3.25 | 3 | CC-MAIN-2018-30 | latest | en | 0.916639 |
https://math.stackexchange.com/questions/828844/the-fundamental-group-of-punctured-surface | 1,563,745,846,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00046.warc.gz | 471,490,986 | 35,061 | the fundamental group of punctured surface
Let $S_{g,m}$ be a surface of genus $g$ with $m$ punctured, we know the fundamental group of $S_{g,0}$ is $$\pi_1(S_{g,0}) = \left\langle a_1, b_1, \dots, a_g, b_g {~\large\mid~} [a_1, b_1] \dots [a_g,b_g] = 1 \right\rangle,$$ then what is the fundamental group of $S_{g,m}$? Thanks in advance.
Hint: Show that $S_{g,m} \simeq \bigvee_{i=1}^{m+2g-1}S^1$ for $m>0$. To do this, use the fundamental polygon of $S_g$; this should get easier as you add more holes!
• In particular, you probably showed this for $m=1$ when originally calculating the fundamental groups of $S_{g,0}$ via Van Kampen's theorem. – Dan Rust Jun 10 '14 at 13:27
• So $\pi_1(S_{g,m})=\mathbb{Z}^{m+2g-1}$, $m>0$? – user151938 Jun 11 '14 at 5:40
• @user151938 Careful: what you wrote generally refers to the free abelian group. $\pi_1(S_{g,m})$ is the free group on $m+2g-1$ generators. – user98602 Jun 11 '14 at 5:47 | 352 | 932 | {"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} | 3.28125 | 3 | CC-MAIN-2019-30 | longest | en | 0.745148 |
https://jp.mathworks.com/matlabcentral/cody/problems/141-solve-the-sudoku-row/solutions/597446 | 1,563,317,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00332.warc.gz | 439,165,748 | 15,413 | Cody
# Problem 141. Solve the Sudoku Row
Solution 597446
Submitted on 17 Mar 2015 by Ze Zhang
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
%% x = [8 3 4 0 6 7 1 2 9]; y_correct = [8 3 4 5 6 7 1 2 9]; assert(isequal(solveSudokuRow(x),y_correct))
2 Pass
%% x = [ 3 5 7 1 6 8 0 2 9 ]; y_correct = ... [ 3 5 7 1 6 8 4 2 9 ]; assert(isequal(solveSudokuRow(x),y_correct))
3 Pass
%% x = [ 2 8 0 7 3 9 6 5 4 ]'; y_correct = [ 2 8 1 7 3 9 6 5 4 ]'; assert(isequal(solveSudokuRow(x),y_correct)) | 258 | 616 | {"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} | 2.796875 | 3 | CC-MAIN-2019-30 | longest | en | 0.48948 |
http://iannonechem.com/Sc/workbookanswers/7.answers.htm | 1,544,667,095,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824338.6/warc/CC-MAIN-20181213010653-20181213032153-00112.warc.gz | 143,430,627 | 48,834 | Unit VI Molarity
Lesson Day Date Topic
# 2
Molarity Worksheet # 1
1. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.
15.8 g x 1 mole
Molarity = 74.6 g = 0.941 M
0.225 L
2. Calculate the mass of KCl required to prepare 250. mL of 0.250 M solution.
0.250 L x 0.250 moles x 74.6 g = 4.66 g
1 L 1 mole
3. Calculate the volume of 0.30 M KCl solution that contains 6.00 g of KCl.
6.00 g x 1 mole x 1 L = 0.27 L
74.6 g 0.30 mol
4. Calculate the volume of 0.250 M H2SO4 that contains 0.250 g H2SO4.
0.250 g H2SO4 x 1 mole x 1 L = 0.0102 L
98.12 g 0.250 mole
5. 1.50 g of NaCl is dissolved in 100.0 mL of water. Calculate the concentration.
1.50 g x 1 mole
Molarity = 58.5 g = 0.256 M
0.1000 L
6. How many moles of NaCl are in 250. mL of a 0.200 M solution?
0.250 L x 0.200 mole = 0.0500 moles
1 L
7. How many litres of a 0.200 M KCl solution contain 0.250 moles?
0.250 moles x 1 L = 1.25 L
0.200 moles
8. How many millilitres of 0.200 M H2SO4 are required to completely neutralize 250. mL of 0.250 M NaOH?
H2SO4 + 2NaOH Na2SO4 + 2HOH
? mL 0.250 L
0.250 L NaOH x 0.250 mole x 1 mole H2SO4 x 1 L x 1000 mL = 156 mL
1 L 2 mole NaOH 0.200 mole 1 L
9. Calculate the mass of CuSO4.5H2O required to prepare 100.0 mL of 0.100 M solution.
0.100 L x 0.100 mole x 249.7 g = 2.50 g
1 L 1 mole
10. Calculate the mass of Cu(NO3)2.6H2O required to prepare 100.0 mL of 0.200 M solution.
5.91 g
11. Calculate the mass of CoCl3.6H2O required to prepare 500.0 mL of a 0.200 M solution.
27.4 g
12. 50.0 g of NaCl is dissolved in 200.0 mL of water, calculate the molarity.
4.27 M
13. 25.0 g of CuSO4.8H2O is dissolved in 25.0 mL of water, calculate the molarity.
3.29 M
14. Calculate the mass of NaCl required to prepare 500.0 mL of a 0.500 M solution.
14.6 g
15. Calculate the volume of 0.500 M NaCl solution required to contain 0.0500 g of NaCl.
0.00171 L
16. Calculate the volume of 0.200 M NaCl solution required to contain
0.653 g of NaCl.
0.0558L
17. Calculate the mass of NaCl required to prepare 256 mL of a 0.35 M solution.
5.2 g
18. 25.2 g of NaCl is dissolved in 365 mL of water, calculate the molarity.
1.18 M
19. 56.3 g of CuSO4.8H2O is dissolved in 30. mL of water, calculate the molarity.
6.2 M
Worksheet # 2 Molarity
1. Calculate the mass of CuSO4.6H2O required to prepare 200.0 mL of 0.300 M solution.
0.200 L x 0.300 moles x 267.72 g = 16.1 g
1 L 1 mole
2. Calculate the mass of CoCl3.8H2O required to prepare 300.0 mL of a 0.520 M solution.
0.300 L x 0.520 moles x 309.56 g = 48.3 g
1 L 1 mole
3. 150.0 g of NaCl is dissolved in 250.0 mL of water, calculate the molarity.
150.0 g x 1 mole
Molarity = 58.5 g (3sig figs) = 10.3 M
0.250 L
4. 25.2 g of CuSO4.6H2O is dissolved in 28.0 mL of water, calculate the molarity.
25.2 g x 1 mole
Molarity = 267.72 g = 3.36 M
0.0280 L
5. Calculate the mass of NaCl required to prepare 565.0 mL of a 0.450 M solution.
0.5650 L x 0.450 moles x 58.5 g = 14.9 g
1 L 1 mole
6. Calculate the volume of 0.250 M NaCl solution required to contain
0.0300 g of NaCl.
0.0300 g NaCl x 1 mole x 1 L = 0.00205 L
58.5 g 0.250 mole
7. Calculate the volume of 0.500 M NaCl solution required to contain 0.52 g of NaCl.
0.52 g NaCl x 1 mole x 1 L = 0.018 L
58.5 g 0.500 mole
8. Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M solution.
0.3600 L x 0.35 moles x 58.5 g = 7.4 g
1 L 1 mole
9. 55.6 g of NaCl is dissolved in 562 mL of water, calculate the molarity.
55.6 g x 1 mole
Molarity = 58.5 g = 1.69 M
0.562 L
10. 78.9 g of CuSO4.8H2O is dissolved in 500.0 mL of water, calculate the molarity.
78.9 g x 1 mole
Molarity = 303.76 g = 0.519 M
0.5000 L
Stoichiometry Worksheet # 3
1. Excess sodium hydroxide solution is added to 20.0 mL of 0.184 M ZnCl2, calculate the mass of zinc hydroxide that will precipitate.
NaOH(aq) + ZnCl2(aq) Zn(OH)2(s) + 2NaCl(aq)
0.0200 L ? g
0.0200 L ZnCl2 x 0.184 mole x 1 mole Zn(OH)2 x 99.42 g = 0.366 g Zn(OH)2
1 L 1 mole ZnCl2 1 mole
2. How many millilitres of 1.09 M HCl are required to react with a solution formed by dissolving 0.775 g of sodium carbonate?
Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
0.775g ? mL
0.775 g Na2CO3 x 1 mole x 2 mole HCl x 1 L x 1000 mL = 13.4 mL
106 g 1 mole Na2CO3 1.09 moles 1 L
3. Calculate the number of grams barium carbonate that can be precipitated by adding 50.0 mL of 0.424 MBa(NO3)2.
Ba(NO3)2(aq) + K2CrO4(aq) BaCrO4(s) + 2KNO3(aq)
0.0500 L Ba(NO)2 x 0.424 mole x 1 mole BaCO3 x 253.3 g = 5.37 g BaCrO4
1 L 1 mole Ba(NO)2 1 mole
4. Determine the number of millilitres of 0.246 M AgNO3 required to precipitate all the phosphate ion in a solution containing 2.10 g of sodium phosphate.
3AgNO3(aq) + Na3PO4(aq) Ag3PO4(s) + 3NaNO3(aq)
156 mL
5. How many grams of silver nitrate must be used in the preparation of 150. mL of 0.125 M solution.
3.19 g
6. What volume of SO2 is generated by the complete reaction of 35.0 mL of 0.924 M Na2SO3?
Na2SO3(aq) + 2NaOH(aq) 2NaCl(aq) + H2O(l) + SO2(g)
0.724 L
7. How many milliliters of 6.2 M NaOH must react to liberate 2.4 L of hydrogen at STP?
2Al(s) + 6NaOH(aq) 2Na3AlO3(aq) + 3H2(g)
35 mL
8. Calculate the weight of H2C2O4.2H2O required to make 750.0 mL of a 0.480 M solution.
45.4 g
9. 25.4 L of HCl gas at STP are dissolved in 2.5 L of water to produce an acid solution. What volume of 0.200 M Ba (OH) 2 will this solution neutralize?
2HCl + Ba(OH)2 BaCl2 + 2HOH
25.4 L HCl x 1 mole x 1mole Ba (OH) 2 x 1L = 2.8 L
22.4 L 2 mole HCl 0.200 mol
10. 8.25 L at STP of HCL gas is dissolved in 500 ml of water to produce an acid solution. What volume of 0.200 M Ca (OH) 2 will this solution neutralize?
Ca (OH) 2 + 2HCl CaCl2 + 2H2O
8.25 L HCL x 1 mole x 1 mole Ca (OH) 2 x 1 L = 0.921L
22.4 L 2 mole HCL 0.200 mol
11. 250 mL of water is added to 100 mL of 0.0200M H2SO4. What volume of 0.100M KOH will it neutralize?
H2SO4 + 2KOH 2K2SO4 + 2HOH
0.1 L H2SO4 x 0.0200mol x 2 mole KOH x 1 L x 1000 mL
1L 1 mole H2SO4 0.100 mol 1 L
= 40ml
12. What volume of 0.924 M Na2SO3 is required for the production of 350.0 mL of SO2 at STP?
2HCl + Na2SO3(aq) + 2NaOH(aq) → 2NaCl(aq) + H2O(l) + SO2(g)
1.69 x 10-2L
13. How many milliliters of hydrogen at STP can be generated by 500.0 mL 6.2 M NaOH completely reacting with excess Al.
2Al(s) + 6NaOH(aq) → 2Na3AlO3(aq) + 3H2(g)
3.5 x 104 mL
14. 64.5 L of HCl gas at STP are dissolved in water to produce an acid solution. What volume of 0.200 M Ba (OH)2 will this solution neutralize?
7.20 L
Titration Calculations Worksheet # 4
1. In a titration 12.5 mL of 0.200 M NaOH ia required to neutralize 10.0 mL of H2SO4.
Calculate the concentration of the acid ?
H2SO4 + 2NaOH Na2SO4 + 2HOH
0.0100 L 0.0125 L
? M 0.200 M
Molarity = 0.0125 L NaOH x 0.200 moles x 1 mole H2SO4
1 L 2 mole NaOH
0.0100 L
= 0.125 M
2. In a titration 22.5 mL of 0.100 M HCl ia required to neutralize 20.0 mL of Ba(OH)2 .
Calculate the concentration of the base ?
2HCl + Ba(OH)2 BaCl2 + 2HOH
0.0225 L 0.0200 L
0.100 M ? M
Molarity = 0.0225 L HCl x 0.100 moles x 1 mole Ba(OH)2
1 L 2 mole HCl
0.0200 L
= 0.0563 M
3. A burette filled with 1.52 M nitric acid solution reads 33.10 mL initially. After titrating a 25.00 mL sample of
barium hydroxide the endpoint was reached and the burette showed 46.30 mL. What is the barium hydroxide
concentration?
46.30 - 33.10 = 13.20 mL = 0.01320 L
2HNO3 + Ba(OH)2 Ba(NO3)2 + 2HOH
0.01320 L 0.02500 L
1.52 M ? M
0.0132L HNO3 x 1.52 mole x 1 mole Ba(OH)2
Molarity = 1 L 2 mole HNO3 = 0.402 M
0.02500 L
4. A burette filled with 2.557 M sodium hydroxide solution reads 15.62 mL initially. After titrating a 25.00 mL sample of phosphoric acid the endpoint was reached and the burette now showed 39.22 mL. What is the [phosphoric acid]?
39.22 - 15.62 = 23.60 mL = 0.02360 L
3NaOH + H3PO4 Na3PO4 + 3H2O
0 .02360 L 0.02500 L
2.557 M ? M
0.02360 L NaOH x 2.557 mole x 1 mole H3PO4
Molarity = 1 L 3 mole NaOH = 0.8046 M
0.02500 L
5. A 10.00 mL sample of 2.120 M sodium hydroxide solution is placed in a 250.0 mL Erlenmeyer flask. An indicator called bromothymol blue is added to the solution. The solution is blue. Hydrochloric acid was added from a burette until there was a green color (endpoint had been reached). Determine the concentration of hydrochloric acid given the following burette readings:
Burette final = 22.04 mL
Burette initial - 12.08 mL
Difference = 9.96 mL Beware subtraction!!!! One sig fig is lost!!!
NaOH + HCl NaCl + H2O
.01000 L .00996 L
2.120 M ? M
0.01000 L NaOH x 2.1220 mole x 1 mole HCl
Molarity = 1 L 1 mole NaOH = 2.13 M HCl
0.00996 L
6. The following data was obtained during the titration of 1.0097 M sodium hydroxide with a 25.00 mL aliquot of hydrofluoric acid:
Trial 1 Trial 2 Trial 3
Burette Final Reading 34.56 mL 39.42 mL 44.20 mL
Burette Initial Reading 14.94 mL 19.86 mL 24.66 mL
Use the above information to determine the concentration of the acid.
NaOH + HF NaF + H2O
0 .01955 L 0 .0250 L
1.0097 M ? M
0.01955 L NaOH x 1.0097 mole x 1 mole HF
[HF] = 1 L 1 mole NaOH = 0.7896 M
0.0250 L
7. The following data was obtained during the titration of 0.0998 M sodium hydroxide with a
10.00 mL aliquot of sulphuric acid:
Trial 1 Trial 2 Trial 3
Burette Final Reading 26.05 mL 48.52 mL 33.78 mL
Burette Initial Reading 2.46 mL 34.94 mL 20.22 mL
Use the above information to determine the concentration of the acid.
6.77 x 10-2 M
8. The following data was obtained during the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:
Trial 1 Trial 2 Trial 3
Burette Final Reading 22.92 mL 25.32 mL 41.30 mL
Burette Initial Reading 0.06 mL 2.58 mL 18.54 mL
Use the above information to determine the concentration of the barium hydroxide.
0.9352 M
9. What volume of 2.549 M NaOH is needed to fully titrate 50.0 mL of 1.285 M HCl solution ?
HCl + NaOH →
0.0500 L HCl x 1.285 mol x 1 mole NaOH x 1L = 0.0252 L
1 L 1 mole HCl 2.549 mole
10. What volume of 1.146 M KOH is needed to fully titrate 20.8 mL of 0.557 M H2SO4 solution ?
H2SO4 + 2KOH →
0.0208 L H2SO4 x 0.557 mol x 2 mole KOH x 1L = 0.0202 L
1 L 1 mole H2SO4 1.146 mole
Titrations Worksheet # 5
1. Calculate the mass of H2C2O4.2H2O required to prepare 500.0 mL of a 0.200M solution.
12.6 g
2. Calculate the mass of Cu2SO4.6H2O required to prepare 200.0 mL of a 0.300M solution.
19.9 g
3. In a titration 0.200 M NaOH is used to neutralize 10.0 mL of H2SO4. In three runs, the following data was collected. Calculate the concentration of the acid.
Volume of 0.20 M NaOH (mL)
Initial Burette Reading 12.90 15.70 18.50
Final Burette Reading 15.70 18.50 21.50
0.028 M
4. In a titration 0.250 M KOH is used to neutralize 25.0 mL of H3PO4. In three runs, the following data was collected. Calculate the concentration of the acid.
Volume of 0.250 M KOH (mL)
Initial Burette Reading 2.90 15.70 28.70
Final Burette Reading 15.70 28.70 42.70
12.80 13.00 14.00
Use 12.90 mL
0.0430 M
5. Calculate the volume of 0.500M H3PO4 required to neutralize 25.0 mL of
0.200M NaOH.
0.00333L
6. Calculate the volume of 0.50 M NaOH required to neutralize 35.0 mL of
0.100M H2C2O4.
0.014L
7. In a titration 35.7 mL of 0.250 M H3PO4 is used to neutralize 25.0 mL of KOH. Calculate the molarity of the base.
2.08 M
8. In a titration 35.2 mL of 0.20 M H2C2O4 is used to neutralize 10.0 mL of NaOH. Calculate the molarity of the base.
1.4 M
9. 2 Al + 3 I2 2 AlI3
Initial 12.0 mol 15.0 mol 0
Change:
End: 2.0 mol 0mol 10.0 mol
10. C + 2Cl2 CCl4
Initial 16.0 mol 34.0 mol 0
Changes:
End: 0 mol 2.0 mol 16.0 mol
11. 4 Fe + 3 O2 2 Fe2O3
Initial 12.0 mol 8.0 mol 0
Change:
End: 1.3 mol 0 mol 5.33 mol
12. 2 NO + O2 2 NO2
100. g x 1 mole 100. g x 1 mole 0
30.0 g 32.0 g
Init: 3.333 3.125 3.333
Change: 3.333 1.667 3.333
End: 0 1.458 mol x 32.0 g 3.333 x 46.0 g
1 mole 1 mole
Grams: 0 g 46.7 g 153 g
13. Calculate the volume of H2 gas produced at STP by the reaction of 300. mL of 0.500 M HCl with excess Zn.
Zn + 2HCl H2 + ZnCl2
0.300 L x 0.500 moles x 1 mole H2 x 22.4 L = 1.68 L
1 L 2 mole HCl 1 mole
14. Calculate the volume of 0.30 M KCl solution that contains 9.00 g of KCl.
0.40 L
Dilutions Worksheet # 6
1. 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.
M1V1 = M2V2
(20.0)(0.200) = M2(100.0)
M2 = 0.0400 M
2. 15.0 mL of a solution of NaOH is diluted to a final volume of 250.0 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.
M1V1 = M2V2
(15.0) M1 = (250.0) (0.0500)
M1 = 0.833 M
3. 50.0 mL of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate the new molarity.
V2 = 50.0 mL + 150.0 mL = 200.0
M1V1 = M2V2
(0.025)(50.0) = M2 (200.0)
M2 = 0.0063M
4. 45.0 mL of a solution of NaOH is diluted by adding 250.0 mL of water to produce a new molarity of 0.0500 M. Calculate the molarity of the base.
0.328 M
5. A 0.125 M solution is concentrated by evaporation to a reduced final volume of 100.0 mL and a molarity of 0.150 M. Calculate the original volume.
120. mL
6. 850.0 mL of 0.280 M KOH solution is diluted to a final volume of 1000.0 mL, calculate the new concentration.
0.238 M
7. 95.0 mL of a solution of NaOH is diluted to a final volume of 135 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.
0.0711 M
Molarity Review # 7
1. Convert 250. g AgNO3 to formula units and then to atoms of O.
2.66 x 1024 at O
2. Convert 5.9 x1025 H2 molecules to grams.
2.0 x 102 g H2
3. Calculate the percentage composition of MgSO4.
20.2 % Mg 26.7 % S 53.2 % O
4. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular formula.
C3H5N2O C15H25N10O5
5. How many grams O2 are required to consume 56.3 g Al?
4Al + 3O2 2Al2O3
50.0 g O2
6. 25.5 mL of 0.100 M HCl reacts with excess Zn to produce 25.3 mL of H2 gas at STP. Calculate the theoretical yield in mL and the percentage yield of H2 gas.
Zn + 2HCl H2 + ZnCl2.
0.0255 L x 0.100 mole x 1 mole H2 x 22.4 L = 0.0286 L 88.6 %
1 L 2 mole HCl 1 mole
7. Calculate the energy produced by the complete reaction of 150. g H2.
2H2 + O2 2H2O + 130. KJ
4.83 x 103 KJ
8. 84.0 g of Al reacts with 122 g O2 to produce Al2O3. How many grams of Al2O3 are produced? Determine the mass of the reactant in excess and the limiting reactant.
4Al + 3O2 2Al2O3
84.0 g x 1 mole 122 g x 1 mole
27 g 32 g
I 3.111 mole 3.8125 mole 0
C 3.111 mole 2.333 mole 1.5555 mole
E 0 1.4795 mole 1.5555 mole
47.3 g 159 g
9. 15.2 g of Al reacts with 14.3 g O2 to produce Al2O3. How many grams of Al2O3 are produced? Determine the mass of the reactant in excess and the limiting reactant.
4Al + 3O2 2Al2O3
15.2 g x 1 mole 14.3 g x 1 mole
27 g 32 g
I 0.5630 mole 0.4469 mole 0
C 0.5630 mole 0.4222 mole 0.2815 mole
E 0 0.247 mole 0.2815 mole
0.790 g 28.7 g
10. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.
[KCl] = 15.8 g x 1 mole = 0.941 M
74.6 g
0.225 L
11. Calculate the mass of KCl required to prepare 250.0 mL of 0.250 M solution.
0.2500L x 0.250 mole x 74.6 g = 4.66 g KCl
1 L 1 mole
12. Calculate the volume of 0.30 M BaCl2 solution that contains 6.00 g of KCl.
6.00 g x 1 mole x 1 L = 0.27 L
74.6 g 0.30 mole
13. Calculate the volume of 0.250 M H3PO4 that contains 0.250 g H2SO4.
0.250 g x 1 mole x 1 L = 0.0102 L
98.03 g 0.250 mole
14. 1.5 g of BaCl2 is dissolved in 100.0 mL of water. Calculate the concentration.
0.072 M
15. How many moles of BaCl2 are in 250.0 mL of a 0.200 M solution?
0.0500 moles
16. How many litres of a 0.200 MBaCl2 solution contain 0.250 moles?
1.25 L
17. Calculate the volume of H2 gas produced at STP by the reaction of 400.0 mL of 0.800 M HCl with excess Zn.
Zn + 2HCl H2 + ZnCl2
3.58 L
18. Calculate the volume of 0.250 M H3PO4 required to neutralize 25.5 mL of 0.200 M NaOH.
H3PO4 + 3NaOH ® Na3PO4 + 3HOH
? L 0.0255 L
0.0255 L NaOH x 0.200 mole x 1 mole H3PO4 x 1 L = 0.00680 L
1 L 3 mole NaOH 0.250 mole
19. Calculate the volume of 0.500 M KOH required to neutralize 45.3 mL of 0.320 M H2SO4 .
H2SO4 + 2KOH ® K2SO4 + 2HOH
0.0453 L ? L
0.0453 L H2SO4 x 0.320 mole x 2 mole KOH x 1 L = 0.0580 L
1 L 1 mole H2SO4 0.500 mole
20. Calculate the mass of CoCl3.6H2O required to prepare 500.0 mL of a 0.200 M solution.
27.4 g
Worksheet # 8 Dilutions and Molarity
1. 40.0 mL of 0.400 M NaOH solution is diluted to a final volume of 200.0 mL, calculate the new concentration.
M1V1 = M2V2
(0.400)(40.0) = M2(200.0)
M2 = 0.0800 M
2. 85.0 mL of a solution of NaOH is diluted to a final volume of 290.0 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.
M1V1 = M2V2
M1(85.0) = (0.0500)(290.0)
M1 = 0.171 M
3. 150.0 mL of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate the new molarity.
M1V1 = M2V2
(0.025)(150.0) = M2(300.0)
M2 = 0.013 M
4. 220.0 mL of a solution of NaOH is diluted by adding 250.0 mL of water to produce a new molarity of 0.0500 M. Calculate the molarity of the base.
M1V1 = M2V2
M1(220.0) = (0.0500)(470.0)
M1 = 0.107 M
5. A 0.350 M solution is concentrated by evaporation to a reduced final volume of 100.0 mL and a molarity of 0.825 M. Calculate the original volume.
V1 = 236 mL
6. 850.0 mL of 0.280 M KOH solution is diluted to a final volume of 1000.0 mL, calculate the new concentration.
M2 = 0.238 M
7. 28 g of KCl is dissolved in 225 mL of water, calculate the molarity.
1.7 M
8. Calculate the mass of KCl required to prepare 125 mL of 0.450 M solution.
4.20 g
9. Calculate the volume of 0.40 M KCl solution that contains 8.00 g of KCl.
0.27 L
10. Calculate the volume of 0.400 M H2SO4 required to neutralize 25.0 mL of 0.200 M NaOH.
0.00625 L
11. Calculate the volume of H2 gas produced at STP by the reaction of 250.0 mL of 0.600 M HCl with excess Zn. Zn + 2HCl H2 + ZnCl2
1.68 L
12. 8.5 L of HCl gas at STP is dissolved in 325 mL of water, calculate the molarity of the acid solution.
1.2 M
13. How many moles of NaCl are in 350.0 mL of a 0.400 M solution?
0.140M
14. How many litres of a 0.300 M KCl solution contain 0.350 moles?
1.17 L
15. Calculate the mass of 8.25 x 105 mL of H2 gas at STP.
74.4 g
16. Calculate the number of formula units of KCl in 200.0 mL of 0.300 M solution.
3.61 x 1022 FU
Worksheet # 9 Ion Concentration
1. What is the concentration of each ion in a 10.5 M sodium sulphite solution?
Na2SO3 2 Na+ + SO32-
10.5 M 21.0M 10.5M
2. What is the concentration of each ion in the solution formed when 94.5 g of nickel (III) sulphate is dissolved into 850.0 mL of water?
Ni2 (SO4) 3 2Ni3+ + 3S042-
0.274 M 0.548M 0.822M
Molarity = 94.5 g x 1 mole
405.7 g = 0.274M
0.8500L
3. If 3.78 L of 0.960 M calcium fluoride solution is added to 6.36 L of water, what is the resulting concentration of each ion?
M1V1 = M2V2 CaF2 Ca2+ + 2F-
(0.960) (3.78) = M2 (10.14) 0.358 M 0.358 M 0.716 M
M2= 0.358 M
4. What is the concentration of each ion in a 5.55 M zinc phosphate solution?
Zn3 (PO4) 2 3Zn 2+ + 2PO43-
5.55 M 16.7 11.1 M
5. What is the concentration of each ion in the solution formed when 94.78 g of iron (III) sulphate
is dissolved into 550.0 mL of water?
Fe2 (SO4)3 2Fe3+ + 3SO42-
0.4309 M 0.8619 M 1.293 M
94.78 g x 1 mol
[Fe2 (SO4) 3] = 399.9 g = 0.4309 M
0.5500 L
6. If 6.25 L of 0.560 M sodium bromide solution is added to 3.45 L of water, what is the resulting
Concentration of each ion?
M1V1 = M2V2 NaBr Na + + Br -
(0.560) (6.25) = M2 (9.70) 0.361 M 0.361 M 0.361 M
M2= 0.361 M
7. 50.0 mL of 0.200 M Na3PO4 solution is mixed with 150.0 mL of 0.400 M Na2CO3.
Calculate all ion concentrations.
Na3PO4 3 Na+ + PO43-
50.0 0.200 M 0.150 M 0.0500 M
200.0
Na2PO4 → 2 Na+ + CO32-
150.0 0.400 M 0.600 M 0.300 M
200.0
[Na+] = 0.600 M + 0.150 M = 0.750 M
8. What is the concentration of each ion in the solution formed when 16.5 g of Aluminum sulphate is
dissolved into 600.0 mL of water?
16.5 g x 1 mol
[Al2 (SO4) 3] = 342.3 g = 0.0803 M
0.600 L
Al2 (SO4) 3 2Al3+ + 3SO42-
0.0803 M 0.161 M 0.241M
9. If 1.78 L of 0.420 M barium fluoride solution is added to 2.56 L of water, what is the resulting
concentration of each ion?
M1V1 = M2V2 BaF2 Ba2+ + 2F-
(0.420)(1.78) = M2 (4.34) 0.172 M 0.172 M 0.345 M
M2 = 0.172 M
10. What is the concentration of each ion in a 1.22 M zinc acetate solution?
Zn (CH3COO) 2 → Zn 2+ + 2CH3COO-
1.22 M 1.22M 2.44M
11. What is the concentration of each ion in the solution formed when 94.78 g of cobalt (III) sulphate
is dissolved into 400.0 mL of water?
[Co2(SO4)3] = 94.78 g x 1 mole
406.1 g = 0.5835 M
0.4000 L
Co2(SO4)3 2Co3+ + 3SO42-
0.5835 M 1.167 M 1.750 M
12. If the chloride concentration in 2.00 L of solution is 0.0900 M, calculate the [Al3+] and the molarity of the AlCl3 solution.
AlCl3 Al3+ + 3Cl-
0.0300 M 0.0300 M 0.0900 M
13. If the [Ga3+] concentration in 2.00 L of solution is 0.0300 M, calculate the [SO42-] and the molarity of the Ga2(SO4)3 solution.
Ga2(SO4)3 2Ga3+ + 3SO42-
0.0150 M 0.0300 M 0.0450 M
14. In a titration 12.5 mL of 0.200 M NaOH is needed to neutralize 10.0 mL of H3PO4, calculate the acid concentration.
H3PO4 + 3NaOH Na3PO4 + 3HOH
0.0100 L 0.0125 L
? M 0.200 M
[H3PO4] = 0.0125 L NaOH x 0.200 mole x 1 moles H3PO4
1 L 3 mole NaOH = 0.0833 M
0.0100 L
15. What volume of 0.200 M H2SO4 is required to neutralize 25.0 mL of 0.300 M NaOH?
H2SO4 + 2NaOH Na2SO4 + 2HOH
? L 0.0250 L
0.200 M 0.300 M
0.0250 L NaOH x 0.300 mole x 1 moles H2SO4 x 1 L = 0.0188 L
1 L 2 NaOH 0.200 mole
16. The [Cl-] = 0.600 M in 100.0 mL of a AlCl3 solution. How many grams AlCl3 are in the solution?
AlCl3 ® Al3+ + 3Cl-
0.200 M 0.600 M
0.1000 L x 0.200 mole x 133.5 g = 2.67 g
1 L 1 mole
17. The [SO42-] = 0.600 M in 100.0 mL of a Al2(SO4)3 solution. How many grams Al2(SO4)3 are in the solution?
6.85 g
Worksheet # 10 Molarity Unit Review # 1
1. 200.0 mL of 0.200 M H2SO4 reacts with 250.0 mL of 0.40 M NaOH, calculate the concentration of the excess base.
H2SO4 + 2NaOH Na2SO4 + 2HOH
0.2000 L x 0.200 mole 0.250 L x 0.40 mole
1 L 1 L
I 0.0400 mole 0.100 mole
C 0.0400 mole 0.0800 mole
E 0 0.020 Note the loss of one sig fig!
[NaOH] = 0.020 mole = 0.044 M
0.4500 L Note that the final volume is 250.0 + 200.0 mL
2. 100.0 mL of 0.100 M H2SO4 reacts with 50.0 mL of 0.20 M NaOH, calculate the concentration of the excess acid.
H2SO4 + 2NaOH Na2SO4 + 2HOH
0.1000 L x 0.100 mole 0.050 L x 0.20 mole
1 L 1 L
I 0.0100 mole 0.010 mole
C 0.0050 mole 0.010 mole
E 0.0050 mole 0
[H2SO4] = 0.0050 mole = 0.033 M
0.1500 L
Note that the final volume is 100.0 + 50.0 mL
3. 500.0 mL of 0.100 M H2SO4 reacts with 400.0 mL of 0.400 M NaOH, calculate the concentration of the excess base.
H2SO4 + 2NaOH Na2SO4 + 2HOH
0.5000 L x 0.100 mole 0.4000 L x 0.40 mole
1 L 1 L
I 0.0500 mole 0.160 mole
C 0.0500 mole 0.100 mole
E 0 mole 0.060 mole
Note the loss of sig figs!
[NaOH] = 0.06 mole = 0.067 M
0.900 L
Note that the final volume is 500.0 + 400.0 mL
4. 100.0 mL of 0.200 M MgCl2 reacts with 300.0 mL of 0.400 M AlCl3, calculate all ion concentrations.
MgCl2 Mg2+ + 2Cl-
100.0 0.200 M 0.0500 M 0.100 M
400.0
AlCl3 → Al3+ + 3Cl-
300.0 0.400 M 0.300 M 0.900 M
400.0
[Cl-] = 0.100 M + 0.900 M = 1.000 M
5. Change 2.66 moles of H2O to molecules.
1.60 x 1024 molecules
6. Change 9.7x1019 atoms Fe to moles.
1.6 x 10-4 mole
7. Convert 88.3 g AgNO3 to formula units and then to atoms of O.
88.3 g x 1 mol x 6.02 x 1023 FU x 3 atoms O = 9.39 x 1023 atoms O
169.9 g 1 mol 1 FU
8. Convert 3.8 x 1025 H2 molecules to grams.
1.3 x 102 g
9. Calculate the empirical formula of a compound that is 62.2 % Pb, 8.454 % N, and 28.8 % O.
Is this compound ionic or covalent?
Pb (NO3) 2
10. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and 18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular formula.
C15H25N10O5
11. How many moles of Al2O3 are produced by the reaction 200. g Al?
4Al + 302 2Al2O3
3.70 mole
12. How many moles Al are required to produce 300. g Al2O3?
4Al + 302 2Al2O3
5.88 mole
13. 100. g Al reacts with excess O2 to produce 150. g Al2O3 according to
Calculate the theoretical and percentage yield. 4Al + 302 2 Al2O3.
79.4 %
14. Calculate the energy produced by the complete reaction of 150. g H2.
2H2 + O2 2H2O + 130. KJ
4.83 x 103 kJ
15. How many grams of H2 would be needed to produce 260. KJ of energy?
2H2 + O2 2H2O + 130. KJ
8.08 g
16. 20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant.
8 g H2 , 2.9 x 102 g H2O
17. How many litres of O2 gas are required to produce 100. g Al2O3?
4Al + 302 2Al2O3
32.9 L
18. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.
0.941 M
19. Calculate the mass of KCl required to prepare 250.0 mL of 0.250 M solution.
4.66 g
20. Calculate the volume of 0.30 M KCl solution that contains 6.00 g of KCl.
0.27 L
21. Calculate the volume of 0.250 M H2SO4 required to neutralize 20.0 mL of 0.100 M NaOH.
0.00400 L
22. Calculate the volume of H2 gas produced at STP by the reaction of 150.0 mL of 0.500 M HCl with excess Zn.
Zn + 2HCl H2 + ZnCl2
0.840 L
23. 1.5 L of HCl gas at STP is dissolved in 225 mL of water, calculate the molarity of the acid solution.
0.30 M
24. How many moles of NaCl are in 250. mL of a 0.200 M solution?
0.0500 moles
25. How many litres of a 0.200 M KCl solution contain 0.250 moles?
1.25 L
26. Calculate the mass of 2.25 x 105 mL of H2 gas at STP.
20.3 g
27. Calculate the number of formula units of KCl in 100.0 mL of 0.200 M solution.
1.20 x 1022 FU
28. 40.6g of KBr is dissolved in 500.0 mL of water, calculate the molarity.
0.682 M
29. Calculate the mass of KBr required to prepare 450.0 mL of 0.350 M solution.
18.7 g
30. Calculate the volume of 0.50 M KCl solution that contains 3.00 g of KCl.
0.080 L
31. Calculate the volume of 0.250 M H3PO4 required to neutralize 25.5 mL of
0.200 M NaOH.
0.00680 L
32. In a titration 22.5 mL of 0.200 M H3PO4 is required to neutralize 10.0 mL of KOH. What is the molarity of the base?
1.35 M
Worksheet # 11 Molarity Unit Review # 2
1. 100.0 mL of 0.200 M HCl, 200.0 mL of 0.100 M HBr, and 175 mL of 0.100 M Ba(OH)2.
Calculate the concentration of the excess acid or base.
0.1000 L x 0.200 mole = 0.0200 mole HCl
1L
0.2000 L x 0.100 mole = 0.0200 mole HBr
1L
= 0.0400 mole HX
0.175L x 0.100 mole = 0.0175 mole Ba(OH)2
1L
2HX + Ba(OH)
I 0.0400 mole 0.0175 mole
C 0.0350 mole 0.0175 mole
E 0.0050 mole 0.0000
Total Volume = 100.0 mL + 200.0 mL + 175 mL = 475 mL
[HX] = 0.0050 mole
0.475 L
[HX] = 0.011 M
2. 150.0 mL of 0.200 M HCl and 250 mL of 0.300 M HNO3 react with excess CaCO3
Calculate the theoretical yield of CO2. Start by writing an equation.
HCl + CaCO3 ® CO2 + CaCl2 + H2O
0.1500 L HCl x 0.200 mole x 1 mole CO2 x 44.0 g = 0.660 g
1 L 2 mole HCl 1 mole
HNO3 + CaCO3 ® CO2 + Ca(NO3)2 + H2O
0.2500 L HNO3 x 0.300 mole x 1 mole CO2 x 44.0 g = 1.65 g
1 L 2 mole HCl 1 mole
Total 2.31 g
3. Calculate the percentage composition of Al2(SO4)3 to three significant figures.
2 Al 54.0 %Al = 54.0 x 100% = 15.8 %
342.3
3 S 96.3 %S = 96.3 x 100% = 28.1 %
342.3
12 O 192.0 % O = 192.0 x 100% = 56.1%
342.3 342.3
4. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and and18.8 % O and has a molecular mass of 850.5g/mol. Calculate the empirical and molecular formula.
42.3 g C x 1 mol = 3.525 mol = 3 C3H5N2O C30H50N20O10
12.0 g
5.94 g H x 1 mol = 5.881 mol = 5
1.01 g
32.9 g N x 1 mol = 2.350 mol = 2
14.0 g
18.8 g O x 1 mol = 1.175 mol = 1
16.0 g
5. How many grams of 02 are required to consume 56.3 g Al?
4Al + 302 2Al2O3
56.3 g Al x 1 mol x 3 mol O2 x 32.0 g = 50.0 g O2
27.0 g 4 mol Al 1 mol
6. 15.8 g of AlCl3 is dissolved in 225 mL of water, calculate the molarity.
Molarity = 15.8 g x 1 mol
133.5 g = 0.526 M
0.225 L
7. Calculate the mass of AlCl3 required to prepare 250.0 mL of 0.250 M solution.
0.250 L x 0.250 mol x 133.5 g = 8.34 g
L 1 mol
8. Calculate the volume of 0.30 M AlCl3 solution that contains 6.00 g of AlCl3.
6.00 g x 1 mol x 1 L = 0.15 L
33.5 g 0.30 mol
9. Calculate the volume of 0.450 M H2SO4 required to neutralize 25.0 mL of
0.200 M NaOH.
H2SO4 + 2NaOH Na2SO4 + 2HOH
? L 0.025 L
0.025 L NaOH x 0.200 mol x 1 mol H2SO4 x L = 0.00556 L
L 2 mol NaOH 0.450 mol
10. Calculate the volume of H2 gas produced at STP by the reaction of 350.0 mL of 0.600 M HCl with excess Zn.
Zn + 2HCl H2 + ZnCl2
0.350 L HCl x 0.600 mol x 1 mol H2 x 22.4 L = 2.35 L
L 2 mol HCl 1 mol
11. 2.9 L of HCl gas at STP is dissolved in 225 mL of water, calculate the molarity of the acid solution.
Molarity = 2.9 L x 1 mol = 0.58 M
22.4 L
0.225 L
12. How many moles of NaCl are in 500.0 mL of a 0.300 M solution?
0.500 L x 0.300 mol = 0.150 mol
L
13. How many litres of a .2300 M KCl solution contain 0.250 moles?
0.250 mol x L = 1.09 L
0.2300 mol
14. Calculate the mass of 560. mL of CO2 gas at STP.
0.560 L x 1 mol x 44.0 g = 1.10g
22.4 L 1 mol
15. Calculate the number of formula units of NaCl in 100.0 mL of 0.200 M solution.
0.100 L x 0.200 mol x 6.02 x 1023 FU = 1.20 x 1022 FU
L 1 mol
16. 25.5 mL of 0.100 M HCl reacts with excess Zn to produce 25.3 mL of H2 gas at STP.
Calculate the theoretical yield in mL and the percentage yield of H2 gas.
Zn + 2HCl H2 + ZnCl2.
0.0255 L x 0.100 mol x 1 mol H2 x 22.4 L = 0.0286 L
L 2 mol HCL 1 mol
% Yield = 25.3 x 100 % = 88.6 %
28.6
17. Calculate the energy produced by the complete reaction of 150. g H2.
2H2+O2 2H2O + 130KJ
150 g H2 x 1mol x 130 KJ = 4.83 x 103 KJ
2.02 g 2 mol
18. 84.0 g of Al reacts with 122g O2 to produce Al2O3. How many grams of Al2O3 are produced? Determine the mass of the reactant in excess and the limiting reactant.
4 Al + 3O2 2Al2O3
84.0 g x 1 mol 122g x 1mol
27.0 g 32.0 g
I 3.111 mol 3.813 mol 0
C 3.111 mol 2.333 mol 1.556
E 0 1.48 mol 1.556
Limiting 47.4 g O2 excess 159 g
19. Calculate the percentage composition of Na2SO4.
2 Na 46.0 g % Na = 32.4 %
1 S 32.1 g % S = 22.6 %
4 O 64.0 g % O = 45.0 %
142.1
20. How many litres of O2 gas are required to produce 100. g Al2O3?
4Al + O2 2Al2O3
100. g Al2O3 x 1 mole x 3 mole O2 x 22.4 L = 32.9 L
102 g 2 mole Al2O3 1 mole
21. Calculate the molar mass of a gas that weighs 19.43 g and has a STP volume of 9.894 L.
If the gas is a very funny one containing nitrogen and used by the dentist, determine the
molar mass and molecular formula for the gas.
9.894 L x 1 mole = 0.4417 mole
22.4 L
Molar Mass = 19.43 g = 44.0 g/mole N2O
0.4417 mole
Write a balanced formula equation, complete ionic equation, and net ionic equation for each reaction. There are two no reactions.
22. Zn(s) + 2AgNO3(aq) 2Ag(s) + Zn(NO3)2(aq)
Zn(s) + 2Ag+ + 2NO3- 2Ag(s) + Zn2+ + 2NO3-
Zn(s) + 2Ag+ 2Ag(s) + Zn2+
23. BaS (aq) + 2KOH(aq) Ba(OH)2(s) + K2S(aq)
Ba2+ + S2- + 2K+ + 2OH- Ba(OH)2(s) + 2K+ + S2-
Ba2+ + 2OH- Ba(OH)2(s)
24. 2NaCl(aq) + F2(g) 2NaF(aq) + Cl2(g)
2Na+ + 2Cl- + F2(g) 2Na+ + 2F- + Cl2(g)
2Cl- + F2(g) 2F- + Cl2(g)
25. Sr(OH)2 (aq) + CuSO4(aq) Cu(OH)2 (s) + SrSO4(s)
Sr2+ + 2OH- + Cu2+ + SO42- Cu(OH)2 (s) + SrSO4(s)
Sr2+ + 2OH- + Cu2+ + SO42- Cu(OH)2 (s) + SrSO4(s)
26. NaCl(aq) + Cu(NO3)2(aq)
No reaction, both possible products have high solubility
27. NaCl(aq) + ZnF2(aq)
No reaction, both possible products have high solubility
28. 100.0 g of an aqueous compound that is 45.49 % Pb, 12.31 % N, and 42.20 % O reacts with another compound that is 28.16 % N, 8.13 % H, 20.79 % P, and 42.91 % O. If the actual yield of the product containing lead is 60.0 g, calculate the percentage yield.
60.0 g Actual Yield
3Pb(NO3)4(aq) + 4(NH4)3PO4(aq) Pb3(PO4)4(s) + 12NH4NO3(aq)
100.0 g ? g Theoretical Yield
100.0 g Pb(NO3)4 x 1 mole x 1 mole Pb3(PO4)4 x 1001.6 g = 75.3 g
455.2 g 3 mole 3Pb(NO3)4 1 mole
% yield = 60.0 g x 100% = 81.8 %
29. The following data was obtained during the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:
Trial 1 Trial 2 Trial 3
Burette Final Reading 22.92 mL 25.32 mL 41.30 mL
Burette Initial Reading 0.06 mL 2.58 mL 18.54 mL
Vol. of Acid Added 22.86 mL 22.74 mL 22.76 mL
Use the above information to determine the concentration of the barium hydroxide.
2HCl + Ba(OH)2
0.02275 L 0.02500 L
2.0554 M ? M
Molarity Ba(OH)2 = 0.02275 L x 2.0554 mole x 1 mole Ba(OH)2
1 L 2 mole HCl
0.02500 L
= 0.9352 M
30. 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.
M1V1 = M2V2
(0.200)(20.0) = M2(100)
M2 = 0.0400 M
31. 20.0 mL of 0.300 M AlCl3 is mixed with 20.0 mL of 0.300 CaCl2, calculate all ion concentrations.
ACl3 Al3+ + 3Cl-
20.0 0.300 M 0.150 M 0.450 M
40.0
CaF2 → Ca2+ + 2Cl-
20.0 0.300 M 0.150 M 0.300 M
40.0
[Cl-] = 0.450 M + 0.300 M = 0.750 M
32. A burette filled with 2.000 M sodium hydroxide solution reads 20.20 mL initially. After titrating a
25.00 mL sample of phosphoric acid the endpoint was reached and the burette now showed 40.20 mL. What is the [phosphoric acid]?
H3PO4 + 3NaOH
0.02500 L 0.02000 L
? M 2.000 M
Molarity H3PO4 = 0.02000 L x 2.000 mole x 1 mole H3PO4
1 L 3 mole NaOH
0.02500 L
= 0.5333 M
33. Calculate the volume of 0.500M KOH required to neutralize 45.0 mL of 0.320 M H2SO4.
H2SO4 + 2KOH
0.0450 L ? L
0.0450 L H2SO4 x 0.320 moles x 2 mole KOH x 1 L = 0.0576 L
1 L 1 mole H2SO4 0.500 mole
34. 100.0 mL of 0.200 M H2SO4 reacts with 150.0 mL of 0.40 M NaOH, calculate the concentration of the excess base.
H2SO4 + 2NaOH
0.1000 L x 0.200 mole 0.1500 L x 0.40 mole
1 L 1L
I 0.0200 mole 0.060 mole
C 0.0200 mole 0.0400 mole
E 0 0.020 mole Note the loss of one sig fig
[NaOH] = 0.020 mole = 0.080 M
0.250 L
35. 200.0 mL of 0.10 M H2SO4 reacts with 100.0 mL of 0.20 M NaOH, calculate the concentration of the excess acid.
H2SO4 + 2NaOH
0.2000 L x 0.100 mole 0.1000 L x 0.20 mole
1 L 1L
I 0.0200 mole 0.020 mole
C 0.010 mole 0.020 mole
E 0.010 loss of one sig fig 0
[H2SO4] = 0.010 mole = 0.033 M
0.300 L
36. 250.0 mL of 0.100 M H2SO4 reacts with 100.0 mL of 0.400 M NaOH and
200.0 mL of 0.200 M KOH, calculate the concentration of the excess base.
H2SO4 + 2XOH
0.2500 L x 0.100 mole 0.1000 L x 0.400 mole + 0.2000 L x 0.200 mole
1 L 1L 1L
I 0.0250 mole 0.0800 mole
C 0.0250 mole 0.0500 mole
E 0 0.0300 mole
[NaOH] = 0.0300 mole = 0.0545 M
0.550 L
Worksheet # 12 Chemistry 11 Calculations Practice Test # 1
Pick two formulas that match each classification:
1. a b Acid a) HCl e) KOH
2. c d Covalent Nonacid b) CH3COOH f) NH4Cl
3. h f Salt c) CH4 g) Ba(OH)2
4. e g Base d) HOH h) AgNO3
5. Calculate the molarity of the solution formed when 200 g of NaCl is dissolved in
100 mL of H2O.
Molarity = 200g x 1 mole
58.5g = 34.2 M
0.100 L
6. How many grams of AgCl are required to prepare 150 mL of 0.200 M solution?
0.150L x 0.200 mole x 143.4 g = 4.30 g
1 L 1 mole
7. How many litres of 0.200 M AgCl are needed to provide 50 g of AgCl?
50g x 1 mole x 1 L = 1.7 L
143.4g 0.200 mole
8. 100 g of AlCl3 is dissolved in 200 mL H2O, calculate [Al3+] and [Cl-].
100 g x 1 mole
Molarity = 133.5 g = 3.745 M AlCl3 Al3+ + 3Cl-
0.200 L
3.745 M 3.75 M 11.2M
9. In three runs of a titration 36.9, 34.4 and 34.3 mL of 0.200 M NaOH was required to neutralize a 25.0 mL sample of H2CO3. Calculate the molarity of the acid.
H2SO4 + 2NaOH Na2SO4 + 2HOH
0.0250 L 0.3435 L
? M 0.200 M
[ H2SO4] = 0.03435 L x 0.200 mole x 1 mole H2SO4 = 0.137 M
1 L 2 mole NaOH
0.0250 L
= 0.137 M
10. Calculate the [NaOH] due to excess NaOH in the new solution produced by mixing 100. mL
0.200 M HCl and 100. mL 0.300 M NaOH.
HCL + NaOH NaCl + HOH
0.100L x 0.200 mole = 0.0200 mol 0.100L x 0.300 mol = .030 mole
1 L 1 L
I 0.0200 mole 0.0300 mole
C 0.0200 mole 0.0200 mole
E 0 mole 0.0100 mole
Total Volume = 200 mL = 0.200 L Molarity = 0.0100 mole = 0.0500 M
0.200 L
11. A empty beaker has a mass of 29.86 g. The same beaker is filled with 0.250 L with a solution of CaCl2 and weighs 87.26 g. The solution is evaporated to dryness and the mass of the beaker and solid is 62.31 g. Calculate the molarity of the solution.
Mass of CaCl2 = 62.31 – 29.86 = 32.45g
Molarity = 32.45g x 1 mole
111.1g = 1.17 M
.250 L
12. Complete the reaction equations.
i) Formula Equation/Chemical Equation
2AgNO3 (aq) + Na2SO4 (aq) Ag2SO4(s) + 2NaNO3(aq)
ii) Total Ionic Equation
2Ag+(aq) + 2NO3- + 2Na+(aq) + SO42- Ag2SO4(s) + 2Na+(aq) + 2NO3-(aq)
iii) Net Ionic Equation
2Ag+(aq) + SO42- Ag2SO4(s)
13. Complete the formula equation:
2H3PO4(aq) + 3Sr(OH)2(aq) Sr3(PO4)2(s) + 6HOH(l)
Complete the complete ionic equation:
6H+(aq) + 2PO43- + 3Sr2+(aq) + 6OH- Sr3(PO4)2(s) + 6HOH(l)
Complete the net ionic equation:
6H+(aq) + 2PO43- + 3Sr2+(aq) + 6OH- Sr3(PO4)2(s) + 6HOH(l)
14. Complete the formula equation:
Fe3(PO4)2(aq) + 3Zn(s) 3Fe(s) + Zn3(PO4)2(s)
Complete the complete ionic equation:
3Fe2+(aq) + 2PO43- + 3Zn(s) 3Fe(s) + Zn3(PO4)2(s)
Complete the net ionic equation:
3Fe2+(aq) + 2PO43- + 3Zn(s) 3Fe(s) + Zn3(PO4)2(s)
Worksheet # 13 Chemistry Calculations Practice Test # 2
1. Calculate the number of formula units in 250. g CaCl2.
1.35 x 1024 FU
2. Calculate the mass of 2.35 x 1020 molecules of CO2.
0.0172 g
3. Calculate the STP volume of 10.0 g of CO2 gas.
5.09 L
4. Calculate the number of grams CaCl2 in 350. mL of a 0.250M solution.
9.72g
5. Calculate the volume of 0.250 M NaCl solution that would contain 0.17 g NaCl.
0.012L
6. 1.26 g of AlCl3 are dissolved in 160.0 ml of water. Calculate the molarity of the solution.
0.0590M
7. 12.5 ml of CO2 gas at STP are dissolved in 250.0 ml of water. Calculate the molarity of the solution.
0.00223M
8. 10.0 g of Al2(SO4)3 is dissolved in 155 ml of water. Calculate the two ion concentrations.
0.377 M
0.565 M
9. 200.0 ml of 0.200M H3PO4 reacts with 200.0 ml of 0.300M KOH. Calculate the molarity of the excess acid in the new solution formed.
0.0500M
10. 16 g of Ca react with water. Calculate the volume of H2 gas produced at STP.
Ca + 2H2O ® H2 + Ca(OH)2
8.9L
11. In a titration 0.200 M NaOH is used to neutralize 10.0 mL of H2SO4. In three runs the following data was collected. Calculate the concentration of the acid.
Volume of 0.200 M NaOH 25.3 mL 25.8 mL 25.6 mL
0.256 M
12. 60.0 g of Al react with 60.0 g of O2. Calculate the amount of excess reactant.
4Al + 3O2 2Al2O3
6.66 g O2
13. Calculate the percentage composition of the elements in Ga2 (SO4)3 to three significant digits.
32.6% , 22.5% , 44.9%
14. What volume of 0.300 M solution must be diluted to a final volume of 1200.0 mL and have a molarity of 0.2500M.
1.00L
15. Calculate the number of grams NaCl produced by the complete reaction of
520 g Cl2. 2Na + Cl2 2NaCl
857g
16. If the actual yield of NaCl in the last question was 200. g, calculate the percentage yield of NaCl.
23.3%
17. 200.0 ml 0.200 M HCl reacts with 400.0 ml 0.150M NaOH. Calculate the molarity of excess base.
HCl + NaOH NaCl + H2O
0.0333M
18. 100.0 mL of 0.250 M HCl solution is diluted by adding 250.0 mL of water, calculate the new concentration.
0.0714M
19. 65.5 mL of 0.300 M is diluted to a new molarity of 0.0600 M, how much water was added?
262mL
20. 56.0 mL of 0.100 M HCl reacts with 0.250 M Ba (OH) 2, calculate the volume of base required to completely neutralize the acid.
0.0112L
21. Write the formula, complete, and net ionic equation for each.
H3PO4 (aq) and NaOH (aq).
H3PO4 (aq) + 3NaOH (aq) Na3PO4 (aq) + 3HOH (l)
3H+ (aq) + PO4-3 (aq) + 3Na+ (aq) + 3OH- (aq) 3Na+ (aq) + PO4-3 (aq) + 3HOH (l)
H+ (aq) + OH- (aq) HOH (l)
22. Write the formula, complete, and net ionic equation for each.
Na3PO4 (aq) and Ca (NO3) 2(aq).
2Na3PO4 (aq) + 3Ca(NO3) 2 (aq) Ca3 (PO4) 2 (s) + 6NaNO3 (aq)
Na+ (aq) + 2PO43- (aq) +3Ca2+ (aq) + 6NO3- (aq) Ca3 (PO4) 2 (s) + 6Na+ (aq)+ 6NO3- (aq)
3Ca2+ (aq) + 2PO43- (aq) Ca3 (PO4) 2 (s)
23. Write the formula, complete, and net ionic equation for each.
Cu (NO3) 2(aq) and Ag(s).
Ag (s) + Cu (NO3) 2 (aq) No Reaction
24. A empty beaker has a mass of 25.86 g. The same beaker is filled with 0.250 L with a solution of Cl2 and weighs 87.26 g. The solution is evaporated to dryness and the mass of the beaker and solid is 36.31 g. Calculate the molarity of the solution.
0.376 M
25. 125.0 g of an aqueous compound that is 3.091 % H, 31.62 % P, and 65.29 % O reacts with another compound that is 80.14 % Ba, 18.68 % O, and 1.179 % H. If the actual yield of the solid product is 350. g, calculate the percentage yield of the solid.
91.2%
26. 0.0250 M
27. 0.0091 M
28. [Ca2+] = 0.0714 M [Al3+] = 0.193 M [Cl-] = 0.722M
29 [Na+] = 0.333 M [PO43-] = 0.0667 M [SO42-] = 0.0667 M
30. 0.0827 M
31. 150.0 mL of 0.200 M HCl and 250.0 mL of 0.300 M HNO3 react with excess CaCO3.
Calculate the theoretical yield of CO2. Start by writing an equation.
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Give me an example of two Riemann-integrable functions $f,g:[0,1]\to[0,1]$ such that $g\circ f$ isn't integrable! I already know the following example: f(x)=\begin{cases} 0, & \text{if ... | 1,656 | 6,005 | {"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} | 3.3125 | 3 | CC-MAIN-2014-41 | latest | en | 0.852459 |
https://us.metamath.org/mpeuni/lighneallem3.html | 1,685,698,412,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00421.warc.gz | 643,947,703 | 20,452 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > lighneallem3 Structured version Visualization version GIF version
Theorem lighneallem3 42545
Description: Lemma 3 for lighneal 42549. (Contributed by AV, 11-Aug-2021.)
Assertion
Ref Expression
lighneallem3 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) ∧ ((2↑𝑁) − 1) = (𝑃𝑀)) → 𝑀 = 1)
Proof of Theorem lighneallem3
Dummy variables 𝑘 𝑗 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 oveq2 6930 . . . . . . . . . . . . 13 (𝑁 = 1 → (2↑𝑁) = (2↑1))
2 2cn 11450 . . . . . . . . . . . . . 14 2 ∈ ℂ
3 exp1 13184 . . . . . . . . . . . . . 14 (2 ∈ ℂ → (2↑1) = 2)
42, 3ax-mp 5 . . . . . . . . . . . . 13 (2↑1) = 2
51, 4syl6eq 2830 . . . . . . . . . . . 12 (𝑁 = 1 → (2↑𝑁) = 2)
65oveq1d 6937 . . . . . . . . . . 11 (𝑁 = 1 → ((2↑𝑁) − 1) = (2 − 1))
7 2m1e1 11508 . . . . . . . . . . 11 (2 − 1) = 1
86, 7syl6eq 2830 . . . . . . . . . 10 (𝑁 = 1 → ((2↑𝑁) − 1) = 1)
98adantl 475 . . . . . . . . 9 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑁 = 1) → ((2↑𝑁) − 1) = 1)
109eqeq1d 2780 . . . . . . . 8 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑁 = 1) → (((2↑𝑁) − 1) = (𝑃𝑀) ↔ 1 = (𝑃𝑀)))
11 eldifi 3955 . . . . . . . . . . . . 13 (𝑃 ∈ (ℙ ∖ {2}) → 𝑃 ∈ ℙ)
12 prmnn 15793 . . . . . . . . . . . . 13 (𝑃 ∈ ℙ → 𝑃 ∈ ℕ)
13 nnnn0 11650 . . . . . . . . . . . . 13 (𝑃 ∈ ℕ → 𝑃 ∈ ℕ0)
1411, 12, 133syl 18 . . . . . . . . . . . 12 (𝑃 ∈ (ℙ ∖ {2}) → 𝑃 ∈ ℕ0)
1514nn0zd 11832 . . . . . . . . . . 11 (𝑃 ∈ (ℙ ∖ {2}) → 𝑃 ∈ ℤ)
16 iddvdsexp 15412 . . . . . . . . . . 11 ((𝑃 ∈ ℤ ∧ 𝑀 ∈ ℕ) → 𝑃 ∥ (𝑃𝑀))
1715, 16sylan 575 . . . . . . . . . 10 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → 𝑃 ∥ (𝑃𝑀))
18 breq2 4890 . . . . . . . . . . . . 13 (1 = (𝑃𝑀) → (𝑃 ∥ 1 ↔ 𝑃 ∥ (𝑃𝑀)))
1918adantl 475 . . . . . . . . . . . 12 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 1 = (𝑃𝑀)) → (𝑃 ∥ 1 ↔ 𝑃 ∥ (𝑃𝑀)))
20 dvds1 15448 . . . . . . . . . . . . . . 15 (𝑃 ∈ ℕ0 → (𝑃 ∥ 1 ↔ 𝑃 = 1))
2114, 20syl 17 . . . . . . . . . . . . . 14 (𝑃 ∈ (ℙ ∖ {2}) → (𝑃 ∥ 1 ↔ 𝑃 = 1))
22 eleq1 2847 . . . . . . . . . . . . . . . 16 (𝑃 = 1 → (𝑃 ∈ ℙ ↔ 1 ∈ ℙ))
23 1nprm 15797 . . . . . . . . . . . . . . . . 17 ¬ 1 ∈ ℙ
2423pm2.21i 117 . . . . . . . . . . . . . . . 16 (1 ∈ ℙ → 𝑀 = 1)
2522, 24syl6bi 245 . . . . . . . . . . . . . . 15 (𝑃 = 1 → (𝑃 ∈ ℙ → 𝑀 = 1))
2611, 25syl5com 31 . . . . . . . . . . . . . 14 (𝑃 ∈ (ℙ ∖ {2}) → (𝑃 = 1 → 𝑀 = 1))
2721, 26sylbid 232 . . . . . . . . . . . . 13 (𝑃 ∈ (ℙ ∖ {2}) → (𝑃 ∥ 1 → 𝑀 = 1))
2827ad2antrr 716 . . . . . . . . . . . 12 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 1 = (𝑃𝑀)) → (𝑃 ∥ 1 → 𝑀 = 1))
2919, 28sylbird 252 . . . . . . . . . . 11 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 1 = (𝑃𝑀)) → (𝑃 ∥ (𝑃𝑀) → 𝑀 = 1))
3029ex 403 . . . . . . . . . 10 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (1 = (𝑃𝑀) → (𝑃 ∥ (𝑃𝑀) → 𝑀 = 1)))
3117, 30mpid 44 . . . . . . . . 9 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (1 = (𝑃𝑀) → 𝑀 = 1))
3231adantr 474 . . . . . . . 8 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑁 = 1) → (1 = (𝑃𝑀) → 𝑀 = 1))
3310, 32sylbid 232 . . . . . . 7 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑁 = 1) → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1))
3433ex 403 . . . . . 6 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (𝑁 = 1 → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1)))
3534com23 86 . . . . 5 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (((2↑𝑁) − 1) = (𝑃𝑀) → (𝑁 = 1 → 𝑀 = 1)))
3635a1d 25 . . . 4 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → ((¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) → (((2↑𝑁) − 1) = (𝑃𝑀) → (𝑁 = 1 → 𝑀 = 1))))
37363adant3 1123 . . 3 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) → (((2↑𝑁) − 1) = (𝑃𝑀) → (𝑁 = 1 → 𝑀 = 1))))
38373imp 1098 . 2 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) ∧ ((2↑𝑁) − 1) = (𝑃𝑀)) → (𝑁 = 1 → 𝑀 = 1))
39 neqne 2977 . . . . . . . . . . . 12 𝑁 = 1 → 𝑁 ≠ 1)
4039anim2i 610 . . . . . . . . . . 11 ((𝑁 ∈ ℕ ∧ ¬ 𝑁 = 1) → (𝑁 ∈ ℕ ∧ 𝑁 ≠ 1))
41 eluz2b3 12069 . . . . . . . . . . 11 (𝑁 ∈ (ℤ‘2) ↔ (𝑁 ∈ ℕ ∧ 𝑁 ≠ 1))
4240, 41sylibr 226 . . . . . . . . . 10 ((𝑁 ∈ ℕ ∧ ¬ 𝑁 = 1) → 𝑁 ∈ (ℤ‘2))
43 oddge22np1 15477 . . . . . . . . . 10 (𝑁 ∈ (ℤ‘2) → (¬ 2 ∥ 𝑁 ↔ ∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁))
4442, 43syl 17 . . . . . . . . 9 ((𝑁 ∈ ℕ ∧ ¬ 𝑁 = 1) → (¬ 2 ∥ 𝑁 ↔ ∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁))
45443ad2antl3 1195 . . . . . . . 8 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ ¬ 𝑁 = 1) → (¬ 2 ∥ 𝑁 ↔ ∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁))
46 oveq2 6930 . . . . . . . . . . . . . . . . . 18 (𝑁 = ((2 · 𝑗) + 1) → (2↑𝑁) = (2↑((2 · 𝑗) + 1)))
4746oveq1d 6937 . . . . . . . . . . . . . . . . 17 (𝑁 = ((2 · 𝑗) + 1) → ((2↑𝑁) − 1) = ((2↑((2 · 𝑗) + 1)) − 1))
4847eqcoms 2786 . . . . . . . . . . . . . . . 16 (((2 · 𝑗) + 1) = 𝑁 → ((2↑𝑁) − 1) = ((2↑((2 · 𝑗) + 1)) − 1))
492a1i 11 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → 2 ∈ ℂ)
50 2nn0 11661 . . . . . . . . . . . . . . . . . . . . . . 23 2 ∈ ℕ0
5150a1i 11 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → 2 ∈ ℕ0)
52 nnnn0 11650 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → 𝑗 ∈ ℕ0)
5351, 52nn0mulcld 11707 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → (2 · 𝑗) ∈ ℕ0)
5449, 53expp1d 13328 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → (2↑((2 · 𝑗) + 1)) = ((2↑(2 · 𝑗)) · 2))
5551, 53nn0expcld 13352 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → (2↑(2 · 𝑗)) ∈ ℕ0)
5655nn0cnd 11704 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → (2↑(2 · 𝑗)) ∈ ℂ)
5756, 49mulcomd 10398 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → ((2↑(2 · 𝑗)) · 2) = (2 · (2↑(2 · 𝑗))))
5854, 57eqtrd 2814 . . . . . . . . . . . . . . . . . . 19 (𝑗 ∈ ℕ → (2↑((2 · 𝑗) + 1)) = (2 · (2↑(2 · 𝑗))))
5958oveq1d 6937 . . . . . . . . . . . . . . . . . 18 (𝑗 ∈ ℕ → ((2↑((2 · 𝑗) + 1)) − 1) = ((2 · (2↑(2 · 𝑗))) − 1))
60 npcan1 10800 . . . . . . . . . . . . . . . . . . . . . . 23 ((2↑(2 · 𝑗)) ∈ ℂ → (((2↑(2 · 𝑗)) − 1) + 1) = (2↑(2 · 𝑗)))
6156, 60syl 17 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → (((2↑(2 · 𝑗)) − 1) + 1) = (2↑(2 · 𝑗)))
6261eqcomd 2784 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → (2↑(2 · 𝑗)) = (((2↑(2 · 𝑗)) − 1) + 1))
6362oveq2d 6938 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → (2 · (2↑(2 · 𝑗))) = (2 · (((2↑(2 · 𝑗)) − 1) + 1)))
64 peano2cnm 10689 . . . . . . . . . . . . . . . . . . . . . 22 ((2↑(2 · 𝑗)) ∈ ℂ → ((2↑(2 · 𝑗)) − 1) ∈ ℂ)
6556, 64syl 17 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → ((2↑(2 · 𝑗)) − 1) ∈ ℂ)
66 1cnd 10371 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → 1 ∈ ℂ)
6749, 65, 66adddid 10401 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → (2 · (((2↑(2 · 𝑗)) − 1) + 1)) = ((2 · ((2↑(2 · 𝑗)) − 1)) + (2 · 1)))
6863, 67eqtrd 2814 . . . . . . . . . . . . . . . . . . 19 (𝑗 ∈ ℕ → (2 · (2↑(2 · 𝑗))) = ((2 · ((2↑(2 · 𝑗)) − 1)) + (2 · 1)))
6968oveq1d 6937 . . . . . . . . . . . . . . . . . 18 (𝑗 ∈ ℕ → ((2 · (2↑(2 · 𝑗))) − 1) = (((2 · ((2↑(2 · 𝑗)) − 1)) + (2 · 1)) − 1))
7049, 65mulcld 10397 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → (2 · ((2↑(2 · 𝑗)) − 1)) ∈ ℂ)
71 ax-1cn 10330 . . . . . . . . . . . . . . . . . . . . . 22 1 ∈ ℂ
722, 71mulcli 10384 . . . . . . . . . . . . . . . . . . . . 21 (2 · 1) ∈ ℂ
7372a1i 11 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → (2 · 1) ∈ ℂ)
7470, 73, 66addsubassd 10754 . . . . . . . . . . . . . . . . . . 19 (𝑗 ∈ ℕ → (((2 · ((2↑(2 · 𝑗)) − 1)) + (2 · 1)) − 1) = ((2 · ((2↑(2 · 𝑗)) − 1)) + ((2 · 1) − 1)))
75 2t1e2 11545 . . . . . . . . . . . . . . . . . . . . . . 23 (2 · 1) = 2
7675oveq1i 6932 . . . . . . . . . . . . . . . . . . . . . 22 ((2 · 1) − 1) = (2 − 1)
7776, 7eqtri 2802 . . . . . . . . . . . . . . . . . . . . 21 ((2 · 1) − 1) = 1
7877a1i 11 . . . . . . . . . . . . . . . . . . . 20 (𝑗 ∈ ℕ → ((2 · 1) − 1) = 1)
7978oveq2d 6938 . . . . . . . . . . . . . . . . . . 19 (𝑗 ∈ ℕ → ((2 · ((2↑(2 · 𝑗)) − 1)) + ((2 · 1) − 1)) = ((2 · ((2↑(2 · 𝑗)) − 1)) + 1))
8074, 79eqtrd 2814 . . . . . . . . . . . . . . . . . 18 (𝑗 ∈ ℕ → (((2 · ((2↑(2 · 𝑗)) − 1)) + (2 · 1)) − 1) = ((2 · ((2↑(2 · 𝑗)) − 1)) + 1))
8159, 69, 803eqtrd 2818 . . . . . . . . . . . . . . . . 17 (𝑗 ∈ ℕ → ((2↑((2 · 𝑗) + 1)) − 1) = ((2 · ((2↑(2 · 𝑗)) − 1)) + 1))
8281ad2antlr 717 . . . . . . . . . . . . . . . 16 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((2↑((2 · 𝑗) + 1)) − 1) = ((2 · ((2↑(2 · 𝑗)) − 1)) + 1))
8348, 82sylan9eqr 2836 . . . . . . . . . . . . . . 15 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ ((2 · 𝑗) + 1) = 𝑁) → ((2↑𝑁) − 1) = ((2 · ((2↑(2 · 𝑗)) − 1)) + 1))
8483eqeq1d 2780 . . . . . . . . . . . . . 14 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ ((2 · 𝑗) + 1) = 𝑁) → (((2↑𝑁) − 1) = (𝑃𝑀) ↔ ((2 · ((2↑(2 · 𝑗)) − 1)) + 1) = (𝑃𝑀)))
85143ad2ant1 1124 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → 𝑃 ∈ ℕ0)
86 nnnn0 11650 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑀 ∈ ℕ → 𝑀 ∈ ℕ0)
87863ad2ant2 1125 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → 𝑀 ∈ ℕ0)
8885, 87nn0expcld 13352 . . . . . . . . . . . . . . . . . . . . 21 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑃𝑀) ∈ ℕ0)
8988nn0cnd 11704 . . . . . . . . . . . . . . . . . . . 20 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑃𝑀) ∈ ℂ)
9089adantr 474 . . . . . . . . . . . . . . . . . . 19 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (𝑃𝑀) ∈ ℂ)
91 1cnd 10371 . . . . . . . . . . . . . . . . . . 19 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → 1 ∈ ℂ)
9270adantl 475 . . . . . . . . . . . . . . . . . . 19 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (2 · ((2↑(2 · 𝑗)) − 1)) ∈ ℂ)
9390, 91, 923jca 1119 . . . . . . . . . . . . . . . . . 18 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → ((𝑃𝑀) ∈ ℂ ∧ 1 ∈ ℂ ∧ (2 · ((2↑(2 · 𝑗)) − 1)) ∈ ℂ))
9493adantr 474 . . . . . . . . . . . . . . . . 17 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((𝑃𝑀) ∈ ℂ ∧ 1 ∈ ℂ ∧ (2 · ((2↑(2 · 𝑗)) − 1)) ∈ ℂ))
95 subadd2 10626 . . . . . . . . . . . . . . . . 17 (((𝑃𝑀) ∈ ℂ ∧ 1 ∈ ℂ ∧ (2 · ((2↑(2 · 𝑗)) − 1)) ∈ ℂ) → (((𝑃𝑀) − 1) = (2 · ((2↑(2 · 𝑗)) − 1)) ↔ ((2 · ((2↑(2 · 𝑗)) − 1)) + 1) = (𝑃𝑀)))
9694, 95syl 17 . . . . . . . . . . . . . . . 16 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((𝑃𝑀) − 1) = (2 · ((2↑(2 · 𝑗)) − 1)) ↔ ((2 · ((2↑(2 · 𝑗)) − 1)) + 1) = (𝑃𝑀)))
97 nncn 11383 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑃 ∈ ℕ → 𝑃 ∈ ℂ)
9811, 12, 973syl 18 . . . . . . . . . . . . . . . . . . . . . 22 (𝑃 ∈ (ℙ ∖ {2}) → 𝑃 ∈ ℂ)
99983ad2ant1 1124 . . . . . . . . . . . . . . . . . . . . 21 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → 𝑃 ∈ ℂ)
10099, 87pwm1geoser 15004 . . . . . . . . . . . . . . . . . . . 20 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑃𝑀) − 1) = ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
101100adantr 474 . . . . . . . . . . . . . . . . . . 19 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → ((𝑃𝑀) − 1) = ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
102101eqeq1d 2780 . . . . . . . . . . . . . . . . . 18 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (((𝑃𝑀) − 1) = (2 · ((2↑(2 · 𝑗)) − 1)) ↔ ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = (2 · ((2↑(2 · 𝑗)) − 1))))
103102adantr 474 . . . . . . . . . . . . . . . . 17 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((𝑃𝑀) − 1) = (2 · ((2↑(2 · 𝑗)) − 1)) ↔ ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = (2 · ((2↑(2 · 𝑗)) − 1))))
10499ad2antrr 716 . . . . . . . . . . . . . . . . . . . . 21 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 𝑃 ∈ ℂ)
105 1cnd 10371 . . . . . . . . . . . . . . . . . . . . 21 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 1 ∈ ℂ)
106104, 105subcld 10734 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (𝑃 − 1) ∈ ℂ)
107 fzfid 13091 . . . . . . . . . . . . . . . . . . . . . . 23 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (0...(𝑀 − 1)) ∈ Fin)
10885adantr 474 . . . . . . . . . . . . . . . . . . . . . . . . 25 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → 𝑃 ∈ ℕ0)
109 elfznn0 12751 . . . . . . . . . . . . . . . . . . . . . . . . . 26 (𝑘 ∈ (0...(𝑀 − 1)) → 𝑘 ∈ ℕ0)
110109adantl 475 . . . . . . . . . . . . . . . . . . . . . . . . 25 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → 𝑘 ∈ ℕ0)
111108, 110nn0expcld 13352 . . . . . . . . . . . . . . . . . . . . . . . 24 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → (𝑃𝑘) ∈ ℕ0)
112111nn0zd 11832 . . . . . . . . . . . . . . . . . . . . . . 23 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → (𝑃𝑘) ∈ ℤ)
113107, 112fsumzcl 14873 . . . . . . . . . . . . . . . . . . . . . 22 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℤ)
114113zcnd 11835 . . . . . . . . . . . . . . . . . . . . 21 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℂ)
115114ad2antrr 716 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℂ)
116106, 115mulcld 10397 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ∈ ℂ)
11756ad2antlr 717 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (2↑(2 · 𝑗)) ∈ ℂ)
118117, 105subcld 10734 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((2↑(2 · 𝑗)) − 1) ∈ ℂ)
119 2rp 12142 . . . . . . . . . . . . . . . . . . . . 21 2 ∈ ℝ+
120119a1i 11 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 2 ∈ ℝ+)
121120rpcnne0d 12190 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (2 ∈ ℂ ∧ 2 ≠ 0))
122 divmul2 11037 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ∈ ℂ ∧ ((2↑(2 · 𝑗)) − 1) ∈ ℂ ∧ (2 ∈ ℂ ∧ 2 ≠ 0)) → ((((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = ((2↑(2 · 𝑗)) − 1) ↔ ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = (2 · ((2↑(2 · 𝑗)) − 1))))
123116, 118, 121, 122syl3anc 1439 . . . . . . . . . . . . . . . . . 18 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = ((2↑(2 · 𝑗)) − 1) ↔ ((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = (2 · ((2↑(2 · 𝑗)) − 1))))
124 div23 11052 . . . . . . . . . . . . . . . . . . . . 21 (((𝑃 − 1) ∈ ℂ ∧ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℂ ∧ (2 ∈ ℂ ∧ 2 ≠ 0)) → (((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
125106, 115, 121, 124syl3anc 1439 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
126125eqeq1d 2780 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = ((2↑(2 · 𝑗)) − 1) ↔ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1)))
12751nn0zd 11832 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑗 ∈ ℕ → 2 ∈ ℤ)
128 2nn 11448 . . . . . . . . . . . . . . . . . . . . . . . . . 26 2 ∈ ℕ
129128a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . 25 (𝑗 ∈ ℕ → 2 ∈ ℕ)
130 id 22 . . . . . . . . . . . . . . . . . . . . . . . . 25 (𝑗 ∈ ℕ → 𝑗 ∈ ℕ)
131129, 130nnmulcld 11428 . . . . . . . . . . . . . . . . . . . . . . . 24 (𝑗 ∈ ℕ → (2 · 𝑗) ∈ ℕ)
132 iddvdsexp 15412 . . . . . . . . . . . . . . . . . . . . . . . 24 ((2 ∈ ℤ ∧ (2 · 𝑗) ∈ ℕ) → 2 ∥ (2↑(2 · 𝑗)))
133127, 131, 132syl2anc 579 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑗 ∈ ℕ → 2 ∥ (2↑(2 · 𝑗)))
134133notnotd 141 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → ¬ ¬ 2 ∥ (2↑(2 · 𝑗)))
13555nn0zd 11832 . . . . . . . . . . . . . . . . . . . . . . 23 (𝑗 ∈ ℕ → (2↑(2 · 𝑗)) ∈ ℤ)
136 oddm1even 15471 . . . . . . . . . . . . . . . . . . . . . . 23 ((2↑(2 · 𝑗)) ∈ ℤ → (¬ 2 ∥ (2↑(2 · 𝑗)) ↔ 2 ∥ ((2↑(2 · 𝑗)) − 1)))
137135, 136syl 17 . . . . . . . . . . . . . . . . . . . . . 22 (𝑗 ∈ ℕ → (¬ 2 ∥ (2↑(2 · 𝑗)) ↔ 2 ∥ ((2↑(2 · 𝑗)) − 1)))
138134, 137mtbid 316 . . . . . . . . . . . . . . . . . . . . 21 (𝑗 ∈ ℕ → ¬ 2 ∥ ((2↑(2 · 𝑗)) − 1))
139138ad2antlr 717 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ¬ 2 ∥ ((2↑(2 · 𝑗)) − 1))
140 breq2 4890 . . . . . . . . . . . . . . . . . . . . . . . 24 ((((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1) → (2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ 2 ∥ ((2↑(2 · 𝑗)) − 1)))
141140notbid 310 . . . . . . . . . . . . . . . . . . . . . . 23 ((((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1) → (¬ 2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ ¬ 2 ∥ ((2↑(2 · 𝑗)) − 1)))
142141adantl 475 . . . . . . . . . . . . . . . . . . . . . 22 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1)) → (¬ 2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ ¬ 2 ∥ ((2↑(2 · 𝑗)) − 1)))
143 fzfid 13091 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (0...(𝑀 − 1)) ∈ Fin)
144112ad4ant14 742 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → (𝑃𝑘) ∈ ℤ)
145 elnn0 11644 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (𝑘 ∈ ℕ0 ↔ (𝑘 ∈ ℕ ∨ 𝑘 = 0))
146 eldifsn 4550 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 (𝑃 ∈ (ℙ ∖ {2}) ↔ (𝑃 ∈ ℙ ∧ 𝑃 ≠ 2))
147 simpr 479 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 ((𝑃 ∈ ℙ ∧ 𝑃 ≠ 2) → 𝑃 ≠ 2)
148147necomd 3024 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 ((𝑃 ∈ ℙ ∧ 𝑃 ≠ 2) → 2 ≠ 𝑃)
149146, 148sylbi 209 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 (𝑃 ∈ (ℙ ∖ {2}) → 2 ≠ 𝑃)
150149adantl 475 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → 2 ≠ 𝑃)
151150neneqd 2974 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ¬ 2 = 𝑃)
152 2prm 15810 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 2 ∈ ℙ
15311adantl 475 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → 𝑃 ∈ ℙ)
154 simpl 476 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → 𝑘 ∈ ℕ)
155 prmdvdsexpb 15832 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((2 ∈ ℙ ∧ 𝑃 ∈ ℙ ∧ 𝑘 ∈ ℕ) → (2 ∥ (𝑃𝑘) ↔ 2 = 𝑃))
156152, 153, 154, 155mp3an2i 1539 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → (2 ∥ (𝑃𝑘) ↔ 2 = 𝑃))
157151, 156mtbird 317 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 ((𝑘 ∈ ℕ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ¬ 2 ∥ (𝑃𝑘))
158157ex 403 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 (𝑘 ∈ ℕ → (𝑃 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ (𝑃𝑘)))
159 n2dvds1 15496 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ¬ 2 ∥ 1
160 oveq2 6930 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 (𝑘 = 0 → (𝑃𝑘) = (𝑃↑0))
16198exp0d 13321 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 (𝑃 ∈ (ℙ ∖ {2}) → (𝑃↑0) = 1)
162160, 161sylan9eq 2834 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 ((𝑘 = 0 ∧ 𝑃 ∈ (ℙ ∖ {2})) → (𝑃𝑘) = 1)
163162breq2d 4898 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ((𝑘 = 0 ∧ 𝑃 ∈ (ℙ ∖ {2})) → (2 ∥ (𝑃𝑘) ↔ 2 ∥ 1))
164159, 163mtbiri 319 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 ((𝑘 = 0 ∧ 𝑃 ∈ (ℙ ∖ {2})) → ¬ 2 ∥ (𝑃𝑘))
165164ex 403 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 (𝑘 = 0 → (𝑃 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ (𝑃𝑘)))
166158, 165jaoi 846 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 ((𝑘 ∈ ℕ ∨ 𝑘 = 0) → (𝑃 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ (𝑃𝑘)))
167145, 166sylbi 209 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (𝑘 ∈ ℕ0 → (𝑃 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ (𝑃𝑘)))
168167, 109syl11 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑃 ∈ (ℙ ∖ {2}) → (𝑘 ∈ (0...(𝑀 − 1)) → ¬ 2 ∥ (𝑃𝑘)))
1691683ad2ant1 1124 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑘 ∈ (0...(𝑀 − 1)) → ¬ 2 ∥ (𝑃𝑘)))
170169ad2antrr 716 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (𝑘 ∈ (0...(𝑀 − 1)) → ¬ 2 ∥ (𝑃𝑘)))
171170imp 397 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → ¬ 2 ∥ (𝑃𝑘))
172 nnm1nn0 11685 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 (𝑀 ∈ ℕ → (𝑀 − 1) ∈ ℕ0)
173 hashfz0 13533 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 ((𝑀 − 1) ∈ ℕ0 → (♯‘(0...(𝑀 − 1))) = ((𝑀 − 1) + 1))
174172, 173syl 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (𝑀 ∈ ℕ → (♯‘(0...(𝑀 − 1))) = ((𝑀 − 1) + 1))
175 nncn 11383 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 (𝑀 ∈ ℕ → 𝑀 ∈ ℂ)
176 1cnd 10371 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 (𝑀 ∈ ℕ → 1 ∈ ℂ)
177175, 176npcand 10738 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (𝑀 ∈ ℕ → ((𝑀 − 1) + 1) = 𝑀)
178174, 177eqtr2d 2815 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (𝑀 ∈ ℕ → 𝑀 = (♯‘(0...(𝑀 − 1))))
1791783ad2ant2 1125 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → 𝑀 = (♯‘(0...(𝑀 − 1))))
180179adantr 474 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → 𝑀 = (♯‘(0...(𝑀 − 1))))
181180breq2d 4898 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (2 ∥ 𝑀 ↔ 2 ∥ (♯‘(0...(𝑀 − 1)))))
182181biimpa 470 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 2 ∥ (♯‘(0...(𝑀 − 1))))
183143, 144, 171, 182evensumodd 15519 . . . . . . . . . . . . . . . . . . . . . . . . . 26 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 2 ∥ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘))
184183olcd 863 . . . . . . . . . . . . . . . . . . . . . . . . 25 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (2 ∥ ((𝑃 − 1) / 2) ∨ 2 ∥ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
185152a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → 2 ∈ ℙ)
186 oddn2prm 15921 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (𝑃 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑃)
187 oddm1d2 15488 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (𝑃 ∈ ℤ → (¬ 2 ∥ 𝑃 ↔ ((𝑃 − 1) / 2) ∈ ℤ))
18815, 187syl 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (𝑃 ∈ (ℙ ∖ {2}) → (¬ 2 ∥ 𝑃 ↔ ((𝑃 − 1) / 2) ∈ ℤ))
189186, 188mpbid 224 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (𝑃 ∈ (ℙ ∖ {2}) → ((𝑃 − 1) / 2) ∈ ℤ)
190189adantr 474 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → ((𝑃 − 1) / 2) ∈ ℤ)
191 fzfid 13091 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (0...(𝑀 − 1)) ∈ Fin)
19214ad2antrr 716 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → 𝑃 ∈ ℕ0)
193109adantl 475 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → 𝑘 ∈ ℕ0)
194192, 193nn0expcld 13352 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → (𝑃𝑘) ∈ ℕ0)
195194nn0zd 11832 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) ∧ 𝑘 ∈ (0...(𝑀 − 1))) → (𝑃𝑘) ∈ ℤ)
196191, 195fsumzcl 14873 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℤ)
197185, 190, 1963jca 1119 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ) → (2 ∈ ℙ ∧ ((𝑃 − 1) / 2) ∈ ℤ ∧ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℤ))
1981973adant3 1123 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (2 ∈ ℙ ∧ ((𝑃 − 1) / 2) ∈ ℤ ∧ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℤ))
199 euclemma 15829 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 ((2 ∈ ℙ ∧ ((𝑃 − 1) / 2) ∈ ℤ ∧ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘) ∈ ℤ) → (2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ (2 ∥ ((𝑃 − 1) / 2) ∨ 2 ∥ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘))))
200198, 199syl 17 . . . . . . . . . . . . . . . . . . . . . . . . . 26 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ (2 ∥ ((𝑃 − 1) / 2) ∨ 2 ∥ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘))))
201200ad2antrr 716 . . . . . . . . . . . . . . . . . . . . . . . . 25 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) ↔ (2 ∥ ((𝑃 − 1) / 2) ∨ 2 ∥ Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘))))
202184, 201mpbird 249 . . . . . . . . . . . . . . . . . . . . . . . 24 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → 2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)))
203202pm2.24d 149 . . . . . . . . . . . . . . . . . . . . . . 23 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (¬ 2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) → 𝑀 = 1))
204203adantr 474 . . . . . . . . . . . . . . . . . . . . . 22 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1)) → (¬ 2 ∥ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) → 𝑀 = 1))
205142, 204sylbird 252 . . . . . . . . . . . . . . . . . . . . 21 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ (((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1)) → (¬ 2 ∥ ((2↑(2 · 𝑗)) − 1) → 𝑀 = 1))
206205ex 403 . . . . . . . . . . . . . . . . . . . 20 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1) → (¬ 2 ∥ ((2↑(2 · 𝑗)) − 1) → 𝑀 = 1)))
207139, 206mpid 44 . . . . . . . . . . . . . . . . . . 19 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((((𝑃 − 1) / 2) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = ((2↑(2 · 𝑗)) − 1) → 𝑀 = 1))
208126, 207sylbid 232 . . . . . . . . . . . . . . . . . 18 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → ((((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) / 2) = ((2↑(2 · 𝑗)) − 1) → 𝑀 = 1))
209123, 208sylbird 252 . . . . . . . . . . . . . . . . 17 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((𝑃 − 1) · Σ𝑘 ∈ (0...(𝑀 − 1))(𝑃𝑘)) = (2 · ((2↑(2 · 𝑗)) − 1)) → 𝑀 = 1))
210103, 209sylbid 232 . . . . . . . . . . . . . . . 16 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((𝑃𝑀) − 1) = (2 · ((2↑(2 · 𝑗)) − 1)) → 𝑀 = 1))
21196, 210sylbird 252 . . . . . . . . . . . . . . 15 ((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) → (((2 · ((2↑(2 · 𝑗)) − 1)) + 1) = (𝑃𝑀) → 𝑀 = 1))
212211adantr 474 . . . . . . . . . . . . . 14 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ ((2 · 𝑗) + 1) = 𝑁) → (((2 · ((2↑(2 · 𝑗)) − 1)) + 1) = (𝑃𝑀) → 𝑀 = 1))
21384, 212sylbid 232 . . . . . . . . . . . . 13 (((((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) ∧ 2 ∥ 𝑀) ∧ ((2 · 𝑗) + 1) = 𝑁) → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1))
214213exp31 412 . . . . . . . . . . . 12 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (2 ∥ 𝑀 → (((2 · 𝑗) + 1) = 𝑁 → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1))))
215214com23 86 . . . . . . . . . . 11 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝑗 ∈ ℕ) → (((2 · 𝑗) + 1) = 𝑁 → (2 ∥ 𝑀 → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1))))
216215rexlimdva 3213 . . . . . . . . . 10 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁 → (2 ∥ 𝑀 → (((2↑𝑁) − 1) = (𝑃𝑀) → 𝑀 = 1))))
217216com34 91 . . . . . . . . 9 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁 → (((2↑𝑁) − 1) = (𝑃𝑀) → (2 ∥ 𝑀𝑀 = 1))))
218217adantr 474 . . . . . . . 8 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ ¬ 𝑁 = 1) → (∃𝑗 ∈ ℕ ((2 · 𝑗) + 1) = 𝑁 → (((2↑𝑁) − 1) = (𝑃𝑀) → (2 ∥ 𝑀𝑀 = 1))))
21945, 218sylbid 232 . . . . . . 7 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ ¬ 𝑁 = 1) → (¬ 2 ∥ 𝑁 → (((2↑𝑁) − 1) = (𝑃𝑀) → (2 ∥ 𝑀𝑀 = 1))))
220219com24 95 . . . . . 6 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ ¬ 𝑁 = 1) → (2 ∥ 𝑀 → (((2↑𝑁) − 1) = (𝑃𝑀) → (¬ 2 ∥ 𝑁𝑀 = 1))))
221220ex 403 . . . . 5 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (¬ 𝑁 = 1 → (2 ∥ 𝑀 → (((2↑𝑁) − 1) = (𝑃𝑀) → (¬ 2 ∥ 𝑁𝑀 = 1)))))
222221com25 99 . . . 4 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (¬ 2 ∥ 𝑁 → (2 ∥ 𝑀 → (((2↑𝑁) − 1) = (𝑃𝑀) → (¬ 𝑁 = 1 → 𝑀 = 1)))))
223222impd 400 . . 3 ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) → (((2↑𝑁) − 1) = (𝑃𝑀) → (¬ 𝑁 = 1 → 𝑀 = 1))))
2242233imp 1098 . 2 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) ∧ ((2↑𝑁) − 1) = (𝑃𝑀)) → (¬ 𝑁 = 1 → 𝑀 = 1))
22538, 224pm2.61d 172 1 (((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (¬ 2 ∥ 𝑁 ∧ 2 ∥ 𝑀) ∧ ((2↑𝑁) − 1) = (𝑃𝑀)) → 𝑀 = 1)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 198 ∧ wa 386 ∨ wo 836 ∧ w3a 1071 = wceq 1601 ∈ wcel 2107 ≠ wne 2969 ∃wrex 3091 ∖ cdif 3789 {csn 4398 class class class wbr 4886 ‘cfv 6135 (class class class)co 6922 ℂcc 10270 0cc0 10272 1c1 10273 + caddc 10275 · cmul 10277 − cmin 10606 / cdiv 11032 ℕcn 11374 2c2 11430 ℕ0cn0 11642 ℤcz 11728 ℤ≥cuz 11992 ℝ+crp 12137 ...cfz 12643 ↑cexp 13178 ♯chash 13435 Σcsu 14824 ∥ cdvds 15387 ℙcprime 15790 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1839 ax-4 1853 ax-5 1953 ax-6 2021 ax-7 2055 ax-8 2109 ax-9 2116 ax-10 2135 ax-11 2150 ax-12 2163 ax-13 2334 ax-ext 2754 ax-rep 5006 ax-sep 5017 ax-nul 5025 ax-pow 5077 ax-pr 5138 ax-un 7226 ax-inf2 8835 ax-cnex 10328 ax-resscn 10329 ax-1cn 10330 ax-icn 10331 ax-addcl 10332 ax-addrcl 10333 ax-mulcl 10334 ax-mulrcl 10335 ax-mulcom 10336 ax-addass 10337 ax-mulass 10338 ax-distr 10339 ax-i2m1 10340 ax-1ne0 10341 ax-1rid 10342 ax-rnegex 10343 ax-rrecex 10344 ax-cnre 10345 ax-pre-lttri 10346 ax-pre-lttrn 10347 ax-pre-ltadd 10348 ax-pre-mulgt0 10349 ax-pre-sup 10350 This theorem depends on definitions: df-bi 199 df-an 387 df-or 837 df-3or 1072 df-3an 1073 df-tru 1605 df-fal 1615 df-ex 1824 df-nf 1828 df-sb 2012 df-mo 2551 df-eu 2587 df-clab 2764 df-cleq 2770 df-clel 2774 df-nfc 2921 df-ne 2970 df-nel 3076 df-ral 3095 df-rex 3096 df-reu 3097 df-rmo 3098 df-rab 3099 df-v 3400 df-sbc 3653 df-csb 3752 df-dif 3795 df-un 3797 df-in 3799 df-ss 3806 df-pss 3808 df-nul 4142 df-if 4308 df-pw 4381 df-sn 4399 df-pr 4401 df-tp 4403 df-op 4405 df-uni 4672 df-int 4711 df-iun 4755 df-br 4887 df-opab 4949 df-mpt 4966 df-tr 4988 df-id 5261 df-eprel 5266 df-po 5274 df-so 5275 df-fr 5314 df-se 5315 df-we 5316 df-xp 5361 df-rel 5362 df-cnv 5363 df-co 5364 df-dm 5365 df-rn 5366 df-res 5367 df-ima 5368 df-pred 5933 df-ord 5979 df-on 5980 df-lim 5981 df-suc 5982 df-iota 6099 df-fun 6137 df-fn 6138 df-f 6139 df-f1 6140 df-fo 6141 df-f1o 6142 df-fv 6143 df-isom 6144 df-riota 6883 df-ov 6925 df-oprab 6926 df-mpt2 6927 df-om 7344 df-1st 7445 df-2nd 7446 df-wrecs 7689 df-recs 7751 df-rdg 7789 df-1o 7843 df-2o 7844 df-oadd 7847 df-er 8026 df-en 8242 df-dom 8243 df-sdom 8244 df-fin 8245 df-sup 8636 df-inf 8637 df-oi 8704 df-card 9098 df-cda 9325 df-pnf 10413 df-mnf 10414 df-xr 10415 df-ltxr 10416 df-le 10417 df-sub 10608 df-neg 10609 df-div 11033 df-nn 11375 df-2 11438 df-3 11439 df-n0 11643 df-z 11729 df-uz 11993 df-rp 12138 df-fz 12644 df-fzo 12785 df-fl 12912 df-mod 12988 df-seq 13120 df-exp 13179 df-hash 13436 df-cj 14246 df-re 14247 df-im 14248 df-sqrt 14382 df-abs 14383 df-clim 14627 df-sum 14825 df-dvds 15388 df-gcd 15623 df-prm 15791 This theorem is referenced by: lighneal 42549
Copyright terms: Public domain W3C validator | 19,948 | 30,967 | {"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} | 3.328125 | 3 | CC-MAIN-2023-23 | longest | en | 0.158174 |
https://issuu.com/examzone/docs/computer-graphics24 | 1,529,928,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00460.warc.gz | 643,134,436 | 21,509 | Texture Mapping & Other Fun Stuff
MIT EECS 6.837, Durand and Cutler
The Problem: • Don't want to represent all this detail with geometry
MIT EECS 6.837, Durand and Cutler
Procedural Solid (3D) Textures • Write a function: f(x,y,z) → color • non-intuitive • difficult to match existing texture Image removed due to copyright considerations.
Ken Perlin, SIGGRAPH '85. MIT EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping – Perspective Correct Interpolation – Illumination – Texture Mapping Difficulties – Projective Texturing
• Other Mapping Techniques
MIT EECS 6.837, Durand and Cutler
Photo-textures
Courtesy of Leonard McMillan, Computer Science at the University of North Carolina in Chapel Hill. Used with permission.
MIT EECS 6.837, Durand and Cutler
Texture Mapping • Like wallpapering or gift-wrapping with stretchy paper • Curved surfaces require extra stretching or cutting
MIT EECS 6.837, Durand and Cutler
Texture Coordinates • Specify a texture coordinate (u,v) at each vertex • Canonical texture coordinates (0,0) → (1,1) • Often the texture size is a power of 2 (but it doesn't have to be) • How can we tile this texture? (0,1)
(0,0)
MIT EECS 6.837, Durand and Cutler
(1,0)
Tiling Texture (0,3)
(0,0)
(0,3)
tiles with visible seams
(3,0)
(0,0)
MIT EECS 6.837, Durand and Cutler
seamless tiling (repeating)
(3,0)
Texture Coordinates • Specify a texture coordinate (s,t) at each vertex • Canonical texture coordinates (0,0) → (1,1) • Can we just linearly interpolate the values in screen space? (0,1)
(0,0)
MIT EECS 6.837, Durand and Cutler
(1,0)
What Goes Wrong?
texture source
what we get
MIT EECS 6.837, Durand and Cutler
what we want
Looking at One Edge • Consider one edge from a given triangle. This edge and its projection onto our viewport lie in a single common plane illustrated below:
Courtesy of Leonard McMillan, Computer Science at the University of North Carolina in Chapel Hill. Used with permission.
MIT EECS 6.837, Durand and Cutler
Visualizing the Problem Let's assume that the viewport is located 1 unit away from the center of projection.
Notice that uniform steps on the image plane do not correspond to uniform steps along the edge. MIT EECS 6.837, Durand and Cutler
How do we fix it? • We can reduce the perceived artifacts of nonperspective correct interpolation by subdividing the model into smaller triangles. Why? • However, sometimes the errors become obvious – At "T" joints – When switching between levels-of-detail representations (mipmapping... next time)
MIT EECS 6.837, Durand and Cutler
Subdivision
MIT EECS 6.837, Durand and Cutler
Subdivision
texture source
what we get MIT EECS 6.837, Durand and Cutler
what we want
Linear Interpolation in Screen Space
Compare linear interpolation in screen space to interpolation in 3-space
MIT EECS 6.837, Durand and Cutler
Perspective Correct Interpolation We need a mapping from t values to s values:
Solve for s in terms of t:
Unfortunately, at this point in the pipeline (after projection) we no longer have z. However, we do have w1= 1/z1 and w2 = 1/z2, so:
MIT EECS 6.837, Durand and Cutler
Questions?
MIT EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping – Perspective Correct Interpolation – Illumination – Texture Mapping Difficulties – Projective Texturing
• Other Mapping Techniques
MIT EECS 6.837, Durand and Cutler
Texture Mapping & Illumination • Texture mapping can be used to alter some or all of the constants in the illumination equation: as the final color for the pixel, or as the diffuse color, or to alter the normal, ... the possibilities are endless! (e.g. GL_DECAL, GL_MODULATE, GL_BLEND, ...)
Phong's Illumination Model
Constant Diffuse Color
Diffuse Texture Color
Texture used as Label
Texture used as Diffuse Color
Courtesy of Leonard McMillan, Computer Science at the University of North Carolina in Chapel Hill. Used with permission.
MIT EECS 6.837, Durand and Cutler
2D Texture Mapping • Increases the apparent complexity of simple geometry • Requires perspective projection correction • Specifies variations in shading within a primitive: – Illumination – Surface Reflectance Courtesy of Leonard McMillan, Computer Science at the University of North Carolina in Chapel Hill. Used with permission.
MIT EECS 6.837, Durand and Cutler
Texture Mapping Difficulties • Tedious to specify texture coordinates for every triangle • Easier to model variations in reflectance than illumination • Acquiring textures is surprisingly difficult – Texture image can't have projective distortions – Seamless tiling – Non-repeating textures Courtesy of Leonard McMillan, Computer Science at the University of North Carolina in Chapel Hill. Used with permission.
MIT EECS 6.837, Durand and Cutler
Common Texture Coordinate Mappings • • • •
Orthogonal Cylindrical Spherical Perspective Projection • Texture Chart MIT EECS 6.837, Durand and Cutler
Projective Textures • Treat the texture as a light source (a slide projector) • No need to specify texture coordinates explicitly • A good model for shading variations due to illumination • A fair model for reflectance (can use pictures) MIT EECS 6.837, Durand and Cutler
Projective Texture Example • Modeling from photograph • Using input photos as textures
Images removed due to copyright considerations
Figure from Debevec, Taylor & Malik http://www.debevec.org/Research MIT EECS 6.837, Durand and Cutler
Texture Chart • Pack triangles into a single image
MIT EECS 6.837, Durand and Cutler
Questions?
MIT EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping • Other Mapping Techniques: – Projective Shadows and Shadow Maps – Bump Mapping – Displacement Mapping – Environment Mapping (for Reflections)
MIT EECS 6.837, Durand and Cutler
Images removed due to copyright considerations.
Figure from Moller & Haines “Real Time Rendering� MIT EECS 6.837, Durand and Cutler
Figure from FoleyMIT et al. “Computer Graphics Principles and Practice” EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping • Other Mapping Techniques: – Projective Shadows and Shadow Maps – Bump Mapping – Displacement Mapping – Environment Mapping (for Reflections)
MIT EECS 6.837, Durand and Cutler
What's Missing? • What's the difference between a real brick wall and a photograph of the wall texture-mapped onto a plane? • What happens if we change the lighting or the camera position?
MIT EECS 6.837, Durand and Cutler
Remember Phong Normal Interpolation? • Instead of using the normal of the triangle, interpolate an averaged normal at each vertex across the face
MIT EECS 6.837, Durand and Cutler
Bump Mapping • Textures can be used to alter the surface normal of an object. • This does not change the actual shape of the surface -we are only shading it as if it were a different shape!
Sphere w/Diffuse Texture
Swirly Bump Map
Sphere w/Diffuse Texture & Bump Map
Courtesy of Leonard McMillan. Used with permission.
MIT EECS 6.837, Durand and Cutler
Bump Mapping • The texture map is treated as a single-valued height function. • The partial derivatives of the texture tell us how to alter the true surface normal at each point to make the object appear as if it were deformed by the height function.
Courtesy of Leonard McMillan. Used with permission.
MIT EECS 6.837, Durand and Cutler
Another Bump Map Example
Bump Map
Cylinder w/Diffuse Texture Map
Cylinder w/Texture Map & Bump Map
MIT EECS 6.837, Durand and Cutler
Another Bump Map Example
MIT EECS 6.837, Durand and Cutler
Questions?
MIT EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping • Other Mapping Techniques: – Projective Shadows and Shadow Maps – Bump Mapping – Displacement Mapping – Environment Mapping (for Reflections)
MIT EECS 6.837, Durand and Cutler
What's Missing? • What does a texture- & bumpmapped brick wall look like as you move the viewpoint? • What does the silhouette of a bump-mapped sphere look like?
MIT EECS 6.837, Durand and Cutler
Displacement Mapping • Use the texture map to actually move the surface point. How is this different than bump mapping? • The geometry must be displaced before visibility is determined. Is this easily done in the graphics pipeline? In a ray-tracer?
MIT EECS 6.837, Durand and Cutler
Displacement Mapping Image from: Geometry Caching for Ray-Tracing Displacement Maps by Matt Pharr and Pat Hanrahan.
Courtesy of Matt Pharr. Used with permission.
note the detailed shadows cast by the stones MIT EECS 6.837, Durand and Cutler
Displacement Mapping
Courtesy of Ken Musgrave. Used with permission.
MIT EECSKen 6.837,Musgrave Durand and Cutler
Questions?
MIT EECS 6.837, Durand and Cutler
Today • 2D Texture Mapping • Other Mapping Techniques: – Projective Shadows and Shadow Maps – Bump Mapping – Displacement Mapping – Environment Mapping (for Reflections)
MIT EECS 6.837, Durand and Cutler
Environment Maps • We can simulate reflections by using the direction of the reflected ray to index a spherical texture map at "infinity". • Assumes that all reflected rays begin from the same point.
MIT EECS 6.837, Durand and Cutler
What's the Best Chart?
Image removed due to copyright considerations.
MIT EECS 6.837, Durand and Cutler
Questions?
Image by Henrik Wann Jensen Environment by Paul MIT EECSmap 6.837, Durand and CutlerDebevec
Textures can Alias • Aliasing is the under-sampling of a signal, and it's especially noticeable during animation
point sampling
mipmaps & linear interpolation MIT EECS 6.837, Durand and Cutler
Next Time:
Aliasing, Anti-Aliasing & Mipmaps for Texturing
MIT EECS 6.837, Durand and Cutler
computer graphics24
Texture Mapping & Other Fun Stuff MIT EECS 6.837, Durand and Cutler •Don't want to represent all this detail with geometry MIT EECS 6.83...
computer graphics24
Texture Mapping & Other Fun Stuff MIT EECS 6.837, Durand and Cutler •Don't want to represent all this detail with geometry MIT EECS 6.83... | 2,622 | 10,176 | {"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} | 3.25 | 3 | CC-MAIN-2018-26 | latest | en | 0.728612 |
https://proxies-free.com/tag/shortest/ | 1,600,971,235,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00369.warc.gz | 590,113,474 | 12,116 | ## How many shortest paths are there from source to destination for Dijkstra’s algorithm?
Given an acyclic, directed, and weighted graph.
Now how can I determine how many shortest paths are there from source to destination, if i use Dijkstra’s algorithm?
## Solving shortest path problem with Dijkstra’s algorithm for n negative-weight edges and no negative-weight cycle
Although many texts state Dijkstra’s algorithm does not work for negative-weight edges, the modification of Dijkstra’s algorithm can. Here is the algorithm to solve a single negative-weight edge without negative-weight edges.
Let $$d_s(v)$$ be the shortest distance from source vertex s to vertex v.
Suppose the negative edge $$e$$ is $$(u, v)$$
First, remove the negative edge $$e$$, and run Dijkstra from the source vertex s.
Then, check if $$d_s(u) + w(u, v) leq d_s(v)$$. If not, we are done. If yes, then run Dijkstra from $$v$$, with the negative edge still removed.
Then, $$forall t in V$$, $$d(t) = min(d_s(t), d_s(u) + w(u, v) + d_v(t))$$
Given the above algorithm, I want to modify the above algorithm again to solve n negative-weight edges and no negative weight cycle. Any hint?
## weighted graphs – Why is DFS not suited for shortest path problem?
I am sorry for the repetition of the question. I understand that this question has already been answered before by the community, but most answers tend to focus on unweighted graphs. I want to know Can DFS we used to find the shortest path for weighted graphs? I know that Dijkstra’s algorithm is used to find the shortest path for weighted graphs. But, what I want to know is what is fundamentally different in using DFS for unweighted graphs compared to Dijkstra (which is BFS + priority queue/set) and why can’t we create DFS + priority queue/set implementation for shortest path problem?
Ref link: Shortest Path using DFS on weighted graphs,
Why can’t DFS be used to find shortest paths in unweighted graphs?
Any help would be really appreciated.
Thanks!
## algorithms – Shortest Path in a Directed Acyclic Graph with two types of costs
I am given a directed acyclic graph $$G = (V,E)$$, which can be assumed to be topologically ordered (if needed). The edges in G have two types of costs – a nominal cost $$w(e)$$ and a spiked cost $$p(e)$$.
The goal is to find a path from a node $$s$$ to a node $$t$$ that minimizes the following cost: $$sum_e w(e) + max_e {p(e)},$$ where the sum and maximum are taken over all edges of the path.
Standard dynamic programming methods show that this problem is solvable in $$O(E^2)$$ time. Is there a more efficient way to solve it? Ideally, an $$O(Ecdot operatorname{polylog}(E,V))$$ algorithm would be nice.
## shortest path algorithm – why backtrack from the end node instead of starting from the starting node?
You want to find the shortest paths $$S(s,v)$$ to all nodes $$v$$ (31:30 in the video). In other words, we are interested in computing a function $$f(v) = S(s,v)$$, and, as explained in lecture, you can write it as $$f(v) = min_{(u,v) in E} f(u) + w(u,v)$$. In the end, it allows you to compute all such values in one go.
If you instead consider an equation $$S(s,v) = min_{(s,u) in E} w(s,u) + S(u,v)$$, you do can find $$S(s,v)$$ this way. However, you won’t be able to find $$S(s,u)$$ for all other $$u$$ in the process, and you’ll have to run the same algorithm again for other end vertices.
## Graph search or shortest path algorithm for graph with multiple “goals”?
I did a project in a class using A* search to solve an 8-puzzle.
But what about a puzzle with multiple ‘solved’ states? For example, and 8 puzzle with some repeated tiles.
I’m not sure whether something like A* search could still work or not. I have trouble fathoming how a heuristic might be designed.
Are their other shortest path algorithms or search algorithms that are better suited for this kind of problem?
## shortest path – Bellman Ford Dynamic Programming
Dynamic programming doesn’t necessarily mean that your solution will be efficient. It just means that your problem can be defined using a recursive function, for which you can use memoization.
To understand what this function is, take a look at the proof. The invariant is “after $$i$$ iterations, we’ve found all shortest paths of length at most $$i$$“. Therefore, we can define our function $$f(v, ell)$$ – the length of the shortest path to vertex $$v$$ with length at most $$ell$$ – in the following way:
$$f(v, ell) = min_{u in in(v)} (f(u, ell-1) + d(u,v))$$
What Bellman-Ford does is slightly different (e.g. it can compute all shortest paths in a single iteration, if our order is lucky), but it can only help us, so we don’t mind.
## shortest path – Removing any arbitrary vertex from a directed graph?
I came upon this particular question which I do not understand from Jeff E. Algorithms, Chapter 9, ex. 8. https://jeffe.cs.illinois.edu/teaching/algorithms/book/09-apsp.pdf
How can we remove any arbitrary vertex from a directed graph, with weighted edges that can be positive negative or zero, without changing the shortest paths distance between all other pairs of vertices, in $$O(V^2)$$.
Let $$G=(V,E)$$ be a directed graph with weighted edges; edge weights could be positive, negative, or zero.
a) How would we delete an arbitrary vertex $$v$$ from this graph, without changing the shortest-path distance between any other pair of vertices? Describe an algorithm that constructs a directed graph $$G’=(Vsetminus{v},E’)$$ with weighted edges, such that the shortest-path distance between any two vertices in $$G’$$ is equal to the shortest-path distance between the same two vertices in $$G$$, in $$O(V2)$$ time.
I have an idea would be to just consider the incoming edges $$u rightarrow v$$ and outgoing edges $$v rightarrow t$$ and for each $$u$$, add an edge $$u rightarrow t$$ with the weight being the sum of the weights of $$u rightarrow v$$ and $$v rightarrow t$$, for all $$t$$. The second part builds upon the first, but now I am confused about how to go about it.
Now suppose we have already computed all shortest-path distances in $$G’$$.Describe an algorithm to compute the shortest-path distances in the original graph G from v to every other vertex, and from every othervertex tov, all in $$O(V^2)$$ time
Is there a way to do this now that I have deleted the node?
## graphs and networks – How to choose the shortest route?
Someone needs to start from home and complete the three tasks of going to the post office to send a letter, going to the bookstore to buy books, and going to the supermarket to buy food. He can walk through some nodes repeatedly. How should he choose the route to make the path the shortest?
``````Graph[{Home [UndirectedEdge] School,
Home [UndirectedEdge] Supermarket,
Home [UndirectedEdge] PostOffice,
PostOffice [UndirectedEdge] Home,
PostOffice [UndirectedEdge] Bookstore,
PostOffice [UndirectedEdge] Supermarket,
Bookstore [UndirectedEdge] PostOffice,
Bookstore [UndirectedEdge] Supermarket,
Supermarket [UndirectedEdge] Bookstore,
Supermarket [UndirectedEdge] PostOffice,
Supermarket [UndirectedEdge] Home,
Supermarket [UndirectedEdge] School,
School [UndirectedEdge] Supermarket,
School [UndirectedEdge] Home},
EdgeWeight -> {410, 510, 218, 218, 75, 329, 75, 440, 440, 329, 510,
125, 125, 410}, VertexLabels -> "Name",
VertexCoordinates -> {Home -> {0, 0}, School -> {1, 0},
PostOffice -> {0.2, 1}, Supermarket -> {1.2, 0.8},
Bookstore -> {0.4, 1.7}}]
``````
If possible, I hope the respondents can provide as many methods as possible to solve this problem, such as neural network algorithm, genetic algorithm, or built-in function solution, etc.
## algorithm – A system that figure out the fastest (not shortest) navigation method between points for 2D spaceship with 2 axes propulsion
I am working on a space themed game with space being a 2D pane. The player issue order by clicking on the coordinate for his/her ship to navigate to a certain point on the pane.
The player ship can:
1. Propel forward and backward
2. Strafe starboard and port
3. Yaw starboard and port
The ship’s performance on each direction varies greatly depending on configuration. Some may move faster on sides and back directions than they do forward direction.
I have combined A* and PID control for navigation but couldn’t help but noticed that in many occasions where the ship could have accelerate to the full speed and turn towards the target to reach the target faster, it instead turns in place while strafing slowly until it faces the destination before accelerating.
I would like to add another calculation that let the ship knows which navigation approach it should take to minimize the time to reach the target but I have no idea which field of study I should look into or if there is a quick win scenario I could use to achieve an expected outcome. | 2,129 | 8,862 | {"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": 46, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-40 | latest | en | 0.885572 |
http://www.ni.com/documentation/en/labview/1.0/analysis-node-ref/voronoi-diagram/ | 1,534,485,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211719.12/warc/CC-MAIN-20180817045508-20180817065508-00542.warc.gz | 536,728,513 | 9,677 | # Voronoi Diagram (G Dataflow)
Version:
Computes the Voronoi diagram of specified points in a plane.
## x
X-coordinates of the points in the plane. You must specify at least three elements in this input.
## y
Y-coordinates of the points in the plane. You must specify the same number of elements in y and x.
## error in
Error conditions that occur before this node runs.
The node responds to this input according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
Default: No error
## Voronoi x
X-coordinates of the Voronoi points.
## Voronoi y
Y-coordinates of the Voronoi points.
## Voronoi edges
An n-by-2 array of indexes of the Voronoi points, where n is the number of edges in the Voronoi diagram. Each row of Voronoi edges represents an edge of the Voronoi diagram defined by the indexes.
## error out
Error information.
The node produces this output according to standard error behavior.
Standard Error Behavior
Many nodes provide an error in input and an error out output so that the node can respond to and communicate errors that occur while code is running. The value of error in specifies whether an error occurred before the node runs. Most nodes respond to values of error in in a standard, predictable way.
error in does not contain an error error in contains an error
If no error occurred before the node runs, the node begins execution normally.
If no error occurs while the node runs, it returns no error. If an error does occur while the node runs, it returns that error information as error out.
If an error occurred before the node runs, the node does not execute. Instead, it returns the error in value as error out.
## Voronoi Diagrams
For n given points in the plane, the Voronoi diagram divides the plane into n Voronoi cells. Each given point belongs to one Voronoi cell. The point in a Voronoi cell is closer to the corresponding given point than to any of the other given points.
The following figure shows an example of a Voronoi diagram of eight given points.
Where This Node Can Run:
Desktop OS: Windows
FPGA: This product does not support FPGA devices | 618 | 2,840 | {"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-2018-34 | latest | en | 0.755151 |
http://www.acmerblog.com/POJ-2769-Reduced-ID-Numbers-blog-867.html | 1,477,120,230,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00220-ip-10-171-6-4.ec2.internal.warc.gz | 295,820,202 | 12,088 | 2013
11-12
# Reduced ID Numbers
T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 106-1. T. Chur finds this range of SINs too large for identification within her groups. For each group, she wants to find the smallest positive integer m, such that within the group all SINs reduced modulo m are unique.
On the first line of the input is a single positive integer N, telling the number of test cases (groups) to follow. Each case starts with one line containing the integer G (1 ≤ G ≤ 300): the number of students in the group. The following G lines each contain one SIN. The SINs within a group are distinct, though not necessarily sorted.
For each test case, output one line containing the smallest modulus m, such that all SINs reduced modulo m are distinct.
2
1
124866
3
124866
111111
987651
1
8
//* @author [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
import java.util.Hashtable;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Hashtable< Integer,Integer> table;
Scanner scan=new Scanner(System.in);
int M[]=new int[300];
int n=scan.nextInt(),k=0,m=1;
boolean band=false;
for(int i=0;i< n;i++){
k=scan.nextInt();
band=true;
table=new Hashtable< Integer, Integer>();
for(int j=0;j< k;j++){
M[j]=scan.nextInt();
}
m=k;
while(true){
band=true;
for(int j=0;j< k;j++){
if(table.get((M[j]%m))==null){
table.put((M[j]%m), 0);
}
else {
band=false;
break;
}
}
if(band==true) break;
table.clear();
m++;
}
System.out.println(m);
}
}
} | 576 | 2,079 | {"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} | 2.84375 | 3 | CC-MAIN-2016-44 | longest | en | 0.511828 |
https://www.jiskha.com/display.cgi?id=1244939744 | 1,516,532,336,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00656.warc.gz | 952,572,731 | 3,600 | # Algebra
posted by .
Let g(x) = 10-x/2+x
a. Find g (–1)
b. State the domain of the function
g(x)= 10-x/2+x
c. Find g (t + 1) and simplify as much as possible. Show work.
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More Similar Questions | 624 | 1,829 | {"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} | 3.84375 | 4 | CC-MAIN-2018-05 | latest | en | 0.841481 |
https://brainly.com/question/245620 | 1,484,667,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279923.28/warc/CC-MAIN-20170116095119-00511-ip-10-171-10-70.ec2.internal.warc.gz | 801,581,207 | 9,190 | # There is a 750 ml bottle of concentrate orange squash it is enough to make fifteen 250 ml glasses of diluted orange drink how much water is needed to make 10 litre of this drink
1
by alontje66
2015-01-08T15:32:51-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Fifteen 250 ml glasses contain 3750 ml of the drink, and 750 ml of orange squash.
There are 3000 ml of water in 3750 ml of the drink.
3000 ml of water - 3750 ml of the drink
x l of water - 10 l of the drink
8 litres of water are needed to make 10 litres of this drink. | 206 | 803 | {"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-2017-04 | latest | en | 0.942737 |
https://au.mathworks.com/matlabcentral/profile/authors/25586676 | 1,669,991,402,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00197.warc.gz | 150,723,035 | 22,195 | Community Profile
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http://us.metamath.org/mpeuni/islnopp.html | 1,657,181,212,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00234.warc.gz | 55,802,232 | 6,909 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > islnopp Structured version Visualization version GIF version
Theorem islnopp 25852
Description: The property for two points 𝐴 and 𝐵 to lie on the opposite sides of a set 𝐷 Definition 9.1 of [Schwabhauser] p. 67. (Contributed by Thierry Arnoux, 19-Dec-2019.)
Hypotheses
Ref Expression
hpg.p 𝑃 = (Base‘𝐺)
hpg.d = (dist‘𝐺)
hpg.i 𝐼 = (Itv‘𝐺)
hpg.o 𝑂 = {⟨𝑎, 𝑏⟩ ∣ ((𝑎 ∈ (𝑃𝐷) ∧ 𝑏 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑎𝐼𝑏))}
islnopp.a (𝜑𝐴𝑃)
islnopp.b (𝜑𝐵𝑃)
Assertion
Ref Expression
islnopp (𝜑 → (𝐴𝑂𝐵 ↔ ((¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
Distinct variable groups: 𝐷,𝑎,𝑏 𝐼,𝑎,𝑏 𝑃,𝑎,𝑏 𝑡,𝐴 𝑡,𝐵 𝑡,𝑎,𝑏
Allowed substitution hints: 𝜑(𝑡,𝑎,𝑏) 𝐴(𝑎,𝑏) 𝐵(𝑎,𝑏) 𝐷(𝑡) 𝑃(𝑡) 𝐺(𝑡,𝑎,𝑏) 𝐼(𝑡) (𝑡,𝑎,𝑏) 𝑂(𝑡,𝑎,𝑏)
Proof of Theorem islnopp
Dummy variables 𝑢 𝑣 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 islnopp.a . . 3 (𝜑𝐴𝑃)
2 islnopp.b . . 3 (𝜑𝐵𝑃)
3 eleq1 2828 . . . . . 6 (𝑢 = 𝐴 → (𝑢 ∈ (𝑃𝐷) ↔ 𝐴 ∈ (𝑃𝐷)))
43anbi1d 743 . . . . 5 (𝑢 = 𝐴 → ((𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ↔ (𝐴 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷))))
5 id 22 . . . . . . . 8 (𝑢 = 𝐴𝑢 = 𝐴)
65oveq1d 6830 . . . . . . 7 (𝑢 = 𝐴 → (𝑢𝐼𝑣) = (𝐴𝐼𝑣))
76eleq2d 2826 . . . . . 6 (𝑢 = 𝐴 → (𝑡 ∈ (𝑢𝐼𝑣) ↔ 𝑡 ∈ (𝐴𝐼𝑣)))
87rexbidv 3191 . . . . 5 (𝑢 = 𝐴 → (∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣) ↔ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝑣)))
94, 8anbi12d 749 . . . 4 (𝑢 = 𝐴 → (((𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣)) ↔ ((𝐴 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝑣))))
10 eleq1 2828 . . . . . 6 (𝑣 = 𝐵 → (𝑣 ∈ (𝑃𝐷) ↔ 𝐵 ∈ (𝑃𝐷)))
1110anbi2d 742 . . . . 5 (𝑣 = 𝐵 → ((𝐴 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ↔ (𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷))))
12 oveq2 6823 . . . . . . 7 (𝑣 = 𝐵 → (𝐴𝐼𝑣) = (𝐴𝐼𝐵))
1312eleq2d 2826 . . . . . 6 (𝑣 = 𝐵 → (𝑡 ∈ (𝐴𝐼𝑣) ↔ 𝑡 ∈ (𝐴𝐼𝐵)))
1413rexbidv 3191 . . . . 5 (𝑣 = 𝐵 → (∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝑣) ↔ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵)))
1511, 14anbi12d 749 . . . 4 (𝑣 = 𝐵 → (((𝐴 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝑣)) ↔ ((𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
16 hpg.o . . . . 5 𝑂 = {⟨𝑎, 𝑏⟩ ∣ ((𝑎 ∈ (𝑃𝐷) ∧ 𝑏 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑎𝐼𝑏))}
17 simpl 474 . . . . . . . . 9 ((𝑎 = 𝑢𝑏 = 𝑣) → 𝑎 = 𝑢)
18 eqidd 2762 . . . . . . . . 9 ((𝑎 = 𝑢𝑏 = 𝑣) → (𝑃𝐷) = (𝑃𝐷))
1917, 18eleq12d 2834 . . . . . . . 8 ((𝑎 = 𝑢𝑏 = 𝑣) → (𝑎 ∈ (𝑃𝐷) ↔ 𝑢 ∈ (𝑃𝐷)))
20 simpr 479 . . . . . . . . 9 ((𝑎 = 𝑢𝑏 = 𝑣) → 𝑏 = 𝑣)
2120, 18eleq12d 2834 . . . . . . . 8 ((𝑎 = 𝑢𝑏 = 𝑣) → (𝑏 ∈ (𝑃𝐷) ↔ 𝑣 ∈ (𝑃𝐷)))
2219, 21anbi12d 749 . . . . . . 7 ((𝑎 = 𝑢𝑏 = 𝑣) → ((𝑎 ∈ (𝑃𝐷) ∧ 𝑏 ∈ (𝑃𝐷)) ↔ (𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷))))
23 oveq12 6824 . . . . . . . . 9 ((𝑎 = 𝑢𝑏 = 𝑣) → (𝑎𝐼𝑏) = (𝑢𝐼𝑣))
2423eleq2d 2826 . . . . . . . 8 ((𝑎 = 𝑢𝑏 = 𝑣) → (𝑡 ∈ (𝑎𝐼𝑏) ↔ 𝑡 ∈ (𝑢𝐼𝑣)))
2524rexbidv 3191 . . . . . . 7 ((𝑎 = 𝑢𝑏 = 𝑣) → (∃𝑡𝐷 𝑡 ∈ (𝑎𝐼𝑏) ↔ ∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣)))
2622, 25anbi12d 749 . . . . . 6 ((𝑎 = 𝑢𝑏 = 𝑣) → (((𝑎 ∈ (𝑃𝐷) ∧ 𝑏 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑎𝐼𝑏)) ↔ ((𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣))))
2726cbvopabv 4875 . . . . 5 {⟨𝑎, 𝑏⟩ ∣ ((𝑎 ∈ (𝑃𝐷) ∧ 𝑏 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑎𝐼𝑏))} = {⟨𝑢, 𝑣⟩ ∣ ((𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣))}
2816, 27eqtri 2783 . . . 4 𝑂 = {⟨𝑢, 𝑣⟩ ∣ ((𝑢 ∈ (𝑃𝐷) ∧ 𝑣 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝑢𝐼𝑣))}
299, 15, 28brabg 5145 . . 3 ((𝐴𝑃𝐵𝑃) → (𝐴𝑂𝐵 ↔ ((𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
301, 2, 29syl2anc 696 . 2 (𝜑 → (𝐴𝑂𝐵 ↔ ((𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
311biantrurd 530 . . . . 5 (𝜑 → (¬ 𝐴𝐷 ↔ (𝐴𝑃 ∧ ¬ 𝐴𝐷)))
32 eldif 3726 . . . . 5 (𝐴 ∈ (𝑃𝐷) ↔ (𝐴𝑃 ∧ ¬ 𝐴𝐷))
3331, 32syl6bbr 278 . . . 4 (𝜑 → (¬ 𝐴𝐷𝐴 ∈ (𝑃𝐷)))
342biantrurd 530 . . . . 5 (𝜑 → (¬ 𝐵𝐷 ↔ (𝐵𝑃 ∧ ¬ 𝐵𝐷)))
35 eldif 3726 . . . . 5 (𝐵 ∈ (𝑃𝐷) ↔ (𝐵𝑃 ∧ ¬ 𝐵𝐷))
3634, 35syl6bbr 278 . . . 4 (𝜑 → (¬ 𝐵𝐷𝐵 ∈ (𝑃𝐷)))
3733, 36anbi12d 749 . . 3 (𝜑 → ((¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷) ↔ (𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷))))
3837anbi1d 743 . 2 (𝜑 → (((¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵)) ↔ ((𝐴 ∈ (𝑃𝐷) ∧ 𝐵 ∈ (𝑃𝐷)) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
3930, 38bitr4d 271 1 (𝜑 → (𝐴𝑂𝐵 ↔ ((¬ 𝐴𝐷 ∧ ¬ 𝐵𝐷) ∧ ∃𝑡𝐷 𝑡 ∈ (𝐴𝐼𝐵))))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 383 = wceq 1632 ∈ wcel 2140 ∃wrex 3052 ∖ cdif 3713 class class class wbr 4805 {copab 4865 ‘cfv 6050 (class class class)co 6815 Basecbs 16080 distcds 16173 Itvcitv 25556 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1989 ax-6 2055 ax-7 2091 ax-9 2149 ax-10 2169 ax-11 2184 ax-12 2197 ax-13 2392 ax-ext 2741 ax-sep 4934 ax-nul 4942 ax-pr 5056 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2048 df-eu 2612 df-mo 2613 df-clab 2748 df-cleq 2754 df-clel 2757 df-nfc 2892 df-rex 3057 df-rab 3060 df-v 3343 df-dif 3719 df-un 3721 df-in 3723 df-ss 3730 df-nul 4060 df-if 4232 df-sn 4323 df-pr 4325 df-op 4329 df-uni 4590 df-br 4806 df-opab 4866 df-iota 6013 df-fv 6058 df-ov 6818 This theorem is referenced by: islnoppd 25853 oppne1 25854 oppne2 25855 oppne3 25856 oppcom 25857 oppnid 25859 opphllem1 25860 opphllem3 25862 opphllem4 25863 opphllem5 25864 opphllem6 25865 oppperpex 25866 outpasch 25868 lnopp2hpgb 25876 hpgerlem 25878 colopp 25882 colhp 25883 lmiopp 25915 trgcopyeulem 25918
Copyright terms: Public domain W3C validator | 3,867 | 5,006 | {"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} | 3.65625 | 4 | CC-MAIN-2022-27 | latest | en | 0.270494 |
https://www.freecodecamp.org/forum/t/basic-algorithm-scripting-confirm-the-ending/222468 | 1,539,701,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510754.1/warc/CC-MAIN-20181016134654-20181016160154-00207.warc.gz | 973,343,405 | 5,545 | # Basic Algorithm Scripting: Confirm the Ending
Basic Algorithm Scripting: Confirm the Ending
0
#1
Hello…
my solution to this…is this okay? it works.
Summary
function confirmEnding(str, target) {
// “Never give up and good luck will find you.”
// – Falcor
var noob = str.length; //get string length
var noob2 = target.length; //get length of target
var result = str.substring(noob-noob2); //here subtract lengths of string and target to extract equivalent end of string
var match = target.substring(); //here outputting target
console.log(match);
console.log(result);
if (match == result){ //matching both sub string outputs
return true;
}
else{
return false;
}
}
confirmEnding(“Bastian”, “n”);
#2
Looks good to me. You could certainly make it shorter in some ways.
For example, this
if (match == result) {
return true;
} else {
return false;
}
could just be
return match == result;
#3
Hi,
It gets the job done, so congrats!
The useless line is var match = target.substring(); because substring without any arguments will always return the target string, so that line means var match = target and you already have a variable with that value: target.
This is your code without that line:
function confirmEnding(str, target) {
var noob = str.length;
var noob2 = target.length;
var result = str.substring(noob - noob2);
if (target == result) {
return true;
} else {
return false;
}
}
You can improve it the way @michaelfoland just told you:
function confirmEnding(str, target) {
var noob = str.length;
var noob2 = target.length;
var result = str.substring(noob - noob2);
return target == result;
}
Have you seen regular expressions already? I think this example is a good fit for that:
function confirmEnding(str, target) {
return new RegExp( target + "\\$" ).test( str );
}
Happy coding! | 440 | 1,810 | {"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-2018-43 | longest | en | 0.562396 |
https://brainmass.com/business/bond-valuation/pg27 | 1,596,454,289,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735810.18/warc/CC-MAIN-20200803111838-20200803141838-00335.warc.gz | 244,218,587 | 14,881 | Explore BrainMass
# Bond Valuation
### Price of a Bond: Bond Valuation for Complex Systems
Complex Systems has an outstanding issue of \$1,000-par value bonds with a 12% coupon interest rate. The issue pays interest annually and has 16 years reamining to its maturity date. a. If bond of similar risk are currently earning a 10% rate of return, how much should the Complex Systems bond sell for today? b. Describe th
### Solve: Default Risk Premium on a Corporate Bond
Question: A treasury bond that matures in ten years has a yield of 6 percent. A 10 year corporate bond has a yield of 8 percent. Assume that the liquidity premium on the corporate bond is .5 percent. What is the default risk premium on the corporate bond?
### Current Yield, Yield to Maturity & Effective Annual Yield
The 7 percent coupon bonds of Firm X are selling for 102 percent of par value. The bonds mature in 6 years and pay interest semiannually. What is the current yield, yield to maturity and effective annual yield? 1. 6.86; 6.59; 6.70 2. 6.86; 6.23; 6.53 3. 6.37; 6.23; 6.53 4. 6.37; 6.59; 6.86 5. 6.59; 6.65;
### Current & Effective Yield of a Bond
What is the current and effective annual yield for the following? A bond with a yield to maturity of 7.12 percent, a 6.5 percent coupon rate, a face value of \$1,000, a market price of \$961.85, and semiannual interest payments.
### Implied Treasury bond rate
Looking at the Wall Street Journal you observe that the settlement price on a hypothetical 15-year, semiannual payment, 6% coupon bond is 109 9-32. If the bond has a \$1,000 par value, what is the implied Treasury bond rate? 5.11% 5.55% 5.91% 6.35% 6.79%
### FINANCE calculations: NPV, ROE, MRP, CAPM, WACC, IRR, Beta and others
Please see attach document. 1. The primary operating goal of a publicly-owned firm interested in serving its stockholders should be to _________. (Points: 10) maximize its expected total corporate income maximize its expected EPS minimize the chances of losses maximize the stock price per
### Bonds and Stocks in a Market
Bond No. Maturity Coupon Price Yield to Maturity 1 2 years \$50 \$992 ? 2 3 years \$45 ? 5.52% 3 4 years \$60 \$1,015 ? 4 6 years \$54 ? 5.82% (a) Compute the yi
### Methods of Retiring Bonds
What are methods that a company may use to retire its bonds?
### Interest payments a bondholder
Face Value of \$1000 bond has a coupon rate of 8% with 9 years until maturity and sells at a yield to maturity of 9%. What interest payments do Bondholders receive each year and at what price does the bond sell (assume annual interest payments)? If I'm able to calculate the total present value of interest payments how do I get
### Present value or price of bond
I have a \$1000 bond which matures in 10 years with current market interest rate of 8%.If I know that the bond pays \$30 every six months, then how can I find out what the yield to maturity is, expressed on a semiannual basis. I need the yield to maturity rate in order to calculate the present value of interest payments and the p
### Yield to maturity calculations for bond
Consider this: Zero coupon money multiplier notes of 2008. Bonds were issued on July 1,1990 for \$100. Interest is paid every July 1 and the bond matures on July 1, 2008. Determine the yield to maturity if the bonds are purchased at the: a. issue price in 1990 b. Market price as of July 1, 2004, of \$750 c. Explain why th
### Calculating returns for stock and bonds - Set of 14 problems:
Please show all work and complete in excel. Problem Set #1 Calculating Returns: 1. a) Assume you bought 1000 shares of stock at an initial price of \$25 per share. The stock paid a dividend of \$0.50 per share during the following year, and the share price at the end of the year when you sold it, was \$35. Compute y
### Valuation of a Stock and Bond
With that in mind, you decide to put an Excel spreadsheet together that values the firm's stock and bonds. The company's stock trades for US\$35 per share, with an annual dividend payment of US\$1.50, expected to grow to US\$1.58 next year. The required return on stocks is 10%, and the dividend is expected to increase by 6% for the
### Three Theories used to Describe the Shape of the Yield Curve
Compare and contrast the three theories used to describe the shape of the yield curve. Explain how each theory might impact interest rates.
### Coupon rate, coupon yield, and yield to maturity
A firm issues a bond at par value. Shortly thereafter, interest rates fall. If you calculated the coupon rate, coupon yield, and yield to maturity for this bond after the decline in interest rates, which of the three value would be highest and which would be lowest? Explain
### Derivatives-practice problems with index and interest rate futures
1. A portfolio manager controls \$5 million in common stocks. In anticipation of a stock market decline, the manager decides to hedge the portfolio using S&P 500 futures contract. The portfolio beta is 1.20, and the current value of the S&P 500 futures contract selected is 1438.50. a) Calculate the number of futures contracts
### Stocks and Bonds
How do stocks and bonds differ? What are the key differences between them with respect to ownership rights, claims on income and assets, maturity, risks, and tax treatment? Why might an organization choose one versus the other as a long-term financing instrument?
### Dividend Valuation Model and Wealth Maximization
4) Dividends, Retained Earnings, and Yield - Springsteen Music Company earned \$820 million last year and paid out 20 percent of earnings in dividends. a) By how much did the company's retained earnings increase? b) With 100 million shares outstanding and a stock price of \$50, what was the dividend yield? (Hint: First compute
### Bond Valuation Questions
1. Bond valuation Callaghan Motors' bonds have 10 years remaining to maturity. Interest is paid annually; they have a \$1,000 par value; the coupon interest rate is 8 percent; and the yield to maturity is 9 percent. What is the bond's current market price? 2. Current yield and yield to maturity A bond has a \$1,000 par value,
### Risk and Return and Stock Valuation
Please assist me with this problem: Go to Yahoo! Finance at http://finance.yahoo.com/ and look up the stock information for a publicly traded company of your using. If you already know your publicly traded company's stock symbol, all you have to do is enter the letters in the "Get Quotes" field near the top of the page and th
### Bond Valuation
You are evaluating two different \$1,000 maturity corporate bonds to buy. The ABC Company bond has a 7 percent annual coupon with 7 years remaining while the XYZ Company bond has a 10 percent annual coupon with 5 years remaining. You could also buy a newly issued 10-year bond from Widget Company of America that has a 12 percent c
### Price of bond
Burns Fire and Casualty Company has \$1,000 par value bonds outstanding at 11 percent interest. The bonds will mature in 20 years. Compute t he current price of the bonds if the present yield to maturity is: a. 6 percent. b. 8 percent. c. 12 percent.
### Price of bond
Bonds mature in 10 yrs, par value of 1,000, annual coupon payment of \$80, interest rate for bonds is 9%. What is price of bonds?
### Npminal yield to call
Outstanding bond with 12 yrs to maturity, 9% coupon paid semiannually, 1,000 par value, bond has 7% nominal yield to maturity, but it can be called in 3 yrs at a price of 1,045. What is bond's nominal yield to call?
### Implied annual yield, Forward rate
A Treasury bond futures contract has a settlement price of 89-8. What is the implied annual yield? Suppose that 1 Swiss franc could be purchased in the foreign exchange market for 60 U.S. cents today. If the franc depreciated 10% tomorrow against the dollar, how many francs would a dollar buy tomorrow?
### Bond Prices and Returns
One bond has a coupon rate of 8 percent, another a coupon rate of 12 percent. Both bonds have 10-year maturities and sell at a yield to maturity of 10 percent. If their yields to maturity next year are still 10 percent, what is the rate of return on each bond? (Show your work proving your answers). Does the higher coupon
### Risk Analysis in Investments
1) Graph and explain the risk profile for the following: Risk Expected Return 0.10 0.07 0.14 0.10 0.20 0.15 0.30 0.25 2) Given the following two investment options, explain what an investor would choose and why: Investment 1, an investmen
### Bonds And Shares Valuation
Please find the attached file for the problem. Many thanks in advance ? Anna Hegg has been considering investing in the bonds of Atilier Industries. The bonds were issued 5 years ago at their \$1000 par value and have exactly 25 years remaining until they mature. They have an 8% coupon interest rate, are convertible into 50 s
### Calculating Yield to Maturity and Current Yield
4. Bond Pricing. A 6-year Circular File bond pays interest of \$80 annually and sells for \$950. What are its coupon rate, current yield, and yield to maturity? 5. Bond Pricing. If Circular File (see question 4) wants to issue a new 6-year bond at face value, what coupon rate must the bond offer? 6. Bond Yields. A bond has 1
### Bond Amortization for Bishop Company
On January 1, 2006, Bishop Company issued 11% bonds dated January 1, 2006, with a face amount of \$10 million. The bonds mature in 2015 (10 years). For bonds of similar risk and maturity, the market yield is 13%. Interest is paid semiannually on June 30 and December 31. (1.) Determine the price of the bonds at January 1, | 2,368 | 9,611 | {"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} | 3.25 | 3 | CC-MAIN-2020-34 | latest | en | 0.881358 |
https://www.physicsforums.com/threads/unsolvable-exponental.261115/ | 1,521,901,065,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650685.77/warc/CC-MAIN-20180324132337-20180324152337-00535.warc.gz | 831,351,377 | 17,520 | # Unsolvable exponental?
1. Oct 2, 2008
### soandos
Is it possible to solve x = y^y for y?
2. Oct 2, 2008
### CRGreathouse
There's a closed form in Lambert's W, or you could use Newton's method. (Actually, depending on how fast you can evaluate logarithms and exponentials, the secant method is probably faster.)
3. Oct 2, 2008
### SW VandeCarr
I'm sure I'm missing something here, but isn't the solution simply y equals the yth root of x?.
4. Oct 2, 2008
### mathman
Since y is unknown, the yth root of x is unknown.
5. Oct 2, 2008
### soandos
Just some clarification:
is there a way to get an exact value, or just a numeric approximation?
also, i do not really understand the Lambert W function. the link just says that it is the inverse function, but not how.
also, how is the following true (or is it false?):
Sqrt[x]^Sqrt[x]==1/x^Sqrt[x]
Last edited: Oct 2, 2008
6. Oct 3, 2008
### WarPhalange
False. 1/x = x^(-1), so you'd have [x^(1/2)]^[x^(1/2)] = [x^(-1)]^[x^(1/2)]
Also, a quick numerical check can show you it's false. Take 4, for example. Square root 4 = 2, so you have 2^2 = (1/4)^2, so that's false.
7. Oct 3, 2008
### CRGreathouse
Is there a way to get an exact value for $\sqrt 2$, or just a numeric approximation? Because Newton's method is used for square roots as well.
This is a philosophical question more than a mathematical one, I suspect.
8. Oct 3, 2008
### Irrational
*erroneous post*
9. Oct 3, 2008
### soandos
to get an exact answer, i think the only real way is with a continuous fraction. could someone please explain to me exactly what newtons method is?
10. Oct 3, 2008
### CRGreathouse
Last edited by a moderator: May 3, 2017
11. Oct 4, 2008
### soandos
I am not sure that i understand. if the root is negative, then it will never work, as the derivatives of the function are x^x (1 + \text {Log}[x]) for the first derivative, and x^{-1 + x} + x^x (1 + \text {Log}[x])^2 for the second derivative. i believe that both of these are non-continuous at any negative. is this right, or am is making a mistake?
12. Oct 4, 2008
### CRGreathouse
You didn't sy you were looking at x < 0. You'll need to use the secant method (as I recommended in my first post!) or something similar instead of Newton in that case. Otherwise you could use Newton's method or Halley's method to compute W and get the answer that way.
I generally recommend Newton's method only because it is what is taught. The secant method is often better. Of course you can always fall back onto bisection if you have trouble... or don't want to use something complex like regula falsi.
13. Oct 4, 2008
### soandos
i am not sure that i understand. i thought from what i read that the secant method has the same kind of problem that newton's method does, namely that if there is no continuous second derivative, then there is no guarantee that it will converge.
also, from what i understand of Halley's function, is that it is dependent on newtons method, so for -x it will also fail. in fact, i belive that all of these methods are for continuous functions only.
is there a non-iterative way of doing this, or at least one that will work for negative x?
14. Oct 5, 2008
### CRGreathouse
I suggested Halley's method for solving W, not x^x.
Your function is continuous, what's the problem?
15. Oct 6, 2008
### soandos
i thought that it is not continious at all negative x.
16. Oct 6, 2008
### CRGreathouse
Well, just take absolute values and find the magnitude, then see if a solution is possible with a sign change.
17. Oct 6, 2008
### soandos
how could that work if the magnitude of the absolute value is a different expresion? even with a sign change, the absolute value is changing more than that.
18. Oct 7, 2008
### soandos
also, how could someone compute W using halleys mehtod?
19. Oct 7, 2008
### CRGreathouse
20. Oct 7, 2008
### soandos
thanks. but what about the previous question? | 1,139 | 3,945 | {"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} | 3.71875 | 4 | CC-MAIN-2018-13 | latest | en | 0.914744 |
http://forum.twbts.com/viewthread.php?action=printable&tid=16006 | 1,566,699,802,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00407.warc.gz | 76,051,424 | 2,038 | [attach]22969[/attach]
[attach]22970[/attach]
1. Function 大寫轉數字(xNumber\$)
2. Dim N1%, N2%, NN, T, X%, i%
3. For i = Len(xNumber) To 1 Step -1
4. T = Mid(xNumber, i, 1)
5. X = InStr("Ｐ十拾百佰千仟", T): If X > 1 Then N1 = Int(X / 2): GoTo 101
6. X = InStr("萬億兆", T): If X > 0 Then N2 = X * 4: N1 = 0: GoTo 101
7. X = InStr("Ｐ一壹二貳三參四肆五伍六陸七柒八捌九玖", T)
8. If X > 1 Then NN = NN + Int(X / 2) * (10 ^ N1) * (10 ^ N2)
9. 101: Next
10. 大寫轉數字 = NN
11. End Function
[attach]22971[/attach]
歡迎光臨 麻辣家族討論版版 (http://forum.twbts.com/) | 286 | 527 | {"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-2019-35 | longest | en | 0.288117 |
https://community.ptc.com/t5/Mathcad-Tips/tkb-p/tkb_mathcadtips/sort-string/topicPublishDate | 1,680,325,331,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00087.warc.gz | 223,065,802 | 44,516 | cancel
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# Mathcad Tips
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## What's New In PTC Mathcad Prime 9
What's new in PTC Mathcad Prime 9 blog, interview, and webinar. Out now!
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## PTC Mathcad Community Challenge Index and Guidelines
PTC Mathcad Community Challenge is a bimonthly event where we pose a math problem for you to collaborate and solve using Mathcad Prime.
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## PTC Community Spotlight: Tetsuro Tokoro (ttokoro)
Professor Tetsuro Tokoro (ttokoro on the PTC Community) is a member of the Institute of Electrical Engineers of Japan (IEEJ), and he has dedicated much of his time to teaching and studying electric circuit design for his college students at Japan’s National Institute of Technology (KOSEN), Gifu College. A true man of math, ttokoro finds helping community members solve their problems to be very fun. He seeks out math-based brain teasers and shares them with the Mathcad community to gain the insights of his kind and very intelligent fellow Mathcad users. These include resistor meshes (including in 3D), perfect squared squares, and a puzzle where you must identify the minimum radius on a circle with n number of points where all distances between points must be an integer. Sometimes, he looks for problems on rosettacode.org to solve. According to ttokoro, many of these problems are unique and won’t be found in a textbook (and many math textbooks in use were written before math software came to the scene) but demonstrate the value of math software. Ttokoro ventures to say that a lot of these problems cannot be solved by hand, or at least it would be unviable to do so, especially for students. Instead, Mathcad and its programming operators must be used to solve these problems. Montage of resistor meshes and other fun problems ttokoro has introduced Beginning in 2021, ttokoro has also become a prolific uploader of Japanese PTC Mathcad content on YouTube. Many of these videos share tips or are tie-ins to the puzzles he or others post on the PTC Community, including the Mathcad Community Challenges. Ttokoro’s best resistor 3D mesh electric circuit. (Each edge is made by one ohm. Find the resistance between Node [0,0,0] and [1,1,2].) Ttokoro first started using Mathcad with Mathcad 12, and his biggest requests for Mathcad Prime features are transparent colour availability and for the Animation tool to return. You’ll notice that in his brain teasers, as well as in his submissions to the Mathcad Community Challenges, ttokoro likes to show results with plots because it helps convey the meaning better than with math results alone and having more tools to show the results via images helps with that. This sort of thinking carries over to his academic work as well. Ttokoro wearing his PTC Champion Jacket and PTC Mathcad T- shirt! Thank you for your enthusiasm for math and contributions to the PTC Community, ttokoro!
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## Registration Open for the Mathcad Virtual Conference 2022!
Registration has opened for the Mathcad Virtual Conference 2022! Happening September 13, 2022, 10:00 Eastern time. This will be our third annual Mathcad conference, and this year we're promising to talk only about information with demos that you will find valuable and interesting. You'll hear from... Brian Thompson, Mathcad general manager, introducing the show and discussing the current state of Mathcad Andrew McGough, Mathcad product manager, presenting an updated Mathcad Prime roadmap, along with world-premiere demos of some of the new functionality coming to next year's Mathcad Prime 9 Anji Seberino, Mathcad application engineer, showcasing Mathcad Prime's powerful design of experiments functionality Anna Novikova, Mathcad software engineer, deeply detailing the improvements made to Mathcad Prime 8's numeric and symbolic engines (including some enhancements that were not documented in Mathcad Help) A Mathcad customer and how Mathcad Prime powers his innovative structural engineering start-up Throughout the conference, you'll get to share your feedback to PTC about the sessions. We'll also be manning a text-based Q&A box. And if you are within the first 500 registrants, you will be able to have your chance to win a raffle to win one of 25 brand-new 100% cotton Mathcad T-shirts. PTC hasn't had new Mathcad merchandise in about a decade, but now we do, and these shirts are quite wonderful. The raffle numbers will be drawn using a Mathcad Prime program during the conference. We hope to see you there! And if you aren't able to make it on the day of, we are recording the sessions and will send you the recordings afterward. We really hope that you'll find this to be the best Mathcad conference that we've ever hosted. (view in My Videos)
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## Mathcad Scripted Control design Working Group
PTC has a Mathcad Scripted Control design Working Group for implementing scripted controls in Mathcad Prime 10.
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## Mathcad Community Challenge September 2022 - Resistors in Parallel
This month’s challenge is related to electrical engineering. We have a simple circuit with an electrical potential of 220 Volts. Initially we have a single 10 Ohm resistor. Then we add a second resistor in parallel, with 10% higher resistance. Then we add a third resistor in parallel, with 10% higher resistance than the previous resistor. And so on. Calculate the current in the circuit for the single resistor case. Calculate the resistance of each additional resistor and current through each resistor for 2, 3, 4, 5, and 10 resistors in parallel. Can you write a function or program that calculates the resistance of each resistor and current through each resistor for n resistors in parallel? These calculations are fairly straightforward, so it will be interesting to see what tools – vectors, matrices, loops, plots, charts, etc. – that you use to solve the problem. As always, how you document your calculations is important as your worksheet will be visible to the community. Here is an example of three resistors in parallel, as drawn in Creo Schematics: Find the Mathcad Community Challenge Guidelines here!
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## Mathcad Community Challenge July 2022 - Area of a Spherical Triangle
This month's challenge is another geometry-based challenge - but it's non-Euclidean geometry. Given the following three points on a circle of radius 10 units, calculate the area of the spherical triangle: (5.339, 6.948, 4.819) (-4.018, -3.703, 8.375) (6.455, -6.679, -3.705) (Note that the point coordinates are rounded to three decimal places.) Optional: Can you write a program or function that calculates the area for any three points on the surface? Is there a symbolic evaluation for three generic points (a,b,c), (d,e,f), and (g,h,k)? Here is a 3D model in Creo 9 of the sphere and points: Find the Mathcad Community Challenge Guidelines here! Edit: the coordinates of the second point were incorrect. Wildly incorrect. I have no idea how I wrote down the wrong numbers.
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## Mathcad Community Challenge May 2022 - An Isoperimetric Geometry Problem
The first two challenges were biased towards mechanical engineering. May’s challenge pertains to geometry. Create a worksheet in which you calculate (1) the diameter of a circle and (2) the length of a side of a square that yields the minimum combined area for a combined perimeter of 1 meter. This is an optimization problem. What tools within Mathcad can provide you with a result? Optional: How can you depict the results? Can you use a 2D plot or Chart Component to visualize the answer? Find the Mathcad Community Challenge Guidelines here!
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## Transferring Mathcad Single User or Student license from an existing Host ID to a new Host ID
Solution: Mathcad Single User and Student Licenses can be transferred by using the link. Select the version and enter the product code Enter the new HOST ID onto which the license needs to be transferred - To correctly identify the Host ID on new machine, go to Windows Start > Programs > Accessories > Command Prompt > type ipconfig/all (ENTER) and with the Physical Address below the Ethernet adapter Local Area Connection there stands the Host ID - Click on #Submit Request Below is the sample email triggered with new HOST ID license file (attachment) Please save this license file to a secure location on your computer.
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## Why Mathcad Prime 8: A Chat with Andy McGough
PTC Mathcad Prime 8 has now released! @CatMcchad a quick interview with PTC Mathcad's product manager, @amcgough, for this blog article about what's new and exciting in Mathcad Prime 8, what new functionality users and veterans will look forward to using, and how the Mathcad team decides what new functionality to put into each release. Much of that is customer-driven, and Andy cited several delivered requests from the Mathcad Ideas board on this very community. Read the interview here: https://www.mathcad.com/en/blogs/why-prime8-andy-mcgough Andy will also be available next week (March 22 at 10:00 Eastern) for a webinar about what's new in Mathcad Prime 8, featuring demos of the new functionality, plus a live Q&A portion with Andy that yours truly will be moderating. Register for that here (or catch the replay-on-demand after the fact): https://www.mathcad.com/en/resources/webcast/whats-new-mathcad-8 Happy calculating~
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## Mathcad Community Challenge March 2022 - Planetary Gear Calculator
Let’s face it, planetary gears are cool. They are used in all kinds of mechanisms, including transmissions, motors, and turbines. Your challenge, should you choose to accept it, is to use Mathcad to build a worksheet that calculates the gear ratios and output speeds of a planetary gear under different configurations. The math behind the gear ratios is straightforward. But users should be able to select: The sun or planet carrier as the input; The planet carrier or ring as the output; and, The ring, sun, or planet carrier as the stationary component. How can the user select the desired component? How can you add intelligence to the worksheet to prevent an incorrect combination of input, output, and stationary? What inputs do you want for your calculator? You will have to make some choices. Your choices could include: Number of planets Number of teeth Component diameters Module / pitch Tools that you might consider include Combo boxes, programs, and functions involving strings. Here are some technical references to get you started: KHK Gears Instructables How Stuff Works Once again, this is a great project for students to work on in teams. Make the worksheet as simple or as complicated as you want. Others can build on your work, and you can build on others. Let’s see what the community can create! Good luck! Find the Mathcad Community Challenge Guidelines here!
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## For the Love of Mathcad ❤
(Made in PTC Mathcad Prime 7)
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## What's in Mathcad Prime 8? SNEAK PEEK
Hi all! I'm pleased to announce that while Mathcad Prime 8 is releasing March 2022, we at PTC have just published a blog article letting you all know in advance what is being included! And PTC Community are the first ones to know. Check it out: https://www.mathcad.com/en/blogs/whats-mathcad-prime8 We'll post another announcement when Mathcad Prime 8 is actually released. In the meantime, we've opened up registration for our accompanying What's New in Mathcad Prime 8 webinar coming March 22, 2022. I plan to moderate it, with Mathcad product manager Andrew McGough demoing the functionality in much more detail. The webinar will end with a live question and answer session. Mathcad Prime 8 Webinar March 22 (Note that I'm not able to answer your Mathcad Prime 8 questions on this post, but Andy will certainly be able to in that webinar.) Thanks!
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## Mathcad Community Challenge January 2022
Mathcad Contest Idea - January 2022 Plane Truss Mechanics Can you use Mathcad to solve this problem? This comes from page 88 of “An Introduction to the Mechanics of Solids,” by Stephen H. Crandall, Norman C. Dahl, and Thomas J. Lardner. This was the text for the first mechanical engineering class (2.01 Mechanics) that I took my sophomore year at MIT way back in 1989. You can find the book for as little as \$5 on Amazon. It shows three different solutions for the problem, including: closed-form solutions involving equilibrium of forces and the beam-deflection equation a computer program called IBM STRESS Castigliano’s theorem which solves via elastic energy. Any introductory mechanics text or a Schaum’s Outline should guide you to a solution. As with the other contests that will follow this one, the point is not the answer to the problem, but your execution of the solution. Some ideas you may consider including in your worksheet: Allowing the user to change the material via a Combo Box Input Controls. Depicting the results with a Chart Component, such as the truss in the deformed shape, or the deflection at D due to changing input loads. Making the problem more open-ended, such as using matrices and programming for different geometry and loading. Note that these are just ideas; I have not tried any of these. Maybe you can explore different approaches. The problem is simply a starting point. This is an excellent problem for a team to solve, especially for civil and mechanical engineering students. How would you tackle this problem in Mathcad? Find the Mathcad Community Challenge Guidelines here!
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## PTC Community Spotlight: Professor Valery Ochkov
A Mathcad user for 25 years, Professor Valery (‘Val’) Ochkov takes to the mountains when not involved in developing training and educational software for fossil and nuclear power plants. He’s shown below in the Chimbulak region of Kazakstahn. Val has long been involved in the Mathcad community! See the photo of the “Mathcad Clock.” This is from a post back in 2010! Beyond answering questions, Valery is also known for using math equations to create designs and authoring Mathcad trivia challenges for other community members to participate in. Some of the more entertaining challenges might be the Hare and Snell’s law and the Problem of Cockroach Races. The Bicycle Post where Valery quizzes the community on how to solve “How many strokes with a bicycle pump need to be done to inflate the tire to 5atm” also inspired engagement. When asked for hobbies that don’t involve hiking boots, Val assured us his hobbies really are writing books and articles about Mathcad. He’s now deeply interested in using Mathcad for STEM education, the subject of his latest book, 2⁵ Problems for STEM Education 2 (2020). He says collaborations with community members were critical to developing the material. Way to go Mathcad Community! Keep Collaborating! Keep engaging and great things like a book is born!
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## "My Top 3 Engineering Nightmares" Interview
https://www.engineering.com/story/my-top-3-engineering-nightmares-expert-shares-the-truth-on-software-selection Anji Seberino, the head of our PTC Mathcad Application Engineer team, recently took some time to interview with engineering.com about her top 3 engineering nightmares accumulated from her own experience as an engineer. From missing units to debugging custom code to trying to reverse engineer IP stored in crazy, inherited spreadsheets, you too have probably faced these issues in your time in the industry when not everyone is using the right software and processes. Let us know what you think of the interview!
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## "For Power Mathematics, Retire the Spreadsheet" with Brian Thompson
I'm sharing a recent long-form interview that PTC's CAD/engineering calculations general manager, Brian Thompson, had with engineering.com: https://www.engineering.com/story/video/for-power-mathematics-retire-the-spreadsheet Alongside with the video interview, there is also a written transcript. You don't have to share any of your own information to access it. For Power Mathematics, Retire the Spreadsheet Brian talked through a wide variety of topics, including why engineers and research scientists across many various industries choose PTC Mathcad and how those customers use it, the old and painful days of trying to perform complex calculations in spreadsheets, and some glimpses into the future of Mathcad.
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## Official PTC Mathcad 15 and Prime 1–6 End-of-Sale Announcement
Hi Mathcad community members, For those of you who don't know me (some do), my name is Andrew McGough and I'm the product manager for Mathcad here at PTC. I have talked to some of you in the past individually, but I don't get to participate as much as I'd like on the community. I hope this announcement is the start of me being able to be more active here. I would like to share with you that PTC is announcing the end-of-sale for the following versions of PTC Mathcad, effective December 31, 2021: Mathcad 15 Mathcad Prime 1.0 – Prime 6.0 Effective January 1, 2022, the only version of PTC Mathcad available for subscription will be PTC Mathcad Prime 7 and future versions of PTC Mathcad Prime. After December 31, 2021, PTC will no longer have the right to sell or distribute software versions moving to end of sale due to a third-party component contained in those versions. I appreciate the passion of our Mathcad userbase, so I would like to pass along some additional information on a helpful offer. If you currently have a subscription license and wish to continue using your current version of PTC Mathcad while you evaluate Mathcad Prime 7, we are offering a multi-year renewal option that will give you the ability to use your current version for up to four additional years. If you decide to take advantage of this offer, it is important to place the order prior to December 7, 2021 to ensure no issues in booking the order or the subscription start date. PTC may decline or be unable to process orders received after that date. Please contact your PTC account representative or PTC partner for more information. Please bear in mind that the use of old versions of software does introduce implicit risk in areas of security, hardware, and software platform support. PTC will continue to provide technical support access, phone assistance and expert input for versions of Mathcad that have reached end-of-sale. PTC will not be able to issue new licenses or provide access to software downloads for end-of-sale versions. Our customers in English-speaking countries received the End-of-Sale announcement on February 17, 2021. Our global customers are receiving the localized communication now. We have put together multiple resources for you to answer your questions. Please refer to our FAQ for more information. We know this is a big change so the Mathcad team will be monitoring this thread for feedback. Best regards, Andrew McGough, Product Manager, PTC
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## Converting .mcd to .mcdx
We received an email inquiry which I had forwarded to a Mathcad product manager. I thought it may be useful for others, too, so instead of just emailing the response to the customer, I am posting it here as well. ------- We need to convert .mcd to .mcdx...We have Prime 6.0.0.0 What do we need and how much does the program cost to do this conversion. ------- The .mcdx file format is Mathcad Prime. The .mcd file format is for earlier versions of the legacy software, later versions of the legacy software use the .xmcd file format. If this customer uses Mathcad Prime 6.0 there is a converter that comes with the software they can use to convert all .xmcd and most .mcd files to .mcdx - ‘most’ .mcd as we only go back as far as something like legacy Mathcad 7, which was late 90’s I think. From the Input/Output tab in Mathcad Prime 6.0 they can find an ‘xmcd/mcd converter’ button - clicking that will open the converter. They will need the latest version of Mathcad 15.0 (M050) for that to work, and they can download that currently from the support software download site (Prime 7.0 converter does not need Mathcad 15.0 for the converter to work). -------- Hope this helps! For additional questions, don't hesitate to open a Support ticket!
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## GET A PTC MATHCAD EDUCATION: Fundamentals of Data Analysis
Are you making the most of your engineering calculation software? Join a free webcast exploring the fundamentals of data analysis. Presented by Anji Seberino, PTC’s leading Mathcad expert, the online event takes place Thursday, June 25 at 10 am PDT and 1 pm EDT. REGISTER NOW! Anji will focus the session on the basic principles of analyzing data in Mathcad. You’ll learn about Interpolating data Filtering data Fitting data to a curve This webcast is part of the weekly education series for engineers who want to learn more about PTC Mathcad as a powerful alternative for engineering calculations. Visit the registration page to view weekly topics and sign up today to join in.
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## GET A PTC MATHCAD EDUCATION: Solving and Optimization
Are you making the most of your engineering calculation software? REGISTER NOW for a free webcast that explores solving and optimization. Presented by Anji Seberino, PTC’s leading Mathcad expert, the online event takes place Thursday, June 18 at 10 am PDT and 1 pm EDT. Anji will focus the session on solving linear and nonlinear systems. Plus, you’ll learn how to set up and solve optimization problems. This webcast is part of the weekly education series for engineers who want to learn more about PTC Mathcad as a powerful alternative for engineering calculations. Visit the registration page to view weekly topics and sign up today to join in. REGISTER NOW!
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Announcements | 4,918 | 21,924 | {"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-2023-14 | latest | en | 0.944576 |
http://www.nachi.org/forum/f20/rule-thumb-c-sizing-question-practice-exam-4672/ | 1,419,226,902,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802774464.128/warc/CC-MAIN-20141217075254-00165-ip-10-231-17-201.ec2.internal.warc.gz | 668,629,332 | 16,952 | InterNACHI Inspection Forum Rule of Thumb A/C sizing question from Practice Exam
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Inspecting HVAC Systems Topics include heating, venting, and air conditioning inspections.
#1
5/23/06, 11:04 AM
Gregory A. Liebig InterNACHI Member Join Date: May 2006 Location: Sheboygan, WI Posts: 787
Rule of Thumb A/C sizing question from Practice Exam
I have a question regarding a question on the practice test. On page 11 there is a question that seems to conflict with the information I was looking up. The correct answer is 700 sq ft but some of the research I did supported 500 sq ft per ton. What is the correct answer? Wisconsin requires passing the National Exam and I’m doing everything I can to get ready for it within 2 weeks.
Greg Liebig
NACHI06051095
A rule of thumb for air conditioning is a ton of air conditioning is needed for:
• 700 square ft
• 500 square ft
• A one story house
• A hot day
References:
http://www.healthgoods.com/education/healthy_home_information/Space_Heating_and_Cooling/sizing_heat_and_ac.htm
The contractor asks you how many square feet of living space there are in your house. He (or she) then tells you what size unit you need. This is called "sizing by square footage" and is the most commonly used inaccurate method of sizing. A typical value used for air conditioners is one ton (12,000 Btu/hour) per 500 square feet (46 m2). This does not take into account differences among houses in design, construction, or energy efficiency.
Table 1: Capacity rules of thumb for room air conditioners
Rules of thumb for estimating air conditioner size vary by manufacturer. This example, from Carrier, includes the following suggestions for adjustments: If a room is heavily shaded, reduce capacity by 10 percent; if the room is very sunny, increase by 10 percent; and if using the unit in a kitchen, increase capacity by 4,000 Btu per hour.
Room area Capacity
(ft²) (Btu/h)
100 to 150 5,000
150 to 250 6,000
250 to 300 7,000
300 to 350 8,000
350 to 400 9,000
400 to 450 10,000
450 to 550 12,000
550 to 700 14,000
700 to 1,000 18,000
Source: Platts
#2
5/23/06, 12:17 PM
rdawes Active Poster Join Date: Nov 2004 Location: Plano, TX Posts: 425
Please Note: rdawes is a non-member guest and is in no way affiliated with InterNACHI or its members.
Re: Rule of Thumb A/C sizing question from Practice Exam
That's the rule-of-thumb we used for the Texas TREC exam. Of course, Texas is a bit warmer than some climates.
#3
5/23/06, 10:37 PM
jwilliams7 InterNACHI Member Join Date: Apr 2006 Location: Springdale, AR Posts: 36
Re: Rule of Thumb A/C sizing question from Practice Exam
I passed the National Exam about a month and a half ago, I used 500 sq feet/ton.
#4
5/23/06, 10:48 PM
Gary Reecher Active Poster Join Date: Aug 2004 Posts: 273
Please Note: Gary Reecher is a non-member guest and is in no way affiliated with InterNACHI or its members.
Re: Rule of Thumb A/C sizing question from Practice Exam
Rules of thumb are dumb.
Using Air Conditioning Contractors of America (ACCA) Manual J heat loss/gain calculation method is more precise with less chance of oversizing equipment and reducing comfort.
#5
5/24/06, 12:14 AM
Joseph Burkeson, CMI InterNACHI Member Join Date: May 2003 Location: Greater Tampa Bay Posts: 22,004
Re: Rule of Thumb A/C sizing question from Practice Exam
Quote:
Originally Posted by Gary Reecher Rules of thumb are dumb. Using Air Conditioning Contractors of America (ACCA) Manual J heat loss/gain calculation method is more precise with less chance of oversizing equipment and reducing comfort.
You Betcha!
IRC - M1401.3 Sizing.
Heating and cooling equipment shall be sized based on building loads calculated in accordance with ACCA Manual J or other approved heating and cooling calculation methodologies.
Yes, I'd personally choose a CMI every time over every other inspector. ~ Nick Gromicko, InterNACHI Founder
Certified Master Inspector - 2010
Florida Licensed Home Inspector - HI176
Florida Licensed Mold Assessor - MRSA208
Square-One Inspection "Assurance begins here"
#6
5/24/06, 8:10 AM
Michael W. Gault InterNACHI Member Join Date: Apr 2005 Location: Mount Pleasant, SC Posts: 1,093
Please Note: mgault is a non-member guest and is in no way affiliated with InterNACHI or its members.
Re: Rule of Thumb A/C sizing question from Practice Exam
Quote:
Originally Posted by Gary Reecher Rules of thumb are dumb. Using Air Conditioning Contractors of America (ACCA) Manual J heat loss/gain calculation method is more precise with less chance of oversizing equipment and reducing comfort.
Doesn't negate the fact that it's on the NHIE...
#7
5/24/06, 8:33 PM
rdawes Active Poster Join Date: Nov 2004 Location: Plano, TX Posts: 425
Please Note: rdawes is a non-member guest and is in no way affiliated with InterNACHI or its members.
Re: Rule of Thumb A/C sizing question from Practice Exam
Rules of thumb may be dumb but if I inspect a house that's a ton undersized based on that rule and we are here in hot Texas then I will call for a HVAC guy to be consulted so he can do those calcs and tell the buyer if the AC is properly sized. I won't say it is not properly sized but I will say that it should be looked into.
#8
5/24/06, 10:56 PM
Michael W. Gault InterNACHI Member Join Date: Apr 2005 Location: Mount Pleasant, SC Posts: 1,093
Please Note: mgault is a non-member guest and is in no way affiliated with InterNACHI or its members.
Re: Rule of Thumb A/C sizing question from Practice Exam
What Ronald said!
#9
5/25/06, 2:36 AM
Gregory A. Liebig InterNACHI Member Join Date: May 2006 Location: Sheboygan, WI Posts: 787
Re: Rule of Thumb A/C sizing question from Practice Exam
I also agree that one should defer if they cannot give the client the best decsion. That still doesn't take away the facts of the question on the exam.
#10
5/25/06, 2:38 AM
Gregory A. Liebig InterNACHI Member Join Date: May 2006 Location: Sheboygan, WI Posts: 787
Re: Rule of Thumb A/C sizing question from Practice Exam
Thanks Mike! That's my point exactly... | 1,622 | 6,277 | {"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-2014-52 | longest | en | 0.929485 |
http://libros.duhnnae.com/2017/jul8/150138580968-Sur-les-plus-grands-facteurs-premiers-dentiers-consecutifs.php | 1,542,369,051,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743011.30/warc/CC-MAIN-20181116111645-20181116133645-00317.warc.gz | 200,458,881 | 7,824 | # Sur les plus grands facteurs premiers dentiers consécutifs
Sur les plus grands facteurs premiers dentiers consécutifs - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
1 Analyse IECL - Institut Élie Cartan de Lorraine
Abstract : Let §P^+ n §denote the largest prime factor of the integer §n§ and §P ^+ y n§ denote the largest prime factor §p§ of §n§ which satisfies §p≤ y§. In this paper, firstly we show that the triple consecutive integers with the two patterns §P^+ n − 1 > P^+ n < P ^+ n + 1§ and §P^+ n−1 < P^+ n > P^+ n+1§ have a positive proportion respectively. More generally, with the same methods we can prove that for any §J ∈ Z, J\geq3§, the J−tuple consecutive integers with the two patterns §P^+ n + j 0 = \min {0 \leq j \leq J−1} P^+ n + j and §P^+ n + j 0 = max {0\leq j \leqJ−1} also have a positive proportion respectively. Secondly for §y = x^θ§ with §0 < θ ≤1§ we show that there exists a positive proportion of integers §n§ such that §P y^+n < P y^+n+ 1.§ Specially, we can prove that the proportion of integers §n§ such that §P^+n < P^+n + 1§ is larger than 0.1356, which improves the previous result -0.1063- of the author.
Autor: Zhiwei Wang -
Fuente: https://hal.archives-ouvertes.fr/
DESCARGAR PDF | 408 | 1,297 | {"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} | 2.625 | 3 | CC-MAIN-2018-47 | longest | en | 0.586915 |
https://oneclass.com/class-notes/ca/utsg/soc/soc-202h1/147621-soc202-sampling-distributions-confidence-intervals.en.html | 1,571,824,916,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00437.warc.gz | 617,085,281 | 95,658 | Class Notes (1,100,000)
CA (630,000)
UTSG (50,000)
SOC (3,000)
Lecture
# SOC202H1 Lecture Notes - Sampling Distribution, Standard Error
Department
Sociology
Course Code
SOC202H1
Professor
Scott Schieman
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February 5th
Sampling Distributions & Confidence Intervals
-sample to sample variability
how much variation do you have from sample to sample
calculate standard deviation; little consensus if there is a large SD
less spread around the mean = smaller SD = more consensus
-large standard error = less reliability in sample
standard deviation is numerator for formula of standard error
denominator is sample size
larger denominator = better because it spreads out more; therefore a
decreases standard error
-sampling distribution of means: taking samples over and over again and plotting the means
-large error term: large sample to sample variability and wide confidence intervals
-larger CI = size of error term
*Analytical Thinking Exercises
1) Standard error would increase; greater sample to sample variability
Size of error term would increase
Width of CI would widen
Value of upper critical limit would increase and value of lower critical limit would
decrease
3) Standard error stays the same (does not depend on CI)
Size of error term increases
Value of critical Z-score increases
Width of CI widens
Value of upper critical limit would increase and value of lower critical limit would
decrease
2.58 = 99% CI
1.96 = 95% CI
4) Standard error increases
Error term increases
CI widens
Value of upper critical limit would increase and value of lower critical limit would
decrease
Unlock to view full version | 382 | 1,700 | {"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} | 3.28125 | 3 | CC-MAIN-2019-43 | latest | en | 0.88689 |
https://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-12th-edition/chapter-2-summarizing-and-graphing-data-2-2-frequency-distributions-basic-skills-and-concepts-page-52/14 | 1,537,429,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156418.7/warc/CC-MAIN-20180920062055-20180920082055-00224.warc.gz | 749,766,943 | 12,268 | ## Elementary Statistics (12th Edition)
Published by Pearson
# Chapter 2 - Summarizing and Graphing Data - 2-2 Frequency Distributions - Basic Skills and Concepts - Page 52: 14
2, 6, 12, 12.
#### Work Step by Step
The lower two frequencies are the same as the first two frequencies, so they are 2 and 6. Now we have to find the rest two frequencies, which we can assume are peak frequencies. This can simply be done by subtracting the sum of the frequencies we know (2+6+6+2=16) from the total (40) and dividing it by two. Therefore the peak frequency is 40-16=24, 24/2=12, therefore the two peaks are 12.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 201 | 771 | {"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} | 4.15625 | 4 | CC-MAIN-2018-39 | longest | en | 0.920725 |
https://digitalcommons.kennesaw.edu/undergradsymposiumksu/2018/Posters/1/ | 1,718,204,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00250.warc.gz | 181,792,079 | 8,487 | # Improved Regional Sports Scheduling using SAS Optgraph
## Disciplines
Statistics and Probability
## Abstract (300 words maximum)
My project improves upon the tmanual cluster method of state interleague scheduling used by the Georgia State Soccer Association by use of an automated multi-round linear assignment algorithm based on SAS Optgraph.
Currently, interleague schedules are formed by manually grouping teams according to placement on a map. Typically, ten teams are assigned to a group and each team plays all of the others once per season (nine games). This approach has two flaws: 1) distance on a map does not always correlate with travel time, and 2) assigning teams to clusters precludes teams from playing teams from other groups. The second flaw becomes particularly apparent when the ideal boundaries between clusters are not obvious, but they must be drawn based on a ten-team cluster.
The improved system uses travel time between venues as the edge weight and multiple rounds of linear-assignment mapping to generate a full schedule. As Optgraph only supports bipartite linear mapping, the nodes are randomly partitioned between an A and a B subgraph, which is shuffled after each round. This approach is also compared to a non-bipartite matching algorithm written in Python.
## Academic department under which the project should be listed
CCSE - Data Science and Analytics
Dr. Joe DeMaio
## Share
COinS
Improved Regional Sports Scheduling using SAS Optgraph
My project improves upon the tmanual cluster method of state interleague scheduling used by the Georgia State Soccer Association by use of an automated multi-round linear assignment algorithm based on SAS Optgraph.
Currently, interleague schedules are formed by manually grouping teams according to placement on a map. Typically, ten teams are assigned to a group and each team plays all of the others once per season (nine games). This approach has two flaws: 1) distance on a map does not always correlate with travel time, and 2) assigning teams to clusters precludes teams from playing teams from other groups. The second flaw becomes particularly apparent when the ideal boundaries between clusters are not obvious, but they must be drawn based on a ten-team cluster.
The improved system uses travel time between venues as the edge weight and multiple rounds of linear-assignment mapping to generate a full schedule. As Optgraph only supports bipartite linear mapping, the nodes are randomly partitioned between an A and a B subgraph, which is shuffled after each round. This approach is also compared to a non-bipartite matching algorithm written in Python. | 517 | 2,656 | {"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-2024-26 | latest | en | 0.934061 |
https://www.coursehero.com/file/6616728/7-Ch-24-College-Physics-ProblemCH24-Wave-Optics/ | 1,524,438,894,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00539.warc.gz | 762,357,698 | 57,404 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
7_Ch 24 College Physics ProblemCH24 Wave Optics
# 7_Ch 24 College Physics ProblemCH24 Wave Optics - Wave...
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Wave Optics 309 24.5 (a) From d sin m θ λ = , the angle for the 1 m = maximum for the sound waves is 1 1 1 354 m s 1 sin 36.2 0.300 m 2 000 Hz sound v m m d d = = sin θ λ = = sin f ° (b) For 3.00-cm microwaves, the required slit spacing is ( )( ) 1 3.00 cm 5.08 cm sin sin36.2 m d λ θ = = = ° (c) The wavelength is sin d m θ λ = ; and if this is light, the frequency is ( ) ( ) ( ) 8 14 -6 1 3.00 10 m s 5.08 10 Hz sin 1.00 10 m sin36.2 mc c f d λ θ × = = = = × × ° 24.6 The position of the first order bright fringe for wavelength λ is 1 L y d λ = Thus, ( ) ( ) ( ) 9 1 3 700 400 10 m 1.5 m 1.5 1 m 1.5 mm 0.30 10 m L y d λ
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{[ snackBarMessage ]} | 360 | 950 | {"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} | 3.078125 | 3 | CC-MAIN-2018-17 | latest | en | 0.624806 |
https://twistedone151.wordpress.com/2007/12/ | 1,553,169,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202523.0/warc/CC-MAIN-20190321112407-20190321134407-00392.warc.gz | 641,144,472 | 14,464 | ## Archive for December, 2007
### Monday Math 1: The Isoperimetric Problem
December 31, 2007
Isoperimetric Problem
The isoperimetric problem is the problem of finding the closed plane curve with a given perimeter that encloses the greatest area. Below, I will use the calculus of variations in a proof of the long-known solution, the circle.
We let our enclosed region be D, and the enclosing curve ∂D, with orientation on ∂D chosen so that D is to it's left (counterclockwise). We thus want to minimize subject to the constraint . Note that these are quite different kinds of integrals, and cannot be combined directly via the multiplier method of including constraints.
However, using Green's theorem, which says that over a plane region D with (counterclockwise oriented) boundary ∂D we have:
By choosing , and using the fact that the area of D is , we obtain
So now we can combine the minimized quantity and the constraint using a multiplier. As both are integral constraints, the multiplier is a constant λ, so parametrizing our curve ∂D as (x(t),y(t)), so
and we want to find the extrema of
The integrand is , so we have the Euler-Lagrange equation for x:
Now, , so , and
Similarly for y:
So we have and
Squaring and summing these, we obtain:
This is the equation of a circle with radius λ and and center (C2,C1). From our original constraint , we obtain λ=p/2π.
### The Pain Beam
December 29, 2007
No, it's not the twisted creation of a mad scientist in a grade B movie, it's here. While the concept and it's execution are pretty cool, scientifically speaking; and while I'm in favor of developing non-lethal weaponry; the abuse potential of something like this is rather worrying.
### Orexin A: Substitute for Sleep…
December 29, 2007
This is a very interesting result, but it will be years, if ever, before we see medical use of these results; not to mention that we still don't know what the long-term side-effects may be. Still, anything that might help people with sleep disorders is a good thing, in my opinion.
### Pocket Veto
December 29, 2007
Here is an analysis of the constitutionality of President Bush's current “pocket veto.” Look also at the first comment and at this Overlawyered post for info on the President's key objection to the bill.
### testing again
December 29, 2007
これは日本語です。
### Physics Friday 1
December 28, 2007
Let us consider a thin hoop of radius R. On this hoop is strung a small bead of mass m which can slide along the hoop. The bead-hoop interaction is frictionless. If we place the hoop on end, and rotate it about the vertical with a constant angular velocity , what is the equation of motion for the movement of the bead on the hoop? What are the equilibrium positions, and are they stable or unstable equilibria? We decribe the position of the bead on the hoop by the angle from the bottom of the hoop.
Rotating Reference Frame Approach:
We work in the reference frame of the hoop. As this is a rotating reference frame, it is non-inertial and we have fictitious forces to consider, namely the centrifugal and Coriolis forces.
As our bead’s motion is constrained to the plane of the hoop, and the axis of the frame’s rotation is also in that plane, the Coriolis force is perpendicular to the plane, and can be ignored (as it will be entirely canceled by a normal force of the hoop on the bead, and will not affect the motion at all). Thus we have as the only relevant fictitious force the centrifugal force straight away from the axis.
We need only consider the component of net force tangential to the hoop, as only motion along the hoop is possible. The component of gravity tangential to the hoop is, and the component of the centrifugal force is , so
The equation of motion:
so:
If we define , then the above can be written as
Equilibrium angles are those where
which occurs when , so , so long as
or
Note when , we have only the point , as the two coincide.
Stability: Look at the sign of near the zeroes
1. For , we have only the one equilibrium at . We can see that for <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?0<theta, 0″> and <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?\omega^2\cos\theta-\omega_0^2<\omega^2-\omega_0^2, so <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?\ddot{\theta}, and the equilibrium is stable.
2. For \omega_0″>, we have two equilibria, at and at . We can see that for <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?0<\theta, 0″>. For <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?0<\theta, 0″>, so 0″>; for <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?\theta_0<\theta, <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?\omega^2\cos\theta-\omega_0^2, so <img src=”http://www.forkosh.dreamhost.com/mimetex.cgi?\ddot{\theta}.
and thus the equilibrium at is unstable, and that at is stable.
For a 50 cm diameter hoop, we have
or approximately one revolution a second.
### Nintendo should hire this guy…
December 28, 2007
This is very cool:
It demonstrates how powerful the motion-tracking of the Wiimote really is.
### This is an interesting video
December 28, 2007
### Testing 2…
December 10, 2007 | 1,336 | 5,223 | {"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": 52, "/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} | 3.375 | 3 | CC-MAIN-2019-13 | latest | en | 0.940278 |
https://www.stuffmakesmehappy.com/knowledgebase/can-anything-outrun-a-black-hole | 1,708,638,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.45/warc/CC-MAIN-20240222193722-20240222223722-00252.warc.gz | 1,041,847,464 | 17,401 | # Can anything outrun a black hole?
The gravitational pull of a black hole is so strong that nothing, not even light, can escape once it gets too close. However, there is one way to escape a black hole — but only if you're a subatomic particle.
## Is there anything that can escape a black hole?
Black holes are some of the universe's most fascinating objects. Black holes are dark, dense regions in space where the pull of gravity is so strong that nothing can escape. Not even light can get out of these regions.
## How fast do you need to go to outrun a black hole?
At a certain point, escape velocity is greater than the speed of light, or 186,282 miles/second (299,792 kilometers/second). For comparison, the Earth's escape velocity is about 25,000 mph (40,270 km/h) at the surface. Since nothing can go faster than light, that means nothing can escape a black hole.
## Is black hole faster than speed of light?
Using NASA's Chandra X-ray Observatory, astronomers have seen that the famous giant black hole in Messier 87 is propelling particles at speeds greater than 99% of the speed of light.
## What can destroy a black hole?
On the other hand, there is nothing we could throw at a black hole that would damage it, not even another black hole would do it, they will simply merge into a larger one (figure 2). However, there may be a way to destroy these objects, all we need to do is to wait.
## What is black hole weakness?
Its mass, M, is its sensitivity to gravity. So Q > M means gravity is the weaker of the two. From their assumption that black holes ought to be able to decay, the four physicists made a more sweeping conjecture that gravity must be the weakest force in any viable universe.
## What happens if two black holes collide?
It is possible for two black holes to collide. Once they come so close that they cannot escape each other's gravity, they will merge to become one bigger black hole. Such an event would be extremely violent.
## What is the fastest thing in the universe?
It's more honest to say that the fastest 'physical' thing in the Universe is light itself (or in fact the entire electromagnetic spectrum). Of course, the Universe has a self-imposed speed limit – the speed of light, which is 299,792.458km/s. Nothing moves faster than this.
## What is the closest black hole to Earth?
The two black holes are being called Gaia BH1 and Gaia BH2 and are the closest to Earth of any black holes found so far, ESA said. Gaia BH1 is situated only 1,560 light-years away - about thrice as close as the previous record holder. Gaia BH2 is situated 3,800 light-years away.
## What is inside a black hole?
Don't let the name fool you: a black hole is anything but empty space. Rather, it is a great amount of matter packed into a very small area - think of a star ten times more massive than the Sun squeezed into a sphere approximately the diameter of New York City.
## How long is 1 minute in a black hole?
Black hole news: Standing on edge of black hole would cause 700 years to pass in 1 minute.
## Do wormholes exist?
While researchers have never found a wormhole in our universe, scientists often see wormholes described in the solutions to important physics equations. Most prominently, the solutions to the equations behind Einstein's theory of space-time and general relativity include wormholes.
## Has anyone lost in a black hole?
Fortunately, this has never happened to anyone — black holes are too far away to pull in any matter from our solar system.
## What's beyond a black hole?
Beyond the event horizon lies a truly minuscule point called a singularity, where gravity is so intense that it infinitely curves space-time itself. This is where the laws of physics, as we know them, break down, meaning all theories about what lies beyond are just speculation.
## Does matter ever leave a black hole?
When matter falls into or comes closer than the event horizon of a black hole, it becomes isolated from the rest of space-time. It can never leave that region. For all practical purposes the matter has disappeared from the universe.
## How many black holes are in the Milky Way?
Since the Milky Way contains over 100 billion stats, our home galaxy must harbor some 100 million black holes. Though detecting black holes is a difficult task and estimates from NASA suggest there could be as many as 10 million to a billion stellar black holes in the Milky Way.
## What is the largest black hole in the Universe?
Known as TON 618, it is the most massive black hole observed so far in the Universe. NASA has revealed that it tips the scales at 66 billion times the Sun's mass!
## Is black hole reaching Earth?
Despite their abundance, there is no reason to panic: black holes will not devour Earth nor the Universe. It is incredibly unlikely that Earth would ever fall into a black hole.
## How fast is the speed of darkness?
Darkness travels at the speed of light. More accurately, darkness does not exist by itself as a unique physical entity, but is simply the absence of light. Any time you block out most of the light – for instance, by cupping your hands together – you get darkness.
## What is the oldest thing in the universe?
They made observations via the European Space Agency's (ESA) Hipparcos satellite and estimated that HD140283 — or Methuselah as it's commonly known — was a staggering 16 billion years old. Such a figure was rather baffling.
## What is the strongest thing in the universe?
These super-intense beams of gamma radiation — the gamma-ray bursts — are created by the most powerful type of stellar explosion: a hypernova.
## How long do black holes last?
In pure general relativity, with no other modifications or considerations of other physics, they remain black for eternity. Once one forms, it will just hang out there, being a black hole, forever.
## Who created dark matter?
The term dark matter was coined in 1933 by Fritz Zwicky of the California Institute of Technology to describe the unseen matter that must dominate one feature of the universe—the Coma Galaxy Cluster.
## How loud is a black hole collision?
Normally this is about 413.3 Pa s/m in 20 degree air, which would give us -4.2 dB for the stellar mass black holes. Which is below the threshold of hearing. | 1,365 | 6,292 | {"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-2024-10 | latest | en | 0.943189 |
http://www.instructables.com/id/simple-alternate-LED-flashing-using-555-timer/ | 1,531,782,365,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00325.warc.gz | 477,792,962 | 10,365 | Simple Alternate LED Flashing Using 555 Timer
10,022
16
5
Published
Introduction: Simple Alternate LED Flashing Using 555 Timer
here is the circuit diagram.!
Step 1: Components Required
1) 2 led of different colours
2) 100uf capacitor
3) 555 timer ic
4) wires
5)9v battery
6) battery connector
7) 1 k ohm resistor 1
8) 10 k ohm resistor 1
9) 220 ohm resistor 2
Step 2: Connecting the Components
1) mount the 555 ic in the middle of the bread board
2)connect pin no 7 and 8 with 1 k ohm resistor
3) connect pin 6and 7 with 10k ohm resistor
4)join 2nd and 6th pin
5) join 4th and 8th pin
6) connect pin 1 and 2 with the capacitor
7) take a separate line from the pin 3 and connect one of the 220 ohm resistor to it, and connect the anode(+) of one of the led to the other end of the resistor and connect the cathode(-) of led to pin no 2.
8) connect the anode of another led to pin 4, then connect the another 220 ohm resistor series with it
9)connect the another end of the resistor to pin no 3
Step 3: Testing
connect the positive end of the battery to pin no 8 and negative to pin no 1.
You will see the LEDs flashing alternatingly... plz try this @ home it will cost below 2\$.
hope you guys like this..
thank you..! meet you all soon with another circuit..
kathiravan...!
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5 Discussions
Could this circuit be connected to a 12v supply?
And can this circuit be connected to the main (220v) voltage using resistors and diodes without a transforme?
What are the polarity of the capacitor (to what pin we connect the capacitor +ve pole ?
What is the voltage of the 100uf capacitor and the wattage of each resistor of the circuit?
this is what ive been looking for, what is the time interval of the two leds blinking? and if i want to slower the time should i raise the amm=ount of resistor? to what cvalue? thanks | 523 | 1,886 | {"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-2018-30 | latest | en | 0.842907 |
http://eschooltoday.com/science/electricity/what-is-a-parallel-circuit.html | 1,495,973,897,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609817.29/warc/CC-MAIN-20170528120617-20170528140617-00036.warc.gz | 144,007,374 | 5,988 | RECOMMENDED LESSONS:
Parallel circuits In a parallel circuit, there is more than one resistor (bulb) and they are arranged on many paths. This means electricity (electrons) can travel from one end of the cell through many branches to the other end of the cell. Look at the illustration below involving two resistors in a parallel circuit: You will notice from the above that there is more than one path: PATH 1: A-B-C-D-E-F and back to A. PATH 2: A-B-C-G-H-D-E-F and back to A. From the above, it is clear that electricity from the cell can take either path A or Path B to return to the cell. The great thing about parallel circuits is that, even when one resistor (bulb) burns out, the other bulbs will work because the electricity is not flowing through one path. Think of all the light bulbs in your home. If one bulb burns out, the other bulbs in the rooms still work. Another great thing is that the bulbs in a parallel circuit do not dim out like the case in series circuits. This is because the voltage across one branch is the same as the voltage across all other branches.
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All Rights Reserved | 465 | 1,694 | {"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} | 3.015625 | 3 | CC-MAIN-2017-22 | longest | en | 0.844832 |
https://onlinejudge.org/board/viewtopic.php?p=21378 | 1,610,907,528,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00559.warc.gz | 486,379,765 | 13,389 | ## 160 - Factors and Factorials
All about problems in Volume 1. If there is a thread about your problem, please use it. If not, create one with its number in the subject.
Moderator: Board moderators
problem
New poster
Posts: 27
Joined: Mon Nov 10, 2003 12:40 am
### PE why ?help me.
[c]
PE help me plz.here is my source code.
/* @JUDGE_ID: xxxxxx 160 C++ */
#include<stdio.h>
#include<string.h>
int main()
{
long int i,j,n,k,p=0;
static int x[500];
memset(x,0,500);
scanf("%ld",&n);
while(n>0)
{ for(i=2;i<=n;i++)
{ j=i;
for(k=2;k<=i;k++)
{ while(j%k==0)
{ ++x[k];
j/=k;
}
}
}
printf("%2ld! = ",n);
for(i=0;i<=99;i++)
{
if(x)
{
p++;
if(p==16&&n==100)
{
printf("\n ");
p=0;
}
if(p==16)
{
printf("\n ");
p=0;
}
printf("%2d ",x);
}
/* if(n>=53&&p>=15)
{
if(n==100)
printf("\n ");
else
printf("\n ");
p=0;
} */
}
memset(x,0,500);
printf("\n");
scanf("%ld",&n);
p=0;
}
return 0;
}
[\c]
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
### 160 why WA?i think my output is exactly as sample:(
[cpp]
#include "stdio.h"
int main(int argc, char* argv[])
{
int num,temp,point,done,max,change;
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
int numOfPrime[25];
while(scanf("%d",&num))
{
max=0;
if(num==0) break;
for(temp=0;temp<25;temp++)
numOfPrime[temp]=0;
for(done=2;done<=num;done++)
{
temp=done;
while(temp!=1)
{
for(point=0;point<25 && prime[point]<=temp;point++)
{
if(temp%prime[point]==0)
{
temp/=prime[point];
numOfPrime[point]++;
if(point>max) max=point;
}
}
}
}
printf("%3d! =",num);
for(temp=0,change=0;temp<=max;temp++)
{
if(numOfPrime[temp]!=0)
{
change++;
printf("%3d",numOfPrime[temp]);
if(change==15)
{
change=0;
printf("\n ");
}
}
}
if(change!=15) printf("\n");
}
return 0;
}
[/cpp]
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
Frostina
New poster
Posts: 23
Joined: Mon Dec 15, 2003 5:21 am
why it gets wa ? plz..
[c]#include <stdio.h>
#include <math.h>
int prime[40];
int is_prime(int n) {
int i;
for (i=0;prime*prime<=n;i++)
if (!(n%prime))
return 0;
return 1;
}
void make_table(void) {
int i, index = 1;
prime[0] = 2;
for (i=3;i<=150;i+=2)
if (is_prime(i))
prime[index++] = i;
}
int main(void) {
int n, i, t, j;
make_table();
while (scanf("%d",&n)==1) {
if (!n) break;
printf("%3d! =",n);
for (i=t=0;prime<=n;i++,t=0) {
for (j=1;;j++) {
if ((int)pow(prime*1.0,j*1.0)>n)
break;
t+=n/(int)pow(prime*1.0,j*1.0);
}
printf(" %3d",t);
if (!(i%14)&&i) printf("\n ");
}
putchar('\n');
}
return 0;
}
[/c]
Thanks for your help !
junbin
Experienced poster
Posts: 174
Joined: Mon Dec 08, 2003 10:41 am
### Re: 160 why WA?i think my output is exactly as sample:(
I tested your code again my AC program and I get the following:
100! = 97 48 24 16 9 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1
My output:
100! = 95 48 22 16 9 7 5 5 4 3 3 2 2 2 2
1 1 1 1 1 1 1 1 1 1 1
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
Thanks so much,i'll recode it
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
hey,it's interesting,but seems ur AC code is wrong,here's some data,1st is the number who can be divided by 2 in 100!,2nd is the times that it can be divided by 2:
2 1
4 2
6 1
8 3
10 1
12 2
14 1
16 4
18 1
20 2
22 1
24 3
26 1
28 2
30 1
32 5
34 1
36 2
38 1
40 3
42 1
44 2
46 1
48 4
50 1
52 2
54 1
56 3
58 1
60 2
62 1
64 6
66 1
68 2
70 1
72 3
74 1
76 2
78 1
80 4
82 1
84 2
86 1
88 3
90 1
92 2
94 1
96 5
98 1
100 2
[cpp]
#include"iostream.h"
int main()
{
int a,b,c;
a=1;
c=0;
while(a!=0)
{
cin>>a>>b;
c+=b;
cout<<c<<endl;
}
}[/cpp]
copy the data and as the input of this program,u will find 97 should be right answer
????????
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
junbin
Experienced poster
Posts: 174
Joined: Mon Dec 08, 2003 10:41 am
oops.. you're right... when I determine primes, I used an auto generator to count up to 99.. so 100 is flagged as prime by mistake. :p
Now we know 100 is not in the data set..
Anyway, the reason is very obvious if you make a input file with all 100 numbers and pipe the output to a file and then edit the file.
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
then it will be another problem maybe difficult
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
so anyone can provide ur AC code to me to let me compare the output with mine?Please mail to MorningCX@hotmail.com
thanks a lot
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
junbin
Experienced poster
Posts: 174
Joined: Mon Dec 08, 2003 10:41 am
Morning wrote:so anyone can provide ur AC code to me to let me compare the output with mine?Please mail to MorningCX@hotmail.com
thanks a lot
You don't need someone else's code... just genereate a test data file from 2 to 100. Run your program with this and output to a FILE (not to standard io). Edit the file and look carefully from top to bottom.. you cannot miss the mistake.
Morning
Experienced poster
Posts: 134
Joined: Fri Aug 01, 2003 2:18 pm
Location: Shanghai China
I've done it but i can't find the mistake,for the out-put code algrithm is the same as the WA code.So i should have another AC code.
Have u correct ur code?
"Learning without thought is useless；thought without learning is dangerous."
"Hold what you really know and tell what you do not know -this will lead to knowledge."-Confucius
junbin
Experienced poster
Posts: 174
Joined: Mon Dec 08, 2003 10:41 am
Morning wrote:I've done it but i can't find the mistake,for the out-put code algrithm is the same as the WA code.So i should have another AC code.
Have u correct ur code?
I've corrected my code.. it's still AC.. the judge does not test 100.
Anyway, take a very careful look at your output.. it should be around 30-60.. the answers are correct, at least the maths part. it's the OUTPUT that is wrong.
aakash_mandhar
New poster
Posts: 38
Joined: Thu Dec 11, 2003 3:40 pm
Location: Bangalore
### 160: Factors and factorials..... WA Help Please... (Now AC)
What is wrong with my code.. I guess allignement.. It seems to work fine and logic is simple and infact it also works for inputs greater than 100 but i go WA... Please can anyone help me or tell me in which case it fails..
Thx a lot in advance
Aakash
[cpp]
Code removed. Never mind i got it accepted. The problem was that you can print only 15 digits on each line.. Silly me :) This might help others..
[/cpp]
...I was born to code...
UFP2161
A great helper
Posts: 277
Joined: Mon Jul 21, 2003 7:49 pm
Contact:
These should be right justified in fields of width 3 and each line (except the last of a block, which may be shorter) should contain fifteen numbers. Any lines after the first should be indented. Follow the layout of the example shown below exactly.
Ghust_omega
Experienced poster
Posts: 115
Joined: Tue Apr 06, 2004 7:04 pm
Location: Venezuela
### NEEEEDDDD HELP PROBLEM 160 pleaseeeeeeeeee
[java]
hi!
someone please help me with the problem 160 i have compile error in JAVA and I dont know why, if someone can tell me whats wrong, I apriciate that here my source code
import java.io.*;
import java.util.*;
class Main
{
static String ReadLn (int maxLg) // utility functito read from stdin
{
byte lin[] = new byte [maxLg];
int lg = 0, car = -1;
String line = "";
try
{
while (lg < maxLg)
{
if ((car < 0) || (car == '\n')) break;
lin [lg++] += car;
}
}
catch (IOException e)
{
return (null);
}
if ((car < 0) && (lg == 0)) return (null); // eof
return (new String (lin, 0, lg));
}
public static void main (String args[]) // entry point from OS
{
Main myWork = new Main(); // create a dinamic instance
myWork.Begin(); // the true entry point
}
void Begin()
{
String input;
StringTokenizer idata;
int a=0;
Vector numeros= new Vector();
while ((input = Main.ReadLn (255)) != null)
{
idata = new StringTokenizer (input);
a = Integer.parseInt (idata.nextToken());
if(a==0){
break;
}
if(a==1){System.out.println(" "+a+"!"+" = "+1);}
else {
Vector v= new Vector();
int [] a1= new int [a];
int j=2;
for(int x=0;x!=a-1;x++){
a1[x]=j;
j++;
}
int i=0,m=0,l=0;
while(i!=a-1){
j=i+a1;
if(a1==1) {i++; }
else { Double k= new Double((a-1)/a1);
m= k.intValue();
do{
if(j<=a-1&&a1[j]!=0){
if(a1[j]%a1==0){
a1[j]=a1[j]/a1;
Integer aux= (Integer)v.lastElement();
int u = aux.intValue();
v.remove(v.lastElement());
u++;
}
else{
l++;
j=j+a1;
}
}
else break;
}
while(j<=a-1);
i++;
}
}
if(new Integer(a).toString().length()==1)System.out.print(" "+a+"!"+" =");
if(new Integer(a).toString().length()==2)System.out.print(" "+a+"!"+" =");
if(new Integer(a).toString().length()==3)System.out.print(a+"!"+" =");
for(int p=0;p!=v.size();p++){
System.out.print("\n");
System.out.print(" ");
}
if(v.get(p).toString().length()==1)System.out.print(" "+v.get(p));
if(v.get(p).toString().length()==2)System.out.print(" "+v.get(p));
if(v.get(p).toString().length()==3)System.out.print(v.get(p)); | 3,200 | 9,680 | {"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-2021-04 | latest | en | 0.459137 |
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slide2
# slide2 - Chapter 2 Organizing and Graphing Data Bin Wang...
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Chapter 2: Organizing and Graphing Data Bin Wang [email protected] Department of Mathematics and Statistics University of South Alabama Mann E7 1/33
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2.1 Raw data Definition (Raw Data (source data, ungrouped data)) Data recorded in the sequence in which they are collected and before they are processed or ranked are called raw data . Mann E7 2/33
2.1 Raw data Definition (Raw Data (source data, ungrouped data)) Data recorded in the sequence in which they are collected and before they are processed or ranked are called raw data . Mann E7 2/33
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2.1 Raw data Definition (Raw Data (source data, ungrouped data)) Data recorded in the sequence in which they are collected and before they are processed or ranked are called raw data . Mann E7 2/33
2.2 Organizing and Graphing Qualitative (Categorical) Data Definition A frequency distribution for categorical data lists all categories and the number of elements that belong to each of the categories. Mann E7 3/33
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2.2 Organizing and Graphing Qualitative (Categorical) Data Definition A frequency distribution for categorical data lists all categories and the number of elements that belong to each of the categories. Example Table 2.3 (p. 28) Type of employment students intend to engage In Mann E7 3/33
2.2 Organizing and Graphing Qualitative (Categorical) Data Example Table 2.4 (p. 29) Frequency Distribution of Stress on Job Mann E7 4/33
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2.2.2 Relative Frequency and Percentage Distributions Definition The relative frequency for a particular category is the fraction or proportion of the time that the category appears in the data set. It is calculated as relative frequency = frequency of the category sum of all frequencies . Mann E7 5/33
2.2.2 Relative Frequency and Percentage Distributions Definition The relative frequency for a particular category is the fraction or proportion of the time that the category appears in the data set. It is calculated as relative frequency = frequency of the category sum of all frequencies . When the table includes relative frequencies, it is sometimes referred to as a relative frequency distribution. Mann E7 5/33
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2.2.2 Relative Frequency and Percentage Distributions Definition The relative frequency for a particular category is the fraction or proportion of the time that the category appears in the data set. It is calculated as relative frequency = frequency of the category sum of all frequencies .
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http://www.jiskha.com/display.cgi?id=1307312412 | 1,498,277,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320215.92/warc/CC-MAIN-20170624031945-20170624051945-00642.warc.gz | 538,010,417 | 3,920 | # Physics
posted by .
You hear a beat produced by 2 sources: 342.6 and 351.8 Hz. What is the period of the resulting beat tone? -- is there a formula for this? If so what is it and how do you solve the question?
• Physics -
beats occur in nonlinear mediums, you get the sum, and the difference frequency. Often, we just look at the difference frequency as the beat...here it is 351.8-342.6=9.2hz
period then is 1/9.2 seconds
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Question
# In figure, if AB∥CD,∠APQ=50o and ∠PRD=127o, find x and y.
A
x=50o
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B
x=77o
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C
y=50o
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D
y=77o
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Solution
## The correct options are A x=50o C y=77oGiven, AB∥CD,∠APQ=50ox=∠APQ (Corresponding angles) x=50o x+y=127o (Triangle's angle sum rule) y=127o−x=127o−50o=77o
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=Maplets/Examples | 1,462,076,277,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860114285.32/warc/CC-MAIN-20160428161514-00071-ip-10-239-7-51.ec2.internal.warc.gz | 650,655,861 | 24,605 | Overview of the Maplets[Examples] Package - Maple Help
Home : Support : Online Help : Programming : Maplets : Examples : Maplets/Examples
Overview of the Maplets[Examples] Package
Calling Sequence Maplets[Examples][command](arguments) command(arguments)
Description
• The Maplets[Examples] package contains commands that display Maplet applications and shows how they can be used.
• The Using the Example Maplets worksheet demonstrates the use of the Maplets[Examples] commands.
• Each command in the Maplets[Examples] package can be accessed by using either the long form or the short form of the command name in the command calling sequence.
As the underlying implementation of the Maplets[Examples] package is a module, it is also possible to use the form Maplets:-Examples:-command to access a command from the package. For more information, see Module Members.
List of Maplets[Examples] Package Commands
The following is a list of available commands.
To display the help page for a particular Maplets[Examples] command, see Getting Help with a Command in a Package.
Sample Maplet Application Worksheets
These worksheets demonstrate how to write Maplet applications that behave similarly to those produced by the commands in the Maplets[Examples] package.
List of Maplets[Examples][LinearAlgebra] Package Commands
• The Maplets[Examples][LinearAlgebra] package provides visual interfaces to some routines in the LinearAlgebra package.
The following is a list of available commands.
• For more information on the commands in this subpackage, see Overview of the Maplets[Examples][LinearAlgebra].
Sample Maplet Application Worksheets
These worksheets demonstrate how to write Maplet applications that behave similarly to those produced by the commands in the Maplets[Examples][LinearAlgebra] package.
Examples
> $\mathrm{with}\left(\mathrm{Maplets}[\mathrm{Examples}]\right):$
> $\mathrm{GetEquation}\left(\right)$
${{x}}^{{2}}{-}{4}{}{x}{+}{3}{=}{0}$ (1) | 448 | 1,990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 2.53125 | 3 | CC-MAIN-2016-18 | longest | en | 0.73337 |
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## Goal
There is a building with N floors. You have X identical eggs.
One of the floors (from 0 to N) is the highest in the building you can drop an egg from without breaking it. If you drop the egg from a higher floor, the egg will break. If you drop the egg from this floor or below, the egg will not be broken. If an egg is not broken after the drop, it keeps its same physical condition and you can use it again.
What is minimal amount of egg drops required in worst case to find out this highest floor from which the egg will not be broken?
It is recommended to solve "The Two Egg problem" first.
Input
Line 1: Integer N - number of floors in a building.
Line 2: Integer X - number of available eggs.
Output
Minimal amount of egg drops in worst case required to find out the highest floor from which the egg will not be broken.
Constraints
1 ≤ N ≤ 1000000
1 ≤ X ≤ 20
Example
Input
100
1
Output
100
A higher resolution is required to access the IDE
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# The Mathematics Behind the Biochemistry of DNA
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The Mathematics Behind the Biochemistry of DNAProblems concerning the understanding of scientific evidence in forensic science are investigated with reference to measures of improbability related with the presentation of such evidence in an adversarial perspective. The investigation includes the use of probabilistic arguments associated with expert scientific testimony in the courts. In Scotland, forensic medicine students were given problems presented in court in order for them to learn about uncertainties.
These problems were supported by scientific evidences for (Taroni and Aiken 169).
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How is mathematics a useful tool in understanding such problems? As mentioned above, probability is being used to interpret the data which derived from the scientific analysis, usually done in the laboratory with forensic experts or chemists, with the application of established chemical reactions and powerful methods of chemistry and biotechnology. The theories of probability are proven by the laws of mathematics.The likelihood of a specific occasion to happen is probability is the chance for an event to happen.
It can be written in fractions or as decimals between 0 (an impossible event) and 1 (a sure event). Decimals can be further written as percentages. The probability can be calculated as:The probability of getting heads in one toss is: p(heads) = 1/(1 + 1) = 1?2.Craven mentioned in his Primer on Probability for Discrete Variables that the Bayes theorem is an effective tool in estimating the likelihood of events.
In theories of statistics, some conditions are hard to estimate for its likelihood, but the Bayes theorem solves it. An example of events which are hard to estimate for its probability is the case of the DNA matching. This paper will only give the importance of the Bayes theorem to somehow appreciate the application of mathematics in the filed of statistics.The DNA fingerprint is the identity of a human being, in which he can be differentiated from other humans.
It is commonly encountered on movies concerning the real identity of an individual, to whom must he really belong, who is the real mother of this individual, whatsoever. The last resort to finally end the arguments is the DNA matching. DNA matching has been extensively done since DNA sequencing and matching was discovered.Let us have a comparison of the evidence and a reference or the sample which is of the known source, such as a hair sample from a suspect.
Considering a rape case, we could get seminal fluid from the rape victim or from the crime scene to compare with the DNA present in the blood of the suspect or from the fresh semen obtainable from the suspect if he would allow it.If the two samples produce an indistinguishable DNA profile, then it is strongly evidenced that the unknown and the known sample came from a common source, and the contributor of the unknown sample is the suspect.Are we confident of the results? From the Forensic Mathematics of DNA Matching by Brenner, when a DNA profile includes rarely a mixture of traits, we can conclude that the suspect is the contributor. It would be better for us to say that the questionable probably be isolated from the wrongdoing and that he matches only by probability.
Deoxyribonucleic acid or more commonly known as DNA is composed of nucleic acid sequences. The first evidence that DNA is a hereditary material in the body was provided in 1944 by scientist Oswald Avery, Colin MacLoed and Maclyn McCarty. Other features of the DNA like its double helical structure were revealed in 1953 by James D. Watson and Francis H.
C. Crick with the help of Rosalind Franklin, an X-ray crystallographer. Watson and Crick had the Nobel Prize in Physiology or Medicine in 1962 with Maurice H.F.
Wilkins (). These discoveries led to the extensive research of the applications of DNA in terms of its forensic value.Instead of knowing the whole sequence of the DNA, a percentage of each allele present in the sample is taken into consideration. This is useful if there are many alleles to consider and there are a lot of suspects to observe.
The frequency of each allele as observed in the sample is tallied and divided to the total number of alleles present, not considering what alleles there are. The ratios will be compared to what was seen in the crime scene; then the investigator will have an idea of who the suspect more likely is.The mathematics behind this is very easy. In this manner we could appreciate chemistry, and especially the simple mathematics underneath.
Here presented are simple to complicated mathematical applications which make easy the tasks of biochemists and forensic experts in their fields. Numerical representations will help in concretizing the application of math in DNA matching.Profile probability of the DNAA match would occur by chance. The following table will give us the data to support the previous statement.
It is easiest to illustrate by example how the probability is determined: Allele 10 at the locus CSF1PO was found 109 times out of 432 alleles which is equivalent to 216 people. There is enough reason to say that there is a p=0.25 estimated chance that a CSF1PO allele selected at random would be a 10. Furthermore, a 0.
31 chance (q) of a random CSP1PO allele to be 11 can be estimated. Before we type the suspect, if we take into account that he is not the source of the evidence, then we may think of him as someone who randomly received a CSF1PO allele from his parents. The chance of receiving either 10 from his mother and 11 from his father is the product of p and q (pq). In this, since the chance for the mother and father can be interchanged since they are just mere chances; 2pq formula arises, which will give genotype frequency locus probability.
When the numbers of alleles observed are the same, a formula of p2 is adapted. The locus TPOX illustrates this. The term pp or p2 shows the probability of getting allele 8 from each parent combined. Twenty-eight percent of people have similar TPOX genotype as that of the evidence.
There is about 4% chance that a person will have a combined genotype of the two loci. It is derived by multiplying 0.28 by 0.16 or 28% of 16%.
The chance for a multiple-locus genotype is achieved by multiplication – by getting the product of the occurrence of the per-locus genotypes and by multiplying by 2 for the event of a heterozygous locus.Same calculations applied for THO1 and vWA loci, and taking these alleles into account makes a person chosen in random to have the combined genotype from 4% down to approximately 1/7000 (the actual precise value is 1/7142.86).The example given distrusts the notion of the DNA tests being the best way to identify a person in the crime scene.
This is not to say that the technique is disqualified from all applications but it showed us clearly that this technique has a flaw and is not perfect at all.The overall profile frequency for this case is 0.00014 or about 1/7000. Therefore, we can say that either the suspect contributed the evidence, or an improbable coincidence happened – the once-in-7000 coincidence that an unrelated person would by chance have the same DNA profile as that obtained from the evidence.
The Prosecutor’s Misleading NotionWriters in newspapers incorrectly interpret on their own the DNA matching results; in this case, that there is only 1 chance out of 7000 that the semen was left by a person other than the suspect. The prosecutor’s fallacy is being committed this way. It is not true for the reason that it acts as if the DNA evidence alone can be an enough criterion to declare that the suspect might be the donor with great probability, which unreasonably tells us that extra substantiation in the case cannot cause any change.Logically, we cannot rely solely on DNA evidence alone as verification – we still need additional information, which may be very small as needed.
We can assume that the suspect was jailed, or he was an ex-convict, before his DNA analysis, which can be a proof already against him.The Legal Representative’s Way OutBrenner also mentioned into his article that the defense sometimes tries to lessen the influence of 7000 to one matching. A nice argument would be that there are many men in the city which could probably have that kind of profile, so there is a small chance that the client is the source of the semen. This is to consider that every man in the city have same chances of being on the place of the crime.
Errors in processing the sample or interpreting the outcome may be a source of forensic DNA testing false positives. A model such as the Bayesian model show how the potential for a false positive affects the factual value of DNA evidence and the satisfactoriness of DNA evidence to meet sufficiently the traditional legal standards for conviction. The Bayesian analysis is contradictory with the “false positive fallacy,” an instinctively appealing but flawed alternative interpretation. The results show the relevance of having accurate data regarding both the random match probability and the false positive probability when looking at the DNA evidence.
It is emphasized that disregarding or undervaluing the potential for a false positive can lead to serious errors of interpretation, chiefly when the suspect is identified through a “DNA dragnet” or by searching the database, and that ignoring the true rate of error imposes an important element of improbability about the correctness of the DNA evidence .SummaryOne of the applications of mathematics is the computation of the probability for an event to occur. Probability is used on the other hand to venture on the applications of DNA matching. It is the common notion that DNA matching is a perfect technique in forensic investigations.
This paper gives light to the small, yet probable mistake that this technique could bring. This paper will make people appreciate the little application of mathematics in this field, yet really important one. Mathematics as a language of science is being revealed because it is not that emphasized when analysis are made. This will somehow give credit to those biologists and chemists went through in studying the DNA, and especially the mathematicians and statisticians who devoted their lives in this endeavour to widen the horizon for the possibilities of the human mind.
Works CitedBrenner, Charles H. “Forensic mathematics of DNA matching”. 23 April 2008. <http://dna-view.
com/profile.htm>Craven, Mark. “Primer on Probability for Discrete Variables.” September 2007.
24 April 2008. <http://www.biostat.wisc.
edu/bmi576/lectures/intro-probability.pdf>Franco Taroni, Silvia Bozza and Alex Biedermann. “Probabilistic reasoning in the law. Part 1: Assessment of probabilities and explanation of the value of DNA evidence.” Journal of Forensic Sciences 38(3) (1998): 165-77.
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# MIT Sloan School of Management
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17 Jun 2010, 20:35
I'm interested in working in a tech consulting firm after my MBA such as Deloitte, Accenture, Booz Allen, and so on. I was wondering how much the presence in campus of those companies is.
From what I've seen in employment reports most tech companies at MIT are Google, Microsoft style.
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13 Jun 2011, 10:36
MIT ambassadors, is this topic stile alive? I'd like to ask a few questions from students with concentration in Supply Chain / Operations
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13 Jun 2011, 13:36
Likewise would like to know if this thread is still alive. I would love to learn about entrepreneurship track and also technology innovation side of things at MIT.
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https://number.academy/1000047 | 1,702,251,908,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00708.warc.gz | 479,332,884 | 11,848 | # Number 1000047 facts
The odd number 1,000,047 is spelled 🔊, and written in words: one million and forty-seven, approximately 1.0 million. The ordinal number 1000047th is said 🔊 and written as: one million and forty-seventh. The meaning of the number 1000047 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1000047. What is 1000047 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,000,047 in other units
The decimal (Arabic) number 1000047 converted to a Roman number is (M)XLVII. Roman and decimal number conversions.
#### Weight conversion
1000047 kilograms (kg) = 2204703.6 pounds (lbs)
1000047 pounds (lbs) = 453618.3 kilograms (kg)
#### Length conversion
1000047 kilometers (km) equals to 621401 miles (mi).
1000047 miles (mi) equals to 1609421 kilometers (km).
1000047 meters (m) equals to 3280955 feet (ft).
1000047 feet (ft) equals 304819 meters (m).
1000047 centimeters (cm) equals to 393719.3 inches (in).
1000047 inches (in) equals to 2540119.4 centimeters (cm).
#### Temperature conversion
1000047° Fahrenheit (°F) equals to 555563.9° Celsius (°C)
1000047° Celsius (°C) equals to 1800116.6° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1000047 seconds equals to 1 week, 4 days, 13 hours, 47 minutes, 27 seconds
1000047 minutes equals to 2 years, 3 weeks, 1 day, 11 hours, 27 minutes
### Codes and images of the number 1000047
Number 1000047 morse code: .---- ----- ----- ----- ----- ....- --...
Sign language for number 1000047:
Number 1000047 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1000047
### Multiplications
#### Multiplication table of 1000047
1000047 multiplied by two equals 2000094 (1000047 x 2 = 2000094).
1000047 multiplied by three equals 3000141 (1000047 x 3 = 3000141).
1000047 multiplied by four equals 4000188 (1000047 x 4 = 4000188).
1000047 multiplied by five equals 5000235 (1000047 x 5 = 5000235).
1000047 multiplied by six equals 6000282 (1000047 x 6 = 6000282).
1000047 multiplied by seven equals 7000329 (1000047 x 7 = 7000329).
1000047 multiplied by eight equals 8000376 (1000047 x 8 = 8000376).
1000047 multiplied by nine equals 9000423 (1000047 x 9 = 9000423).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1000047
Half of 1000047 is 500023,5 (1000047 / 2 = 500023,5 = 500023 1/2).
One third of 1000047 is 333349 (1000047 / 3 = 333349).
One quarter of 1000047 is 250011,75 (1000047 / 4 = 250011,75 = 250011 3/4).
One fifth of 1000047 is 200009,4 (1000047 / 5 = 200009,4 = 200009 2/5).
One sixth of 1000047 is 166674,5 (1000047 / 6 = 166674,5 = 166674 1/2).
One seventh of 1000047 is 142863,8571 (1000047 / 7 = 142863,8571 = 142863 6/7).
One eighth of 1000047 is 125005,875 (1000047 / 8 = 125005,875 = 125005 7/8).
One ninth of 1000047 is 111116,3333 (1000047 / 9 = 111116,3333 = 111116 1/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
1000047
#### Is Prime?
The number 1000047 is not a prime number. The closest prime numbers are 1000039, 1000081.
#### Factorization and factors (dividers)
The prime factors of 1000047 are 3 * 333349
The factors of 1000047 are 1, 3, 333349, 1000047.
Total factors 4.
Sum of factors 1333400 (333353).
#### Powers
The second power of 10000472 is 1.000.094.002.209.
The third power of 10000473 is 1.000.141.006.627.103.872.
#### Roots
The square root √1000047 is 1000,0235.
The cube root of 31000047 is 100,001567.
#### Logarithms
The natural logarithm of No. ln 1000047 = loge 1000047 = 13,815558.
The logarithm to base 10 of No. log10 1000047 = 6,00002.
The Napierian logarithm of No. log1/e 1000047 = -13,815558.
### Trigonometric functions
The cosine of 1000047 is -0,886323.
The sine of 1000047 is 0,463068.
The tangent of 1000047 is -0,52246.
### Properties of the number 1000047
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1000047 in Computer Science
Code typeCode value
1000047 Number of bytes976.6KB
Unix timeUnix time 1000047 is equal to Monday Jan. 12, 1970, 1:47:27 p.m. GMT
IPv4, IPv6Number 1000047 internet address in dotted format v4 0.15.66.111, v6 ::f:426f
1000047 Decimal = 11110100001001101111 Binary
1000047 Decimal = 1212210210210 Ternary
1000047 Decimal = 3641157 Octal
1000047 Decimal = F426F Hexadecimal (0xf426f hex)
1000047 BASE64MTAwMDA0Nw==
1000047 MD5e0b4754cb5395a920dcd421aaa193079
1000047 SHA2248bb8a82abc9b83e6eb217116752c5ed3e88d686aebc93aa4d8ac4ca3
1000047 SHA256be982fa6e35f84580045ba88bd5eb707e88d0bfc25ce78e848831744b79c412e
1000047 SHA384c140b83ac622e0d9063c8e2159a03ef90f2febb64bf087f2eaf5123e1c89e9ddc7389436ece186a5874b1b9b8ab9265e
More SHA codes related to the number 1000047 ...
If you know something interesting about the 1000047 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1000047
### Character frequency in the number 1000047
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 4 4 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1000047, the numbers 1+0+0+0+0+4+7 = 1+2 = 3 are added and the meaning of the number 3 is sought.
## № 1,000,047 in other languages
How to say or write the number one million and forty-seven in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.000.047) un millón cuarenta y siete German: 🔊 (Nummer 1.000.047) eine Million siebenundvierzig French: 🔊 (nombre 1 000 047) un million quarante-sept Portuguese: 🔊 (número 1 000 047) um milhão e quarenta e sete Hindi: 🔊 (संख्या 1 000 047) दस लाख, सैंतालीस Chinese: 🔊 (数 1 000 047) 一百万零四十七 Arabian: 🔊 (عدد 1,000,047) واحد مليون و سبعة و أربعون Czech: 🔊 (číslo 1 000 047) milion čtyřicet sedm Korean: 🔊 (번호 1,000,047) 백만 사십칠 Dutch: 🔊 (nummer 1 000 047) een miljoen zevenenveertig Japanese: 🔊 (数 1,000,047) 百万四十七 Indonesian: 🔊 (jumlah 1.000.047) satu juta empat puluh tujuh Italian: 🔊 (numero 1 000 047) un milione e quarantasette Norwegian: 🔊 (nummer 1 000 047) en million og førti-syv Polish: 🔊 (liczba 1 000 047) milion czterdzieści siedem Russian: 🔊 (номер 1 000 047) один миллион сорок семь Turkish: 🔊 (numara 1,000,047) birmilyonkırkyedi Thai: 🔊 (จำนวน 1 000 047) หนึ่งล้านสี่สิบเจ็ด Ukrainian: 🔊 (номер 1 000 047) один мільйон сорок сім Vietnamese: 🔊 (con số 1.000.047) một triệu lẻ bốn mươi bảy Other languages ...
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Jeffrey1992 June 23, 2012 10:14
Mixture and Decay of a tracer in a Water Tank
Hello :)
I have a problem that i cannot answer yet.
My simulation is a water tank, circular, with Diameter of 10m and a inlet at a center-bottom, with diameter of 0.125m.
There is a mixture of water+tracer entering at inlet, with concentration of 3mg/l. (MixtureInlet)
But, in the water tank, there is another mixture of water+tracer, with a concentration of 4mg/l. (MixtureInside)
Initially, the water lever of the mixture that was alredy in the tank was 9,85m (my tank has 10m high), the water rises until 10m, and then, water goes out until the water level is 3,75m.
The two mixtures will mix, and i will analyse the decay of the tracer with time.
My problem is, the mass fraction of inlet mixture is constant,so, I created a mixture fixed values (Mass fraction of tracer = 2x10^-6).
The mixture INSIDE the tank will mix and decay, so, i created a mixture with variable values.
The Mass fraction, initialy at the tank, before entering any water, is 3x10^-6.
So, i will create a equation that simulates the decay.
But, for that , i will use C= Co. e^-kt
I am trying now, to put the initial value of mass fraction of mixture in tank, 3x10^-6, PLUS the mass fraction of water entering in the water tank.
But it is not working ;/
I tried (Tracer.MixtureInlet.Volume Fraction + 3x10^-6).e^-kt, but it is not working.
Any ideas?
ghorrocks June 24, 2012 07:49
I do not completely understand your question - do you want the mass fraction of the incoming material to decay over time?
This sounds straight forward. Please post the output file which shows the error.
Felggv July 27, 2012 13:31
Hello,
First of all, are you familiar with the CEL language? Do you know the variables you're allowed to use? Do you know how to write them properly?
Shouldn't you use Mass Fraction instead of Volume Fraction?
Describe how your materials are set! Do you have water + tracer as a variable composition mixture? Is water set as constraint?
Give us more details of how did you set your parameters, so help will come easier. :)
Good luck,
Felipe.
All times are GMT -4. The time now is 00:19. | 644 | 2,422 | {"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-2015-35 | longest | en | 0.904087 |
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# Equation of a Circle
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How the equation of a circle is derived given that the circle has centre O (0, 0) and O (a, b)
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### Equation of a Circle
1. 1. Equation of Circle Additional Mathematics
2. 2. Distance between Two Points 5 y 4 3 2 1 x-5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 By Phytagoras Theorem: -3 -4 -5
3. 3. Equation of Circle with Centre (0, 0) 6 y 5 4 3 2 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 Distance -2 -3 -4 -5 -6 What can you conclude from this pattern?
4. 4. Equation of Circle with Centre (0, 0) 6 y By applying Pythagoras theorem, 5 Distance 4 3 2 As the centre of the circle (0, 0) 1 and the radius OD, then the x equation of the circle:-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6
5. 5. Find an equation of the circle with center at (0, 0) thatpasses through the point (-1, -4).Since the center is at (0, 0) well have 6 y 2 2 2x y r 5 4The point (-1, -4) is on the circle then 3we may substitute the coordinate on 2the equation to find the radius of the 1circle x -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 2 2 2 -1 1 4 r -2 1 16 17 -3 -4 Subbing this in for r2 we have: -5 -6 2 2 x y 17
6. 6. Equation of Circle with Centre (a, b) 6 y 5 4 3 2 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 Distance -2 -3 -4 -5 -6
7. 7. Equation of Circle with Centre (a, b) By applying distance formula, 6 y Distance 5 4 3 As the centre of the circle (a, b) 2 and the radius OD, then the equation of the circle: 1 x-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6
8. 8. Find an equation of the circle with center at (2, -1) andradius 4.Substitute the center and the radius into 6the general equation y 5 4 2 -1 4 3 2Expand and simply the equation to 1 xgive alternative form: -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 | 796 | 2,081 | {"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} | 4.15625 | 4 | CC-MAIN-2018-13 | latest | en | 0.762706 |
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# Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
Sal solves the system of equations 4x - 2y = 5 and 2x - y = 2.5 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• What is the point of standard form? It seems to me that all it is good for is solving systems of equations like this. But can't you just use slope-int form?
• Plus when you get to matrices, standard form is the best/only way to fromulate them
• I cannot figure this out at all:
-5x +2y = 22
9x + 8y= 30
• The answer above was not correct... you take -5x+2y=22 and multiply the whole equation by four to get: -20x+8y=88, and then you can subtract the other original equation: 9x+8y=30 from it... and you get -29x=58, so you divide -29 on both sides, which gives you x=-2... then to find y you simply plug this into the equation: 9(-2)+8y=30 and you get y=6... hope that helps sorry it's 9 months late though!;)
• what if the problem is a ratio
• At , instead of changing 4x - 2y = 5 to slope-intercept form in order to graph it, how (if you even can) would you graph the equation while keeping it in standard form?
• It is possible to do that but it is harder. So just put it into slope intercept form.
(1 vote)
• if -3x-6y=-63, how do you know if the "6y" is negative?
(1 vote)
• You always take the sign that is before the number, for example you can separate the expression into -3x and -6y.
• if the answer is o=o is there any solutions?
(1 vote)
• If you end up with 0 = 0, you will have infinite solutions.
• What is the solution of 4x+y=57 and 7x-2y=21?
(1 vote)
• First at all,we need to simplify both equations. The first one we will isolated the y, and we get y=-4x+57 and the second one y=3.5x -10.5.And we need to multiply -1 to any one of the equations.I choose to do with the second one, then -y=-3.5x+10.5,and add this to the first one,we get 0=-7.5x+67.5,and you can solve for x. x=9 Then substitute x to any of this equation.You will get y=21.You can check it out.
Remember this method!
• How do i solve:
8x-7y=21
2x=3y+4
using addition/elimination? can u help me with the steps?
• 8x-7y=21 2x=3y+4
So. with the first equation i'll multiply it by 3, the second, i'll multiply it by 7, that should cancel the Ys out when i add the two equations.
8x-7y=21 = 24x-21y=63 2x=3y+4 = 14x=21y+28 Now i'll add the equations together.
24x-21y=63 +
28+21y=14x =
24x+28=14x+63 Now i'll subtract 14x from both sides, then 28 from both sides.
24x+28=14x+63
-14x -14x =
10x+28=63
-28 -28
10x=35 i'll divide by 10 and then if i haven't messed up we'll have your x value.
10x=35
____
10 =
x=3.5 So now i'll substitute 3.5 for x to solve for y.
7=3y+4 That's your second equation, with 3.5 for x, therefore 7 for 2x. i'll subtract 4.
-4 -4 =
3=3y Divide by 3 and...
__
3 =
1=y Now i'll see if that checks out with you first equation (x=3.5, and y=1)
+7 +7
28=28 Yes it does, there's your answer and i hope that helps make things a bit clearer.
Good fortune problem solving!
(1 vote)
• what are the rules to follow when calculating simultaneous equations
(1 vote)
• Can you always use both adding and subtraction to do elimination?
(1 vote)
• Yes, in the end you get the same answer either way.
(1 vote)
## Video transcript
We're told to solve and graph the solution for the system of equations right here. And the first thing that jumps out at me, is that we might be able to eliminate one of the variables. And if we just focus on the x, we have a 4x here and we have a 2x right here. If we were to just add them right now, we would get a 6x. So that wouldn't eliminate it. But if we can multiply this 2x by negative 2, it'll become a negative 4x, and then when you add it, they would cancel out. So let's multiply this equation, this second equation, by negative 2. So I'm going to multiply both sides of this equation by negative 2. And the whole motivation is so that this 2x becomes a negative 4x. And, of course, I can't just multiply only the 2x. Anything I do to the left-hand side of the equation I have to do to every term, and I have to do to both sides of the equation. So the second equation becomes negative 4x-- that's negative 2 times 2x-- plus-- we have negative 2 times negative y-- which is plus 2y is equal to 2.5 times negative 2, is equal to negative 5. I just rewrote the second equation, multiplying both sides by negative 2. Now, this top equation-- I'll write it on the bottom now-- we have 4x minus 2y is equal to positive 5. And now we can eliminate it. We can say, hey, look, the negative 4x and the positive 4x should cancel out, or they will cancel out. So let's add these two equations. Let's add the left side to the left side, the right side to the right side, and we can do that because these two things are equal. We're doing the same thing to both sides of the equation. So what do we get? If we take our negative 4x plus our 4x, well, those cancel out. So you're left with nothing. Maybe I could write a 0 there. 0x if you want. And then you have your plus 2y and your negative 2y. Those also cancel out. So you're also left with 0y. And then that equals negative 5 plus 5 is equal to 0. So this just simplifies to 0 equals 0, which is true, but it's kind of bizarre. We had all these x's and y's. Everything canceled out. So let's explore this a little bit more. Let's graph it and see what this 0 equals 0 is telling us when we try to solve this system of equations. So let me graph this top guy. I'll do it in blue. So right now it's in standard form. Let's put it in slope-intercept form. So we have 4x minus 2y is equal to 5. Let's subtract 4x from both sides. I want the x terms on the right-hand side. So then I'm left with negative 2y is equal to negative 4x plus 5. Now we can divide both sides by negative 2. And we are left with y is equal to positive 2x, right, that's positive 2x, minus 2.5. So let's graph that. The y-intercept is negative 2.5. So negative 2.5 right there, and then it has a slope of 2. So if we move up 1, if we move up in the x-direction, if we move to the right 1 in the positive x-direction, we will move up 2. So 1, 2. Right there. And if we were to do it again, we move up 1, 2. Just like that. So the line's going to look something like this. I'll try my best to draw a straight line. This is the hardest part about a lot of these problems. There you go. So that's the top equation. Now, let me draw the bottom equation. Let me draw and I'll do it in this green color. So this bottom equation was 2x minus y is equal to 2.5. And we can subtract 2x from both sides. The left-hand side becomes negative y is equal to 2x plus-- or is equal to negative 2x plus 2.5. Now let's multiply or divide both sides by negative 1. And you get y is equal to positive 2x minus 2.5. And let's try to graph this, and you already might notice something interesting about these two equations. You try to graph this, the y-intercept is at negative 2.5, right there. The slope is 2. So it's going to be this exact same line. And you saw that algebraically. I didn't have to graph it. These two lines have the exact same equation when you put them in slope-intercept form. That's the first equation. That's the second equation. So what this 0 equals 0 is telling us is actually that these are the same line. That these actually have an infinite number of solutions. Any point on this line, which is both of those lines, will satisfy both of these equations. You give me an arbitrary y, solve for x in the top equation, that x and y will also satisfy the bottom equation. So this actually has an infinite number of solutions. These are the same line. | 2,265 | 7,881 | {"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} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.928505 |
https://thedifferenceengine.eu/engineer/understanding-the-deflection-of-beams.html | 1,653,116,919,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662538646.33/warc/CC-MAIN-20220521045616-20220521075616-00267.warc.gz | 627,842,881 | 18,553 | Understanding the Deflection of Beams
Thanks to Brilliant for sponsoring this video. Beams are used to carry loads in a whole range of different structures, and it’s important for engineers to be able to predict how much a beam will deflect when loads are applied to it. Allow a beam to deflect too much and it could cause damage to other parts of the structure, feel unsafe to the user, or fail to meet its intended function, even if stresses in the beam are low and there’s no risk of failure.
Construction codes typically define what the maximum allowable deflection is, and so it’s up to the engineer to figure out how much the beam will deflect. There are quite a few different approaches that can be used to do this. I’ll cover five of the more common methods in this video. All of these methods make the same important assumptio
n, which is that deflections are small and stresses remain in the elastic region. For this reason, the deformed shape of the beam is called the elastic curve. We can define x as the distance along the axis of the undeformed beam and y as the vertical deflection of the beam.
The objective is to figure out how y varies along the length of the beam. We will assume that there’s no displacement along the x axis. We’re also interested in the angle of rotation of the beam, Theta. Let’s take a closer look at the angle. We can define the slope of the beam as dy divided by dx. Since we’re limiting ourselves to scenarios where deflections are small, we can apply the small angle approximation, and so we can assume that the angle of rotation is equal to the slope of the beam. In other words, the angle of rotation is equal to the first derivative of y with respect to x. We can start thinking about how beams deflect under load by reviewing the moment-curvature equation. It allows us to calculate the curvature of a beam for a given bending moment.
M is the bending moment, R is the radius of curvature of the deformed shape, and El is the flexural rigidity of the beam, which defines how much the beam will resist bending due to the stiffness of its material, and due to its cross-sectional shape. The term 1/R is the curvature of the beam. The moment-curvature equation is the first step towards being able to calculate the deflection of a beam, since it gives us information about how the beam will deform for an applied bending moment. It turns out there’s a mathematical equation that can be used to calculate the curvature at any point along a curve. We can combine this with the moment-curvature equation to get an expression that relates the deflection of the beam and the bending moment along it.
But this equation is still quite complicated, so let’s try and simplify it. Since we’re limiting ourselves to elastic deformation and assuming that the slope of the elastic curve is small, the dy/dx term will be small. And since it’s squared it will be even smaller, and so can be neglected. This gives us a much simpler differential equation defining the deflection of the beam. All we need to do to get an expression for the deflection along a beam is integrate the equation twice.
This is called the double integration method. It’s easiest to understand how the method works using an example. Here we have a cantilever with a load applied at the end. We’ll start by drawing the free body diagram and applying equilibrium to determine the reaction forces. Next we want to calculate the bending moment along the beam. If we draw a free body diagram of a slice of the beam we can see that as we move along the beam, the bending moment increases linearly with the distance from the concentrated force P. So the bending moment M is equal to -Px. We can insert this expression for M into the deflection differential equation and apply the double integration method to figure out how the beam will deflect. E and I are constant along the length of the beam, so we can pull them outside of the integral, and then integrate the equation for M with respect to x. Remember that whenever you integrate you need to add a constant of integration at the end. We’ll call it C1. It’s an unknown for now.
Then we integrate a second time, remembering to add a second constant of integration. And so we now have an equation that defines the deflection y of the beam. The only problem is that we have two unknown constants. Fortunately we can use what we know about the deflection and the slope of the beam at the support to calculate C1 and C2. At the fixed end the beam cannot rotate, so the slope dy/dx must be equal to zero. This allows us to calculate C1.
Understanding Laminar and Turbulent Flow
And at that same location the deflection y must also be zero, which gives us C2. And so we’ve solved the deflection curve for this beam. In this case it’s obvious that the maximum deflection will occur at the end of the beam. For cases where the location of maximum deflection is less obvious, like for this simply supported beam, we can observe that the slope of the beam will be zero at the location of maximum deflection. So by setting the slope equation to be equal to zero we can figure out the location of ymax. Boundary conditions are key to applying the double integration method. We just saw that at a fixed support the displacement and the slope must both be zero. At a pinned support the beam is free to rotate, so the slope can be non-zero and only the boundary condition for the displacement applies. The same is true for a roller support, which is also free to rotate. The double integration method is quite easy to apply, but even for some quite simple configurations like this one the bending moment won’t be defined by a single expression along the full length of the beam.
To apply the double integration method in situations like this we need to split the beam into sections, and determine separate expressions for the bending moment in each section. We then apply the double integration method to the different sections, and we end up with a lot of different constants of integration. By considering the boundary conditions at the supports we can calculate some of these constants. To solve the remaining constants we need to consider what’s called the continuity condition between the different sections of the beam. Since the beam remains continuous across the transition between these two sections, the displacement and the slope must be the same at the right of Section 1 and at the left of Section 2.
This gives us a system of equations we can solve to determine the remaining constants of integration. Having to split the beam into sections and solve a system of equations complicates things and makes it more difficult to solve the deflections of the beam, particularly for cases where the beam needs to be split into a large number of sections. Fortunately there’s a powerful mathematical trick that allows us to define a bending moment profile like this one using a single expression, so that we can apply the double integration method without having to split the beam into sections. It uses singularity functions to make solving this type of beam much easier. They are defined using this special notation, and follow specific rules.
When the expression inside the bracket gives a positive number, the output is the expression, but when it is negative, the output is zero. These are called Macaulay brackets, and applying them to calculate the deflection of beams is called Macaulay’s Method. Their behaviour allows us to essentially turn different terms in the bending moment equation on and off as we move along a beam. We can work through the beam we looked at earlier to understand how. Let’s see how the bending moment changes as we move along the beam. In the first half of the beam only the reaction force contributes to the bending moment, and the moment arm is equal to x. But once we get past halfway, we need to include the effect of the concentrated load as well, and so the bending moment expression has to include an additional term. The moment arm is equal to x minus half the length of the beam. If we replace the brackets in this expression with Macaulay brackets, the second term in the expression will be “switched off” in the first half of the beam, since x – L/2 will give a negative number.
And so this expression using Macaulay brackets defines the bending moment along the entire length of the beam. The same thing works for distributed loads, although it’s slightly different. When using Macaulay brackets for a uniformly distributed load, the bracket has to appear twice – once in the magnitude of the load and once in the moment arm. The moment arm is the distance to the centre of the distributed load, so is equal to the length over which the load is applied, divided by 2. And so we get this expression, where the Macaulay bracket is squared and the distributed load is divided by 2. This expression is only valid if the distributed load extends all the way to the end of the beam. If it doesn’t, we have to adjust the expression to balance the part of the load that’s missing. Here’s how you can apply Macaulay brackets for a few other load types.
Understanding Young's Modulus
Macaulay brackets also have special rules for integration. They are integrated in a similar way to polynomial functions, like this, which makes it easy to apply the double integration method when using them. Let’s go back to our beam with a concentrated load and solve the deflections using Macaulay’s method. Since we have a single expression that defines M along the full length of the beam, we no longer need to split the beam into sections to solve it, and we can just apply the double integration method as we normally would. The double integration method can be quite time consuming to apply, even when using Macaulay’s method. To save us from having to go through the integration process every time we want to calculate the deflection of a beam, reference texts often include tables which list the deflection equations for a set of typical load and support configurations.
This is more powerful than it might seem at first glance, because the principle of superposition can be used to apply this small set of solutions to a much wider range of problems. It allows us to combine different load configurations on top of one another, and to calculate the overall deflection caused by the combined loading by summing up the contribution of each load to the total deflection. This is a really powerful technique that’s not just used for the analysis of beams, but also more widely in structural analysis. For the superposition method to be applicable, deflections must vary linearly with the applied loads. And deflections need to be small, since large deflections could mean that the applied loads don’t act independently.
Sometimes when analysing a beam we might only be interested in calculating the angle of rotation or the displacement at a single location, instead of along the entire length of the beam. There’s another method we can use for cases like this, which is called the moment-area method because it’s based on the area under the bending moment diagram. If we take two points, 1 and 2, and we integrate both sides of the deflection differential equation between those two points, we end up with this equation. The term on the left is the change in angle between both points, which we can visualise by extending lines tangent to the elastic curve at the two points.
And the term on the right is just the area under the bending moment diagram between both points, A, divided by the flexural rigidity EI, assuming that both E and I are constant along the length of the beam. So to calculate the change in angle between two points along the elastic curve all we have to do is divide the area under the bending moment diagram between the two points by EI. This is the first moment-area theorem. If we rewind and multiply both sides of the deflection differential equation by x before integrating, we can get another useful equation. This time the integration is a bit more complicated, since we need to use integration by parts. But we end up with this expression, which is the second moment-area theorem. It tells us that the vertical distance between point 2 and the tangent line at point 1 is equal to the first moment of the area of the bending moment diagram, divided by EI. The first moment of area is just the area under the diagram, multiplied by the horizontal distance from point 2 to the centroid of the area This vertical distance is called the tangential deviation. It might not seem that useful, since it doesn’t directly give us the deflection at point 2. But if we calculate the tangential deviation at the other end of the beam we can then use trigonometry to figure out the actual deflection at point 2. The moment-area method is particularly useful when there’s a point along the beam where we know that the slope of the elastic curve will be equal to zero, since in that case calculating the tangential deviation gives us the deflection directly. This is why the moment-area method is often applied for cantilever problems. Let’s look at an example. We’re interested in the deflection and angle of rotation at the free end, so let’s define point 1 at the fixed support and point 2 at the free end. Because the angle and deflection are zero at point 1, the moment-area method will give us the deflection and angle at point 2 directly. We can define x as the horizontal distance from the free end, and draw the bending moment diagram. To apply the moment-area method we need to calculate the area under the bending moment diagra
An Introduction to Stress and Strain
m between points 1 and 2, and the distance between point 2 and the centroid of the area. It’s easy to do this by splitting the diagram into two triangles and a rectangle. The areas are easy to calculate. And for the centroids we just need to remember that the centroid of a right angled triangle is one third of the way down and one third of the way across from the right angle. Then we can just plug the numbers into the first and second theorems to calculate the angle and deflection at point 2. There’s one last approach to analysing the deflection of beams that I’d like to cover, which is an energy approach that uses the concept of strain energy. When a body is deformed elastically, energy is stored within it in the form of strain energy. For a beam deformed by bending the strain energy is a function of the bending moment, and it can be calculated using this equation. It’s common to neglect any shear deformation and assume all of the strain energy is coming from pure bending. Castigliano’s theorem states that the deflection at the location of an applied load along the line of action of the load is equal to the partial derivative of the strain energy with respect to that load. The deflection can be a displacement or an angle of rotation. Because the bending moment term is squared in the strain energy equation, it’s usually easier to move the partial derivative inside of the integral, like this. Let’s look at an example. The bending moment along the beam shown here is defined by this expression. We can calculate the partial derivative of M with respect to P.
Using Castigliano’s theorem, and integrating between zero and L, we get an expression for the displacement at point A. The powerful thing about Castigliano’s thoerem is that the load we consider doesn’t even have to be a real load. If we want to determine deflections at a location where no load is applied, we can just apply a fictitious force. We include the fictitious force in the bending moment equation, and then once we’ve calculated the partial derivative of the strain energy with respect to that force, we can set it to zero to obtain an equation for the deflection at that location. The fictitious load can be either a moment or a force. Applying a moment allows us to calculate the angle of rotation, and applying a force allows us to calculate the deflection y of the beam. Applying a fictitious moment at Point A allows us to calculate the angle of rotation at that point. This time we take the partial derivative of M with respect to M_A.
Once we have an expression for the angle of rotation we can set the fictitious moment M_A to be equal to zero. Although it might be slightly more difficult to conceptualise than the other methods we’ve looked at, Castigliano’s theorem can be very powerful when applied properly, and it has uses that extend well beyond the analysis of beams. With experience you’ll be able to select the method that will be easiest or most appropriate to apply, based on the specifics of the situation you are analysing. But one inescapable fact when it comes to applying any of these methods is that they rely heavily on having a solid understanding of calculus. Mastering the fundamentals of calculus is the best way to make beam deflection problems feel much less daunting, and this video’s sponsor Brilliant is a math and science learning platform that will help you do exactly that. It has all of the steps covered, from the basics of what differentiation and integration actually are, all the way up to complex topics like partial differential equations and Laplace transforms. The thing I love most about Brilliant is how easy it makes it to learn. | 3,558 | 17,356 | {"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-2022-21 | latest | en | 0.946482 |
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# At what positions is the speed of a simple harmonic oscillator half its maximum That is
ISBN: 9781938168932 372
## Solution for problem 39 Chapter 16
College Physics for AP® Courses | 1st Edition
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Problem 39
At what positions is the speed of a simple harmonic oscillator half its maximum? That is, what values of x / X give v = vmax / 2 , where X is the amplitude of the motion?
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THE STRANGE NEW SCIENCE OF CHAOS ● Chaos = errors or deviations from expected results ○ Scientists try to minimize or eliminate it ● Linear system = the output is directly proportional to the input ● Examples of chaotic systems ○ Water bubbling down a stream ○ Smoke rising from a cigarette ○ Water dripping from a tap ○ The weather ○ Our nervous systems ● Chaotic systems are sensitive to initial conditions = a small change in the starting condition leads to a big change in the result ● Weather is hard to predict ○ The turbulent atmosphere cannot be described by simple linear equations ● Attractor = a preferred state towards which a chaotic system may tend ● Turbulence = much of the universe is fluids of one kind or another in motion ● Chaos does not mean randomness, but patterns within what appear to be random ● Our brains depend on chaos, but chaotic behavior is characterized of a heart attack
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##### ISBN: 9781938168932
College Physics for AP® Courses was written by and is associated to the ISBN: 9781938168932. The full step-by-step solution to problem: 39 from chapter: 16 was answered by , our top Physics solution expert on 03/09/18, 08:07PM. Since the solution to 39 from 16 chapter was answered, more than 259 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: College Physics for AP® Courses, edition: 1. The answer to “At what positions is the speed of a simple harmonic oscillator half its maximum? That is, what values of x / X give v = vmax / 2 , where X is the amplitude of the motion?” is broken down into a number of easy to follow steps, and 37 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 34 chapters, and 2282 solutions.
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A228506 T(n,k)=Number of nXk binary arrays with top left value 1 and no two ones adjacent horizontally, vertically, diagonally or antidiagonally. 11
%I #7 Sep 01 2013 17:48:59
%S 1,1,1,2,1,2,3,3,3,3,5,5,12,5,5,8,11,29,29,11,8,13,21,88,87,88,21,13,
%T 21,43,239,358,358,239,43,21,34,85,684,1252,2002,1252,684,85,34,55,
%U 171,1909,4749,9528,9528,4749,1909,171,55,89,341,5392,17285,49101,59839,49101
%N T(n,k)=Number of nXk binary arrays with top left value 1 and no two ones adjacent horizontally, vertically, diagonally or antidiagonally.
%C Table starts
%C ..1...1....2.....3.......5........8........13.........21..........34
%C ..1...1....3.....5......11.......21........43.........85.........171
%C ..2...3...12....29......88......239.......684.......1909........5392
%C ..3...5...29....87.....358.....1252......4749......17285.......64235
%C ..5..11...88...358....2002.....9528.....49101.....243118.....1228036
%C ..8..21..239..1252....9528....59839....413786....2724191....18387032
%C .13..43..684..4749...49101...413786...3862849...34229311...311423874
%C .21..85.1909.17285..243118..2724191..34229311..405580157..4951454523
%C .34.171.5392.64235.1228036.18387032.311423874.4951454523.81304395949
%H R. H. Hardin, <a href="/A228506/b228506.txt">Table of n, a(n) for n = 1..1740</a>
%F Empirical for column k:
%F k=1: a(n) = a(n-1) +a(n-2)
%F k=2: a(n) = a(n-1) +2*a(n-2)
%F k=3: a(n) = 2*a(n-1) +3*a(n-2) -2*a(n-3)
%F k=4: a(n) = 2*a(n-1) +7*a(n-2) -2*a(n-3) -3*a(n-4)
%F k=5: a(n) = 2*a(n-1) +16*a(n-2) +a(n-3) -27*a(n-4) +a(n-5) +4*a(n-6)
%F k=6: [order 8]
%F k=7: [order 14]
%e Some solutions for n=4 k=4
%e ..1..0..0..0....1..0..0..0....1..0..0..1....1..0..0..0....1..0..0..1
%e ..0..0..0..0....0..0..0..0....0..0..0..0....0..0..0..0....0..0..0..0
%e ..0..0..0..0....1..0..0..0....0..0..0..0....0..0..0..1....1..0..0..0
%e ..1..0..0..0....0..0..0..0....0..0..0..0....0..0..0..0....0..0..1..0
%Y Column 1 is A000045
%Y Column 2 is A001045
%K nonn,tabl
%O 1,4
%A _R. H. Hardin_ Aug 23 2013
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Last modified June 24 00:02 EDT 2024. Contains 373661 sequences. (Running on oeis4.) | 1,064 | 2,509 | {"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-2024-26 | latest | en | 0.571154 |
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What are some similarity postulates?
If two of the angles are the same, the third angle is the same and the triangles are similar. If the three sides are in the same proportions, the triangles are similar. If two sides are in the same proportions and the included angle is the same, the triangles are similar.
Is there ss test of similarity?
Explain. While two pairs of sides are proportional and one pair of angles are congruent, the angles are not the included angles. This is SSA, which is not a similarity criterion. Therefore, you cannot say for sure that the triangles are similar.
Why is SSA not a similarity theorem?
What about SSA (Side Side Angle) theorem? The ASS Postulate does not exist because an angle and two sides does not guarantee that two triangles are congruent. If two triangles have two congruent sides and a congruent non included angle, then triangles are NOT NECESSARILLY congruent.
What are the 3 similarity postulates?
These three theorems, known as Angle – Angle (AA), Side – Angle – Side (SAS), and Side – Side – Side (SSS), are foolproof methods for determining similarity in triangles.
Is AAA a postulate?
In Euclidean geometry, the AA postulate states that two triangles are similar if they have two corresponding angles congruent. (This is sometimes referred to as the AAA Postulate—which is true in all respects, but two angles are entirely sufficient.) The postulate can be better understood by working in reverse order.
Is AAS same as SAA?
A variation on ASA is AAS, which is Angle-Angle-Side. Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
Is AAA a similarity theorem?
may be reformulated as the AAA (angle-angle-angle) similarity theorem: two triangles have their corresponding angles equal if and only if their corresponding sides are proportional. Two similar triangles are related by a scaling (or similarity) factor s: if the first triangle has sides a, b, and c, then the second…
Can SSA prove triangles similar?
Given two sides and non-included angle (SSA) is not enough to prove congruence. You may be tempted to think that given two sides and a non-included angle is enough to prove congruence. But there are two triangles possible that have the same values, so SSA is not sufficient to prove congruence.
Why is there no AAA postulate?
Knowing only angle-angle-angle (AAA) does not work because it can produce similar but not congruent triangles. Because there are 6 corresponding parts 3 angles and 3 sides, you don’t need to know all of them.
How do you know if its AAS or ASA?
While both are the geometry terms used in proofs and they relate to the placement of angles and sides, the difference lies in when to use them. ASA refers to any two angles and the included side, whereas AAS refers to the two corresponding angles and the non-included side.
What is the AAS Theorem?
Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.
What is SSS for similarity?
SSS for Similarity is a way of determining whether two triangles are similar or not. Similarity means the triangles are identical in shape, but not in size. SSS stands for side-side-side. If you think two triangles are similar, you look at the ratios of the matching sides for each triangle.
Does SAS prove similarity?
SAS is one of the three tests for similarity of triangles. Its says:- if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angle between them is the same then the triangles are similar.
What is the definition of SAS similarity therom?
SAS means side, angle, side, and refers to the fact that two sides and the included angle of a triangle are known. The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.
How do you prove SSS theorem in similar triangles?
Part 3 of 4: Using the Side-Side-Side Theorem Define the Side-Side-Side (SSS) Theorem for similarity. Two triangles would be considered similar if the three sides of both triangles are of the same proportion. Measure the sides of each triangle. Using a ruler, measure all three sides of each triangle. Calculate the proportions between the sides of each triangle. | 1,014 | 4,659 | {"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-2022-27 | latest | en | 0.930339 |
https://calculatorbeast.com/math/cube-root-calculator | 1,603,896,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00173.warc.gz | 253,460,215 | 23,194 | ##### Formula and Execution
Number This is the value whose cube root or nth root is required. In this section, you will substitute the number to find its nth root. Degree of Root The degree of root is the number which, when multiplied by the same times as the root number, generates the original number. Like square root has root 2, and when it is multiplied 2 times, it generates the original number. Example: √x * √x = x. In this section, you substitute the number to which the root is required. For example: for square root, you will substitute 2; for cube root, you substitute 3, similarly, for fourth root, 4, and so on. Root This is the result section for all your answers!
## Cube Root Calculator
Smart, handy, quick, precise, and free! Sounds like a complete package, right? Well, that is because it is. Our online cube root calculator computes all your cube root calculations instantly making your life a lot easier than you thought. And guess what? We are not just limited to cube roots, but to roots of any degree. So basically, it is an nth Root Calculator!
Just type in your values, and as the second you feed the calculator up, the second it generates results.
Besides, also check out full list of all the calculators that are all complete packages of all the good attributes just like this one, to make your math ventures a lot easier and fun for you.
### How to Calculate Cube Root or nth Root by Using Calculator Beast
Substituting numbers in the right section is all it takes. Let us calculate the cube root of 5832. Just substitute in 5832 in the ‘Number’ section, 3 in the ‘Degree of Root’ section, and see the result in the ‘Root’ section! Here is the screenshot from our calculator for better illustrations.
To calculate the nth root of any number, we again don’t have to deal with climbing up the mountains. Follow the same steps as above, just instead of 3, type the number to which the root needs to be calculated. I have taken the example of 6th root for the number 1000.
Another good news is that our cube root/nth root calculator works both ways. You can also substitute a number in the root section, and the degree of root to which the number is raised. And see your results in the number section. Brilliant, right?
### Concept of Cube Root
Before you read further, I would appreciate it if you take a glance at the concepts described in our online square root calculator. It is not mandatory, but it just makes things easy, you know. And who doesn’t like easy?
#### Cubes
We all know what cubes are. The famous Rubik’s Cube is an example from our daily lives. So, when a number is raised to the power of 3, we say it is cubed. Why? Because it forms a cube in a 3-dimensional plane, as shown in the figure.
We discussed in the square root calculator that squaring a number generates the area of a square. Similarly, cubing a number generates the area of a cube. The figure above shows 33 plotted in a 3-dimensional plane.
#### Cube Root
Just as we saw that the square roots are the inverse of the squares, the cube roots, in the same way, are inverse of the cubes. We use the radical ‘√’ sign, the same way we use for the square roots. The difference is that we write a small 3 with the radical sign like ∛. Mathematically, we define cube roots as,
“If a number y can be expressed as y = x3, we say that y is the cube of and x and x is the cube root of y.”
∛y = x ó y = x3
Cube roots can also be written in the form of exponents as ∛y = y1/3.
Examples:
1. Calculate the cube root of 9261.
Solution
The calculator at your left-hand side will make all your cube roots calculations a piece of cake! Substitute 9261 in the number section, 3 in the degree of root section, and I hope you have got 21 as a result in the root section.
This means when you multiply 18 thrice, you get 5832. So, we say that the cube root of 5832 in 18.
1. Calculate the number whose cube root is 1.396.
Solution
So, by reading the problem, you already know that this time, we are going the other way round. Our calculator not only gives results for the cube roots, but it also generates cubes when roots are added to it.
So, substitute 1.396 in the root section, 3 in the degree of root section, and what you got? 2.72, right? Well, that’s an approximation up to two decimal places of the value of the exponential constant ‘e. So, now we know that the cube root of e is 1.396.
#### Perfect Cubes
If we see the examples discussed above, the concept of perfect cubes will become easy to understand. In example 1, 9261 is the perfect cube of 21. And in the second example, 1.396 is not the perfect cube of the exponential constant.
Were you able to draw any conclusions from above?
Well, if we calculate the cube root of any number, and the result is a whole number, then the original number is said to be the perfect cube of the result. And if the result on calculating the cube is not a whole number, then the original number is not the perfect cube of any number.
#### Some Commonly Used Values of Cubes and Cube Roots
Calculators are a very handy tool when it comes to calculating the cubes and cube roots of bigger and irrational numbers. But we definitely would not recommend you to completely rely on these computation tools.
One must learn the basic values of the cubes and cube roots
#### How to Calculate Cube Roots by Prime Factorization?
Prime Factorization is something we all learn at elementary school levels. I assume you already know how to calculate prime factors; if not, our online GCF calculator is always there to teach you prime factorization.
We list down the prime factors of 1728,
1728 = 2.2.2.2.2.2.3.3.3 (here dot represents multiplication)
To calculate the cube root of 1728, we take cube root on both sides of the above equation
∛1728 = ∛(2.2.2.2.2.2.3.3.3)
= ∛{(2.2.2)(2.2.2)(3.3.3)}
= ∛{(2)3(2)3(3)3}
We already discussed that the cube roots could also be written in the form of exponents. So we write the above expression as,
∛1728 = [(2)3(2)3(3)3]1/3
= 2.2.3 = 12
So the cube root of 1728 is 12.
#### How to Estimate Cube Roots?
To estimate the cube root of any number, you first need to learn the cube root table written above. Once those values are at your fingertips, then you can easily leave your friends in awe by this trick.
Notice the values in the table. Notice how the cubes for each number ends. Did you see anything observable? Well, the values of the cubes for the numbers 1,4,5,6,9, and 0, end with the same digits respectively. Like the cube root of 1 ends in 1, the cube root of 4 ends in 4, and the same goes for the rest of the numbers.
While the cubes for 2 and 8, and 3 and 7, flip. Like cube root of 2 is 8, that is, it ends in 8, and the cube root of 8 is 512, i.e., it ends in 2. The same goes for 3 and 7.
This pattern follows for all numbers. Once you have mastered this pattern, then as you see any number whose cube root is required, the first thing you do is to simply write the last number.
Let us suppose the number 46,656. The last digit is 6, so we know that our last number of the cube root would be 6. Now ignore the last three numbers, i.e., 656. Consider 46 and run through your brain the list you just memorized and pick the nearest cube. The nearest cube is 27, whose cube root is 3. Simply place this 3 before 6. So our cube root for 46,656 is 36.
Let us dive into another example. If we have to find the cube root of 314,432, we see the last digit. It is 2, so we know that the cube root would end in 8. Now our last digit is known to us. Ignoring 432, we see the cube nearest to 314 from our list. It is 216, and its cube root is 6, so I guess you already know the cube root of 314,432. Yes, it is 68. No big a deal, right? Well, it might seem time-requiring in the beginning, but the more you practice, the more you ace at it.
#### Graph of a Cube Root Function
The cube root function is odd i.e., it fulfills the condition f(-x) = -f(x). Being an odd function makes it symmetric about the origin. The graph of a cube root function also has a vertical tangent at x=0.
#### The Derivative of a Cube Root Function
The formula for carrying out basic differentiation is,
f(x) = xn ó f’(x) = nx(n-1)
If we consider the cube root function f(x) = ∛x, then by using the above formula, the derivative of the cube root function would be,
f’(x) = 1/3 (x)1/3-1
f’(x) = 1/3 (x)-2/3
f’(x) = 1/3 . 1/x2/3 (dot implies multiplication)
f’(x) = 1/(3∛x2)
#### Nth Root
We studied square roots and cube roots. We also discussed why they are termed as square and cube. Now the math is not limited to the roots of these numbers. It, of course, goes on to infinity.
We cannot name the root of every single number individually because no one would be able to remember infinite names, and it just makes things complex rather than easy. So, when we take the roots of numbers except for the square root and cube root, we simply say, we are taking the nth root of that number. In general notation, we also write n as the index of the root, like n√x. If we are to calculate the 4th root of any number, we say 4th root and write it as ∜x. The same goes on for the 5th, 6th, 7th, and so on for roots up to infinity.
Our nth root calculator helps you solve your problems related to roots to any number. Just type in the number in the root section, nth root in the degree of root section, and get your results instantly in the root section. The screenshot below shows the 7th root of 356,897.
### FAQs
#### How do I calculate the cube root by a scientific calculator?
If you are using a scientific calculator, then calculating either the cube root or the nth root is no big a deal. Find out the square root button. In most of the calculators, the square root button has cube root written above it, on the base of the calculator. This function works after pressing the shift button. So, press the shift button and now press the square root button. Type in the number whose cube root is required, press = sign, and bingo!
When it comes to calculating the nth root, find the key in your calculator that has a box in its index place. Since the calculator comes in different versions, so this button may exist at different places in different versions. Find out where it exists in your calculator. Press the button, now type the root in the index box, and the number under the radical by using the arrow keys to shift from index to under the radical sign. Press = sign and get your results for the nth root of any number.
#### How do I calculate the cube root on a regular calculator?
Well, the steps to calculate the cube root on a regular calculator are something you will have to memorize, but it is not something impossible, and also you can verify your answers during your exams if cube root is required to be calculated manually.
• The first step is to type the number in the calculator.
• The second step is to press the root sign twice.
• Now press the multiply sign and again press root sign for four times.
• Again press multiply sign and this time press the radical/root sign for eight times. Make sure you do not miscount.
• This is the last time you have to press the multiply sign. Now press the radical sign twice.
• Press = and get your results.
To memorize the steps, simply learn this list: type number, then press root sign two times, four times, eight times, and two times. During every comma, use multiplication operation. Hope this makes things easier to memorize.
Note: Slight variation in the answer may exist due to rounding off issues. Otherwise, the results are worth using.
#### Do cube roots of negative numbers exists?
Yes, cube roots of negative numbers exist. And calculating them is easy in comparison to the square roots. In square roots, we studies about how complex numbers get involved when we calculate the square root of negative numbers. But in this case, things don’t get complex.
Why? Because when we multiply a negative number three times, it gives us a negative result. So, what do you think would be the cube root of -216? -6, of course, because -6*-6*-6=-216.
#### We have two results for square roots, positive and negative, what about cube roots and nth roots?
The easiest way to know about the number of results existing while calculating the roots is simply to see the degree of root. If the degree is 2, results would be 2, if it is 3, we will have three results as its roots, if it’s 4, we have 4 roots, and if it’s n, we have n roots.
So you must be thinking, why do you get 1 result when you calculate the cube root? Well, the answer because that one result is the real result, the other two results are complex numbers, i.e., numbers in the form of a+ib, where i is the imaginary number equal to √-1.
The imaginary numbers are usually not dealt with very often. So, most of the time, we ignore those roots until there is a need. And yes, although they are termed as imaginary, the numbers do exhibit great use in the real world, and that is why we have them. The mathematicians, physicists, and other scientists know where and how to use them.
Our calculator deals with real values only. The imaginary values take some time to calculate, so we won’t be going into the details of calculating them, but we will prove our point that they exist, by an example.
The cube roots of 27 are 3, -1.5+1.5(√3)i, and -1.5-1.5(√3)i. To prove these are the cube roots of 27, we simply take a cube of these numbers. If the result is 27, they are the cube roots.
• For 3, (3)3 = 3*3*3 = 27
• For -1.5+1.5(√3)i, we use the multiplication formula (a+b)3 = a3 + b3 + 3a2b + 3ab2 and we keep in mind that i2 = -1 and i3 = -i.
We consider a=-1.5 and b=1.5(√3)i
(-1.5+1.5(√3)i)3 = (-1.5)3 + (1.5(√3)i)3 + 3(-1.5)2(1.5(√3)i) + 3(-1.5)( 1.5(√3)i)2
= -3.375 – (81√3)i/8 + (81√3)i/8 -30.375
= 27
You can carry out the calculation for the third value and see if you get 27 as the answer. | 3,559 | 13,946 | {"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-2020-45 | longest | en | 0.904427 |
https://grahamhancock.com/phorum/read.php?1,1131508,1218341 | 1,619,112,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00572.warc.gz | 397,466,612 | 14,407 | Mysteries : The Official GrahamHancock.com forums
For serious discussion of the controversies, approaches and enigmas surrounding the origins and development of the human species and of human civilization. (NB: for more ‘out there’ posts we point you in the direction of the ‘Paranormal & Supernatural’ Message Board).
Hi, David
Thanks for the valuable feedback
DavidK Wrote:
-------------------------------------------------------
> Thinkitover
>
> I have a side length value in my data for the
> "Zangkunchong" pyramid of "29.34 m" (I'm still
> trying to figure out if it may be a misplaced
> diagram of the "Tomb of the General").
>
> i would interpret this as follows
>
> 29.33333r metres is the actual intended measure.
>
> Because your metre is 39.37 it has distorted the
> analysis.
>
> using 39.375 x 29.333r = 1155 inches and this is
> surely the intended number because it is base 7
> and base 11
Good observation. I haven't had the easiest time working with the ~1155 range because of the apparent diversity - typical "trying to find the one but there's three of them or maybe five" kind of stuff - but happy to see that is it's still classic 260 / 225 = 1.155555555 in some form or another.
> 1155 / 11 = 105 this is 9 egyptian feet of 11.666r
> inches
>
> 1155 / 7 = 165 The imperial rod (10 off)
>
> So 1155 x 22/7 = 3630 / 20.625 = 176
More good observations. I hadn't actually starting taking my own measures seriously yet, but I did want to point out some of the things the inquiry led to.
> If you read Berriman you will realise the chisnese
> system was based on root 2 as a measure on the
> ground.
If I take Berriman at face value, you mean, lol.
> if you are getting numbers such as 101.818181818r
> then they will divide by 99/100
>
> 101.818181r / 100 x 99 = 100.8
>
>
> 101.81818181r x 22/7 is 320 and the hekat is being
> expressed.
>
> 101.81818181r is the sum of a proven Egyptian
> series summed proving the use of 22/7.
>
> 20.61818181818181818r / 100 x 99 = 20.412
>
> Everything can be resolved into whole numbers
> because all the whole numbers have been
> fractionalised using Egyptian methods.
>
> 132000000 / 3600 = 36666.666r
The big thing I'm looking for is the 360 / Lunar Year that goes with a third set of calendar numbers, and in spite of being a great suggestion, 360 / (~320 / Pi) doesn't seem to be it. As things presently stand, I have 1.017532310, one of a great many new numbers since a renewed looked at Stonehenge, that I've barely begun to explore.
Still, I'd probably be wise to explore your suggestion and mine further. I've had to revise a few planetary values already, they seem to be tricky.
Interestingly, if I try to fit your suggestion with a couple of workhorses from Stonehenge ((51.95151515 x 10) / 2) / 224.8373808 = 1155.31311 inches. Converted to feet, that still seems to be trying to head the same place
1155.31311 in / 12 = 96.27609250 = 1 / 103.8679462, but I'm still not sure if I've gotten that right or wrong.
I still don't know some of these numbers well enough, it's only been the past several weeks I've watched Stonehenge explode like that. At least I have the very good fortune of having Stockdale and Harris making me think I'm right about a lot of things before I've had to work out all of it for myself, and all I've read from them so far is Harris' paper from that thread you linked to.
I really wish I had more data on the Asian pyramids, though - I still don't even know where those diagrams originated - but I'm a lot more comfortable with the idea of having more parts to cross-reference to each other to help keep things on the right track before getting too invested in interpreting them. That might help me to get a feel for the reliability of the measurements also - I was a long time trusting even Petrie as to accuracy.
A value of 1152.833215 in isn't impossible, and might be more elegantly simple, but I think I'll try to continue to explore the complexities because I'm hoping they might further illuminate the way forward with the planetary and lunar numbers.
Cheers!
Subject Views Written By Posted
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Persuasive Evidence of the Modern Manipulation of Fibonacci 597 Dr. Troglodyte 15-Jan-18 19:00
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 613 magisterchessmutt 15-Jan-18 19:18
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 562 magisterchessmutt 15-Jan-18 19:37
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 87 DavidK 12-Jan-20 20:18
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 529 DavidK 15-Jan-18 20:10
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 539 DavidK 15-Jan-18 19:26
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 491 molder 15-Jan-18 23:35
Another Canonical alternative 775 cloister 16-Jan-18 00:57
Re: Another Canonical alternative 493 molder 16-Jan-18 11:35
Re: Another Canonical alternative 490 cloister 16-Jan-18 12:23
Another perspective???????????????? 495 Sirfiroth 16-Jan-18 18:09
Re: Another perspective???????????????? 521 DavidK 17-Jan-18 14:07
Re: Another perspective???????????????? 459 cloister 17-Jan-18 14:31
Re: Another perspective???????????????? 518 Sirfiroth 17-Jan-18 17:02
Re: Another perspective???????????????? 479 Dr. Troglodyte 17-Jan-18 17:41
Re: Another perspective???????????????? 484 cloister 17-Jan-18 21:55
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 93 DavidK 12-Jan-20 20:25
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 499 Harte 17-Jan-18 22:40
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 571 Sirfiroth 18-Jan-18 01:29
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 468 DavidK 18-Jan-18 10:11
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 424 Sirfiroth 18-Jan-18 14:10
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 415 DavidK 18-Jan-18 14:40
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 469 Sirfiroth 18-Jan-18 15:16
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 475 DavidK 18-Jan-18 15:48
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 511 Sirfiroth 19-Jan-18 18:38
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 438 DavidK 19-Jan-18 21:55
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 495 Sirfiroth 20-Jan-18 01:15
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 508 DavidK 20-Jan-18 09:56
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 439 Sirfiroth 20-Jan-18 16:12
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 564 rodz111 20-Jan-18 16:20
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 566 molder 20-Jan-18 11:31
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 473 DavidK 20-Jan-18 16:57
Re: CALL ROYAL CUBITS INCHES 475 molder 20-Jan-18 20:35
Re: CALL ROYAL CUBITS INCHES 459 molder 20-Jan-18 21:09
Re: CALL ROYAL CUBITS INCHES 495 Sirfiroth 21-Jan-18 03:58
Re: CALL ROYAL CUBITS INCHES 453 DavidK 21-Jan-18 10:31
Re: CALL ROYAL CUBITS INCHES 511 DavidK 21-Jan-18 10:57
Re: CALL ROYAL CUBITS INCHES 457 DavidK 21-Jan-18 11:22
Re: CALL ROYAL CUBITS INCHES 459 DavidK 21-Jan-18 11:39
Re: CALL ROYAL CUBITS INCHES 554 Sirfiroth 21-Jan-18 01:29
Persuasive Evidence of the Ancient Manipulation of Phi? 480 Dr. Troglodyte 18-Jan-18 14:53
Re: Persuasive Evidence of the Ancient Manipulation of Phi? 483 Sirfiroth 18-Jan-18 18:49
Re: Persuasive Evidence of the Ancient Manipulation of Phi? 526 Dr. Troglodyte 18-Jan-18 20:10
Re: Persuasive Evidence of the Ancient Manipulation of Phi? 579 DavidK 18-Jan-18 20:51
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 495 Harte 18-Jan-18 17:27
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 658 rodz111 21-Jan-18 12:00
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 431 DavidK 08-Feb-18 23:40
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 445 DavidK 08-Feb-18 23:50
is this for real? 473 DavidK 09-Feb-18 00:21
Linking different versions of Pi and the canon of ancient measure 445 DavidK 09-Feb-18 12:28
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 692 rodz111 11-Feb-18 10:17
Rind papyrus 50 is not Pi as 256/81 557 DavidK 09-Feb-18 15:31
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 508 DavidK 13-Feb-18 10:01
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 516 DavidK 13-Feb-18 16:28
Pi as 3.1416 526 DavidK 08-Mar-18 09:39
3 versions of Pi to get to 2912 at Stonehenge 521 DavidK 12-Mar-18 12:23
Fibonacci Pi, 7920 miles and the canon of ancient measure 611 DavidK 21-Mar-18 11:18
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 509 DavidK 09-Jun-18 10:28
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 380 DavidK 28-Jun-18 10:40
Finding real Pi at the Aubrey circle 390 DavidK 28-Jun-18 10:56
Re: Finding real Pi at the Aubrey circle 399 rodz111 28-Jun-18 12:06
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 393 molder 28-Jun-18 11:23
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 398 DavidK 28-Jun-18 12:01
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 380 molder 28-Jun-18 12:13
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 419 molder 28-Jun-18 21:37
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 434 DavidK 29-Jun-18 09:52
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 386 molder 29-Jun-18 21:23
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 362 DavidK 30-Jun-18 09:48
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 425 molder 30-Jun-18 10:23
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 374 DavidK 30-Jun-18 12:08
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 433 molder 30-Jun-18 12:18
Pi at Stonehenge and 2912 478 DavidK 07-Jul-18 06:47
the simplicity of using pi as 3.1416 at Stonehenge 410 DavidK 12-Aug-18 07:32
Re: the simplicity of using pi as 3.1416 at Stonehenge 353 DavidK 13-Aug-18 06:09
Fibonacci Pi the saros and 3.1416 at Stonehenge 331 DavidK 13-Aug-18 07:40
Astronomically produced pi 374 DavidK 13-Aug-18 07:43
variable pi using Michell's canon 375 DavidK 14-Aug-18 06:36
Re: variable pi using Michell's canon 309 cloister 14-Aug-18 09:48
Re: variable pi using Michell's canon 356 DavidK 14-Aug-18 10:10
Link between pi and root 2 418 DavidK 10-Aug-18 11:14
Re: Link between pi and root 2 362 DavidK 18-Aug-18 22:08
The Eclipse!? 374 PB Bytes 18-Aug-18 22:20
Re: The Eclipse!? 394 DavidK 18-Aug-18 22:41
Re: The Eclipse!? 417 PB Bytes 18-Aug-18 23:16
Re: The Eclipse!? 437 molder 19-Aug-18 11:12
Re: The Eclipse!? 459 DavidK 20-Aug-18 19:08
Jim Wakefield's fibonnacci Pi 375 DavidK 14-Dec-18 14:32
Re: Persuasive Evidence of the Ancient Manipulation of Pi? 300 DavidK 07-Jan-19 16:49
The hidden unit and Pi and real root 2 293 DavidK 09-Apr-19 12:30
Re: The hidden unit and Pi and real root 2 - Berriman had it 123 DavidK 03-Jan-20 12:51
Ellifino, As Usual 122 thinkitover 03-Jan-20 21:40
Re: Ellifino, As Usual 129 DavidK 03-Jan-20 23:11
Re: Ellifino, As Usual 104 thinkitover 13-Jan-20 06:22
Re: Ellifino, As Usual 135 DavidK 13-Jan-20 13:55
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# rightmost non-zero digit
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What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8
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24 Aug 2009, 18:04
I am getting 6.
I got 8 initially but now I am getting 6...
I am wondering if 6 or 8 is right or neither...lol
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24 Aug 2009, 22:53
tejal777 wrote:
What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8
ps-rightmost-nonzero-digit-product-of-factors-82401.html?view-post=617960#p617960
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Re: rightmost non-zero digit [#permalink] 24 Aug 2009, 22:53
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### Course: Calculus, all content (2017 edition)>Unit 3
Lesson 12: Planar motion
# Motion along a curve: finding velocity magnitude
Given that a particle moves along the implicit curve xy=16 and given its rate of change with respect to y at some point, Sal finds the magnitude of the particle's velocity vector using implicit differentiation.
## Want to join the conversation?
• I am still really confused about doing d/dx or dy/dx or dx/dt and so on and so forth.
I know that "d/dx" = "take the derivative of everything in the equation that has an 'x'"
but if it doesn't have an 'x' you just place a "d(whatever)/dx"? is that correct? and then somehow later find a way to find that "d(whatever)/dx"?
For example. I am confused where t comes in this equation. t was never given in the original equation, so how can you just add it in?
Thank you.
2nd half of question!
okay how is
d/dt of x = 1*dx/dt?
Why isn't it just one? or dx/dt? in other words, why do you use the chain rule for doing a d/dt of x?
(9 votes)
• This website uses notation that differs from my calculus textbook, which also differs (on occasion) from the notation in my calculus class! There are lots of derivative notations floating around. It is confusing and a video explaining them all might be helpful. I'll try my best here.
Jelle, you are correct, the notation "d/dx" can also be stated as "the derivative with respect to x." We can also write it as "dy/dx," aka "the derivative of y (the function) with respect to x." Of course, we can name a function pretty much anything other than y, hence notations such as "db/dt" or "dr/dt" and the list goes on (you'll use such notation in related rate problems). Notation "d/dt" or "dx/dt" or even "dy/dt" can be stated as "the derivative (in general, or of x or y) with respect to t," with t being a third parameter. Again, regarding related rate problems, or acceleration or velocity problems, t usually stands for time, but it can also stand for some third parameter when differentiating.
To clarify (x)(dx/dt) = 1(dx/dt), we treat it similar to the way we treat y in implicit differentiation: We take the derivative of x, which is 1, and multiply it by x', similar to how 2y(dy/dt) = 2(dy/dt) or 2y'. Since we are differentiating every variable in such equations with respect to a third parameter, this explains why every variable has its own derivative notation.
To try and sum the ones I know:
f'(x) = [letter representing function name, usually y]' = d[letter representing function name, again usually y]/dx (or dt, if a third parameter, such as time, applies). Of course the symbol ' is read as "prime."
More explicitly: f'(x) = y' = dy/dx or dy/dt.
I hope this is helpful and accurate. :)
(4 votes)
• Beginning at , Mr. Khan is trying to explain how to solve for dx/dt. He states: "We're going to use a little bit of the product rule and a little bit of the chain rule here." Then he proceeds to apply the product rule, which is justified by the nature of the left hand side of the equation x·y, but he never uses the chain rule for the obvious reason that it's inapplicable in this problem. Why did he say we will be using it? this kind of error is really confusing; and, frankly, it's unacceptable that no correction has been made.
(2 votes)
• There is no error. Sal uses the chain rule to compute the derivative of x with respect to t and the derivative of y with respect to t.
d/dt[x]=dx/dx•dx/dt by chain rule. But dx/dx=1, which is why he wrote in the multiplication by 1 in his calculation.
(10 votes)
• Wish he'd graph the thing so I could understand what's actually going on
(4 votes)
• If you can use Google to do searches, you can do this yourself!
Just enter `16/x`
Of course that won't show you motion over time ...
(0 votes)
• Is there anywhere on Khan U where Sal provides a summary of the various "rules" (i.e. the power rule, the chain rule, the product rule, etc.)? If not, I think it would be beneficial to all of us students to publish a video with this information.
(2 votes)
• Hi everyone! I get the set up and the computation. But, what exactly does "the magnitude of our velocity vector" mean in this context? What's the difference between saying "the velocity is 2 square root of 2 units per minute" vs, "the magnitude of the velocity is 2 square-root of 2 units per minute"? I'm probably missing something. Any thoughts/clarification is appreciated.
(2 votes)
• "the velocity is 2 square root of 2 units per minute" is an incorrect statement, as velocity isn't a scalar. Instead something like "the velocity is 2 square root of 2 units per minute towards North" would be correct.
In some cases, we don't need the direction. Hence, we talk about the magnitude of velocity (also called the speed) and there, "the magnitude of the velocity is 2 square-root of 2 units per minute" would be the correct statement to use.
(3 votes)
• I want to visualise it graphically. Any ideas?
(2 votes)
• so what would the position vector be defined as? Since velocity is derived from postition
(1 vote)
• Super helpful!! Just wanted to say thanks
(1 vote)
• it seems like the initial path function is not a function of t--it would not make sense
to have x(t)*y(t)=16. So this is really different than anything done in the courses
up to this point, but there is no explanation. It seems the t parameter is being
introduced when differentiating with respect to t.
(1 vote)
## Video transcript
- [Voiceover] A particle moves along the curve xy equals 16, so that the y coordinate is increasing, we underline this, the y coordinate is increasing at a constant rate of two units per minute. That means that the rate of change of y with respect to t is equal to two. What is the magnitude, in units per minute, of the particle's velocity vector when the particle is at the point four comma four, so when x is four, and y is four. So let's see what's going on. So, let's first just remind ourselves what a velocity vector, what the velocity vector will look like. So our velocity is going to be a function of time. And it's going to have two components, it's going to be what is the rate of change in the x direction, and the rate of change in the y direction. So the rate of change in the x direction is going to be dx dt, and the rate of change in the y direction is going to be dy dt. And they tell us that this is, that dy dt is a constant two units per minute. But they're not even just asking us for just the velocity vector for its components, they're asking for the magnitude, they're asking for the magnitude of the particle's velocity vector. Well, if I have some vector, let me do a little bit of a side here, if I have some vector, let's say a, that has components b and c, well then the magnitude of my vector, sometimes you'll see it written like that, sometimes you'll see it written with double bars like that, the magnitude of my vector, and this comes straight out of the Pythagorean theorem, this is going to be the square root of b squared plus c squared. The square root of the x component squared plus the y component squared. So, if we wanted the magnitude of our velocity vector, the magnitude of the particle's velocity vector, well I could write that as the magnitude of v, I could even write it as a function of t, it's going to be equal to the square root of the x component squared, so that's the rate of change of x with respect to time, squared, plus the y component squared, which in this case is the rate of change of y, with respect to t, squared. So how do we figure out these, how do we figure out these two things? Well we already know the rate of change of y with respect to t, they say that's a constant rate of two units per minute. So we already know that this is gonna be two, or that this whole thing right over here's gonna be four, but how do we figure out the rate of change of x with respect to t? Well, we could take our original equation that describes the curve, we could take the derivative of both sides with respect to t, and then that's going to give us an equation that involves xy and dx dt and dy dt, so let's do that. So we have xy is equal to 16, I'm gonna to take the derivative with respect to t, of both sides, we do that in a different color, just for a little bit of variety, so the derivative with respect to t of the left hand side, derivative with respect to t of the right hand side, now the left hand side, we view this as a product of two functions, if we say, look, x is a function of t, and y is also a function of t, this is we're gonna do a little bit of the product rule and a little bit of the chain rule here. And so this is going to be equal to, derivative of the first function, which is, so, we'll first say the derivative of x with respect to x is one, times the derivative of x with respect to t. Remember, taking the derivative with respect to t, not with respect to x, times the second function, so times, times the second function, so times y, times y, plus, the first function, which is just x, times the derivative of the second function with respect to t. So first what's the derivative of y with respect to y, well that's just one, and then what's the derivative of y with respect to t, well that's dy dt, and that is going to be equal to, that is going to be equal to, derivative of a constant is just zero. So let's see, what does this simplify to, this simplifies to, in fact we don't even have to simplify it more, we can actually plug in the values to solve for dx dt. We know that dy dt is a constant two, and we want the magnitude of the particle's velocity vector when the particle's at the point four comma four, so when x is equal to four, so when x is equal to four, and y is equal to four, and y is equal to four. So now, it's a little messy right now, but this right here is an equation we can solve for, there's only one unknown here, the rate of change of x with respect to t, right when we are at the point four comma four, and so if we're able to figure that out, we can substitute that in here and figure out the magnitude of our velocity vector. So let us write it out, so this gives us four, four dx dt, plus what is this four times two, plus eight is equal to zero, and so we have four dx dt is equal to negative eight, just subtracted eight from both sides, divide both sides by four, you get dx dt, scroll down, is equal to negative two. So when all this stuff is going on the rate of change of x with respect to t is negative two, and then you square it, you get a four, right over here, and so the magnitude of our velocity vector is going to be equal to the square root of four plus four, which is equal to eight, which is the same thing as four times two, so this is going to be two square root of two, units per minute, so that's the magnitude of the velocity vector. | 2,673 | 11,072 | {"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-2024-22 | latest | en | 0.960108 |
https://math.answers.com/math-and-arithmetic/What_is_your_grade_if_you_miss_three_out_12 | 1,726,862,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00331.warc.gz | 341,801,844 | 47,958 | 0
Updated: 9/26/2023
Wiki User
8y ago
Providing that the other nine were correct then it is 9/12 which is 75%
Wiki User
8y ago
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Related questions
### What is your grade if you miss 1 out of 12?
A 92%. you have an A-
83%
### What is your grade if you miss three out of twelve?
To calculate your grade, you divide the number you got correct (9) by the total number of questions (12), which would give you 75%.
58% or F
### What is your grade if you miss 12 out of 20 questions?
You would have a 60%.
83% or B
### What grade will you get if you miss 1 out of 12?
Well, it is a good grade 92%, you are sill the A student in class. Good Job!
### What is your grade if you miss 12 out of 43?
If you missed 12, then you got 31 correct [43-12=31]. If all questions are weighted equally, then 31/43 = 0.721 or 72.1 %, so your grade is 72.
### How long has Burris Ewell been going to the first grade?
Burris Ewell has been going to the first grade for three years.
### Who is scouts first grade teacher?
Miss Caroline Fisher is her first grade teacher.
### What is the score to miss 12 out of 30?
The "raw score" is 60%. The letter grade earned by that score is entirely up to the teacher. We hope it wasn't a math quiz, and frankly, we wouldn't expect a high grade.
### How many are four sets of three?
why are you asking this if you went to 3rd grade you would know this but it is 12 | 401 | 1,444 | {"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-2024-38 | latest | en | 0.969906 |
https://www.exceldemy.com/uniform-probability-distribution-excel/ | 1,696,390,565,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00561.warc.gz | 834,444,994 | 59,780 | # How to Model Uniform Probability Distribution in Excel (3 Ways)
Get FREE Advanced Excel Exercises with Solutions!
Looking for ways to model uniform probability distribution in Excel? Then, this is the right place for you. Here, you will find 3 suitable ways to model uniform probability distribution in Excel.
## What Is Uniform Probability Distribution?
Uniform Probability Distribution is a type of distribution where the probability of all the events in the distribution is almost equal. If we try to plot a graph using the value of probability for this kind of distribution, we will get a rectangular form that represents the equality between the probability values.
## 3 Ways to Model Uniform Probability Distribution in Excel
Here, we will show you 3 different ways to model uniform probability distribution in Excel. Suppose, we have a dataset containing the Dice Values from 1 to 6. Now, we will model a uniform probability distribution using random values.
### 1. Apply Excel RAND & TRUNC Functions to Model Uniform Probability Distribution
In the first method, we will apply the RAND and TRUNC functions to generate random data to model a uniform probability distribution in Excel.
Here are the steps.
Steps:
• Firstly, select Cell B5 and insert the following formula.
`=TRUNC(RAND()*6+1)`
• After that, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells.
In the formula, we first created some random data using the RAND function. We multiplied the data with 6 and added 1 to get those data between 1 to 6. Then, we used the TRUNC function to remove the decimal part from the data.
• Thus, some random data will be created from 1 to 6.
• Then, select Cell E5 and insert the following formula.
`=COUNTIF(B5:B14,D5:D10)`
• Next, press Enter to get all the Count values.
Here, in the COUNTIF function, we inserted cell range B5:B14 as the range and cell range D5:D10 as the criteria.
• Next, select Cell E11 and insert the following formula.
`=SUM(E5#)`
• After that, press Enter.
In the SUM function, we added the values of the cell range E5:E10.
• Further, select Cell F5 and insert the following formula.
`=E5/\$E\$11`
• Then, press Enter and drag down the Fill Handle tool to AutoFill the formula for the rest of the cells.
Here, in the formula, we divided Cell E5 by the value of the Total number of Dice Data in Cell E11 (fixed this value as it is constant for all the following Cells).
• Finally, you will get all the Probability values for each Dice Value.
• Next, select Cell F11 and insert the following formula.
`=SUM(F5:F10)`
• Then, press Enter.
In the SUM function, we added the values of the cell range F5:F10.
### 2. Illustrate Uniform Probability Distribution Using RANDBETWEEN Function
You can also use the RANDBETWEEN function to illustrate a uniform probability distribution. Now, we will create some random data between 1 to 6 using this function to model uniform probability distribution.
Steps:
• To start with, select Cell B5 and insert the following formula.
`=RANDBETWEEN(1,6)`
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells.
In the RANDBETWEEN function, we inserted 1 as the bottom and 6 as the top value to get some random data between 1 to 6.
• Thus, you can create some random data using the RANDBETWEEN function.
• Finally, model a uniform probability distribution going through the same steps shown above.
### 3. Insert Simple Formula to Get Probability in Uniform Distribution in Excel
In the last method, we will show you how you can get the probability of some specific data in a uniform distribution by inserting a simple formula. Suppose you have some parcel having a uniformly distributed weight from 5kg to 25kg. Now, you want to know the probability of parcels having 10kg to 15kg weight in that distribution. You can solve this problem by going through the steps given below.
Steps:
• Firstly, select Cell C10 and insert the following formula.
`=(C7-C8)/(C5-C6)`
• Then, press Enter to get the probability.
Here, in the formula, we divided the value of the difference between Cell C7 and Cell C8 by the value of the difference between Cell C5 and Cell C6.
## Practice Section
In the article, you will find an Excel workbook like the image given below to practice on your own.
## Conclusion
So, in this article, we have shown you 3 ways to model uniform probability distribution in Excel. I hope you found this article interesting and helpful. If something seems difficult to understand, please leave a comment. Please let us know if there are any more alternatives that we may have missed. And, visit ExcelDemy for many more articles like this. Thank you!
## Related Articles
Arin Islam
Hello, I'm Arin. I graduated from Khulna University of Engineering and Technology (KUET) from the Department of Civil Engineering. I am passionate about learning new things and increasing my data analysis knowledge as well as critical thinking. Currently, I am working and doing research on Microsoft Excel and here I will be posting articles related to it.
We will be happy to hear your thoughts
Advanced Excel Exercises with Solutions PDF | 1,155 | 5,194 | {"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} | 4.09375 | 4 | CC-MAIN-2023-40 | latest | en | 0.847303 |
https://numbermatics.com/n/32305196/ | 1,571,215,211,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00495.warc.gz | 623,201,993 | 6,370 | 32305196
32,305,196 is an even composite number composed of five prime numbers multiplied together.
What does the number 32305196 look like?
This visualization shows the relationship between its 5 prime factors (large circles) and 48 divisors.
32305196 is an even composite number. It is composed of five distinct prime numbers multiplied together. It has a total of forty-eight divisors.
Prime factorization of 32305196:
22 × 7 × 11 × 53 × 1979
(2 × 2 × 7 × 11 × 53 × 1979)
See below for interesting mathematical facts about the number 32305196 from the Numbermatics database.
Names of 32305196
• Cardinal: 32305196 can be written as Thirty-two million, three hundred five thousand, one hundred ninety-six.
Scientific notation
• Scientific notation: 3.2305196 × 107
Factors of 32305196
• Number of distinct prime factors ω(n): 5
• Total number of prime factors Ω(n): 6
• Sum of prime factors: 2052
Divisors of 32305196
• Number of divisors d(n): 48
• Complete list of divisors:
• Sum of all divisors σ(n): 71850240
• Sum of proper divisors (its aliquot sum) s(n): 39545044
• 32305196 is an abundant number, because the sum of its proper divisors (39545044) is greater than itself. Its abundance is 7239848
Bases of 32305196
• Binary: 11110110011110000001011002
• Base-36: J8EUK
Squares and roots of 32305196
• 32305196 squared (323051962) is 1043625688598416
• 32305196 cubed (323051963) is 33714532420806794169536
• The square root of 32305196 is 5683.7660050357
• The cube root of 32305196 is 318.4863278415
Scales and comparisons
How big is 32305196?
• 32,305,196 seconds is equal to 2 days, 21 hours, 39 minutes, 56 seconds.
• To count from 1 to 32,305,196 would take you about twenty-eight weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 32305196 cubic inches would be around 26.5 feet tall.
Recreational maths with 32305196
• 32305196 backwards is 69150323
• The number of decimal digits it has is: 8
• The sum of 32305196's digits is 29
• More coming soon! | 683 | 2,398 | {"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} | 3.25 | 3 | CC-MAIN-2019-43 | latest | en | 0.862961 |
https://codepoints.net/U+2124?lang=en | 1,716,206,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00724.warc.gz | 148,995,472 | 7,866 | Home: go to the homepage U+2100 to U+214F Letterlike Symbols
# U+2124Double-Struck Capital Z
U+2124 was added in Unicode version 1.1 in 1993. It belongs to the block U+2100 to U+214F Letterlike Symbols in the U+0000 to U+FFFF Basic Multilingual Plane.
This character is a Uppercase Letter and is commonly used, that is, in no specific script. The character is also known as the set of integers.
The glyph is a font version of the glyph Latin Capital Letter Z. It has no designated width in East Asian texts. In bidirectional text it is written from left to right. When changing direction it is not mirrored. The word that U+2124 forms with similar adjacent characters prevents a line break inside it. The glyph can be confused with one other glyph.
An integer is the number zero (0), a positive natural number (1, 2, 3, etc.) or a negative integer (−1, −2, −3, etc.). The negative numbers are the additive inverses of the corresponding positive numbers. The set of all integers is often denoted by the boldface Z or blackboard bold $Z$.
The set of natural numbers $N$ is a subset of $Z$, which in turn is a subset of the set of all rational numbers $Q$, itself a subset of the real numbers $R$. Like the set of natural numbers, the set of integers $Z$ is countably infinite. An integer may be regarded as a real number that can be written without a fractional component. For example, 21, 4, 0, and −2048 are integers, while 9.75, 5+1/2, and 2 are not.
The integers form the smallest group and the smallest ring containing the natural numbers. In algebraic number theory, the integers are sometimes qualified as rational integers to distinguish them from the more general algebraic integers. In fact, (rational) integers are algebraic integers that are also rational numbers.
## Representations
System Representation
8484
UTF-8 E2 84 A4
UTF-16 21 24
UTF-32 00 00 21 24
URL-Quoted %E2%84%A4
HTML hex reference ℤ
Wrong windows-1252 Mojibake ℤ
HTML named entity ℤ
HTML named entity ℤ
alias the set of integers
LATEX \mathbb{Z}
## Complete Record
Property Value
Age (age) 1.1 (1993)
Unicode Name (na) DOUBLE-STRUCK CAPITAL Z
Unicode 1 Name (na1) DOUBLE-STRUCK Z
Block (blk) Letterlike Symbols
General Category (gc) Uppercase Letter
Script (sc) Common
Bidirectional Category (bc) Left To Right
Combining Class (ccc) Not Reordered
Decomposition Type (dt) font
Decomposition Mapping (dm) Latin Capital Letter Z
Lowercase (Lower)
Simple Lowercase Mapping (slc) Double-Struck Capital Z
Lowercase Mapping (lc) Double-Struck Capital Z
Uppercase (Upper)
Simple Uppercase Mapping (suc) Double-Struck Capital Z
Uppercase Mapping (uc) Double-Struck Capital Z
Simple Titlecase Mapping (stc) Double-Struck Capital Z
Titlecase Mapping (tc) Double-Struck Capital Z
Case Folding (cf) Double-Struck Capital Z
ASCII Hex Digit (AHex)
Alphabetic (Alpha)
Bidi Control (Bidi_C)
Bidi Mirrored (Bidi_M)
Composition Exclusion (CE)
Case Ignorable (CI)
Changes When Casefolded (CWCF)
Changes When Casemapped (CWCM)
Changes When NFKC Casefolded (CWKCF)
Changes When Lowercased (CWL)
Changes When Titlecased (CWT)
Changes When Uppercased (CWU)
Cased (Cased)
Full Composition Exclusion (Comp_Ex)
Default Ignorable Code Point (DI)
Dash (Dash)
Deprecated (Dep)
Diacritic (Dia)
Emoji Modifier Base (EBase)
Emoji Component (EComp)
Emoji Modifier (EMod)
Emoji Presentation (EPres)
Emoji (Emoji)
Extender (Ext)
Extended Pictographic (ExtPict)
FC NFKC Closure (FC_NFKC) Latin Small Letter Z
Grapheme Cluster Break (GCB) Any
Grapheme Base (Gr_Base)
Grapheme Extend (Gr_Ext)
Hex Digit (Hex)
Hyphen (Hyphen)
ID Continue (IDC)
ID Start (IDS)
IDS Binary Operator (IDSB)
IDS Trinary Operator and (IDST)
IDSU (IDSU) 0
ID_Compat_Math_Continue (ID_Compat_Math_Continue) 0
ID_Compat_Math_Start (ID_Compat_Math_Start) 0
Ideographic (Ideo)
InCB (InCB) None
Indic Mantra Category (InMC)
Indic Positional Category (InPC) NA
Indic Syllabic Category (InSC) Other
Jamo Short Name (JSN)
Join Control (Join_C)
Logical Order Exception (LOE)
Math (Math)
Noncharacter Code Point (NChar)
NFC Quick Check (NFC_QC) Yes
NFD Quick Check (NFD_QC) Yes
NFKC Casefold (NFKC_CF) Latin Small Letter Z
NFKC Quick Check (NFKC_QC) No
NFKC_SCF (NFKC_SCF) Latin Small Letter Z
NFKD Quick Check (NFKD_QC) No
Other Alphabetic (OAlpha)
Other Default Ignorable Code Point (ODI)
Other Grapheme Extend (OGr_Ext)
Other ID Continue (OIDC)
Other ID Start (OIDS)
Other Lowercase (OLower)
Other Math (OMath)
Other Uppercase (OUpper)
Prepended Concatenation Mark (PCM)
Pattern Syntax (Pat_Syn)
Pattern White Space (Pat_WS)
Quotation Mark (QMark)
Regional Indicator (RI)
Sentence Break (SB) Upper
Soft Dotted (SD)
Sentence Terminal (STerm)
Terminal Punctuation (Term)
Unified Ideograph (UIdeo)
Variation Selector (VS)
Word Break (WB) Alphabetic Letter
White Space (WSpace)
XID Continue (XIDC)
XID Start (XIDS)
Expands On NFC (XO_NFC)
Expands On NFD (XO_NFD)
Expands On NFKC (XO_NFKC)
Expands On NFKD (XO_NFKD)
Bidi Paired Bracket (bpb) Double-Struck Capital Z
Bidi Paired Bracket Type (bpt) None
East Asian Width (ea) neutral
Hangul Syllable Type (hst) Not Applicable
ISO 10646 Comment (isc)
Joining Group (jg) No_Joining_Group
Joining Type (jt) Non Joining
Line Break (lb) Alphabetic
Numeric Type (nt) none
Numeric Value (nv) not a number
Simple Case Folding (scf) Double-Struck Capital Z
Script Extension (scx)
Vertical Orientation (vo) R | 1,577 | 5,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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-2024-22 | latest | en | 0.876601 |
http://eel.is/c++draft/rand.adapt.ibits | 1,547,756,690,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00249.warc.gz | 78,049,362 | 4,436 | # 25 Numerics library [numerics]
## 25.6 Random number generation [rand]
An independent_bits_engine random number engine adaptor combines random numbers that are produced by some base engine e, so as to produce random numbers with a specified number of bits w.
The state x of an independent_bits_engine engine adaptor object x consists of the state e of its base engine e; the size of the state is the size of e's state.
The transition and generation algorithms are described in terms of the following integral constants:
• Let and .
• With n as determined below, let , , , and .
• Let if and only if the relation holds as a result.
Otherwise let .
[Note
:
The relation always holds.
end note
]
The transition algorithm is carried out by invoking e() as often as needed to obtain values less than and values less than .
The generation algorithm uses the values produced while advancing the state as described above to yield a quantity S obtained as if by the following algorithm:
S = 0;
for (k = 0; k≠n0; k += 1) {
do u = e() - e.min(); while (u≥y0);
S = 2w0⋅S+umod2w0;
}
for (k = n0; k≠n; k += 1) {
do u = e() - e.min(); while (u≥y1);
S = 2w0+1⋅S+umod2w0+1;
}
template<class Engine, size_t w, class UIntType>
class independent_bits_engine {
public:
// types
using result_type = UIntType;
// engine characteristics
static constexpr result_type min() { return 0; }
static constexpr result_type max() { return 2w−1; }
// constructors and seeding functions
independent_bits_engine();
explicit independent_bits_engine(const Engine& e);
explicit independent_bits_engine(Engine&& e);
explicit independent_bits_engine(result_type s);
template<class Sseq> explicit independent_bits_engine(Sseq& q);
void seed();
void seed(result_type s);
template<class Sseq> void seed(Sseq& q);
// generating functions
result_type operator()(); | 467 | 1,832 | {"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} | 2.9375 | 3 | CC-MAIN-2019-04 | longest | en | 0.720397 |
https://customelitewriters.com/index.php/2018/11/06/discuss-the-steps-to-take-to-increase-the-security-of-information-on-your-computer-and-online/ | 1,601,146,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244353.70/warc/CC-MAIN-20200926165308-20200926195308-00037.warc.gz | 319,325,580 | 14,483 | # Discuss the steps to take to increase the security of information on your computer and online
Discuss the steps to take to increase the security of information on your computer and online
INVITATION TO
Computer Science 11
Chapter 8 Information Security
Objectives
After studying this chapter, students will be able to:
• Describe the steps to take to increase the security of
information on your computer and online
• Explain how passwords are encrypted using a hash
function on many systems
• Describe cyber-attacks, including viruses, worms,
Trojan horses, DOS attacks, and phishing, and explain
how they differ from each other
• Encrypt and decrypt messages using simple Caesar
ciphers and matrix-based block ciphers
Invitation to Computer Science, 6th Edition 2
Objectives (continued)
After studying this chapter, students will be able to:
• Describe the overall process used by symmetric
encryption algorithms such as DES
• Compare symmetric versus asymmetric (public key)
encryption
• Describe the overall process used by RSA encryption
• Explain why web transmission protocols such as SSL
and TLS use multiple forms of encryption to secure
data transfer over the web
• Explain the importance of considering computer
security for networked embedded systems
Invitation to Computer Science, 6th Edition 3
Introduction
• Information security
– Keep information safe
• Physical security – lock doors, maintain control of
devices
• Online security
– Secure assembly language
– Secure operating system
– Secure network
Invitation to Computer Science, 7th Edition 4
Threats and Defenses
• Authentication: establishing identity
• Secure a password file with a hash function;
one-way encryption
1. Replace letters by numbers: 2 1 4 2 15 25 2
3. Remainder of sum/7: 51 mod 7 = 2
4. Add 1 and multiply by 9: (2+1)*9 = 27
5. Reverse digits and convert to letters: 72 = gb
Invitation to Computer Science, 7th Edition 5
Threats and Defenses (cont’d.)
stored
– Hash input password and compare
• More secure method
– Identical passwords won’t hash to identical values
Invitation to Computer Science, 7th Edition 6
Threats and Defenses (cont’d.)
• Guess password, brute force or from knowledge
– Try personal references (e.g., pet name)
– Try all possible passwords (computationally difficult)
software
– Tries words and word combinations, millions of
• Social engineering: get a person to tell password
Invitation to Computer Science, 7th Edition 7
Threats and Defenses (cont’d.)
Other authentication methods
• Biometric information (fingerprint or retinal scans)
– User enters ID and a partial password
– System or user device generates last half of the
– Last half of the password is good for only a few
seconds
Invitation to Computer Science, 7th Edition 8
Threats and Defenses (cont’d.)
• Authorization: set of permitted actions for each
authorized person
• Operating system maintains access control lists
– Write access (modify a file)
– Execute access (run a program)
– Delete access (remove a file
• System administrator or superuser has universal
access and sets up authorization
Invitation to Computer Science, 7th Edition 9
Threats and Defenses (cont’d.)
• Malware: malicious software arriving from the
network
– Virus: program embedded within another program
or file, replicates itself and attacks other files
– Worm: program that can send copies of itself to
other nodes on the network
– Trojan horse: program that seems beneficial, but
hides malicious code within it
• Keystroke logger: records all keys typed
Invitation to Computer Science, 7th Edition 10
Threats and Defenses (cont’d.)
• Denial-of-service (DOS) attack: many computers
try to access the same URL at the same time
– Clogs the network, prevents legitimate access, and
causes the server to crash
– Distributed DOS uses thousands of computers
• Uses a zombie army (botnet): many innocent
computers infected with malware
• Phishing: obtain sensitive information by
impersonating legitimate sources
– Many emails; just a few “bites” are enough
Invitation to Computer Science, 7th Edition 11
Encryption
• Cryptography: science of secret writing
• Encryption and decryption (inverse operations)
– Convert from plaintext to ciphertext and back again
• Symmetric encryption algorithm
– A secret key shared by the sender and the receiver
– Same key is used to encrypt and decrypt
• Asymmetric encryption algorithm (public key)
– Uses two keys, public and private
– Use public key (generally known) to encrypt
– Use private key (known only to receiver) to decrypt
Invitation to Computer Science, 7th Edition 12
Encryption (cont’d.)
• Caesar cipher (shift cipher)
– Map characters to others a fixed distance away in
alphabet
– Example: AE, BF, CG…UY, VZ, WA
– Stream cipher: encode each character as it comes
• Substitution cipher: similar, but implement other
mappings
• Pros: easy and fast, can do character by character
• Cons: letter frequency, double letters, still pertain,
makes it easy to break
Invitation to Computer Science, 7th Edition 13
Encryption (cont’d.)
Block cipher
• Block of plaintext encoded into a block of ciphertext
• Each character contributes to multiple characters
• Matrix-based block cipher
– Group characters into blocks n characters long
– Find invertible n by n matrix, M, and its inverse, M’
as keys
– Map characters to letters A1, B2, etc.
– Wrap values 26 and above back to zero: 260,
271, etc.
Invitation to Computer Science, 7th Edition 14
Invitation to Computer Science, 7th Edition 15
Encryption (cont’d.)
Example: use 2 by 2 matrices:
M = M′ =
Encrypt block GO
• Convert to vector V = [7 15]
• Matrix multiplication:
V x M = [7*3 + 15*2 7*5 + 15*3]
= [51 80] = [25 2]
• Convert to string: YB
Invitation to Computer Science, 7th Edition 16
3 5
2 3
23 5
2 23
Encryption (cont’d.)
Example: Use 2 by 2 matrices:
M = M′ =
Decrypt block YB
• Convert to vector V2 = [25 2]
• Matrix multiplication:
V2 x M′ = [25*23 + 2*2 5*25 + 23*2]
= [579 171] = [7 15]
• Convert to string: YB
Invitation to Computer Science, 7th Edition 17
3 5
2 3
23 5
2 23
Encryption (cont’d.)
DES (Data Encryption Standard)
• Symmetric encryption algorithm
• Designed for digital data: plaintext is a binary string
• Uses 64-bit binary key (56 bits actually used)
• Sixteen rounds of the same series of manipulations
Invitation to Computer Science, 7th Edition 18
Encryption (cont’d.)
DES (Data Encryption Standard)
• Decryption uses the same algorithm; keys in
reverse
• Fast and effective, but requires shared key
– 56 bits is too small for modern technology
• AES (Advanced Encryption Standard) uses a
similar approach; longer keys
Invitation to Computer Science, 7th Edition 19
Encryption (cont’d.)
DES manipulations
• Split string
• Duplicating some bits
• Omit some bits
• Permute bit order
• Combine bit strings with
XOR (exclusive or)
Invitation to Computer Science, 7th Edition 20
Invitation to Computer Science, 7th Edition 21
Encryption (cont’d.)
Public-key systems: RSA key creation
• Pick 2 large prime numbers: p and q
• Compute n = p×q, and m = (p-1)×(q-1)
• Choose large number e at random, so that e and m
are relatively prime (no common factors except 1)
• Find unique value d, between 0 and m, such that
(e×d) modulo m = 1
• Public key = (n, e), Private key = d
Invitation to Computer Science, 7th Edition 22
Encryption (cont’d.)
RSA key creation, example:
• p = 7, q = 13
• n = 7×13 = 91, and m = 6×12 = 72
• Let e = 77 (72 = 2 * 2 * 2 * 3 * 3, 77 = 7 * 11)
• d = 29
• Public key = (91, 25), Private key = 29
Invitation to Computer Science, 7th Edition 23
Encryption (cont’d.)
RSA encryption:
Given public key (n, e)
• Convert message to integer P
• Calculate C = Pe modulo n
RSA decryption:
Given private key d
• Calculate Cd modulo n
Invitation to Computer Science, 7th Edition 24
Encryption (cont’d.)
RSA encryption, example:
Given public key (91, 25)
• Convert message to integer P = 37
• Calculate C = 3725 modulo 91 = 46
RSA decryption:
Given private key 29
• Calculate 4629 modulo 91 = 37
Invitation to Computer Science, 7th Edition 25
Web Transmission Security
• Ecommerce requires secure transmission of
names, passwords, and credit card numbers
• Web protocols: SSL (Secure Sockets Layer) and
TLS (Transport Layer Security)
– Client-server applications
– Server provides certificate of authentication and
server’s public key
– Client sends its DES key, encrypted using RSA
– Data is sent encrypted by the (now shared) DES key
Invitation to Computer Science, 7th Edition 26
Invitation to Computer Science, 7th Edition 27
Embedded Computing
• Embedded computers: special-purpose, limited
computers in other systems
• Examples: automobiles, smart appliances, remote
controls, and patient monitoring systems
• New trend: connect embedded computers to a
network
• Targeting embedded systems could cause chaos
– Change thermostats, disrupt patient care, or disable
aircraft or automobiles
Invitation to Computer Science, 7th Edition 28
Summary
• Internet and web are meant to promote information
exchange, so information security is hard
• Online attacks include viruses, worms, Trojan
horses, DOS attacks, and phishing, among others
• Data security involves encrypting sensitive data
before transmitting or storing in unsecured location
• Symmetric encryption requires a shared key
• Asymmetric encryption uses public and private
keys
Invitation to Computer Science, 7th Edition 29
Summary (cont’d.)
• Caesar cipher is a simple symmetric encryption,
substitution ciphers are similar
• Block ciphers combine blocks of plaintext symbols
into blocks of ciphertext
• DES and AES are strong symmetric encryption
algorithms
• RSA is the most common asymmetric algorithm
• Secure web transmission requires protocols:
SSL/TLS
• Embedded systems are the next problem to solve
Invitation to Computer Science, 7th Edition 30 | 2,513 | 10,089 | {"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} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.847484 |
https://www.mathworks.com/matlabcentral/cody/problems/32-most-nonzero-elements-in-row/solutions/213984 | 1,511,102,389,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805649.7/warc/CC-MAIN-20171119134146-20171119154146-00622.warc.gz | 822,892,760 | 11,645 | Cody
# Problem 32. Most nonzero elements in row
Solution 213984
Submitted on 7 Mar 2013 by Heather
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
%% a = [ ... 1 2 0 0 0 0 0 5 0 0 2 7 0 0 0 0 6 9 3 3]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
2 Pass
%% a = [ ... 1 2 0 0 0 0 5 0 0 6 9 -3 2 7 0 0 0 0 0 0]; r_correct = 3; assert(isequal(fullest_row(a),r_correct))
3 Pass
%% a = [ ... 1 0 0 0 0 0 0 0 0 0 0 0 0 2 3]; r_correct = 5; assert(isequal(fullest_row(a),r_correct))
4 Pass
%% a = [ ... 0 0 0 -3 0 0]; r_correct = 4; assert(isequal(fullest_row(a),r_correct)) | 303 | 711 | {"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} | 2.796875 | 3 | CC-MAIN-2017-47 | latest | en | 0.586748 |
http://www.ltu.edu/courses/lowry/techcalc/mod4-3.htm | 1,472,500,358,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982965886.67/warc/CC-MAIN-20160823200925-00028-ip-10-153-172-175.ec2.internal.warc.gz | 547,525,068 | 2,462 | Unit 4 - Objective 3 - Graphing Rational Functions
Asymptotes: If the graph of y = f(x) gets closer and closer to the line y = L or x = a, then these lines are asymptotes.
Definition: If f(x) = L, then y = L is a Horizontal Asymptote.
y = 1 is a horizontal asymptote for each graph below
Note: To find the horizontal asymptotes for f(x), take the limit as of the given funtion. If this limit is a number, L, then y = L is a horizontal asymptote for the graph.
!!An abbreviated limit notation will be used in this review!!
AbbreviationMeaning
as "y approaches 4 as x approaches 1"
and means limit of the y values, as x goes to 1, is 4
"x increases without bounds" and we say "x goes to infinity"
"x decreases without bounds"
"x approaches 5 from the right"
x only has values greater than 5
"x approaches 5 from the left"
x less than 5
Definition: If as or as then x = a is a Vertical Asymptote. In other words, the y values increase or decrease without bound as the x values get closer to the line x = a.
x = 2 is a Vertical Asymptote for each of the graphs below.
Note: To find the vertical asymptote, reduce f(x) then if x = a makes the denominator zero (and not the numerator), x = a is a vertical asymptote for the graph. The graph will NEVER cross a vertical asymptote. The function is undefined there.
Given: find the vertical and horizontal asymptotes. Then, sketch the graph.
To find the horizontal asymptotes, take the limit as
To find the vertical asymptotes, factor demonimator (after reducing if possible)
denominator = 0 (numerator not 0)when x = 3 or x = -3
Therefore: x = 3 and x = -3 are Vertical Asymptotes
To sketch by hand:
1. Draw all of the asymptotes with dashed lines.
2. Plot points in between the vertical asymptotes.
xy
-432/-7 -4.5
00
4-32/7 -4.5
Now you may plot more points or you can investigate the behavior of y = f(x) as the x values get close to the vertical asymptotes. If then the y values become larger and larger positive and the graph goes up. If then the y values become more and more negative and the graph goes down. The graph never crosses a vertical asymptote.
Using the points above, sketch information about the horizontal asymptotes, as, , and as , (behavior at left and right ends) Usually the outside branches of the graph approach thevertical asymptotes, or plot the y values for x close to the vertical asymptotes like x = 3.1, -3.1 In the middle of the graph, plot points inside the asymptotes, like x = 2.9, -2.9, or check to see if the graph crosses the horizontal asymptotes by setting y = -2 and see if you can solve for x.
[Unit 4 Outline] [Course Outline] [Home Page] | 683 | 2,644 | {"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} | 4.84375 | 5 | CC-MAIN-2016-36 | latest | en | 0.865861 |
http://www.qacollections.com/How-to-Get-Your-Creative-Juices-Flowing | 1,545,190,948,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00075.warc.gz | 458,669,344 | 5,877 | # How to Get Your Creative Juices Flowing?
Have you ever been stumped for an idea? It can be really frustrating and counterproductive when the right idea just won't come to you. So, no matter what type of idea you are looking for, whether i... Read More »
http://www.wikihow.com/Get-Your-Creative-Juices-Flowing
Top Q&A For: How to Get Your Creative Juices Flowing
## Why is there no water flowing from the kitchen faucet but everywhere else including the dishwasher there is water flowing?
Sound level Lp = 64 dBSPL equals what sound pressure p (rms)?The sound pressure is: 0.0317 pascals (Pa) = 31.7 millipascals (mPa)The reference sound pressure is: p0 = 20 μPa = 2 · 10-5 Pa (0 dB)T... Read More »
## Anyone up for a rating of Juices On a rating of 0-10 how do you rate the Juices on this list?
1) Grapefruit Juice-92) Orange Juice.-103) Apple Juice-104) Pineapple Juice.-95) Vegetable Juice-I'm not sure. I've only tried one and it was good.6) Pear Juice.-97) Grape Juice-108) Prune Juice-19... Read More »
## How to Get Ink Flowing in Gel Pens?
Do you have any gel pens? Do you want them to have ink flowing? You've come to the right place.
http://www.wikihow.com/Get-Ink-Flowing-in-Gel-Pens
## How to Write Free Flowing Lyrics?
If you want to write song lyrics in a manner unencumbered by rules about writing lyrics but you just don't know where to start, this article has some suggestions to follow.
http://www.wikihow.com/Write-Free-Flowing-Lyrics
Related Questions | 408 | 1,487 | {"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} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.895651 |
https://www.physicsforums.com/threads/work-function.311581/ | 1,508,353,096,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823067.51/warc/CC-MAIN-20171018180631-20171018200631-00526.warc.gz | 939,833,732 | 16,762 | # Work function
1. May 3, 2009
### itryphysics
1. The problem statement, all variables and given/known data
Red light of wavelength 670 nm produces photoelectrons from a certain material. Green light of wavelength 522 nm produces photoelectrons from the same material with 1.2 times the previous maximum kinetic energy. What is the material's work function? answer in units of eV
2. Relevant equations
hf = KEmax + Wo
3. The attempt at a solution
(6.63e^-34)(3e^8) / 522e-9 = 1.2 + W
the answer i get after punching this into the calculator is 1.2 since the left side of the equation virtually equals zero. Please help me solve this. Thanks!
2. May 3, 2009
### dx
You just have to write the equation hf = KEmax + W0 for both the cases and solve them as simultaneous equations. If KEmax for red is A, then KEmax for green is (1.2)A.
3. May 3, 2009
### itryphysics
Yes I did that. But its the work function that I need. everytime i had 1.2 on the left side of the equation , the work function equalled 1.2
Please, can you provide any further help?
4. May 3, 2009
### dx
Write out your steps and I'll tell you where you went wrong.
5. May 3, 2009
### itryphysics
6.63e^-34 (3e^8) / (522e^-9) = 1.2 + W
7.8 e^-19 = 1.2 + W
7.8 e ^-19 - 1.2 = W
W= 1.2 (since 7.8 e^-19 is almost equal to zero)
help =(
6. May 3, 2009
### dx
1.2 multiplies KEmax of red. You seem to have it by itself added to W, and I don't really get what you did.
First write out the equations in symbols. What are the two equations in symbols?
7. May 3, 2009
### itryphysics
hf = KE + W and hf = 1.2KE + W
???
8. May 3, 2009
### itryphysics
I dont have the KE of red. So wht do i multiply 1.2 by?
9. May 3, 2009
### dx
The equations you just wrote are correct. Now there are two variables, KE and W in these equations, and you have two equations, so you can eliminate KE, and solve for W.
10. May 3, 2009
### itryphysics
sorry still a bit confused. Do I not take KE into account at all? wht do i do with the 1.2?
11. May 3, 2009
### dx
Do you know how to solve simultaneous equations?
12. May 3, 2009
### itryphysics
eeeek ...not sure .. can you get me started at least , please.
13. May 3, 2009
### dx
Ok. Take the first equation, and solve it for KE. You will get it in terms of fred and W. Now substitute this in the second equation, and you will get an equation without KE in it.
14. May 3, 2009
### itryphysics
So....
KE = hf - W
then after substitution ..
hf = 1.2 (hf - W) + W
hf = 1.2 hf -1.2W + W
-0.2 hf = -0.2 W
since f=c/lambda
0.2 hc / lambda =0.2 W
right?
15. May 3, 2009
### dx
Almost, but you made one mistake. The f in the first equation should be fred = 670 nm and the f in the second equation should be fgreen = 522 nm. They're different.
16. May 3, 2009
### itryphysics
ok thanks!
but my answer is coming out to be a negative ..
(6.63 e^-34)(670E-9)= 1.2 (6.63e^-34)(522e^-9) - 0.2W
4.4421e^-40 = 4.15303 e^-40 -0.2W
2.8907e^-41 = -0.2W this makes W negative.. | 1,000 | 3,005 | {"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-2017-43 | latest | en | 0.92277 |
http://mathhelpforum.com/calculus/10570-hard-derivative.html | 1,524,436,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00286.warc.gz | 207,902,399 | 9,554 | 1. Hard derivative
Differentiate the fnction
$\displaystyle f(x) = \frac{(x^{-1} + x^{2})^{-1}}{(x^{3} - 2x^{-2})^{-2}}$
Differentiate the fnction
$\displaystyle f(x) = \frac{(x^{-1} + x^{2})^{-1}}{(x^{3} - 2x^{-2})^{-2}}$
Someone wants to give you an ulcer, don't they?
Simplify first:
$\displaystyle f(x) = \frac{(x^{-1} + x^{2})^{-1}}{(x^3 - 2x^{-2})^{-2}}$
$\displaystyle f(x) = \frac{(x^3 - 2x^{-2})^2}{x^{-1} + x^2}$
$\displaystyle f(x) = \frac{\left ( x^3 - \frac{2}{x^2} \right )^2}{\frac{1}{x} + x^2}$
I'm going to expand the numerator so my next step will be a bit clearer:
$\displaystyle f(x) = \frac{x^6 - \frac{4x^3}{x^2} + \frac{4}{x^4}}{\frac{1}{x} + x^2}$
$\displaystyle f(x) = \frac{x^6 - 4x + \frac{4}{x^4}}{\frac{1}{x} + x^2}$
Now mulitply the numerator and denominator by x^4. We would normally need to be wary of doing this, but the domain of the original function excludes x = 0 anyway in this case.
$\displaystyle f(x) = \frac{x^{10} - 4x^5 + 4}{x^3 + x^6}$
You should be able to take the derivative now.
-Dan | 413 | 1,043 | {"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} | 4.4375 | 4 | CC-MAIN-2018-17 | latest | en | 0.679847 |
http://mathbitsnotebook.com/Algebra1/StatisticsData/STCenterPractice.html | 1,656,328,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00490.warc.gz | 39,861,851 | 5,306 | Measures of Center and Shapes of Distributions Terms of Use Contact Person: Donna Roberts
Directions: Read carefully and choose the best answers.
1. The number of snowboarding accidents reported weekly during one winter season at the Fun Mountain Resort: 12, 15, 6, 8, 12, 17, 10, 8, 7, 13, 14, 16, 8, 18, 11
(a) Find the mean.
Choose: 11.7 12 12.7 12.8
(b)
Find the median.
Choose: 11 12 13 14
(c)
Find the mode.
Choose: 8 11 12 no mode
(d)
A teenager wants to emphasize the safety of snowboarding. Which measure of central tendency should the teen use?
Choose: mean median mode
(e)
A parent wants to emphasize the dangers of snowboarding. Which measure of central tendency should the parent use?
Choose: mean median mode
2. A company is releasing two new types of knee socks: argyle and plaid. The table below shows the sales numbers for the first week.
Day of Week Mon Tues Wed Thurs Fri Sat Sun Argyle 8 10 1 10 2 10 1 Plaid 14 6 8 6 18 12 6
(a)
Find the mean, median and mode of the
argyle knee socks sales.
Choose: 6, 8, 10 10, 8, 6 8, 6, 10 10, 6, 8
(b)
Find the mean, median and mode of the
plaid knee socks sales.
Choose: 6, 8, 10 10, 8, 6 8, 10, 6 10, 6, 8
(c)
Which measure of central tendency would you use to show that the
argyle knee socks are selling better?
Choose: mean median mode
(d)
Which measure of central tendency would you use to show that the
plaid knee socks are selling better?
Choose: mean median mode
(e)
Which measure of central tendency would you use to show that there is no difference between the sales of the two types of knee socks?
Choose: mean median mode
(f)
Which knee socks sales had the larger spread of sales?
Choose: argyle plaid both were the same
3. The box plot shown at the right is described as being a ____________.
Choose:
symmetric distribution skewed left distribution skewed right distribution
4. Compare mean and median in differently shaped data distributions. For each type of distribution stated below, decide what is "typically" the relationship between the mean and the median.
Type of distribution:
mean larger median larger mean = median
symmetric, bell-shaped distribution
distribution skewed right
distribution skewed left
uniform distribution
5. The best measure of center to use when data is skewed is the ____________.
Choose:
mean median mode
6. Which of the following statements is false?
Choose:
In a skewed distribution, the mean typically gets pulled toward the tail. The graph of a distribution with one clear peak is called a unimodal distribution. The graph clearly showing the median value is the box plot. The median can also be thought of as a balance point.
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | 747 | 2,877 | {"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-2022-27 | latest | en | 0.888482 |
https://hyperleap.com/topic/Sectional_curvature | 1,611,480,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00664.warc.gz | 381,768,116 | 18,473 | # Sectional curvature
curvaturecurvature tensorsmanifolds with constant sectional curvaturenegatively curvednon-positive sectional curvaturenon-positively curvedsectional
In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds.wikipedia
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### Space form
space forms
In mathematics, a space form is a complete Riemannian manifold M of constant sectional curvature K.
### Hyperbolic space
hyperbolic 3-spacehyperbolic planehyperbolic 4-space
Hyperbolic n-space, denoted H n, is the maximally symmetric, simply connected, n-dimensional Riemannian manifold with a constant negative sectional curvature.
### Exponential map (Riemannian geometry)
exponential mapexponential map of this Riemannian metricExponential map, Riemannian geometry
It is the Gaussian curvature of the surface which has the plane σ p as a tangent plane at p, obtained from geodesics which start at p in the directions of σ p (in other words, the image of σ p under the exponential map at p).
The exponential map is also useful in relating the abstract definition of curvature to the more concrete realization of it originally conceived by Riemann himself—the sectional curvature is intuitively defined as the Gaussian curvature of some surface (i.e., a slicing of the manifold by a 2-dimensional submanifold) through the point p in consideration.
### Gaussian curvature
Gauss curvaturecurvatureLiebmann's theorem
It is the Gaussian curvature of the surface which has the plane σ p as a tangent plane at p, obtained from geodesics which start at p in the directions of σ p (in other words, the image of σ p under the exponential map at p).
In 1928, Élie Cartan proved the Cartan–Hadamard theorem: if M is a complete manifold with non-positive sectional curvature, then its universal cover is diffeomorphic to a Euclidean space.
In mathematics, the Cartan–Hadamard theorem is a statement in Riemannian geometry concerning the structure of complete Riemannian manifolds of non-positive sectional curvature.
### Riemann curvature tensor
Riemann tensorcurvature tensorcurvature
The sectional curvature determines the curvature tensor completely.
The Gaussian curvature coincides with the sectional curvature of the surface.
### Comparison theorem
comparecomparison methodcomparison theorem in Riemannian geometry
If tighter bounds on the sectional curvature are known, then this property generalizes to give a comparison theorem between geodesic triangles in M and those in a suitable simply connected space form; see Toponogov's theorem.
### Toponogov's theorem
Toponogov theoremToponogov's triangle comparison theoremV. Toponogov
If tighter bounds on the sectional curvature are known, then this property generalizes to give a comparison theorem between geodesic triangles in M and those in a suitable simply connected space form; see Toponogov's theorem. Toponogov's theorem affords a characterization of sectional curvature in terms of how "fat" geodesic triangles appear when compared to their Euclidean counterparts.
Let M be an m-dimensional Riemannian manifold with sectional curvature K satisfying
### Soul theorem
soul conjecturesouls
The soul theorem implies that a complete non-compact non-negatively curved manifold is diffeomorphic to a normal bundle over a compact non-negatively curved manifold.
In mathematics, the soul theorem is a theorem of Riemannian geometry that largely reduces the study of complete manifolds of non-negative sectional curvature to that of the compact case.
### Preissman's theorem
Preissman's theorem restricts the fundamental group of negatively curved compact manifolds.
In Riemannian geometry, a field of mathematics, Preissman's theorem is a statement that restricts the possible topology of a negatively curved compact Riemannian manifold M.
### Myers's theorem
Myers theoremMyers' theoremBonnet–Myers theorem
A weaker result, due to Ossian Bonnet, has the same conclusion but under the stronger assumption that the sectional curvatures is bounded below by k.
### Curvature
curvednegative curvatureextrinsic curvature
The curvature of a curve can naturally be considered as a kinematic quantity, representing the force felt by a certain observer moving along the curve; analogously, curvature in higher dimensions can be regarded as a kind of tidal force (this is one way of thinking of the sectional curvature).
### Riemannian geometry
Riemannianlocal to global theoremsRiemann geometry
In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds.
### Curvature of Riemannian manifolds
curvatureabstract definition of curvaturecurvature of a Riemannian manifold
In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds.
### Tangent space
tangent planetangenttangent vector
The sectional curvature K(σ p ) depends on a two-dimensional plane σ p in the tangent space at a point p of the manifold.
### Surface (topology)
surfaceclosed surfacesurfaces
It is the Gaussian curvature of the surface which has the plane σ p as a tangent plane at p, obtained from geodesics which start at p in the directions of σ p (in other words, the image of σ p under the exponential map at p).
### Geodesic
geodesicsgeodesic flowgeodesic equation
It is the Gaussian curvature of the surface which has the plane σ p as a tangent plane at p, obtained from geodesics which start at p in the directions of σ p (in other words, the image of σ p under the exponential map at p).
### Grassmannian
GrassmanniansGrassmann manifold Grassmannian manifolds
The sectional curvature is a smooth real-valued function on the 2-Grassmannian bundle over the manifold.
### Fiber bundle
structure grouplocal trivializationtrivial bundle
The sectional curvature is a smooth real-valued function on the 2-Grassmannian bundle over the manifold.
### Riemannian manifold
Riemannian metricRiemannianRiemannian manifolds
Given a Riemannian manifold and two linearly independent tangent vectors at the same point, u and v, we can define
### Linear independence
linearly independentlinearly dependentlinear dependence
Given a Riemannian manifold and two linearly independent tangent vectors at the same point, u and v, we can define
### Tangent vector
tangent vectorstangent directionstangent
Given a Riemannian manifold and two linearly independent tangent vectors at the same point, u and v, we can define
### Orthonormality
orthonormalorthonormal setorthonormal sequence
In particular, if u and v are orthonormal, then
### Euclidean space
EuclideanspaceEuclidean vector space
In 1928, Élie Cartan proved the Cartan–Hadamard theorem: if M is a complete manifold with non-positive sectional curvature, then its universal cover is diffeomorphic to a Euclidean space. | 1,529 | 6,872 | {"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-2021-04 | latest | en | 0.884436 |
http://stackoverflow.com/questions/804268/moving-particles-in-c-and-opengl/804322 | 1,435,635,583,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375091587.3/warc/CC-MAIN-20150627031811-00290-ip-10-179-60-89.ec2.internal.warc.gz | 240,438,575 | 20,697 | Moving particles in C and OpenGL
I want to be able to move a particle in a straight line within a 3D environment but I can't think how to work out the next location based on two points within a 3D space?
I have created a struct which represents a particle which has a location and a next location? Would this be suitable to work out the next location to move too? I know how to initially set the next location using the following method:
``````// Set particle's direction to a random direction
void setDirection(struct particle *p)
{
float xnm = (p->location.x * -1) - p->velocity;
float xnp = p->location.x + p->velocity;
float ynm = (p->location.y * -1) - p->velocity;
float ynp = p->location.y + p->velocity;
float znm = (p->location.z * -1) - p->velocity;
float znp = p->location.z + p->velocity;
struct point3f nextLocation = { randFloat(xnm, xnp), randFloat(ynm, ynp), randFloat(znm, znp) };
p->nextLocation = nextLocation;
}
``````
The structs I have used are:
``````// Represents a 3D point
struct point3f
{
float x;
float y;
float z;
};
// Represents a particle
struct particle
{
enum TYPES type;
float velocity;
struct point3f location;
struct point3f nextLocation;
struct point3f colour;
};
``````
here's all my code http://pastebin.com/m469f73c2
-
The other answer is a little mathish, it's actually pretty straight forward.
You need a "Velocity" which you are moving. It also has x, y and z coordinates.
In one time period, to move you just add the x velocity to your x position to get your new x position, repeat for y and z.
On top of that, you can have an "Acceleration" (also x,y,z) For instance, your z acceleration could be gravity, a constant.
Every time period your velocity should be recalcualted in the same way, Call velocity x "vx", so vx should become vx + ax, repeat for y and z (again).
It's been a while since math, but that's how I remember it, pretty straight forward unless you need to keep track of units, then it gets a little more interesting (but still not bad)
-
The tricky part is to figure out the correct direction to move in. After that it should be cake. – Tim Lin May 3 '09 at 17:58
Actually, the tricky part is that in reality, there is a hole because this is done in finite "steps". To get real-world movement, you have to make the steps very close together. When you make them infinitely close together, you have a true model. There is an entire branch of math based on this idea of making the steps perfectly continuous (Calculus), but doing it in finite steps like this is a good approximation and simplifies the formulas a LOT. – Bill K May 11 '09 at 17:10
I'd suggest that a particle should only have one location member -- the current location. Also, the velocity should ideally be a vector of 3 components itself. Create a function (call it `move`, `displace` whatever) that takes a `particle` and a time duration `t`. This will compute the final position after `t` units of time has elapsed:
``````struct point3f move(struct *particle, int time) {
particle->location->x = particle->velocity->x * t;
/* and so on for the other 2 dimensions */
return particle->location;
}
``````
-
I would recomend two things:
1. read an article or two on basic vector math for animation. For instance, this is a site that explains 2d vectors for flash.
2. start simple, start with a 1d point, ie a point only moving along x. Then try adding a second dimension (a 2d point in a 2d space) and third dimension. This might help you get a better understanding of the underlying mechanics. hope this helps
-
Think of physics. An object has a position (x, y, z) and a movement vector (a, b, c). Your object should exist at its position; it has a movement vector associated with it that describes its momentum. In the lack of any additional forces on the object, and assuming that your movement vector describes the movement over a time period t, the position of your object at time x will be (x + (a*t), y + (b*t), z + (c*t)).
In short; don't store the current position and the next position. Store the current position and the object's momentum. It's easy enough to "tick the clock" and update the location of the object by simply adding the momentum to the position.
-
Store velocity as a struct point3f, and then you have something like this:
``````void move(struct particle * p)
{
p->position.x += p->velocity.x;
p->position.y += p->velocity.y;
p->position.z += p->velocity.z;
}
``````
Essentially the velocity is how much you want the position to change each second/tick/whatever.
-
You want to implement the vector math `X_{i+1} = X_{i} + Vt`. For the `X`s and `V` vectors representing position and velocity respectively, and `t` representing time. I've parameterized the distance along the track by time because I'm a physicist, but it really is the natural thing to do. Normalize the velocity vector if you want to give track distance (i.e. scale `V` such that `V.x*V.x + V.y*V.y + V.z*V.z = 1`).
Using the `struct` above makes it natural to access the elements, but not so convenient to do the addition: arrays are better for that. Like this:
``````double X[3];
double V[3];
// initialize
for (int i=0; i<3 ++1){
X[i] = X[i] + V[i]*t;
}
``````
With a union, you can get the advantages of both:
``````struct vector_s{
double x;
double y;
double z;
}
typedef
union vector_u {
struct vector_s s; // s for struct
double a[3]; // a for array
} vector;
``````
If you want to associate both the position and the velocity of with the particle (a very reasonable thing to do) you construct a structure that support two vectors
``````typedef
struct particle_s {
vector position;
vector velocity;
//...
} particle_t;
``````
and run an update routine that looks roughly like:
``````void update(particle *p, double dt){
for (int i=0; i<3 ++i){
p->position.a[i] += p->velocity.a[i]*dt;
}
}
``````
-
Would you care to explain this in a bit more detail? I am a unsure of what you mean exactly. – Malachi Apr 29 '09 at 21:11
Afaik, there are mainly two ways on how you can calculate the new position. One is like the other have explaint to use an explicit velocity. The other possibility is to store the last and the current position and to use the Verlet integration. Both ways have their advantages and disadvantages. You might also take a look on this interresting page.
-
If you are trying to move along a straight line between two points, you can use the interpolation formula:
``````P(t) = P1*(1-t) + P2*t
``````
P(t) is the calculated position of the point, t is a scalar ranging from 0 to 1, P1 and P2 are the endpoints, and the addition in the above is vector addition (so you apply this formula separately to the x, y and z components of your points). When t=0, you get P1; when t=1, you get P2, and for intermediate values, you get a point part way along the line between P1 and P2. So t=.5 gives you the midpoint between P1 and P2, t=.333333 gives you the point 1/3 of the way from P1 to P2, etc. Values of t outside the range [0, 1] extrapolate to points along the line outside the segment from P1 to P2.
Using the interpolation formula can be better than computing a velocity and repeatedly adding it if the velocity is small compared to the distance between the points, because you limit the roundoff error.
- | 1,870 | 7,301 | {"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} | 3.171875 | 3 | CC-MAIN-2015-27 | latest | en | 0.898984 |
https://fr.mathworks.com/matlabcentral/cody/problems/191-proper-factors/solutions/536157 | 1,579,397,241,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00460.warc.gz | 452,754,020 | 15,606 | Cody
# Problem 191. Proper Factors
Solution 536157
Submitted on 27 Nov 2014 by Zikobrelli
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
%% assert(isequal(pfactors(5),[]))
ans = Empty matrix: 1-by-0 ans = []
2 Pass
%% assert(isequal(pfactors(10),[2 5]))
ans = 2 5
3 Pass
%% assert(isequal(pfactors(12),[2 3 4 6]))
ans = 2 3 4 6
4 Pass
%% assert(isequal(pfactors(15432),[2 3 4 6 8 12 24 643 1286 1929 2572 3858 5144 7716]))
ans = Columns 1 through 8 2 3 4 6 8 12 24 643 Columns 9 through 14 1286 1929 2572 3858 5144 7716 | 253 | 656 | {"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-2020-05 | latest | en | 0.531622 |
https://whomadewhat.org/how-many-cards-are-in-a-donkey/ | 1,660,181,251,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00673.warc.gz | 555,661,070 | 14,602 | Playing Donkey Players slide the unwanted cards one at a time to the player to their left. As they receive a card from the player on their right they look at it quickly and decide whether to keep it or pass it along. Play continues until a player has 4 cards of the same rank.
Subsequently, How do you play Donkey Card Game?
Playing Donkey Players slide the unwanted cards one at a time to the player to their left. As they receive a card from the player on their right they look at it quickly and decide whether to keep it or pass it along. Play continues until a player has 4 cards of the same rank.
Also, How do you play round pass donkey?
All players choose a card from their hand and pass to the next player. Keep swapping until someone gets a set of 4 matching cards. As soon as a player gets 4 matching donkey cards they shout DONKEY and everybody grabs for a carrot. If you didn’t get a carrot, you lift up one of your donkey ears.
How do you play Ash in cards?
Playing Donkey Players slide the unwanted cards one at a time to the player to their left. As they receive a card from the player on their right they look at it quickly and decide whether to keep it or pass it along. Play continues until a player has 4 cards of the same rank.
Last Review : 18 days ago.
## How do you play pass round donkey card game?
– Each player will end up with 4 cards. …
– Play continues until a player has 4 cards of the same rank. …
– The rest of the players upon noticing grab an object from the table. …
– The player who loses the round shuffles the cards and deals them out to the players one at a time face down and the next round begins.
## How do you play Magarac?
In Magarac, the loser of each hand gets one letter of the word Magarac (which means Jackass, which conveniently also has 7 letters). The game continues until a player has lost 7 times, spelling out the whole word. This player loses the whole game and is mockingly called jackass by other players.
## What are the rules for playing hearts?
At the end of each hand, players count the number of hearts they have taken as well as the queen of spades, if applicable. Hearts count as one point each and the queen counts 13 points. The aggregate total of all scores for each hand must be a multiple of 26. The game is usually played to 100 points (some play to 50).
## How do you play higher or lower card game?
You simply have to decide whether the second card is searched for higher or lower than the first card. As you progress, you will be shown a new card. Choose higher or lower to decide whether it is searched for more or less than the previous card. The objective is to get the most right in a row.
## How do you play Donkey card game in India?
Donkey card game is played in India in every household at family get-together and parties. Objective of the game is to empty your cards before your opponents. The player who is left with the maximum number of cards in the end of the game is crowned as ‘DONKEY’.
## How do you play don’t be a donkey?
All players choose a card from their hand and pass to the next player. Keep swapping until someone gets a set of 4 matching cards. As soon as a player gets 4 matching donkey cards they shout DONKEY and everybody grabs for a carrot. If you didn’t get a carrot, you lift up one of your donkey ears.
## Is Hearts a game of skill or luck?
Hearts is a game of skill — to a certain extent. You rely on luck to get good cards dealt to you, but strategic playing and a good memory make an enormous difference in this game. Keeping track of the cards played in each suit helps you to master this game, and practice and experience have no substitute.
## What is the object of the card game hearts?
Hearts is a trick taking game that requires 4 players and a standard 52 playing card deck with Aces high and 2’s low. The objective of the game is to have the fewest points when someone reaches 100 points. For other trick taking games, see our guides for Spades and Euchre.
## Is Hearts a gambling game?
Hearts is a trick-taking card game that belongs to the “whist” family of card games. … Yes, it’s still gambling if you bet on it, but if your skills are better than your opponents’, you can be playing a game with a positive expectation.
## What is the aim of the game Hearts?
Hearts is a trick taking game that requires 4 players and a standard 52 playing card deck with Aces high and 2’s low. The objective of the game is to have the fewest points when someone reaches 100 points. For other trick taking games, see our guides for Spades and Euchre.
## How do you play the card game Rummikub?
A player cannot lay down a tile he has just picked; he must wait until his next turn. Play continues until one player empties his rack and calls, “Rummikub®”. That ends the round and players tally their points (see Scoring). When the pouch is empty, play continues until no more plays can be made; that ends the round.
## How do u play high low card game?
– Place your bet on the table.
– A card is dealt.
– Choose whether the next card will be higher or lower.
– The next card is dealt.
– If you guessed incorrectly, your stake is lost, and the game round ends.
– You must place another bet to play again.
## How do you play hearts for beginners?
Watch the video on YouTube
## Can you break hearts on the first trick?
A player can’t lead with a heart until a heart has been “broken,” or played in the game. Hearts are broken when a player lacks a card in the suit that has been led, so he throws down a heart instead. Many people play that hearts can’t be broken on the first trick.
Spread the word ! Don’t forget to share. | 1,301 | 5,658 | {"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} | 3.5 | 4 | CC-MAIN-2022-33 | longest | en | 0.976053 |
https://codedump.io/share/Yt48RZesEntL/1/calculate-mean-value-of-column-if-multiple-occurrences-in-another-column | 1,524,772,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948464.28/warc/CC-MAIN-20180426183626-20180426203626-00525.warc.gz | 602,145,368 | 9,209 | user2697374 - 2 years ago 115
Perl Question
# Calculate mean value of column if multiple occurrences in another column
This is part of a tab separated file that I generated:
Sample Gene RPKM
MK_001_27 HPSE2 17.3767266340978
MK_003_11 HPSE2 51.1152373602497
MK_002_5 HPSE2 16.5372913024845
MK_001_23 HPSE2 25.8985857481585
MK_001_23 HPSE2 21.6045197173757
MK_001_27 HPSE2 139.450963428357
MK_001_23 HPSE2 36.7603351866179
MK_003_9 HPSE2 25.2860867858956
MK_001_22 HPSE2 100.915250867745
MK_003_9 HPSE2 35.078964327254
MK_003_12 HPSE2 34.078813048573
MK_003_9 HPSE2 13.5865540939141
Any gene is present in at least one sample. It may be present multiple times in the same sample.
I want to generate the mean of the Reads Per Kilobase per Million (RPKM) value and use it to replace multiple value for any given gene from the same sample
For example
MK_003_9 HPSE2 35.078964327254
MK_003_9 HPSE2 13.5865540939141
MK_003_9 HPSE2 25.2860867858956
becomes
MK_003_9 HPSE2 24.650535069
s there a way that you know of that could resolve this in Unix or in Perl?
Pseudocode
-For each line, combine column 0 (sample) and column 1 (gene) to create a "key"
though not always unique
-Run through the file to see if this "key" is present somewhere else
-Count the number of times this "key" is present
-If present > 1 time, calculate the mean of the RPKM values by sum()/count
-Create occurrence with this "key" and the new RPKM value
-Delete(?) the other corresponding "keys"
This is quite straightforward if you accumulate the contents of your data file into a Perl hash. You don't make it clear whether there may be multiple genes in a single file, so I've coded this for multiple samples and multiple genes
The program expects the input file as a parameter on the command line, and prints its output to STDOUT which may be redirected as normal
use strict;
use warnings qw/ all FATAL /;
use List::Util 'sum';
print scalar <>; # Copy the header line
my %mean_rpkm;
while ( <> ) {
my (\$sample, \$gene, \$rpkm) = split;
push @{ \$mean_rpkm{\$gene}{\$sample} }, \$rpkm;
}
for my \$gene ( sort keys %mean_rpkm ) {
for my \$sample ( sort keys %{ \$mean_rpkm{\$gene} } ) {
my \$rpkm = \$mean_rpkm{\$gene}{\$sample};
my \$mean = sum(@\$rpkm) / @\$rpkm;
printf "%s\t%s\t%.3f\n", \$sample, \$gene, \$mean;
}
}
### output
Sample Gene RPKM
MK_001_22 HPSE2 100.915
MK_001_23 HPSE2 28.088
MK_001_27 HPSE2 78.414
MK_002_5 HPSE2 16.537
MK_003_11 HPSE2 51.115
MK_003_12 HPSE2 34.079
MK_003_9 HPSE2 24.651
### Update
The output from my solution is essentially unordered. I have sorted the genes and samples lexically, but you may want them in the same order as they appear in the input file. If so then you should say so
A useful intermediate solution is to install Sort::Naturally (it's not a core module) and add
use Sort::Naturally 'nsort';
to the top of the above program. If you then replace both occurrences of sort with nsort then you will get this output. It may not be idea, but it's an improvement because it sorts MK_003_9 before MK_003_11 which a simple lexical sort doesn't do
Sample Gene RPKM
MK_001_22 HPSE2 100.915
MK_001_23 HPSE2 28.088
MK_001_27 HPSE2 78.414
MK_002_5 HPSE2 16.537
MK_003_9 HPSE2 24.651
MK_003_11 HPSE2 51.115
MK_003_12 HPSE2 34.079
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 1,125 | 3,444 | {"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-2018-17 | latest | en | 0.786038 |
https://study.com/academy/topic/oae-middle-grades-math-rates-ratios-proportions.html | 1,542,672,111,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746171.27/warc/CC-MAIN-20181119233342-20181120015342-00367.warc.gz | 689,761,789 | 35,735 | # Ch 17: OAE Middle Grades Math: Rates, Ratios & Proportions
As you study for the OAE Middle Grades Mathematics test, review these lessons to improve your understanding ratios, rates, and proportions. Our study guide also show you how they can be applied to real-life situations.
## OAE Middle Grades Math: Rates, Ratios & Proportions - Chapter Summary
Use the lesson videos and assessments in this chapter to prepare for questions on the Ohio Assessments for Educators (OAE) Middle Grades Mathematics test asking about ratios and proportions. Start by watching our professional instructors in a series of videos as they explain the procedures used to solve problems with rates, ratios and proportions. Once you've completed this chapter you should be better prepared to:
• Calculate and find unit rates
• Solve one-step problems involving ratios and rates
• Calculate ratios and proportions
• Assess proportional relationships
• Graph proportional relationships
• Use proportional relationships to solve multi-step ratio and percent problems
• Apply these skills to solve problems with money, time, and other real life problems
Be sure to ask our instructors any questions you may come up with so that they can help you master the material presented in these lessons. After viewing the lessons, reinforce your retention of the information by completing the quizzes that accompany them. Use your quiz results to determine what topics you should review further, and organize this review with video tags that can take you straight to the locations of the lesson that cover your areas of weakness.
### OAE Middle Grades Math: Rates, Ratios & Proportions Chapter Objectives
The state of Ohio uses the OAE Middle Grades Mathematics to certify middle grade math teachers. This computer-administered certification exam is made of 150 multiple-choice questions and is administered in a four hours and 15 minutes testing session. On test day you, will see that 17% of these questions belong to the domain of Number Sense and Operations. Prepare for some of these question to ask you to solve problems with ratios and proportions by completing the videos and quizzes in this chapter.
16 Lessons in Chapter 17: OAE Middle Grades Math: Rates, Ratios & Proportions
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### Earning College Credit
Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 571 | 2,871 | {"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-2018-47 | longest | en | 0.899911 |
http://www.lucky-name-numerology.com/calculate-my-birth-numbers-name-numerology.html | 1,537,361,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156224.9/warc/CC-MAIN-20180919122227-20180919142227-00331.warc.gz | 362,861,303 | 5,955 | # Calculate My Birth Numbers Name Numerology
by Leslie
I just sent in a question about my name numerology for Leslie Abbagliato, 11/18/1975, but I just thought of something. On calculating my birthday instead of it adding up to 33/6, can you add:
1+1 =2
(18) 1+8 =9
(1975) 1+9+7+5 =22
Then
2+9+4=15
1+5 =6 or
2+9=11
1+1=2
2+4=6
to avoid it equaling to 33 for 3+3=6, since 3 is not compatible with 6 or does it not matter any when you add it? Will the 3's still affect my 6 life path number? Thanks! | 183 | 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} | 3.296875 | 3 | CC-MAIN-2018-39 | latest | en | 0.827366 |
https://www.exceltip.com/mathematical-functions/how-to-use-the-permut-function-in-excel.html | 1,618,372,404,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00484.warc.gz | 860,065,781 | 17,712 | # How to use the PERMUT Function in Excel
In this article, we will learn about how to use the PERMUT function in Excel.
Permutation is a mathematical operator for getting the number of possible permutations having number (n) & number chosen (r). Order is significant in permutations.
Pnr= FACT(n) / FACT(r)
OR
Cnr= (n)! / (r)!
Factorial function in excel is called FACT function and its mathematical symbol is ! .
The PERMUT function returns the value which is the number of possible permutations. Order is significant in permutations.
Syntax:
=PERMUT (number, number_chosen)
Number : total number of elements or values
Number_chosen : number of values chosen from the total values.
All of this might be confusing so let’s Gear up and understand the PERMUT function using it in an example.
If there are 10 numbers stated 1 to 10. And you get the total number ways selecting any one number from 1 to 10 can be calculated using the PERMUT function.
There are 10 ways i.e. 10 possible ways to select one number from the PERMUTation of 10. Order is significant in permutations but here one value is selected so there is no ordered pairing.
Possible values :
1 2 3 4 5 6 7 8 9 10
Now Let’s consider vowels . There are 5 vowels in the alphabet which are a, i, e, o & u.
And we need to find the ordered pair of two vowels
So we need to use the formula to get the Permutations of two vowels.
=PERMUT(5, 2)
5 : five vowels
2 : two vowels PERMUTation required
There are 20 possible ways which are
ae ai ao au ea = 5
ei eo eu ia ie = 5
io iu oa oe oi = 5
ou ua ue ui uo = 5
Total = 20
Order of vowels is significant when using the PERMUT function. So ae & ea are counted as different.
Now take a few examples to get the possible permutation having bigger numbers.
There are 970200 possible permutations when three different numbers chosen & set in order in a total of 100 numbers.
Note:
• Arguments that contain decimal values are truncated to integers.
• A permutation is a group of values in which order/sequence matters.
• The PERMUT function returns a #VALUE! error value if either argument is not numeric.
• If number is less than number_chosen, the function returns #NUM! error.
As you can get from the above examples that the PERMUT function returns the total number of possible permutations of choosing & then ordering numbers out of total numbers.
Hope you understood how to use PERMUT function in Excel. Explore more articles on Excel COMBIN function
here. Please feel free to state your query or feedback for the above article.
Related Article:
How to use the COMBIN Function in Excel
How to use the FACT Function in Excel
Popular Articles:
50 Excel Shortcuts to Increase Your Productivity
How to use the VLOOKUP Function in Excel
How to use the COUNTIF in Excel 2016
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# What is the Mode?
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### What is the Mode?
1. 1. Mode Conceptual Explanation What is the Mode?
2. 2. Mode What is the Mode?
3. 3. Mode So here is a technical definition of the “mode”: What is the Mode?
4. 4. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” What is the Mode?
5. 5. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” Data Set 1 2 2 2 2 3 What is the Mode?
6. 6. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” Data Set 1 2 2 2 2 3 What is the Mode?
7. 7. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 Data Set 1 2 2 2 2 3 What is the Mode?
8. 8. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 22 Data Set 1 2 2 2 2 3 What is the Mode?
9. 9. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 2 22 Data Set 1 2 2 2 2 3 What is the Mode?
10. 10. Data Set 1 2 2 2 2 3 Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 2 2 22 What is the Mode?
11. 11. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 2 2 2 22 Data Set 1 2 2 2 2 3 What is the Mode?
12. 12. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” Data Set 1 2 2 2 2 3 1 2 3 2 2 2 What is the Mode?
13. 13. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 2 2 2 2 Data Set 1 2 2 2 2 3 3 What is the Mode?
14. 14. Mode So here is a technical definition of the “mode”: “The mode is simply the most frequently occurring observation in the distribution.” 1 2 2 2 2 Data Set 1 2 2 2 2 3 3 The Mode What is the Mode?
15. 15. And What is the Mode?
16. 16. And “It is a rough estimate of central tendency” What is the Mode?
17. 17. And “A distribution may have multiple modes which confuses the concept of central tendency” What is the Mode?
18. 18. And “A distribution may have multiple modes which confuses the concept of central tendency” What is the Mode?
19. 19. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” What is the Mode?
20. 20. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 What is the Mode?
21. 21. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 Mode = 3 and 7 What is the Mode?
22. 22. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 Mode = 3 and 7 What is the Mode?
23. 23. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 Mode = 3 and 7 the same What is the Mode?
24. 24. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 Mode = 3 and 7 the same What is the Mode?
25. 25. But “can reveal attributes of the distribution which might not be observed by considering only the mean or median” Normal Distribution Mean = 5 Median = 5 Mode = 5 Mean = 5 Median = 5 Mode = 3 and 7 the same different What is the Mode? | 1,234 | 4,556 | {"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} | 3.078125 | 3 | CC-MAIN-2020-10 | latest | en | 0.904509 |
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# Combined Filters
Topic closed. 11 replies. Last post 7 years ago by user044.
Page 1 of 1
Tx
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May 4, 2004
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Posted: February 21, 2010, 1:28 pm - IP Logged
It is an error for lottery softwares not to have and make use of what I guess could be called combined filters or some other name(s).
When you have 2 choices it is either one or the other, few choices of filter patterns often lead to filter failure when using what I call "The Negative Prediction" technique.
Some examples of 2 choice Boxed or straight filters are: (Mostly already includes All)
High or Low (Mostly)
Even or Odd (Mostly)
In or Out (Mostly)
How can statistics tell you which of the 2 patterns will come out next and which one won't?
Your chances of being right and or wrong might be 50 percent either way, that is 50% chance of being wrong and or right, that is not good, but is bad.
If you have any 10 equal patterns and if you filter out only 1 of the 10, your chances of being right might be about 90%, but if you filter out 5 of them, you might lower your chances to maybe about 50%.
There should be a balance between amount of filtration and accuracy of the prediction, accuracy being the chance of the filtration not failing.
Too much filtration used on each filter lowers the accuracy of the prediction and makes for more filters failures.
But too little filtration leaves too many combinations that need to be bought and played and as we know, the lottery pays too little.
If the game is fair and honest, then we will only be dealing with what they might call "Random Factors", then more filters and patterns for each filter might be needed.
Having the wrong filters-patterns might be the same as having no filters-patterns as then they might be useless.
Winning by division (Filtering Out) and by numbers (Having many filters-Patterns) is what "Negative Prediction" is about.
-----------
An example:
We do see Hot Patterns here, but it is a very risky thing:
11/25/2009;4;5;0;Mostly low
11/26/2009;0;4;3;All low
11/27/2009;7;2;9;Mostly high
11/28/2009;7;3;4;Mostly low
11/30/2009;4;9;5;Mostly high
12/1/2009;0;5;5;Mostly high
12/2/2009;0;4;5;Mostly low
12/3/2009;8;0;8;Mostly high
12/4/2009;8;6;1;Mostly high
12/5/2009;6;5;8;All high
12/7/2009;9;1;1;Mostly low
12/8/2009;4;6;4;Mostly low
12/9/2009;2;5;7;Mostly high
12/10/2009;0;8;1;Mostly low
12/11/2009;6;0;1;Mostly low
12/12/2009;1;8;6;Mostly high
12/14/2009;8;4;2;Mostly low
12/15/2009;2;3;1;All low
12/16/2009;1;1;0;All low
12/17/2009;3;0;9;Mostly low
12/18/2009;3;8;3;Mostly low
12/19/2009;9;1;0;Mostly low
12/21/2009;6;4;4;Mostly low
12/22/2009;0;4;7;Mostly low
12/23/2009;6;5;7;All high
12/24/2009;4;3;6;Mostly low
12/25/2009;9;9;9;All high
This listings format is uneven, the date should have 2 digits for the Day and also 2 for the Month, such as:
The patterns themselves could just be represented by equal length symbols such as either digits or letters, but I know that that would confuse some or many people:
For example:
0 = Mostly Low
1 = Mostly High
2 = All Low
3 = All High
Then it would be easier to see and also count each of the 4 patterns there, but there could also be a choice of just:
Mostly Low and Mostly High, just 2 equal patterns, then:
0 = Mostly Low
1 = Mostly High
Each pattern then equals 1/2 of the total combinations.
But as I said, it might be best in some or many cases to instead use many patterns for each filter and if each of the patterns are more or less equal then even better yet.
----------
Well, by combined filters what I have in mind is kind of like these examples:
Low-Even = LE
Low-Odd = LO
High-Even = HE
High-Odd = HO
Not exactly equal patterns, but close enough.
Those patterns in some way apply oe might apply to other numeric filters and or digits-numbers.
-----------
As to Hot-Cold filters-patterns, one should first setup the basic rules of their use, with them their being many more variables and also it might be best if the Hot-Cold counting is not done in the regular way, but in a weighted way, I also have posted about that before.
I might later today explain what I mean by that.
"Ten measures of beauty descended to the world, nine were taken by Jerusalem."
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Posted: February 21, 2010, 1:40 pm - IP Logged
Well, by combined filters what I have in mind is kind of like these examples:
Low-Even = LE
Low-Odd = LO
High-Even = HE
High-Odd = HO
(Well it's nice too see someone else has discovered this little gem besides me,I use it for selecting lead numbers for Positions 1-2)
Tx
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May 4, 2004
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Posted: February 21, 2010, 6:23 pm - IP Logged
Well, by combined filters what I have in mind is kind of like these examples:
Low-Even = LE
Low-Odd = LO
High-Even = HE
High-Odd = HO
(Well it's nice too see someone else has discovered this little gem besides me,I use it for selecting lead numbers for Positions 1-2)
Yes, but it is very old stuff, known about it for many years, but lottery programs are very far behind.
"Ten measures of beauty descended to the world, nine were taken by Jerusalem."
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Posted: February 21, 2010, 7:10 pm - IP Logged
True but there are some very good programs out there that have an excellent array of filtering options
Tx
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Posted: February 21, 2010, 7:27 pm - IP Logged
The regular Hot-Cold count:
Boxed:
Pick a particular number of past draws.
Aways use the same number of past draws.
There are 10 digits from 0 to 9 to be counted.
Count all Zeroes on all 3 positions on all those draws that you are going to use.
Do the same for all the other 9 digits.
Each of the 10 boxed digits would have come out a given number of times.
Some digits more often than others.
And some the same as others.
The digits that came out more times will be hotter than the others that came out fewer times.
The problem might be that either always or sometimes, some digits will come out the same number of times as other digits.
-------
It is the same for straight Hot-Cold, but the digits counted are done by position and not boxed for all positions.
--------
The first rule in any way is that always the very same number of past draws needs to be used, but it does not have to be the same number of past draws for boxed counting as it is used for straight counting.
Maybe for boxed counting fewer draws could be used than for straight counting.
----------
A maybe better way:
If 10 past draws were used:
12/15/2009 2 3 1 X 1
12/16/2009 1 1 0 X 2
12/17/2009 3 0 9 X 3
12/18/2009 3 8 3 X 4
12/19/2009 9 1 0 X 5
12/21/2009 6 4 4 X 6
12/22/2009 0 4 7 = X 7 Times
12/23/2009 6 5 7 = X 8 Times
12/24/2009 4 3 6 Digits here = X 9
12/25/2009 9 9 9 Newest and last draw. Digits here coiunt X 10
The digits are weighted by each past draw used in that kind of way, that could be rule # 2.
If 5 past draws were used then they would be weighted from 5 to 1, if 15 past draws then from 15 to 1.
It is possible that even sometimes or often or always some digits might have the same count, so 1 or more rules might be needed.
For example, if 2 digits or more have the same count then the digit(s) that has 1 or the most digits closer to the very last draw that came out is hotter than the other, that would maybe break the tie(s).
Other rules could be possible, but I think that only those might be needed.
As to how to use those Hot-Cold counts, that is for another post sometime soon maybe.
"Ten measures of beauty descended to the world, nine were taken by Jerusalem."
Tx
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Posted: February 21, 2010, 9:11 pm - IP Logged
?I made that post just a moment ago and something happened as it does sometimes when I am posting, the post was "lost" and I was out of LP all at once, it happens every so often along with other "Weird things"
I can't again make the post today, it will have to wait for some other time.
You have no idea how many past posts I have lost in some very weird ways, a few sometimes I can recover and or repost, but many just get lost-deleted even before they get posted and don't get reposted due to some reasons, like their being too hard to make and or too long and or just can't remember exactly what I had made.
"Ten measures of beauty descended to the world, nine were taken by Jerusalem."
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Posted: February 21, 2010, 9:41 pm - IP Logged
You're unlikely to find any off-the-shelf lottery software that does exactly what you want.
That leaves you with a) putting bits and pieces of different filters and workouts together, b) having something custom written, or c) making your own code.
Filters that are 50/50 are best left for when you've exausted all the ones that are more tilted one way or the other.
In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.
Chicago
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Posted: February 21, 2010, 9:43 pm - IP Logged
?I made that post just a moment ago and something happened as it does sometimes when I am posting, the post was "lost" and I was out of LP all at once, it happens every so often along with other "Weird things"
I can't again make the post today, it will have to wait for some other time.
You have no idea how many past posts I have lost in some very weird ways, a few sometimes I can recover and or repost, but many just get lost-deleted even before they get posted and don't get reposted due to some reasons, like their being too hard to make and or too long and or just can't remember exactly what I had made.
Whenever I'm on the "beast monster" aka my mini-notebook computer (wireless) it seems to have a mind of it's own. One minute I'm in a forum and the next I'm at a previous website I visited. I don't have this problem when I'm on my desk PC.
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Posted: February 22, 2010, 10:59 am - IP Logged
"Time Treat"
Your exactly right that's why I have an arsenal of GOOD programs and usually combine 2 or 3 (if necessary)that have the best analysis features and filtering options and so far it's worked quite well,I use P34Lotto and LotsoftPro for the Play 4 and for the Cash 3 I use P34Lotto and Pro-Filter 3.
Upstate New York
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Posted: February 22, 2010, 12:06 pm - IP Logged
"Time Treat"
Your exactly right that's why I have an arsenal of GOOD programs and usually combine 2 or 3 (if necessary)that have the best analysis features and filtering options and so far it's worked quite well,I use P34Lotto and LotsoftPro for the Play 4 and for the Cash 3 I use P34Lotto and Pro-Filter 3.
Expert lotto has a decent array of good filters some of which I still dont understand after using it for a year. Sadly it only works with the pick 5 and pick6 lotteries. I have looked at p34 lotto and it does have some very nice features.
I have googled Pro-Filter 3 and cannot find a web site that offers this software. Could you please advise me as to where to find this program? Many thanks in advance.
jeff
Ambition is a poor excuse for not having sense enough to be lazy.
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Posted: February 22, 2010, 12:48 pm - IP Logged
Do a search for "IntelBet Lottery Tools" once you get too the site look at the left side under MENU towards the bottom you will see a listing for "Pro-Filter 3"
Upstate New York
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Posted: February 22, 2010, 1:15 pm - IP Logged
Do a search for "IntelBet Lottery Tools" once you get too the site look at the left side under MENU towards the bottom you will see a listing for "Pro-Filter 3"
Found it. Thank you very much again!
Ambition is a poor excuse for not having sense enough to be lazy.
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This page is about coal calculator to convert gcv arb gcv adb,, coal gcv adb to arb conversion formula, (CV 2250 kcal/kg NAR*). saber más. conversion of gcv arb to gcv adb The difference between and gross,coal gcv adb to arb conversion formula The what is the difference between, what is the difference between gcv on arb and adb gcv adb to .
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Calculator Adb To Arb theasters . This page is about coal calculator to convert gcv arb gcv adb,, coal gcv adb to arb conversion formula, (CV 2250 kcal/kg NAR*): . saber más. conversion of gcv arb to gcv adb The difference between and gross,coal gcv adb to arb conversion formula The what is the difference between, what is the difference between gcv on arb and adb gcv adb to | 1,859 | 7,640 | {"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-2021-49 | latest | en | 0.845285 |
https://questions.examside.com/past-years/jee/question/let-overrightarrow-a--widehat-i-alpha-widehat-j--jee-main-mathematics-trigonometric-functions-and-equations-7ym2snsxjntamwws | 1,713,153,198,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00716.warc.gz | 426,238,242 | 42,579 | 1
JEE Main 2021 (Online) 27th July Evening Shift
Numerical
+4
-1
Out of Syllabus
Let $$\overrightarrow a = \widehat i - \alpha \widehat j + \beta \widehat k$$, $$\overrightarrow b = 3\widehat i + \beta \widehat j - \alpha \widehat k$$ and $$\overrightarrow c = -\alpha \widehat i - 2\widehat j + \widehat k$$, where $$\alpha$$ and $$\beta$$ are integers. If $$\overrightarrow a \,.\,\overrightarrow b = - 1$$ and $$\overrightarrow b \,.\,\overrightarrow c = 10$$, then $$\left( {\overrightarrow a \, \times \overrightarrow b } \right).\,\overrightarrow c$$ is equal to ___________.
2
JEE Main 2021 (Online) 27th July Morning Shift
Numerical
+4
-1
Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b$$ and $$\overrightarrow c = \widehat j - \widehat k$$ be three vectors such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c$$ and $$\overrightarrow a \,.\,\overrightarrow b = 1$$. If the length of projection vector of the vector $$\overrightarrow b$$ on the vector $$\overrightarrow a \times \overrightarrow c$$ is l, then the value of 3l2 is equal to _____________.
3
JEE Main 2021 (Online) 25th July Evening Shift
Numerical
+4
-1
If $$\left( {\overrightarrow a + 3\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 5\overrightarrow b } \right)$$ and $$\left( {\overrightarrow a - 4\overrightarrow b } \right)$$ is perpendicular to $$\left( {7\overrightarrow a - 2\overrightarrow b } \right)$$, then the angle between $$\overrightarrow a$$ and $$\overrightarrow b$$ (in degrees) is _______________.
4
JEE Main 2021 (Online) 25th July Morning Shift
Numerical
+4
-1
Let $$\overrightarrow p = 2\widehat i + 3\widehat j + \widehat k$$ and $$\overrightarrow q = \widehat i + 2\widehat j + \widehat k$$ be two vectors. If a vector $$\overrightarrow r = (\alpha \widehat i + \beta \widehat j + \gamma \widehat k)$$ is perpendicular to each of the vectors ($$(\overrightarrow p + \overrightarrow q )$$ and $$(\overrightarrow p - \overrightarrow q )$$, and $$\left| {\overrightarrow r } \right| = \sqrt 3$$, then $$\left| \alpha \right| + \left| \beta \right| + \left| \gamma \right|$$ is equal to _______________. | 706 | 2,188 | {"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} | 3.546875 | 4 | CC-MAIN-2024-18 | latest | en | 0.489189 |
https://naif.jpl.nasa.gov/pub/naif/toolkit_docs/IDL/icy/cspice_wnsumd.html | 1,566,053,952,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313428.28/warc/CC-MAIN-20190817143039-20190817165039-00081.warc.gz | 567,434,788 | 3,085 | cspice_wnsumd
Abstract I/O Examples Particulars Required Reading Version Index_Entries
```
```
#### Abstract
```
CSPICE_WNSUMD summarizes the contents of a double precision window.
For important details concerning this module's function, please refer to
the CSPICE routine wnsumd_c.
```
#### I/O
```
Given:
window the scalar SPICE window containing zero or more intervals.
You must create 'window' as a window structure via a
cspice_celld call.
the call:
cspice_wnsumd, window, meas, avg, stddev, shortest, longest
returns:
meas the double precision scalar total measure of
the intervals in the input window. This is just
the sum of the measures of the individual intervals.
avg the double precision scalar average measure
of the intervals in the input window.
stddev the double precision scalar standard deviation
of the measures of the intervals in the input window.
shortest,
longest the integer scalar indices of the left endpoint
of, respectively, the shortest and longest intervals
in the data contained in 'window'.
The following algorithm describes the relation of
'shortest' and 'longest' to the window data:
data_start = window.data
data_end = data_start + window.card - 1
array = window.base[ data_start:data_end]
The shortest interval:
[ array[shortest], array[shortest+1] ]
The longest interval:
[ array[longest], array[longest+1] ]
```
#### Examples
```
Any numerical results shown for this example may differ between
platforms as the results depend on the SPICE kernels used as input
and the machine specific arithmetic implementation.
;;
;; Create a cell containing a double precision 12-vector.
;;
win1 = cspice_celld( 12 )
;;
;; Define an array representing a window with six intervals.
;; The values in 'darray' have correct order for a
;; SPICE window.
;;
darray = [ [ 1.d, 3.], \$
[ 7., 11.], \$
[ 18., 18.], \$
[ 23., 27.], \$
[ 30., 69.], \$
[ 72., 80.] ]
;;
;; Add the 'darray' data to the cell.
;;
count = size( darray, /n_elements)/2l
for i=0l, count -1 do begin
cspice_wninsd, darray[0,i], darray[1,i], win1
endfor
;;
;; Calculate the summary for 'win1'.
;;
cspice_wnsumd, win1, \$
meas, \$
avg, \$
stddev, \$
shortest,\$
longest
;;
;; 'shortest' and 'longest' refer to the indices of
;; the 'cell' data array.
;;
intrvl_short= shortest/2l
intrvl_long = longest/2l
print, 'Measure : ', meas
print, 'Average : ', avg
print, 'Standard Dev : ', stddev
print, 'Index shortest : ', shortest
print, 'Index longest : ', longest
print, 'Interval shortest : ', intrvl_short
print, 'Interval longest : ', intrvl_long
data_start = win1.data
data_end = data_start + win1.card - 1
array = win1.base[ data_start:data_end]
print, 'Shortest interval : ', [ array[shortest], array[shortest+1] ]
print, 'Longest interval : ', [ array[longest], array[longest+1] ]
IDL outputs:
Measure : 57.000000
Average : 9.5000000
Standard Dev : 13.413302
Index shortest : 4
Index longest : 8
Interval shortest : 2
Interval longest : 4
Shortest interval : 18.000000 18.000000
Longest interval : 30.000000 69.000000
```
#### Particulars
```
This routine provides a summary of the input window, consisting
of the following items:
- The measure of the window.
- The average and standard deviation of the measures
of the individual intervals in the window.
- The indices of the left endpoints of the shortest
and longest intervals in the window.
All of these quantities are zero if the window contains no
intervals.
```
```
ICY.REQ
CELLS.REQ
WINDOWS.REQ
```
#### Version
```
-Icy Version 1.0.1, 04-DEC-2008, EDW (JPL)
the code example.
-Icy Version 1.0.1, 12-SEP-2006, EDW (JPL)
Correct Required Reading citation CELL.REQ to CELLS.REQ.
-Icy Version 1.0.0, 08-AUG-2004, EDW (JPL)
```
#### Index_Entries
```
summary of a d.p. window
```
`Wed Apr 5 17:58:05 2017` | 1,137 | 4,022 | {"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} | 2.578125 | 3 | CC-MAIN-2019-35 | latest | en | 0.67925 |
https://petrowiki.spe.org/Ideal_gases | 1,723,128,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00882.warc.gz | 365,228,547 | 10,912 | You must log in to edit PetroWiki. Help with editing
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# Ideal gases
While no ideal gases exist, many gases behave like ideal gases under certain conditions. The concept of an ideal gas is useful for understanding gas behavior and simplifying the calculation of gas properties. This page describes an ideal gas, and develops the ideal gas law and the gas law constant.
## Ideal gas
The kinetic theory of gases postulates that a gas is composed of a large number of very small discrete particles. These particles can be shown to be identified with molecules. For an ideal gas, the volume of these particles is assumed to be so small that it is negligible compared with the total volume occupied by the gas. It is assumed also that these particles or molecules have neither attractive nor repulsive forces between them. The average energy of the particles or molecules can be shown to be a function of temperature only. Thus, the kinetic energy, Ek, is independent of molecule type or size. Because kinetic energy is related to mass and velocity by Ek = 1/2 mv2, it follows that small molecules (less mass) must travel faster than large molecules (more mass) when both are at the same temperature.
## Boyle's law and Charles' law
Molecules are considered to be moving in all directions in a random manner as a result of frequent collisions with one another and with the walls of the containing vessel. The collisions with the walls create the pressure exerted by the gas. Thus, as the volume occupied by the gas is decreased, the collisions of the particles with the walls are more frequent, and an increase in pressure results. It is a statement of Boyle’s law that this increase in pressure is inversely proportional to the change in volume at constant temperature:
....................(1)
where:
• p is the absolute pressure
• V is the volume.
Further, if the temperature is increased, the velocity of the molecules and, therefore, the energy with which they strike the walls of the containing vessel will be increased, resulting in a rise in pressure. To maintain the pressure constant while heating a gas, the volume must be increased in proportion to the change in absolute temperature. This is a statement of Charles’ law,
....................(2)
where:
• T is the absolute temperature
• p is a constant
From a historical viewpoint, the observations of Boyle and Charles in no small degree led to the establishment of the kinetic theory of gases, rather than vice versa. It follows from this discussion that, at zero degrees absolute, the kinetic energy of an ideal gas, as well as its volume and pressure, would be zero. This agrees with the definition of absolute zero, which is the temperature at which all the molecules present have zero kinetic energy.
Because the kinetic energy of a molecule depends only on temperature, and not on size or type of molecule, equal molecular quantities of different gases at the same pressure and temperature would occupy equal volumes. The volume occupied by an ideal gas therefore depends on three things: temperature, pressure, and number of molecules (moles) present. It does not depend on the type of molecule present.
## Ideal gas law
The ideal gas law, which is actually a combination of Boyle’s and Charles’ laws, is a statement of this fact:
....................(3)
where:
• p = pressure
• V = volume
• n = number of moles
• R = gas-law constant
• T = absolute temperature.
## Gas law constant
The gas law constant, R, is a proportionality constant that depends only on the units of p, V, n, and T. Tables 1A through 1C present different values of R for the various units of these parameters. The value of the gas constant is experimental, and more-accurate values are reported occasionally. The values in Tables 1A through 1C are based on the values reported by Moldover et al.[1] Their value was determined from measurements of the speed of sound in argon as a function of pressure at the temperature of the triple point of water. Note that because pV has the units of energy, the value of R is typically given in units of energy per mole per absolute temperature unit [e.g., the appropriate SI value for R is 8.31447 J/(g mol-K), and the appropriate British gravitational (sometimes called the American customary units) value for R is 1,545.35 ft-lbf/(lb-mol°R]. However, sometimes pressure and volume units are more appropriate, such as R = 10.7316 (psia-ft3)/ (lb mol-°R).
## Nomenclature
Ek = kinetic energy, J m = mass, kg n = number of moles p = absolute pressure, Pa R = gas-law constant, J/(g mol-K) T = absolute temperature, K v = velocity, m/s V = volume, m3
## References
1. Moldover, M.R., Trusler, J.P.M., Edwards, T.J. et al. 1988. Measurement of the Universal Gas Constant R Using a Spherical Acoustic Resonator. J. Res. Nat. Inst. Stand. Technol. 93 (2): 85.
## Noteworthy papers in OnePetro
Use this section to list papers in OnePetro that a reader who wants to learn more should definitely read | 1,135 | 5,187 | {"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} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.927997 |
http://itknowledgeexchange.techtarget.com/itanswers/listbox-4/ | 1,454,978,245,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701154682.35/warc/CC-MAIN-20160205193914-00060-ip-10-236-182-209.ec2.internal.warc.gz | 113,260,257 | 20,666 | ## Listbox in VB.NET
110 pts.
Tags:
VB.NET
VB.NET 2005
VB.NET 2008
I load list box with file data of 5 elements (k2,k2,k3,k4,k5) with up to 10000 results. I get the sum of these elements to be some integer (3,6,7,8,9,11). I write a code to eliminate some results like this:
```If (k1+k2+k3+k4+k5) <> 3 or
(k1+k2+k3+k4+k5) <> 6 or
If (k1+k2+k3+k4+k5) <> 7 or
(k1+k2+k3+k4+k5) <> 8 or
If (k1+k2+k3+k4+k5) <> 9 or
(k1+k2+k3+k4+k5) <> 11 then
k2 " ;" & _ etc```
Is it possible to write a code that reflects all these integer (3,6,7,8,9,11) at ones?
Thanks. We'll let you know when a new response is added.
First, I would like to bring to your attention the fact that when using OR conditions the final result will be true if at least one of the conditions is met (all other conditions don’t even need to be evaluated).
If you were doing something like this
```If (k1+k2+k3+k4+k5) <b><> 3 AND
(k1+k2+k3+k4+k5) <b><> 6 AND
(k1+k2+k3+k4+k5) <b><> 7 AND```
or this:
```If (k1+k2+k3+k4+k5) <b>= 3 OR
(k1+k2+k3+k4+k5) <b>= 6 OR
(k1+k2+k3+k4+k5) <b>= 7 OR```
You could use an array and the IndexOf function.
Something like this (you might need to change the “<> -1” to “= -1” below):
```Dim myArray() As Integer = New Integer() {3, 6, 7, 8, 9, 11}
...
If Array.IndexOf(myArray, k1+k2+k3+k4+k5) <> -1 Then
End If```
-CarlosDL | 526 | 1,331 | {"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-2016-07 | longest | en | 0.708555 |
http://www.jiskha.com/display.cgi?id=1206740984 | 1,498,189,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00273.warc.gz | 564,615,706 | 3,796 | # Algebra
posted by on .
Does this relation define a function?
y^2=3x
3x=y^2|0|3|12|27
y |0|1| 2| 3
y^2=3x is a function because the coordinates I chose are not duplicated.
• Algebra - ,
y^2 = 3 x
y = + sqrt(3x)
BUT
y = - sqrt (3x)
for example if x = 12
y = sqrt(36) = 6
BUT
y = -sqrt(36) = -6
works
so there are two values of y for every x
NOT a function | 145 | 360 | {"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-2017-26 | latest | en | 0.699456 |
http://www.tokenrock.com/explain-golden-ratio-177.html | 1,493,598,281,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917126237.56/warc/CC-MAIN-20170423031206-00004-ip-10-145-167-34.ec2.internal.warc.gz | 690,615,201 | 9,480 | # Golden Ratio
DESCRIPTION
The golden ratio, also known as the golden proportion, golden mean, golden section, golden number, and divine proportion is the division of a given unit of length into two parts such that the ratio of the shorter to the longer equals the ratio of the longer part to the whole or, when a line is divided such that the ratio of the longer part of the line to the whole is exactly the same ratio as the shorter part of the line is to the longer part.
It is a number often encountered when taking the ratios of distances in simple geometric figures, such as the pentagon, pentagram, decagon and dodecahedron. It is a ratio or proportion defined by the number Phi = 1.618033988749895... It is an irrational number, meaning it is a number that cannot be written as a simple fraction - the decimal goes on forever without repeating. Phi, like Pi, is a ratio defined by a geometric construction.
Just as Pi is the ratio of the circumference of a circle to its diameter, Phi is simply the ratio of the line segments that result when a line is divided in one very special and unique way.
The figure of a golden section illustrates the geometric relationship that defines this constant. Expressed algebraically:
This equation has as its unique positive solution the algebraic irrational number
Shapes proportioned according to the golden ratio have long been considered aesthetically pleasing throughout many cultures, and is still used frequently in art and design, suggesting a natural balance between symmetry and asymmetry.
The ancient Pythagoreans, who defined numbers as expressions of ratios (and not as units as is common today), believed that reality is numerical and that the golden ratio expressed an underlying truth about existence.
The Golden Ratio seems to get its name from the Golden Rectangle, a rectangle whose sides are in the proportion of the Golden Ratio.
The theory of the Golden Rectangle is an aesthetic one, that the ratio is an aesthetically pleasing one and so can be found spontaneously or deliberately turning up in a great deal of art. The front of the Parthenon can be comfortably framed with a Golden Rectangle. Additional classic subdivisions of the rectangle align perfectly with major architectural features of the structure.
The Golden Rectangle can be used to create a spiral, the Golden Spiral. Starting with one Golden Rectangle, a second Golden Rectangle can be attached to the first using the longest side of the rectangle, side A as the shortest side B of the next rectangle. To this end the second rectangle is constructed 90 degrees perpendicular to the first rectangle. If this process is continued, called the spiraling of the Golden Rectangle, a curved line can be drawn through the corners of the rectangles creating the Golden Mean spiral. The spiraling of the Golden Mean spiral continues indefinitely in inward and outward directions, it's getting smaller and smaller spiraling inwards and getting bigger and bigger spiraling outwards.
Another connection of the Golden Ratio to partial symmetries in nature is through the Fibonacci Numbers. This is a number series where each member is simply the sum of the previous two numbers. Fibonacci spirals and Golden Mean ratios appear everywhere in the universe. The spiral is the natural flow form of water when it is going down the drain. It is also the natural flow form of air in tornadoes and hurricanes. Here's another beautiful example of a Fibonacci spiral in nature, it's the Nautilus shell and every book about sacred geometry contains one:
The Great Pyramid of Giza's dimensions are also based on the Golden Ratio. If we take a cross section of the Great Pyramid, we get a right triangle, the so-called Egyptian Triangle. The ratio of the slant meta-height of the pyramid (hypotenuse of the triangle) to the distance from ground center (half the base dimension) is 1.61804 ... which differs from phi by only one unit in the fifth decimal place. If we let the base dimension be 2 units, then the sides of the right triangle are in the proportion 1:sqrt(phi):phi and the pyramid has a meta-height of sqrt(phi).
Leonardo Da Vinci exhibited the golden ratio in many of his paintings and illustrations calling it the De Divina Proportione or 'divine proportion'. He conducted an entire exploration of the human body and the ratios of the lengths of various body parts.
## Have Additional Questions or Insight?
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Avia Venefica discusses the symbolic meaning of numbers as a tool! | 987 | 4,895 | {"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-2017-17 | longest | en | 0.959346 |
https://ask.csdn.net/questions/234225 | 1,624,417,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488528979.69/warc/CC-MAIN-20210623011557-20210623041557-00165.warc.gz | 111,458,483 | 18,397 | 弹指间
2016-01-24 10:29 阅读 1.4k
以特定格式输出两个数的和
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
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2条回答默认 最新
• 点赞 评论 复制链接分享
• AgoniAngel 2016-01-24 12:39
//一看就是ACM题，贴上AC代码
``````#include<stdio.h>
#include<string.h>
int main()
{
char a[1000],b[1000],c[1001];
int i,j=1,p=0,n,n1,n2;
scanf("%d",&n);
while(n--)
{
scanf("%s %s",a,b);
printf("Case %d:\n",j);
printf("%s + %s = ",a,b);
n1=strlen(a)-1;
n2=strlen(b)-1;
for(i=0;n1>=0||n2>=0;i++,n1--,n2--)
{
if(n1>=0&&n2>=0){c[i]=a[n1]+b[n2]-'0'+p;}
if(n1>=0&&n2<0){c[i]=a[n1]+p;}
if(n1<0&&n2>=0){c[i]=b[n2]+p;}
p=0;
if(c[i]>'9'){c[i]=c[i]-10;p=1;}
}
if(p==1) printf("%d",p);
while(i--)
printf("%c",c[i]);
j++;
if(n!=1) printf("\n\n");
else printf("\n");
}
}
``````
点赞 评论 复制链接分享 | 590 | 1,507 | {"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} | 2.875 | 3 | CC-MAIN-2021-25 | latest | en | 0.530595 |
https://oeis.org/A319712/internal | 1,632,107,283,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00551.warc.gz | 483,849,504 | 2,905 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A319712 Sum of A034968(d) over divisors d of n, where A034968 gives the sum of digits in factorial base. 4
%I
%S 1,2,3,4,4,5,3,6,6,8,5,9,4,7,10,10,6,11,5,14,10,11,7,12,6,7,9,12,5,17,
%T 4,13,11,11,11,18,5,10,11,21,7,19,6,18,19,14,8,18,6,13,12,13,6,17,12,
%U 18,12,11,7,29,6,10,19,19,14,23,7,19,16,25,9,24,5,10,17,17,13,19,6,30,15,14,8,31,15,13,14,27,9,35,13,23,14,17,17
%N Sum of A034968(d) over divisors d of n, where A034968 gives the sum of digits in factorial base.
%C Inverse Möbius transform of A034968.
%H Antti Karttunen, <a href="/A319712/b319712.txt">Table of n, a(n) for n = 1..40320</a>
%H <a href="/index/Fa#facbase">Index entries for sequences related to factorial base representation</a>
%F a(n) = Sum_{d|n} A034968(d).
%F a(n) = A319711(n) + A034968(n).
%o (PARI)
%o A034968(n) = { my(s=0, b=2, d); while(n, d = (n%b); s += d; n = (n-d)/b; b++); (s); };
%o A319712(n) = sumdiv(n,d,A034968(d));
%Y Cf. A034968, A319711, A319715.
%K nonn
%O 1,2
%A _Antti Karttunen_, Oct 02 2018
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Last modified September 19 22:06 EDT 2021. Contains 347576 sequences. (Running on oeis4.) | 617 | 1,530 | {"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-2021-39 | latest | en | 0.538783 |
http://news.datascience.org.ua/2018/12/26/inverse-reinforcement-learning/ | 1,568,800,659,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573264.27/warc/CC-MAIN-20190918085827-20190918111827-00396.warc.gz | 132,233,467 | 8,187 | # Inverse Reinforcement Learning
Once we have the right reward function, the problem is reduced to finding the right policy, and can be solved with standard reinforcement learning methods.The main problem when converting a complex task into a simple reward function is that a given policy may be optimal for many different reward functions..That is, even though we have the actions from an expert, there exist many different reward functions that the expert might be attempting to maximize.In other words, our goal is to model an agent taking actions in a given environment..We therefore suppose that we have a state space S (the set of states the agent and environment can be in), an action space A (the set of actions the agent can take), and a transition function T(s′|s,a), which gives the probability of moving from state s to state s′ when taking action a..For instance, for an AI learning to control a car, the state space would be the possible locations and orientations of the car, the action space would be the set of control signals that the AI could send to the car, and the transition function would be the dynamics model for the car..The tuple of (S,A,T) is called an MDP∖R, which is a Markov Decision Process without a reward function..(The MDP∖R will either have a known horizon or a discount rate γ but we’ll leave these out for simplicity.)The inference problem for IRL is to infer a reward function R given an optimal policy π∗:S→A for the MDP∖R..We learn about the policy π∗ from samples (s,a) of states and the corresponding action according to π∗ (which may be random)..Typically, these samples come from a trajectory, which records the full history of the agent’s states and actions in a single episode:In the car example, this would correspond to the actions taken by an expert human driver who is demonstrating desired driving behaviour (where the actions would be recorded as the signals to the steering wheel, brake, etc.).Given the MDP∖R and the observed trajectory, the goal is to infer the reward function R..In a Bayesian framework, if we specify a prior on R we have:The likelihood P(ai|si,R) is just πR(s)[ai], where πR is the optimal policy under the reward function R..Note that computing the optimal policy given the reward is in general non-trivial; except in simple cases, we typically approximate the policy using reinforcement learning..Due to the challenges of specifying priors, computing optimal policies and integrating over reward functions, most work in IRL uses some kind of approximation to the Bayesian objective.Reward SignalIn most reinforcement learning tasks there is no natural source for the reward signal..Instead, it has to be hand-crafted and carefully designed to accurately represent the task.Often, it is necessary to manually tweak the rewards of the RL agent until desired behavior is observed.. More details | 604 | 2,872 | {"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-2019-39 | latest | en | 0.935059 |
https://stats.stackexchange.com/questions/tagged/generative-models | 1,657,123,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00190.warc.gz | 570,007,247 | 74,808 | # Questions tagged [generative-models]
A probabilistic or statistical model thought about as describing how the values in a sample is actually generated, and not only as a description or approximation.
248 questions
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1 vote
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• 113
1 vote
23 views
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21 views
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1 vote
33 views
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• 111
91 views
### Should I be using batchnorm and/or dropout in a VAE or GAN?
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1 vote
54 views
### Initialization of GAN discriminator
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• 133
52 views
### Restricted Boltzmann Machine: W matrix visualization results after training MNIST images and Pseudo-log-likelihood
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• 21
1 vote
44 views
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33 views
### A Generative model for binary classification (Modeling, Network architecture
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11 views
### Problem for training Wasserstein GAN
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• 437
32 views
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• 169
29 views
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1 vote
39 views
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8 views
### Conventional autoencoder training instability [duplicate]
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83 views
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69 views
### I need help understanding the meaning of the loss values of a WGAN with Gradient Penalty
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1 vote
30 views
### What should I expect if I train a Variational Autoencoder (VAE) with a dataset composed of identical images?
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• 95
1 vote
388 views
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14 views
### Naive Bayes for data generation
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101 views
### loss function that penalizes empirical CDF
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• 165
1 vote
109 views
### Issues with GAN and VAE models
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27 views
### Do GANs compute a posterior distribution, and if not how do they have such good results with just MLE?
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• 179
1 vote
31 views
• 215
30 views
### Can a Simple ANN be Generative?
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325 views
• 2,206 | 2,113 | 9,423 | {"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} | 2.5625 | 3 | CC-MAIN-2022-27 | longest | en | 0.910145 |
https://docs.pymor.org/2023-1-0/autoapi/pymor/algorithms/riccati/index.html | 1,709,501,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00470.warc.gz | 212,962,809 | 7,507 | # pymor.algorithms.riccati¶
## Module Contents¶
pymor.algorithms.riccati.solve_pos_ricc_lrcf(A, E, B, C, R=None, S=None, trans=False, options=None, default_dense_solver_backend=_DEFAULT_RICC_LRCF_DENSE_SOLVER_BACKEND)[source]
Compute an approximate low-rank solution of a positive Riccati equation.
Returns a low-rank Cholesky factor $$Z$$ such that $$Z Z^T$$ approximates the solution $$X$$ of a (generalized) positive continuous-time algebraic Riccati equation:
• if trans is False
$A X E^T + E X A^T + (E X C^T + S^T) R^{-1} (C X E^T + S) + B B^T = 0.$
• if trans is True
$A^T X E + E^T X A + (E^T X B + S) R^{-1} (B^T X E + S^T) + C^T C = 0.$
If E is None, it is taken to be identity, and similarly for R. If S is None, it is taken to be zero.
If the solver is not specified using the options argument, a solver backend is chosen based on availability in the following order:
1. pymess (see pymor.bindings.pymess.solve_pos_ricc_lrcf),
2. slycot (see pymor.bindings.slycot.solve_pos_ricc_lrcf),
3. scipy (see pymor.bindings.scipy.solve_pos_ricc_lrcf).
Parameters
A
The non-parametric Operator A.
E
The non-parametric Operator E or None.
B
The operator B as a VectorArray from A.source.
C
The operator C as a VectorArray from A.source.
R
The matrix R as a 2D NumPy array or None.
S
The operator S as a VectorArray from A.source or None.
trans
Whether the first Operator in the positive Riccati equation is transposed.
options
The solver options to use. See:
default_dense_solver_backend
Default dense solver backend to use (pymess, slycot, scipy).
Returns
Z
Low-rank Cholesky factor of the positive Riccati equation solution, VectorArray from A.source.
pymor.algorithms.riccati.solve_ricc_dense(A, E, B, C, R=None, S=None, trans=False, options=None, default_solver_backend=_DEFAULT_RICC_DENSE_SOLVER_BACKEND)[source]
Compute the solution of a Riccati equation.
Returns the solution $$X$$ of a (generalized) continuous-time algebraic Riccati equation:
• if trans is False
$A X E^T + E X A^T - (E X C^T + S^T) R^{-1} (C X E^T + S) + B B^T = 0.$
• if trans is True
$A^T X E + E^T X A - (E^T X B + S) R^{-1} (B^T X E + S^T) + C^T C = 0.$
If E is None, it is taken to be identity, and similarly for R. If S is None, it is taken to be zero.
We assume:
• A, E, B, C, R, S are real NumPy arrays,
• E is nonsingular,
• (E, A, B, C) is stabilizable and detectable,
• R is symmetric positive definite, and
• $$B B^T - S^T R^{-1} S$$ ($$C^T C - S R^{-1} S^T$$) is positive semi-definite if trans is False (True).
If the solver is not specified using the options argument, a solver backend is chosen based on availability in the following order:
1. slycot (see pymor.bindings.slycot.solve_ricc_dense)
2. scipy (see pymor.bindings.scipy.solve_ricc_dense)
Parameters
A
The matrix A as a 2D NumPy array.
E
The matrix E as a 2D NumPy array or None.
B
The matrix B as a 2D NumPy array.
C
The matrix C as a 2D NumPy array.
R
The matrix B as a 2D NumPy array or None.
S
The matrix S as a 2D NumPy array or None.
trans
Whether the first matrix in the Riccati equation is transposed.
options
The solver options to use. See:
default_solver_backend
Default solver backend to use (slycot, scipy).
Returns
X
Riccati equation solution as a NumPy array.
pymor.algorithms.riccati.solve_ricc_lrcf(A, E, B, C, R=None, S=None, trans=False, options=None, default_sparse_solver_backend=_DEFAULT_RICC_LRCF_SPARSE_SOLVER_BACKEND, default_dense_solver_backend=_DEFAULT_RICC_LRCF_DENSE_SOLVER_BACKEND)[source]
Compute an approximate low-rank solution of a Riccati equation.
Returns a low-rank Cholesky factor $$Z$$ such that $$Z Z^T$$ approximates the solution $$X$$ of a (generalized) continuous-time algebraic Riccati equation:
• if trans is False
$A X E^T + E X A^T - (E X C^T + S^T) R^{-1} (C X E^T + S) + B B^T = 0.$
• if trans is True
$A^T X E + E^T X A - (E^T X B + S) R^{-1} (B^T X E + S^T) + C^T C = 0.$
If E is None, it is taken to be identity, and similarly for R. If S is None, it is taken to be zero.
We assume:
• A and E are real Operators,
• B, C and S are real VectorArrays from A.source,
• R is a real NumPy array,
• E is nonsingular,
• (E, A, B, C) is stabilizable and detectable,
• R is symmetric positive definite, and
• $$B B^T - S^T R^{-1} S$$ ($$C^T C - S R^{-1} S^T$$) is positive semi-definite if trans is False (True).
For large-scale problems, we additionally assume that len(B) and len(C) are small.
If the solver is not specified using the options argument, a solver backend is chosen based on availability in the following order:
Parameters
A
The non-parametric Operator A.
E
The non-parametric Operator E or None.
B
The operator B as a VectorArray from A.source.
C
The operator C as a VectorArray from A.source.
R
The matrix R as a 2D NumPy array or None.
S
The operator S as a VectorArray from A.source or None.
trans
Whether the first Operator in the Riccati equation is transposed.
options
The solver options to use. See:
default_sparse_solver_backend
Default sparse solver backend to use (pymess, lrradi).
default_dense_solver_backend
Default dense solver backend to use (pymess, slycot, scipy).
Returns
Z
Low-rank Cholesky factor of the Riccati equation solution, VectorArray from A.source. | 1,610 | 5,303 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.642401 |
http://www.dovov.com/androidmatrix-preconcatpostconcat.html | 1,566,091,751,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313536.31/warc/CC-MAIN-20190818002820-20190818024820-00013.warc.gz | 245,836,822 | 7,914 | # Android：Matrix – > preconcat和postconcat有什么区别？
• postRotate
• preRotate
• setRotate
### 3 Solutions collect form web for “Android：Matrix – > preconcat和postconcat有什么区别？”
``[320] [240] [ 1]` `
` `[1.5 0 0] [ 0 1.5 0] [ 0 0 1]` `
` `[480] [360] [ 1]` `
` `[1 0 -320] [0 1 -240] [0 0 1]` `
` `[320*1 + 1*-320] [0] [240*1 + 1*-240] = [0] [ 1*1 ] [1]` `
` `matrix: float[] values ={1.2f,0.5f,30,0.5f,1.2f,30,0,0,1}; //as we all know, the basic value in matrix,means no transformation added matrix2: float[] values2 ={1f,0,0,0,1f,0,0,0,1}; Let's say our matrix values are the values above.` `
1，当我们做如下的转换：
` `matrix.preTranslate(-50, -50); is equals to do sequence transformation to matrix2 above like below: matrix2.postTranslate(-50, -50); matrix2.postSkew(0.5f/1.2f,0.5f/1.2f);// note here matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
2，当我们做如下的转换：
` `matrix.preRotate(50); is equals to do sequence transformation to matrix2 like below: matrix2.postRotate(50); matrix2.postSkew(0.5f/1.2f,0.5f/1.2f); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
3，当我们做如下的变换：
` `matrix.preScale(1.3f,1.3f); is equals to do sequence transformation to matrix2 like below: matrix2.postScale(1.3f,1.3f); matrix2.postSkew(0.5f/1.2f,0.5f/1.2f); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
4，当我们做如下的转换：
` ` matrix.preSkew(0.4f,0.4f);` `
` ` matrix2.postSkew(0.4f,0.4f); matrix2.postSkew(0.5f/1.2f,0.5f/1.2f); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
` `matrix: float[] values ={1.2f,0,30,0,1.2f,30,0,0,1}; matrix2: float[] values2 ={1f,0,0,0,1f,0,0,0,1};` `
1. 当我们做如下的转换：
` `matrix.preTranslate(-50, -50);` `
等同于对上面的matrix2进行序列转换，如下所示：
` `matrix2.postTranslate(-50, -50); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
2. 当我们做如下的转换：
` `matrix.preRotate(50);` `
等同于对matrix2进行如下的序列转换：
` `matrix2.postRotate(50); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
3. 当我们做如下的转换：
` `matrix.preScale(1.3f,1.3f);` `
等同于对matrix2进行如下的序列转换：
` `matrix2.postScale(1.3f,1.3f); matrix2.postScale(1.2f, 1.2f); matrix2.postTranslate(30, 30);` `
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• 为什么MATLAB在matrix乘法中如此之快？ | 1,137 | 2,455 | {"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-2019-35 | longest | en | 0.297908 |
http://hismath.blogspot.com/2010/02/jsh-random-distributions-and-prime.html | 1,506,313,306,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690318.83/warc/CC-MAIN-20170925040422-20170925060422-00588.warc.gz | 149,801,742 | 7,233 | ## JSH: Random distributions and prime numbers
It's important to highlight the debate about randomness and prime numbers as for years now the side which claims that random cannot be found with primes has been winning, when all the evidence actually says that they can.
And it's not a minor issue. If primes can give random distributions then random may possibly defined through prime numbers. Random in our reality may BE about prime numbers.
It's an opportunity to answer one of the biggest questions in our reality: what exactly is random?
So I presented a rather simple mathematical axiom:
Prime residue axiom: Given differing primes p_1 and p_2, there is no preference for any particular residue of p_2 for p_1 mod p_2 over any other. (And I'll note that I don't consider 0 to be a residue. )
The axiom indicates then that by residue, there should be random behavior. Here is an example mod 3.
Here is what you get with the first 23 primes greater than 3:
5 mod 3 = 2, 7 mod 3 = 1, 11 mod 3 = 2, 13 mod 3 = 1, 17 mod 3 = 2, 19 mod 3 = 1, 23 mod 3 = 2, 29 mod 3 = 2, 31 mod 3 = 1, 37 mod 3 = 1, 41 mod 3 = 2, 43 mod 3 = 1, 47 mod 3 = 2, 53 mod 3 = 2, 59 mod 3 = 2, 61 mod 3 = 1, 67 mod 3 = 1, 71 mod 3 = 2, 73 mod 3 = 1, 79 mod 3 = 1, 83 mod 3 = 2, 89 mod 3 = 2, 97 mod 3 = 1
There are some mathematical details which have to be handled though before you rush to higher primes, as the maximum gap between primes is roughly p+1, where p^2 is the smallest integer. So to look mod 101, for instance, you'd need to start at 101^2, before you use primes, so you'd take the residue modulo primes greater than p^2.
So I need to clip the first two and start at 11 mod 3.
So the sequence is
2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1
and by the prime residue axiom, it is random.
Primes could be used to label random sequences. As imagine the sequence above were to be labeled, then it could be, residues mod 3, from 11 through 23. For different random sequences, you could just look for them in prime residues, and use the primes themselves to label in the same way.
The max gap isn't a complicated thing to handle. So if you wish to test this idea out, you can program it easily enough, and just look at the distribution with the standard methods to determine randomness.
But notice, if you did not know about the max gap issue, and did so, you could convince yourself that the sequence is not random as you'd have a tendency towards smaller residues until you broke through the barrier.
For those still skeptical consider now twin primes. The prime residue axiom would indicate that for twin primes—two primes in a row separated only by 2, for instance 11, 13, or 17, 19—the probability calculation for their occurrence is actually very easy.
For example, between 5^2 and 7^2, there are 6 primes. The probability then is given by:
prob = ((5-2)/(5-1))*((3-2)/(3-1) = (3/4)*(1/2) = 0.375
(That calculation is fairly straightforward probability.)
And 6*0.375 = 2.25 so you expect 2 twin primes in that interval.
The primes are 29, 31, 37, 41, 43, 47 and you'll notice, two twin primes as predicted: 29,31 and 41, 43.
However, there is an issue which shifts the probability slightly.
If you go into the actual residues it jumps out at you:
29, 31, 37, 41, 43, 47
mod 3: 2, 1, 1, 2, 1, 2
mod 5: 4, 1, 2, 1, 3, 2
Here all the residues for 5 were in evidence so the count came out right, but for random it should have been possible for ALL the residues mod 5 to be 4, but it's not because with 6 primes there isn't enough space in the interval—4*5 = 20, but 49-25=24, where only 12 are odd and only 6 are primes. So the probability is actually off! A scenario where all residues are 4 is precluded by the size of the interval for the larger prime.
Which is an issue like the max gap problem.
That will tend to over-count because the higher residues are less likely to occur because they cannot fit. Easy explanation that jumps out at you with even a small example. Easy.
For the smaller primes it's not an issue as if the prime is greater than interval/(prime count in interval) then that prime isn't affected and its residues can have purely random behavior. For instance, for 3 between 25 and 49, you have 24/(6) = 4, and as that is greater than 3, there is no clipping for 3.
And that's it.
You have all the information needed to see randomness with prime numbers.
For the residue of one prime relative to another, you have to go beyond the max gap. I've hypothesized that's just a matter of going to primes greater than p^2, to get a random sequence of numbers using that prime's residues. For instance, again for p = 101, you'd use primes greater than 101^2.
I've also shown how you can see the count of twin primes following the predictions from random, with a slight over in the expectation value given by difficulty in fitting in higher residues of the larger primes.
Physicists who are curious who are good with their probability and statistics can test out distributions to see if they now look random, and should consider why they believed before that primes did not give us random.
Primes may have been the key all along. The answer to random. By using them fully we may be able to greatly enhance our understanding of random in our own world.
Who knows? Random in our everyday lives may just be about prime behavior. | 1,474 | 5,376 | {"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-2017-39 | longest | en | 0.913471 |
https://poetrywithmathematics.blogspot.com/2015/01/probability-and-coincidence.html | 1,539,750,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510969.31/warc/CC-MAIN-20181017023458-20181017044958-00277.warc.gz | 771,600,088 | 35,393 | Sunday, January 18, 2015
Probability and Coincidence
On page 26 of my copy of the latest New Yorker is a poem by Lia Purpura entitled "Probability." In her brief poem Purpura renders with poetic power the astonishment each of us feels when meeting a long-ago classmate at an out-of-town super market or some other unexpected event. Take time to follow the link and read this poem.
Recently several friends have shared with me their amazement at unexpected coincidences and I have been tempted to illustrate -- perhaps with the birthday paradox -- how likely to happen unexpected events may be.
With more than 23 persons in a room the chances are more than 50-50
that two of them will share a birthday (same day, maybe different years).
Many websites offer explanation of this "birthday paradox" -- here is one.
I offer this link to some thought-provoking quotes about coincidence -- here is one of them, from Isaac Asimov:
People are entirely too disbelieving
of coincidence.
They are far too ready to dismiss it
and to build arcane structures of extremely rickety substance
in order to avoid it.
I, on the other hand, see coincidence
everywhere as an inevitable consequence
of the laws of probability,
according to which having no unusual coincidence
is far more unusual than any coincidence
could possibly be.
These words are found in Asimov's essay, "The Planet that Wasn't," originally published in The Magazine of Fantasy and Science Fiction (May 1975).
Mathematician Joe Mazur is working on a book (FLUKE) about the nature of randomness and coincidences.
1. if there exist more trees
in the world than there are
leaves on any one tree
then surely we find two
trees that are bound to have
the same number of leaves
probably more
than two as is
probable and
almost certain
1. Thanks for your "square" comment! | 416 | 1,825 | {"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-2018-43 | longest | en | 0.961682 |
https://byjus.com/question-answer/in-equilibrium-reaction-x-moles-of-the-reactant-a-decompose-to-give-1-mole-each-1/ | 1,721,500,638,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00239.warc.gz | 120,731,600 | 27,635 | 1
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Question
# In equilibrium reaction, x moles of the reactant A decompose to give 1 mole each of C and D. If the fraction of A decomposed at equilibrium is independent of initial concentration, then the value of x would be:
A
1
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Solution
## The correct option is A 2The fraction of A decomposed at equilibrium is independent of initial concentration.This implies that the concentration term is not present in the equilibrium constant expression.The equilibrium reaction is xA⇌C+D.Let the inital concentration of A be c M.The equilibrium concentrations of A,C and D are c(1−a),cαx and cαx respectively.The expression for the equilibrium constant is KC=[C][D][A]x=(cαx)×(cαx)[c(1−α)]x=(cα)2x2[c(1−α)]x.When x=2, the equilibrium constant expression will be free from the concentration term.Hence, KC=c2α24[c(1−α)]2=α24(1−α)2.Thus, option B is correct.
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Join BYJU'S Learning Program | 406 | 1,372 | {"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-2024-30 | latest | en | 0.821388 |
https://kr.mathworks.com/matlabcentral/cody/problems/2020-area-of-an-isoceles-triangle/solutions/690329 | 1,579,974,808,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00437.warc.gz | 508,977,218 | 15,509 | Cody
# Problem 2020. Area of an Isoceles Triangle
Solution 690329
Submitted on 23 Jun 2015 by Kirti Raikar
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
%% x = 5; y = 8; A_correct = 12; tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
2 Pass
%% x = 2; y = 2; A_correct = sqrt(3); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance)
3 Pass
%% x = 10; y = 2; A_correct = sqrt(99); tolerance = 1e-12; assert(abs(isocelesArea(x,y)-A_correct)<tolerance) | 210 | 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} | 2.875 | 3 | CC-MAIN-2020-05 | latest | en | 0.651491 |
https://www.skillsworkshop.org/index.php/ict?search=&f%5B0%5D=ict_resource_type%3A2183&f%5B1%5D=ict_resource_type%3A2190&f%5B2%5D=subjects_ict_area%3A954 | 1,581,983,762,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00201.warc.gz | 909,402,995 | 10,709 | # ICT Resources
Displaying 1 - 10 of 37 resources:
## ITQ Level 1 Cult Film PowerPoint Challenge
Completing these challenges will help students to create a basic PowerPoint that runs automatically and provides evidence through printed handouts.
ITQ Learning outcomes covered: 1.1 1.3 1.6 2.3 2.4 3.2 3.3
Editor's note
I think the learning outcomes quoted above relate to the OCR ICT Award (or similar) but this beautifully clear, step by step task sheet is ideal for any Level 1-2 ICT course.
Level
L1
L2
ICT
Finding and selecting information
Developing, presenting and communicating information
Level 1 and 2 CLAIT / ECDL
Context
Art Film Media Music TV
## ICT (ITQ) Microsoft Word Task Sheet
An exercise in Microsoft Word built around the content of a HR article from:
https://www.consultancy.uk/services/hr-consulting
Editor’s note
Three separate step by step activities cover paragraph formatting, working with tables, and pictures/objects.
Level
L2
L1
ICT
Level 1 and 2 CLAIT / ECDL
Developing, presenting and communicating information
Functional ICT
Context
## Fatal Affraction
This is a set of worksheets in a MS Excel workbook that deals with fractions.
- Sheet one: a set of pictures (pie charts) that shows fractions from halves to tenths
- Sheet two: equivalent fractions with two pie charts, learners can input fractions and see if they are equivalent by looking at the shape of the pie charts
- Sheet three: starting to look at fractions being equivalent to decimals with two pie charts one for fractions and one for decimals
Level
GCSE L1-5
L2
L1
Maths
N2/L2.3
N2/L2.1
GCSE N10
Functional Maths - numbers and the number system
ICT
Using ICT
## Presenting information L1 ICT task sheet
A task sheet for L1 Functional Skills ICT. It focuses on Task 3 of the Edexcel exam paper which involves being creative and producing a document that uses clear presentation features and is fit for purpose.
Includes two tasks (a restaurant menu and a flyer for a student production) plus a short extension.
Level
L1
ICT
Developing, presenting and communicating information
General ICT resources
Context
Catering Food Nutrition
Art Film Media Music TV
## Haunted Portsmouth
Designed for Entry 3 – Level 2 Functional Skills English.
Students loved this lesson. Can be designed around other areas but covers:
Writing persuasively.
Fact and opinion.
Research write a factual article
Create a poster using appropriate language.
Level
L2
L1
E3
English
Functional Skills English
ICT
Finding and selecting information
Context
History
## Working comfortably and safely
A word processing activity which involves typing and formatting and also contains useful information.
The idea is that the learners digest the text that they are working with and learn from it. I would put it at about level E2/3.
Level
E3
E2
ICT
Functional ICT
## International Talk Like a Pirate Day
Literacy – aimed at Level 1-2 and links to imaginative writing at GCSE.
Quirky fun starter – find your pirate name. This can be done on mobile phones if there are no PCs.
After finding the King’s shilling, write descriptive letter home.
Level
L2
L1
English
Functional English - writing
ICT
Finding and selecting information
Context
History
## Littergram renamed? L1-2 Functional English pack
This is a literacy resource based around a recent news story. There are a range of tasks relating to an anti-litter app called ‘littergram’ and legal challenges made to the use of its name, with the potential to extend tasks into other aspects of littering and how the litter problem might be resolved. A final extension task asks students to take their own photographs of local examples of litter as supporting evidence for a writing activity.
Level
L2
L1
English
Functional English - writing
Functional English - speaking, listening & communicating
General literacy / English
ICT
Finding and selecting information
Context
News, Politics & Government items
Electrical, Electronics & Technology
## Plan a Christmas Fete E1-E2
Students are to plan and advertise a Christmas fete.
The resource covers writing lists and sentences, proof reading and using basic adjectives.
Students could use the Internet to find some ideas for Christmas stalls.
Level
E2
E1
English
Functional English - writing
Wt/E2.1
Wt/E1.1
ICT
Finding and selecting information
Context
Voluntary & Charity
## Car Service L2 Functional ICT
Set of resources created by myself and colleague, David Bayne , to help students use Word, Excel and Access. We wanted students to practise for the Level 2 ICT Functional Skills exams, using a method that would be familiar to them. They all drive cars and need to get them serviced, so this seemed a suitable area to use.
Level
L2
ICT
Functional ICT
Context
Motor vehicles & Transport | 1,103 | 4,780 | {"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} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.867249 |
https://zakladkadrugs.xyz/?post=3338 | 1,627,961,427,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00309.warc.gz | 1,107,044,995 | 5,184 | # Can dust particles be charged electronically
### Particle adhesion due to static electricity
When particles move in currents of air caused by air conditioners or similar devices, they take on a positive or negative charge through contact with various objects. Positively charged particles are attracted to negatively charged objects and vice versa. The more particles there are in the air, the more particles adhere to the objects in the room. If the particle sources (usually people and clothing) are also electrically charged, the particles emanating from these sources are also electrically charged. This attractive force generated by electrostatic charges is known as the Coulomb force.
The Coulomb force is expressed using the following formula:
F 3338 is the Coulomb force, Q1 and Q2 [C] are electrostatic charges, d [m] is the distance between the charges and ε0 [F / m] is the electrical constant (ε0 = 8.85 × 10-12 F / m). The greater the electrostatic charge of the objects, the greater the Coulomb force and the weaker the greater the distance between the objects. If the distance is multiplied by 2, the Coulomb force is 1/4 of its previous value. If the distance is multiplied by 10, the Coulomb force has 1/100 of its previous value. This means that the force acting between two objects with a distance of 1 mm is reduced to 1/100 of its previous strength when the distance between the objects increases to 10 mm.
### Attachment of particles to particles
We have calculated the distance at which two highly charged objects begin to adhere to one another due to the Coulomb force.
The distance is shown on the horizontal axis and the Coulomb force on the vertical axis. The three lines indicate three examples with different particle sizes. The three different particles have a diameter of 0.5 μm, 5 μm and 50 μm. Let us concentrate on points A, B and C. They mark the distances at which the Coulomb force acting between the particles and the gravitation acting on the particles are in equilibrium. This means that the particles stick to each other when the distance between them decreases because the Coulomb force is greater than gravity. If the distance is greater than indicated by the dots, gravity exceeds the Coulomb force and the particles fall to the ground without sticking to each other. The 5 μm particles attract each other at a distance of less than 0.1 mm. If the distance is 0.1 mm or more, the particles fall to the ground without sticking to each other. It follows that particles only adhere to one another as a result of electrostatic charging if the distance between them is extremely small.
### Attachment of particles to objects
The size of the Coulomb force also changes with the strength of the electrostatic charge. In other words, the size of the Coulomb force depends on what kind of object is electrostatically charged. Since particles are only weakly electrostatically charged, they only adhere to one another if they are very close. Let's look at the case where both a particle and an object are electrostatically charged. Since the object is significantly larger than a particle, the amount of static electricity is greater. It is also true that the attenuation of the Coulomb force in a large object is no longer inversely proportional to the square of the distance. We left out the detailed calculations, but if the object is electrostatically charged, it can attract particles from a distance of 20, 30 and more centimeters.
Next, let's look at the case where the object is not electrostatically charged, but the particles are. In this case, there is another aspect to consider: the conductivity of the object. First, let's consider a ladder. When electrostatically charged particles approach the conductor, static electricity is generated on the surface of the conductor, which is opposite to the charge on the particles. The reason for this is the movement of the electrical charges in the conductor. This phenomenon is known as electrostatic induction.
Grounding the object is ineffective as a countermeasure for this phenomenon. This does not prevent an opposite charge from building up as a result of the Coulomb force. The only possible countermeasure is the electrostatic discharge of the particles. Since the electrostatic charges of a conductor and the particles are opposite to each other, the Coulomb force occurs.
As before, we calculated the distance at which a particle is attracted to an object.
For particles with a diameter of 5 μm, the Coulomb force becomes so great that a particle adheres to the object at a distance of 0.1 mm or less. Particles do not adhere to the object if the distance between the two is at least a few millimeters.
However, only the particles are electrostatically charged and the charge is small, so that the Coulomb force is weak and the distance at which the particles adhere to the object is only a few millimeters.
Next, let's look at the case where the object is an uncharged insulator and only the particles have an electrostatic charge. Non-conductors have an insulating effect on electricity. Therefore there is no electrostatic induction. In the above formula for the Coulomb force, since the charge Q1 is zero, the Coulomb force is also zero. In other words, the particle does not adhere to the object. | 1,110 | 5,316 | {"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} | 3.640625 | 4 | CC-MAIN-2021-31 | latest | en | 0.941701 |
https://www.environmentalistsforeurope.org/what-is-a-5-generation-family-tree/ | 1,720,839,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00065.warc.gz | 603,010,098 | 11,679 | ## What is a 5 generation family tree?
This blank 5 generation family tree includes the parents of the child, the grandparents on both sides, the great grandparents on 4 sides and up till 3rd generation great great grandparents.
## How many 5 generation relatives are there?
But a 10 generation list of ancestors contains 1,022 because this is the total number of ancestors for the 2nd to the 10th generation inclusive….Ancestral Reference Numbering System.
2nd generation 2 ancestors = 2 ancestors in total
5th generation 16 ancestors = 30 ancestors in total
What are 5 generations?
Baby Boomers: 1946 – 1964. Generation X: 1965 – 1979. Millennial: 1980 – 1996. Gen Z: 1997 – 2015.
How do you label a generation on a pedigree?
As you add generations, label each generation using Roman numerals (I, II, III, and so on). The original set of parents, located at the top or far left, are generation I. Continue the sequential numbering to their children, grandchildren, great grandchildren and so forth.
### How far back is 5 generations?
A Familial generation is used based on when an actual birth took place within a family. A few centuries ago, the accepted and often used value was about 20 years, meaning that as many as five familial generations could exist in a single century.
### How do you calculate generation of family?
To start, you and your siblings and cousins make one generation, and your parents and their siblings from the next. Your grandparents and their siblings form the third generation, and so on. The top-level of any family tree is the first generation, next down is their children, making up the second generation.
How do you make a generational family tree?
1. Gather information about your family. Write down what you know, ask family members to fill in the gaps, and find pictures and documents.
2. Draft a family tree outline. Compile all of the information you have and create an outline.
3. Add information to each leaf.
4. Distribute your family tree diagram.
How many years is 5 generations?
## How do you categorize generations?
Generation names explained
1. The Lost Generation — born 1883-1900.
2. The Greatest Generation — born 1901-1924.
3. The Silent Generation — born 1925-1945.
4. Baby Boomer Generation — born 1946-1964.
5. Generation X — born 1965-1980.
6. Generation Y — born 1981-1996.
7. Generation Z — born 1997-2012.
8. Generation Alpha — born 2013-2025.
## What is 3 generation performance pedigree package?
Displays the names, titles, and coat colors of up to 14 ancestors. Also includes the number of pups each ancestor has produced and a summary of the titles earned by those pups. Before a pedigree can be ordered, the dog must be permanently registered.
How do you count generations in a family?
How many generations are shown in the pedigree?
(7.4) Pedigree flashcards
A B
How many generations are shown on this pedigree?, 3 generations are shown.
Which individual is a female affected by the trait of interest?, I-2 is the affected female.
Which individual is a male affected by the trail of interest?, III-2 is the affected male.
### How many generations is 5 ethnicity?
Concepts – Calculating Ethnicity Percentages
Generation # You Have Approximate Percentage of Their DNA That You Have Today
2 4 25%
3 8 12.5%
4 16 6.25%
5 32 3.12% | 791 | 3,317 | {"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} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.929886 |
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