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https://www.teacherspayteachers.com/Product/Coordinate-Graphing-Ordered-Pairs-Mystery-Pictures-Bird-Butterfly-Fish-etc-1525225 | 1,537,815,852,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160641.81/warc/CC-MAIN-20180924185233-20180924205633-00279.warc.gz | 888,205,237 | 21,224 | # Coordinate Graphing Ordered Pairs Mystery Pictures: Bird, Butterfly, Fish etc
Subject
Resource Type
Common Core Standards
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14 MB|10 pictures x 2 levels
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Product Description
Practice plotting ordered pairs with these fun beginner level coordinate graphing mystery pictures! The pictures are simplified and only have a few shapes to make it easy for students who have just begun learning about plotting coordinates. These are also helpful for students who need extra practice.
This activity is easy to differentiate by choosing either the first quadrant (positive whole numbers) or the four quadrant (positive and negative whole numbers) worksheet. All points are represented by whole numbers, there are no fractions or decimals. This activity is perfect for math centers, early finishers or homework.
Students who have a good understanding of coordinate graphing will enjoy my more advanced mystery pictures.
This bundle includes 10 mystery pictures: apple, bird, boat, butterfly, fish, kite, pencil, smiley face, star and tree.
Graphing paper, coordinates worksheets and answer keys are included.
Instructions:
Students plot the ordered pairs and draw connecting straight lines as they plot. When the word “STOP” is reached, the student should NOT connect the last point with the first in the group.
Each picture comes in a separate PDF file.
BUNDLE DISCOUNT! Save 50% when you purchase this product as part of my Coordinate Graphing Mystery Pictures All Year Mega Bundle.
If you like this activity then you may also like my other Mystery Pictures worksheets.
Coordinate Graphing:
- Thanksgiving Graphing
- Halloween Graphing
- Superhero Kids Graphing
- Graphing Coordinates Mystery Pictures for Beginners - Bundle of 10
- Back to School Animals Graphing
- Summer Ice Cream, Sun, Popsicle, Flip Flops Graphing
- Easter bunny and chicks graphing
- Spring Bee, Butterfly, Ladybug, Flower
- Mother's Day Graphing
- Earth Day Graphing
- Superhero
- Christmas
- Spring
- Monsters
- St. Patrick's Day
and many more!
Search filter: CGMP EOYAct SumAct BTSAct
Total Pages
10 pictures x 2 levels
N/A
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N/A
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\$3.99 | 473 | 2,202 | {"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-2018-39 | latest | en | 0.857346 |
http://www.xbiz.com/features/1513/understanding-conversion-ratios | 1,501,074,259,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426161.99/warc/CC-MAIN-20170726122153-20170726142153-00022.warc.gz | 573,430,118 | 12,676 | educational
# Understanding Conversion Ratios
Stephen Yagielowicz
I've discussed many aspects of sponsor programs in the past, and today I'm going to delve into one of the most mysterious: the correlation of payouts, conversion ratios, and earnings. If you don't understand this, then you better read this article now!
One of the most confusing aspects of working with adult Webmaster affiliate (or "sponsor") programs is knowing which will be the most profitable for you. Notice that I did not say "highest paying" but carefully stated "most profitable." The difference is not merely semantic, but could be the deciding factor between your success and failure.
There are many types of sponsorship programs wooing Webmaster's for their surfer traffic; all singing the siren song of "Highest paying program on the 'Net!" or some such foolishness: and why not? Most newbies gravitate to the sponsor touting the biggest numbers in their ads, while rarely looking beyond the hype to the math underneath the offer, or the fine print behind it.
On the face of it, the choice between one sponsor offering a payout of \$25 per trial membership, and another offering a payout of \$50 per trial membership is a no-brainer. After all, since one sponsor will pay you twice as much as the other, then you should choose that one, and make twice as much money, right? Perhaps, but is that all there is to the equation? Not when you factor conversion ratios into the equation:
Converting Traffic Into Cash
Conversion ratios (or "conversions") are probably the single biggest factor influencing your actual earnings, yet so many Webmasters know nothing about them or even consider them in their exit traffic marketing plan. Simply put, a sponsor site's conversion ratio is the ratio of sales made per number of unique visitors you sent to the sponsor's site. In other words, if 500 surfers on YOUR site click a banner to visit your sponsor, and ONE of those folks then buys a trial membership to the sponsor's site, that sponsor site converted at 1:500 for you.
With all things being equal, and given the two payout examples above, you would receive either \$25 or \$50 for that sale, depending upon the sponsor you choose. The problem is that all things ARE NOT equal, and a site's conversion ratio is not a static value, but can fluctuate throughout a single day, let alone throughout the life of the site. Let's take a look at another example:
Let's say that sponsor "A" paying \$25 per is converting at 1:250 for you, and sponsor "B" paying \$50 per is converting at 1:500. You're now receiving an equal rate of pay (your actual earnings) from both sponsors, since "A" converts twice as well as "B" (1:250 vs. 1:500). This example illustrates the fact that the amount you will actually EARN is not the same thing as what the sponsor is offering to PAY.
Consider also that just because a program offers a higher payout rate, it does not mean that it will convert any worse than a program offering a smaller payout rate. In another example, sponsor "A" could convert at 1:500, earning you \$25 for the 500 surfers you sent, while sponsor "B" converts at 1:250, earning you \$100 from those same 500 surfers.
Factors Affecting Conversions
When evaluating conversions, realize that a sponsor's claims about their sites are all but meaningless to you. This is not to say that they are lying about how well (or poorly) their sites convert, but that conversions are a very subjective issue, and tossing out your best number is not the same as your average ratio, and certainly not representative of what YOU may expect.
Conversions are most heavily influenced by two factors: the quality of your traffic, and the quality of the sponsor's offer. The quality of your traffic is determined by its source (TGP traffic will convert differently from Search Engine traffic, for instance) and by how well you target it to the sponsor. For example, if your free site is focused on a Mature Ebony niche, your sales will be disappointing if you send all your traffic to "Cracker Teens" or "Asian Lolitas." Two "real world" examples of notable conversion ratios: I used to pull from 1:5 to 1:125 sending my AVS traffic to CyberErotica, while I pulled 1:3500+ sending blind TGP traffic to UltraTeen. In this case, traffic quality, rather than sponsor quality, was likely the main difference. Perhaps the "join page" was too slow, unintuitive, or simply wasn't available in the prospect's native language.
Sponsor "quality" can and does make a difference however. Some factors that will influence the sponsor's ability to convert the traffic you send in to sales are once again its quality, including the way in which it was sent: Blind link traffic will perform more poorly than traffic clicking on an approved banner, for instance. A sponsor's ability to "make the sale" is also quite vital, and not just the product of a decent tour that "pulls." Perhaps his credit card processor "scrubs" a bit too hard, declining potential members due to a poor credit rating; and no secondary processor was available to "cascade" the prospect's application into. Perhaps the "join page" was too slow, unintuitive, or simply wasn't available in the prospect's native language. There are many factors that come into play here, and almost all of them are beyond your control.
Since your traffic mix is YOUR traffic mix, and particular to your blend of sites and sources, you must find the sponsor and program that does the best for YOU, and it may not be the sponsor that does the best for everyone else. This isn't bad, or a problem, just an acknowledgement that your traffic has unique aspects that will prove more profitable at one site than at another, and the only way for you to know which is best is to test! Good Luck! ~ Stephen
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Steve Hamilton · | 1,424 | 6,405 | {"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-2017-30 | longest | en | 0.956062 |
https://statshelponline.com/linear-modeling-survival-analysis-assignment-help-19150 | 1,519,204,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813602.12/warc/CC-MAIN-20180221083833-20180221103833-00625.warc.gz | 770,546,269 | 14,197 | Select Page
## Linear Modeling Survival Analysis Assignment Help
Introduction
Generalized linear modeling and Survival Analysis is one of the modern-day subjects in data and generally consists of ANOVA, R shows, optimum probability estimate and associated principles. Our Stats specialists and Stats online tutors being skilled in these contemporary ideas can cater to whole selection of your requirements in generalized linear modeling and Survival Analysis such as research help, project help, argumentation help, quizzes preparation help and so on circulation flow survival time. In this paper, we focus on those styles that can be established in terms of the probability of default by using survival analysis techniques.
It is not unexpected that R needs to be abundant in survival analysis functions. CRAN's Survival Analysis Job View, a curated list of the finest pertinent R survival analysis bundles and functions, is undoubtedly powerful. It is a wonderful building that offers some concept of the substantial contributions R designers have actually made both to the theory and practice of Survival Analysis. As efficient as it is, nevertheless, I picture that even survival analysis professionals require some time to discover their method around this job view. Survival analysis styles components that impact the time to an event. Routine least squares regression strategies fall short due to the reality that the time to event is typically not typically distributed, and the style can not deal with censoring, exceptionally normal in survival info, without modification.
The Survival platform fits a single Y that represents time to event (or time to failure). Make use of the Survival platform to take an appearance at the flow of the failure times. Survival analysis assists the scientist examine if, any why, specific people are exposed to a greater threat of experiencing an occasion of interest, such as death, device failure, drug regression, and so on. This is likewise described as occasion history analysis. Survival analysis includes a wide array of strategies that help the scientist examine time-to-event designs. Survival analysis can be utilized to study lots of things and is incredibly practical in studying the reason for deaths and births. It can likewise be utilized by the scientist in order to comprehend the cause( s) of divorces and marital relationships, as well as the reason for wars and transformations.
The very first action in survival analysis is for the scientist to develop exactly what occasion is to be evaluated. The term 'occasion' is a shift from one state to another. When the occasion has actually happened for people or group of people, the next action is to make sure that the information consists of a longitudinal record of. Utilizing the example above, this might consist of ages and marital relationship dates, in addition to other explanatory variables (e.g., gender, earnings, and so on). The reliant variable can be survival time or shift rate, and analyses can be carried out to explain the differing percentages of making it through cases at various times or to examine the relationship in between survival time and a set of covariates/predictors. The survival time information in survival analysis has 2 crucial unique qualities. The survival time information in survival analysis is not unfavorable and is typically favorably manipulated.
The survival time, which is the item of research study in survival analysis, must be separated from the calendar time. The survival time in survival analysis ought to constantly be determined associated to some suitable time origin. It presumes you currently have the fundamental tools of logistic and linear regression, parametric and semi-parametric survival analysis in your well-stocked analytical tool box which you obtained in graduate school. The concern this book addresses is how do you utilize those regression tools effectively. Our Data professionals and Stats online tutors being proficient in these modern-day principles can cater to whole variety of your requirements in generalized linear modeling and Survival Analysis such as research help, project help, argumentation help, quizzes preparation help and so on
circulation flow survival time. CRAN's Survival Analysis Job View, a curated list of the finest pertinent R survival analysis plans and functions, is certainly powerful. Not just is the plan itself abundant in functions, however the item produced by the Surv() function, which consists of failure time and censoring info, is the fundamental survival analysis information structure in R. Dr. Terry Therneau, the plan author, started working on the survival plan in 1986. The survival and risk functions are important concepts in survival analysis for discussing the blood circulation of celebration times. The survival time information in survival analysis has 2 crucial unique qualities.The survival plan is the foundation of the whole R survival analysis building. Not just is the bundle itself abundant in functions, however the item produced by the Surv() function, which includes failure time and censoring details, is the standard survival analysis information structure in R. Dr. Terry Therneau, the bundle author, started working on the survival bundle in 1986. Afterwards, the bundle was included straight into Plus, and consequently into R.
Generalized linear direct designs supply common typical technique a broad range variety response modeling problemsIssues The GLM can be fitted utilizing a typical treatment and a system for hypothesis screening is readily available. Diagnostics utilizing deviance residuals offer a method to inspect that picked designs are sufficient. When the information about their survival time is inadequate; the most usually come throughout kind is perfect censoring, observations are called censored. The survival time for this person is thought of to be a minimum of as long as the duration of the research study. Censoring is a necessary issue in survival analysis, representing a particular kind of losing out on info. The survival and threat functions are necessary concepts in survival analysis for describing the flow of event times. The survival function supplies, for each time, the possibility of withstanding (or not experiencing the event) approximately that time. While these are normally of direct interest, great deals of other quantities of interest (e.g., suggest survival) may subsequently be estimated from comprehending either the hazard or survival function.
https://youtu.be/LXltTeSe594 | 1,187 | 6,589 | {"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-2018-09 | longest | en | 0.919249 |
https://www.magicalapparatus.com/pencil-reading/my-word-by-corinda.html | 1,548,318,489,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584519757.94/warc/CC-MAIN-20190124080411-20190124102411-00524.warc.gz | 863,906,948 | 10,144 | ## My Word by Corinda
Before I describe the effect I would like to say that this is a creditable example of simplicity and boldness. You might think it too simple to work— but if you do as I have done, try it out, you will find that the "obvious is not always apparent!" You cannot judge any effect from paper. It must be performed before you can really tell whether or not is is any good. Tricks that look like miracles in print—sometimes flop badly in practice. This is the reverse; it looks silly in print and works like a charm.
The Effect. A spectator is given a sheet of paper which bears three words. "Page, Line and Position". He is asked to imagine that he has a book in his hands and that he opens it to a page and sees the number at the top. He is handed a pencil and told to write the page number alongside the heading "Page". He then imagines that he counts down to any line—and fills the number of lines alongside "Line". Finally, he counts along the line to any word and fills in the Position number. All this is done "in imagination". Whilst this is going on, you hand a book to another spectator and tell him. to wait a moment. You now take the numbered sheet from the first person and hand it to the spectator with the book. He opens the book at the selected page number, counts down to the chosen line and finds the word at the given position. He calls out that word. You proceed to turn round a large slate which has been on show all the time—and on the back is seen written the chosen word!
The Method. Take a piece of paper and in ink write at the top "Page "
below this write "Line " and below this write "Position now take a good thick book and open it anywhere in the middle. With a PENCIL and in DIFFERENT HANDWRITING fill in that page number on your form. Count down seven lines and put Line 7 on the form and finally count along three words and fill in a three alongside Position Prepare a second form with the three headings written in ink—but leave it blank. Look to be quite sure what word is at the third position on the selected page and write that in bold letters on your slate.
Turn the slate with the blank side towards the audience so that the word is not seen until the climax. Fold the prepared sheet once each way to reduce it to a convenient size for palming. Hand the unprepared slip to a spectator near the BACK and be sure that the book goes to someone near the front. Choose two people that are seated as far apart as possible. Have the person with the slip fill in his numbers, telling him to show no one for a moment and to fold his paper when done. You take it and fold once more and then whilst walking from the back to the front (the greater the distance—the more time you have) simply exchange the two slips—handing YOUR prepared paper to the book man. He does not know what the other person chose and until afterwards, does not know what is going to happen. You tell him to open the book "at the chosen page"—count down to the "Chosen Line" and call out the "Chosen Word". The other spectator does not know that his numbers are in your pocket. That is all there is to it—excepting the switch. You have so much time to perform this essential move, that there is no call for complicated switching of papers—simply EXCHANGE them and put the other one in your pocket. Be sure that you hand a Pencil to the spectator who decides upon the numbers.
If you don't mind hard work, look carefully through many books and sooner or later you will find the same word in the same position in three different books. You may then give the second spectator a choice of any of three books which adds to the effect. The occurrence of three words the same in identical positions is not as unlikely as it sounds—you can find "and, of, the, etc.," jn many books—but try and find something better than anything so commonplace as these words.
Should any of my readers feel that the trick may be discovered, perhaps they would care to try the same effect another way:—Have the numbers called out loud and the page found as the number is called—finally when' the word is called write it on a card with a Swami Gimmick. (Step One will tell you how to do this!)
## Practical Mental Influence
Unlock the Powers of Mental Concentration to Influence Other People and to Change Situations. Learn How to mold the mind one-pointed, until you have focused and directed a mighty degree of Mental Influence toward the desired object.
Get My Free Ebook | 976 | 4,475 | {"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-2019-04 | longest | en | 0.967746 |
http://calculatorpack.com/micro-to-base-calculator/ | 1,721,290,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00002.warc.gz | 5,684,550 | 11,287 | # Micro to Base Calculator
Have you ever struggled with converting microvalues to their respective base values or vice versa? Introducing the Micro to Base Calculator! This handy tool simplifies the process by allowing you to easily convert microvalues to their base equivalents or base values to their microequivalents. Our calculator is user-friendly and easy to use, even for those who are not adept at calculations. With just a few clicks, this calculator will save you time and frustration in trying to figure out complicated conversions. Whether you are in the healthcare industry, scientific research, or any field that deals with microvalues, the Micro to Base Calculator is a must-have tool. Try it out today and simplify your conversion needs!
## Micro to Base Calculator
Convert micro to base value
Micro to Base Calculator Results
Micro Value:0
Conversion Factor:0
Base Unit:0
Base Value:0
converting between micro units can be useful in various contexts. Our micro to base calculator simplifies this conversion. To explore calorie-related calculations and understand nutritional values, consider linking it with our micro calorie calculator. This combination offers a comprehensive resource for converting and analyzing nutritional data.
## How to Use the "Micro to Base Calculator"
The "Micro to Base Calculator" is a useful tool that converts micro values to their corresponding base values. This calculator is significant for scientists, researchers, and individuals working in technical fields that frequently use micro values. It simplifies the calculation process and eliminates the possibility of errors that may occur when converting values manually.
The primary applications of the "Micro to Base Calculator" include converting micro values to their corresponding base values for measurements such as length, weight, and volume. The calculator is also useful for converting micro values to base values in chemistry and physics equations.
## Instructions for Utilizing the Calculator
To use the "Micro to Base Calculator," users must input three values: the micro value, conversion factor, and base unit. The micro value is the value to be converted to a base value. The conversion factor is the multiplier that converts the micro value to a base value. The base unit is the unit of measurement for the resulting base value.
After inputting the values, the calculator will automatically calculate and display the resulting base value. The output fields include the micro value, conversion factor, base unit, and base value. The micro value, conversion factor, and base unit will be displayed as entered by the user, while the base value will be automatically calculated and displayed by the calculator.
## Formula
The "Micro to Base Calculator" formula is simple. It multiplies the micro value by the conversion factor to get the base value. Mathematically, the formula can be expressed as:
Base Value = Micro Value x Conversion Factor
## Illustrative Example
Suppose a researcher wants to convert a micro value of 500 micrometers to millimeters. The researcher knows that one millimeter is equivalent to 1000 micrometers, making the conversion factor 0.001. To use the "Micro to Base Calculator," the researcher will input a micro value of 500, a conversion factor of 0.001, and a base unit of millimeters. The calculator will automatically calculate the base value as 0.5 millimeters.
## Illustrative Table Example:
Micro Value Conversion Factor Base Unit Base Value 1000 0.001 meters 1 500 0.001 millimeters 0.5 750 0.00001 liters 0.0075 5000 0.0001 grams 0.5 200 0.01 seconds 2
The "Micro to Base Calculator" simplifies the conversion of micro values to base values. By following the outlined steps and formula, users can easily calculate base values from micro values. The calculator saves time and eliminates the possibility of human error that may occur when converting values manually. It is a handy tool that is essential for scientists, researchers, and individuals working in technical fields that require frequent conversions of micro values. | 813 | 4,095 | {"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} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.854495 |
http://www.statmodel.com/discussion/messages/23/23895.html?1491876517 | 1,581,942,323,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875142323.84/warc/CC-MAIN-20200217115308-20200217145308-00245.warc.gz | 245,825,288 | 4,671 | Output for theta and delta parameteri...
Message/Author
Richard T. Campbell posted on Wednesday, March 22, 2017 - 2:57 pm
I ran example 3.13 which uses WLMV with the theta parameterization. The output shows scale factors for the endogenous variables. When I change to the delta I see residual variances. I think I understand the distinction between the two, but I am puzzled by the output which seems to contradict the explanation in the user manual.
Bengt O. Muthen posted on Wednesday, March 22, 2017 - 7:04 pm
Are you adding to the output requests for 3.13?
Could you send the output for your run to Support?
James Algina posted on Saturday, April 08, 2017 - 7:42 pm
Hi:
If the theta parameterization is used in a two-group one-factor factor model and the scalar model is believed to be an adequate model for the data, is the following correct? The expressions from page 495 in the Mplus manual
P (u = 0 | x) = F (t1 - b*x),
P (u = 1 | x) = F (t2 - b*x) - F (t1 - b*x),
P (u = 2 | x) = F (- t2 + b*x),
with b = the factor loading and x = the factor, can be used to calculate category probabilities on item i for the group in which the residual variance equals one, but have to be modified for the other group by dividing each term by the residual standard deviation for item i.
Thanks,
Jamie
Bengt O. Muthen posted on Monday, April 10, 2017 - 7:08 pm
These formulas are ok if you want the probabilities conditional on specific factor values x. | 390 | 1,453 | {"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-2020-10 | latest | en | 0.90265 |
https://numberdividedby.com/short-division-worksheet/ | 1,669,946,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00221.warc.gz | 480,572,354 | 20,558 | Ace division by practicing them through a short division worksheet. In the primary ages a child absorbs most of what is taught to them, so try to make them learn and understand basic topics of mathematics like division using worksheets so that the topics are properly understood by them which can help them in solving difficult ones in the upcoming years.
Short worksheets are easy to solve and are not that time consuming. They will boost your kid’s confidence and help in keeping them engaged in studies in a fun and competitive way. Worksheets help you understand where the kid is lacking and where there needs to be some improvement.
## Short Division Worksheet
PDF
We all need to test ourselves and check whether we understood the topic clearly or not. So do the children that are studying in primary classes like 2nd,3rd and 4th. Once the basic topics are clearly understood in these classes there is no tension for the future. They will grasp things very easily afterwards. To make sure your kid has learned the topic without any confusions, you can use these short division worksheets that are very much like a short test, covering almost all the necessary topics.
### Division Table Charts Worksheet
PDF
When we talk about division, division tables are also necessary to learn for the clarification of the whole topic. Division tables are the tabular form of division equations which clearly shows what will be the outcome when two numbers are divided. Learning these tables will help them solve division equations quickly and division concepts are easily understood by them. Students who are interested in clarifying the concept of division and division tables in depth can take a short kind of a test with these worksheets to know what they so far have learned and on what topics they need to focus more and which tables are hard for them to learn.
### Division Table Worksheet for Kids
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There are numerous division tables, but primary level students learn only the starting easy ones only because first of all they don’t learn the big ones in the early ages and second of all that would create a lot of burden for them. But on the other side high schools need to learn the tougher and big division tables for their higher studies and competitive exams. Therefore there are division tables worksheets for small kids, that are easy comparatively and more fun for the kids take more interest in these worksheets.
### Division Table Worksheet Free
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Worksheets of any subject are important to be solved. Till the time you don’t test yourself or put your theoretical knowledge into practical work it won’t be considered as a topic clearly understood and learned. Mathematics is a subject of practical knowledge only. The more you practice, the perfect you’ll get in it. So keep on practicing with division tables and division worksheets that are available free and ace the subject and concept for the future exams and competitive entrances. Small kids can also find these interesting and playful if you try to indulge them in solving by using the correct concept and formulas. They will even get excited after solving them correctly. | 592 | 3,158 | {"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-2022-49 | latest | en | 0.95869 |
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# It is as difficult to prevent crimes against property as
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14 Jul 2005, 20:11
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441.It is as difficult to prevent crimes against property as those that are against a person.(A) those that are against a
(B) those against a
(C) it is against a
(D) preventing those against a
(E) it is to prevent those against a
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15 Jul 2005, 10:09
Picked E.
It is as difficult to X ....as it is to Y.
HMTG.
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15 Jul 2005, 15:51
C is my choice..
It is as difficult to prevent crimes against property as it is against a person.
(A) those that are against a
- OUT verbose - those and that
(B) those against a
- OUT need a subject and a verb after as
(C) it is against a
- Good keep this
(D) preventing those against a
- OUT participle use. no parallelism
(E) it is to prevent those against a
- wordy, can be shorter
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15 Jul 2005, 18:48
folks....b, d, e and A are all out...
C it is...
you have to make this statment parallel...
it is ....it is...
so we are down to C and E...
E I dont what those is referring to? plus the construction is overly wordy...C is short and sweet!
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16 Jul 2005, 16:03
I vote for C) it is against a
simple and concise. it is.. it is parallel .. E is verbose and no sure what does the 'those' refer to?
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16 Jul 2005, 18:20
E is correct, but why can't we assume B as
(it is to prevent) those against a ...
with the part in paraenthesis implied ?
Not quite clear about the rule of omission ...
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16 Jul 2005, 18:50
folks whats wrong with C?....
I can see "those" in E to be problamatic....no one has given a good reason why E should be correct? on the other hand, ritesh and my explanation on C sounds pretty OK to me...
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16 Jul 2005, 19:42
those in E refer to crimes.
E is correct but wordy from my point of view... Ofcourse in GMAT, they want a precise and best answer among the choices.
I think it is time for perezhan to post OA and OE
I am more interested in OE.
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18 Jul 2005, 23:14
OA is E
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18 Jul 2005, 23:14
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# It is as difficult to prevent crimes against property as
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,332 | 4,316 | {"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} | 3.46875 | 3 | CC-MAIN-2017-47 | latest | en | 0.952164 |
https://stat.ethz.ch/pipermail/r-sig-finance/2012q1/009623.html | 1,669,564,573,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00498.warc.gz | 574,374,284 | 2,835 | # [R-SIG-Finance] correlation based time series clustering?
Murali.Menon at avivainvestors.com Murali.Menon at avivainvestors.com
Thu Feb 23 10:47:52 CET 2012
Folks,
There's been quite a bit of work on clustering in finance:
http://www.mendeley.com/research/correlation-based-hierarchical-clustering-in-financial-time-series/
Very simply, though, you can calculate a correlation matrix given your time series of returns, and then apply, say, a medoid-clustering scheme (e.g. the pam function in package 'cluster') and see what happens. According to Mantegna, if you do this on the components of the Dow Jones, the various industrial types very naturally form individual clusters.
Hope this helps.
Murali
-----Original Message-----
From: r-sig-finance-bounces at r-project.org [mailto:r-sig-finance-bounces at r-project.org] On Behalf Of julien cuisinier
Sent: 23 February 2012 09:29
To: comtech.usa at gmail.com; r-sig-finance at stat.math.ethz.ch
Subject: Re: [R-SIG-Finance] correlation based time series clustering?
Hi Michael,
A very general question here with little input from you...I am not surprised to see little feedback
I have been looking for something similar & same result so I do not think it exist yet. I am a complete newbie in clustering but looking around there are plenty of R function available, nothing that I could find as simple as using correlation per se.
Thinking about it Im not sure how it would work & anything I can think of would be quite sensitive to the starting point (e.g. calculate pair-wise correls within a market, then start by one stock & cluster with it all other stocks with corrells higher than a certain threshold?) May be some recursive function trying many different starting points? But then what to do with the resulting different cluster structure?
Could you share with the list what reference (not in R) you found on the topic? That would be great if you could share / bring something to the list as well & then see if we can build that in? (very very ambitious of me here =)
Thanks & regards,
Julien
> Date: Tue, 21 Feb 2012 15:05:59 -0600
> From: comtech.usa at gmail.com
> To: r-sig-finance at stat.math.ethz.ch
> Subject: [R-SIG-Finance] correlation based time series clustering?
>
> Hi all,
>
> I am looking for a function for correlation based time-series clustering in
> R... I have googled for quite a while and couldn't find any in R...
>
>
> Thanks a lot!
>
> [[alternative HTML version deleted]]
>
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-- Also note that this is not the r-help list where general R questions should go. | 750 | 3,110 | {"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-2022-49 | latest | en | 0.91255 |
http://oeis.org/A339040 | 1,618,213,379,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00524.warc.gz | 76,401,048 | 3,707 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A339040 Number of unlabeled connected simple graphs with n edges rooted at two noninterchangeable vertices. 6
1, 3, 10, 35, 125, 460, 1747, 6830, 27502, 113987, 485971, 2129956, 9591009, 44341610, 210345962, 1023182861, 5100235807, 26035673051, 136023990102, 726877123975, 3970461069738, 22156281667277, 126234185382902, 733899631974167, 4351500789211840 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS FORMULA G.f.: f(x)/g(x) - r(x)^2 where f(x), g(x) and r(x) are the g.f.'s of A339063, A000664 and A339039. PROG (PARI) \\ See A339063 for G. seq(n)={my(A=O(x*x^n), g=G(2*n, x+A, [])); Vec(G(2*n, x+A, [1, 1])/g - (G(2*n, x+A, [1])/g)^2)} CROSSREFS Cf. A000664, A002905, A303832, A304074, A339039, A339041, A339042, A339044, A339063. Sequence in context: A224509 A026026 A047037 * A201058 A318113 A216710 Adjacent sequences: A339037 A339038 A339039 * A339041 A339042 A339043 KEYWORD nonn AUTHOR Andrew Howroyd, Nov 20 2020 STATUS approved
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Last modified April 12 03:14 EDT 2021. Contains 342912 sequences. (Running on oeis4.) | 522 | 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} | 2.96875 | 3 | CC-MAIN-2021-17 | latest | en | 0.58826 |
https://www.mrsmactivity.co.uk/downloads/year-2-counting-forwards-and-backwards-within-50-lesson-presentation/ | 1,656,419,043,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00543.warc.gz | 965,957,871 | 52,281 | # Year 2 | Counting Forwards and Backwards Within 50 Lesson Presentation
## Year 2 Place Value Resources
This lesson focuses on counting forwards and backwards between 0 and 50. It recaps Year 1 objectives at the start of Year 2.
Aligned with the maths mastery approach, this Year 2 | Counting Forwards and Backwards Within 50 Lesson Presentation is fully editable, and is designed for the Year 2 maths curriculum covering the following maths objectives for the autumn term:
Topic/Block: Place Value
Small Steps: Counting forwards and backwards within 50.
NC Links: Read and write numbers to at least 100 in numerals and in words. • Identify, represent and estimate numbers using different representations including the number line.
Ready-to-progress criteria: Year 1 Conceptual Prerequisites: Know the 10 ones are equivalent to 1 ten. Know that multiples of 10 are made up from a number of tens.
2-NPV-1 Recognise the place value of each digit in two-digit numbers.
Year 1 Conceptual Prerequisites: Count forwards and backwards to and from 100.
2NPV-2 Reason about the location of any two digit number in the linear number system.
TAF Statements: Working towards – Read and write numbers in numerals up to 100.
Working at – Read scales in divisions of ones.
There are teacher prompts and notes included in the “note” section of each slide.
Aligned with the order of teaching of the White Rose Maths scheme of work, use this to help your class get to grips with each mathematical concept. This lesson presentation also includes varied fluency activities, problem solving, and mathematical discussion questions too.
Explore our other year 2 place value resources. | 359 | 1,677 | {"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} | 3.671875 | 4 | CC-MAIN-2022-27 | longest | en | 0.896601 |
https://mathematica.stackexchange.com/questions/66168/how-to-plot-curves-in-ternary-plot-triangular-plot | 1,571,072,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00268.warc.gz | 617,731,087 | 28,846 | # How to plot curves in ternary plot (triangular plot) [duplicate]
I need to plot the following curve in a triangular (ternary) plot: $$|x_{BB}-x_{AA}|=(\sigma_c-2x_{AB}+1)^{\gamma}, \qquad (1)$$ subjected to the condition (this condition must satisfy for a ternary plot) $$x_{AA}+x_{AB}+x_{BB}=1, \qquad (2)$$
where $\sigma_c$ is a positive constant, say 1, and $\gamma=0.35$.
My attempt is to express $x_{AB}$ in terms of the other two variables from the equation (2) and substitute the resulting expression in (1), then I obtain the following:
Abs[b - a] == (2 - 2 (1 - a - b))^0.35
Questions:
a) Is this expression equivalent to the equations (1) and (2)?
b) How do I plot the curve (1) subjected to (2) in a ternary form?
## marked as duplicate by bobthechemist, Kuba♦, Öskå, Karsten 7., gpapNov 21 '14 at 13:25
• – Szabolcs Nov 20 '14 at 18:52
• Indeed, I believe István Zachar's answer in particular would work perfectly for this question. – Rahul Nov 20 '14 at 18:57
• Your example seems to have no solutions, by the way. ( I think you need sc<1 .. ) – george2079 Nov 20 '14 at 19:32
• Try to leverage the power of ContourPlot3D. – Silvia Nov 22 '14 at 8:58
make use of ContourPlot to find solutions..
dat = Table[ {#[[1]], #[[2]], 1 - #[[1]] - #[[2]]} & /@
Select[
Cases[ ContourPlot[
Abs[ b - a] == ( sig - 2 (1 - b - a ) + 1 )^.35 ,
{a, 0, 1}, {b, 0, 1}] ,
List[a_Real, b_Real] :> { a, b } ,
Infinity] , #[[1]] + #[[2]] <= 1 & ] ,
{sig, -1, 1, .25}];
then transform to a ternary figure:
Graphics[{Line[{{0, 0}, {1, 0}, {1/2, Sqrt[3]/2}, {0, 0}}],
Point[{#[[2]] + #[[3]]/2 , Sqrt[3]/2 #[[3]] } & /@ #] & /@ dat}]
Edit -- same thing preserving the lines form ContourPlot
tercp[cp_Graphics] :=
Quiet@Cases[ Normal@First@Cases[cp, _GraphicsComplex, Infinity] ,
Line[x_] :> Line[{
1 - #[[1]] + #[[2]],
Sqrt[3] (1 - #[[1]] - #[[2]])}/2 & /@
Select[x, Total[#] <= 1 &] ], Infinity]
Graphics[{Line[{{0, 0}, {1, 0}, {1/2, Sqrt[3]/2}, {0, 0}}],
{Dashed,
Table[ tercp [ ContourPlot[(1 - b - a) == ci , {a, 0, 1}, {b, 0, 1}] ],{ci, .1, .9, .1}],
Table[tercp [ ContourPlot[a == ci , {a, 0, 1}, {b, 0, 1}] ], {ci, .1, .9, .1}],
Table[ tercp [ ContourPlot[b == ci , {a, 0, 1}, {b, 0, 1}] ] , {ci, .1, .9, .1}]},
Table[ {Hue[RandomReal[]],
tercp [ ContourPlot[Abs[b - a] == (sig - 2 (1 - b - a) + 1)^.35,
{a, 0, 1}, {b, 0, 1}] ] }, {sig, -1, 1, .2}]}] | 951 | 2,371 | {"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} | 3.46875 | 3 | CC-MAIN-2019-43 | latest | en | 0.718738 |
https://math.stackexchange.com/questions/644327/how-unique-are-u-and-v-in-the-singular-value-decomposition-a-udv-dagger/644397 | 1,721,668,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00299.warc.gz | 327,037,122 | 47,522 | # How unique are $U$ and $V$ in the singular value decomposition $A=UDV^\dagger$?
According to Wikipedia:
A common convention is to list the singular values in descending order. In this case, the diagonal matrix $$\Sigma$$ is uniquely determined by $$M$$ (though the matrices $$U$$ and $$V$$ are not).
My question is, are $$U$$ and $$V$$ uniquely determined up to some equivalence relation (and what equivalence relation)?
• Do you understand why $U$ and $V$ are not uniquely determined? Think about what happens if $\Sigma$ is not regular, which means that some eigenvalues are zero. Or what happens if you have multiple eigenvalues.
– Marc
Commented Jan 19, 2014 at 23:29
• @Marc *singular values. ;-) Commented Jan 20, 2014 at 0:25
• Oops, of course. Thanks for the correction.
– Marc
Commented Jan 20, 2014 at 0:35
• The two posts address the phase ambiguities explicitly: math.stackexchange.com/questions/2287795/…, math.stackexchange.com/questions/1805191/… Commented Jun 10, 2017 at 23:07
Let $A = U_1 \Sigma V_1^* = U_2 \Sigma V_2^*$. Let us assume that $\Sigma$ has distinct diagonal elements and that $A$ is tall. Then
$$A^* A = V_1 \Sigma^* \Sigma V_1^* = V_2 \Sigma^* \Sigma V_2^*.$$
From this, we get
$$\Sigma^* \Sigma V_1^* V_2 = V_1^* V_2 \Sigma^* \Sigma.$$
Notice that $\Sigma^* \Sigma$ is diagonal with all different diagonal elements (that's why we needed $A$ to be tall) and $V_1^* V_2$ is unitary. Defining $V := V_1^* V_2$ and $D := \Sigma^* \Sigma$, we have
$$D V = V D.$$
Now, since $V$ and $D$ commute, they have the same eigenvectors. But, $D$ is a diagonal matrix with distinct diagonal elements (i.e., distinct eigenvalues), so it's eigenvectors are the elements of the canon basis. That means that $V$ is diagonal too, which means that
$$V = \operatorname{diag}(e^{{\rm i}\varphi_1}, e^{{\rm i}\varphi_2}, \dots, e^{{\rm i}\varphi_n}),$$
for some $\varphi_i$, $i=1,\dots,n$.
In other words, $V_2 = V_1 V$. Plug that back in the formula for $A$ and you get
$$A = U_1 \Sigma V_1^* = U_2 \Sigma V_2^* = U_2 \Sigma V^* V_1^* = U_2 V^* \Sigma V_1^*.$$
So, $U_2 = U_1 V$ if $\Sigma$ (and, in extension, $A$) is square nonsingular. Other options, somewhat similar to this, are possible if $\Sigma$ has zeroes on the diagonal and/or is rectangular.
If $\Sigma$ has repeating diagonal elements, much more can be done to change $U$ and $V$ (for example, one or both can permute corresponding columns).
If $A$ is not thin, but wide, you can do the same thing by starting with $AA^*$.
So, to answer your question: for a square, nonsingular $A$, there is a nice relation between different pairs of $U$ and $V$ (multiplication by a unitary diagonal matrix, applied in the same way to the both of them). Otherwise, you get quite a bit more freedom, which I believe is hard to formalize.
• I believe that if there are repeated singular values, you just simply use "small" unitary matrices instead of the factors $e^{\mathrm{i}\varphi}$, that is, $V$ is block diagonal instead (of course, assuming that the singular values are properly ordered). Commented Jan 20, 2014 at 10:38
• @AlgebraicPavel Yes, you are right. Such small matrices then correspond to linear combinations that keep the unitarity. These can be random, and completely different between $U$ and $V$. Commented Jan 20, 2014 at 12:28
• very good answer, thanks! So I meant to think this through on my own, but I haven't gotten around to it, so I will ask: Is it just that any such pairs are related in this way, OR can you multiply any U and V by ANY unitary diagonal matrix and get another valid U1, V1 that would SVD the original matrix A? Commented Jan 25, 2014 at 19:56
• If you have an SVD of $A$, you can multiply its $U$ and $V$ by a diagonal unitary matrix (the same one for each!). However, sometimes you can do even more, as discussed in my answer and Pavel's comment. Commented Jan 25, 2014 at 22:00
I'm going to provide a full characterisation of the set of SVDs for a given matrix $$A$$, using two different (but of course ultimately equivalent) kinds of formalisms. First a standard matrix formalism, and then using dyadic notation.
The TL;DR is that if $$A=UDV^\dagger=\tilde U \tilde D\tilde V^\dagger$$ for $$U,\tilde U,V,\tilde V$$ isometries and $$D,\tilde D>0$$ squared strictly positive diagonal matrices, then we can safely assume $$D=\tilde D$$ by trivially rearranging the basis for the underlying space, and furthermore that $$\tilde V=V W$$ with $$W$$ a unitary block-diagonal matrix that leaves invariant certain subspaces directly determined by $$A$$ (further details below). This characterises all and only the possible SVDs. So given any SVD, the freedom in the choice of other SVDs corresponds to the freedom in choosing these unitaries $$W$$ (how much freedom is that, depends in turn on the degeneracy of the singular values of $$A$$).
## Regular notation
Consider the SVD of a given matrix $$A$$, in the form $$A=UDV^\dagger$$ with $$D>0$$ a diagonal squared matrix with strictly positive entries, and $$U,V$$ isometries (this writing is general: the SVD is often written in a way that makes $$U,V$$ unitaries and $$D$$ not necessarily squared, but you can have instead $$D>0$$ squared if you allow $$U,V$$ to be isometries).
The question is, if you have $$A = UDV^\dagger = \tilde U \tilde D \tilde V^\dagger,$$ with $$U,\tilde U,V,\tilde V$$ isometries, and $$D,\tilde D>0$$ diagonal squared matrices, what does this imply for $$\tilde U,\tilde D,\tilde V$$? And more specifically, can we somehow find an explicit relation between them?
1. The first easy observation is that you must have $$D=\tilde D$$, modulo trivial rearrangements of the basis elements. This follows from $$AA^\dagger=UD^2 U^\dagger = \tilde U \tilde D^2 \tilde U^\dagger$$, which means that $$D^2$$ and $$\tilde D^2$$ both contain the eigenvalues of $$AA^\dagger$$. Because the spectrum is determined algebraically from $$AA^\dagger$$, the set of eigenvalues must be identical, and the singular values are by definition positive reals, we must have $$D=\tilde D$$.
2. The above reduces the question to: if $$UDV^\dagger=\tilde U D\tilde V^\dagger$$, with $$U,\tilde U,V,\tilde V$$ isometries, what can we say about $$\tilde U,\tilde V$$? To this end, we observe that the freedom in the choice of $$U,V$$ amounts to the different possible ways to decompose the subspaces associated to each distinct singular value.
More precisely, consider the subspace $$V^{(d)}\equiv \{v: \, \|Av\|=d\}$$ corresponding to a singular value $$d$$. We can then uniquely decompose the matrix as $$A=\sum_d A_d$$ where $$A_d\equiv A \Pi_d$$ and $$\Pi_d$$ is the projection onto $$V^{(d)}$$, and the sum is over all nonzero singular values $$d$$ of $$A$$. We can now observe that any and all SVDs of $$A$$ correspond to a choice of orthonormal basis for each $$V^{(d)}$$. Namely, for any such basis $$\{\mathbf v_k\}$$ we associate the partial isometry $$V_d\equiv \sum_k \mathbf v_k \mathbf e_k^\dagger$$. The corresponding basis for the image of $$A_d$$ is then determined as $$\mathbf u_k= A \mathbf v_k$$, and we then define the partial isometry $$U_d\equiv \sum_k \mathbf u_k \mathbf e_k^\dagger$$. Here, $$\mathbf e_k$$ denotes an orthonormal basis spanning the elements of $$D$$ corresponding to the singular value $$d$$. This procedure provides a decomposition $$A_d= U_d D V_d^\dagger$$, and therefore an SVD for $$A$$ itself by summing these. Any SVD can be constructed this way.
In conclusion, the freedom in choosing an SVD is entirely in the choice of bases $$\{\mathbf v_k\}$$ above. We can summarise this freedom concisely by saying that given any SVD $$A=UDV^\dagger$$, any other SVD can be written as $$A=UW D (VW)^\dagger$$ for some unitary $$W$$ such that $$[W,D]=0$$. This commutation property is a concise way to state that $$W$$ is only allowed to mix vectors corresponding to the same singular value, that is, to the same eigenspace of $$D$$.
#### Toy example #1
Let's work out a simple toy example to illustrate the above results.
Let $$H \equiv \begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$ This is a somewhat trivial example because $$H$$ is Hermitian, but still illustrates some aspects. A standard SVD reads $$H = \underbrace{\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix} }_{\equiv U} \underbrace{\begin{pmatrix}\sqrt2 & 0\\0&\sqrt2\end{pmatrix}}_{\equiv D} \underbrace{\begin{pmatrix}0&1\\1&0\end{pmatrix}}_{\equiv V^\dagger}.$$ In this case, we have two identical singular values. According to our discussion above, this means that we can apply a(ny) unitary transformation to the columns of $$V$$ and still obtain another SVD. That is, given any unitary $$W$$, $$\tilde V\equiv V W$$ gives another SVD for $$H$$. In this simple case, you can also observe this directly, as $$D=\sqrt2 I$$, and therefore $$H = UDV^\dagger= UD W \tilde V^\dagger = (UW) D V^\dagger,$$ hence $$\tilde U\equiv UW$$, $$\tilde V\equiv UV$$ give the alternative SVD $$H=\tilde U D\tilde V^\dagger$$, and all SVDs have this form.
#### Toy example #2
Consider a simple non-squared case. Let $$A \equiv \begin{pmatrix}1&1\\1 & \omega \\ 1 & \omega^2\end{pmatrix}, \qquad \omega\equiv e^{2\pi i/3}.$$ This is again almost trivial because $$A$$ is an isometry, up to a constant. Still, we can write its SVD as $$A = \underbrace{\frac{1}{\sqrt3}\begin{pmatrix}1&1\\1&\omega\\1&\omega^2\end{pmatrix}}_{\equiv U} \underbrace{\begin{pmatrix}\sqrt3 & 0 \\0&\sqrt3\end{pmatrix}}_{\equiv D} \underbrace{\begin{pmatrix}0&1\\1&0\end{pmatrix}}_{\equiv V^\dagger}.$$ Notice that now $$U,V$$ are isometries, but not unitaries, and that $$D>0$$ is squared. Are per our results above, any SVD will have the form $$A=\tilde U D \tilde V^\dagger$$ with $$\tilde V=V W$$ for some unitary $$W$$.
for example, taking $$W=V$$ (we can do this, because here $$V$$ is also unitary), we get the alternative SVD $$A=\tilde U D \tilde V^\dagger$$ with $$\tilde V=VW=VV=I$$ and $$\tilde U= U W^\dagger=UV=\frac1{\sqrt3}\begin{pmatrix}1&\omega&\omega^2\\1&1&1\end{pmatrix}^T$$, that is, $$A = \underbrace{\frac{1}{\sqrt3}\begin{pmatrix}1&1\\\omega&1\\\omega^2&1\end{pmatrix}}_{\equiv \tilde U} \underbrace{\begin{pmatrix}\sqrt3 & 0 \\0&\sqrt3\end{pmatrix}}_{\equiv D} \underbrace{\begin{pmatrix}1&0\\0&1\end{pmatrix}}_{\equiv \tilde V^\dagger}.$$
#### Toy example #3
Let's do an example with non-degenerate singular values. Let $$A = \begin{pmatrix}1& 2 \\ 1 & 2\omega \\ 1 & 2\omega^2\end{pmatrix}, \qquad \omega\equiv e^{2\pi i/3}.$$ This time the singular values are $$\sqrt3$$ and $$2\sqrt3$$. One SVD is easily derived as $$A = \underbrace{\frac{1}{\sqrt3}\begin{pmatrix}1&1\\1&\omega\\1&\omega^2\end{pmatrix}}_{\equiv U} \underbrace{\begin{pmatrix}\sqrt3 & 0 \\0&2\sqrt3\end{pmatrix}}_{\equiv D} \underbrace{\begin{pmatrix}1&0\\0&1\end{pmatrix}}_{\equiv V^\dagger}.$$ However, in this case there is much less freedom in choosing other SVDs, because these must correspond to $$\tilde V=VW$$ where $$W$$ only mixes columns of $$V$$ corresponding to the same values of $$D$$. In this case $$D$$ is non-degenerate, thus $$W$$ must be diagonal, and therefore the full set of SVDs must correspond to $$W=\begin{pmatrix}e^{i\alpha}&0\\0&e^{i\beta}\end{pmatrix}$$, that is, $$A = \underbrace{\frac{1}{\sqrt3}\begin{pmatrix}e^{-i\alpha}&e^{-i\beta}\\e^{-i\alpha}&\omega e^{-i\beta}\\e^{-i\alpha}&\omega^2 e^{-i\beta}\end{pmatrix}}_{\equiv \tilde U} \underbrace{\begin{pmatrix}\sqrt3 & 0 \\0&2\sqrt3\end{pmatrix}}_{\equiv D} \underbrace{\begin{pmatrix}e^{i\alpha}&0\\0&e^{i\beta}\end{pmatrix}}_{\equiv \tilde V^\dagger}.$$ All SVDs will look like this, for some $$\alpha,\beta\in\mathbb{R}$$. Although note that by permuting the elements of $$D$$, we obtain SVDs which look different, although are ultimately equivalent to the above.
#### SVD in dyadic notation removes "trivial" redundancies
The SVD of an arbitrary matrix $$A$$ can be written in dyadic notation as $$A=\sum_k s_k u_k v_k^*,\tag A$$ where $$s_k\ge0$$ are the singular values, and $$\{u_k\}_k$$ and $$\{v_k\}_k$$ are orthonormal sets of vectors spanning $$\mathrm{im}(A)$$ and $$\ker(A)^\perp$$, respectively. The connection between this and the more standard way of writing the SVD of $$A$$ as $$A=UDV^\dagger$$ is that $$u_k$$ is the $$k$$-th column of $$U$$, and $$v_k$$ is the $$k$$-th column of $$V$$.
#### Global phase redundancies are always present
If $$A$$ is nondegenerate, the only freedom in the choice of vectors $$u_k,v_k$$ is their global phase: replacing $$u_k\mapsto e^{i\phi}u_k$$ and $$v_k\mapsto e^{i\phi}v_k$$ does not affect $$A$$.
#### Degeneracy gives more freedom
On the other hand, when there are repeated singular values, there is additional freedom in the choice of $$u_k,v_k$$, similarly to how there is more freedom in the choice of eigenvectors corresponding to degenerate eigenvalues. More precisely, note that (A) implies $$AA^\dagger=\sum_k s_k^2 \underbrace{u_k u_k^*}_{\equiv\mathbb P_{u_k}}, \qquad A^\dagger A=\sum_k s_k^2 \mathbb P_{v_k}.$$ This tells us that, whenever there are degenerate singular values, the corresponding set of principal components is defined up to a unitary rotation in the corresponding degenerate eigenspace. In other words, the set of vectors $$\{u_k\}$$ in (A) can be chosen as any orthonormal basis of the eigenspace $$\ker(AA^\dagger-s_k^2)$$, and similarly $$\{v_k\}_k$$ can be any basis of $$\ker(A^\dagger A-s_k^2)$$.
However, note that a choice of $$\{v_k\}_k$$ determines $$\{u_k\}$$, and vice-versa (otherwise $$A$$ wouldn't be well-defined, or injective outside its kernel).
#### Summary
A choice of $$U$$ uniquely determines $$V$$, so we can restrict ourselves to reason about the freedom in the choice of $$U$$. There are twe main sources of redundancy:
1. The vectors can be always scaled by a phase factor: $$u_k\mapsto e^{i\phi_k}u_k$$ and $$v_k\mapsto e^{i\phi_k}v_k$$. In matrix notation, this corresponds to changing $$U\mapsto U \Lambda$$ and $$V\mapsto V\Lambda$$ for an arbitrary diagonal unitary matrix $$\Lambda$$.
2. When there are "degenerate singular values" $$s_k$$ (that is, singular values corresponding to degenerate eigenvalues of $$A^\dagger A$$), there is additional freedom in the choice of $$U$$, which can be chosen as any matrix whose columns form a basis for the eigenspace $$\ker(AA^\dagger-s_k^2)$$.
Finally, we should note that the former point is included in the latter, which therefore encodes all of the freedom allowed in choosing the vectors $$\{v_k\}$$. This is because multiplying the elements of an orthonormal basis by phases does not affect its being an orthonormal basis.
I will complete the proof of @Vedran for the case when there exist repeating eigenvalues, which would justify what @glS have said. Let $$A = U \Sigma V^T = U^{'} \Sigma {V^{'}}^T$$ be a matrix with real entries - the case with complex entries is similar. Then, $$A^T A = V \Sigma^T \Sigma V^{T} = V^{'} \Sigma^T \Sigma {V^{'}}^T.$$
From this, we get $$\Sigma^T \Sigma V^T V^{'} = V^T V^{'} \Sigma^T \Sigma.$$ Defining the square matrix $$Q$$ as $$Q = V^T V^{'}$$, we have $$Q^T Q = (V^T V^{'})^T V^T V^{'} = I$$, and similarly, $$Q Q^T = I.$$ Hence, $$Q$$ is an orthogonal matrix that satisfies the Sylvester equation $$Q\Sigma^T\Sigma - \Sigma^T \Sigma Q = 0.\tag{1}$$
Aiming to simplify the Sylvester equation (1) a little, counting the multiplicities, we can write $$\Sigma^T \Sigma$$ $$= \sigma_1^2 I_{n_1} \oplus \sigma_2^2 I_{n_2} \oplus \cdots$$ $$\oplus \sigma_{k}^2 I_{n_k}$$ $$= \text{diag}(\sigma_1^2 I_{n_1}, \sigma_2^2 I_{n_2},$$ $$\cdots, \sigma_{k}^2 I_{n_k}),$$ with $$\sigma_i=\sigma_j$$ iff $$i=j$$ and $$n_i$$ the multiplicity of $$\sigma_{i}$$.
Now, writing $$Q$$ in blocks conformally to $$\Sigma^T \Sigma$$, i.e., with $$Q \Sigma^T \Sigma$$ and $$\Sigma^T \Sigma Q$$ making sense, we have a new system of Sylvester equations $$\sigma_i^2 Q_{ij} I_{n_i} - \sigma_j^2 I_{n_j} Q_{ij}=0,$$ for each $$1\leq i,j\leq k$$. This means that, since both matrices $$\sigma_i^2 I_{n_i}$$ and $$\sigma_j^{2}I_{n_j}$$ do not share any of its eigenvalues for $$i\not=j,$$ by the result discussed here, if $$i\not= j$$, we have necessarily that $$Q_{ij} = 0$$. This means that $$V^T V^{'} = \text{diag} (Q_{11}, Q_{22},\cdots, Q_{kk})$$ for orthogonal matrices $$Q_{ii}$$, meaning that $$V' = V \text{diag}(Q_{11}, Q_{22},\cdots, Q_{kk})=V Q.\tag{2}$$
We can repeat the argument above for the matrix $$A A^T$$ and conclude that there exist an orthogonal matrix $$\bar{Q}$$ such that $$\bar{Q} = \text{diag}(\bar{Q}_{11}, \bar{Q}_{22},\cdots, \bar{Q}_{kk})\tag{3}$$ and $$U' = U \bar{Q},$$ with the matrices $$\bar{Q}_{ii}$$ of the same size of $$Q_{ii}$$ whenever $$\sigma_i\not=0$$ - this is possible because the matrices $$A^T A$$ and $$A A^T$$ have the same eigenvalues with the same multiplicity, except possibly the null eigenvalue. Taking into account that $$A = U \Sigma V^T = U^{'} \Sigma {V^{'}}^T$$, we would be able to conclude $$\Sigma = U^T U^{'} \Sigma {V^{'}}^T V = U^T U \bar{Q} \Sigma Q^T V^T V,$$ which simplifies to $$\Sigma = \bar{Q} \Sigma Q^T.$$ Lastly, considering the nonzero blocks of $$\Sigma$$ associated with the matrices $$Q_{ii}$$ and $$\bar{Q}_{ii}$$, we are able to conclude that $$\bar{Q}_{ii} = Q_{ii}$$ in the expression (3), whenever $$\sigma_i\not=0$$.
$$\newcommand\R{\mathbb{R}}$$I see that this question has been asked a few times. I'd like to provide a simple answer to this.
The singular values $$s_1, \dots, s_k$$ of an $$n$$-by-$$m$$ matrix $$M$$ are the square roots of the positive eigenvalues of $$M^*M$$. Let $$d_1, \dots, d_k$$ be the respective dimensions of the eigenspaces, and $$d_{k+1}$$ be the dimension of the kernel. Observe that $$r = d_1+\cdots + d_k$$ is the rank of $$M$$.
If $$M = U\Sigma V^*$$ is a singular decomposition, then $$M^*M = V\Sigma^2V^*$$, and therefore the columns of $$V$$ form a unitary basis of eigenvectors $$(v_1, \dots, v_m)$$. $$V$$ is clearly unique up to unitary transformations of each eigenspace.
Now, for each $$1 \le j \le k$$, let $$w_j = Mv_j$$ and observe that for any $$1 \le i,j \le r$$, $$\langle w_i,w_j\rangle = \langle Mv_i,Mv_j\rangle = \langle v_i,M^*Mv_j\rangle = \lambda_j\langle v_i,v_j\rangle = \lambda_j\delta_{ij},$$ where the $$\lambda_j$$ are the singular values listed with the appropriate multiplicities. Therefore, $$u_j = \frac{w_j}{\lambda_j},\ 1 \le j \le r$$ are a unitary basis of the image of $$M$$. They can be extended to a unitary basis $$(u_1, \dots, u_n)$$ of $$\R^n$$. It is now easy to show that the matrix $$U$$ whose columns are $$u_1, \dots, u_n$$ satisfies the equation $$M = U\Sigma V^*.$$ It is also easy to show that conversely, if $$M = U\Sigma V^*,$$ then, for each $$1 \le j \le r$$, then $$\lambda_j u_j = Mv_j,$$ and $$u_{r+1}, \dots, u_n$$ are a unitary basis of the orthogonal complement of the image of $$M$$. It follows that given any $$V$$ satisfying the conditions above, the first $$r$$ columns of $$U$$ are uniquely determined and the last $$n-r$$ columns can be any unitary basis of $$(\operatorname{image} M)^\perp$$.
From all this, we see that $$U$$ and $$V$$ are unique up to unitary transformations of the eigenspaces of $$M^*M$$ and unitary transformations of $$(\operatorname{image} M)^\perp$$. | 6,115 | 19,301 | {"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": 238, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.834083 |
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SICP 2.19: Counting Change Revisited
From SICP section 2.2.1 Representing Sequences
Exercise 2.19 asks us to modify the change counting procedure from section 1.2.2 (which we looked at in depth in exercise 1.14). We need to modify the program so that it takes a list of coins to be used instead of having the denominations hard coded as they were originally. The following lists of coins are provided:
(define us-coins (list 50 25 10 5 1))(define uk-coins (list 100 50 20 10 5 2 1 0.5))
We're also provided with the following modified version of the cc procedure:
(define (cc amount coin-values) (cond ((= amount 0) 1) ((or (< amount 0) (no-more? coin-values)) 0) (else (+ (cc amount (except-first-denomination coin-values)) (cc (- amount (first-denomination coin-values)) coin-values)))))
All that's left for us to do is define the procedures first-denomination, except-first-denomination, and no-more? in terms of primitive list operations. These procedures are very straightforward.
; return true if passed an empty list(define (no-more? coin-values) (if (null? coin-values) true false)); return all coin values except the first(define (except-first-denomination coin-values) (cdr coin-values)); return only the first coin value(define (first-denomination coin-values) (car coin-values))> (cc 100 us-coins)292> (cc 100 uk-coins)104561
Finally, we're asked if the order of the list coin-values affects the answer produced by cc. We can find out quickly enough by experiment.
; reversed list of us coins(define su-coins (list 1 5 10 25 50))> (cc 100 us-coins)292> (cc 100 su-coins)292
We can see that the order doesn't matter. This is because the procedure recursively evaluates every sub-list after subtracting the value of the first coin from the target amount. It doesn't matter if that value is the smallest, largest, or even if the values are shuffled.
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
Monday, December 27, 2010
SICP 2.18: Reversing a List
From SICP section 2.2.1 Representing Sequences
Exercise 2.18 asks us to define a procedure reverse that takes a list as its argument and returns a list with the same elements in reverse order.
We'll be using the append procedure given in the text which takes two lists and appends the second list to the end of the first:
(define (append list1 list2) (if (null? list1) list2 (cons (car list1) (append (cdr list1) list2))))
We can reverse a list by appending the car of the list to the reverse of the cdr of the list. (See how easy it is to start thinking in Scheme?) In other words, we just move the first item to the end of the list after reversing the remaining items. This means reverse will make a recursive call to itself, so we also need a base case to make the recursion stop. We can just stop when we reach an empty list.
(define (reverse items) (if (null? items) items (append (reverse (cdr items)) (list (car items)))))
Once again, we can test it out with a few example lists from the text.
> (reverse (list 1 2 3 4))(4 3 2 1)> (reverse (list 1 4 9 16 25))(25 16 9 4 1)> (reverse (list 1 3 5 7))(7 5 3 1)> (reverse (list 23 72 149 34))(34 149 72 23)
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
SICP 2.17: Last Item in a List
From SICP section 2.2.1 Representing Sequences
Exercise 2.17 asks us to define a procedure last-pair that returns a list that contains only the last element of a given (non-empty) list.
Earlier in SICP section 2.2.1 we saw that a list is represented in Scheme as a chain of pairs.
Our new procedure needs to return a list, so we need to be careful not to simply return the last value in the list, or even the last pair.
There are several correct ways to do this, but here's a recursive solution (based on the length example from the text) that uses the method of "cdring down" the list.
(define (last-pair items) (if (null? (cdr items)) (list (car items)) (last-pair (cdr items))))
The first line checks to see if the cdr of the list of items passed in is nil. If it is, the second line creates a new list from the car of the items. If not, the last line makes a recursive call to last-pair with the remaining items in the list.
We can verify that it works by testing it with several of the example lists from the chapter.
> (last-pair (list 1 2 3 4))(4)> (last-pair (list 1 4 9 16 25))(25)> (last-pair (list 1 3 5 7))(7)> (last-pair (list 23 72 149 34))(34)
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
Saturday, December 18, 2010
SICP 2.12 - 2.16: Extended Exercise: Interval Arithmetic (Part 2)
From SICP section 2.1.4 Extended Exercise: Interval Arithmetic
The solutions in the first half of this extended exercise, SICP 2.7 - 2.11: Extended Exercise: Interval Arithmetic (Part 1), are only one possible way to approach the problem of implementing interval arithmetic. Another way to represent intervals is as a center value and an added tolerance. The following alternate constructor and selectors are supplied in the text:
; Represent intervals as a center value and a width(define (make-center-width c w) (make-interval (- c w) (+ c w)))(define (center i) (/ (+ (lower-bound i) (upper-bound i)) 2))(define (width i) (/ (- (upper-bound i) (lower-bound i)) 2))
Here's a test case to show how it works:
> (define a (make-center-width 5 1))> a(4 . 6)> (center a)5> (width a)1
A third way of representing intervals, often used by engineers, is as a center value and a percent tolerance measured as the ratio of the width of the interval to its center value.
Exercise 2.12 asks us to define a constructor make-center-percent that takes a center and a percent tolerance and creates the desired interval. We'll also need to define a selector percent that extracts the percentage tolerance for a given interval.
We can define our new constructor in terms of make-center-width. We'll make our procedure take a whole number percentage p.
(define (make-center-percent c p) (make-center-width c (* c (/ p 100.0))))
Since the percent tolerance of an interval is the ratio of the interval's width to its center, we can implement percent in terms of the provided width and center procedures.
(define (percent i) (* 100.0 (/ (width i) (center i))))
Here's how we'd define the same interval as before with the new constructor:
> (define a (make-center-percent 5 20))> a(4.0 . 6.0)> (center a)5.0> (width a)1.0> (percent a)20.0
Exercise 2.13 asks us to show that there is, for small tolerances, a simple formula for the approximate percentage tolerance of the product of two intervals in terms of the tolerances of the factors. We can assume that all numbers are positive.
Recall the old definition of interval multiplication that we used was:
[a,b] × [c,d] = [min (ac, ad, bc, bd), max (ac, ad, bc, bd)]
Since we're assuming all positive numbers for this exercise, we can change the definition to a simpler form:
[a,b] × [c,d] = [ac, bd]
Now we need to complicate things again and look at what it means to represent an interval in terms of its center (ci) and percent tolerance (pi).
i = [ci - ci(pi/100), ci + ci(pi/100)]
i = [ci(1 - pi/100), ci(1 + pi/100)]
Multiplying two intervals x and y together would be:
xy = [cxcy(1 - px/100)(1 - py/100), cxcy(1 + px/100)(1 + py/100)]
We can use the FOIL method to combine some of the terms above.
(1 - px/100)(1 - py/100) = (1 - py/100 - px/100 + pxpy/10,000)
= 1 - (px + py)/100 + pxpy/10,000
(1 + px/100)(1 + py/100) = 1 + py/100 + px/100 + pxpy/10,000
= 1 + (px + py)/100 + pxpy/10,000
Inserting those back into the interval notation we get:
xy = [cxcy(1 - (px + py)/100 + pxpy/10,000), cxcy(1 + (px + py)/100 + pxpy/10,000)]
Since the exercise lets us assume that both px and py are small, and because we're only looking for an approximation, we can ignore the pxpy/10,000 terms because they'll be very small.
xy = [cxcy(1 - (px + py)/100), cxcy(1 + (px + py)/100)]
Now we have things back in terms of an interval's center and percent tolerance. The center of the product of the two intervals is cxcy, and the percent tolerance is (px + py). The approximate percentage tolerance of the product of the two intervals is the sum of the tolerances of the two factors.
We can check this with a quick example.
> (define a (make-center-percent 5 2))> (define b (make-center-percent 10 3))> (define c (mul-interval a b))> (percent c)4.997001798920647
Electrical resistor values are often expressed as a center value and a percent tolerance, so an engineer writing programs that work with resistances would be likely to make use of an interval arithmetic library. The formula for parallel resistors can be written in two algebraically equivalent ways:
and
The following two programs implement the two different computations:
(define (par1 r1 r2) (div-interval (mul-interval r1 r2) (add-interval r1 r2)))(define (par2 r1 r2) (let ((one (make-interval 1 1))) (div-interval one (add-interval (div-interval one r1) (div-interval one r2)))))
Exercise 2.14 asks us to demonstrate that the two programs above give different answers for the same inputs by investigating the behavior of the system on a variety of arithmetic expressions.
First we'll make some intervals and look at the results of interval division in terms of center and percent.
> (define a (make-center-percent 100 5))> (define b (make-center-percent 200 2))> (define aa (div-interval a a))> aa(0.9047619047619049 . 1.1052631578947367)> (define ab (div-interval a b))> ab(0.46568627450980393 . 0.5357142857142857)> (center aa)1.0050125313283207> (center ab)0.5007002801120448> (percent aa)9.97506234413964> (percent ab)6.993006993006991
What's notable here is that the center of the result of dividing an interval by itself is not 1, but just an approximation to it. This will be important when we explain what's wrong with our interval system in the next exercise. For now we're just trying to show that something isn't right.
We can use the same intervals that we defined above to illustrate the difference between the two parallel resistance procedures.
> (define apb1 (par1 a b))> (define apb2 (par2 a b))> apb1(60.25889967637541 . 73.6082474226804)> apb2(63.986254295532646 . 69.32038834951456)> (define apa1 (par1 a a))> (define apa2 (par2 a a))> apa1(42.97619047619048 . 58.026315789473685)> apa2(47.5 . 52.49999999999999)
This verifies that there is a significant difference between the two procedures.
Exercise 2.15 points out that a formula to compute with intervals using the library we've been developing will produce tighter error bounds if it can be written in such a form that no variable that represents an uncertain number (an interval) is repeated. The conclusion is that par2 is a "better" program for parallel resistances than par1. We need to explain if this is correct, and why.
First, this is a good time to show that the two formulas used to develop par1 and par2 are algebraically equivalent. We're trying to show that
We'll start with the formula on the right hand side and derive the formula on the left. To do so, all we really need to do is multiply by R1/R1 and R2/R2. Since both of these fractions equal 1, these are valid transformations.
So the formulas are algebraically equivalent, but they don't give the same answer. Why could that be? The answer lies in the trick we used just now to show equivalence. We used the ratios R1/R1 and R2/R2 to change the formula and said that it was okay because that's just like multiplying by 1. But R1 and R2 represent resistor values, which are intervals, and we saw in exercise 2.14 that dividing an interval by itself doesn't equal 1, it just approximates it. Transforming the equation in this way introduces error. That's why the observation that we can get tighter error bounds if we avoid repeating variables that represent uncertain numbers is correct.
Exercise 2.16 asks us to explain why equivalent algebraic expressions may lead to different answers. It goes on to ask if we can devise an interval-arithmetic package that does not have this shortcoming, or if the task is impossible.
As the text points out, this is a very difficult problem. Wikipedia to the rescue.
The so-called dependency problem is a major obstacle to the application of interval arithmetic. Although interval methods can determine the range of elementary arithmetic operations and functions very accurately, this is not always true with more complicated functions. If an interval occurs several times in a calculation using parameters, and each occurrence is taken independently then this can lead to an unwanted expansion of the resulting intervals.
...
In general, it can be shown that the exact range of values can be achieved, if each variable appears only once. However, not every function can be rewritten this way.
In short, no we cannot design an interval arithmetic package that does not have this shortcoming in the general case. The best we can do, as was indicated in the previous exercise, is to try and write formulas that avoid repeating variables that represent intervals. This is not always possible.
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.
Saturday, December 4, 2010
SICP 2.7 - 2.11: Extended Exercise: Interval Arithmetic (Part 1)
From SICP section 2.1.4 Extended Exercise: Interval Arithmetic
Section 2.1.4 is a small project that has us design and implement a system for working with intervals (objects that represent the range of possible values of an inexact quantity). We need to implement interval arithmetic as a set of arithmetic operations for combining intervals. The result of adding, subtracting, multiplying, or dividing two intervals is itself an interval, representing the range of the result. We're provided with the procedures for adding, multiplying, and dividing two intervals.
The minimum value the sum could be is the sum of the two lower bounds and the maximum value it could be is the sum of the two upper bounds:
(define (add-interval x y) (make-interval (+ (lower-bound x) (lower-bound y)) (+ (upper-bound x) (upper-bound y))))
The product of two intervals is computed by finding the minimum and the maximum of the products of the bounds and using them as the bounds of the resulting interval.
(define (mul-interval x y) (let ((p1 (* (lower-bound x) (lower-bound y))) (p2 (* (lower-bound x) (upper-bound y))) (p3 (* (upper-bound x) (lower-bound y))) (p4 (* (upper-bound x) (upper-bound y)))) (make-interval (min p1 p2 p3 p4) (max p1 p2 p3 p4))))
We can divide two intervals in terms of the mul-interval procedure by multiplying the first interval by the reciprocal of the second. Note that the bounds of the reciprocal interval are the reciprocal of the upper bound and the reciprocal of the lower bound, in that order.
(define (div-interval x y) (mul-interval x (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y)))))
Exercise 2.7 gives us the implementation for make-interval and asks us to complete the interval abstraction by implementing the selectors.
; by convention the bounds are added in lower-upper order(define (make-interval a b) (cons a b))(define (upper-bound intrvl) (cdr intrvl))(define (lower-bound intrvl) (car intrvl))
Now that we have most of our definitions in place, we can start testing some of the given procedures.
> (define a (make-interval 5 10))> (define b (make-interval 10 20))> (add-interval a b)(15 . 30)> (mul-interval a b)(50 . 200)> (div-interval a b)(0.25 . 1.0)
Exercise 2.8 asks us to implement a procedure for computing the difference of two intervals, sub-interval, using reasoning similar to that used to implement add-interval. The smallest value possible when subtracting interval y from interval x is to subtract the upper bound of y from the lower bound of x. The largest result of subtracting interval y from interval x is to subtract the lower bound of y from the upper bound of x.
; interval subtraction: [a,b] − [c,d] = [a − d, b − c](define (sub-interval x y) (make-interval (- (lower-bound x) (upper-bound y)) (- (upper-bound x) (lower-bound y))))
We can test this with a few different intervals to show that it works whether the intervals overlap or not.
> (define a (make-interval 1 10))> (define b (make-interval 50 100))> (define c (make-interval 5 20))> (sub-interval b a)(40 . 99)> (sub-interval a b)(-99 . -40)> (sub-interval a c)(-19 . 5)> (sub-interval c a)(-5 . 19)
Exercise 2.9 defines the width of an interval as half the difference between its upper and lower bounds. For some arithmetic operations the width of the result of combining two intervals is a function only of the widths of the argument intervals, whereas for others the width of the combination is not a function of the widths of the argument intervals. We are to show that the width of the sum (or difference) of two intervals is a function only of the widths of the intervals being added (or subtracted).
I think this is best proven mathematically by showing that the width of the sum of two intervals is the same as the sum of the widths of two intervals. We start with a couple of definitions for interval addition and computing the width of an interval:
[a,b] + [c,d] = [a + c, b + d]
width([x, y]) = (y - x) / 2
Now we can combine these to come up with a formula for the width of the sum of two intervals.
width([a, b] + [c, d]) = width([a + c, b + d])
= ((b + d) - (a + c)) / 2
Finally, we can derive a formula for the sum of the widths of two intervals.
width([a, b]) + width([c, d]) = (b - a) / 2 + (d - c) / 2
= ((b - a) + (d - c)) / 2
= ((b + d) - (a + c)) / 2
This is the same formula we derived above. This proves that if we add any two intervals with widths x and y, the width of the resulting interval will always be z, no matter what the bounds of the intervals are.
To prove that this is not the case for interval multiplication we can use a simple counterexample.
> (define a (make-interval 2 4))> (define b (make-interval 5 10))> (define c (make-interval 10 15))> (mul-interval a b)(10 . 40)> (mul-interval a c)(20 . 60)
The intervals b and c have the same width, but when we multiply each of them by interval a, the resulting intervals have different widths. This means that the width of the product of two intervals cannot be a function of only the widths of the operands.
Exercise 2.10 points out that it is not clear what it means to divide by an interval that spans zero. We need to modify the provided div-interval routine to check for this condition and signal an error if it occurs.
Before we make the required modifications, let's take a closer look at the div-interval procedure to see what the problem is.
(define (div-interval x y) (mul-interval x (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y)))))
If the interval y spans zero, then there's a small problem with this portion of the code:
(make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y))
If y is the interval [-10, 10], for example, then the snippet above would produce the interval [0.1, -0.1], which has a lower bound that is higher than its upper bound.
To fix the problem we can just do as the exercise suggests and signal an error if a zero-spanning interval is found in the divisor.
; Exercise 2.10(define (spans-zero? y) (and (<= (lower-bound y) 0) (>= (upper-bound y) 0)))(define (div-interval x y) (if (spans-zero? y) (error "Error: The denominator should not span 0.") (mul-interval x (make-interval (/ 1.0 (upper-bound y)) (/ 1.0 (lower-bound y))))))
Run an example in your interpreter to verify.
> (define a (make-interval 2 5))> (define b (make-interval -2 2))> (div-interval a b). . Error: The denominator should not span 0.
Exercise 2.11 suggests that by testing the signs of the endpoints of the intervals, we can break mul-interval into nine cases, only one of which requires more than two multiplications. We are to rewrite mul-interval using this suggestion.
The suggestion is based on the result of multiplication of two numbers with the same or opposite signs. For each interval there are three possibilities, both signs are positive, both are negative, or the signs are opposite. (Note that an interval with the signs [+, -] is not allowed, since the lower bound would be higher than the upper bound.) Since there are two intervals to check, that makes nine possibilities. All nine possibilities are listed below.
[+, +] * [+, +]
[+, +] * [-, +]
[+, +] * [-, -]
[-, +] * [+, +]
[-, +] * [-, +]
[-, +] * [-, -]
[-, -] * [+, +]
[-, -] * [-, +]
[-, -] * [-, -]
For most of the combinations above, we can see directly which pairs need to be multiplied to form the resulting interval. For example, if all values are positive, then multiplying the two upper bounds and two lower bounds are the only two products we need to find. The only case where we need to do more than two multiplications is in the case [-, +] * [-, +], since the product of the two lower bounds could possibly be larger than the product of the two upper bounds.
Note that the following code is much less readable and would be harder to maintain than the original procedure. In addition to that, without benchmarking both procedures we aren't even sure which one is faster. The trade-off in maintainability certainly doesn't seem to be worth any potential savings you might get from eliminating two multiplications from most cases. The developer who suggested this enhancement should probably have their commit access revoked (and be shot out of a cannon).
; Exercise 2.11(define (mul-interval x y) (let ((xlo (lower-bound x)) (xhi (upper-bound x)) (ylo (lower-bound y)) (yhi (upper-bound y))) (cond ((and (>= xlo 0) (>= xhi 0) (>= ylo 0) (>= yhi 0)) ; [+, +] * [+, +] (make-interval (* xlo ylo) (* xhi yhi))) ((and (>= xlo 0) (>= xhi 0) (<= ylo 0) (>= yhi 0)) ; [+, +] * [-, +] (make-interval (* xhi ylo) (* xhi yhi))) ((and (>= xlo 0) (>= xhi 0) (<= ylo 0) (<= yhi 0)) ; [+, +] * [-, -] (make-interval (* xhi ylo) (* xlo yhi))) ((and (<= xlo 0) (>= xhi 0) (>= ylo 0) (>= yhi 0)) ; [-, +] * [+, +] (make-interval (* xlo yhi) (* xhi yhi))) ((and (<= xlo 0) (>= xhi 0) (<= ylo 0) (>= yhi 0)) ; [-, +] * [-, +] (make-interval (min (* xhi ylo) (* xlo yhi)) (max (* xlo ylo) (* xhi yhi)))) ((and (<= xlo 0) (>= xhi 0) (<= ylo 0) (<= yhi 0)) ; [-, +] * [-, -] (make-interval (* xhi ylo) (* xlo ylo))) ((and (<= xlo 0) (<= xhi 0) (>= ylo 0) (>= yhi 0)) ; [-, -] * [+, +] (make-interval (* xlo yhi) (* xhi ylo))) ((and (<= xlo 0) (<= xhi 0) (<= ylo 0) (>= yhi 0)) ; [-, -] * [-, +] (make-interval (* xlo yhi) (* xlo ylo))) ((and (<= xlo 0) (<= xhi 0) (<= ylo 0) (<= yhi 0)) ; [-, -] * [-, -] (make-interval (* xhi yhi) (* xlo ylo))))))
We'll need several test cases to make sure we get good coverage of all possible conditions.
> (define a (make-interval 2 4))> (define b (make-interval -2 4))> (define c (make-interval -4 -2))> (mul-interval a a)(4 . 16)> (mul-interval a b)(-8 . 16)> (mul-interval a c)(-16 . -4)> (mul-interval b a)(-8 . 16)> (mul-interval b b)(-8 . 16)> (mul-interval b c)(-16 . 8)> (mul-interval c a)(-16 . -4)> (mul-interval c b)(-16 . 8)> (mul-interval c c)(4 . 16)
Related:
For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge. | 6,532 | 24,560 | {"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} | 3.640625 | 4 | CC-MAIN-2016-30 | longest | en | 0.846452 |
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how to calculate magnetic quantum number? explain clearly with an example
6 years ago
419 Points
Dear Student
The magnetic quantum number has values between -l and +l. When l =1, for example, m can have three values: -1, 0, and +1. Because you know that the subshell designation for l =1 is "p", you now know that the p orbital has three components px, py, and pz. Notice how the subscripts are related to a three-dimensional coordinate system, x, y, and z. The chart below shows a summary of the quantum numbers:
Principal Quantum Number (n) Azimuthal Quantum Number (l) Subshell Designation Magnetic Quantum Number (m) Number of orbitals in subshell
1
0
1s
0
1
2
01
2s2p
0-1 0 +1
13
3
012
3s3p3d
0-1 0 +1-2 -1 0 +1 +2
135
4
0123
4s4p4d4f
0-1 0 +1-2 -1 0 +1 +2-3 -2 -1 0 +1 +2+3
1357
All the best.
AKASH GOYAL
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
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6 years ago
SAGAR SINGH - IIT DELHI
879 Points
Dear student,
To describe the magnetic quantum number m you begin with an atomic electron's angular momentum, L, which is related to its quantum number $l\,$ by the following equation:
$\mathbf{L} = \hbar\sqrt{l(l+1)}$
where $\hbar = h/2\pi$ is the reduced planck constant.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Sagar Singh
B.Tech, IIT Delhi
6 years ago
Aakash Dutta
29 Points
this is very simple
6 years ago
Think You Can Provide A Better Answer ?
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# qgis / QGIS
fix QgsDualEdgeTriangulation
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vcloarec authored and nyalldawson committed Sep 20, 2020
1 parent 8b03293 commit 5e0a14fca6b91970b8f3d1242492fc2297d1947d
Showing with 45 additions and 4 deletions.
1. +45 −4 src/analysis/interpolation/qgsdualedgetriangulation.cpp
@@ -155,11 +155,11 @@ int QgsDualEdgeTriangulation::addPoint( const QgsPoint &p ) //test, if it is the same point as the first point if ( p.x() == mPointVector[0]->x() && p.y() == mPointVector[0]->y() ) { //QgsDebugMsg( QStringLiteral( "second point is the same as the first point, it thus has not been inserted" ) ); //second point is the same as the first point QgsPoint *p = mPointVector[1]; mPointVector.remove( 1 ); delete p; return -100; return 0; } unsigned int edgeFromPoint0ToPoint1 /* 2 */ = insertEdge( -10, -10, 1, false, false );//edge pointing from point 0 to point 1 @@ -190,6 +190,7 @@ int QgsDualEdgeTriangulation::addPoint( const QgsPoint &p ) mEdgeOutside = firstEdgeOutSide(); if ( mEdgeOutside < 0 || mHalfEdge[mEdgeOutside]->getPoint() < 0 || mHalfEdge[mHalfEdge[mEdgeOutside]->getDual()]->getPoint() < 0 ) return -100; double leftOfNumber = MathUtils::leftOf( p, mPointVector[mHalfEdge[mHalfEdge[mEdgeOutside]->getDual()]->getPoint()], mPointVector[mHalfEdge[mEdgeOutside]->getPoint()] ); if ( fabs( leftOfNumber ) <= leftOfTresh ) { @@ -204,7 +205,22 @@ int QgsDualEdgeTriangulation::addPoint( const QgsPoint &p ) int point1 = mHalfEdge[mEdgeOutside]->getPoint(); int point2 = mHalfEdge[mHalfEdge[mEdgeOutside]->getDual()]->getPoint(); double distance1 = p.distance( *mPointVector[point1] ); if ( distance1 <= leftOfTresh ) // point1 == new point { QgsPoint *pt = mPointVector.last(); mPointVector.removeLast(); delete pt; return point1; } double distance2 = p.distance( *mPointVector[point2] ); if ( distance2 <= leftOfTresh ) // point2 == new point { QgsPoint *pt = mPointVector.last(); mPointVector.removeLast(); delete pt; return point2; } double edgeLength = mPointVector[point1]->distance( *mPointVector[point2] ); if ( distance1 < edgeLength && distance2 < edgeLength ) @@ -427,7 +443,27 @@ int QgsDualEdgeTriangulation::addPoint( const QgsPoint &p ) //the point is exactly on an existing edge (the number of the edge is stored in the variable 'mEdgeWithPoint'--------------- else if ( number == -20 ) { //QgsDebugMsg( "point exactly on edge" ); //point exactly on edge; //check if new point is the same than one extremity int point1 = mHalfEdge[mEdgeWithPoint]->getPoint(); int point2 = mHalfEdge[mHalfEdge[mEdgeWithPoint]->getDual()]->getPoint(); double distance1 = p.distance( *mPointVector[point1] ); if ( distance1 <= leftOfTresh ) // point1 == new point { QgsPoint *pt = mPointVector.last(); mPointVector.removeLast(); delete pt; return point1; } double distance2 = p.distance( *mPointVector[point2] ); if ( distance2 <= leftOfTresh ) // point2 == new point { QgsPoint *pt = mPointVector.last(); mPointVector.removeLast(); delete pt; return point2; } int edgea = mEdgeWithPoint; int edgeb = mHalfEdge[mEdgeWithPoint]->getDual(); @@ -558,7 +594,12 @@ int QgsDualEdgeTriangulation::baseEdgeOfPoint( int point ) int QgsDualEdgeTriangulation::baseEdgeOfTriangle( const QgsPoint &point ) { unsigned int actEdge = mEdgeInside;//start with an edge which does not point to the virtual point (usually number 3) unsigned int actEdge = mEdgeInside;//start with an edge which does not point to the virtual point if ( mHalfEdge[actEdge]->getPoint() < 0 ) actEdge = mHalfEdge[mHalfEdge[mHalfEdge[actEdge]->getDual()]->getNext()]->getDual();//get an real inside edge if ( mHalfEdge[mHalfEdge[actEdge]->getDual()]->getPoint() < 0 ) actEdge = mHalfEdge[mHalfEdge[actEdge]->getNext()]->getDual(); int counter = 0;//number of consecutive successful left-of-tests int nulls = 0;//number of left-of-tests, which returned 0. 1 means, that the point is on a line, 2 means that it is on an existing point int numInstabs = 0;//number of suspect left-of-tests due to 'leftOfTresh'
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### Incorrect total in Power Bi
Hi, I have a decimal column in Power Bi with positive and negative values, when I put that column (by itself) to a graph or a table I get a total of -2 billion, ...
| 2 Replies | 515 Views | Created by RachelM.A - Sunday, February 18, 2018 6:09 AM | Last reply by RachelM.A - Sunday, February 18, 2018 6:33 AM
### How to calculate conditional running total in power query (Excel 2016)?
Is it possible to calculate conditional based running total in power query? I need to add numbers(cumulative sum) based on the group. Please see the image attached. ...
| 4 Replies | 4655 Views | Created by Jithin.RG - Friday, March 23, 2018 4:16 AM | Last reply by Jithin.RG - Monday, March 26, 2018 9:06 AM
### cant runnig
run script User's domain is mydomian. Total estimated token size is 1304. For access to DCs and delegatable resources the ...
| 14 Replies | 604 Views | Created by hamed_forum - Wednesday, January 30, 2019 12:31 PM | Last reply by hamed_forum - Sunday, February 3, 2019 5:30 AM
### Power Query and Power View
Yes, currently Power Query does not support SSAS Tabular as a data source.
| 2 Replies | 1231 Views | Created by garyandadriel - Thursday, August 21, 2014 8:22 PM | Last reply by Ehren - MSFT - Thursday, October 16, 2014 5:53 PM
### Totals in Power BI
Excel 365 Pro Plus with PowerPivot and Power Query (aka Get & Transform) Instead of DAX-SUMMARIZE() you might try PowerQuery-GROUP(). PQ has ...
| 1 Replies | 837 Views | Created by Dan_dosReis - Wednesday, December 12, 2018 2:17 PM | Last reply by Herbert Seidenberg - Saturday, December 15, 2018 4:43 AM
### Power Q&A - Sorting by total
Hi, I've got a view like below on my Power Q&A, which is produced by asking following question: show day and <fieldname> sorted by ...
| 3 Replies | 3286 Views | Created by Dawid-SC - Monday, December 1, 2014 11:35 AM | Last reply by Ed Price - MSFT - Sunday, July 10, 2016 6:30 AM
### Power Query
Hi, i have Microsoft Office Professional Plus 2010 (64 bit). I am trying to install Power Query and i get a message "you need to have excel 2013 or excel 2010 ...
| 4 Replies | 424 Views | Created by komhs74 - Friday, December 21, 2018 11:28 PM | Last reply by komhs74 - Saturday, December 22, 2018 1:11 AM
### Total query
What is the query to also give the something_cntal per foobar_name? Something like this: Total foobar_name ...
| 5 Replies | 354 Views | Created by aujong - Wednesday, July 27, 2016 4:39 PM | Last reply by --CELKO-- - Wednesday, July 27, 2016 11:32 PM
### RUNNIG A JOB???
using an OLE DB source to retrieve some data from the db... netx step i use this data conversion task because some of the results of the query aren`t supported by mi ...
| 18 Replies | 7131 Views | Created by ruk_walled - Thursday, May 4, 2006 9:10 PM | Last reply by ruk_walled - Friday, March 30, 2007 5:13 PM
### 80/20 rule
ABC Approach in Power Query as well: In my last comment there:
| 11 Replies | 1624 Views | Created by yan_elf - Friday, August 21, 2015 9:46 PM | Last reply by yan_elf - Friday, July 29, 2016 9:35 PM
### Power Pivot & Power Query
| 3 Replies | 1133 Views | Created by SJL6357 - Tuesday, December 2, 2014 12:25 PM | Last reply by George123345 - Thursday, December 4, 2014 1:19 AM
### Lost Totals in Access Queries
Are you consuming this Access data in Power Query, or are you referring only to Access queries? Ehren
| 1 Replies | 204 Views | Created by cFolds - Tuesday, April 11, 2017 1:29 PM | Last reply by Ehren - MSFT - Wednesday, April 12, 2017 5:17 PM
### Power Query : How to insert custom column based on total hours
Hi Rajender, From what I can tell, your desired output should be doable in Power Query. However, I'm still having trouble understanding exactly where you're ...
| 6 Replies | 1049 Views | Created by AskQuery1984 - Friday, August 19, 2016 4:15 PM | Last reply by Ehren - MSFT - Friday, September 2, 2016 10:08 PM | 1,655 | 5,507 | {"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} | 2.703125 | 3 | CC-MAIN-2020-34 | latest | en | 0.909438 |
https://diy.stackexchange.com/questions/169693/is-this-the-right-way-to-ground-fault-protect-both-sides-of-an-mwbc | 1,717,101,009,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00296.warc.gz | 162,153,245 | 40,161 | # is this the right way to ground fault protect both sides of an mwbc?
is this the correct way to ground fault-protect both sides of an mwbc? pigtail the (line) neutral to two separate gfci outlets, and pigtail the (load) neutrals to the neutral that's going outside.
Close, but not quite. From the point of the two GFCIs onward, the neutrals must be kept separate.
Consider: if you had a device drawing power on the black hot circuit from a load farther down the line, and nothing on the red, which GFCI would its neutral return current flow through? Since they're just connected in parallel in your diagram, it would flow through both, and then trip both, since current out is not equal to current in within each GFCI. If you separate the neutrals, you can make sure all the current from the black hot flows back through its proper GFCI, and none crosses over to the other one.
Also note that this scheme will fail if you use any 240V devices on this circuit. The only way to do a mixed 120/240V circuit is with a 2-pole GFCI breaker.
That drawing cannot possibly work. Right off the bat, you are paralleling, providing two paths for neutral to go. That itself is a code violation, nevermind the GFCIs. It won't work because as Nate Strickland describes, neutral won't know which path to take.
GFCIs involve looping all the wires of a circuit through a current transformer, where each wire creates a magnetic field and all the magnetic fields should cancel each other out. Therefore, all the wires of the circuit have to go through ONE current transformer. Any scheme where a GFCI doesn't handle all the wires, can't work by definition.
I gather you have an existing MWBC that stops at several junction boxes, and you need to GFCI protect them all.
## Solution: A 2-pole GFCI
So only a 2-pole GFCI will work. I've never found one offered as a stand-alone GFCI (deadfront). Mainly they seem to appear as 2-pole GFCI+circuit breaker combo's. And those always take 2 full spaces. And they must be listed or classified for your panel, so unless you have a modern GE, Eaton, Sq.D. or Siemens panel (or Challenger which take BR), you could be in trouble. Check Eaton's CL line which fits some panels, but you could be out of luck. In that case, it's "subpanel time".
## Solution: Don't use Load terminals
You're not required to use Load terminals. You certainly can install GFCI receptacles and leave the warning tape on the "Load" terminals. Just then, the downline "rest of the circuit" won't be protected. Place all the wires on "Line" (most GFCIs permit 2 wires per screw, but remember, you already need to pigtail neutral because it's a multi-wire branch circuit!)
This means downline receptacles will not be protected, and you will need separate GFCI receptacles there as well if you need protection there.
• Are you saying that keeping neutrals separate after the GFCI won't work? Or are you assuming that isn't plausible on already-installed wire because there isn't a second neutral? Jul 23, 2019 at 15:42
• @JPhi1618 I'm saying his drawing won't work. Which means yes, if there's a pre-installed situation of 3-wire connecting to a junction box of outlets, then onward as 3-wire to another junction box of outlets, then he is sunk indeed. Jul 23, 2019 at 15:47 | 793 | 3,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.947282 |
http://www.smart-kit.com/s246/short-but-challenging-logic-puzzle/ | 1,498,497,894,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320841.35/warc/CC-MAIN-20170626170406-20170626190406-00252.warc.gz | 670,068,052 | 9,258 | School-Safe Puzzle Games
## Short, but challenging, logic puzzle
You’ll need the full powers of your working memory for this one, courtesy of www.braingle.com:
There is a very special five digit number. The special thing about it is that its first digit describes how many 0s there are in the whole number, the second digit describes how many 1s there are in the whole number, and so one, all the way to the fifth digit describing how many 4s there are in the big number
Tags:
### 9 Comments to “Short, but challenging, logic puzzle”
1. RLP | Profile
The number is 21200.
There are 2 zeroes, 1 one, 2 twos, 0 threes and 0 fours.
2. lowtec | Guest
40000 works also…
3. Nerd | Guest
no, 4000 does not work, because the 5th digit is the number of the 4-s
4. Scott K. | Guest
I feel dumb after that. I was going to say 14111, but it just goes to show, I can’t take a guess based on just the first 2 numbers working, I have to think it the full way through.
5. Kelby B. | Guest
6. Megan W | Guest
Kelby, 31000 doesn’t work because there would have to be a 1 to represent the number of 3s. i.e 31010 – but this then means there are 2 1s. hence making it an incorrect answer.
21200 is the only correct answer.
7. DrNumlock | Guest
Wich is wrong.
Correct it: 4 0 0 0 1
wich is wrong
Correct it: 3 0 0 0 1
the answer is close, but not right.
Correct it: 3 0 0 1 0
I cant find any flaws. so i call it right.
8. Dr.Numnut | Guest
Dr. Numlock there is a flaw in your theory in the # 30010 the 2nd digit would represent how many 1’s are in the 5-digit #,ur # states 0 and u have one #1 in ur 5-digit # i think 21200 is the only correct answer
9. max | Guest
Scott K. (above) is my dad. i am his 13 year old son. he couldnt get the answer no matter how hard he tried. he worked on it for a good 25-30 minutes. i then took it up to my room and figured it out in about 5 minutes. my dad is a very smart lawyer, so this has to just be how your individual brain thinks logically. i probably embarassed him by saying all this, so sorry dad! | 597 | 2,044 | {"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} | 3.546875 | 4 | CC-MAIN-2017-26 | latest | en | 0.953194 |
https://www.coursehero.com/file/5826557/ATTACHMENT231/ | 1,512,967,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00564.warc.gz | 715,598,486 | 40,829 | ATTACHMENT+%231+
ATTACHMENT+%231+ - complete circle of 2 radians in a time...
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ATTACHMENT #1 Point P moves with constant speed vp on the circumference of a circle. Point Q moves such that x coordinates are equal. The motion of pont Q on the diameter is defined as SHM. P Q x axis 0 x x m x m w NOTES: 1. The radius of the circle is x m, the maximum value of x, which is called the amplitude of the motion. 2. The radius line follows P around the circle, rotating one
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Unformatted text preview: complete circle of 2 radians in a time of one period, T. 3. The constant angular velocity of the radius is w= 2 /T. 4. In the above circle, if time begins when point Q is at x m , the angle between the base x and the hypotenuse x m is wt as shown. wt...
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Ask a homework question - tutors are online | 252 | 973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-51 | latest | en | 0.863681 |
http://mathhelpforum.com/algebra/197273-simplifying.html | 1,524,695,091,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947968.96/warc/CC-MAIN-20180425213156-20180425233156-00370.warc.gz | 198,881,741 | 9,541 | # Thread: Simplifying
1. ## Simplifying
How can I simplify the following...
$\displaystyle 14-9/4^2- 2^3$
The test puts it exactly in this format I don't understand if it's to confuse me? Or I simply don't know my math very well.
Here I will use PEMDAS.
My solution
$\displaystyle 14-\frac{9}{16}-8$
$\displaystyle \frac{14}{1}-\frac{9}{16}-8$
=$\displaystyle \frac{224}{16}-\frac{9}{16}-8$
=$\displaystyle \frac{215}{16}-\frac{128}{16}$
=$\displaystyle \frac{87}{16}$
or
=$\displaystyle 5\frac{7}{16}$
2. ## Re: Simplifying
This is correct.
Edit: A simpler way is
$\displaystyle 14-\frac{9}{16}-8=6-\frac{9}{16}=5+\frac{16}{16}-\frac{9}{16}=5\frac{7}{16}$
3. ## Re: Simplifying
Originally Posted by emakarov
This is correct.
Weird my sample test is not providing me the right answer.
Probably the test did not include the "NONE OF THESE" answer. | 289 | 860 | {"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 | 4 | CC-MAIN-2018-17 | latest | en | 0.779451 |
https://www.rdocumentation.org/packages/igraph/versions/1.0.1/topics/which_multiple | 1,580,026,650,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00194.warc.gz | 1,045,895,517 | 5,821 | # which_multiple
0th
Percentile
##### Find the multiple or loop edges in a graph
A loop edge is an edge from a vertex to itself. An edge is a multiple edge if it has exactly the same head and tail vertices as another edge. A graph without multiple and loop edges is called a simple graph.
Keywords
graphs
##### Usage
which_multiple(graph, eids = E(graph))
##### Arguments
graph
The input graph.
eids
The edges to which the query is restricted. By default this is all edges in the graph.
##### Details
which_loop decides whether the edges of the graph are loop edges.
any_multiple decides whether the graph has any multiple edges.
which_multiple decides whether the edges of the graph are multiple edges.
count_multiple counts the multiplicity of each edge of a graph.
Note that the semantics for which_multiple and count_multiple is different. which_multiple gives TRUE for all occurences of a multiple edge except for one. Ie. if there are three i-j edges in the graph then which_multiple returns TRUE for only two of them while count_multiple returns ‘3’ for all three.
See the examples for getting rid of multiple edges while keeping their original multiplicity as an edge attribute.
##### Value
any_multiple returns a logical scalar. which_loop and which_multiple return a logical vector. count_multiple returns a numeric vector.
simplify to eliminate loop and multiple edges.
##### Aliases
• any_multiple
• count.multiple
• count_multiple
• has.multiple
• is.loop
• is.multiple
• which_loop
• which_multiple
##### Examples
# NOT RUN {
# Loops
g <- graph( c(1,1,2,2,3,3,4,5) )
which_loop(g)
# Multiple edges
g <- barabasi.game(10, m=3, algorithm="bag")
any_multiple(g)
which_multiple(g)
count_multiple(g)
which_multiple(simplify(g))
all(count_multiple(simplify(g)) == 1)
# Direction of the edge is important
which_multiple(graph( c(1,2, 2,1) ))
which_multiple(graph( c(1,2, 2,1), dir=FALSE ))
# Remove multiple edges but keep multiplicity
g <- barabasi.game(10, m=3, algorithm="bag")
E(g)$weight <- count_multiple(g) g <- simplify(g) any(which_multiple(g)) E(g)$weight
# }
Documentation reproduced from package igraph, version 1.0.1, License: GPL (>= 2)
### Community examples
Looks like there are no examples yet. | 545 | 2,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-05 | latest | en | 0.811264 |
https://www.wisegeek.net/what-are-the-different-ways-to-determine-rate-of-return.htm | 1,708,758,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00292.warc.gz | 1,080,700,611 | 22,436 | Finance
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What Are the Different Ways to Determine Rate of Return?
Alex Newth
Alex Newth
Investors have many methods to use to determine rate of return on investment vehicles, depending on the factors they want to consider. The stated rate is the most basic method to determine rate of return, because it only considers the percentage of interest paid. Effective rate of return takes into consideration compounding, which is present on most investments. Internal rate of return is based on how much money is in an account or investment vehicle, so the return can dynamically alter. Time-weighted return is the opposite of internal rate of return, because it does not matter how much money is in an account.
Stated rate of return is the most basic method for investors who want to determine rate of return, because no math is required and investors will typically know this number before investing. The stated rate is the interest rate that will be added to an investment. For example, if the interest rate is 15 percent, then this also is the stated rate. Unlike effective and internal rate of return, it does not matter how much money is in the investment vehicle, nor does it matter if compounding — extra interest added on top of the investment’s current value, as opposed to its base value — is used.
The effective rate of return is often used on static bonds and stocks that do not increase or decrease in value based on an investor’s actions, unlike a bank account that allows an investor to add or remove money. Using this measurement to determine rate of return often happens when the investment vehicle has compounding, because effective rate of return is specifically made to account for compounding. If there is no compounding, then effective rate of return can still be used, but it will often turn out to be the same as the stated return.
An internal rate of return is similar to the effective return rate, but this is used for bank accounts or other investment vehicles that allow the investor to add or subtract money from the investment. For example, if the interest rate is 10 percent and an account has \$100 US Dollars (USD), then the return will be \$110 USD; if the account has \$200 USD, then the return will be \$220 USD. This leads many investors to try to add money to their accounts or investment vehicles before the payment time.
With time-weighted return, it does not matter how much money is in an account. To determine rate of return, investors are typically given a flat number or a static interest rate is applied to a base amount of money. For example, if the return is \$10 USD — or it may be 10 percent of a base amount of \$100 USD — it does not matter how much money the investor has in an account, the return will still be \$10 USD. This type of return rate is much safer, because investors are practically guaranteed a payment, regardless of an investment vehicle’s current value. | 639 | 3,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.955479 |
https://www.doubtnut.com/qna/642545581 | 1,709,019,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00585.warc.gz | 752,471,118 | 45,265 | # If→a,→b,→c are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is (A) →a+→b+→c (B) →a|→a|+→b∣∣→b∣∣+→/|→c| (C) →a|→a|2+→b∣∣→b∣∣2+→c|→c|2 (D) |→a|→a−∣∣→b∣∣→b+|→c|→c
A
a+b+c
B
a|a|+bb+c|c|
C
a|a|2+bb2+c|c|2
D
|a|abb+|c|c
Video Solution
Text Solution
Verified by Experts | 184 | 326 | {"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-2024-10 | latest | en | 0.700692 |
https://ontrack-media.net/gateway_updated/gateway/algebra2/g_a2m3l3rs3.html | 1,726,757,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00211.warc.gz | 404,979,026 | 3,456 | In the previous section, you noticed changes to the graph of the rational parent function by changing the parameters a, h, and k. You noticed that changes in a caused a vertical stretch or compression, changes in h caused a horizontal shift, and changes in k caused a vertical shift. The same holds true for square root functions in the following form:
Four functions, four graphs, and four descriptions of transformations of the square root parent function are listed below. Some functions have already been placed in the table. Place the remaining objects next to their corresponding counter parts.
### Conclusion Questions
• What happens to the graph when the value of a is changed?
Interactive popup. Assistance may be required.
As the value of a increases, the graph is vertically stretched. As the value of a gets closer to zero, the graph is reflected horizontally. When the value of a is negative, the graph is reflected across its horizontal asymptote.
• What happens to the graph when you increase or decrease the value of h?
Interactive popup. Assistance may be required.
As the value of h increases, the function shifts to the right. As the value of h decreases, the function shifts to the left.
• What happens to the graph when you increase or decrease the value of k?
Interactive popup. Assistance may be required.
As the value of k increases, the function shifts up. As the value of k decreases, the function shifts down.
### Pause and Reflect
Think about the square root functions, f(x) = √(x + 1) – 2 and f(x) = √(x – 4) – 5. How would you describe the vertical shift from the first function to the second function?
Interactive popup. Assistance may be required.
The second function is translated down 3 units from the first function.
Now think about the rational functions, f(x) = 1 over x + 1 1 (x + 1) – 2 and f(x) = 1 over x - 4 1 (x – 4) – 5. Would the vertical shift between this pair of functions be the same or different as the vertical shift between the square root functions? Explain your answer.
Interactive popup. Assistance may be required.
The vertical shift from the first function to the second function is the same. The vertical shift on a function is the result of the constant being added or subtracted regardless of the parent function.
### Practice
1. Describe the transformation of the graph of the square root parent function, f(x) = √x, to the graph of the square root function f(x) = 1 over 2 - 1 2 x.
Interactive popup. Assistance may be required.
Recall the answer to the conclusion question #1 above.
Interactive popup. Assistance may be required.
The graph of the parent function will be compressed by a factor of 1 over 2 1 2 and reflected across the x-axis.
2. Below are the coordinates of 2 square root functions. Describe the transformation of the graph from Function 1 to Function 2 based on these coordinates.
Function 1 x y 0 1 1 2 4 3 9 4 16 5
Function 2 x y 0 -3 1 -2 4 -1 9 0 16 1
Interactive popup. Assistance may be required.
Find the difference between the y-values of Function 1 and Function 2.
Interactive popup. Assistance may be required.
Function 2 is translated down 4 units from Function 1.
3. If the graph of f(x) = √(x + 2) + 4 were translated 1 unit to the left and 4 units down, what would be g(x), the function representing the transformed graph?
Interactive popup. Assistance may be required.
Think about which part of the function is affected by a horizontal shift and which part is affected by a vertical shift. Refer back to the conclusion questions for additional help.
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g(x) = √(x + 3)
4. Describe the transformation from f1(x) that would generate f2(x) in the graph below.
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Locate the starting point on the graph of f1(x) and count the units to the left and up to the starting point on the graph of f2(x).
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The graph of function f2(x) has been translated 4 units to the left and 4 units up from the graph of function f1(x). | 939 | 4,079 | {"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} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.869248 |
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# Which term of the following A.P $3,\text{ }15,\text{ }27,\text{ }39,\text{ }\ldots$will be 132 more than the 54th term of the A.P.
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360k+ views
Hint: Use the formula for the nth term of A.P ${{a}_{n}}=a+(n-1)d$. Where ‘a’ is the first term and ‘d’ is the common difference to find the 54th term. Then use ${{a}_{n}}=a+(n-1)d$again to find the term which is 132 more than the 54th term. Alternatively, you can use the property that if mth term is p more than nth term then $m=n+\dfrac{p}{d}$.
Complete step-by-step solution -
We know that ${{a}_{n}}=a+(n-1)d\text{ (i)}$
In the given A.P we have a = 3
d = 15-3 = 12
Put n = 54, a = 3 and d = 12 in equation (i), we get
\begin{align} & {{a}_{54}}=3+\left( 54-1 \right)\times 12 \\ & \Rightarrow {{a}_{54}}=3+53\times 12 \\ & \Rightarrow {{a}_{54}}=3+636 \\ & \Rightarrow {{a}_{54}}=639 \\ \end{align}
Let the kth term of A.P be 132 more than 54th term
Hence ${{a}_{k}}={{a}_{54}}+132$
$\Rightarrow a+\left( k-1 \right)d=639+132$
Put a = 3 and d = 12
$\Rightarrow 3+\left( k-1 \right)12=771$
Subtracting 3 from both sides we get
\begin{align} & \Rightarrow 3+\left( k-1 \right)12-3=771-3 \\ & \Rightarrow \left( k-1 \right)12=768 \\ \end{align}
Dividing both sides by 12, we get
\begin{align} & \Rightarrow \dfrac{\left( k-1 \right)12}{12}=\dfrac{768}{12} \\ & \Rightarrow k-1=64 \\ \end{align}
Transposing 1 to RHS we get
k = 64+1 = 65
Hence the 65th term is 132 more than 54th term
Note: [a] Alternatively, let mth term is p more than nth term
Then we have
\begin{align} & {{a}_{m}}={{a}_{n}}+p \\ & \Rightarrow a+\left( m-1 \right)d=a+\left( n-1 \right)d+p \\ \end{align}
Subtracting a from both sides we get
\begin{align} & \Rightarrow a+\left( m-1 \right)d-a=a+\left( n-1 \right)d+p-a \\ & \Rightarrow \left( m-1 \right)d=\left( n-1 \right)d+p \\ \end{align}
Dividing by d on both sides we get
\begin{align} & \dfrac{\left( m-1 \right)d}{d}=\dfrac{\left( n-1 \right)d+p}{d} \\ & \Rightarrow m-1=\dfrac{\left( n-1 \right)d+p}{d} \\ \end{align}
Transposing 1 to RHS we get
\begin{align} & \Rightarrow m=\dfrac{\left( n-1 \right)d+p}{d}+1 \\ & \Rightarrow m=\dfrac{\left( n-1 \right)d}{d}+\dfrac{p}{d}+1 \\ & \Rightarrow m=n-1+\dfrac{p}{d}+1 \\ & \Rightarrow m=n+\dfrac{p}{d} \\ \end{align}
Put n = 54, p = 132 and d = 12 we get
$m=54+\dfrac{132}{12}$
m = 54+11 = 65
[b] The value of the first term was not needed in solving the question through the second method.
[c] Some of the most important formulae in A.P are:
[1] ${{a}_{n}}=a+\left( n-1 \right)d$
[2] ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.
[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$
Last updated date: 25th Sep 2023
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Mathematica script by Chris Parrish,
cparrish@sewanee.edu
Sources and references for some of these problems include
James Stewart, "Calculus: Concepts and Contexts," Second Edition, Brooks/Cole, 2001
Deborah Hughes-Hallett, Andrew M. Gleason, et. al., "Calculus," Second Edition, John Wiley & Sons, 1998
Robert Fraga, ed., "Calculus Problems for a New Century," The Mathematical Association of America, 1993
Population of Switzerland
Hughes-Hallett, Gleason, et al, Exercise 9.7.1, page 536
In[1083]:=
In[1085]:=
Spread of Information by Mass Media
Hughes-Hallett, Gleason, et al, Exercise 9.7.5a, page 537
Let's choose some notation:
population size = m
number of informed people = i
number of people who know initially = i_0
Then, di/dt = k (m - i).
In[1090]:=
?DSolve
In[1091]:=
Clear[i,t,k,m]
DSolve[ i'[t] == k (m - i[t]), i[t],t]
Out[1092]=
We know that i[0] = i_0, so that
i_0 = i[0] = m + c;
thus,
c = i_0 - m.
Choose units so that m = 1 and k = 1 and i_0 = 0.
In[1093]:=
Spread of Information by Word of Mouth
Hughes-Hallett, Gleason, et al, Exercise 9.7.5b, page 537
Let's choose some notation:
population size = m
number of informed people = i
number of people who know initially = i_0
Then, di/dt = k i (m - i).
In[1098]:=
Clear[i,t,k,m]
DSolve[ i'[t] == k i[t] (m - i[t]), i[t],t]
Out[1099]=
Here is a large-scale view of the slope field.
Take k = 1.
In[1100]:=
Clear[f,i,t,k,m,a,b,c,d];
f[i_,t_] := i (m - i);
m = 1;
a = 0; b = 1; (* a <= t <= b *)
c = 0; d = 1; (* c <= i <= d *)
init1 = 0.0;
init2 = 0.05;
init3 = 0.75;
pts = {{0,init1},{0,init2},{0,init3}};
field = PlotVectorField[{1,f[i,t]},
{t,a,b},{i,c,d},
PlotLabel -> "Spread of Information by Word of Mouth",
Axes -> True,
AxesLabel -> {"t","i"},
PlotPoints -> 20,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Let's plot the solutions which pass through the red dots.
Since m = 1, we have
i[t] = 1/(1 + c Exp[-kt])
Hence, c = (1 / i[0] - 1)
In[1110]:=
Clear[i1,i2,i3,t]
m = 1;
k = 1;
c1 = 1/init1 - 1;
c2 = 1/init2 - 1;
c3 = 1/init3 - 1;
i1[t_] := 1/(1 + c1 Exp[-k t])
i2[t_] := 1/(1 + c2 Exp[-k t])
i3[t_] := 1/(1 + c3 Exp[-k t])
plots = Plot[{i1[t],i2[t],i3[t]},{t,a,b},
PlotStyle->Indigo,
PlotRange->{0,1},
DisplayFunction->Identity];
Show[{field,plots},
DisplayFunction->\$DisplayFunction];
In[1122]:=
When is the information spreading the most rapidly?
di/dt is the rate of spread of information
In[1124]:=
i'[t]
Out[1124]=
When is this function a maximum? -- when its derivative is zero.
In[1125]:=
i''[t]
Out[1125]=
Let's simplify this expression.
In[1126]:=
Simplify[i''[t]]
Out[1126]=
When is this zero?
(1) when i0 = 0
(2) when k = 0
(3) when m = i0
(4) when i0 + i0 Exp[kmt] - m = 0
(5) when m = 0
Case (4) is the most interesting
In[1127]:=
Solve[i0 + i0 Exp[k m t] - m == 0, t]
Out[1127]=
Gause's Yeast
Hughes-Hallett, Gleason, et al, Exercise 9.7.6, page 537
In[1128]:=
Clear[data];
data = {{0,0.37},{10,8.87},{18,10.66},{23,12.5},{34,13.27},{42,12.87},{47,12.7}};
dots = ListPlot[data,
PlotStyle->{MarsOrange,PointSize[0.02]}];
Let's take L = 13.
In[1131]:=
L = 13;
Exercise 9.7.6b
Estimate k.
In[1132]:=
Clear[p,k];
p[0] = 0.37;
p[10] = 8.87;
estimatedDerivativeAtZero = (p[10] - p[0])/10;
k = estimatedDerivativeAtZero / (p[0] (1 - p[0]/L))
Out[1136]=
Exercise 9.7.6c
As a check, look at p[18] and compare your answer with the value of 10.66 from the table.
In[1137]:=
k = .28;
a = (L - p[0])/p[0];
p[t_] := L/(1 + a Exp[- k t]);
c = 0;
d = 40;
plot = Plot[p[t],{t,c,d},
PlotStyle->Red,
PlotLabel->"p[t]",
PlotRange->{0,13},
AxesLabel->{"t (hrs)","p (num cells)"}];
p[18]
Out[1144]=
How well does our solution fit the original data?
In[1145]:=
Show[dots,plot];
Threshold Population
Hughes-Hallett, Gleason, et al, Exercise 9.7.17a, page 537
Graph the derivative,dp/dt = deriv[p], against p:
dp/dt = deriv[p] = p (p - 6)
In[1146]:=
Clear[deriv,p,c,d];
deriv[p_] := p (p - 6);
c = -1; d = 8; (* c <= p <= d *)
Plot[deriv[p],{p,c,d},
PlotStyle->Red,
PlotLabel->"dp/dt = p (p - 6)",
AxesLabel->{"p (population)","dp/dt"}];
Hughes-Hallett, Gleason, et al, Exercise 9.7.17b, page 537
Find the general solution.
In[1150]:=
DSolve[p'[t] == p[t] (p[t] - 6), p[t], t]
Out[1150]=
Define it, and then check that this definition satisfies the differential equation.
In[1151]:=
Clear[p,t,c,d,const,p0];
p[t_] := 6/(1 + const Exp[6t]);
Simplify[p'[t] == p[t] (p[t] - 6)]
Out[1153]=
Find find the solution satisfying p[0] = 5:
p0 = p[0] = 6/(1 + const) => const = (6 / p0) - 1
If p0 = 5, take const = -6/5.
In[1154]:=
p[t_] := 6/(1 + const Exp[6t]);
const = 6/p0 - 1;
p0 = 5;
c = 0; d = 1; (* c <= t <= d *)
diesOut = Plot[p[t],{t,c,d},
PlotStyle->Red,
PlotRange->{0,8},
PlotLabel->"soln[t] = 6/(1 + 1/5 Exp[6t])",
AxesLabel->{"t (time)",None}];
Hughes-Hallett, Gleason, et al, Exercise 9.7.17c, page 537
Find soln for generic p[0].
Try it for p[0] = 8; so that const = -1/4.
We have to shorten the domain of the variable t to get a decent picture,
because this function blows up at t = 0.2.
In[1159]:=
Clear[p,t,c,d,const,p0];
p[t_] := 6/(1 - 1/4 Exp[6t]);
c = 0; d = 0.2; (* c <= t <= d *)
grows = Plot[p[t],{t,c,d},
PlotStyle->Red,
PlotRange->{0,30},
PlotLabel->"soln[t] = 6/(1 - 1/4 Exp[6t])",
AxesLabel->{"t (time)",None}];
Hughes-Hallett, Gleason, et al, Exercise 9.7.17d, page 537
In[1163]:=
Needs["Graphics`PlotField`"]
Clear[f,p,t,a,b,c,d];
(* dp/dt = f[p,t] = p (p - 6) *)
f[p_,t_] := p (p - 6);
a = 0; b = 8; (* a <= t <= b *)
c = 0; d = 8; (* c <= p <= d *)
pts = {};
field = PlotVectorField[{1,f[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "dp/dt = p (p - 6)",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 20,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Let's try to plot everything in the same picture.
In[1171]:=
Clear[p1,t,c1,d1,const,field,grows,diesOut];
p1[t_] := 6/(1 + 1/5 Exp[6t]);
c1 = 0; d1 = 1; (* c <= t <= d *)
diesOut = Plot[p1[t],{t,c1,d1},
PlotStyle->Red,
PlotRange->{0,8},
DisplayFunction->Identity];
Clear[p2,t,c2,d2,const];
p2[t_] := 6/(1 - 1/4 Exp[6t]);
c2 = 0; d2 = 0.1; (* c <= t <= d *)
grows = Plot[p2[t],{t,c2,d2},
PlotStyle->Red,
PlotRange->{0,10},
DisplayFunction->Identity];
Clear[f,p,t,a,b,c,d];
f[p_,t_] := p (p - 6);
a = 0; b = 1; (* a <= t <= b *)
c = 0; d = 10; (* c <= p <= d *)
field = PlotVectorField[{1,f[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "dp/dt = p (p - 6)",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 10,
Prolog -> ManganeseBlue,
AspectRatio->0.5,
DisplayFunction->Identity];
Show[{field,grows,diesOut},
DisplayFunction->\$DisplayFunction];
Could do a bit better with those funky arrowheads, but you get the idea.
Sustainable Yield
Hughes-Hallett, Gleason, et al, Exercises 9.7.19-21, pages 540-541
Let's look at the function which describes the rate of growth
of the fish population when there is no fishing.
In[1189]:=
Clear[r,p,c,d,k];
r[p_] := p (2 - k p);
k = 0.01; (* growth rate parameter *)
c = 0; d = 300; (* c <= p <= d *)
rate = Plot[r[p],{p,c,d},
PlotStyle->Red,
PlotLabel->"dp/dt = p (2 - k p)",
AxesLabel->{"p (population size)","dp/dt"}];
In the absence of fishing, the fish population should grow
when there are fewer than 200 fish, and it should decrease
when there are more than 200 fish. Let's see if we can get
a slope field to tell that story.
In[1194]:=
Needs["Graphics`PlotField`"]
Clear[rate,p,t,a,b,c,d];
rate[p_,t_] := p (2 - k p);
k = 0.01; (* growth rate parameter *)
removalRate = 75; (* fish per year *)
a = 0; b = 1; (* a <= t <= b *)
c = 199; d = 200.5; (* c <= p <= d *)
pts = {};
field = PlotVectorField[{1,rate[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "Growth of Fish Population\n dp/dt = p (2 - k p)",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 20,
AspectRatio->1.0,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Now we allow fishing. Fishermen (fisherfolk) remove 75 fish per year,
and this contributes to the rate of change in the growth of the fish population.
In[1203]:=
Clear[dp,p,c,d,k];
dp[p_] := p (2 - k p) - removalRate;
removalRate = 75;
k = 0.01; (* growth rate parameter *)
c = 0; d = 300; (* c <= t <= d *)
rateWithFishing = Plot[dp[p],{p,c,d},
PlotStyle->Red,
PlotLabel->"dp/dt = p (2 - k p) - removalRate",
AxesLabel->{"p (population size)","dp/dt"}];
Oops! -- big effect! Populations that start with fewer than 50 fish will be driven to extinction, while populations with more fish when the fishing season opens will tend towards the stable size of 150 fish -- lower than the 200 we saw earlier.
Here is a view of the associated slope field.
In[1209]:=
Clear[r2,p,t,a,b,c,d];
r2[p_,t_] := p (2 - k p) - removalRate;
k = 0.01; (* growth rate parameter *)
removalRate = 75; (* fish per year *)
a = 0; b = 6; (* a <= t <= b *)
c = 20; d = 160; (* c <= p <= d *)
pts = {{0,40},{0,60},{0,140},{0,160}};
field = PlotVectorField[{1,r2[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "Growth under Fishing Pressure\ndp/dt = p (2 - k p) - removalRate",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 12,
AspectRatio->1.0,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Now let's plot some particular solutions.
Here, p[0] = 40, 60, and 160.
In[1217]:=
Clear[p,t,p0,k,p40,p60,p140,p160,c,d]
p0 = 60; (* p0 = initial population size *)
k = 1 - 100/(p0 - 50);
k60 = 1 - 100/(60 - 50);
k140 = 1 - 100/(140 - 50);
k160 = 1 - 100/(160 - 50);
p60[t_] := 50 + 100/(1 - k60 Exp[-t])
p140[t_] := 50 + 100/(1 - k140 Exp[-t])
p160[t_] := 50 + 100/(1 - k160 Exp[-t])
c = 0; d = 6; (* c <= t <= d *)
plots = Plot[{p60[t],p140[t],p160[t]},{t,c,d},
PlotStyle->Red,
PlotLabel->"p[0] = 60, 140, and 160",
PlotRange->{40,160},
AxesLabel->{"t (time)","p (population size)"}];
Here is an example in which the population dies out; set p[0] = 40.
In[1228]:=
c = 0; d = 1.2; (* c <= t <= d *)
k40 = 1 - 100/(40 - 50);
p40[t_] := 50 + 100/(1 - k40 Exp[-t])
pop40 = Plot[p40[t],{t,c,d},
PlotStyle->Red,
PlotRange->{0,50},
PlotLabel->"p[0] = 40",
AxesLabel->{"t (time)","p (population size)"}];
Can we get all of this into one picture?
In[1232]:=
Show[{field,plots,pop40},
DisplayFunction->\$DisplayFunction];
Now consider the effect of a harvesting rate of H = 100 fish per year.
In[1233]:=
Clear[dp,p,c,d,k];
dp[p_] := p (2 - k p) - removalRate;
removalRate = 100;
k = 0.01; (* growth rate parameter *)
c = 0; d = 200; (* c <= t <= d *)
rateWithFishing = Plot[dp[p],{p,c,d},
PlotStyle->Red,
PlotLabel->"dp/dt = p (2 - k p) - removalRate",
AxesLabel->{"p (population size)","dp/dt"}];
There is a single root at p = 100, and for any other value of p the derivative dp/dt is negative. Thus, if the fishing rate is 100 fish per year, a population with more than 100 fish will approach an equilibrium population size of 100 fish. Any smaller population will die out.
In[1239]:=
Needs["Graphics`PlotField`"]
Clear[rate,p,t,a,b,c,d];
rate[p_,t_] := p (2 - k p) - removalRate;
k = 0.01; (* growth rate parameter *)
removalRate = 100; (* fish per year *)
a = 0; b = 10; (* a <= t <= b *)
c = 90; d = 110; (* c <= p <= d *)
pts = {};
field = PlotVectorField[{1,rate[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "Growth of Fish Population\n dp/dt = p (2 - k p) - 100",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 20,
AspectRatio->1.0,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Finally, we consider the effect of a harvesting rate of H = 200 fish per year.
In[1248]:=
Clear[dp,p,c,d,k];
dp[p_] := p (2 - k p) - removalRate;
removalRate = 200;
k = 0.01; (* growth rate parameter *)
c = 0; d = 200; (* c <= t <= d *)
rateWithFishing = Plot[dp[p],{p,c,d},
PlotStyle->Red,
PlotLabel->"dp/dt = p (2 - k p) - removalRate",
AxesLabel->{"p (population size)","dp/dt"}];
The derivative dp/dt is always negative! Thus, if the fishing rate is 200 fish per year, any population of fish will die out, no matter how large the initial population.
In[1254]:=
Needs["Graphics`PlotField`"]
Clear[rate,p,t,a,b,c,d];
rate[p_,t_] := p (2 - k p) - removalRate;
k = 0.01; (* growth rate parameter *)
removalRate = 200; (* fish per year *)
a = 0; b = 10; (* a <= t <= b *)
c = 90; d = 110; (* c <= p <= d *)
pts = {};
field = PlotVectorField[{1,rate[p,t]},
{t,a,b},{p,c,d},
PlotLabel -> "Growth of Fish Population\ndp/dt = p (2 - k p) - 200",
Axes -> True,
AxesLabel -> {"t","p"},
PlotPoints -> 10,
AspectRatio->1.0,
Prolog -> ManganeseBlue,
Epilog -> {Red,PointSize[0.02],
Map[Point,pts]}];
Created by Mathematica (April 25, 2004) | 4,761 | 12,907 | {"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} | 4.03125 | 4 | CC-MAIN-2017-39 | latest | en | 0.717051 |
https://math.stackexchange.com/questions/2647884/some-questions-about-a-3-times-3-real-skew-symmetric-matrix | 1,596,719,347,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00568.warc.gz | 452,482,650 | 35,715 | # Some questions about a $3 \times 3$ real skew-symmetric matrix
Let $$A = -A^T$$ be a $$3 \times 3$$ real skew-symmetric matrix.
1. Prove that $$A$$ has an eigenvalue $$\lambda = 0$$ and two other pure imaginary eigenvalues, and that the eigenvectors of $$A$$ are orthogonal in $$\mathbb{C}^3$$.
2. Find $$e^A$$, with: $$A = \left[ \begin{matrix} 0 & 1 & 0 \\ -1 &0 & 2 \\ 0 & -2 & 0 \end {matrix} \right ]$$
I've found a similar question here: Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$
In one of the answers it is said: "In general, a real skew-symmetric 3×3 matrix K looks like: $$\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end {matrix} \right ]$$ "
I don't understand why the general form of the matrix has that diagonal of zeros. I would have thought about something like: $$\left[ \begin{matrix} d & a & b \\ -a & e & c \\ -b & -c & f \end {matrix} \right ]$$
Wich of course has much worse determinant & characteristic polynomial...
Note that$$-\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}^T=\begin{pmatrix}-d & a & b \\ -a & -e & c \\ -b & -c & -f\end{pmatrix},$$which will be equal to$$\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}$$if and only if $d=e=f=0$. Can you take it from here?
• That's right! Thank you! I think I can manage to do the first part, I'll have some problems with the exponential of the matrix – Gitana Feb 12 '18 at 20:19
• Sorry I have some problems also with the eigenvectors part. I can't find an easy way to show that they are orthogonal to each other! – Gitana Feb 12 '18 at 21:04
• maybe I've got something: $( A \vec{v} | \vec{w}) = (\vec{v}| A^T \vec{w})=(\vec{v}|-A^T \vec{w})$ then if $\vec{v}, \vec{w}$ are eigenvectors for, respectively, $\lambda_0 = 0$ and $\lambda_1 = 1$, $( \lambda_0 \vec{v} | \vec{w}) = (\vec{v}| - \lambda_1 \vec{w})$, wich means that $(\lambda_0+lambda_1)(\vec{v}|\vec{w})=0$ and that's never true unless $v \perp w$ (the same for the others eigenvalues) – Gitana Feb 12 '18 at 21:10
• @Gitana You'll find a proof of a more general statement here. – José Carlos Santos Feb 12 '18 at 21:16
• @ José Thank you, very useful proof, and to solve the exponential? I was thinking about something like $e^A=Ve^\lambda V^{-1}$ but I have some difficulties in the determination of the eigenvectors. Is there a simpler way? $e^\lambda$ where $\lambda$ is the matrix of the eigenvaules – Gitana Feb 12 '18 at 21:39
$iA$ is a Hermitian matrix.
Hermitian matrices are always diagonalizable and always have real eigenvalues, and have orthogonal eigen-vectors.
The eigenvalues of $A$ are either $0$ or pure imaginary.
The characteristic equation of $A$ will be a degree 3 polynomial with real coefficients. It must have one real root. And it must be $0.$
If you negate every value on the diagonal, you must have the same value. That is only true of zero. Therefore you get 0 on the diagonal.
The eigenvalues of a real matrix must be either real or in complex conjugate pairs. If you have three eigenvalues, one of them must be a real one. Since you can prove, that skew syymetric matrices have all imaginary eigenvalues, you can directly see, that there is only one number, that is both on the real and the imaginary axis. That is 0. | 1,040 | 3,282 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-34 | latest | en | 0.765563 |
https://msuperl.org/wikis/pcubed/doku.php?id=184_notes:e_and_b | 1,627,875,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00426.warc.gz | 414,506,471 | 7,806 | 184_notes:e_and_b
Thus far in this course, we have considered the electric and magnetic fields completely separately, either only looking at the effects of an electric field by itself or a magnetic field by itself. However, there are many real-world contexts where a charge may be moving in a magnetic field and also near other charges. This means the charge would feel both an electric force and a magnetic force. Through Newton's second law (\$\vec{F}_{net}=\vec{F}_1+\vec{F}_2+…), we can think about how the combination of these forces affects individual charges. Using the magnetic and electric force is one way that we can think about combining electric and magnetic fields. Note that this is not a direct relationship between electric field and magnetic field, but rather relies on using force. The notes this week are going to focus on a more fundamental (and direct) relationship between electric and magnetic fields, which hinges on a changing magnetic field rather than a constant magnetic field. So the real question is:
• 184_notes/e_and_b.txt | 217 | 1,057 | {"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-2021-31 | latest | en | 0.95137 |
https://socratic.org/questions/how-do-you-find-the-critical-points-for-f-x-6x-2-3-x-5-3 | 1,701,870,613,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00746.warc.gz | 594,441,311 | 6,221 | # How do you find the critical points for f(x)= 6x^(2/3) + x^(5/3)?
Jan 12, 2016
$x = - \frac{12}{5} , 0$
#### Explanation:
Use the power rule.
$f ' \left(x\right) = \frac{2}{3} \left(6 {x}^{- \frac{1}{3}}\right) + \frac{5}{3} {x}^{\frac{2}{3}}$
$f ' \left(x\right) = \frac{4}{x} ^ \left(\frac{1}{3}\right) + \frac{5}{3} {x}^{\frac{2}{3}}$
$f ' \left(x\right) = \frac{4}{x} ^ \left(\frac{1}{3}\right) + \frac{5 x}{3 {x}^{\frac{1}{3}}}$
$f ' \left(x\right) = \frac{12 + 5 x}{3 {x}^{\frac{1}{3}}}$
Notice that writing the derivative in this form makes finding critical values much simpler.
Critical values occur when the derivative either equals zero (the numerator equals zero) or when the derivative does not exist (the denominator equals zero).
$12 + 5 x = 0$
$x = - \frac{12}{5}$
This is a critical value.
$3 {x}^{\frac{1}{3}} = 0$
$x = 0$
This is the other critical value.
The function graphed:
graph{6x^(2/3) + x^(5/3) [-25.58, 32.16, -10.9, 17.97]}
Notice the maximum at $x = - \frac{12}{5}$ and sharp turn when $x = 0$. | 407 | 1,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-50 | latest | en | 0.524275 |
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TalkPHP » Pixelate algorithm using GD
11-17-2008, 05:10 AM #1 (permalink) The Frequenter Join Date: Sep 2007 Location: Denmark Posts: 352 Thanks: 8 Pixelate algorithm using GD Long time no script for here, so I decided that a function I was working on the other day might be useful for posting here. Its an pixelate algorithm written in PHP using GD, it includes a basic algorithm that works by coloring the pixel block the first color found in a block. The advanced algorithm iterates through all pixels and finds the average color in a block and colors it like that, through meaning more execution time. Source: PHP Code: ``` * @link http://www.tuxxedo.net/ * @copyright 2006+ * @license Microsoft Public License * * ======================================================================== */ function imagepixelate(\$im, \$blocksize, \$advanced = false) { if(!is_resource(\$im) || strtolower(get_resource_type(\$im)) != 'gd') { return; } \$blocksize = (integer) \$blocksize; \$sx = imagesx(\$im); \$sy = imagesy(\$im); if(\$blocksize <= 0) { return(false); } elseif(\$blocksize == 1) { return(true); } if(\$advanced) { for(\$x = 0; \$x < \$sx; \$x += \$blocksize) { for(\$y = 0; \$y < \$sy; \$y += \$blocksize) { \$colors = Array( 'alpha' => 0, 'red' => 0, 'green' => 0, 'blue' => 0, 'total' => 0 ); for(\$cx = 0; \$cx < \$blocksize; ++\$cx) { for(\$cy = 0; \$cy < \$blocksize; ++\$cy) { if(\$x + \$cx >= \$sx || \$y + \$cy >= \$sy) { continue; } \$color = imagecolorat(\$im, \$x + \$cx, \$y + \$cy); imagecolordeallocate(\$im, \$color); \$colors['alpha'] += (\$color >> 24) & 0xFF; \$colors['red'] += (\$color >> 16) & 0xFF; \$colors['green'] += (\$color >> 8) & 0xFF; \$colors['blue'] += \$color & 0xFF; ++\$colors['total']; } } \$color = imagecolorallocatealpha(\$im, \$colors['red'] / \$colors['total'], \$colors['green'] / \$colors['total'], \$colors['blue'] / \$colors['total'], \$colors['alpha'] / \$colors['total'] ); if(!@imagefilledrectangle(\$im, \$x, \$y, (\$x + \$blocksize - 1), (\$y + \$blocksize - 1), \$color)) { return(false); } } } } else { for(\$x = 0; \$x < \$sx; \$x += \$blocksize) { for(\$y = 0; \$y < \$sy; \$y += \$blocksize) { if(!@imagefilledrectangle(\$im, \$x, \$y, (\$x + \$blocksize - 1), (\$y + \$blocksize - 1), imagecolorat(\$im, \$x, \$y))) { return(false); } } } } return(true); }?> ``` (Note that GD 2.0.1 or greater is required for this to work, but the chances that you're using GD 2.0.1 or greater is very high) The code is licensed under the Ms-Pl license which can be found at: http://www.opensource.org/licenses/ms-pl.html Prototype: `imagepixelate(resource \$image, integer \$blocksize [, boolean \$advanced = false])` The \$image parameter must be a resource valid resource, invalid resources NULL is returned. The \$blocksize parameter defines the blocksize in pixels. If a blocksize of 0 or lower is supplied the function will return false. Note that a blocksize of 1 wouldn't have any effect as each block already is one pixel so therefore the function just returns true. The \$advanced parameter is used to specify which algorithm to use, by default its using the simple algorithm, descriptions can be found above. Return values, NULL on invalid resource specified, FALSE on error and TRUE on success. Example code (basic): PHP Code: ``` ``` Example code (advanced): PHP Code: ``` ``` Example output (basic): Example output (advanced): A clear difference is visable, enjoy! :) C Version I've made a version, also written for gd. So in order to get it to work you have to be a minor libgd hacker. The C code is as follows: c Code: `int gdImagePixelate(gdImagePtr im, int blocksize, int advanced){ int x, y; if (blocksize <= 0) { return 0; } else if (blocksize == 1) { return 1; } if (advanced) { int a, r, g, b, c; int total; int cx, cy; for (x = 0; x < im->sx; x += blocksize) { for (y = 0; y < im->sy; y += blocksize) { a = r = g = b = c = total = 0; for (cx = 0; cx < blocksize; cx++) { for (cy = 0; cy < blocksize; cy++) { if (gdImageBoundsSafe(im, (x + cx), (y + cy)) == 0) { continue; } c = gdImageGetPixel(im, (x + cx), (y + cy)); gdImageColorDeallocate(im, c); a += (c >> 24) & 0xFF; r += (c >> 16) & 0xFF; g += (c >> 8) & 0xFF; b += c & 0xFF; total++; } } c = gdImageColorAllocateAlpha(im, r / total, g / total, b / total, a / total); gdImageFilledRectangle(im, x, y, (x + blocksize - 1), (y + blocksize - 1), c); } } return 1; } for (x = 0; x < im->sx; x += blocksize) { for(y = 0; y < im->sy; y += blocksize) { gdImageFilledRectangle(im, x, y, (x + blocksize - 1), (y + blocksize - 1), gdImageGetPixel(im, x, y)); } } return 1;}` Put that in the bottom of gd.c, and in the gd.h header after the other function prototypes then add the following: c Code: `int gdImagePixelate(gdImagePtr im, int blocksize, int advanced);` Its usage are exactly the same as the php version, except for this is on C level and much faster ofcourse. Furthermore for people who also hacks php a little you can alter the gd binding to support this function, I assume people who do this knows how the basic PHPAPI works so I'll only post the PHP_FUNCTION macro's body: c Code: `/* {{{ proto bool imagepixelate(resource im, int blocksize [, bool advanced]) Applies pixelation to the image */PHP_FUNCTION(imagepixelate){ zval *IM; gdImagePtr im; long blocksize; zend_bool advanced = 0; int retval; if (zend_parse_parameters(ZEND_NUM_ARGS() TSRMLS_CC, "rl|b", &IM, &blocksize, &advanced) == FAILURE) { return; } ZEND_FETCH_RESOURCE(im, gdImagePtr, &IM, -1, "Image", le_gd); retval = gdImagePixelate(im, (int) blocksize, advanced); if (retval) { RETURN_TRUE; } else { RETURN_FALSE; }}/* }}} */` I've submitted it as a feature request to GD itself and yet again, enjoy! :) __________________ Last edited by Wildhoney : 11-23-2008 at 04:13 PM.
11-17-2008, 09:02 AM #2 (permalink) The Acquainted Join Date: Nov 2007 Location: Netherlands Posts: 113 Thanks: 11 Nice script, but can't you just scale an image down and then scale it up to get the same effect?...
11-17-2008, 11:49 AM #3 (permalink) The Prestige Join Date: Sep 2007 Location: Sweden, Stockholm Posts: 1,080 Thanks: 115 I'm sorry if this is a weird question, but I really don't understand what this is for :O __________________
11-17-2008, 07:38 PM #4 (permalink) Moderateur Join Date: Apr 2007 Posts: 1,393 Thanks: 5 A similar technique using imagecopyresized/imagecopyresampled (with a block size of 5) would produce something like: Original Non-advanced (uses imagecopyresized) Advanced (uses imagecopyresampled) Good work with the function Kalle, did you first write it in 2006 (looking at the copyright).
11-17-2008, 09:13 PM #5 (permalink) The Frequenter Join Date: Sep 2007 Location: Denmark Posts: 352 Thanks: 8 I initially wrote the simple pixelate effect in late 2006 yea, but the advanced version was written a few weeks ago :) @Tanax im an image effect, could for example be used to style an image dynamiclly, or maybe on an verification image to blur it. @Sjaq in theory I think its possible yea, but I didn't really look at it. __________________
11-17-2008, 09:15 PM #6 (permalink) The Prestige Join Date: Sep 2007 Location: Sweden, Stockholm Posts: 1,080 Thanks: 115 Okey, I understand now what it does. Now I have a new question. Why would you want to do this? I mean, the original image is better quality? :O __________________
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Quote:
Originally Posted by Tanax Okey, I understand now what it does. Now I have a new question. Why would you want to do this? I mean, the original image is better quality? :O
To make it look kool of coarse...
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Quote:
Originally Posted by Tanax Okey, I understand now what it does. Now I have a new question. Why would you want to do this? I mean, the original image is better quality? :O
The point of the effect isn't the quality, its to do what you need. I've personally written a couple of smaller image editing programs in php which enables me to manipulate images. For example if you wanna use pixelization (pixelate censur), you can easily perform this dynamically with php by using the function.
Theres no exact usage for it, as people's need vary :)
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11-23-2008, 04:04 PM #9 (permalink) The Frequenter Join Date: Sep 2007 Location: Denmark Posts: 352 Thanks: 8 Updated the original post with a C version for use with GD, enjoy! :) __________________
06-27-2009, 12:36 AM #10 (permalink) The Frequenter Join Date: Sep 2007 Location: Denmark Posts: 352 Thanks: 8 As an update for this, since PHP 5.3.0 RC1, this pixelation algorithm have been apart of the GD binding for PHP, and its available as of GD 2.1.0. For more information and documentation see: http://php.net/imagefilter __________________
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http://www.osti.gov/eprints/topicpages/documents/record/186/0082241.html | 1,398,020,220,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00171-ip-10-147-4-33.ec2.internal.warc.gz | 601,318,033 | 3,130 | Predicative Categories of Classes Michael Alton Warren Summary: Predicative Categories of Classes Michael Alton Warren October 28, 2004 Contents Abstract iii Acknowledgments iv Introduction v 1 -Pretopoi and Constructive Set Theories 1 1.1 -Pretopoi and Dependent Type Theory . . . . . . . . . . . . 1 1.1.1 -Pretopoi . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1.2 Dependent Type Theory . . . . . . . . . . . . . . . . . 2 1.1.3 Soundness and Completeness . . . . . . . . . . . . . . 4 1.2 Constructive Set Theories . . . . . . . . . . . . . . . . . . . . 6 1.2.1 Notation and Axioms . . . . . . . . . . . . . . . . . . 6 1.2.2 Basic Properties . . . . . . . . . . . . . . . . . . . . . 8 1.2.3 The Category of Sets . . . . . . . . . . . . . . . . . . . 12 1.2.4 Axioms of Infinity . . . . . . . . . . . . . . . . . . . . 14 2 Predicative Categories of Classes 16 2.1 Categories with Class Structure . . . . . . . . . . . . . . . . . 16 2.1.1 Axioms for Categories with Basic Class Structure . . . 16 Collections: Mathematics | 396 | 1,034 | {"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-2014-15 | longest | en | 0.306777 |
https://www.coursehero.com/file/166776/LRC-lab-writeup/ | 1,490,725,551,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189802.72/warc/CC-MAIN-20170322212949-00405-ip-10-233-31-227.ec2.internal.warc.gz | 897,407,673 | 71,597 | LR(C) lab writeup
# LR(C) lab writeup - us ln(V/V ) 1.7 1 4-0.53063 0.5...
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Saad Ghazipura PHY 134 Section 2 LR(C) Circuits Lab Abstract In this lab, the behavior of series LR circuits and LRC circuits will be observed. An LRC circuit is an electrical circuit that contains a resistor (R), inductor (L), and a capacitor. It can either be oriented in series or parallel. This forms a harmonic oscillator. The equations defining the circuit are as such: . The purpose of the 1 Ω resistor is to prevent the variable resistor from the oscilloscope to affect the readings on the oscilloscope. Main Analysis see attached sheets Data: R=100 Voltage(mV) ±.05 mV Time (us) ±2
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Unformatted text preview: us ln(V/V ) 1.7 1 4-0.53063 0.5 12-1.22378 0.2 20-2.14007 0.1 30-2.83321 \ R=200 Voltage(mV) .05 mV Time (us) 2 us ln(V/V ) 1.8 1.0 10-0.2552 7 0.50 20-0.5563 0.30 30-0.7781 5 0.15 40-1.0791 8 0.01 50-2.2552 7 Conclusion The RLC circuit and RC circuit were studied in this lab. In the RLC circuit, a resonant frequency was observed for the RLC circuit. For the RC circuit, the battery was charging giving a negative slope for the graph. Sources of error include uncertainty in reading the oscilloscope and the ruler....
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## This note was uploaded on 04/28/2008 for the course PHY 132 taught by Professor Rijssenbeek during the Spring '04 term at SUNY Stony Brook.
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LR(C) lab writeup - us ln(V/V ) 1.7 1 4-0.53063 0.5...
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The auto industry has experienced one of its most significant trends
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The auto industry has experienced one of its most significant trends [#permalink]
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The auto industry has experienced one of its most significant trends in the last 50 years, which is the migration of motorists from passenger cars to minivans, sport utility vehicles, and pickups.
(A) The auto industry has experienced one of its most significant trends in the last 50 years, which is
(B) Of the trends the auto industry experienced in the last 50 years has been one of the most significant
(C) In the last 50 years, one of the most significant trends that the auto industry has been experiencing has been
(D) One of the most significant trends that the auto industry has experienced in the last 50 years is
(E) In the last 50 years, the auto industry experienced one of the most significant trends that it has had, that of
Originally posted by grad_mba on 15 May 2007, 07:04.
Last edited by Bunuel on 26 Jun 2018, 04:59, edited 2 times in total.
Edited the question.
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Re: The auto industry has experienced one of its most significant trends [#permalink]
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15 May 2007, 07:52
The auto industry has experienced one of its most significant trends in the last 50 years, which is the migration of motorists from passenger cars to minivans, sport utility vehicles, and pickups.
(A) The auto industry has experienced one of its most significant trends in the last 50 years, which is
ambiguous since 'which' refers to years or trends
(B) Of the trends the auto industry experienced in the last 50 years has been one of the most significant
'Of the trends...has been' is wordy and awkward
(C) In the last 50 years, one of the most significant trends that the auto industry has been experiencing has been
'has been expieriencing' means a continuous action, wrong
(D) One of the most significant trends that the auto industry has experienced in the last 50 years is
correct, short and clear
(E) In the last 50 years, the auto industry experienced one of the most significant trends that it has had, that of
awkward and redundant. the meaning changes: the industry expierienced a trend for 50 years.
Director
Joined: 13 Mar 2007
Posts: 532
Schools: MIT Sloan
Re: The auto industry has experienced one of its most significant trends [#permalink]
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15 May 2007, 14:45
Can one of you explain the subject verb agreement in here ?
"one of the trends" doesnt it require a plural verb?
confused here !
Senior Manager
Joined: 19 Feb 2007
Posts: 302
Re: The auto industry has experienced one of its most significant trends [#permalink]
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15 May 2007, 20:38
Can one of you explain the subject verb agreement in here ?
"one of the trends" doesnt it require a plural verb?
confused here !
No.
One of the trends.......requires singular(has)
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Location: India
Re: The auto industry has experienced one of its most significant trends [#permalink]
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25 Jun 2018, 10:01
The auto industry has experienced one of its most significant trends in the last 50 years, which is the migration of motorists from passenger cars to minivans, sport utility vehicles, and pickups.
(A) The auto industry has experienced one of its most significant trends in the last 50 years, which is
(B) Of the trends the auto industry experienced in the last 50 years has been one of the most significant
(C) In the last 50 years, one of the most significant trends that the auto industry has been experiencing has been
(D) One of the most significant trends that the auto industry has experienced in the last 50 years is
(E) In the last 50 years, the auto industry experienced one of the most significant trends that it has had, that of
Thanks,
GyM
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Re: The auto industry has experienced one of its most significant trends [#permalink]
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26 Jun 2018, 04:52
1
Not sure if the "is" is part of the underline or not
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Posts: 51097
Re: The auto industry has experienced one of its most significant trends [#permalink]
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26 Jun 2018, 04:59
Amirfunc wrote:
Not sure if the "is" is part of the underline or not
Yes, it is underlined. Edited. Thank you.
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The auto industry has experienced one of its most significant trends
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,562 | 6,286 | {"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-2018-51 | latest | en | 0.926455 |
https://www.printablemultiplication.com/multiplication-facts-1-12-printable/ | 1,621,101,241,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00097.warc.gz | 1,006,284,483 | 78,613 | # Multiplication Facts 1 12 Printable
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Multiplication worksheets are incredibly useful for mothers and fathers and kids likewise. They will rely on them to discover and practice distinct math specifics and fix troubles daily. They are going to produce math skills whilst having fun. You will certainly be surprised by your kid’s upgrades following with such worksheets. | 648 | 3,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-21 | longest | en | 0.953765 |
https://www.physicsforums.com/threads/norms-question-parallelogram-law.471205/ | 1,701,356,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00103.warc.gz | 1,040,853,778 | 14,756 | # Norms question (parallelogram law)
• tourjete
In summary, the conversation discusses the concept of a norm in the context of a vector space of continuous complex-valued functions. The norm is defined as the maximum absolute value of the function over the interval [a,b]. The conversation also explores the parallelogram law and how it relates to the norm, showing that the norm does not satisfy the parallelogram law and is therefore not an inner-product norm. The attempt at a solution involves using trigonometric functions to define the norm but results in incorrect calculations.
## Homework Statement
Consider the vector space C[a,b] of all continuous complex-valued functions f(x), x$$\in$$ [a,b]. Define a norm ||f|| = max{|f(x)|, x$$\in$$ [a,b]]
a) show that this is a norm
b) Show that this norm does not satisfy the parallelogram law, thereby showing that its not an inner-product norm.
## Homework Equations
Parallelogram law: ||x-y||$$^{2}$$ + ||x+y||$$^{2}$$ = 2||x||$$^{2}$$ + 2||y||$$^{2}$$
## The Attempt at a Solution
I'm mostly having trouble defining the norm. I'm a little unclear on what the concept of a norm is; we only went over inner-product norms in class. I draw the vector going from the origin to the maximum on [a,b] and to define the norm I wrote ||f|| = $$\sqrt{(([vcos])^2 + ([vsin])^2}$$ since that would make it always positive. When I used the parallelogram law, I used x = vcos(theta) and y = vsin(theta). However, I clearly defined the dorm wrong since I got that the two sides of the equation equaled each other.
Is there another way to define a norm? Am I choosing x and y in the parallelogram law wrong?
Last edited:
hi norm!
i don't understand what you're doing
if eg f(x) = x, g(x) = -x, then ||f+g|| = 0, but ||f|| = ||g|| = b | 473 | 1,784 | {"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} | 3.90625 | 4 | CC-MAIN-2023-50 | latest | en | 0.898704 |
https://essaysontime.us/solved-writing-a-code-programming/ | 1,670,518,335,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00646.warc.gz | 257,345,176 | 27,307 | Select Page
Display your results for each question in a new line.Write an explanation of your code for each line using comments. If the explanation is not clear, you will NOT receive full credit. Use the Developer mode of your
browser to access the JavaScript command line. You can edit your code in a
separate file and then just paste it into the command line to run it. You will be submitting the file with
JavaScript.
This is
the link to the first part. There are 6 parts and you can get to other parts
https://medium.com/@cscalfani/so-you-want-to-be-a-…(5 points) Start with an array called inputtable. The array should have numbers between 1 and 10.(30 points) Use inputtable from step 1 to create the following: – Set of multiples of 5 between 1 and 51. Name it fiveTableSet of multiples of 13 between 1 and 131. Name it thirteenTableSet of squares of the numbers in inputtable. Name it squaresTable(10 points) Get the odd multiples of 5 between 1 and 100. 5, 15, … (20 points) Get the sum of even multiples of 7 between 1 and 100. Example, find the multiples and then sum them: 14 + 28+… (15 points) Use currying to rewrite the function below: -(15 points) Use the following code to take advantage of closures to wrap content with HTML tags, specifically show an HTML table consisting of a table row that has at least one table cell/element. You can use the console to output your results.(5 points) Following instructions(Extra credit) Do the generic version of questions 3 and 4, meaning the target multiple must not be hard coded – first odd or even and then the number whose multiples (in range 1 to 100) you want. NOTE: Do NOT use a form of a ‘for’ loop anywhere, including iterators.This is meant to be a functional exercise. function cylinder_volume(r, h){ var volume = 0.0; volume = 3.14 * r * r * h; return volume; } Use r = 5 and h = 10 to call your curried function. makeTag = function(beginTag, endTag){ return function(textcontent){ return beginTag +textcontent +endTag; } } Example output for #6. Note that the
tag is optional. Please do not use this data, but substitute your own values for the contents of the cells.
Firstname Lastname Age
Jill Smith 50
Eve Jackson 94
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https://www.scribd.com/document/11146474/11147327 | 1,566,743,872,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330233.1/warc/CC-MAIN-20190825130849-20190825152849-00445.warc.gz | 963,494,803 | 56,829 | You are on page 1of 5
# Theorems and Postulates 21/11/2008 17:49:00
Postulates:
Postulate 1: Ruler postulate- The points on a line can be paired with real numbers in such a
way that any two points can have the coordinates 0 and 1. Once a coordinate system has
been chosen in this way, the distance between any two points equals the absolute value of
the difference of the coordinates.
Postulate 2: Segment Addition postulate- States that segment of lines can be added.
Postulate 3: Protractor postulate- On AB in a given plane chose any point O between A and
B. Consider OA and OB all the rays that can be drawn from O on one side of AB. These rays
can be paired with real numbers from 0 to 180 in such a way that (1) OA is paired with 0,
and OB with 180 (2) If OP is paired with x and OQ with y, then POQ = |x-y|.
Postulate 5: A line contains at least two points; a plane three noncollinear points; and space
four points noncoplanar.
Postulate 6: Through any two points there is exactly one line.
Postulate 7: Through any three points there is exactly one plane, and through any three
noncollinear points there is exactly one plane.
Postulate 8: If two points are in a plane, then the line that contains those two points is also
in that plane.
Postulate 9: If two planes intersect, their intersection is a line.
Postulate 10: If two parallel lines are cut by a transversal, then the corresponding angles are
congruent.
Postulate 11: If two lines are cut by a transversal, and the corresponding angles are
congruent, the lines are parallel.
Postulate 12: SSS; if three sides of one triangle are congruent to three sides of another
triangle, then the triangles are congruent.
Postulate 13: SAS; if two sides and the included angle of one triangle are congruent to the
other triangles, then the triangles are congruent.
Postulate 14: ASA; If two angles and the included side of one triangle are congruent to the
other triangles, then the two triangles are congruent.
Theorems:
POINTS, LINES, PLANES, AND ANGLES
Theorem 1-1: If two lines intersect, then they intersect in exactly one point.
Theorem 1-2: Through a line and a point not in the line, there is exactly one plane.
Theorem 1-3: If two lines intersect, then exactly one plane contains the lines.
DEDUCTIVE REASONING
Theorem 2-1: (Midpoint Theorem)- If M is the midpoint of AB, then AM = ½ AB and MB = ½
AB.
Theorem 2-2: (Angle Bisector Theorem)- If BX is the bisector of angle ABC, then ABX = ½
ABC and XBC = ½ ABC.
Theorem 2-3: Vertical angles are congruent.
Theorem 2-4: If two lines are perpendicular, then they form congruent adjacent angles.
Theorem 2-5: If two lines form congruent adjacent angles, then the lines are perpendicular.
Theorem 2-6: If the exterior sides of two adjacent acute angles are perpendicular, then the
angles are complementary.
Theorem 2-7: If two angles are supplements of congruent angles (or of the same angle),
then the two angles are congruent.
Theorem 2-8: If two angles are complements of congruent angles (or of the same angle),
then the two angles are congruent.
PARALLEL LINES AND PLANES
Theorem 3-1: If two parallel planes are cut by a third plane, then the lines of intersection are
parallel.
Theorem 3-2: If two parallel lines are cut by a transversal, then AIA are congruent.
Theorem 3-3: If two parallel lines are cut by a transversal, then SSI angles are
supplementary.
Theorem 3-4: If a transversal is perpendicular to one of two parallel lines, then it is
perpendicular to the other line as well.
Theorem 3-5: Two lines are cut by a transversal and the AIA are congruent, then the lines
are parallel.
Theorem 3-6: If two lines are cut by a transversal and SSI angles are supplementary then
the lines are parallel.
Theorem 3-7: In a plane two lines perpendicular to the same line are parallel.
Theorem 3-8: Through a point outside a line, there is exactly one parallel to the given line.
Theorem 3-9: Through a point outside a line, there is exactly one perpendicular to the given
line.
Theorem 3-10: Two lines parallel to a third line are parallel to each other.
Theorem 3-11: The sum of the measures of the angles of a triangle is 180 degrees.
C 1: If two angles of one triangle are congruent to two angles of another triangle, then the
third angles are congruent.
C 2: Each angle of an equiangular triangle has a measure of 60 degrees.
C 3: In a triangle, there can be at most one right angle or obtuse angle.
C 4: The acute angles of a right triangle are complementary.
Theorem 3-12: The measure of an exterior angle of a triangle equals the sum of the
measures of the two remote interior angles.
Theorem 3-13: The sum of the measures of the angles of a convex polygon with n sides is
(n-2)180.
Theorem 3-14: The sum of the measures of the exterior angles of any convex polygon, one
angle at each vertex, is 360 degrees.
CONGRUENT TRIANGLES
Theorem 4-1: If two sides of a triangle are congruent, then the angles opposite those sides
are congruent.
C 1: An equilateral triangle is also equiangular.
C 2: An equilateral triangle has three 60 degree angles.
C 3: The bisector of the vertex angle of an isosceles triangle is perpendicular to the base at
its midpoint.
Theorem 4-2: Converse of theorem 4-1.
Theorem 4-3: AAS; if two angles and a non-included side of one triangle are congruent to
the corresponding parts of another right triangle, then they are congruent.
Theorem 4-4: HL; if the hypotenuse and a leg on a right triangle, are congruent to that of
another, then the two triangles are congruent.
Theorem 4-5: If a point lies on the perpendicular bisector of a segment, then the point is
equidistant from both ends of the segment.
Theorem 4-6: Converse of theorem 4-5.
Theorem 4-7: If a point is equidistant from the endpoints of a segment, then the point is
equidistant from the sides of the angle.
Theorem 4-8: If a point is equidistant from the sides of an angle, then the point lies on the
bisector of the angle.
Theorem 5-1: Opposite sides of a parallelogram are congruent
Theorem 5-2: Opposite angles of a parallelogram are congruent.
Theorem 5-3: Diagonals of a parallelogram bisect each other.
Theorem 5-4: If both pairs of opposite sides of a quadrilateral are congruent, then it is a
parallelogram.
Theorem 5-5: If one pair of opposite sides is both congruent and parallel, then the
Theorem 5-6: If both pairs of opposite angles of a quadrilateral are congruent, then it is a
parallelogram.
Theorem 5-7: If the diagonals of a parallelogram bisect each other, then the quadrilateral is
a parallelogram.
Theorem 5-8: If two lines are parallel, then all points on one line are equidistant from the
other line.
Theorem 5-9: If three parallel lines are cut off into congruent segments on one transversal,
then they cut off congruent segments in every transversal.
Theorem 5-10: A line that contains the midpoint of one side of a triangle and is parallel to
another side passes thought the midpoint of the third side.
Theorem 5-11: The segment that joins the midpoints of two sides of a triangle (1) is parallel
to the third side (2) is half as long as the third side.
Theorem 5-12: The diagonals of a rectangle are congruent.
Theorem 5-13: The diagonals of a rhombus are perpendicular.
Theorem 5-14: Each diagonal of a rhombus bisects two angles of the rhombus.
Theorem 5-15: The midpoint of the hypotenuse of aright triangle is equidistant from the
three vertices.
Theorem 5-16: If an angle of a parallelogram is a right angle, the parallelogram is a
rectangle.
Theorem 5-17: If two consecutive sides of a parallelogram are congruent, then the
parallelogram is a rhombus.
Theorem 5-18: Base angles of an isosceles trapezoid are congruent.
Theorem 5-19: The median of a trapezoid (1) Is parallel to the bases (2) has a length equal
to the average of the base lengths.
INEQUALITIES IN TRIANGLES
Theorem 6-1: The measure of the exterior of a triangle is greater than the measure of either
remote interior angle.
Theorem 6-2: If one side of a triangle is longer than a second side, then the angle opposite
the first side is larger than the angle opposite the second side.
Theorem 6-3: If one angles of a triangle is larger than a second angle, then the side
opposite the first angle is longer than the side opposite the second angle.
C 1: The perpendicular segment from a point to a line is the shortest segment to the
line.
C 2: The perpendicular segment from a point to a plane is the shortest segment to
the plane
Theorem 6-4: Triangle Inequality: The sum of the lengths of two sides of a triangle is always
greater than the length of the third.
Theorem 6-5: If two sides of one triangle are congruent to two sides of another triangle, but
the included angle of the first triangle is larger than the included angle of the second, then
the third side of the first triangle is longer than the third side of the second triangle.
Theorem 6-6: If two sides of one triangle are congruent t two sides of another triangle, but
the third side of the first triangle is longer that the third side of the second, then the
included angle of the first triangle is larger that the included angle of the second.
SIMILAR POLYGONS
Theorem 7-1: SAS ~ Theorem: If an angles of one triangle is congruent to an angle of
another triangle and the sides including those angles are in proportion, then the triangles
are similar.
Theorem 7-2: SSS ~ Theorem: If the sides of two triangles are in proportion, then the
triangles are similar.
Theorem 7-3: Triangle Proportionality Theorem: If a line parallel to one side of a triangle
intersects the other two sides, then it divides those sides proportionally.
Theorem 7-4: Triangle Angle-Bisector Theorem: If a ray bisects an angle of a triangle, hen I
divides the opposite side into segments proportional to the other two sides. | 2,598 | 9,773 | {"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-2019-35 | latest | en | 0.917955 |
http://en.wikipedia.org/wiki/Buffet_(turbulence) | 1,394,181,935,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999638988/warc/CC-MAIN-20140305060718-00086-ip-10-183-142-35.ec2.internal.warc.gz | 60,366,995 | 32,808 | # Stall (flight)
(Redirected from Buffet (turbulence))
For other uses, see stall (disambiguation).
Stall
Airflow separating from an airfoil at a high angle of attack, as occurs at the stall
In fluid dynamics, a stall is a reduction in the lift coefficient generated by a foil as angle of attack increases.[1] This occurs when the critical angle of attack of the foil is exceeded. The critical angle of attack is typically about 15 degrees, but it may vary significantly depending on the fluid, foil, and Reynolds number.
Stalls in fixed-wing flight are often experienced as a sudden reduction in lift as the pilot increases the wing's angle of attack and exceeds its critical angle of attack (which may be due to slowing down below stall speed in level flight). A stall does not mean that the engine(s) have stopped working, or that the aircraft has stopped moving — the effect is the same even in an unpowered glider aircraft. Vectored thrust in manned and unmanned aircraft is used to surpass the stall limit, thereby giving rise to post-stall technology.[2][3]
Because stalls are most commonly discussed in connection with aviation, this article discusses stalls as they relate mainly to aircraft, in particular fixed-wing aircraft. The principles of stall discussed here translate to foils in other fluids as well.
## Formal definition
A stall is a condition in aerodynamics and aviation wherein the angle of attack increases beyond a certain point such that the lift begins to decrease. The angle at which this occurs is called the critical angle of attack. This critical angle is dependent upon the profile of the wing, its planform, its aspect ratio, and other factors, but is typically in the range of 8 to 20 degrees relative to the incoming wind for most subsonic airfoils. The critical angle of attack is the angle of attack on the lift coefficient versus angle-of-attack curve at which the maximum lift coefficient occurs.[citation needed]
Flow separation begins to occur at small angles of attack while attached flow over the wing is still dominant. As angle of attack increases, the separated regions on the top of the wing increase in size and hinder the wing's ability to create lift. At the critical angle of attack, separated flow is so dominant that further increases in angle of attack produce less lift and vastly more drag.[citation needed]
A fixed-wing aircraft during a stall may experience buffeting or a change in attitude. Most aircraft are designed to have a gradual stall with characteristics that will warn the pilot and give the pilot time to react. For example, an aircraft that does not buffet before the stall may have an audible alarm or a stick shaker installed to simulate the feel of a buffet by vibrating the stick fore and aft. The "buffet margin" is, for a given set of conditions, the amount of ‘g’, which can be imposed for a given level of buffet. The critical angle of attack in steady straight and level flight can be attained only at low airspeed. Attempts to increase the angle of attack at higher airspeeds can cause a high-speed stall or may merely cause the aircraft to climb.[citation needed]
Any yaw of the aircraft as it enters the stall regime can result in autorotation, which is also sometimes referred to as a 'spin'. Because air no longer flows smoothly over the wings during a stall, aileron control of roll becomes less effective, whilst simultaneously the tendency for the ailerons to generate adverse yaw increases. This increases the lift from the advancing wing and accentuates the probability of the aircraft to enter into a spin.[citation needed]
Depending on the aircraft's design, a stall can expose extremely adverse properties of balance and control, in particular in a prototype.[citation needed]
An example of the relationship between angle of attack and lift on a cambered airfoil. The exact relationship is usually measured in a wind tunnel and depends on the airfoil section. The relationship for an aircraft wing depends on the planform and its aspect ratio.
## Graph
The graph shows that the greatest amount of lift is produced as the critical angle of attack is reached (which in early-20th century aviation was called the "burble point"). This angle is 17.5 degrees in this case but changes from airfoil to airfoil. In particular, for aerodynamically thick airfoils (thickness to chord ratios of around 10%), the critical angle is higher than with a thin airfoil of the same camber. Symmetric airfoils have lower critical angles (but also work efficiently in inverted flight). The graph shows that, as the angle of attack exceeds the critical angle, the lift produced by the airfoil decreases.
The information in a graph of this kind is gathered using a model of the airfoil in a wind tunnel. Because aircraft models are normally used, rather than full-size machines, special care is needed to make sure that data is taken in the same Reynolds number regime (or scale speed) as in free flight. The separation of flow from the upper wing surface at high angles of attack is quite different at low Reynolds number from that at the high Reynolds numbers of real aircraft. High-pressure wind tunnels are one solution to this problem. In general, steady operation of an aircraft at an angle of attack above the critical angle is not possible because, after exceeding the critical angle, the loss of lift from the wing causes the nose of the aircraft to fall, reducing the angle of attack again. This nose drop, independent of control inputs, indicates the pilot has actually stalled the aircraft.[4][5]
This graph shows the stall angle, yet in practice most pilot operating handbooks (POH) or generic flight manuals describe stalling in terms of airspeed. This is because all aircraft are equipped with an airspeed indicator, but fewer aircraft have an angle of attack indicator. An aircraft's stalling speed is published by the manufacturer (and is required for certification by flight testing) for a range of weights and flap positions, but the stalling angle of attack is not published.
As speed reduces, angle of attack has to increase to keep lift constant until the critical angle is reached. The airspeed at which this angle is reached is the (1g, unaccelerated) stalling speed of the aircraft in that particular configuration. Deploying flaps/slats decreases the stall speed to allow the aircraft to take off and land at a lower speed.
## Aerodynamic description of a stall
### Stalling a fixed-wing aircraft
A fixed-wing aircraft can be made to stall in any pitch attitude or bank angle or at any airspeed but is commonly practiced by reducing the speed to the unaccelerated stall speed, at a safe altitude. Unaccelerated (1g) stall speed varies on different fixed-wing aircraft and is represented by colour codes on the air speed indicator. As the plane flies at this speed, the angle of attack must be increased to prevent any loss of altitude or gain in airspeed (which corresponds to the stall angle described above). The pilot will notice the flight controls have become less responsive and may also notice some buffeting, a result of the turbulent air separated from the wing hitting the tail of the aircraft.
In most light aircraft, as the stall is reached, the aircraft will start to descend (because the wing is no longer producing enough lift to support the aircraft's weight) and the nose will pitch down. Recovery from this stalled state involves the pilot's decreasing the angle of attack and increasing the air speed, until smooth air-flow over the wing is restored. Normal flight can be resumed once recovery from the stall is complete.[6] The maneuver is normally quite safe and if correctly handled leads to only a small loss in altitude (50'-100'). It is taught and practised in order for pilots to recognize, avoid, and recover from stalling the aircraft.[7] A pilot is required to demonstrate competency in controlling an aircraft during and after a stall for certification,[8] and it is a routine maneuver for pilots when getting to know the handling of a new aircraft type. The only dangerous aspect of a stall is a lack of altitude for recovery.
A special form of asymmetric stall in which the aircraft also rotates about its yaw axis is called a spin. A spin can occur if an aircraft is stalled and there is an asymmetric yawing moment applied to it.[9] This yawing moment can be aerodynamic (sideslip angle, rudder, adverse yaw from the ailerons), thrust related (p-factor, one engine inoperative on a multi-engine non-centreline thrust aircraft), or from less likely sources such as severe turbulence. The net effect is that one wing is stalled before the other and the aircraft descends rapidly while rotating, and some aircraft cannot recover from this condition without correct pilot control inputs (which must stop yaw) and loading.[10] A new solution to the problem of difficult (or impossible) stall-spin recovery is provided by the ballistic parachute recovery system.
The most common stall-spin scenarios occur on takeoff (departure stall) and during landing (base to final turn) because of insufficient airspeed during these maneuvers. Stalls also occur during a go-around manoeuvre if the pilot does not properly respond to the out-of-trim situation resulting from the transition from low power setting to high power setting at low speed.[11] Stall speed is increased when the wing surfaces are contaminated with ice or frost creating a rougher surface, and heavier airframe due to ice accumulation.
Stalls do not derive from airspeed and can occur at any speed - but only if the wings have too high an angle of attack. Attempting to increase the angle of attack at 1g by moving the control column back normally causes the aircraft to climb. However, aircraft often experience higher g, for example when turning steeply or pulling out of a dive. In these cases, the wings are already operating at a higher angle of attack to create the necessary force (derived from lift) to accelerate in the desired direction. Increasing the g loading still further, by pulling back on the controls, can cause the stalling angle to be exceeded -even though the aircraft is flying at a high speed.[12] These "high-speed stalls" produce the same buffeting characteristics as 1g stalls and can also initiate a spin if there is also any yawing.
### Symptoms of an approaching stall
One symptom of an approaching stall is slow and sloppy controls. As the speed of the aircraft decreases approaching the stall, there is less air moving over the wing, and, therefore, less air will be deflected by the control surfaces (ailerons, elevator, and rudder) at this slower speed. Some buffeting may also be felt from the turbulent flow above the wings as the stall is reached. The stall warning will sound, if fitted, in most aircraft 5 to 10 knots above the stall speed.[13]
### Stalling characteristics
Different aircraft types have different stalling characteristics. A benign stall is one where the nose drops gently and the wings remain level throughout. Slightly more demanding is a stall in which one wing stalls slightly before the other, causing that wing to drop sharply, with the possibility of entering a spin. A dangerous stall is one in which the nose rises, pushing the wing deeper into the stalled state and potentially leading to an unrecoverable deep stall. This can occur in some T-tailed aircraft wherein the turbulent airflow from the stalled wing can blanket the control surfaces at the tail.[citation needed]
## Stall speed
Flight envelope of a fast airplane.
Left edge is the stall speed curve.
The airspeed indicator is often used to indirectly predict stall conditions.
Stalls depend only on angle of attack, not airspeed. However, the slower an airplane goes, the more angle of attack it needs to produce lift equal to the aircraft's weight. As the speed slows further, at some point this angle will be equal to the critical (stall) angle of attack. This speed is called the "stall speed". An aircraft flying at its stall speed cannot climb, and an aircraft flying below its stall speed cannot stop descending. Any attempt to do so by increasing angle of attack, without first increasing airspeed, will result in a stall.
The actual stall speed will vary depending on the airplane's weight, altitude, configuration, and vertical and lateral acceleration. Guidelines for the case of zero acceleration are provided by the following V speeds:
• VS: The computed stalling speed with flaps retracted at design speed. Often has the same value as VS1.
• VS0: The stall speed in landing configuration (full flaps, landing gear down, spoilers retracted).
• VS1: The stall speed in a "clean" configuration (flaps, landing gear and spoilers all retracted as far as possible).
• VSR: Reference stall speed.[clarification needed]
• VSR0: Reference stall speed in the landing configuration.
• VSR1: Reference stall speed in the clean configuration.
• VSW: Speed at which onset of natural or artificial stall warning occurs.
On an airspeed indicator, the bottom of the white arc indicates VS0 at maximum weight, while the bottom of the green arc indicates VS1 at maximum weight. While an aircraft's VS speed is computed by design, its VS0 and VS1 speeds must be demonstrated empirically by flight testing.[14]
## Accelerated and turning flight stall
Illustration of a turning flight stall, occurring during a co-ordinated turn with progressively increasing angle of bank.
The normal stall speed, specified by the VS values above, always refers to straight and level flight, where the load factor is equal to 1g. However, if the aircraft is turning or pulling up from a dive, additional lift is required to provide the vertical or lateral acceleration, and so the stall speed is higher. An accelerated stall is a stall that occurs under such conditions.[15]
Considering, for example, a banked turn, the lift required is equal to the weight of the aircraft plus extra lift to provide the centripetal force necessary to perform the turn; that is:[16][17]
$L = nW$
where:
$L$ = lift
$n$ = load factor (greater than 1 in a turn)
$W$ = weight of the aircraft
To achieve the extra lift, the lift coefficient, and so the angle of attack, will have to be higher than it would be in straight and level flight at the same speed. Therefore, given that the stall always occurs at the same critical angle of attack,[18] by increasing the load factor (e.g., by tightening the turn) such critical angle - and the stall - will be reached with the airspeed remaining well above the normal stall speed,[16] that is:[19][20][21]
$V_{st} = V_s \sqrt n$
where:
$V_{st}$ = stall speed
$V_s$ = stall speed of the aircraft in straight, level flight
$n$ = load factor
The table that follows gives some examples of the relation between the angle of bank and the square root of the load factor. It derives from the trigonometric relation (secant) between $L$ and $W$.
bank angle $\sqrt n$
30° 1.07
45° 1.19
60° 1.41
For example, in a turn with bank angle of 45°, Vst is 19% higher than Vs.
It should be noted that, according to Federal Aviation Administration (FAA) terminology, the above example illustrates a so-called turning flight stall, while the term accelerated is used to indicate an accelerated turning stall only, that is, a turning flight stall where the airspeed decreases at a given rate.[22]
A notable example of air accident involving a low-altitude turning flight stall is the 1994 Fairchild Air Force Base B-52 crash.
## Dynamic stall
Dynamic stall is a non-linear unsteady aerodynamic effect that occurs when airfoils rapidly change the angle of attack. The rapid change can cause a strong vortex to be shed from the leading edge of the aerofoil, and travel backwards above the wing. The vortex, containing high-velocity airflows, briefly increases the lift produced by the wing. As soon as it passes behind the trailing edge, however, the lift reduces dramatically, and the wing is in normal stall.[23]
Dynamic stall is an effect most associated with helicopters and flapping wings. During forward flight, some regions of a helicopter blade may incur flow that reverses (compared to the direction of blade movement), and thus includes rapidly changing angles of attack. Oscillating (flapping) wings, such as those of insects—including the most famous one, the bumblebee—may rely almost entirely on dynamic stall for lift production, provided the oscillations are fast compared to the speed of flight, and the angle of the wing changes rapidly compared to airflow direction.[23]
Stall delay can occur on airfoils subject to a high angle of attack and a three-dimensional flow. When the angle of attack on an airfoil is increasing rapidly, the flow will remain substantially attached to the airfoil to a significantly higher angle of attack than can be achieved in steady-state conditions. As a result, the stall is delayed momentarily and a lift coefficient significantly higher than the steady-state maximum is achieved. The effect was first noticed on propellers.[24]
## Deep stall
Normal flight
Deep stall condition – T-tail in "shadow" of wing
The deep stall affects aircraft with a T-tail configuration.
A Schweizer SGS 1-36 being used for deep stall research by NASA over the Mojave Desert in 1983
A deep stall (or super-stall) is a dangerous type of stall that affects certain aircraft designs,[25] notably those with a T-tail configuration. In these designs, the turbulent wake of a stalled main wing "blankets" the horizontal stabilizer, rendering the elevators ineffective and preventing the aircraft from recovering from the stall.
Effects similar to deep stall had long been known to occur on many aircraft designs before the term was coined. Gloster Javelin WD808 was lost in a crash on June 11, 1953, to a "locked in" stall[26] and Handley Page Victor XL159 was lost to a "stable stall" on March 23, 1962.[27] The name "deep stall" first came into widespread use after the crash of the prototype BAC 1-11 G-ASHG on October 22, 1963, killing its crew.[28] This led to changes to the aircraft, including the installation of a stick shaker (see below) to clearly warn the pilot of an impending stall. Stick shakers are now a standard part of commercial airliners. Nevertheless, the problem continues to cause accidents; on June 3, 1966, a Hawker Siddeley Trident (G-ARPY)[29] was lost to deep stall; deep stall is suspected to be cause of another Trident (G-ARPI) crash on June 18, 1972; on April 3, 1980, a prototype of the Canadair Challenger business jet entered deep stall during testing, killing one of the test pilots who was unable to leave the plane in time;[30] and on July 26, 1993, a Canadair CRJ-100 was lost in flight test due to a deep stall.[31] It has been reported that a Boeing 727 entered a deep stall in flight test, but the pilot was able to rock the airplane to increasingly higher bank angles until the nose finally fell through and normal control response was recovered.[32] A 727 accident on December 1, 1974, has also been attributed to a deep stall.[33] The crash of West Caribbean Airways Flight 708 in 2005 was also attributed to a deep stall.
Reports on the crash of Air France Flight 447 have stated that the accident involved a deep stall entered at 38,000 ft (11,582 m) and continued for more than three minutes until impact,[34] but this was a steady state conventional stall[35][36] because the aircraft (an Airbus A330) did not have a T-tail.[37]
Canard-configured aircraft are also at risk of getting into a deep stall. Two Velocity aircraft crashed due to locked-in deep stalls.[38] Testing revealed that the addition of leading edge cuffs to the outboard wing prevented the aircraft from getting into a deep stall. The Piper Advanced Technologies PAT-1, N15PT, another canard-configured aircraft, also crashed in an accident attributed to a deep stall.[39] Wind tunnel testing of the design at the NASA Langley Research Center showed that it was vulnerable to a deep stall.[40]
In the early 1980s, a Schweizer SGS 1-36 sailplane was modified for NASA's controlled deep-stall flight program.[41]
## Tip stall
Aircraft with a swept wing suffer from a particular form of stalling behaviour at low speed. At high speed the airflow over the wing tends to progress directly along the chord, but as the speed is reduced a sideways component due to the angle of the leading edge has time to built up. Airflow at the root is affected only by the angle of the wing, but at a point further along the span, the airflow is affected both by the angle as well as any sideways component of the airflow from the air closer to the root. This results in a pattern of airflow that is progressively "sideways" as one moves toward the wingtip.
As it is only the airflow along the chord that contributes to lift, this means that the wing begins to develop less lift at the tip than the root. in extreme cases, this can lead to the wingtip entering stall long before the wing as a whole. In this case the average lift of the wing as a whole moves forward; the inboard sections are continuing to generate lift and are generally in front of the center of gravity (CoG), while the tips are no longer contributing and are behind the CoG. This produces a strong nose-up pitch in the aircraft, which can lead to more of the wing stalling, the lift moving further forward, and so forth. This chain reaction is considered very dangerous and was known as the pitch-up.
Tip stall can be prevented in a number of ways, at least one of which is found on almost all modern aircraft. An early solution was the addition of wing fences to re-direct sideways moving air back towards the rear of the wing. A similar solution is the dog-tooth notch seen on some aircraft, like the Avro Arrow. A more common modern solution is to use some degree of washout.
## Stall warning and safety devices
Fixed-wing aircraft can be equipped with devices to prevent or postpone a stall or to make it less (or in some cases more) severe, or to make recovery easier.
• An aerodynamic twist can be introduced to the wing with the leading edge near the wing tip twisted downward. This is called washout and causes the wing root to stall before the wing tip. This makes the stall gentle and progressive. Since the stall is delayed at the wing tips, where the ailerons are, roll control is maintained when the stall begins.
• A stall strip is a small sharp-edged device that, when attached to the leading edge of a wing, encourages the stall to start there in preference to any other location on the wing. If attached close to the wing root, it makes the stall gentle and progressive; if attached near the wing tip, it encourages the aircraft to drop a wing when stalling.
• A stall fence is a flat plate in the direction of the chord to stop separated flow progressing out along the wing[42]
• Vortex generators, tiny strips of metal or plastic placed on top of the wing near the leading edge that protrude past the boundary layer into the free stream. As the name implies, they energize the boundary layer by mixing free stream airflow with boundary layer flow thereby creating vortices, this increases the inertia of the boundary layer. By increasing the inertia of the boundary layer, airflow separation and the resulting stall may be delayed.
• An anti-stall strake is a leading edge extension that generates a vortex on the wing upper surface to postpone the stall.
• A stick pusher is a mechanical device that prevents the pilot from stalling an aircraft. It pushes the elevator control forward as the stall is approached, causing a reduction in the angle of attack. In generic terms, a stick pusher is known as a stall identification device or stall identification system.[43]
• A stick shaker is a mechanical device that shakes the pilot's controls to warn of the onset of stall.
• A stall warning is an electronic or mechanical device that sounds an audible warning as the stall speed is approached. The majority of aircraft contain some form of this device that warns the pilot of an impending stall. The simplest such device is a stall warning horn, which consists of either a pressure sensor or a movable metal tab that actuates a switch, and produces an audible warning in response.
• An angle-of-attack indicator for light aircraft, the "AlphaSystemsAOA" and a nearly identical "Lift Reserve Indicator", are both pressure differential instruments that display margin above stall and/or angle of attack on an instantaneous, continuous readout. The General Technics CYA-100 displays true angle of attack via a magnetically coupled vane. An AOA indicator provides a visual display of the amount of available lift throughout its slow speed envelope regardless of the many variables that act upon an aircraft. This indicator is immediately responsive to changes in speed, angle of attack, and wind conditions, and automatically compensates for aircraft weight, altitude, and temperature.[citation needed]
• An angle of attack limiter or an "alpha" limiter is a flight computer that automatically prevents pilot input from causing the plane to rise over the stall angle. Some alpha limiters can be disabled by the pilot.
Stall warning systems often involve inputs from a broad range of sensors and systems to include a dedicated angle of attack sensor.
Blockage, damage, or inoperation of stall and angle of attack (AOA) probes can lead to unreliability of the stall warning, and cause the stick pusher, overspeed warning, autopilot, and yaw damper to malfunction.[44]
If a forward canard is used for pitch control, rather than an aft tail, the canard is designed to meet the airflow at a slightly greater angle of attack than the wing. Therefore, when the aircraft pitch increases abnormally, the canard will usually stall first, causing the nose to drop and so preventing the wing from reaching its critical AOA. Thus, the risk of main wing stalling is greatly reduced. However, if the main wing stalls, recovery becomes difficult, as the canard is more deeply stalled and angle of attack increases rapidly.[45]
If an aft tail is used, the wing is designed to stall before the tail. In this case, the wing can be flown at higher lift coefficient (closer to stall) to produce more overall lift.
Most military combat aircraft have an angle of attack indicator among the pilot's instruments, which lets the pilot know precisely how close to the stall point the aircraft is. Modern airliner instrumentation may also measure angle of attack, although this information may not be directly displayed on the pilot's display, instead driving a stall warning indicator or giving performance information to the flight computer (for fly by wire systems).
## Flight beyond the stall
As a wing stalls, aileron effectiveness is reduced, making the plane hard to control and increasing the risk of a spin starting. Post stall, steady flight beyond the stalling angle (where the coefficient of lift is largest) requires engine thrust to replace lift as well as alternative controls to replace the loss of effectiveness of the ailerons. For high-powered aircraft, the loss of lift (and increase in drag) beyond the stall angle is less of a problem than maintaining control. Control can be provided by vectored thrust as well as a rolling stabilator (or taileron), and the enhanced manoeuvering capability by flights at very high angles of attack can provide a tactical advantage for military fighters such as the F-22 Raptor. Short term stalls at 90–120° are sometimes performed at airshows.[46] The highest angle of attack in sustained flight so far demonstrated was 70 degrees in the X-31 at the Dryden Flight Research Center.[47] Post-stall flight is a type of supermaneuverability.
## Spoilers
Except for flight training, airplane testing, and aerobatics, a stall is usually an undesirable event. Spoilers (sometimes called lift dumpers), however, are devices that are intentionally deployed to create a carefully controlled flow separation over part of an aircraft's wing to reduce the lift it generates, increase the drag, and allow the aircraft to descend more rapidly without gaining speed.[48] Spoilers are also deployed asymmetrically (one wing only) to enhance roll control. Spoilers can also be used on aborted take-offs and after main wheel contact on landing to increase the aircraft's weight on its wheels for better braking action.[citation needed]
Unlike powered airplanes, which can control descent by increasing or decreasing thrust, gliders have to increase drag to increase the rate of descent. In high-performance gliders, spoiler deployment is extensively used to control the approach to landing.[citation needed]
Spoilers can also be thought of as "lift reducers" because they reduce the lift of the wing in which the spoiler resides. For example, an uncommanded roll to the left could be reversed by raising the right wing spoiler (or only a few of the spoilers present in large airliner wings). This has the advantage of avoiding the need to increase lift in the wing that is dropping (which may bring that wing closer to stalling).[citation needed]
## History
Otto Lilienthal died while flying in 1896 as the result of a stall. Wilbur Wright encountered stalls for the first time in 1901, while flying his second glider. Awareness of Lilienthal's accident and Wilbur's experience, motivated the Wright Brothers to design their plane in "canard" configuration. This made recoveries from stalls easier and more gentle. The design saved the brothers' lives more than once.[49]
The aircraft engineer Juan de la Cierva worked on his "Autogiro" project to develop a rotary wing aircraft which, he hoped, would be unable to stall and which therefore would be safer than aeroplanes. In developing the resulting "autogyro" aircraft, he solved many engineering problems which made the helicopter possible. Tragically, he was killed in the crash of an airliner, possibly due to a stall at take-off.[citation needed]
Articles
Notable Accidents
## Notes
1. ^ Crane, Dale: Dictionary of Aeronautical Terms, third edition, page 486. Aviation Supplies & Academics, 1997. ISBN 1-56027-287-2
2. ^ Benjamin Gal-Or, "Vectored Propulsion, Supermaneuverability, and Robot Aircraft", Springer Verlag, 1990, ISBN 1990, ISBN 0-387-97161-0, ISBN 3-540-97161-0
3. ^ USAF & NATO Report RTO-TR-015 AC/323/(HFM-015)/TP-1 (2001)
4. ^ Clancy, L.J., Aerodynamics, Sections 5.28 and 16.48
5. ^ Anderson, J.D., A History of Aerodynamics, p 296-311
6. ^ FAA Airplane flying handbook ISBN 978-1-60239-003-4 Chapter4 Page 7
7. ^ 14 CFR part 61
8. ^ Federal Aviation Regulations Part25 section 201
9. ^ FAA Airplane flying handbook ISBN 978-1-60239-003-4 Chapter4 pages 12-16
10. ^ 14 CFR part 23
11. ^ FAA Airplane flying handbook ISBN 978-1-60239-003-4 Chapter4 page 11-12
12. ^ FAA Airplane flying handbook ISBN 978-1-60239-003-4 Chapter4 Page 9
13. ^ Federal Aviation Regulations part 25 section 207
14. ^ Flight testing of fixed wing aircraft. Ralph D. Kimberlin ISBN 978-1-56347-564-1
15. ^ Brandon, John. "Airspeed and the properties of air". Recreational Aviation Australia Inc. Archived from the original on 2008-07-31. Retrieved 2008-08-09.
16. ^ a b Clancy, L.J., Aerodynamics, Section 5.22
17. ^ McCormick, Barnes W. (1979), Aerodynamics, Aeronautics and Flight Mechanics, p.464, John Wiley & Sons, New York ISBN 0-471-03032-5
18. ^ Clancy, L.J., Aerodynamics, Sections 5.8 and 5.22
19. ^ Clancy, L.J., Aerodynamics, Equation 14.11
20. ^ McCormick, Barnes W. (1979), Aerodynamics, Aeronautics and Flight Mechanics, Equation 7.57
21. ^
22. ^ "Part 23 - Airworthiness Standards: §23.203 Turning flight and accelerated turning stalls". Federal Aviation Administration. February 1996. Retrieved 2009-02-18.
23. ^ a b Article about dynamic stall on an aerodynamics web site
24. ^ Burton, Tony; David Sharpe, Nick Jenkins, Ervin Bossanyi (2001). Wind Energy Handbook (digitized online by Google books). John Wiley and Sons. p. 139. ISBN 0-471-48997-2. Retrieved 2009-01-01.
25. ^ "What is the super-stall?". Aviationshop. Retrieved 2009-09-02.
26. ^ ASN Wikibase Occurrence # 20519 Retrieved 4 September 2011.
27. ^ A Tale of Two Victors Retrieved 4 September 2011.
28. ^ ""Report on the Accident to B.A.C. One-Eleven G-ASHG at Cratt Hill, near Chicklade, Wiltshire on 22nd October 1963, Ministry of Aviation C.A.P. 219, 1965
29. ^ "ASN Aircraft accident Hawker Siddeley HS-121 Trident 1C G-ARPY Felthorpe". Aviation-safety.net. 1966-06-03. Retrieved 2013-04-02.
30. ^ "ASN Aircraft accident Canadair CL-600-1A11 Challenger 600 C-GCGR-X Mojave, CA". Aviation-safety.net. Retrieved 2013-04-02.
31. ^ "ASN Aircraft accident Canadair CL-600-2B19 Regional Jet CRJ-100 C-FCRJ Byers, KS". Aviation-safety.net. 1993-07-26. Retrieved 2013-04-02.
32. ^ Robert Bogash. "Deep Stalls". Retrieved 4 September 2011.
33. ^ Accident description Retrieved 4 September 2011.
34. ^ Pew, Glenn (May 2011). "Air France 447 — How Did This Happen?". AvWeb. Retrieved 30 May 2011.
35. ^
36. ^ Bethany Whitfield (May 27, 2011). "Air France 447 Stalled at High Altitude, Official BEA Report Confirms". Flying.
37. ^ Peter Garrison (Jun 01, 2011). "Air France 447: Was it a Deep Stall?". Flying.
38. ^ Cox, Jack, "Velocity... Solving a Deep Stall Riddle", EAA Sport Aviation, July 1991, pp.53-59.
39. ^ ASN Wikibase Occurrence # 10732 Retrieved 4 September 2011.
40. ^ Williams, L.J.; Johnson, J.L. Jr. and Yip, L.P., "Some Aerodynamic Considerations For Advanced Aircraft Configurations", AIAA paper 84-0562, January 1984.
41. ^ Schweizer-1-36 index: Schweizer SGS 1-36 Photo Gallery Contact Sheet
42. ^ Stall fences and vortex generators
43. ^ US Federal Aviation Administration, Advisory Circular 25-7A Flight Test Guide for Certification of Transport Category Airplanes, paragraph 228
44. ^ Harco Probes Still Causing Eclipse Airspeed Problems
45. ^ Airplane stability and control By Malcolm J. Abzug, E. Eugene Larrabee Chapter 17 ISBN 0-521-80992-4
46. ^ Pugachev's Cobra Maneuver
47. ^ X-31 EC94-42478-3: X-31 at High Angle of Attack
48. ^
49. ^ Designing the 1900 Wright Glider | 7,657 | 34,212 | {"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": 11, "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-2014-10 | longest | en | 0.95537 |
https://apps.uc.pt/courses/EN/unit/8658/26/2018-2019?menor=true&type=ram&id=351 | 1,550,748,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504594.59/warc/CC-MAIN-20190221111943-20190221133943-00389.warc.gz | 489,580,746 | 4,149 | # Mathematical Logic
Year
0
Academic year
2018-2019
Code
01001324
Subject Area
Área Científica do Menor
Language of Instruction
Portuguese
Mode of Delivery
Face-to-face
Duration
SEMESTRIAL
ECTS Credits
6.0
Type
Elective
Level
1st Cycle Studies
## Recommended Prerequisites
Basic courses on Analysis, Algebra, Programming.
## Teaching Methods
Classes are expository, with presentation of detailed proofs on the blackboard. The lectures include exercises and voluntary homework.
## Learning Outcomes
The main goal is to familiarize students with classical mathematical logic (symbolic logic) that aims to formalize propositions of informal language, in particular with respect to mathematical propositions and mathematical reasoning. The ambiguities of informal language led to a more and more symbolic language. It is one of the aims of the course to reflect these developments in the chapters on propositional and predicate logic. The third part of the course is an introduction to topics of the theory of computation.
One hopes that the lectures augments the students aptness to formalize propositions and arguments; augments their aptness to discern between fallacious and correct reasoning in mathematics; clarifies the historic and technical rôles of symbolic logic in mathematics and computer science; and conveys, in particular, knowledge on formal deductive systems and Turing machines.
No
## Syllabus
Propositional Logic (CP)
1. Alphabet, valuations, tautologies, contingencies, contradictions. Substitution theorems, associativity, DNF. Boolean algebras and functions
2. Applications: analysis of arguments, electrical and logical circuits
3. Simplification of formulas. Connections with algorithmic complexity
4. Compactness of PC 5. Formal systems of deduction. A formal system (Hilbert) for CP. Deduction and completeness theorems. First Order Logic (FOL) Alphabet, terms, atomic and prime formulas.
Applications. Theorems of FOL. Computability, Decidability, Enumerability
1. Algorithms and Turing machine. Computability
2. Computable functions. Quantification of relations. Theorems of composition, recursion and minimum operator. Computability of common functions. Existence of non-computable functions
3. Decidibility of relations (order and equality relations)
4. Algorithmic indecidibility of the halting problem
5. Listability. Listable but indecidable sets.
## Head Lecturer(s)
Alexander Kovacec
## Assessment Methods
Assessment
Exam (100%) or Midterm exam (70%) + test (30%): 100.0%
## Bibliography
J. Malitz, Introduction to Mathematical Logic, Springer, 1978.
B. Margaris, First Order Logic, Dover, 1970.
E. Mendelsson, Introduction to Mathematical Logic, v. Nostrand, 1966.
A. Kovacec, Lógica (notas de curso), Departamento de Matemática da FCTUC, 2007.
A.J. Franco de Oliveira, Lógica e Aritmética, Gradiva, 2003.
P.J. Davis, R. Hersh, A Experiência Matemática, Gradiva, 1995 | 691 | 2,923 | {"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-2019-09 | longest | en | 0.829759 |
https://plainmath.org/differential-equations/49683-find-the-solution-the-following-differential-equations-frac-xdy-equal | 1,726,248,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00845.warc.gz | 431,025,659 | 30,092 | eiraszero11cu
2021-12-26
Find the solution of the following Differential Equations.
$\frac{xdy-ydx}{{x}^{2}}=0$
Vivian Soares
$\frac{xdy-ydx}{{x}^{2}}=0$
$⇒\frac{x}{{x}^{2}}dy-\frac{y}{{x}^{2}}dx=0$
$⇒-\frac{y}{{x}^{2}}dx+\frac{1}{x}dy=0$
$Mdx+Ndy=0$
$M=\frac{-y}{{x}^{2}},N=\frac{1}{x}$
Then
Clearly $\frac{\partial m}{\partial y}=\frac{\partial N}{\partial x}=\frac{-1}{{x}^{2}}$
The given differential equation is exact
Hence, the general solution is
$\int Mdx+\int Ndy=c$, where c is constant
$\int \frac{-y}{{x}^{2}}dx+0=c$
$⇒-y\cdot \left(\frac{-1}{x}\right)=c$
Answer: $y=cx$ where c is an arbitrary constant.
Hector Roberts
$\frac{xdy-ydx}{{x}^{2}}=0$
$\left(\frac{1}{x}\right)dy-\left(\frac{y}{{x}^{2}}\right)dx=0$
$d\left(\frac{y}{x}\right)=0$
$\frac{y}{x}=C$, C constants
$y=Cx$.
karton | 353 | 805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "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-2024-38 | latest | en | 0.456615 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-7-review-page-557/3 | 1,544,736,045,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825098.68/warc/CC-MAIN-20181213193633-20181213215133-00275.warc.gz | 910,042,760 | 13,543 | ## Algebra: A Combined Approach (4th Edition)
$\dfrac{2-(-2)}{-2+5}=\dfrac{4}{3}$
$\dfrac{2-z}{z+5};$ $z=-2$ Substitute $z$ with $-2$ and simplify: $\dfrac{2-(-2)}{-2+5}=\dfrac{4}{3}$ | 86 | 184 | {"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} | 3.953125 | 4 | CC-MAIN-2018-51 | latest | en | 0.479257 |
https://forums.wolfram.com/mathgroup/archive/2012/Aug/msg00156.html | 1,722,984,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00413.warc.gz | 218,658,152 | 8,586 | Re: How to Extract Conditional Expression?
• To: mathgroup at smc.vnet.net
• Subject: [mg127635] Re: How to Extract Conditional Expression?
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Thu, 9 Aug 2012 03:52:03 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
• References: <jvl43v\$731\$1@smc.vnet.net>
```It can also handle the general case
step1 = x ->
ConditionalExpression[(1/\[Pi]) 105 (-2 ArcTan[\[Sqrt]Root[
1 - 4584 #1 + 525588 #1^2 - 24932888 #1^3 +
643774850 #1^4 - 10134675448 #1^5 + 103523895748 #1^6 -
713980114440 #1^7 + 3417573832943 #1^8 -
11577472180368 #1^9 + 28132792718632 #1^10 -
49450791062896 #1^11 + 63131885232924 #1^12 -
58540019029360 #1^13 + 39271847011624 #1^14 -
18902427543696 #1^15 + 6437818513647 #1^16 -
1518441755144 #1^17 + 239987843524 #1^18 -
24146287608 #1^19 + 1422915970 #1^20 - 42378776 #1^21 +
478484 #1^22 - 1512 #1^23 + #1^24 &, 1]] + 2 \[Pi] C[1]),
C[1] \[Element] Integers];
Assuming[{Element[C[1], Integers]},
FullSimplify[step1] // N]
x -> -1. + 210. C[1]
% /. C[1] -> 3
x -> 629.
Bob Hanlon
On Wed, Aug 8, 2012 at 9:37 PM, djmpark <djmpark at comcast.net> wrote:
> step1 = x ->
> ConditionalExpression[(1/\[Pi]) 105 (-2 ArcTan[\[Sqrt]Root[
> 1 - 4584 #1 + 525588 #1^2 - 24932888 #1^3 +
> 643774850 #1^4 - 10134675448 #1^5 + 103523895748 #1^6 -
> 713980114440 #1^7 + 3417573832943 #1^8 -
> 11577472180368 #1^9 + 28132792718632 #1^10 -
> 49450791062896 #1^11 + 63131885232924 #1^12 -
> 58540019029360 #1^13 + 39271847011624 #1^14 -
> 18902427543696 #1^15 + 6437818513647 #1^16 -
> 1518441755144 #1^17 + 239987843524 #1^18 -
> 24146287608 #1^19 + 1422915970 #1^20 - 42378776 #1^21 +
> 478484 #1^22 - 1512 #1^23 + #1^24 &, 1]] + 2 \[Pi] C[1]),
> C[1] \[Element] Integers];
>
> step1 /. C[1] -> 3 // N
> x -> 629.
>
>
> David Park
> djmpark at comcast.net
> http://home.comcast.net/~djmpark/index.html
>
>
> From: Scott Guthery [mailto:sbg at acw.com]
>
> Thanks for everyone's help but I still can't seem to get the function out of
> the Conditional. Here's the actual ConditionalExpression I'm dealing with:
>
> x -> ConditionalExpression[(1/\[Pi])
> 105 (-2 ArcTan[\[Sqrt]Root[
> 1 - 4584 #1 + 525588 #1^2 - 24932888 #1^3 + 643774850 #1^4 -
> 10134675448 #1^5 + 103523895748 #1^6 - 713980114440 #1^7 +
> 3417573832943 #1^8 - 11577472180368 #1^9 +
> 28132792718632 #1^10 - 49450791062896 #1^11 +
> 63131885232924 #1^12 - 58540019029360 #1^13 +
> 39271847011624 #1^14 - 18902427543696 #1^15 +
> 6437818513647 #1^16 - 1518441755144 #1^17 +
> 239987843524 #1^18 - 24146287608 #1^19 +
> 1422915970 #1^20 - 42378776 #1^21 + 478484 #1^22 -
> 1512 #1^23 + #1^24 &, 1]] + 2 \[Pi] C[1]),
> C[1] \[Element] Integers]
>
> Using the various suggestions the closest I can get is the error statement
> ...
>
> Root::deg: 1-4584 x+525588 x^2-24932888 x^3+643774850 x^4-10134675448
> x^5+103523895748 x^6-713980114440 x^7+3417573832943 x^8-11577472180368
> x^9+28132792718632 x^10-49450791062896 x^11+63131885232924
> x^12-58540019029360 x^13+39271847011624 x^14-18902427543696
> x^15+6437818513647 x^16-1518441755144 x^17+239987843524 x^18-24146287608
> x^19+1422915970 x^20-42378776 x^21+478484 x^22-1512 x^23+x^24 has fewer than
> 1 root(s) as a polynomial in #1. >>
>
> followed by ...
>
> (1/\[Pi])210 (3 \[Pi] -
> ArcTan[\[Sqrt]Root[
> 1 - 4584 x + 525588 x^2 - 24932888 x^3 + 643774850 x^4 -
> 10134675448 x^5 + 103523895748 x^6 - 713980114440 x^7 +
> 3417573832943 x^8 - 11577472180368 x^9 +
> 28132792718632 x^10 - 49450791062896 x^11 +
> 63131885232924 x^12 - 58540019029360 x^13 +
> 39271847011624 x^14 - 18902427543696 x^15 +
> 6437818513647 x^16 - 1518441755144 x^17 + 239987843524 x^18 -
> 24146287608 x^19 + 1422915970 x^20 - 42378776 x^21 +
> 478484 x^22 - 1512 x^23 + x^24 &, 1, 0]])
>
> I can't seem to get rid of the &,1,0 at the end.
> | 1,777 | 4,262 | {"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-2024-33 | latest | en | 0.340666 |
https://edurev.in/course/quiz/attempt/7813_GATE-Mock-Test-Electronics-Engineering--ECE--5/79010688-1772-4d0b-84d3-96918b6f5f92 | 1,653,416,778,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00347.warc.gz | 281,362,219 | 59,615 | # GATE Mock Test Electronics Engineering (ECE)- 5
## 65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | GATE Mock Test Electronics Engineering (ECE)- 5
Description
Attempt GATE Mock Test Electronics Engineering (ECE)- 5 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
### The following question comprises two words that have a certain relationship between them followed by four pairs of words. Select the pair that has same relationship as the original pair of words Fox : Cunning
Solution:
Fox is cunning animal similarly
Ants are Industrious.
The second word shows the quality of the first.
QUESTION: 3
### Replace the phrase printed in bold to make it grammatically correct ? A twenty-first century economy "cannot be held" hostage by power cuts nor travel on nineteenth century roads.
Solution:
QUESTION: 4
Improve the sentence with suitable options by replacing the underlined word.
He "lay" on the grass enjoying the sunshine.
Solution:
QUESTION: 5
Fill in the blanks :
At times, we are all ______ to be mistaken.
Solution:
At times, we are all likely to be mistaken.
QUESTION: 6
There are two solutions of wine and water, the concentration of wine being 0.4 and 0.6 respectively. If five liters of the first solution be mixed with fifteen liters of the second, find the concentration of wine in the resultant solution.
Solution:
Resultant concentration of wine
QUESTION: 7
Rajiv bought some apples at the rate of 25 for Rs. 400 and an equal number at the rate of 30 for Rs. 270. He then sold all at the rate of 10 for Rs. 180. Find his profit/loss percentage
Solution:
Let Rajiv bought x apples of each kind.
Total cost price of Rajiv =
Total selling price of Rajiv =
SP > CP , Hence he makes profit;
Profit percentage =
QUESTION: 8
Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is alteast one graduate among them ?
Solution:
QUESTION: 9
2 men and 3 boys can do a piece of work in 10 days, while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?
Solution:
Let work done by 1 man in 1 day = x units
Let work done by 1 boy in 1 day = y units
2 men and 1 boy can finish work
No. of days taken by 2 men and 1 boy = = 12.5 days
QUESTION: 10
In an exam, 60% of the candidates passed in Maths and 70% candidates passed in English and 10% of the candidates failed in both the subjects, 300 candidates passed in both the subjects. Find the total number of candidates appeared in the exam, if they took test in only two subjects that is Math and English.
Solution:
Let total number of students appear be X.
X = 750
QUESTION: 11
In the figure, the value of resistor R is (I 5) ohms, where I is the current in amperes. The value of current I is
Solution:
150 – IR = 0
150 – I (I +5) = 0
150 – I2 - 5I = 0
I2 +5I – 150 = 0
I2 +15I – 10I - 150 = 0
I (I +15) – 10 (I +15) = 0
(I +15) (I-10) = 0
I = 10A & I = -15A
QUESTION: 12
The velocity of propagation of electromagnetic wave in an underground cable with relative permittivity of 9 is
Solution:
QUESTION: 13
The function shown in the figure can be represented as
Solution:
QUESTION: 14
Consider an LTI system with impulse response h(t) = e-5t u(t). Determine the output of the system for the input x(t) = 5 u(t)
Solution:
y(t) = h(t) * x(t)
= e-5t u(t) * 5u(t)
y(t) = (1 - e-5t) u(t)
QUESTION: 15
The value of integral is
Solution:
= e0 × 0
= 0
QUESTION: 16
A rectangular pulse x(t) is shown in the figure. Determine the value of
Solution:
According to Parswell’s theorem
QUESTION: 17
If and f(t) is as t → ∞ . Determine the value of K
Solution:
⇒K = 2
QUESTION: 18
In the signal flow graph shown in the given figure, the value of C/R ratio is
Solution:
There are three path
QUESTION: 19
What is the open-loop transfer function for an unity feedback system having root locus shown in the following figure ?
Solution:
QUESTION: 20
The digital circuit shown in the figure works as a
Solution:
For D flip-flop
Qn 1 = D
Here, D = X Qn
So Qn 1 = X Qn
So it becomes T flip-flop
QUESTION: 21
The simplified form of Boolean function F(A,B,C) implemented by the given 3 × 8 decoder is
Solution:
QUESTION: 22
The contents of the accumulator in an 8085 microprocessor is altered after the execution of the instruction
Solution:
Compare operation does not affect the accumulator and ORAA does not change the content of accumulator.
QUESTION: 23
Assuming that all the diodes are ideal in figure, the current in the diode D2 is
Solution:
Assume both the diodes are ON
So, I1 = -1A
But current can not flow from n to p side in a diode so D1 will be off and equivalent circuit becomes
Therefore,
QUESTION: 24
Determine the upper and lower threshold voltage of the schmitt trigger shown below :
Solution:
QUESTION: 25
The transistor in the given circuit, should always be in saturation region. Take VCE (sat) = 0V and VBE = 0.7V. the minimum value of RC is
Solution:
QUESTION: 26
A differential amplifier has input at non inverting terminal is 1050 μV and at inverting terminal 950 μV. What is the error in the differential output if CMRR is 1000.
Solution:
QUESTION: 27
Which one of the following type of negative feedback increases the input resistance and decreases the output resistance of an amplifier ?
Solution:
QUESTION: 28
The spectral density and auto correlation function of white noise are respectively.
Solution:
Auto corelation function and PSD are fourier transform pair
Here, PSD = N0
QUESTION: 29
For a random variable x having the probability density function is shown in the figure, what are the variance of random variable.
Solution:
= 4/3
QUESTION: 30
An amplitude modulated signal is given as, S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct, where f fm. and fc is the carrier frequency and fm is the modulating signal frequency. The power efficiency is.
Solution:
S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct
= 100 [1 0.5 Cos 2pfmt] Cos 2pfct
Here modulation index, μ = 0.5
= 11.11%
QUESTION: 31
Signal x(t) = 4Sinc2 100t. Sin 500pt .At what sampling frequency should this signal be sampled to avoid aliasing ?
Solution:
So, fm = f1 f2
= 350 Hz
So, fs = 2fm
= 2 × 350
= 700 Hz
QUESTION: 32
The value of is
Solution:
QUESTION: 33
The solution of differential equation dy = (1-y) dx is
Solution:
dy = (1 – y) dx
y = 1 ce-x
QUESTION: 34
The rank of the matrix A is 2, when K is
Solution:
= 2 {3k} - 3{2k} 4{2´15 - 3´16}
= 0 {for any value of k}
QUESTION: 35
Given If C is a counter-clockwise path in the z-plane such that |z - i| = 1, the value of is
Solution:
Poles are at z = -1, -2
There is no any pole inside the circle, so integration will be zero
QUESTION: 36
In the circuit shown in figure, determine the maximum power absorbed by the load resistance RL
Solution:
Rth = 4 ohms
For Vth
V2 – V3 V2 – V1 12 = 0
But V1 = 2V
So, 2 V2 – V3 – V1 12 = 0
2 V2 – V3 – 2 12 = 0
2 V2 – V= -10…………....2
From equation 1 and 2
V3 – V= 12………………..1
-V3 2V= -10…………….2
V2 = 2 Volt
So V3 = 14 Volt
So Vth = 14V
QUESTION: 37
In the circuit shown below, the switch is closed at t = 0. What is the initial value of the current through the capacitor C.
Solution:
At t = 0-, the equivalent circuit is shown below :
At t = 0 , the equivalent circuit
So, ic = 4 – 3.2 = 0.8A
QUESTION: 38
Determine the z-parameter for the network shown below.
Solution:
Since it is symmetrical ladder network
QUESTION: 39
At which frequency (in rad/sec) the thevenins equivalent impedance looking at terminals AB becomes purely resistive?
Solution:
From the circuit it is clear, that ZAB becomes real when circuit is in resonance and resonance frequency of this circuit is given as
QUESTION: 40
Consider if y(t) = Ae-t for then value of A is
Solution:
QUESTION: 41
The 4-point DFT of sequence x[n] = [1,2,2,1] is
Solution:
QUESTION: 42
The output y(t) of a continuous-time system for the input x(t) is given by . Which one of the following is correct ?
Solution:
It is an integrator that generally an unstable system because integrator produces an unbounded output and for causal
Let at t = 2
It means y(2) is depends on x(4) that means present output is depends on future input, so it is non-causal
QUESTION: 43
For the NMOS and PMOS shown in the figure are matched and = 1mA/V2 and V n = -V p = 1 V. Determine V0
Solution:
For the circuit, it is clear that since QN and QP are perfectly matched and are operating at |VGS| equal values of i.e. are 2.5V, the symmetrical which dictates that V0 = 0V
QUESTION: 44
Determine the region of operation of BJT in the given circuit.
Solution:
Since, VEB = 0.7
Here VB = 0, So, VE = 0.7 V
So, VC = -10 4.6 × 1
= - 5.4 V
Since, VEC = 0.7 5.4 = 6.1V
That is greater than 0.2V So BJT is in active region.
QUESTION: 45
In the silicon BJT circuit shown below. Assume the emitter area of transistor Q2 is two time of area of Q1. The value of Io { Let b of the transistors are very large}
Solution:
= 2mA
Since β is very large,
So, Iref = Ic1
i.e. Ic1 = 2mA
Also the emitter current area of transistor Q2 is times of Q1
So, Ic2 = 2 Ic1
= 2 × 2 mA
= 4 mA
So, I0 = +4 mA
QUESTION: 46
Determine the current io in the given circuit.
Solution:
Circuit can be redrawn
QUESTION: 47
Consider a bar of silicon in which a hole concentration profile describe by P(x) = P0.e-x/Lp. Find the current at x = 0, given P0 = 1016/cm3, Lp = 1 mm and cross-sectional area of the bar is 100 mm2 and Dp = 12 cm2/sec.
Solution:
= 192 A/cm2
So the current Ip,
Ip = Jp × A
= 192 × 10-8
= 192 mA
QUESTION: 48
A system with the open loop transfer function is connected in a negative feedback gain of unity. What is the value of K for which the system is unstable.
Solution:
The characteristic equation
S (S3 8S2 24S 32) K = 0
Routh – Hurwitz array
S4 1 24 K
S3 8 32 O
S2 20 K
S' (20×32-8K) 0
S0 K
So for stable system,
K > 0
and 20×32 - 8K > 0
K < 80
So for unstable system K > 80
i.e. K = 90
QUESTION: 49
In the system shown below x(t) = e-2t u(t). Determine the output y(t) in steady state.
Solution:
Here, x(t) = e-2t u(t)
QUESTION: 50
The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for
Solution:
System is stable in the region 0.1 to 5 on the left side of 10 as number of enrichment there is zero
0.1K < 1 ⇒ K < 10
5K > 1 ⇒ K > 0.2
0.2 < K < 10
Also 1 > 10 K
K < 0.1
So for the stable system 0.2 < K < 10 and K < 0.1
QUESTION: 51
The sensitivity of transfer function with respect to the parameter K is -0.5, Determine the value of K
Solution:
8K2 10K 3 = -4K
8K2 14K 3 = 0
4K (2K 3) 1 (2K 3) = 0
(2K 3) (4K 1) = 0
K = -1.5 and K = -0.25
QUESTION: 52
Given two binary FSK signals i = 1, 2.The minimum frequency separation of these two signals such that they are orthogonal is
Solution:
For the two signals to be orthogonal
QUESTION: 53
A block code is transmitted in NRZ invent-one-one format with voltage levels
6V → ‘1’
-6V → ‘0’
Three code words are sent : 10010, 11110, 01001. Over these three transmitted code words the average voltage is
Solution:
There are total 15 digits so, In which 8 are 1’s and 7 are 0’s
So average value = = 0.4V
QUESTION: 54
A signal is given by,
γ(t) = s(t) n(t)
where s(t) = 20 Cos2π4000t 40 Cos2π2000t and the noise n(t) is white with power spectral density η0 = 1 W/Hz. The total received signal is put through a band pass filter with pass band between 100 to 110 Hz. The value of SNR at the output of filter is
Solution:
Signal power
= 200 800
= 1000
and noise power, Pn = n0 × (BW)
= 1 × 10
= 10W
So, SNR = 1000/10 = 100
In dB = 10log100 = 20dB
QUESTION: 55
The base band signal m(t) is recovered from the DSB-Sc signal s(t) = m(t). Cos2πfct, by multiplying s(t) by the locally generated carrier Cos (2πfct φ). The product is passed through a LPF which rejects the double frequency signal. If the base band signal is band limited to 4KHz, what is the minimum value of fc for which m(t) can be recovered by filtering ?
Solution:
In order to recover m(t) from s(t) by filter method it is necessary that the lowest frequency contained in the first term of s(t) must be greater than the highest frequency contained in the second term that is
So minimum value of fc = 4 KHz
QUESTION: 56
A plane wave in free space is incident normally on a large block of material . which occupies z > 0. If the incident electric field is = 10 Cos (ωt z) V/m, the reflected magnetic field is :
Solution:
QUESTION: 57
A plane wave with is incident normally on a thick plane conductor lying in the X-Y plane. Its conductivity is 6 × 106 s/m and surface impedance is 5 × 10-4
?. The skin depth of conductor is
Solution:
QUESTION: 58
The volume of the paralleopipe whose sides are given by is
Solution:
Volume of paralleopipe
2 {-1} 2 {-1 3} 0 = -2 4 = 2
QUESTION: 59
A number in 4 bit 2 5 complement representation is 1101. This number when stored using 8-bits will beA number in 4 bit 2’5 complement representation is 1101. This number when stored using 8-bits will be
Solution:
The number in 4 bit 1101
So original number, 0011
Now the original number in 8-bit,
00000011
So, in 2s complement, 11111101
QUESTION: 60
Four variable K-map is shown in the figure. The essential prime implicants are
Solution:
QUESTION: 61
An 8085 microprocessor is interfaced to a 8KB RAM as shown below. The address range of the RAM is
Solution:
The RAM will be active when is 0
when, A13 A14 A15 = 011 as decoder is active low.
So the range of RAM address is from COOOH to DFFFH
QUESTION: 62
If z = x – iy and = p iq, then is
Solution:
= P iq
Z = P3 (iq)3 3P (iq) (P iq)
x – iy = P3 – 3Pq2 i (3P2q – q3)
X = P3 – 3Pq2
QUESTION: 63
The Newton – Raphson method is used to find the roots of the equation f(x) = x - cosπx If the initial guess for the root is 0.5, then the value of x after the first iteration is
Solution:
f(x) = X - Cosπx
f(0.5) = 0.5 – Cos 0.5π
= 0.5
= 0.38
QUESTION: 64
The eign value of the matrix are,
Solution:
QUESTION: 65
The value of is
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code | 4,452 | 14,787 | {"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-2022-21 | latest | en | 0.913057 |
https://joyanswer.org/calculating-cost-per-thousand-a-simple-guide | 1,716,755,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00141.warc.gz | 276,458,845 | 8,992 | # Calculating Cost Per Thousand: A Simple Guide
January 17, 2024 by JoyAnswer.org, Category : Finance
How to calculate cost per thousand? Learn how to easily calculate cost per thousand with this straightforward guide. Gain insights into the formula and factors involved, empowering you to make informed financial decisions.
## How to calculate cost per thousand?
Cost Per Thousand (CPM) is a common metric used in advertising to measure the cost of 1,000 impressions or views of an advertisement. The formula for calculating CPM is:
$CPM = \frac{Cost}{Impressions} \times 1000$
Here's a step-by-step guide on how to calculate Cost Per Thousand:
1. Determine the Cost: Find out how much you spent on the advertising campaign. This could be the total cost of placing the ad, production costs, and any other related expenses.
2. Determine the Number of Impressions: Calculate the total number of times the ad was displayed or viewed. Impressions represent the exposure of the ad to potential viewers.
3. Apply the Formula: Use the formula mentioned above to calculate CPM. Divide the total cost by the number of impressions, and then multiply the result by 1000 to get the cost per thousand impressions.
$CPM = \frac{Cost}{Impressions} \times 1000$
For example, if you spent $500 on an ad campaign that generated 100,000 impressions, the CPM would be calculated as follows: $CPM = \frac{\500}{100,000} \times 1000 = \5$ So, in this case, the Cost Per Thousand impressions is$5.
Keep in mind that CPM is a useful metric for comparing the relative cost-effectiveness of different advertising channels or campaigns, especially in the context of display advertising or online platforms.
## Mastering the Art of Cost per Thousand Calculation
Cost per thousand (CPM) is a crucial metric in advertising, representing the cost of reaching 1,000 impressions. Understanding and mastering its calculation is essential for optimizing your advertising campaigns. Here's a breakdown:
A. Definition:
• CPM stands for Cost per Mille, with "Mille" being the Latin word for thousand.
• It measures the cost of reaching 1,000 potential customers with your ad.
B. Calculation:
• Formula: CPM = (Total Ad Spend / Number of Impressions) x 1,000
• Example: If you spend $100 and receive 50,000 impressions, your CPM would be$2.
C. Importance:
• CPM helps compare the cost-effectiveness of different advertising channels and campaigns.
• It allows you to allocate your budget efficiently by focusing on channels with the best reach at the lowest cost.
D. Mastering the Art:
• Go beyond basic CPM: Consider viewable CPM (vCPM) which focuses on impressions actually seen by users.
• Analyze CPM trends: Track fluctuations over time to identify patterns and optimize spend.
• Benchmark against industry averages: Understand your CPM performance compared to similar businesses.
## What is CPM and how is it calculated in advertising?
CPM (Cost per Mille) is a metric used in advertising to measure the cost of reaching 1,000 impressions of your ad. It essentially tells you how much you pay for each thousand potential views your ad receives.
Calculation:
1. Total Ad Spend: This is the amount you spend on your advertising campaign.
2. Number of Impressions: This is the total number of times your ad is displayed, regardless of whether it's actually seen by users.
Formula: CPM = (Total Ad Spend / Number of Impressions) x 1,000
Example: Imagine you spend $50 and your ad generates 25,000 impressions. Your CPM would be: CPM = ($50 / 25,000) x 1,000 = $2 This means it costs you$2 to reach every 1,000 potential customers with your ad.
• CPM is commonly used for display advertising, where ads are displayed on websites or apps.
• CPM can vary greatly depending on factors like target audience, ad format, and platform.
## How to optimize advertising strategies based on cost per thousand metrics?
Using CPM effectively can significantly improve your advertising campaigns' efficiency and results. Here are some strategies:
A. Analyze CPM data:
• Track CPM trends: Monitor changes over time to identify patterns and potential optimization areas.
• Compare CPM across channels: See which platforms offer the best reach at the lowest cost.
• Benchmark against industry averages: Understand how your CPM performance compares to others in your field.
B. Optimize targeting:
• Refine your target audience: Ensure your ads reach the right people, leading to higher engagement and lower CPM.
• Utilize exclusion targeting: Exclude demographics or interests unlikely to convert, reducing wasted impressions.
• Test different targeting options: Experiment with various targeting parameters to find the most effective combination.
C. Improve ad quality:
• A/B test different ad creatives: Find the ad formats and messaging that resonate best with your audience.
• Focus on ad relevance: Ensure your ads match the platform and content where they appear, increasing engagement.
• Optimize landing pages: Create landing pages that provide a seamless user experience, leading to conversions.
D. Bidding strategies:
• Manual bidding: Set CPM bids yourself based on your budget and campaign goals.
• Automatic bidding: Utilize platform algorithms to optimize bids for clicks or conversions.
• Experiment with different bidding strategies: Find the approach that best aligns with your campaign objectives.
E. Continuous monitoring and adjustments:
• Regularly monitor your campaign performance and analyze data to identify areas for improvement.
• Make adjustments based on your findings, optimizing targeting, ad formats, and bidding strategies.
• Remember, advertising is an ongoing process. Continuously test and refine your approach for optimal results.
By effectively using CPM data and implementing these optimization strategies, you can significantly improve the reach, efficiency, and return on investment of your advertising campaigns.
Tags Cost Calculation , Financial Planning
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Finance | 1,648 | 7,965 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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} | 3.9375 | 4 | CC-MAIN-2024-22 | latest | en | 0.869409 |
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Depdendent Variable
Number of equations to solve: 23456789
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Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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Posted: Monday 06th of Aug 10:08 You can find the program here http://www.mathenomicon.net/solving-quadratic-equations.html. | 780 | 2,864 | {"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-2018-13 | latest | en | 0.894204 |
http://openstudy.com/updates/5326f18fe4b06a39f8621b64 | 1,448,735,274,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453656.76/warc/CC-MAIN-20151124205413-00277-ip-10-71-132-137.ec2.internal.warc.gz | 175,037,123 | 11,826 | ## nickersia one year ago A bag contains 8 blue marbles, 4 red marbles and x white marbles. A marble is drawn at random and not replaced. A second marble is drawn at random. If the probability that both are white is 5/51, how many white marbles are in the bag?
1. nickersia
What I did is: In bag at the start $8+4+x=12+x$ Probability that first one is white: $\frac{ x }{ 12+x}$ Now in the bag: $8+4+x-1=11+x$ Probability that second one is white: $\frac{ x-1 }{ 11+x }$ So $\frac{ x }{ 12+x } + \frac{ x-1 }{ 11+x } =\frac{ 5 }{ 51 }$ Is that true?
2. rvc
i think its correct not solved but just what u wrote
3. nickersia
I just solved it for x, and I got x1=1.138 and x2=-11.51 I don't think that's correct, cause I guess I should be getting whole number... There can't be 1.138 marbles in the bag :S
4. rvc
it might be 12+(x-1)
5. nickersia
It makes no difference because at some point you'll have to get rid of the brackets :L
6. Mashy
7. TwoPointInfinity
I don't see this being possible. How can both marbles have the same probability if the first marble wasn't replaced?
8. moli1993
I think u should multiply (x/12+x) *(x-1/11+x)=5/51 like this
9. Mashy
You first calculated the probability of picking a white ball then u calculated the probability of picking another white ball PROVIDED you have already picked one white ball (this is a conditional probability)
10. Mashy
Hence probability of both happening is the product, not sum!
11. Mashy
$P(A and B) = P(A).P(B provided A has happened)$
12. nickersia
Oh, I got ye, I'll try it that way
13. moli1993
yeh Mashy is correct , multiply the terms :)
14. nickersia
$\frac{ x ^{2}-x}{ x ^{2}+23x+132 }=\frac{ 5 }{ 51 }$ ... ... ... Leads me to x1=6.18 and x2=-2.57
15. Zarkon
try solving again
16. nickersia
$51x^{2}-51x=5x ^{2}+115x+660$ $46x ^{2}-166x-660=0$ $x _{1}=6.18 >----< x _{2}==2.57$
17. nickersia
Sorry
18. Zarkon | 632 | 1,913 | {"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} | 4.21875 | 4 | CC-MAIN-2015-48 | longest | en | 0.8887 |
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Pick3 Observation: Friends in Every State: Doubles Playing Drawing/Day: Theory...
Published:
I heard that doubles play every 5 or 7 days or 4 days or whatever...This is a Theory...If you had family members, friends, relatives in let´s say 10 or more states like: 15 states...Doubles should play every drawing...4 or 5 times a day...The Law of Averages well not actually the law of averages but the Law of Randomness should tell you that they will play everyday...
Think about it as a long string of: Pick3 and imagine that the number 3 is doubles, well if you look at: 10 pick3´s lined up you will see the number 3 in there once or twice or maybe 4 or 5 times...Is a law of randomness/averages....
So doubles across states should play every single drawing in one state or in a couple of states withing a group of 7 or 10 states...Sometimes is the way you look at it..
If Pick4 was in every state the same thing will occur with Pick4...You are going to see 2 numbers out of the 4 numbers repeat every drawing in 1 or 2 or 3 states in the group of 7 to 10 states...
Is a theory....So if you know which doubles are coming and you was to bet: \$20 per state straight in the group of 7 or 10 states (Pick3): you will lose: \$140 and win: \$10,000-\$20,000 PER DRAWING and \$20,000 TO 40,000 PER DAY...You dont have to bet \$20 you can bet: \$1 per state is up to you....Sometimes is the way you look at it....
i owe this idea to the movie: Angels & Demons movie i looked at it and it gave me the idea...
this is if Every states plays pick3 twice a day...And if it doesnt play every drawing it will play every day, one drawing yes the other no...Remember 2 Pick3´s drawings play every day...
sometimes is the way you look at it...
and remember to see the numbers as:
1 and 6 = 1
2 and 7 = 2
3 and 8 = 3
4 and 9 = 4
5 and 0 = 5
Entry #675 | 551 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-04 | latest | en | 0.901362 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-7-integration-techniques-7-2-integration-by-parts-7-2-exercises-page-521/59 | 1,537,449,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156471.4/warc/CC-MAIN-20180920120835-20180920141235-00496.warc.gz | 757,277,900 | 13,137 | ## Calculus: Early Transcendentals (2nd Edition)
$= 2{e^3}$
$\begin{gathered} f\,\left( x \right) = \int_e^x {\sqrt {{{\ln }^2}t - 1} \,\,dt} \hfill \\ \hfill \\ {\text{use }}ds = \int {\sqrt {1 + \,{{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \hfill \\ \hfill \\ \int_{}^{} {ds} = s = \int_e^{{e^3}} {\sqrt {1 + \,{{\left( {\frac{{df\left( x \right)}}{{dx}}} \right)}^2}} dx} \hfill \\ \hfill \\ then \hfill \\ \frac{{df\,\left( x \right)}}{{dx}} = \sqrt {{{\ln }^2}x - 1} \hfill \\ \hfill \\ s = \int_e^{{e^3}} {\sqrt {1 + {{\ln }^2}x - 1} } \,dx \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \int_e^{{e^3}} {\ln xdx} \hfill \\ \hfill \\ integrating\,\,by\,parts\,\, \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ u = \ln x \hfill \\ v = 1 \hfill \\ \hfill \\ S = \left. {x\ln x} \right|_e^{{e^3}} - \int_e^{{e^3}} {x \cdot \frac{1}{x}dx} \hfill \\ \hfill \\ = \,\left( {3{e^3} - e} \right) - \left. x \right|_e^{{e^3}} \hfill \\ \hfill \\ evaluate\,\,the\,\,{\text{limits}} \hfill \\ \hfill \\ = 3{e^3} - e - {e^3} + e = 2{e^3} \hfill \\ \hfill \\ \end{gathered}$ | 536 | 1,171 | {"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} | 4.15625 | 4 | CC-MAIN-2018-39 | latest | en | 0.348498 |
https://glasp.co/youtube/p/bizarre-spinning-toys | 1,716,330,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00737.warc.gz | 226,158,509 | 81,300 | BIZARRE spinning toys | Summary and Q&A
755.6K views
May 4, 2016
by
Physics Girl
BIZARRE spinning toys
TL;DR
Spinning tops and eggs defy gravity and exhibit complex rotational physics due to changes in kinetic and potential energy, angular momentum, and torque.
Key Insights
• 😥 Spinning tops and eggs exhibit complex rotational physics, which fascinated physicists to the point of conducting extensive analyses.
• 📐 Changes in kinetic and potential energy, angular momentum, and torque are crucial in understanding the behavior of spinning objects.
• 🖐️ Friction plays a significant role in providing the torque needed for a spinning object to stand up against gravity.
Transcript
[MUSIC PLAYING] So I got this fancy top called a PhiTOP, which is just an oblong piece of metal. But watch what happens when you spin it lying down. It stands up. That is, if you start it spinning along its minor axis, it will rise up and spin along its major axis. Now, this toy is shiny and pretty, but that physics trick at the end, you could make... Read More
Q: Why does the center of mass of a spinning top or an egg rise when it stands up?
The rising center of mass is a result of converting kinetic energy into potential energy, due to changes in the object's rotational motion.
Q: What is the role of torque in making a spinning top or an egg stand up?
Torque, caused by applied forces and friction, changes the object's angular momentum, allowing it to stand up against gravity.
Q: Why do spinning tops and eggs behave differently on surfaces with little friction?
Without friction, there is no torque to oppose the motion, causing the objects to not stand up like they do on surfaces with friction.
Q: Why doesn't a raw egg stand up when spun?
The fluid inside the raw egg does not move with the shell, leading to a lack of torque and preventing it from standing up.
Summary & Key Takeaways
• When a spinning top or an egg stands up on its end, the center of mass rises, seemingly defying gravity.
• Energy conservation explains how this is possible, as kinetic energy is converted into potential energy.
• Torque, caused by friction and applied forces, changes the angular momentum of the spinning object, allowing it to stand up. | 489 | 2,243 | {"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} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.917747 |
https://www.physicsforums.com/threads/not-the-element-of-the-set.756197/ | 1,606,417,998,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00177.warc.gz | 802,609,990 | 22,253 | # Not the element of the set?
I am going through section 14 in chapter 3 in Mary Boas' "Mathematical Methods for the Physical Sciences 3rd ed"
Example 1 is clear, but this line is confusing in example 3 (I don't understand why it shows that these are not an element of the set). I have pasted example 1 for reference below.
"Modify Example 1 to consider the set of polynomials of degree ≤ 3 with f(1) = 1. Is this a vector space? Suppose we add two of the polynomials; then the value of the sum at x = 1 is 2, so it is not an element of the set. Thus requirement 1 is not satisfied so this is not a vector space. Note that a subset of the vectors of a vector space is not necessarily a subspace."
For reference, here is Example 1:
Example 1. Consider the set of polynomials of the third degree or less, namely functions of the form f(x) = a0 +a1x+a2x2 +a3x3. Is this a vector space? If so, find a basis. What is the dimension of the space?
We go over the requirements listed above:
1. The sum of two polynomials of degree ≤ 3 is a polynomial of degree ≤ 3 and so is a member of the set.
2. Addition of algebraic expressions is commutative and associative.
3. The “zero vector” is the polynomial with all coefficients ai equal to 0, and adding it to any other polynomial just gives that other polynomial. The additive inverse of a function f(x) is just −f(x), and −f(x) + f(x) = 0 as required for a vector space.
4. All the listed familiar rules are just what we do every time we work with algebraic expressions.
So we have a vector space! Now let’s try to find a basis for it. Consider the set of functions: {1, x, x2, x3}. They span the space since any polynomial of degree ≤ 3 is a linear combination of them. You can easily show (Problem 1) by computing the Wronskian [equation (8.5)] that they are linearly independent. Therefore they are a basis, and since there are 4 basis vectors, the dimension of the space is 4.
Thanks,
Chris Maness
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Matterwave
Gold Member
The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?
The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?
So you are saying it is not a part of the set because it is made up of the set. Correct?
Chris
Matterwave
Gold Member
So you are saying it is not a part of the set because it is made up of the set. Correct?
Chris
What? It is not part of the set because the set is defined not to include this element.
Maybe a simpler example is better. Say you have a set {1,2,3,4}. Are all the sums of members of this set included in this set? What about 4+1? Is 4+1 in this set?
I meant it can't be part of the set because it is made up from the elements of the set, and I see from your clear example that this is the case.
Matterwave
Gold Member
I meant it can't be part of the set because it is made up from the elements of the set, and I see from your clear example that this is the case.
I don't see why this would be so... in the set {1,2,3,4}, 1+2=3 IS part of the set. 4+2 ISN'T part of the set. But whether an element is in a set or not depends on the set, not on whether that element is "made up from elements of the set" whatever you mean by that.
1 person
Ok, I see it is simply by definition. Since it is not defined in the set, it is not part of the set. It seems obvious now.
I also understand the point you made:
The set specifically requires that f(1)=1. Elements f(x) with f(1)=2 are not in the set by definition. Adding two functions who has f(1)=1 will yield a function with f(1)=2 and so the sum is not in the set. Is that clear?
I am really having difficulty with this section in the book. I understood the group theory section right before it, and the whole rest of the LA chapter. I get to this section (which is the last section that is not a review), and it is like hitting a brick wall. What could I be lacking that causes this group/set stuff to make me feel obtuse? I don't normally struggle with math concepts to the point where I feel like I need tutoring. I have actually never used a math tutor, but I feel I might need to hire one for this section.
Thanks,
Chris Maness
Last edited:
It's just a definition of a vector space and subspaces.
In a vector space, you must be able to add two elements of the set, given some restrictions/requirements, and still end up with a member of the set. If you have two polynomials f(x) and g(x), where f(1) = 1 for both of them, and you add them, then that polynomial at 1, p(1) = (f(1) + g(1)) = 1 + 1 = 2. It's as easy as 1 + 1 ;)
So that requirement that f(1) = 1 turns it into something other than a vector space. :)
(I didn't just mess up horribly, right? xD)
1 person
Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.
So that requirement that f(1) = 1 turns it into something other than a vector space. :)
You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.
I think it is starting to slowly soak in. I am not used to being this knee deep in abstract math. Everything I have dealt with thus far has been very physical, and intuitive.
Thanks,
Chris Maness
Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.
You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.
Hi Chris,
Well, 1 is a polynomial, but what they're saying is that they're putting a restriction on the polynomials. ALL polynomials in this vector space must be 1 when evaluated at one. By definition of a vector space, you must be able to add two of the vectors (elements of the set) and get a third element of the set. But if you add two polynomials who evaluate to 1, that is 1 + 1 = 2.
So yes, it doesn't satisfy closure. :)
Matterwave
Gold Member
Thanks, I was getting confuses by the fact that they refer to 1 as a polynomial, but I guess it is.
You mean that it turns it into something that (by definition) is not part of the set we have already defined, but f(x) was g(x) a part of that set. Since g+f=2 our definition fails because it does not satisfy the conditions of a proper vector -- that is, Closure.
I think it is starting to slowly soak in. I am not used to being this knee deep in abstract math. Everything I have dealt with thus far has been very physical, and intuitive.
Thanks,
Chris Maness
I think you might have a slight confusion. f(x) and g(x) are not equal to 1. f(1)=1 and g(1)=1, but f(x) might equal for example f(x)=x^3, and g(x) might be g(x)=x^2. Both satisfy f(1)=g(1)=1. When you add them together you get h(x)=f(x)+g(x)=x^3+x^2 which is still a polynomial of degree 3, but which no longer satisfies h(1)=1 as now h(1)=2.
1 person
Yes it is finally starting to soak in. I am also reading a GPL (Free) Linear algebra text book that is opening this up for me too. So I am correct in saying V:={f(1),g(1),1} (where f(1)=g(1)=1) is all that is a part of that failed vector space? So maybe if I defined V like V:={f(1),g(1),1,2} it would now have closure since 2 is now part of the set.
Thanks,
Chris Maness
Matterwave
Gold Member
What set are you defining?
The set you defined earlier was V={f(x)|f(1)=1 and f(x) is a polynomial of degree 3 or less}
This is read "the set V is the set of all function f(x) such that f(1) is equal to 1 and f(x) is a polynomial of degree 3 or less".
I don't understand what you mean with your notation.
If you meant V={f(x)|f(1)=1 or 2 and f(x) is a polynomial of degree 3 or less}, then no this new set doesn't form a vector space either. You will have, in this set elements where f(1)=2. The sum of this element with any other element g(x) of the set will necessarily have h(1)=f(1)+g(1)>2, and now h(x) is not in your set.
I thought of a example. Say my set is V:={i,j,k} where i,j,k are unit vectors. I need to allow something akin to the cross product to be one of the operations so that I get the other vectors as a product (in the general since) of two of the vectors (e.g. ixj=k). I think I might be missing 0, and +/- in front as well to meet all 10 properties of a proper vector space. However, what I have shown is the property of closure. Correct?
Thanks,
Chris Maness
What set are you defining?
The set you defined earlier was V={f(x)|f(1)=1 and f(x) is a polynomial of degree 3 or less}
This is read "the set V is the set of all function f(x) such that f(1) is equal to 1 and f(x) is a polynomial of degree 3 or less".
I don't understand what you mean with your notation.
If you meant V={f(x)|f(1)=1 or 2 and f(x) is a polynomial of degree 3 or less}, then no this new set doesn't form a vector space either. You will have, in this set elements where f(1)=2. The sum of this element with any other element g(x) of the set will necessarily have h(1)=f(1)+g(1)>2, and now h(x) is not in your set.
We were posting at the same time, so take the my last post as occurring before yours.
If we were going to add g(1)=1 to f(1)=1, would not g(1)=1 need to be a part of the original set?
Sorry, if my notation seems confusing or ill formed. I have never worked on math that is this abstract before, so I am an admitted hack.
Thanks,
Chris Maness
Matterwave
Gold Member
I thought of a example. Say my set is V:={i,j,k} where i,j,k are unit vectors. I need to allow something akin to the cross product to be one of the operations so that I get the other vectors as a product (in the general since) of two of the vectors (e.g. ixj=k). I think I might be missing 0, and +/- in front as well to meet all 10 properties of a proper vector space. However, what I have shown is the property of closure. Correct?
Thanks,
Chris Maness
Are you trying to make the cross product the "addition of vectors" operation? You can't do this because addition must be commutative, while the cross product is not.
[STRIKE]Your set is closed under the operation of cross product. That's about all you can say about your set.[/STRIKE] It's not even a group under the cross product operation since there is no identity element. It doesn't even come close to defining a vector space.
EDIT: Actually, I just realized your set is not even closed under the cross product as jxi=-k, which is not in your set.
Matterwave
Gold Member
We were posting at the same time, so take the my last post as occurring before yours.
If we were going to add g(1)=1 to f(1)=1, would not g(1)=1 need to be a part of the original set?
Sorry, if my notation seems confusing or ill formed. I have never worked on math that is this abstract before, so I am an admitted hack.
Thanks,
Chris Maness
I think you have some deep misunderstanding, but it's hard for me to figure out what, because it's hard for me to decipher your language.
g(1)=1 is not an element, it's an equation. g(x) (the whole function, not it's value at some x) is an element. If g(x) satisfies the equation g(1)=1, then g(x) is in the set you originally identified. If not, then g(x) is not in the set.
I think you have some deep misunderstanding, but it's hard for me to figure out what, because it's hard for me to decipher your language.
g(1)=1 is not an element, it's an equation. g(x) (the whole function, not it's value at some x) is an element. If g(x) satisfies the equation g(1)=1, then g(x) is in the set you originally identified. If not, then g(x) is not in the set.
Yes, that helps. g(1) is part of the set because for our purposes it is identical to f(1).
Returning to my little thought experiment.
1. Expand V to include {+/-i,+/-j,+/-k,0,1} where dividing and taking the modulus of any vector is allowed.
Am I getting closer? If not, how is i,j,k properly defined as a vector space? It seems obvious that it is.
Edit: I added a line above after "Yes, that helps"
Thanks,
Chris Maness
Matterwave
Gold Member
Yes, that helps. g(1) is part of the set because for our purposes it is identical to f(1).
Really, I would NEVER say g(1) is part of the set (although technically it is since g(1)=1 and 1 is in the set as a polynomial of degree 0), this is seriously confusing language. I would always say g(x) is part of the set if g(1)=1.
Returning to my little thought experiment.
1. Expand V to include {+/-i,+/-j,+/-k,0,1} where dividing and taking the modulus of any vector is allowed.
Am I getting closer? If not, how is i,j,k properly defined as a vector space? It seems obvious that it is.
Edit: I added a line above after "Yes, that helps"
Thanks,
Chris Maness
i,j,k do NOT define a vector space, they are the basis of a vector space. The vector space for which they are the basis for is ##\mathbb{R}^3##.
The vector space ##\mathbb{R}^3## consists of not only i,j,k (which is only 3 elements!) but of ALL sums of i,j,k over the real numbers. This means that 3i+2j+4k is in the set, and so is .1i+3.23j-4.3k. In general, all vectors of the form ##\mathbb{R}i+\mathbb{R}j+\mathbb{R}k## are in the set.
There is an uncountably infinite number of elements in this vector space. The elements are closed under vector addition obviously, and they are closed under vector additions over the field of real numbers. Vector addition is commutative and associative. The negative of any element is also in the set (e.g. -i is the negative of i). The identity element is the 0 vector. And all the other properties of vector spaces are satisfied.
The cross product does NOT have anything to do with the vector space nature defined above. It is an ADDITIONAL structure that we added to the vector space. | 3,786 | 14,165 | {"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} | 4.40625 | 4 | CC-MAIN-2020-50 | latest | en | 0.928936 |
https://oeis.org/A065806 | 1,632,038,362,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00487.warc.gz | 479,020,325 | 4,628 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A065806 A pseudo-random sequence. 3
2, 0, 4, 0, 5, 2, 8, 0, 7, 3, 8, 10, 10, 3, 6, 17, 0, 12, 16, 4, 21, 2, 18, 21, 5, 22, 22, 3, 16, 12, 14, 23, 30, 17, 29, 18, 15, 39, 0, 25, 37, 43, 0, 7, 37, 6, 33, 31, 40, 46, 37, 8, 43, 17, 22, 30, 25, 19, 43, 8, 10, 50, 49, 60, 14, 36, 45, 28, 56, 38, 33, 26, 68, 61, 63, 72, 72, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The sequence looks like a realization of a sequence of independent random variables x(n), with probability function P(x(n) = i) = 1/(n+2), i=0..n+1. The sequence (a(n)+1)/(n+3) seems to be a good U(0,1) pseudo-random number generator. Conjectures: -1 < a(n) < n+2; If a(n) = n+1 then a(n+1) = 0; All integers >1 occur in the sequence; For each c in [0,1] there exists a subsequence a(i_j) with (a(i_j)+1)/(i_j+3)-> c, j->infinity. a(A241671(n)) = 0; a(A241887(n)) = n and a(m) <> n for m < A241887(n). - Reinhard Zumkeller, Aug 09 2014 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 Klaus Strassburger, Plot of points [n, A065806(n)], n=1..20000 Klaus Strassburger, Plot of points [n,(A065806(n)+1)/(n+3)], n=1..20000 MAPLE N := 2000: b := array(-N..N, [seq(i, i=-N..N)]): s := 0; for i from 0 to N do; j := (b[i]-b[i-1]); s := s+1; a[s] := j+1; if j > N or j < -N then break; end if; c := b[i]; b[i] := b[j]; b[j] := c; end do: a := [seq(a[l], l=1..s)]; MATHEMATICA n = 2000; Clear[b]; b[i_] := b[i] = i; s = 0; For[i = 0, i <= n, i++, j = b[i] - b[i-1]; s++; a[s] = j+1; If[j > n || j < -n, Break[]]; c = b[i]; b[i] = b[j]; b[j] = c]; Table[a[l], {l, 1, s}] (* Jean-François Alcover, Sep 19 2016, adapted from Maple *) PROG (Haskell) following the Maple program import Data.IntMap (empty, findWithDefault, insert) a065806 n = a065806_list !! (n-1) a065806_list = f 0 empty where f i m = (j + 1) : f (i + 1) (insert i (b j) \$ insert j bi m) where j = bi - b (i - 1) bi = b i b x = findWithDefault x x m -- Reinhard Zumkeller, Aug 09 2014 CROSSREFS Cf. A241671, A241887. Sequence in context: A114402 A035647 A225437 * A331185 A228087 A320582 Adjacent sequences: A065803 A065804 A065805 * A065807 A065808 A065809 KEYWORD easy,nice,nonn AUTHOR Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Nov 21 2001 STATUS approved
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Last modified September 19 03:31 EDT 2021. Contains 347550 sequences. (Running on oeis4.) | 1,118 | 2,782 | {"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} | 3.421875 | 3 | CC-MAIN-2021-39 | latest | en | 0.68103 |
https://www.edplace.com/worksheet_info/maths/keystage2/year4/topic/935/877/collecting-data:-how-many-apples | 1,716,164,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00012.warc.gz | 678,310,743 | 12,482 | # Understand Data in a Pictogram
In this worksheet, students will read information from a pictogram and answer questions on it.
Key stage: KS 2
Curriculum topic: Statistics
Curriculum subtopic: Compare Data Using Charts and Graphs
Difficulty level:
#### Worksheet Overview
Example
Complete the information from the pictogram.
A family has an apple orchard.
The pictogram shows the apples (in kg) picked each month.
represents 4 kg of apples. represents 2 kg of apples. represents 1 kg of apples.
Aug Sep Oct Nov
How many kg of apples were picked in the month of August?
We look at the row of apples for August.
There are three whole apples, one half apple and one quarter apple.
According to the key, this means three lots of 4 kg, one lot of 2 kg and one of 1 kg.
4 + 4 + 4 + 2 + 1 = 15 kg
Are you happy to try some questions now?
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Get started | 291 | 1,254 | {"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-2024-22 | latest | en | 0.930036 |
https://bizfluent.com/info-8396377-meaning-gross-quantity.html | 1,600,797,298,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00513.warc.gz | 290,988,786 | 32,491 | # What Is the Meaning of Gross Quantity?
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Gross, in relation in numbers, can be used as either a unit of measurements or in economics as a term applied to numbers indicating that they are prior to deductions. For example, gross profit is the profit that a business has produced after it has deducted cost of goods sold before it has deducted operating expenses. In regards to gross quantity, it is the number of whatever item is being measured prior to existing deductions that would count against that number.
## Gross as a Unit of Measurement
A gross is sometimes used as an unit of measurement meaning a dozen dozens. For example, a gross of hot cross buns would be 144 hot cross buns. A dozen gross, or 1,728, is called a great gross. In common usage, a gross is most often abbreviated as either "gr" or "gro."
## Gross in Economics
In economics, gross is a label applied to numbers to indicate that possible deductions have yet to be deducted from them. For example, a business might sell eight units of what goods it had on hand but also saw two of those units returned in the same period, in this case, its gross quantities sold would be eight but its actual quantities sold would be six.
## Net in Economics
Net is the term used in economics to indicate numbers that have had deductions taken from them. For example, net profit or net income is gross profit minus operating expenses and all other miscellaneous expenses including interest and taxation. Using the above example, the actual quantities sold by the business after the returns could also be called its net quantities sold.
## Calculation of Gross and Net Sales
Gross and net quantities are most likely to come up during the calculation of a business's revenue from sales. Sales revenue is calculated as the net of gross sales revenue once the cost of goods sold has been deducted. Said cost is largely based on the number of goods sold multiplied by their purchase costs. Gross quantity can describe the total number of goods sold by the business in that period while net quantity could describe the total number of goods sold by the business in that period that were not returned. | 444 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-40 | longest | en | 0.977462 |
http://www.pbs.org/wgbh/aso/resources/guide/phyact1index.html | 1,524,695,315,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947968.96/warc/CC-MAIN-20180425213156-20180425233156-00223.warc.gz | 484,973,834 | 2,707 | Looking Back in TimeOverview: Calculate time for sunlight to reach Earth and consider other time-based distance measurementsLearning Goal: Understand that light takes time to travel Video Link: New View of the Universe How do astronomers discover the history of the universe? What we detect in the night sky is essentially a snapshot of the past. To illustrate this concept, ask students when they think they would see the Sun "go out" if it were extinguished at exactly 12 noon. Remind them that the Sun is 150 million km away and that light travels about 300,000 km per second. (Earthlings would see the Sun vanish at about 12:08 P.M.) To explore why we see a delayed image, brainstorm other examples or analogies of delay in transmission time. For example, after you shut off the faucet, water still pours out of a garden hose for a few seconds. Ask students how far light would travel in a year if it takes eight minutes to go 150 million km. That distance, 9,450,000,000,000 km, is a light-year. Why would people invent this measure of distance? Brainstorm examples from students' experience of using time to measure distance (e.g., a 15-minute walk to school). Invite students to invent other speed-time units to measure distance -- a slug-minute, for example. What problems could this method have? (A particularly energetic slug would throw you off.) Discuss how the constancy of light's speed makes light-years extremely accurate. This exploration of light and distance can help students grasp the magnitude of Edwin Hubble's discoveries, seen in the "New View of the Universe" video segment. Physics and Astronomy Program Contents Atomic Ethics Stranger than Fiction? In-Depth Investigation: Universal Proportions Home | Resources for Educators Menu | Educator's Guide Contents | Help WGBH | PBS Online | Search | Feedback | Shop © 1998 WGBH | 393 | 1,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-17 | latest | en | 0.923586 |
http://en.wikipedia.org/wiki/Monotone_class_theorem | 1,419,711,818,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552814.40/warc/CC-MAIN-20141224185912-00080-ip-10-231-17-201.ec2.internal.warc.gz | 33,673,508 | 11,843 | Monotone class theorem
In measure theory and probability, the monotone class theorem connects monotone classes and sigma-algebras. The theorem says that the smallest monotone class containing an algebra of sets G is precisely the smallest σ-algebra containing G. It is used as a type of transfinite induction to prove many other theorems, such as Fubini's theorem.
Definition of a monotone class
A monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections, i.e. if $A_i \in M$ and $A_1 \subset A_2 \subset \ldots$ then $\cup_{i = 1}^\infty A_i \in M$, and similarly for intersections of decreasing sequences of sets.
Monotone class theorem for sets
Statement
Let G be an algebra of sets and define M(G) to be the smallest monotone class containing G. Then M(G) is precisely the σ-algebra generated by G, i.e. σ(G) = M(G)
Proof
The following was taken from Probability Essentials, by Jean Jacod and Philip Protter.[1] The idea is as follows: we know that the sigma-algebra generated by an algebra of sets G contains the smallest monotone class generated by G. So, we seek to show that the monotone class generated by G is in fact a sigma-algebra, which would then show the two are equal.
To do this, we first construct monotone classes that correspond to elements of G, and show that each equals the M(G), the monotone class generated by G. Using this, we show that the monotone classes corresponding to the other elements of M(G) are also equal to M(G). Finally, we show this result implies M(G) is indeed a sigma-algebra.
Let $\mathcal{B} = M(G)$, i.e. $\mathcal{B}$ is the smallest monotone class containing $G$. For each set $B$, denote $\mathcal{B}_B$ to be the collection of sets $A \in \mathcal{B}$ such that $A \cap B \in \mathcal{B}$. It is plain to see that $\mathcal{B}_B$ is closed under increasing limits and differences.
Consider $B \in G$. For each $C \in G$, $B \cap C \in G \subset \mathcal{B}$, hence $C \in \mathcal{B}_B$ so $G \subset \mathcal{B}_B$. This yields $\mathcal{B}_B = \mathcal{B}$ when $B \in G$, since $\mathcal{B}_B$ is a monotone class containing $G$, $\mathcal{B}_B \subset \mathcal{B}$ and $\mathcal{B}$ is the smallest monotone class containing $G$
Now, more generally, suppose $B \in \mathcal{B}$. For each $C \in G$, we have $B \in \mathcal{B}_C$ and by the last result, $B \cap C \in \mathcal{B}$. Hence, $C \in \mathcal{B}_B$ so $G \subset \mathcal{B}_B$, and so $\mathcal{B}_B = \mathcal{B}$ for all $B \in \mathcal{B}$ by the argument in the paragraph directly above.
Since $\mathcal{B}_B = \mathcal{B}$ for all $B \in \mathcal{B}$, it must be that $\mathcal{B}$ is closed under finite intersections. Furthermore, $\mathcal{B}$ is closed by differences, so it is also closed under complements. Since $\mathcal{B}$ is closed under increasing limits as well, it is a sigma-algebra. Since every sigma-algebra is a monotone class, $\mathcal{B} = \sigma\,(G)$, i.e. $\mathcal{B}$ is the smallest sigma-algebra containing G
Monotone class theorem for functions
Statement
Let $\mathcal{A}$ be a π-system that contains $\Omega\,$ and let $\mathcal{H}$ be a collection of functions from $\Omega$ to R with the following properties:
(1) If $A \in \mathcal{A}$, then $\mathbf{1}_{A} \in \mathcal{H}$
(2) If $f,g \in \mathcal{H}$, then $f+g$ and $cf \in \mathcal{H}$ for any real number $c$
(3) If $f_n \in \mathcal{H}$ is a sequence of non-negative functions that increase to a bounded function $f$, then $f \in \mathcal{H}$
Then $\mathcal{H}$ contains all bounded functions that are measurable with respect to $\sigma(\mathcal{A})$, the sigma-algebra generated by $\mathcal{A}$
Proof
The following argument originates in Rick Durrett's Probability: Theory and Examples. [2]
The assumption $\Omega\, \in \mathcal{A}$, (2) and (3) imply that $\mathcal{G} = \{A: \mathbf{1}_{A} \in \mathcal{H}\}$ is a λ-system. By (1) and the π − λ theorem, $\sigma(\mathcal{A}) \subset \mathcal{G}$. (2) implies $\mathcal{H}$ contains all simple functions, and then (3) implies that $\mathcal{H}$ contains all bounded functions measurable with respect to $\sigma(\mathcal{A})$
Results and Applications
As a corollary, if G is a ring of sets, then the smallest monotone class containing it coincides with the sigma-ring of G.
By invoking this theorem, one can use monotone classes to help verify that a certain collection of subsets is a sigma-algebra.
The monotone class theorem for functions can be a powerful tool that allows statements about particularly simple classes of functions to be generalized to arbitrary bounded and measurable functions.
References
1. ^ Jacod, Jean; Protter, Phillip (2004). Probability Essentials. Springer. p. 36. ISBN 978-3-540-438717.
2. ^ Durrett, Rick (2010). Probability: Theory and Examples (4th ed.). Cambridge University Press. p. 100. ISBN 978-0521765398. | 1,408 | 4,931 | {"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": 64, "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-2014-52 | latest | en | 0.901033 |
https://testbook.com/question-answer/the-average-of-14-numbers-is-16-5-the-average-of--607714fdf7142e56d76c4a3a | 1,679,515,306,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00668.warc.gz | 641,941,792 | 80,375 | The average of 14 numbers is 16.5. The average of the first 5 numbers is 12 and that of the last 10 numbers is 18.5. If the 5th number is eliminated then find the average of the rest 13 numbers.
1. 16.7
2. 17
3. 14.5
4. 15.3
Option 1 : 16.7
Free
CT: General Knowledge (Mock Test)
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Detailed Solution
Given:
The average of 14 numbers = 16.5
The average of the first 5 numbers = 12
The average of the last 10 numbers = 18.5
Formula used:
Sum of all observation = average of all observation × total observation
Calculation:
Sum of 14 numbers = 16.5 × 14
⇒ 231
Sum of first 5 numbers = 12 × 5
⇒ 60
Sum of last 10 numbers = 10 × 18.5
⇒ 185
Sum of 15 numbers = 60 + 185
⇒ 245
Here the 5th number is present in the sum of 5 numbers and as well as in the sum of 10 numbers.
So, 5th number = sum of 5 numbers + sum of 10 numbers - the sum of 14 numbers
⇒ 60 + 185 - 231
⇒ 14
Sum of rest 13 numbers = sum of 14 numbers - 14
⇒ 231 - 14
⇒ 217
Average of rest 13 numbers = sum of rest 13 numbers/13
⇒ 217/13
⇒ 16.7
∴ The average of 13 numbers is 16.7.
The average of 14 numbers = 16.5
The average of the first 5 numbers = 12
Changed value in one = 12 - 16.5 = -4.5
Change in 5 numbers = -4.5 × 5 = -22.5
The average of last 10 numbers = 18.5
Changed value = 18.5 - 16.5 = 2
Change in 10 numbers = 2 × 10 = 20
Total change = -22.5 + 20 = -2.5
So, 5th number = 16.5 - 2.5 = 14
Average of 13 numbers excluding 5th number = 16.5 + (16.5 - 14)/13
⇒ 16.5 + 2.5/13
⇒ 16.5 + 0.2
⇒ 16.7
∴ The average of 13 numbers is 16.7. | 616 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-14 | latest | en | 0.765136 |
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Author Message
Pranjal
New User
Joined: 17 Mar 2006
Posts: 4
Posted: Sat Apr 22, 2006 11:55 am Post subject: Calculating space for COMP variables How much space in terms of bytes will the following variables take and how is it calculated? S9(4) COMP S9(5) COMP S9(8) COMP S9(9) COMP S9(10) COMP
iknow
Active User
Joined: 22 Aug 2005
Posts: 413
Posted: Sat Apr 22, 2006 8:17 pm Post subject: Re: Calculating space for COMP variables Hi, COMP is a binary storage format . The sign stored in a COMP field will be in the most significant bit. Bit is ON if -ve, off if +ve. S9(4) COMP - 2 bytes S9(5) COMP - 2 bytes S9(8) COMP - 4 bytes S9(9) COMP - 4 bytes S9(10) COMP - 5 bytes Hope this is what you had expected.
mmwife
Super Moderator
Joined: 30 May 2003
Posts: 1592
Posted: Sat Apr 22, 2006 9:08 pm Post subject: Hi, The best place to find the ans to ques like yours is in the manuals. From the Zos Pgming Reference Manual: Digits in PICTURE clause Storage occupied 1 through 4 2 bytes (halfword) 5 through 9 4 bytes (fullword) 10 through 18 8 bytes (doubleword) Scroll to the top of this page; clk "manuals". Scroll down to manual selections and clk on your choice.
Pranjal
New User
Joined: 17 Mar 2006
Posts: 4
Posted: Sun Apr 23, 2006 12:33 pm Post subject: Re: Calculating space for COMP variables Thanks for the replies. I must be checking the mauals first.
KS
New User
Joined: 28 Feb 2006
Posts: 91
Location: Chennai
Posted: Tue Apr 25, 2006 3:18 pm Post subject: Hi, The COMP occupies Int(n/2) 4 bytes and COMP-3 occupies Int((n/2)+1) bytes. So using this we get the results as : S9(4) COMP - Int(4/2) = 2 bytes S9(5) COMP - Int(5/2) = 2 bytes S9(8) COMP - Int(8/2) = 4 bytes S9(9) COMP - Int(9/2) = 4 bytes S9(10) COMP - Int(10/2) =5 bytes Thanks, KS
mmwife
Super Moderator
Joined: 30 May 2003
Posts: 1592
Posted: Wed Apr 26, 2006 6:27 am Post subject: KS, The info I offered above was cut & pasted from IBM's eCOBOL Language Reference Manual (SC27-1408-01). Where did you get your info?
KS
New User
Joined: 28 Feb 2006
Posts: 91
Location: Chennai
Posted: Thu Apr 27, 2006 4:08 pm Post subject: HI mmwife, Sorry all , I was wrong on tht.. A binary number with a picture description, - less than 4 or fewer decimal digits will occupy 2 bytes. - 5 to 9 decimal digits will occupy 4 bytes. - 10 to 18 bytes occupies 8 bytes. Regret the inconvienience caused. Thanks, KS
kamran
New User
Joined: 01 May 2005
Posts: 55
Posted: Thu Apr 27, 2006 4:48 pm Post subject: Re: Calculating space for COMP variables
Pranjal wrote: How much space in terms of bytes will the following variables take and how is it calculated? S9(4) COMP S9(5) COMP S9(8) COMP S9(9) COMP S9(10) COMP
Hi ,
I have written a rexx routines named copyflds to compute the lenght and offset of each feild in a cobol copy member.
I think you don't need to change it much but you can customize it upon your name conventions.
One of The systax of calling it is:
TSO COPYFLDS dataset-name copy-member-name
which copy-member-name is the name of the cobol copy member name
and dataset-name is the name of the dataset which the copy members is reside in.
please copy both attach files COPYFLDS and COPYPICS into a dataset which is concatenated in your user sysproc ddname .
because I prepared it just for my site, I didn't prepared it any document so if there is any problem using it please put a message here.
kamran
New User
Joined: 01 May 2005
Posts: 55
Posted: Thu Apr 27, 2006 4:58 pm Post subject: Re: Calculating space for COMP variables Sorry I don't know why the mentioned bellow files didn't attached, So I will try it once!. I think the problem was that I didn't select any extention for the files. So use the attached files to this one! Hi , I have written a rexx routines named copyflds to compute the lenght and offset of each feild in a cobol copy member. I think you don't need to change it much but you can customize it upon your name conventions. One of The systax of calling it is: TSO COPYFLDS dataset-name copy-member-name which copy-member-name is the name of the cobol copy member name and dataset-name is the name of the dataset which the copy members is reside in. please copy both attach files COPYFLDS and COPYPICS into a dataset which is concatenated in your user sysproc ddname . because I prepared it just for my site, I didn't prepared it any document so if there is any problem using it please put a message here.[/quote]
new2cobol
New User
Joined: 04 Jan 2006
Posts: 77
Location: Bangalore
Posted: Sat Apr 29, 2006 12:49 am Post subject: This should help!!!
dneufarth
Active User
Joined: 27 Apr 2005
Posts: 289
Location: Cincinnati OH USA
Posted: Sat Apr 29, 2006 2:18 am Post subject: Pranjal, the official wisdom from IBM http://publibz.boulder.ibm.com/cgi-bin/bookmgr_OS390/BOOKS/IGY3LR30/5.3.17.1?DT=20050714120224
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# You win. Your spending at EE?
Posted on 2004-09-23
270 Views
Two part question:
You win the \$64,000.00 (USD) question. You decide to spend it all to buy points on EE.
Questions:
1) How many points can you buy, total?
2) Would you, now, be the expert with the most points?
All points to first respondent with BOTH answers. Show your work or accept my math.
0
Question by:dhsindy
• 2
• 2
• 2
LVL 44
Expert Comment
2) when you buy points, you are actually purchasing Question Points, not 'EXPERT POINTS' - Question points are the currency by which you evaluate the 'worth' of a question when you ask it , and these are then awared to the answerer when he/she provides an accepted answer to the question. They are then converted to the 'Expert Points' earned by that answerer. However, Question Points and Expert Points are NOT the same thing, and are not interchangeable.
For example, at the present time, I pewrsonally have something in excess of 1,014,000 Expert Points, but only 4203 Question Points. That means that I could (if I were so inclined) ask 8 high value questions (500 points each). I cannot use the Expert Points to ask lots of questions.
So you second question is essentially meaningless.
AW
0
LVL 16
Author Comment
Clarification:
My mistake. For the sake of this question only, assume all points are the same - purchased points are added to your current expert points - disregard available points for asking questions.
0
LVL 44
Accepted Solution
Arthur_Wood earned 100 total points
at a rate to 1000 points for \$12.95, you could purchase 4,942,000 points, and yes that you make you the #1 'expert'.
AW
0
LVL 24
Expert Comment
Well, I not 1st, so before reading other submissions and letting even more respondees get in line ahead of me .............................
1) All
2) No
> with BOTH answers
done
> accept my math
OK, I accept 64K result. Payable in silver or gold?
0
LVL 16
Author Comment
Thanks for everyones input.
Sorry, SunBow. Maybe next time.
NOTE: For those that may be interested CrazyOne has about 3.8 million points at the moment.
0
LVL 24
Expert Comment
Arthur_Wood> question is essentially meaningless
I agree.
Another problem here is in space/Time.
> Show your work
example: I recall pts offered at one for ten. A .1 gets 1. My value assessment was 10 for .01 so (a) no matter how much money was available, I'd spend zero for points and use the rest for something else, such as a nicer vehicle. (b) the price per point is not a constant in either history or any physical reality. What this means is, after purchase of the first point, there can be no validity to presumption that the cost will be same for 2nd point or any other. Just look at stock market for goos sense of price stabilizing over time.
example: There is capability for numerous askers to offer an unlimited number of points. Without any fixed ceiling. What this translates to is that any member who responds has the capability of receiveing an unlimited number of points, and that includes not only me + 64k, but all other participants in this thread.
Concluding, that without other information available, one cannot predict
> the expert with the most points?
throughout space/time, However we can say that we each have a chance to reach a tie with others, having a point total of "unlimited", but no way to tell, or predict either of us going any higher. So while you or I could have the highest total points, since we cannot guarantee that no one else would seize same opportunities, we'd not be only one with highest total.
Total points change with the times, and time is not frozen for competition
> 2) Would you, now, be the expert with the most points?
No.
(barring expunging of recordable materiel)
0
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two point charges of magnitude +3.00 x 10^-6 C and -5.00 x 10^-6 C are places along the x-axis at x = 0cm and x = 40cm, respectively. Where must a third charge, q, be places along the x-axis so that it does not experience any net force because of the other two charges? | 86 | 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} | 2.609375 | 3 | CC-MAIN-2013-48 | longest | en | 0.936617 |
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1. Oct 3, 2009
### colossus__
1. The problem statement, all variables and given/known data
The problem is rather simple. A set of linear equations, in the form V=Z*I, is given to represent a circuit in the frequency domain. The values for the V and I vectors are given and i have the Z(impedance) matrix writen in function of the Z1, Z2, Z3,...,Zn variables?
In a simpler way: How to solve the A*x=B equation, when the values for the x and B vectors are given, and A is writen in function of A1, A2, A3,...,An variables.
In my case, the A matrix looks like this:
|A1 -A2 0 |
|A1 0 -A3|
|1 1 1 |
x vector:
-0.26
0.259 - i0.966
0.259 + i0.966
B vector:
150 + i0.342
-150 + i0.342
0
2. Relevant equations
As simple as stated before, the only equation is the A*x=B. My example with all the complex numbers is not the best to ask for help in this subject, but it's the one that has led me to it.
3. The attempt at a solution
So far i have tried algebraic manipulation of the A*x=B equation multiplying it by A^-1 in the attempt to somehow reduce the matrix into a vector or any of the vectors into a matrix.
I tought maybe eigenvectors and eigenvalues could be involved in the solution of this problem, but since i don't have any linear algebra books around and don't really remember how to use this stuff, i'm hoping you guys could help me with this one.
Last edited: Oct 3, 2009
2. Oct 3, 2009
### Dick
It's not that hard. Just multiply A by the vector x and equate the resulting vector to B. That gives you three linear equations in the three unknowns A1, A2, A3. Actually, I'll save you a little time. The (1,1,1) row of the matrix times x gives (-0.26)+2*0.259. That's not zero but the third component of B is zero. That makes the system inconsistent. Is there a typo in there somewhere?
Last edited: Oct 3, 2009
3. Oct 3, 2009
### colossus__
You pointed it right, there is something wrong with the system. I went back to the question and found that the data given is wrong. The sum of the currents in the node in question is not equal to zero, therefore we have this problem.
I found this question in a test that is now 2 years old, and i'm actually surprised that none of the students who took the test noticed this mistake.
We could try to redefine the x vector to:
-0.52
0.26 - i0.966
0.26 + i0.966
Now the sum is really equal to zero. I just don't know if this will mess up with the balance in the rest of the system.
4. Oct 3, 2009
### colossus__
I tried using your suggestion to solve the problem, it ended up looking like this:
(in a step by step version)
$$\begin{bmatrix} A_1 & -A_2 & 0\\ A_1 & 0 & -A_3\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} B_1\\ B_2\\ B_3 \end{bmatrix}$$
$$\begin{bmatrix} x_1A_1 & -x_2A_2 & 0\\ x_1A_1 & 0 & -x_3A_3\\ x_1 & x_2 & x_3 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} B_1\\ B_2\\ B_3 \end{bmatrix}$$
$$\begin{bmatrix} x_1 & -x_2 & 0\\ x_1 & 0 & -x_3\\ \frac{x_1}{A_1} & \frac{x_2}{A_2} & \frac{x_3}{A_3} \end{bmatrix} \begin{bmatrix} A_1\\ A_2\\ A_3 \end{bmatrix} = \begin{bmatrix} B_1\\ B_2\\ B_3 \end{bmatrix}$$
Using the actual values,
$$\begin{bmatrix} -0.52 & -0.26 + i0.966 & 0\\ -0.52 & 0 & -0.26 - i0.966\\ \frac{-0.52}{A_1} & \frac{0.26 - i0.966}{A_2} & \frac{0.26 + i0.966}{A_3} \end{bmatrix} \begin{bmatrix} A_1\\ A_2\\ A_3 \end{bmatrix} = \begin{bmatrix} 150 + i0.342\\ -150 + i0.342\\ 0 \end{bmatrix}$$
And i still got no clue on how to solve it with those divisions by An that showed up now.
5. Oct 4, 2009
### Dick
You are doing some strange matrix manipulations. If you really want to write this as a matrix equation the last row should really be [0,0,0] and B3 should be 0. There's no A dependency in the last equation. Which brings us to the next problem, now you only have two equations in three unknowns. There are many solutions. | 1,314 | 3,933 | {"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} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.938438 |
https://www.saranextgen.com/homeworkhelp/doubts.php?id=54103 | 1,701,232,592,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00438.warc.gz | 1,092,961,415 | 5,805 | # A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by (a) 3 (b) 5 (c) 9 (d) 11
## Question ID - 54103 :- A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by (a) 3 (b) 5 (c) 9 (d) 11
3537
Let the ten’s digit be and the unit’s digit be
Then, number =
Number obtained by interchanging the digits =
which is divisible by 11.
Next Question :
The sum of the digits of a two-digit number is 9 less than the number. Which of the following digits is unit’s place of the number? (a) 1 (b) 2 (c) 4 (d) Data inadequate | 207 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-50 | longest | en | 0.788269 |
https://www.convertunits.com/from/ml/to/ounce+%5BUS,+liquid%5D | 1,701,755,840,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00170.warc.gz | 799,244,011 | 12,920 | ## Convert milliliter to ounce [US, liquid]
ml ounce [US, liquid]
Did you mean to convert megaliter milliliter to ounce [US, liquid] ounce [UK, liquid]
How many ml in 1 ounce [US, liquid]? The answer is 29.5735296875. We assume you are converting between milliliter and ounce [US, liquid]. You can view more details on each measurement unit: ml or ounce [US, liquid] The SI derived unit for volume is the cubic meter. 1 cubic meter is equal to 1000000 ml, or 33814.022558919 ounce [US, liquid]. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between milliliters and ounces. Type in your own numbers in the form to convert the units!
## Quick conversion chart of ml to ounce [US, liquid]
1 ml to ounce [US, liquid] = 0.03381 ounce [US, liquid]
10 ml to ounce [US, liquid] = 0.33814 ounce [US, liquid]
20 ml to ounce [US, liquid] = 0.67628 ounce [US, liquid]
30 ml to ounce [US, liquid] = 1.01442 ounce [US, liquid]
40 ml to ounce [US, liquid] = 1.35256 ounce [US, liquid]
50 ml to ounce [US, liquid] = 1.6907 ounce [US, liquid]
100 ml to ounce [US, liquid] = 3.3814 ounce [US, liquid]
200 ml to ounce [US, liquid] = 6.7628 ounce [US, liquid]
## Want other units?
You can do the reverse unit conversion from ounce [US, liquid] to ml, or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Millilitre
The millilitre (ml or mL, also spelled milliliter) is a metric unit of volume that is equal to one thousandth of a litre. It is a non-SI unit accepted for use with the International Systems of Units (SI). It is exactly equivalent to 1 cubic centimetre (cm³, or, non-standard, cc).
## Definition: Ounce
Note that this is a fluid ounce measuring volume, not the typical ounce that measures weight. It only applies for a liquid ounce in U.S. measurements.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 636 | 2,352 | {"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} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.839947 |
https://www.cut-the-knot.org/proofs/ShortestIsLongest.shtml | 1,524,664,143,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947803.66/warc/CC-MAIN-20180425115743-20180425135743-00489.warc.gz | 768,858,465 | 12,850 | # The Size is in the Eyes of the Beholder
Six points are given in the space such that the pairwise distances between them are all distinct. Consider the triangles with vertices at these points. Prove that the longest side of one of these triangles is at the same time the shortest side of another [Savchev & Andreescu, p. 146].
Proof
Consider the triangles one at a time and, in each, color red the shortest side. Some of the segments may well be colored repeatedly. Paint the remaining segments in blue. This brings us to a situation of a complete graph with 6 nodes all of whose edges are colored in one of two colors. As was mentioned elsewhere and will be shown shortly, there is always a monochromatic triangle. In our situation it must be all red, because an all-blue triangle still needs to have a shortest side which then would have been painted red. In an all-red triangle there is a longest side of course. But being painted red means that in another triangle it is the shortest.
Now let's prove that a bichromatic complete graph with 6 nodes has a monochromatic triangle. An arbitrary node in a complete graph is linked by the edges to all the remaining nodes - 5 for K6. Out of five edges painted in two colors, there are at least three painted in the same color. For example, assume that there are three red edges incident with the same node. These red edges have three "free" ends that are joined by another triple of edges. There are just two possibilities:
1. All of the three edges are painted blue, or
2. one of them is painted red.
In the first case, there is an all-blue triangle. In the second, there is an all-red triangle.
### References
1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003 | 392 | 1,732 | {"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-2018-17 | longest | en | 0.959456 |
https://dividedby.org/category/divided-by-18 | 1,653,633,065,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00497.warc.gz | 276,139,110 | 17,734 | Home » Divided by 18
# Divided by 18
Welcome to divided by 18, our category containing the posts which explain the division of a particular number x by 18. The number x is called nominator, and 18 is called the denominator of the quotient x / 18. In every article we give you the result of the fraction x over 18 in high accuracy, along with the value in percentage and the properties of the result, e.g. terminating or repeating. In the former case, we state the number of decimal places, whereas in the latter case we provide you with information regarding the repented. Also, for any x divided by 18 like 4965 divided by 18, we review the frequently asked question in the context. What’s more is that all entries in this category come with a unique calculator you will like. Information regarding the use of our search box, directions for further information, and how to ask us a question can always be found as well.
## 998000 Divided by 18
Welcome to 998000 divided by 18, our post which explains the division of nine hundred ninety-eight thousand by eighteen to you. 🙂 The number 998000… Read More »998000 Divided by 18
## 999000 Divided by 18
Welcome to 999000 divided by 18, our post which explains the division of nine hundred ninety-nine thousand by eighteen to you. 🙂 The number 999000… Read More »999000 Divided by 18
## 991000 Divided by 18
Welcome to 991000 divided by 18, our post which explains the division of nine hundred ninety-one thousand by eighteen to you. 🙂 The number 991000… Read More »991000 Divided by 18
## 992000 Divided by 18
Welcome to 992000 divided by 18, our post which explains the division of nine hundred ninety-two thousand by eighteen to you. 🙂 The number 992000… Read More »992000 Divided by 18
## 993000 Divided by 18
Welcome to 993000 divided by 18, our post which explains the division of nine hundred ninety-three thousand by eighteen to you. 🙂 The number 993000… Read More »993000 Divided by 18
## 994000 Divided by 18
Welcome to 994000 divided by 18, our post which explains the division of nine hundred ninety-four thousand by eighteen to you. 🙂 The number 994000… Read More »994000 Divided by 18
## 995000 Divided by 18
Welcome to 995000 divided by 18, our post which explains the division of nine hundred ninety-five thousand by eighteen to you. 🙂 The number 995000… Read More »995000 Divided by 18
## 996000 Divided by 18
Welcome to 996000 divided by 18, our post which explains the division of nine hundred ninety-six thousand by eighteen to you. 🙂 The number 996000… Read More »996000 Divided by 18
## 997000 Divided by 18
Welcome to 997000 divided by 18, our post which explains the division of nine hundred ninety-seven thousand by eighteen to you. 🙂 The number 997000… Read More »997000 Divided by 18
## 982000 Divided by 18
Welcome to 982000 divided by 18, our post which explains the division of nine hundred eighty-two thousand by eighteen to you. 🙂 The number 982000… Read More »982000 Divided by 18 | 765 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | latest | en | 0.925826 |
https://www.quizzes.cc/metric/percentof.php?percent=1&of=894 | 1,618,313,558,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00069.warc.gz | 1,060,976,638 | 4,239 | #### What is 1 percent of 894?
How much is 1 percent of 894? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 1% of 894 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 1% of 894 = 8.94
Calculate another percentage below. Type into inputs
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Calculating one of eight hundred and ninety-four How to calculate 1% of 894? Simply divide the percent by 100 and multiply by the number. For example, 1 /100 x 894 = 8.94 or 0.01 x 894 = 8.94
#### How much is 1 percent of the following numbers?
1% of 894.01 = 894.01 1% of 894.02 = 894.02 1% of 894.03 = 894.03 1% of 894.04 = 894.04 1% of 894.05 = 894.05 1% of 894.06 = 894.06 1% of 894.07 = 894.07 1% of 894.08 = 894.08 1% of 894.09 = 894.09 1% of 894.1 = 894.1 1% of 894.11 = 894.11 1% of 894.12 = 894.12 1% of 894.13 = 894.13 1% of 894.14 = 894.14 1% of 894.15 = 894.15 1% of 894.16 = 894.16 1% of 894.17 = 894.17 1% of 894.18 = 894.18 1% of 894.19 = 894.19 1% of 894.2 = 894.2 1% of 894.21 = 894.21 1% of 894.22 = 894.22 1% of 894.23 = 894.23 1% of 894.24 = 894.24 1% of 894.25 = 894.25
1% of 894.26 = 894.26 1% of 894.27 = 894.27 1% of 894.28 = 894.28 1% of 894.29 = 894.29 1% of 894.3 = 894.3 1% of 894.31 = 894.31 1% of 894.32 = 894.32 1% of 894.33 = 894.33 1% of 894.34 = 894.34 1% of 894.35 = 894.35 1% of 894.36 = 894.36 1% of 894.37 = 894.37 1% of 894.38 = 894.38 1% of 894.39 = 894.39 1% of 894.4 = 894.4 1% of 894.41 = 894.41 1% of 894.42 = 894.42 1% of 894.43 = 894.43 1% of 894.44 = 894.44 1% of 894.45 = 894.45 1% of 894.46 = 894.46 1% of 894.47 = 894.47 1% of 894.48 = 894.48 1% of 894.49 = 894.49 1% of 894.5 = 894.5
1% of 894.51 = 894.51 1% of 894.52 = 894.52 1% of 894.53 = 894.53 1% of 894.54 = 894.54 1% of 894.55 = 894.55 1% of 894.56 = 894.56 1% of 894.57 = 894.57 1% of 894.58 = 894.58 1% of 894.59 = 894.59 1% of 894.6 = 894.6 1% of 894.61 = 894.61 1% of 894.62 = 894.62 1% of 894.63 = 894.63 1% of 894.64 = 894.64 1% of 894.65 = 894.65 1% of 894.66 = 894.66 1% of 894.67 = 894.67 1% of 894.68 = 894.68 1% of 894.69 = 894.69 1% of 894.7 = 894.7 1% of 894.71 = 894.71 1% of 894.72 = 894.72 1% of 894.73 = 894.73 1% of 894.74 = 894.74 1% of 894.75 = 894.75
1% of 894.76 = 894.76 1% of 894.77 = 894.77 1% of 894.78 = 894.78 1% of 894.79 = 894.79 1% of 894.8 = 894.8 1% of 894.81 = 894.81 1% of 894.82 = 894.82 1% of 894.83 = 894.83 1% of 894.84 = 894.84 1% of 894.85 = 894.85 1% of 894.86 = 894.86 1% of 894.87 = 894.87 1% of 894.88 = 894.88 1% of 894.89 = 894.89 1% of 894.9 = 894.9 1% of 894.91 = 894.91 1% of 894.92 = 894.92 1% of 894.93 = 894.93 1% of 894.94 = 894.94 1% of 894.95 = 894.95 1% of 894.96 = 894.96 1% of 894.97 = 894.97 1% of 894.98 = 894.98 1% of 894.99 = 894.99 1% of 895 = 895
1% of 894 = 8.94 2% of 894 = 17.88 3% of 894 = 26.82 4% of 894 = 35.76 5% of 894 = 44.7 6% of 894 = 53.64 7% of 894 = 62.58 8% of 894 = 71.52 9% of 894 = 80.46 10% of 894 = 89.4 11% of 894 = 98.34 12% of 894 = 107.28 13% of 894 = 116.22 14% of 894 = 125.16 15% of 894 = 134.1 16% of 894 = 143.04 17% of 894 = 151.98 18% of 894 = 160.92 19% of 894 = 169.86 20% of 894 = 178.8 21% of 894 = 187.74 22% of 894 = 196.68 23% of 894 = 205.62 24% of 894 = 214.56 25% of 894 = 223.5
26% of 894 = 232.44 27% of 894 = 241.38 28% of 894 = 250.32 29% of 894 = 259.26 30% of 894 = 268.2 31% of 894 = 277.14 32% of 894 = 286.08 33% of 894 = 295.02 34% of 894 = 303.96 35% of 894 = 312.9 36% of 894 = 321.84 37% of 894 = 330.78 38% of 894 = 339.72 39% of 894 = 348.66 40% of 894 = 357.6 41% of 894 = 366.54 42% of 894 = 375.48 43% of 894 = 384.42 44% of 894 = 393.36 45% of 894 = 402.3 46% of 894 = 411.24 47% of 894 = 420.18 48% of 894 = 429.12 49% of 894 = 438.06 50% of 894 = 447
51% of 894 = 455.94 52% of 894 = 464.88 53% of 894 = 473.82 54% of 894 = 482.76 55% of 894 = 491.7 56% of 894 = 500.64 57% of 894 = 509.58 58% of 894 = 518.52 59% of 894 = 527.46 60% of 894 = 536.4 61% of 894 = 545.34 62% of 894 = 554.28 63% of 894 = 563.22 64% of 894 = 572.16 65% of 894 = 581.1 66% of 894 = 590.04 67% of 894 = 598.98 68% of 894 = 607.92 69% of 894 = 616.86 70% of 894 = 625.8 71% of 894 = 634.74 72% of 894 = 643.68 73% of 894 = 652.62 74% of 894 = 661.56 75% of 894 = 670.5
76% of 894 = 679.44 77% of 894 = 688.38 78% of 894 = 697.32 79% of 894 = 706.26 80% of 894 = 715.2 81% of 894 = 724.14 82% of 894 = 733.08 83% of 894 = 742.02 84% of 894 = 750.96 85% of 894 = 759.9 86% of 894 = 768.84 87% of 894 = 777.78 88% of 894 = 786.72 89% of 894 = 795.66 90% of 894 = 804.6 91% of 894 = 813.54 92% of 894 = 822.48 93% of 894 = 831.42 94% of 894 = 840.36 95% of 894 = 849.3 96% of 894 = 858.24 97% of 894 = 867.18 98% of 894 = 876.12 99% of 894 = 885.06 100% of 894 = 894 | 2,589 | 4,862 | {"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} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.788096 |
https://math.stackexchange.com/questions/3845109/natural-deduction-proof-fitch-alternative-using-disjunction-exclusion | 1,701,739,656,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00374.warc.gz | 440,757,097 | 37,181 | # Natural deduction proof (Fitch) - Alternative using disjunction exclusion
I have to build a fitch proof for the negation introduction rule
with some constraints: I cannot use ¬¬E, ¬I, RAA (Reductio ad absurdum) and ¬¬I. There is also another constraint saying that I have to use the law of excluded middle (LEM).
There are two tips provided to solve the problem. First of all, I have to use ¬E (or 0I). Second of all, I have to use the disjunction elimination:
The only way I found how to solve it is the following way:
Does anyone have an idea on how to use the disjunction elimination? As my solution doesn't use it?
EDIT 1: Correction in proof
EDIT 2: Thank you so much to @MauroALLEGRANZA for the help. This is what I came up with.
EDIT 3: Here are the rules I can use: https://drive.google.com/file/d/19WQOPuxyskq2s_eWXvR04sl6qMUG5CKx/view?usp=sharing
• Not clear... In your proof above you have proved something different (and the proof is unnecessary complicated). Now the question is: what are you trying to prove ? $\dfrac {\lnot \phi}{\phi \vdash 0}$ or $\dfrac {\phi \vdash 0}{\lnot \phi}$ ? Sep 30, 2020 at 13:28
• "A mathematical proof is an inferential argument for a mathematical statement, showing that the stated assumptions logically guarantee the conclusion." This holds also in mathematical logic. Sep 30, 2020 at 13:34
• Maybe it will be helpful if you list the rules regarding $\lnot$ and $0$ that you are allowed to use: $(\lnot \text E)$, $(0 \text I)$ and $(0 \text E)$ Sep 30, 2020 at 14:38
• So what rules can you use? Sep 30, 2020 at 17:14
• Also, that second proof is still not a proper proof .. you can't have two assumptions in a subproof ... or at least there is no formal rule I know of that makes any practical use of something like that.... unless you do have such a rule? Again, we need to know the rules that you an use in order to answer your question Sep 30, 2020 at 17:29
$$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$$
$$\fitch{ 1. \phi \vdash \bot \qquad \quad Assumption}{ 2. \phi \lor \neg \phi \qquad \quad \qquad LEM\\ \fitch{3. \phi \qquad \quad \quad Assumption} {4. \bot \qquad \vdash \ Elim \ (?) \ 3\\ 5. \neg \phi \qquad \bot \ Elim \ 4\\}\\ 6. \phi \vdash \neg \phi \vdash \ Intro \ (?) \ 3-5\\ \fitch{7. \neg \phi \qquad \quad \quad Assumption} {8. \neg \phi \qquad Reit \ 6\\}\\ 9. \neg \phi \vdash \neg \phi \vdash \ Intro \ (?) \ 6-7\\ 10. \neg \phi \qquad \qquad \lor \ Elim \ 2,6,9\\ }$$ | 782 | 2,474 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-50 | latest | en | 0.923749 |
http://gmatclub.com/forum/legislation-in-the-canadian-province-of-ontario-requires-of-38445.html | 1,485,122,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281574.78/warc/CC-MAIN-20170116095121-00328-ip-10-171-10-70.ec2.internal.warc.gz | 111,304,110 | 57,229 | Legislation in the Canadian province of Ontario requires of : GMAT Sentence Correction (SC)
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Legislation in the Canadian province of Ontario requires of
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14 Nov 2006, 02:20
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Legislation in the Canadian province of Ontario requires of both public and private employers that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
(A) that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are
(B) that pay for jobs historically held by women should be the same as for a job requiring comparable skills
(C) to pay the same in jobs historically held by women as in jobs of comparable skill that are
(D) to pay the same regardless of whether a job was historically held by women or is one demanding comparable skills
(E) to pay as much for jobs historically held by women as for a job demanding comparable skills
I COULD ELIMINATE.. B....Use of should not correct
and E.....jobs r compared to job....
If you have any questions
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14 Nov 2006, 04:17
A seems right.
B and E are out as they compare "jobs" to "job".
A seems the best since it has "require ....that" construction.
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14 Nov 2006, 05:41
requires of X...to.. seems to be idiomatic.
I would go with C.
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14 Nov 2006, 06:11
AK47 wrote:
Legislation in the Canadian province of Ontario requires of both public and private employers that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
(A) that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are
(B) that pay for jobs historically held by women should be the same as for a job requiring comparable skills
(C) to pay the same in jobs historically held by women as in jobs of comparable skill that are
(D) to pay the same regardless of whether a job was historically held by women or is one demanding comparable skills
(E) to pay as much for jobs historically held by women as for a job demanding comparable skills
I COULD ELIMINATE.. B....Use of should not correct
and E.....jobs r compared to job....
In my opinion, "Requires... that" is a better construction. I would bet on A
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14 Nov 2006, 06:12
I pick C
i'm having trouble figuring out what "That" in A and B is referring to.
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15 Nov 2006, 00:07
If OA is not A, please correct it now
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19 Nov 2006, 05:56
oa is A]
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19 Nov 2006, 06:32
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ak_idc wrote:
requires of X...to.. seems to be idiomatic.
I would go with C.
There are a few other ways to look at this problem.
First - this is a typical question where Subjunctive use is tested.....requires <of X> that <pay be the same ...etc.>
The prepositional phrase <of both public and private employers> is an additional non-restrictive information provided in the sentence. The sentence should remain intact even if this prepositional phrase is removed. In other words, when prepositional phrases act as adjectives they can be removed. Be careful because prepositional phrases can also act as adverbs -- The book was lying under the table
Legislation in the Canadian province of Ontario requires of both public and private employers that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
If you remove the prepositional phrase -
Legislation in the Canadian province of Ontario requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
You can even remove "in the canadian province of ontario" because again it is prepositional and provides additional non-restrictive information about Legislation -- the prepositional phrase acts as an adjective modifying Legislation and like all adjectives can be removed from the sentence without distorting the meaning.
Legislation requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are[/u] usually held by men.
Now see.... requires that pay be the same is a typical subjunctive usage.
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19 Nov 2006, 08:14
dwivedys wrote:
ak_idc wrote:
requires of X...to.. seems to be idiomatic.
I would go with C.
There are a few other ways to look at this problem.
First - this is a typical question where Subjunctive use is tested.....requires <of X> that <pay be the same ...etc.>
The prepositional phrase <of both public and private employers> is an additional non-restrictive information provided in the sentence. The sentence should remain intact even if this prepositional phrase is removed. In other words, when prepositional phrases act as adjectives they can be removed. Be careful because prepositional phrases can also act as adverbs -- The book was lying under the table
Legislation in the Canadian province of Ontario requires of both public and private employers that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
If you remove the prepositional phrase -
Legislation in the Canadian province of Ontario requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
You can even remove "in the canadian province of ontario" because again it is prepositional and provides additional non-restrictive information about Legislation -- the prepositional phrase acts as an adjective modifying Legislation and like all adjectives can be removed from the sentence without distorting the meaning.
Legislation requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are[/u] usually held by men.
Now see.... requires that pay be the same is a typical subjunctive usage.
You nailed it buddy
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19 Nov 2006, 09:34
dwivedys wrote:
ak_idc wrote:
requires of X...to.. seems to be idiomatic.
I would go with C.
There are a few other ways to look at this problem.
First - this is a typical question where Subjunctive use is tested.....requires <of X> that <pay be the same ...etc.>
The prepositional phrase <of both public and private employers> is an additional non-restrictive information provided in the sentence. The sentence should remain intact even if this prepositional phrase is removed. In other words, when prepositional phrases act as adjectives they can be removed. Be careful because prepositional phrases can also act as adverbs -- The book was lying under the table
Legislation in the Canadian province of Ontario requires of both public and private employers that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
If you remove the prepositional phrase -
Legislation in the Canadian province of Ontario requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are usually held by men.
You can even remove "in the canadian province of ontario" because again it is prepositional and provides additional non-restrictive information about Legislation -- the prepositional phrase acts as an adjective modifying Legislation and like all adjectives can be removed from the sentence without distorting the meaning.
Legislation requires that pay be the same for jobs historically held by women as for jobs requiring comparable skill that are[/u] usually held by men.
Now see.... requires that pay be the same is a typical subjunctive usage.
Thanks
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30 Aug 2014, 07:37
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Re: Legislation in the Canadian province of Ontario requires of [#permalink] 30 Aug 2014, 07:37
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Display posts from previous: Sort by | 2,683 | 10,976 | {"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-2017-04 | latest | en | 0.946726 |
https://stats.stackexchange.com/questions/443846/linear-regression-p-values-of-t-test-of-significance-of-regression-coeficients | 1,709,486,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00252.warc.gz | 549,130,326 | 40,675 | # Linear Regression: P-values of t-test of significance of regression coeficients are same as p-values of F-test about submodel
Suppose we have regression problem:
$$Y = \beta_{0}+\beta_{1}X_{1} + \beta_{2}X_{2} + \epsilon \text{, } \epsilon \sim \mathcal{N}(0,1),$$
where $$X_1 \sim U(0,1)$$,$$X_{2} \sim U(1,2)$$, and we suppose our model with regression function
$$m:=E[Y|X_1,X_2] = \beta_{0}+\beta_{1}X_{1} + \beta_{2}X_{2}.$$
We apply OLS estimation of coefficient and receive estimate of regression coeficeint. Let's do this in $$\verb|R|$$:
X1 <- runif(1000,0,1)
X2 <- runif(1000,1,2)
Y <- X1 + X2 + rnorm(1000,0,1)
model <- lm(Y ~ X1 + X2)
summary(model)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.1401 0.1828 0.766 0.444
X1 1.0250 0.1150 8.909 < 2e-16 ***
X2 0.9247 0.1139 8.119 1.37e-15 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.026 on 997 degrees of freedom
Multiple R-squared: 0.1276, Adjusted R-squared: 0.1259
F-statistic: 72.94 on 2 and 997 DF, p-value: < 2.2e-16
Next, we want to test, if the regression coeficient $$\beta_2$$ is significant or not, so here is T-test for this:
$$H_{0}: \beta_{2} = 0$$ $$H_{1}: \beta_{2} \neq 0$$
Aproppiate test statistic is:
$$T_n = \frac{\hat{\beta_{2}}}{S.E(\hat{\beta_{2}})} \sim t_{n-r},$$
and p-value:$$2min\{CDF_{t}(t_{0}), 1 - CDF_{t}(t_0)\}$$, where $$CDF_t$$ is cummulative distribution function o $$t$$-distribution with $$n-r$$, in ou case $$1000-3$$ degrees of freedom. Let's suppose, that we also made test about submodel:
$$H_{0} Y \sim X_{1} \text{ holds} (M^{0})$$ $$H_{1} Y \sim X_{1} + X_{2} \text{ holds} (M).$$
Test statistics is $$F = \frac{\frac{SSe^0 - SSe}{r - r_{0}}}{MSe} \sim F_{r-r_0,n-r}$$ and p-value:$$1-CDF_f(f_0)$$, where $$CDF_f$$ is cummulative distribution function of F-distribution with $$r-r_0$$ and $$n-r$$ degrees of freedom. In $$\verb|R|$$ let's make an test about submodel:
m0 <- lm(Y ~ X1)
anova(m0,model)
Analysis of Variance Table
Model 1: Y ~ X1
Model 2: Y ~ X1 + X2
Res.Df RSS Df Sum of Sq F Pr(>F)
1 998 1118.2
2 997 1048.8 1 69.352 65.926 1.373e-15 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Now there is a question, i know that if $$X \sim t_{n-r}$$, then $$X^2 \sim F_{1,n-r}$$. So if we squared t-value in our summary table, in line where regression coeficient $$\hat{\beta}_2$$ is, we geta f-value in output in test of submodel, so that's okay, but we have also same p-values, but p-value in summary table is computed from t-distribution, not squared distribution, and this numbers implies that $$2min\{CDF_{t}(t_{0}), 1 - CDF_{t}(t_0)\} = 1-CDF_f(f_0)$$, but i think this is not really true. What am i missing, how it is possible, that p-value which are computed from two different distributions are same ? (and this is not only for this case, but for every liner regression problems). Why p-values are same ? Please help.
In all cases i asuume normal linear model, that's why i assume those distribution of test statistics under $$H_0$$ hypothesis.
You should expect that the p-values are the same. If you reject the null hypothesis that $$\beta_2$$ = 0 (using the t-test), you should simultaneously reject the null hypothesis that the model without $$\beta_2$$ is adequate (using the F-test).
The p-value for the t-test is the probability that $$|t_0| > t^{\alpha/2}_{n-r}$$, and the p-value for the F-test is the probability that $$F_0 > F^{\alpha/2}_{1, n-r}$$. But note that these are mathematically identical statements, since squaring both sides of $$|t_0| > t^{\alpha/2}_{n-r}$$, gives you $$F_0 > F^{\alpha/2}_{1, n-r}$$, as you stated. Therefore, the p-values will always be numerically identical in this situation.
It's true that the p-value is computed from two different distributions, but they're found using different test statistics which are related to one another in a way that the p-value will always be identical.
• this would be a good answer if you include how are the two test statistics related to each other and why will they always (or almost always?) produce the same p-value Jan 8, 2020 at 17:13
• Thanks for your feedback @rep_ho. I updated my answer with some more details on how the statistics are related to one another. Jan 8, 2020 at 18:14
• Yes, this is right, thank you for your help. :) Jan 9, 2020 at 13:04 | 1,515 | 4,461 | {"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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-10 | latest | en | 0.664795 |
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• 615 | 2,535 | 10,139 | {"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} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.893249 |
https://math.stackexchange.com/questions/3234934/is-it-possible-for-two-recursive-sequences-to-have-the-same-characteristic-equat | 1,560,878,575,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998808.17/warc/CC-MAIN-20190618163443-20190618185443-00552.warc.gz | 531,571,822 | 35,063 | # Is it possible for two recursive sequences to have the same characteristic equations
If I have a sequence, $$\{t_1, \,t_2, \,t_1+t_2,\, t_1+2t_2,\cdots\}$$ I know that the formula for it is:
$$T_1=t_1, T_2=t_2, T_{n+2}=T_{n+1}+T_n.$$
If there are two sequences, $$A_n, B_n$$ such that $$A_1=1, A_2=0, A_{n+1}+A_n =A_{n+2}, B_1=0, B_2=1, B_{n+2}=B_{n+1}+B_n$$
Then $$A_n=F_{n-1}$$ and $$B_n=F_{n-1}$$ with $$F_n$$ being the Fibonacci numbers and $$F_1=1, F_2=1$$.
So the sequence becomes $$T_n =t_1 F_{n-2} +t_2F_{n-1}$$
The characteristic equation, which comes from $$T_{n+2}=T_{n+1}+T_{n}$$ is $$x^2-x-1=0$$ which has the roots $$\frac{1\pm \sqrt{5}}{2}$$. This is the same characteristic equation as the Fibonacci sequence.
Is it possible for two different sequences to have the same characteristic equation?
• What does $t_1,t_2,t_1+t_2,t_1+2t_2$ mean? – lulu May 21 at 21:08
• that is the sequence given – user130306 May 21 at 21:08
• Well, it doesn't make any sense. What does the notation mean? What, say, does $t_2, t_1$ mean? – lulu May 21 at 21:08
• To your broader question, of course two sequences can have the same characteristic equation. They just need to have different initial conditions. – lulu May 21 at 21:10
• A recurrence of the form $a_n=pa_{n-1}+qa_{n-2}$ is obviously completely determined by any two consecutive values. In fact it is determined by any pair of values. This means that if two sequences match in two places they are identical. The same recurrence can have different initial values and these will generate a different sequence. – Mark Bennet May 21 at 21:22 | 561 | 1,605 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-26 | latest | en | 0.887001 |
https://www.ams.org/publicoutreach/feature-column/fcarc-gallery4 | 1,540,170,863,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514443.85/warc/CC-MAIN-20181022005000-20181022030500-00052.warc.gz | 853,223,424 | 11,344 | ## Diagonals: Part II
4. Entering new territory
Whereas all of the polygons in the quadrilateralization in Figure 4 (in the previous section) are convex, it is possible to decompose this same polygon into quadrilaterals which are convex and non-convex.
There is even a way to do this so that all the quadrilaterals share a single vertex. This vertex is at the end of the internal diagonal that splits the only non-convex quadrilateral present into two triangles. This quadrilateralizaton shows that placing a single guard will suffice to guard the whole polygon. The vertex at which to place the single guard is shown in green.
In the diagram above, two of the quadrilaterals are actually trapezoids, and the vertices of the quadrilaterals are all vertices of the original polygon. In the diagram below, the original polygon's vertices are shown as small white dots but new vertices have been added on the edges of that polygon; they are shown in blue. The blue vertices are determined by extending those horizontal sides of the original polygon which when prolonged enter the interior of the polygon. The blue dots are placed where these extensions meet the vertical boundary lines of the polygon. This procedure can guarantee a subdivision of the original polygon into rectangles.
Had we started with an arbitrary polygon, a procedure such as this could have been used to divide a polygon into trapezoids (some of which might be "degenerate," and, hence, be triangles). These diagrams and examples chart out lots of new territory for mathematical investigation. When, if ever, can an orthogonal polygon be subdivided into rectangles using existing vertices of the polygon? What possibilities are added when one can add new internal vertices to the collection of potential vertices of the subdividing polygons of the original? Problems of this kind, where one is allowed to add new points in addition to those initially present, are often referred to as Steiner point variations on the original problem.
Invariably, when one ponders the details of a mathematical question one is led into new territory. This involves looking into new problems which are "nearby" in the sense that it seems natural to ask them as an outgrowth of what one is investigating. In the current context, for example, in the diagram below a polygon has been divided by the red diagonals into quadrilaterals which are convex.
This is not always possible because we know there exists a non-convex 4-gon. Under what conditions can a simple polygon be decomposed into convex 4-gons? Are there polygons with n vertices (n at least 5) for which one can be sure of convex 5-gons? Are there polygons with n vertices (n at least 5) for which one can be sure there is no convex quadrilateralizaton? (Here, I have in mind that the existing vertices of the polygon are to be used and no others.)
Another interesting variant of this type of problem is not to start with a polygon but with a plane set of points. Clearly, not all plane sets of points are such that the points of the set can serve as vertices of a simple polygon. For example, the points might all be on a straight line. Perhaps surprisingly, if P is a set of points which do not all lie on a straight line, there is a polygonalization of the points, that, is a simple polygon whose vertices coincide with points in the given set. Many questions about polygonalizations can be asked. For a fixed set of n points not all on a line, what is the largest number of different polygonalizations that can be obtained using these points? Alternatively, one might try to minimize the number of reflex angles for any (simple) polygonalization that passes through the points. Researchers have been successful in finding combinatorial bounds on this interesting "measure" associated with a set of non-collinear points.
Questions in the spirit of this discussion extend to planar sets of points. Thus, starting with the planar set of blue points, one can add additional lines to the point set so that the result consists of all quadrilaterals except for the infinite face of the associated graph. Such a decomposition is shown below. Can you see how to select some of the edges in the graph so that we obtain a polygon which together with the remaining edges forms the diagram shown?
One can now adopt the following perspective. Given a simple polygon, when can one pick points (and how few can one pick) in the interior of this polygon so that the original points together with the new points admit a convex quadrilaterization of the original point set? Problems such as these are still giving researchers plenty to think about.
Additional questions arise when one considers lattice polygons. These are polygons whose vertices can be assigned coordinates of the form (a, b) where a and b are integers. An example of such a polygon is shown below.
Some interesting work dealing with lattice point polygons is due to Sándor Fekete. Fekete considered the difficulty of solving some interesting polygonalization problems associated with point sets whose vertices are located at the points of an integer lattice. For example, given a set of size n of such lattice points such as the blue points below, is there a simple polygon that contains no other grid either in the polygon's interior or on its boundary? Fekete calls such a polygon a grid avoiding polygon. A solution to this problem is given by the polygon shown, but finding it is a bit like looking for a needle in a haystack.
By contrast, the set of points below differs from the one above in the omission of a single point. However, one can not find a grid avoiding polygon based on this point set.
Fekete proved that given a set of lattice points, the problem of determining if they give rise to a grid avoiding polygon is NP-complete. This suggests that it is unlikely that a polynomial time algorithm for finding such a polygon will ever be found.
It is natural to try to extend the work on Art Gallery Theorems to non-convex polyhedra in 3 dimensions. It is tempting to mimic the wonderful proof of Fisk, the first step of which would be to subdivide a non-convex 3-dimensional polyhedron into tetrahedra (the 3-dimensional analog of a triangle) using diagonals that lie in the polyhedron's interior. Like being able to triangulate a simple polygon, many find this an intuitively clear result--unfortunately, it is false! In 1911 the American topologist N. J. Lennes showed that there are non-convex 3-dimensional polyhedra with the property that the segment joining every pair of vertices which are joined by an edge of the polyhedron, lies in the exterior of the polyhedron. However, a simpler approach is due to the work of E. Schönhardt. He found that the simplest example of such a polyhedron is combinatorially a triangular antiprism (6 vertices, 8 triangular faces) where the parallel planed triangular faces are "twisted" with respect to one another. A diagram is shown below, with the hidden lines in blue:
The result is that all the edges that are not edges of the polyhedron are all in the exterior of the polyhedron. This provides an example of something that must be seen (or a model made) to be believed. Although one can not extend the Fisk argument directly, some partial results are known for what is true in 3 dimensions.
There are many appealing directions for generalizations of the Art Gallery Theorems. Some of these concern guarding the exterior of simple polygons, guarding the interior and exterior with guards, and using guards with x-ray vision. One particularly intriguing variant of the art gallery problem concerns the notion of guarded guards. We have seen that using vertex guards is sufficient to guard any n-sided polygon. However, what happens if one is nervous that guards may not be diligent? One way to deal with this is to make sure that every guard is visible by some other guard. This is the notion of guarded guards.
Two guards are required to guard the polygon below. For example, guards at u and x will do the job; however, guards at z and w guard each other and the rest of the polygon.
We now have the question of determining what is the number of guards which are sometimes necessary and always sufficient to guard an n-sided polygon, where we require that the guard set have the property that for any guard g there is some other guard h who can see g. This problem was solved very recently (2003) by T. Michael and V. Pinciu. For n-sided simple polygons with at least 5 sides there is a "guarded guard" set of points with points and for orthogonal polygons with at least 6 sides with points (this last result is due to G. Hernández-Peñalver).
We'll continue the discussion of diagonals in a later column.
Welcome to the
Feature Column!
These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics. | 1,875 | 8,935 | {"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-2018-43 | latest | en | 0.916083 |
https://www.physicsforums.com/threads/how-many-electron-charges-are-there-on-the-drop-millikan.890676/ | 1,607,006,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727782.88/warc/CC-MAIN-20201203124807-20201203154807-00580.warc.gz | 612,094,192 | 23,040 | # How many electron charges are there on the drop? (Millikan)
## Homework Statement
In a measurement of the electron charge by Millikan's method, a potential difference of 1.5 kV can be applied between horizontal parallel metal plates 12 mm apart. With the field switched off, a drop of oil of mass 10-14 kg is observed to fall with constant velocity 400 μm s-1. When the field is switched on, the drop rises with constant velocity 80 μm s-1. How many electron charges are there on the drop? (You may assume that the air resistance is proportional to the velocity of the drop, and that air buoyancy may be neglected.)
(The electronic charge = 1.6 * 10-19 C, the acceleration due to gravity = 10 m s-2.)
2. The attempt at a solution
As I understand I need to find Q?
In my book I have this formula: Q = 6 π r η v / E, where η is the coefficient of viscosity of air. The formula is applicable "when an electric field has been applied such that the drop is stationary." There are also formulas when there is no electric field, but I think in this case we have electric field.
But we don't have the radius and have the distance between plates instead.
I am also not sure whether I need to find Q, or some other value. Also how to find η?
Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor
As I understand I need to find Q?
In my book I have this formula: Q = 6 π r η v / E, where η is the coefficient of viscosity of air. The formula is applicable "when an electric field has been applied such that the drop is stationary." There are also formulas when there is no electric field, but I think in this case we have electric field.
But we don't have the radius and have the distance between plates instead.
I am also not sure whether I need to find Q, or some other value. Also how to find η?
Yes you need to find Q. In particular you need to find out how many elementary charges it takes to total up to Q.
You are not given enough information to calculate the drag force directly. However, you are told to assume that it is proportional to velocity. Can you determine one value for the drag force at a particular velocity (other than zero velocity, of course)? Can you then scale it for other velocities?
moenste
Yes you need to find Q. In particular you need to find out how many elementary charges it takes to total up to Q.
You are not given enough information to calculate the drag force directly. However, you are told to assume that it is proportional to velocity. Can you determine one value for the drag force at a particular velocity (other than zero velocity, of course)? Can you then scale it for other velocities?
Q = (6 π η / E) * (9 η v / 2 ρo g)1 / 2 v
E = V / d = 1500 / 0.012 = 125 000 V m-1.
Now η (coefficient of viscosity of air) and ρo (density of oil) are required.
Update
As I understand when there is no electric field velocity is 400 μm s-1. Weight = Upthrust due to air + Viscous drag. Upthrust is zero. So: (4/3) π r3 po g = 6 π r η v. Simplify them to find radius: r = √ 4.5 η v / ρo g.
Then we find radius and turn the field on. We already found E = 125 000 V m-1. And then we use Q = (4/3) π r3 ρo g / E. Q = 4 π (√ 4.5 η v / ρo g)3 ρo g / 3 E.
Only need to find η and ρo somehow...
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gneill
Mentor
You can forget the formula for drag and all of its constants that you don't know. Just find the force (in Newtons) that it creates when the oil drop is falling at a constant speed without the electric field in place.
moenste
You can forget the formula for drag and all of its constants that you don't know. Just find the force (in Newtons) that it creates when the oil drop is falling at a constant speed without the electric field in place.
Hm, is it F = m g? We have the mass present.
gneill
Mentor
Hm, is it F = m g? We have the mass present.
Yes. The drop is falling with constant velocity so the net force acting is zero.
moenste
Yes. The drop is falling with constant velocity so the net force acting is zero.
F = m g
E = V / d
m g = Q (V / d)
Q = m g d / V = 10-14 * 10 * 0.012 / 1500 = 8 * 10-19 C.
gneill
Mentor
F = m g
E = V / d
You want to specify that that force F above is the force due to drag for that one velocity only.
m g = Q (V / d)
Q = m g d / V = 10-14 * 10 * 0.012 / 1500 = 8 * 10-19 C.
Can you explain in words what the first equation above represents? I can see an electric force, I can see the gravitational force, but I don't see the drag accounted for.
moenste
You want to specify that that force F above is the force due to drag for that one velocity only.
Can you explain in words what the first equation above represents? I can see an electric force, I can see the gravitational force, but I don't see the drag accounted for.
Weigth = Upthrust + Electric force
Weight, as you say, shouldn't be 4/3 π r3 ρo g but F = m g.
Upthrust is zero.
Electric force = Q E, Q = charge on the drop, E = the electric field strength.
gneill
Mentor
There will be a different drag force acting when the drop is moving upwards because its moving at a different speed than when it was falling. You need to account for that drag force. Use the given assumption about the drag force and velocity.
moenste
There will be a different drag force acting when the drop is moving upwards because its moving at a different speed than when it was falling. You need to account for that drag force. Use the given assumption about the drag force and velocity.
What is drag force?
gneill
Mentor
What is drag force?
Air resistance, or viscous drag. You calculated a value for the drag force for the falling oil drop in posts #5 and #7.
moenste
Air resistance, or viscous drag. You calculated a value for the drag force for the falling oil drop in posts #5 and #7.
I'm sorry, I don't get it.
I need to find Q. Assuming everything I did is not correct. What do I need to do then?
Update
I think I understand what you mean. If we have no electric field then the drop is falling and therefore weight = m g. And therefore we have weight = upthrust due to air + viscous drag (in contrast to when there is an electric field and weight = upthrust + electric force). I used to formula for the situation when an electric field is present.
So now we have weight = viscous drag. m g = 6 π r η v. Though I don't see how it helps -- we need the radius and the coefficient of viscosity of air η.
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gneill
Mentor
Yes, you need to find Q. You do that by finding out what force that charge experiences due to the known electric field. But there are several forces acting at once and you need to factor out their contributions. Gravity is easy because you have the mass of the oil drop. Air resistance also acts on the drop whenever it is moving through air. It's a type of friction.
The idea is to establish what this friction force is for the upward moving drop so that you can "remove it" from the sum of forces, just as you will "remove" the gravitational force.
That's why you went to the trouble of finding what the air resistance force is for the drop while it was falling at constant speed. The only two forces acting then were gravity and this friction force.
Now you need to find the friction force acting on the upward moving drop so that you can account for it in the net force sum.
moenste
Yes, you need to find Q. You do that by finding out what force that charge experiences due to the known electric field. But there are several forces acting at once and you need to factor out their contributions. Gravity is easy because you have the mass of the oil drop. Air resistance also acts on the drop whenever it is moving through air. It's a type of friction.
The idea is to establish what this friction force is for the upward moving drop so that you can "remove it" from the sum of forces, just as you will "remove" the gravitational force.
That's why you went to the trouble of finding what the air resistance force is for the drop while it was falling at constant speed. The only two forces acting then were gravity and this friction force.
Now you need to find the friction force acting on the upward moving drop so that you can account for it in the net force sum.
Not sure whether you saw this:
Update
I think I understand what you mean. If we have no electric field then the drop is falling and therefore weight = m g. And therefore we have weight = upthrust due to air + viscous drag (in contrast to when there is an electric field and weight = upthrust + electric force). I used to formula for the situation when an electric field is present.
So now we have weight = viscous drag. m g = 6 π r η v. Though I don't see how it helps -- we need the radius and the coefficient of viscosity of air η.
Is this correct logic?
gneill
Mentor
Is this correct logic?
Yes the logic is fine as far as it goes. But you don't need to know any of the constants for the air resistance formula; it's not needed here since you can use the value for the force at a particular velocity to find the force for other velocities thanks to the given assumption about how the air resistance varies with velocity.
moenste
Yes the logic is fine as far as it goes. But you don't need to know any of the constants for the air resistance formula; it's not needed here since you can use the value for the force at a particular velocity to find the force for other velocities thanks to the given assumption about how the air resistance varies with velocity.
OK, so no electric field: Weight = Upthrust (zero) + Air resistance → m g = Air resistance.
With electric field: Weight = Upthrust (zero) + Electric force → 4 / 3 π r3 ρo g = Q E → 4 / 3 π r3 ρo g = Q V / d.
I think this should be correct.
gneill
Mentor
You're still plugging in the useless air resistance formula. You need to use the first air resistance force to find a value for the second. Then write your force balance using that.
moenste
You're still plugging in the useless air resistance formula. You need to use the first air resistance force to find a value for the second. Then write your force balance using that.
In that case I'm missing some force.
This is what I have my book:
From book:
No EF: Weight = Upthrust due to air + Viscous drag.
EF: Weight = Upthrust + Electric force.
So:
No EF: Viscous drag = 10-14 * 10 = 10-13 N.
EF: Q = m g / (V / d) = 10-13 / (1500 / 0.012) = 8 * 10-19 C.
So we have the answer Q = 8 * 10-19 C. I don't understand why we need all the other numbers given in the problem and why we need to calculate the no EF force.
gneill
Mentor
In the book's figure the drop on the right hand side is stationary, so no air resistance force, just the electric force balancing the gravitational force.
In the present problem the oil drop is moving in both cases.
moenste
In the book's figure the drop on the right hand side is stationary, so no air resistance force, just the electric force balancing the gravitational force.
In the present problem the oil drop is moving in both cases.
Aaaah, it even says so.
So, in that case:
No EF: Air resistance = 10-14 * 10 = 10-13 N.
EF: Weight = Upthrust + Electric force + Air resistance → 10-13 N = zero + Q E + 10-13 N → Q E = 0?
gneill
Mentor
I don't see where you've determined the air resistance force for the new oil drop speed. The speed is different than the first case so the air resistance must be different. Use the first case result to find the value for the new speed.
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moenste
I don't see where you've determined the air resistance force for the new oil drop speed. The speed is difference than the first case so the air resistance must be different. Use the first case result to find the value for the new speed.
Maybe something like Air resistanceFalling * SpeedFalling = ARRising * SpeedRising → 10-13 * (400 * 10-6) = ARRising * (80 * 10-6) → ARRising = 5 * 10-13 N?
Weight = Upthrust + EF + ARRising
10-13 N = zero + Q E + 5 * 10-13 N
Q E = -4 * 10-13 N
Q = -4 * 10-13 / 125 000 = -3.2 * 10-18 Q.
gneill
Mentor
Let's concentrate on the air resistance for a moment.
In the first case you have:
v1 = 400 (micrometers per second)
F1 = Mg = 10-13 N
Now you are given that air resistance is proportional to velocity. Mathematically: F ∝ v
Write that as an equality by inserting a proportionality constant and solve for that constant. Then you can determine the F for any given v.
moenste
Let's concentrate on the air resistance for a moment.
In the first case you have:
v1 = 400 (micrometers per second)
F1 = Mg = 10-13 N
Now you are given that air resistance is proportional to velocity. Mathematically: F ∝ v
Write that as an equality by inserting a proportionality constant and solve for that constant. Then you can determine the F for any given v.
v1 = 4 * 10-4 m s-1.
F1 = m g = 10-14 * 10 = 10-13 N.
v2 = 8 * 10-5 m s-1.
F2 = ?
F1 = k v1, where k = F1 / v1 = 10-13 / 4 * 10-4 = 2.5 * 10-10.
So F2 = k v2 = 2.5 * 10-10 * 8 * 10-5 = 2 * 10-14 N.
---
Weight = Upthrust + EF + ARRising or F2
10-13 N = zero + Q E + 2 * 10-14 N
Q E = 8 * 10-14 N
Q = 8 * 10-14 / 125 000 = 6.4 * 10-19 Q.
Then 6.4 * 10-19 / 1.6 * 10-19 = 4, not 6 as in the answer. What's wrong? I recalculated everything and checked the book answer, it's 6. So, don't think that it's the calculation.
Last edited: | 3,507 | 13,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | latest | en | 0.953839 |
http://www.nag.com/numeric/MB/manual64_24_1/html/F01/f01lhf.html | 1,394,275,068,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654315/warc/CC-MAIN-20140305060734-00038-ip-10-183-142-35.ec2.internal.warc.gz | 451,131,311 | 7,835 | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_matop_real_gen_blkdiag_lu (f01lh)
## Purpose
nag_matop_real_gen_blkdiag_lu (f01lh) factorizes a real almost block diagonal matrix.
## Syntax
[a, pivot, tol, kpivot, ifail] = f01lh(n, blkstr, a, tol, 'nbloks', nbloks, 'lena', lena)
[a, pivot, tol, kpivot, ifail] = nag_matop_real_gen_blkdiag_lu(n, blkstr, a, tol, 'nbloks', nbloks, 'lena', lena)
## Description
nag_matop_real_gen_blkdiag_lu (f01lh) factorizes a real almost block diagonal matrix, A$A$, by row elimination with alternate row and column pivoting such that no ‘fill-in’ is produced. The code, which is derived from ARCECO described in Diaz et al. (1983), uses Level 1 and Level 2 BLAS. No three successive diagonal blocks may have columns in common and therefore the almost block diagonal matrix must have the form shown in the following diagram:
Figure 1
This function may be followed by nag_linsys_real_blkdiag_fac_solve (f04lh), which is designed to solve sets of linear equations AX = B$AX=B$ or ATX = B${A}^{\mathrm{T}}X=B$.
## References
Diaz J C, Fairweather G and Keast P (1983) Fortran packages for solving certain almost block diagonal linear systems by modified alternate row and column elimination ACM Trans. Math. Software 9 358–375
## Parameters
### Compulsory Input Parameters
1: n – int64int32nag_int scalar
n$n$, the order of the matrix A$A$.
Constraint: n > 0${\mathbf{n}}>0$.
2: blkstr(3$3$,nbloks) – int64int32nag_int array
Information which describes the block structure of A$A$ as follows:
• blkstr(1,k)${\mathbf{blkstr}}\left(1,k\right)$ must contain the number of rows in the k$k$th block, k = 1,2,,nbloks$k=1,2,\dots ,{\mathbf{nbloks}}$;
• blkstr(2,k)${\mathbf{blkstr}}\left(2,k\right)$ must contain the number of columns in the k$k$th block, k = 1,2,,nbloks$k=1,2,\dots ,{\mathbf{nbloks}}$;
• blkstr(3,k)${\mathbf{blkstr}}\left(3,k\right)$ must contain the number of columns of overlap between the k$k$th and (k + 1)$\left(k+1\right)$th blocks, k = 1,2,,nbloks1$k=1,2,\dots ,{\mathbf{nbloks}}-1$. blkstr(3,nbloks)${\mathbf{blkstr}}\left(3,{\mathbf{nbloks}}\right)$ need not be set.
The following conditions delimit the structure of A$A$:
• blkstr(1,k),blkstr(2,k) > 0, k = 1,2,,nbloks${\mathbf{blkstr}}\left(1,k\right),{\mathbf{blkstr}}\left(2,k\right)>0\text{, }k=1,2,\dots ,{\mathbf{nbloks}}$,
• blkstr(3,k)0, k = 1,2,,nbloks1${\mathbf{blkstr}}\left(3,k\right)\ge 0\text{, }k=1,2,\dots ,{\mathbf{nbloks}}-1$,
(there must be at least one column and one row in each block and a non-negative number of columns of overlap);
• blkstr(3,k1) + blkstr(3,k)blkstr(2,k), k = 2,3,,nbloks1${\mathbf{blkstr}}\left(3,k-1\right)+{\mathbf{blkstr}}\left(3,k\right)\le {\mathbf{blkstr}}\left(2,k\right)\text{, }k=2,3,\dots ,{\mathbf{nbloks}}-1$,
(the total number of columns in overlaps in each block must not exceed the number of columns in that block);
• blkstr(2,1)blkstr(1,1)${\mathbf{blkstr}}\left(2,1\right)\ge {\mathbf{blkstr}}\left(1,1\right)$,
• blkstr(2,1) + k = 2 j [blkstr(2,k)blkstr(3, k 1 )] k = 1 j blkstr(1,k) ${\mathbf{blkstr}}\left(2,1\right)+\sum _{k=2}^{j}\left[{\mathbf{blkstr}}\left(2,k\right)-{\mathbf{blkstr}}\left(3,k-1\right)\right]\ge \sum _{k=1}^{j}{\mathbf{blkstr}}\left(1,k\right)$, j = 2,3,,nbloks1 $j=2,3,\dots ,{\mathbf{nbloks}}-1$,
• k = 1j[blkstr(2,k)blkstr(3,k)]k = 1jblkstr(1,k), j = 1,2,,nbloks1$\sum _{k=1}^{j}\left[{\mathbf{blkstr}}\left(2,k\right)-{\mathbf{blkstr}}\left(3,k\right)\right]\le \sum _{k=1}^{j}{\mathbf{blkstr}}\left(1,k\right)\text{, }j=1,2,\dots ,{\mathbf{nbloks}}-1$,
(the index of the first column of the overlap between the j$j$th and (j + 1)$\left(j+1\right)$th blocks must be $\le$ the index of the last row of the j$j$th block, and the index of the last column of overlap must be $\ge$ the index of the last row of the j$j$th block);
• k = 1 nbloks blkstr(1,k) = n $\sum _{k=1}^{{\mathbf{nbloks}}}{\mathbf{blkstr}}\left(1,k\right)=n$,
• blkstr(2,1) + k = 2 nbloks [blkstr(2,k)blkstr(3, k 1 )] = nk ${\mathbf{blkstr}}\left(2,1\right)+\sum _{k=2}^{{\mathbf{nbloks}}}\left[{\mathbf{blkstr}}\left(2,k\right)-{\mathbf{blkstr}}\left(3,k-1\right)\right]=nk$,
(both the number of rows and the number of columns of A$A$ must equal n$n$).
3: a(lena) – double array
lena, the dimension of the array, must satisfy the constraint lena k = 1 nbloks [blkstr(1,k) × blkstr(2,k)] ${\mathbf{lena}}\ge \sum _{k=1}^{{\mathbf{nbloks}}}\left[{\mathbf{blkstr}}\left(1,k\right)×{\mathbf{blkstr}}\left(2,k\right)\right]$.
The elements of the almost block diagonal matrix stored block by block, with each block stored column by column. The sizes of the blocks and the overlaps are defined by the parameter blkstr.
If ars${a}_{rs}$ is the first element in the k$k$th block, then an arbitrary element aij${a}_{ij}$ in the k$k$th block must be stored in the array element:
a(pk + (j − r)mk + (i − s) + 1) $a(pk+(j-r)mk+(i-s)+1)$
where
k − 1 pk = ∑ blkstr(1,l) × blkstr(2,l) l = 1
$pk=∑l= 1 k- 1blkstr1l×blkstr2l$
is the base address of the k$k$th block, and
mk = blkstr(1,k) $mk=blkstr1k$
is the number of rows of the k$k$th block.
4: tol – double scalar
A relative tolerance to be used to indicate whether or not the matrix is singular. For a discussion on how tol is used see Section [Further Comments]. If tol is non-positive, then tol is reset to 10ε$10\epsilon$, where ε$\epsilon$ is the machine precision.
### Optional Input Parameters
1: nbloks – int64int32nag_int scalar
Default: The dimension of the array blkstr.
n$n$, the total number of blocks of the matrix A$A$.
Constraint: 0 < nbloksn$0<{\mathbf{nbloks}}\le {\mathbf{n}}$.
2: lena – int64int32nag_int scalar
Default: The dimension of the array a.
The dimension of the array a as declared in the (sub)program from which nag_matop_real_gen_blkdiag_lu (f01lh) is called.
Constraint: lena k = 1 nbloks [blkstr(1,k) × blkstr(2,k)] ${\mathbf{lena}}\ge \sum _{k=1}^{{\mathbf{nbloks}}}\left[{\mathbf{blkstr}}\left(1,k\right)×{\mathbf{blkstr}}\left(2,k\right)\right]$.
None.
### Output Parameters
1: a(lena) – double array
The factorized form of the matrix.
2: pivot(n) – int64int32nag_int array
Details of the interchanges.
3: tol – double scalar
Unchanged unless tol0.0${\mathbf{tol}}\le 0.0$ on entry, in which case it is set to 10ε$10\epsilon$.
4: kpivot – int64int32nag_int scalar
If ${\mathbf{ifail}}={\mathbf{2}}$, kpivot contains the value k$k$, where k$k$ is the first position on the diagonal of the matrix A$A$ where too small a pivot was detected. Otherwise kpivot is set to 0$0$.
5: ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).
## Error Indicators and Warnings
Errors or warnings detected by the function:
Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.
ifail = 1${\mathbf{ifail}}=1$
On entry, n < 1${\mathbf{n}}<1$, or nbloks < 1${\mathbf{nbloks}}<1$, or ${\mathbf{n}}<{\mathbf{nbloks}}$, or lena is too small, or illegal values detected in blkstr.
W ifail = 2${\mathbf{ifail}}=2$
The factorization has been completed, but a small pivot has been detected.
## Accuracy
The accuracy of nag_matop_real_gen_blkdiag_lu (f01lh) depends on the conditioning of the matrix A$A$.
Singularity or near singularity in A$A$ is determined by the parameter tol. If the absolute value of any pivot is less than tol × amax${\mathbf{tol}}×{a}_{\mathrm{max}}$, where amax${a}_{\mathrm{max}}$ is the maximum absolute value of an element of A$A$, then A$A$ is said to be singular. The position on the diagonal of A$A$ of the first of any such pivots is indicated by the parameter kpivot. The factorization, and the test for near singularity, will be more accurate if before entry A$A$ is scaled so that the $\infty$-norms of the rows and columns of A$A$ are all of approximately the same order of magnitude. (The $\infty$-norm is the maximum absolute value of any element in the row or column.)
## Example
```function nag_matop_real_gen_blkdiag_lu_example
n = int64(18);
blkstr = [int64(2),4,5,3,4;4,7,8,6,5;3,4,2,3,0];
a = [-1;
-1;
-0.98;
0.25;
-0.79;
-0.87;
-0.15;
0.35;
0.78;
-0.82;
-0.83;
-0.21;
0.31;
0.12;
-0.98;
-0.93;
-0.85;
-0.01;
-0.58;
-0.84;
0.89;
0.75;
0.04;
0.37;
-0.69;
0.32;
0.87;
-0.94;
-0.98;
-1;
0.38;
-0.96;
-0.76;
-0.53;
-1;
-1;
-0.99;
-0.87;
-0.93;
0.85;
0.17;
-0.91;
-0.14;
-0.91;
-0.39;
-1.37;
-0.28;
-1;
0.1;
0.79;
1.29;
0.9;
-0.59;
-0.89;
-0.71;
-1.59;
0.78;
-0.99;
-0.68;
0.39;
1.1;
-0.93;
0.21;
-0.09;
-0.99;
-1.63;
-0.76;
-0.73;
-0.58;
-0.12;
-1.01;
0.48;
-0.48;
-0.21;
-0.75;
-0.27;
0.08;
-0.67;
-0.24;
0.61;
0.56;
-0.41;
0.54;
-0.99;
0.4;
-0.41;
0.16;
-0.93;
0.16;
-0.16;
0.7;
-0.46;
0.98;
0.43;
0.71;
-0.47;
-0.25;
0.89;
-0.97;
-0.98;
-0.92;
-0.94;
-0.6;
-0.73;
-0.52;
-0.54;
-0.3;
0.07;
-0.46;
-1;
0.18;
0.04;
-0.58;
-0.36];
tol = 0;
[aOut, pivot, tolOut, index, ifail] = nag_matop_real_gen_blkdiag_lu(n, blkstr, a, tol)
```
```
aOut =
-1.0000
1.0000
-0.9800
1.2300
-0.7900
-0.0800
-0.1500
0.5000
0.6341
-0.6667
-0.6748
-0.1707
-1.0340
-0.3489
-0.0645
0.9126
-0.2426
-1.2517
-0.2458
0.4260
0.3800
-0.8474
-1.1838
0.7989
0.8700
-0.3865
0.2811
-1.7939
0.0400
0.9040
0.9748
-0.8304
-1.0000
-1.1089
-0.8671
1.0778
0.2365
0.8447
-0.0845
-0.6673
-1.0897
0.5443
0.2104
0.4940
0.1128
0.5929
1.7246
-0.8806
-0.4456
0.9244
-0.2536
-3.1739
-2.8793
0.3112
-0.5292
0.7990
1.1000
-0.0213
1.2768
-0.4998
-0.3007
-1.6300
-1.2254
-1.2751
-0.7689
-0.1200
-0.2700
-0.7178
0.5830
-0.5889
0.3997
-1.0100
-1.6194
-0.7062
-0.3962
0.1979
1.5262
1.4011
-1.0258
-0.9473
0.2344
0.0391
-0.9900
0.7720
0.5682
0.9800
-0.6897
0.7836
-0.1600
0.1975
0.5940
0.1600
-0.4475
-0.6820
-0.7657
-0.9316
-0.6636
-0.6891
-0.5306
0.8039
0.9708
0.9909
0.4200
-1.4430
0.1524
0.7753
-1.0000
0.8739
0.5376
-0.2729
-0.3600
0.3294
0.4793
-0.3479
pivot =
1
2
3
3
6
5
5
2
5
4
8
2
2
6
4
2
4
4
tolOut =
1.1102e-15
index =
0
ifail =
0
```
```function f01lh_example
n = int64(18);
blkstr = [int64(2),4,5,3,4;4,7,8,6,5;3,4,2,3,0];
a = [-1;
-1;
-0.98;
0.25;
-0.79;
-0.87;
-0.15;
0.35;
0.78;
-0.82;
-0.83;
-0.21;
0.31;
0.12;
-0.98;
-0.93;
-0.85;
-0.01;
-0.58;
-0.84;
0.89;
0.75;
0.04;
0.37;
-0.69;
0.32;
0.87;
-0.94;
-0.98;
-1;
0.38;
-0.96;
-0.76;
-0.53;
-1;
-1;
-0.99;
-0.87;
-0.93;
0.85;
0.17;
-0.91;
-0.14;
-0.91;
-0.39;
-1.37;
-0.28;
-1;
0.1;
0.79;
1.29;
0.9;
-0.59;
-0.89;
-0.71;
-1.59;
0.78;
-0.99;
-0.68;
0.39;
1.1;
-0.93;
0.21;
-0.09;
-0.99;
-1.63;
-0.76;
-0.73;
-0.58;
-0.12;
-1.01;
0.48;
-0.48;
-0.21;
-0.75;
-0.27;
0.08;
-0.67;
-0.24;
0.61;
0.56;
-0.41;
0.54;
-0.99;
0.4;
-0.41;
0.16;
-0.93;
0.16;
-0.16;
0.7;
-0.46;
0.98;
0.43;
0.71;
-0.47;
-0.25;
0.89;
-0.97;
-0.98;
-0.92;
-0.94;
-0.6;
-0.73;
-0.52;
-0.54;
-0.3;
0.07;
-0.46;
-1;
0.18;
0.04;
-0.58;
-0.36];
tol = 0;
[aOut, pivot, tolOut, index, ifail] = f01lh(n, blkstr, a, tol)
```
```
aOut =
-1.0000
1.0000
-0.9800
1.2300
-0.7900
-0.0800
-0.1500
0.5000
0.6341
-0.6667
-0.6748
-0.1707
-1.0340
-0.3489
-0.0645
0.9126
-0.2426
-1.2517
-0.2458
0.4260
0.3800
-0.8474
-1.1838
0.7989
0.8700
-0.3865
0.2811
-1.7939
0.0400
0.9040
0.9748
-0.8304
-1.0000
-1.1089
-0.8671
1.0778
0.2365
0.8447
-0.0845
-0.6673
-1.0897
0.5443
0.2104
0.4940
0.1128
0.5929
1.7246
-0.8806
-0.4456
0.9244
-0.2536
-3.1739
-2.8793
0.3112
-0.5292
0.7990
1.1000
-0.0213
1.2768
-0.4998
-0.3007
-1.6300
-1.2254
-1.2751
-0.7689
-0.1200
-0.2700
-0.7178
0.5830
-0.5889
0.3997
-1.0100
-1.6194
-0.7062
-0.3962
0.1979
1.5262
1.4011
-1.0258
-0.9473
0.2344
0.0391
-0.9900
0.7720
0.5682
0.9800
-0.6897
0.7836
-0.1600
0.1975
0.5940
0.1600
-0.4475
-0.6820
-0.7657
-0.9316
-0.6636
-0.6891
-0.5306
0.8039
0.9708
0.9909
0.4200
-1.4430
0.1524
0.7753
-1.0000
0.8739
0.5376
-0.2729
-0.3600
0.3294
0.4793
-0.3479
pivot =
1
2
3
3
6
5
5
2
5
4
8
2
2
6
4
2
4
4
tolOut =
1.1102e-15
index =
0
ifail =
0
``` | 5,531 | 11,990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 77, "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-2014-10 | longest | en | 0.520302 |
http://mathoverflow.net/questions/15800/calculating-6j-symbols-aka-racah-wigner-coefficients-for-quantum-groups/15825 | 1,469,829,454,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257832385.29/warc/CC-MAIN-20160723071032-00298-ip-10-185-27-174.ec2.internal.warc.gz | 152,522,037 | 17,772 | # Calculating 6j-symbols (aka Racah-Wigner coefficients) for quantum groups
Which $6j$-symbols for quantised enveloping algebras are known explicitly?
The quantum $6j$-symbols for $sl(2)$ are well-known. The references are Masbaum and Vogel and Frenkel and Khovanov.
What is known for other simple Lie algebras?
In case this seems somewhat vague here is a precise question. The data for a $6j$-symbol starts with a tetrahedral graph with edges labelled by highest weights. Then usually there is additional information needed at the vertices which I want to avoid. Take the example of $sl(n)$ and use partitions instead of highest weights. Label two opposite edges by a partition of the form $1^k$ (corresponding to an exterior power of the vector representation) and label the other four edges by partitions. Then associated to this data is a scalar.
Then I would expect this function to be determined by linear recurrence relations (i.e. D-finite or holonomic). Is this correct? and if so can you give recurrence relations?
Ideally we would also regard $n$ as an indeterminate.
Background In general $6j$-symbols arise for any semisimple abelian category which is also monoidal. They are the components of the natural transformation $(\otimes)(\otimes \times 1)\cong (\otimes)(1\times \otimes)$.
In more down to earth terms. If you know the Grothendieck ring of a semisimple abelian monoidal category and you attempt to construct this then the information you are missing is the $6j$-symbols. You can construct the abelian category and you can construct the tensor product functor but you don't have the associator.
For example it is well-known that the character table of a finite group does not determine the group. It does determine the category of representations as a semisimple abelian category. The $6j$-symbols are needed to make this a monoidal category.
For further discussion see:
Character table does not determine group Vs Tannaka duality
-
I think that the result for sl(2) is due to Kirillov and Reshetikhin, and that these other references are later, although they may give new arguments. – Greg Kuperberg Feb 19 '10 at 14:43
I agree. Kirillov and Reshetikhin stated the result. Masbaum and Vogel proved it by showing both sides satisfied the same linear recurrence relations. Frenkel and Khovanov derive the result. The Frenkel and Khovanov method is the only one which looks as though it could be applied in other examples. – Bruce Westbury Feb 19 '10 at 15:22
In SL(2) the decomposition of 2-fold tensor products is unique, prompting one to compare the decomposition of $(A \otimes B) \otimes C$ with $A \otimes (B\otimes C)$. But for other groups the decomposition isn't unique, without some canonical basis input perhaps. So it's never been clear to me how to define 6j symbols, much less compute them. – Allen Knutson Feb 20 '10 at 15:05
Allen, The decomposition of an exterior power with any representation is unique. The question was phrased to avoid this problem. I defined the 6j-symbols as the components of a functor. In general this would mean specifying bases in vector spaces and there is no consensus on how to do this (but there are several ways it can be done). You put a vector at each vertex of the tetrahedron graph to define a scalar. – Bruce Westbury Feb 22 '10 at 9:06
Allen: Yes, you can define a numerical 6j-symbol given any system of canonical bases of $\text{Inv}(A \otimes B \otimes C)$, where "canonical" is taken in the informal sense. More formally, you can (a) use Lusztig's dual canonical bases; or (b) simply ask for any self-consistent set of formulas. A consistent set of formulas would define the bases with respect to which they are correct. – Greg Kuperberg Feb 23 '10 at 21:21
## 1 Answer
We can also use partitions ($k$) (symmetric powers) instead of ($1^k$) on one or two of the edges. This still gives just scalars, but includes the full story for sl(2).
This problem seems to be equivalent to the problem of computing the exchange operator in the tensor product of two (quantum) symmetric or exterior powers of the vector representation of the quantum sl(n), e.g. $S^kV\otimes S^mV$, in the standard basis of the symmetric (resp., exterior) powers (about exchange operators see my paper with Varchenko arXiv:math/9801135 and my ICM talk arXiv:math/0207008). This can be computed if you know the fusion operator for these representations, which can be computed efficiently using the ABRR (Arnaudon-Buffenoir-Ragoucy-Roche) equation, see e.g. the appendix to arXiv:math/9801135. I am not sure if the answer is completely worked out anywhere, but there are at least some answers. For instance, see the paper arXiv:q-alg/9704005 where something is done even in the elliptic setting (which relates to elliptic 6j symbols of Frenkel-Turaev). What they do is compute the matrix elements for $m=1$, but the general $m$ can be obtained using the fusion procedure. This should be a really nice computation with a nice answer of the type you are expecting. In particular, in a special case you'll get coefficients of Macdonald's difference operators attached to symmetric powers. In the exterior powers case (or a product of a symmetric and an exterior power) the answer will be simpler, since $k$ cannot get larger than $n$ (in this case you should perhaps get a pure product, and it should be completely covered in the above paper).
EDIT: Remark. Let $V,W$ be representations of the quantum group with 1-dimensional zero weight spaces. Then a natural basis in ${\rm Hom}(L_\lambda, (V\otimes L_\mu)\otimes W)$ is the compositions of intertwiners in one order, while a natural basis of ${\rm Hom}(L_\lambda, V\otimes (L_\mu\otimes W))$ is the composition of intertwiners in the other order. Thus, the 6j-symbol matrix (which is by definition the transition matrix between these two bases) is the exchange operator for intertwiners.
-
It has taken me a while to respond as I had not come across fusion operators or exchange operators. I have found a formula which I almost understand which says that the 6j-symbols can be calculated from 3j-symbols using the fusion operator. I have not seen a relation between the 6j-symbols and the exchange operator. – Bruce Westbury Mar 4 '10 at 21:17
I edited my answer to explain the relation between 6j symbols and exchange operators. – Pavel Etingof Mar 4 '10 at 22:13
I need to clear my desk so I can work through this. Meanwhile; if you use Lusztig's tensor product then the fusion operators are identity maps. This means the 6j-symbols are equal to the 3j-symbols (with respect to this tensor product). Then I think the exchange operator is the R_0-matrix. I am not sure how this helps calculate 6j-symbols. – Bruce Westbury Mar 5 '10 at 8:04
I am not sure what you mean by "Lusztig's tensor product". By fusion operator I mean the one defined in my paper with Varchenko "Exchange dynamical quantum groups", and I don't see how it can be equal to the identity map. Also the fusion and exchange operators depend on the dynamical parameter $\lambda$ which is one of the six j-labels. – Pavel Etingof Mar 5 '10 at 14:30 | 1,774 | 7,128 | {"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} | 2.8125 | 3 | CC-MAIN-2016-30 | latest | en | 0.924395 |
http://cn.metamath.org/mpeuni/19.45.html | 1,642,883,532,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00521.warc.gz | 12,132,861 | 3,098 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 19.45 Structured version Visualization version GIF version
Theorem 19.45 2105
Description: Theorem 19.45 of [Margaris] p. 90. See 19.45v 1911 for a version requiring fewer axioms. (Contributed by NM, 12-Mar-1993.)
Hypothesis
Ref Expression
19.45.1 𝑥𝜑
Assertion
Ref Expression
19.45 (∃𝑥(𝜑𝜓) ↔ (𝜑 ∨ ∃𝑥𝜓))
Proof of Theorem 19.45
StepHypRef Expression
1 19.43 1808 . 2 (∃𝑥(𝜑𝜓) ↔ (∃𝑥𝜑 ∨ ∃𝑥𝜓))
2 19.45.1 . . . 4 𝑥𝜑
3219.9 2070 . . 3 (∃𝑥𝜑𝜑)
43orbi1i 542 . 2 ((∃𝑥𝜑 ∨ ∃𝑥𝜓) ↔ (𝜑 ∨ ∃𝑥𝜓))
51, 4bitri 264 1 (∃𝑥(𝜑𝜓) ↔ (𝜑 ∨ ∃𝑥𝜓))
Colors of variables: wff setvar class Syntax hints: ↔ wb 196 ∨ wo 383 ∃wex 1702 Ⅎwnf 1706 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1720 ax-4 1735 ax-5 1837 ax-6 1886 ax-7 1933 ax-12 2045 This theorem depends on definitions: df-bi 197 df-or 385 df-ex 1703 df-nf 1708 This theorem is referenced by: eeor 2169
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Homework 4 Solution
# Homework 4 Solution - EE 351K PROBABILITY RANDOM PROCESSES...
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EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi [email protected] Homework 4 Solution Problem 1 There are n multiple-choice questions in an exam, each with 5 choices. The student knows the correct answer to k of them, and for the remaining n - k guesses one of the 5 randomly. Let C be the number of correct answers, and W be the number of wrong answers. (a) What is the PMF of W ? Is W one of the common random variables we have seen in class? (b) What is the PMF of C ? What is its mean, E [ C ] ? Sol : (a) The student guesses one of the 5 choices randomly, probability that a question is guessed correctly = 1 5 . Since the student knows k answers correctly, the number of wrong answers W [0 , n - k ] . Therefore P W ( w ) = { ( n - k w ) ( 4 5 ) w ( 1 5 ) n - k - w w [0 , n - k ] 0 otherwise which is a binomial random variable. (b) Similarly the PMF of C is P C ( c ) = { ( n - k c - k ) ( 1 5 ) c - k ( 4 5 ) n - c c [ k, n ] 0 otherwise Then E [ C ] = n c = k c ( n - k c - k ) ( 1 5 ) c - k ( 4 5 ) n - c = k + n - k 5 . Problem 2 Fischer and Spassky play a sudden-death chess match whereby the first player to win a game wins the match. Each game is won by Fischer with probability p , by Spassky with probability q , and is a draw with probability 1 - p - q .
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{[ snackBarMessage ]} | 500 | 1,709 | {"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 | 4 | CC-MAIN-2018-17 | latest | en | 0.871476 |
https://www.kidsacademy.mobi/printable-worksheets/easy/age-3-7/math/graphs-worksheets/ | 1,722,740,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00812.warc.gz | 669,912,102 | 56,097 | # Easy Graphs Worksheets for Ages 3-7
Check out this Trial Lesson on Easy Graphs Worksheets for Ages 3-7!
Organize and Categorize Data #2
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Interactive
• 3-7
• Graphs
• Easy
## Graphs: Thanksgiving Treats Worksheet
Enjoy Thanksgiving treats with this fun PDF worksheet! Kids will count the treats and choose which graph accurately shows the number of each item. From turkey to cranberry sauce to pie slices, children will develop visual discrimination and graphing skills while assigning value to pictures.
Graphs: Thanksgiving Treats Worksheet
Worksheet
## Fruit Math: Picture Graphs Worksheet
Help your child learn to draw graphs by having them look at pictures. Ask them to identify a graph that accurately portrays the data and have them explain why. This will prepare them to easily draw more complex graphs in the future. For example, have them look at the picture with the number of fruits in the colorful worksheet and explain why the graph is accurate.
Fruit Math: Picture Graphs Worksheet
Worksheet
## Ticket for the Train Worksheet
It's important to use fun scenarios with familiar objects when teaching kids. Try discussing train rides and the accompanying picture graph. Ask questions about it, then help kids check the correct answers. Let them enjoy the learning process!
Ticket for the Train Worksheet
Worksheet
## Julia's Shopping Graph Worksheet
This fun PDF worksheet provides little mathematicians with practice using pictographs. They will answer questions about Julia's graph, helping to assess their understanding of reading and interpreting data. Counting and choosing the correct answer is sure to be an enjoyable experience!
Julia's Shopping Graph Worksheet
Worksheet
Learning Skills | 344 | 1,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.876352 |
https://www.noisebridge.net/wiki/Talk:DreamTeam/Brainduino | 1,653,575,491,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00197.warc.gz | 1,044,457,761 | 9,051 | # Talk:DreamTeam/Brainduino
## March 21 2017
Q&A with Masahiro!
<< We are making progress on our Brainduino V0.1 assembly. We think we isolated the abnormality to IC3, the low-pass filters. We observe 10,000x gain out of the first two rounds of amplification. Our understanding is that 10,000x gain amplifies a +/-50uV brain powered signal to a +/-0.5V signal, within the operational range of the Arduino ACD. >>
Maybe not easy understanding of so many option setup.
We use option A. OPA2111 set gain X 100. AD8422 use as Gain X1 ( no need put R5 & R6 to AD8422 , because we use option A )
How to get X100 : R23 (100K) / R27 (2K) = 50 R24 (100K) / R27 (2K) = 50 50 + 50 = 100
Then IC3D set gain X250
How to get X250 : 1 + (R16 (1M) / R15 (4.02K) ) = 250 (249.75622)
Total : X 100 X 250 = X 25,000
It is not X 10,000.
C15 : 1 µF makes 0.16Hz low cut ( default ) C15 : 0.1 µF makes 1.6Hz low cut.
C35 : 47 pF makes 3200 Hz high cut ( default ) C35 : 1 nF makes 160 Hz high cut
Need setup for other channel too.
<<Coming out of the first and second round of filters in IC3 on both channels we're seeing our signal clipping at +/-12V. Is this expected? >>
From IC3 output, max clip is +-12V. Usually not clip to +-12V. In case clip continue then not working. Make sure solder jumper : SJ8 : 2-3 connect is default.
C46 1µF is option. default is put 0 ohm.
## March 08 2017
Debugging the brainduino v0.1_foo ...
One of these resistors is not like the other.
R15 is upside down ? A 4.02 KOhm resistor. (John wonders if this is a correct value for setting op-amp gain anyway, given the corresponding 1 MOhm resistor in that part of the circuit.)
R17 the same ... maybe not upside down afterall. Jade observes that neither R15 or R17 have any visible printing on the side facing up, unlike any of the many many other resistors on the board. Hence the initial speculation as to its potential upside-downness ...
Breaking out the multimeter now ... Here we go ... What's going on with this multimeter ?
Now reads 4.01 KOhm on R15 same on R17, very good.
The feedback resistors R16 and R19: R16: 1.001 MOhm R19: 1.001 MOhm
C35 difficult to get consistent reading of capacitance (maybe due to rest of circuit not isolated) C36 11.85 nanoFarads ? Also not consistent .007 nanoFarads ... 0.4 ... 2. ... 1. ... 0.006 ok, we want 0.047 nanoFarads ie 47 picoFarads as specified in schematic but we never see that ... but that's okay, because measuring capacitance in circuit is a fool's errand.
backing up a bit ... R3 tests within 2% of 1 MegaOhm despite being GROSSLY OFF CENTER R4 1.000 MegaOhm feels more solid than R3
should we resolder R3 ?
John would (if not feeling adventurous) read documentation and study IC 1b and IC 1a ...
Dan, feeling adventurous, reads Masahiro's instructables article which is full of win, and yields A CLUE as per corner frequency of the EIGHT POLE BUTTERWORTH FILTER: "1/100 of clock is filter high cut frequency."
## December 21 2016
Understanding the brainduino v0.1
Power supplies, voltage/current regulator. voltage reference
LT1761 - 100mA, Low Noise, LDO Micropower Regulators (IC7) (5 Volts for DC offset, 2.5 V after voltage divider, feeds + signal inputs of 2 different op amps, also provides 5 V reference for Arduino)
RB-0512 - +/- 12 V for op amp (RB-0512 D/P -- recom 12v dc/dc converter) (drives all op amps, also feeds LT1761)
Voltage divider (cf LT1761) built around SJ10, SJ11 a voltage divider for each channel ... also linked with capacitors (two 10 microFarad on input, 0.1 microFarad for each output channel)
Op Amps
• for amplification (instrumentation amp setup)
• for filtering (capacitors C15 and C11
Butterworth filter
Arduino (ADC, communication with Bluetooth module)
... examining the hardware: question about IC2 - funky solder, corrosion? First amplifier of channel one (IC2A and IC2B)
verified IC5, IC6 not populated ...
still looking for IC7! found it. It has some funky solder right next to it. IC7 is tiny. Is it actually IC7? Unlabeled solder blob next to it has white rectangle printed around it ... so maybe solder jumper
IC7 connects to C32
SJ16 is one step closer to bringing voltage divider into effect.
Can measure on blob for 5 Volts ...
There may be some test points near U\$7
Some resistors from divider network identified? R39 is 4.3 K Ohms R34 and R35 are each 10 K Ohms ? both contact R29 which is 2.15 K Ohms to ground
## December 14 2016
what is value of c33? 50 pF or 1 nF ? determines high-cut frequency either 160 Hz (1 nF) or 3200 Hz (50 pF)
Op amps ...
OPA2111 - what gain? expecting ~ 100 ?
what values for R27, R23, R24 ? (also R28, R25, R26)
R27 should be 2K ohms R23, R24 -> 100K ohms
cf instrumentation amp (figure 9 of http://www.ti.com/lit/ds/symlink/opa2111.pdf) 1 + (R23 + R24) / R27 ... so 1 + (200 K / 2 K) = 101
so if input signal is on order of 100 microvolts then output of first amp will be 10.100 millivolts (really more like -50 to +50 microvolts -> -5 to +5 millivolts)
AD8422 R5 and R6 are 200 ohms G = 1 + (19.8 kΩ/RG) = 1 + 99 = 100
??? also see fig. 56 on page 20 of http://www.analog.com/media/en/technical-documentation/data-sheets/AD8422.pdf
now consider if input to AD8422 is on order of 10 millivolts output will be 1.000 volts (or if no offset voltage and input is actually ~ -5 to +5 millivolts, then output will be -0.5 to +0.5 volts)
(however, there is a +2.5 Volt DC offset impinging on the AD2177 part of the instrumentation amp setup ... we think this is ok to say ... But is option A grounded -- no offset -- or 2.5 volts ?
where might DC offset voltage come into play?
2nd Stage AD8422 Instrumentation amp with gain resistor 2(10 kΩ + RREF)/(20 kΩ + RREF) http://www.analog.com/media/en/technical-documentation/data-sheets/AD8422.pdf In figure 56, you will see that an op amp is used to input to the VREF pin.
3rd stage Low-cut (aka high-pass ) filter and DC offset to 3rd Stage OP2177ARMZ ( IC1B Ch1 and IC1A Ch2 )
4th Stage IC3D>IC3C Ch1, and IC3A>IC3B, Potentially considered a two-pole high-cut(aka low-pass)
Next week: verify 101 x gain, 100 x gain, DC offset ! | 1,857 | 6,139 | {"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-2022-21 | latest | en | 0.831644 |
https://justaaa.com/statistics-and-probability/995144-your-friend-wants-to-play-a-game-with-a-pair-of | 1,717,099,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971670239.98/warc/CC-MAIN-20240530180105-20240530210105-00534.warc.gz | 282,085,472 | 9,381 | Question
# Your friend wants to play a game with a pair of dice. You win if the...
Your friend wants to play a game with a pair of dice. You win if the sum of the top faces is 5, 6, 7, or 8. The winner pays the loser \$1 each game. Should you play?
Total number of ways = 36
Number of ways of of getting sum 2 = 1 i.e(1,1)
Number of ways of of getting sum 3 = 2 i.e (1,2) (2,1)
Number of ways of of getting sum 4 = 3 i.e (1,3) (2,2) (3,1)
Number of ways of of getting sum 9 = 4 i.e (4,5) (5,4) (6,3) (3,6)
Number of ways of of getting sum 10 = 3 i.e (6,4) (4,6) (5,5)
Number of ways of of getting sum 11 = 2 i.e (6,5) (5,6)
Number of ways of of getting sum 12 =1 i.e (6,6)
Number of unfavourable cases = 16
Number of favourable cases = 20
Probability of winning \$1 = 20/36
Probability of losing \$1 = 16/36
E(X) = 1*(20/36) - 1*(16/36) = 1/9
since expected value is positive
we should play the game
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 346 | 976 | {"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-2024-22 | latest | en | 0.854579 |
https://www.apsense.com/article/4-ways-to-make-mathematics-assignments-more-interesting.html | 1,675,563,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00571.warc.gz | 667,669,422 | 6,251 | # 4 Ways to Make Mathematics Assignments More Interesting
Math is one of the most dreaded subject areas for a majority of the students. This explains why a lot of students prefer to ask the online experts to “Do my math homework" when they are assigned solve to a tricky math problem. However, solving math problems can be interesting if you allow it to. This blog discusses 4 such ways in which you can make math more interesting for yourself.
1. Treat math problems like puzzles:
Most of the students find the math problems to be intimidating as they fear they can’t solve it. That’s the reason why most people in their higher studies in math require research paper writing service. However, their approach towards a puzzle isn't the same. Instead of looking at math homework like a complicated problem, consider it a puzzle, which it actually is. A changed perception of math problems can make a lot of difference when you are solving the problem.
2. Learn alternative methods:
A math problem can be solved in more number of ways than just one. The problem with the academic curriculum is that they want you to learn only the ones that are mentioned in the syllabus. If you want, you can learn alternative ways to solve math problems. Just go to Google.com and search "Vedic Mathematics" You can thank this blog later.
3. Learn the real-life application of the concepts:
Solving math problems can be a lot more interesting when you start associating real-life problems with the ones you find on your math homework. If you can calculate the time you will need to reach your grandma’s home by using the car’s average speed and the distance of the starting points and destination, you will start enjoying mathematics for sure.
4. Explore math outside the curriculum:
Apart from learning the alternative methods to solve the problems you find in your homework, you should also look for new concepts and formulas to learn outside your curriculum. Math can be a lot of exciting when you find out the cool parts of the subject area. You can do a random search on the internet and see what new you can find about math.
In conclusion,
As mentioned, the problem with math homework is your approach towards solving it. When you start practicing these aforementioned things, you can find math solving a lot more interesting than you do now. Good luck. | 472 | 2,353 | {"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-2023-06 | longest | en | 0.936211 |
http://amcairlines.com/9g6rmcv/kmvmjq.php?785db7=oxidation-state-of-en-ligand | 1,632,714,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00672.warc.gz | 3,789,422 | 6,706 | 2. There are four of them, so we will use the name "tetrahydroxo". +2 means that ⦠A ligand, or complexing agent, is a polar molecule or an ion bonded to a central metal ion. There is only one monodentate ligand, hydroxide. B) What is the coordination number of the central metal in [Co(en) 3]Cl 2? No charge transfer between ligands was observed when Ru was in a 2+ oxidation state. The en (NH 2 CH 2 CH 2 NH 2) ligand can be classified as:_____ (monodentate, bidentate ,tridentate, tetradentate, pentadentate, hexadentate). By doing some simple algebra and using the -1 oxidation state of chloro ligand and the overall charge of -4, we can figure out that the oxidation state of copper is +2 charge. Let the oxidation state of Co = x â´ x + 2(-1)+2(0)= +2 x = +2+2 =+4. In the vanadium redox battery (VRB), each half-cell is composed of a vanadium redox couple. The ease with which the different oxidation states of vanadium can be interconverted has led to its usage in a vanadium flow battery. Metals may exhibit multiple oxidation states 3. In its non-ionized state, copper has the following electron distribution: [Ar]4s 1 3d 10 . Let's assume oxidation state of Cobalt = x and, that of C l â = â 1 Then, Each pyridine has one nitrogen that possesses a set of unpaired electrons that it can âdonateâ to a metal to form a dative covalent (co-ordinate) bond. Oxidation state ambiguity is the central aspect to metal complexes involving redox non-innocent ligands Jørgenson, 1966: "Ligands are innocent when they allow the oxidation state of the central atoms to be defined" Chirik, 2010: "Redox-active, or 'noninnocent,' ligands have more energetically accessible levels that allow redox In the dioxidized state, the spectroelectrochemistry of the disubstituted complex shows a ligand-to-ligand charge transfer at 1425 nm, with a transition moment of 1.25 Å and an effective two-state coupling of 1200 cm â1. Reactivity includes: A) Ligand exchange processes: i) Associative (S. N. 2; expanded coordination no.) 4. Potassium has an oxidation number of +1, giving an overall charge of +2. Multiple Oxidation States. Sometimes, the oxidation states can also be written as a superscripted number to the right of the element symbol (Fe 3+ ). The oxidation number for metals that can have more than one oxidation state is represented by a Roman numeral. Because the compound is neutral and 2(1)+(Cr)+4(-2)=0, chromium must have an oxidation number of +6. The oxidation number for metals that can have more than one oxidation state is represented by a Roman numeral. Metals may exhibit paramagnetism dependent on metal oxidation state and on ligand field. A complex ion is a polyatomic species consisting of a central metal ion surrounded by several ligands. a) Each Cl ligand has a charge of â1, so 4 x â1 = â4 16. Calculate the oxidation state of the metal and the number of d electrons in the following coordination complexes: a) [CoCl4] 2â ; b) [Fe(bpy) 3] 3+; c) [Cu(ox) 2] 2â; d) [Cr(CO) 6] Ans. A) What is the oxidation state of the central metal in [Ni(en) 3]SO 4? C) What is the oxidation state of the central metal in [Ni(en) 3](NO 3) 2? Immediately we know that this complex is an anion. As the name suggests, the molecule comprises two connected pyridines. The oxidation number is placed in parentheses after the name of the element (iron(III)). The oxidation state of the metal is 3 (x+(-1)4=-1). At the anode VO2+ ions are converted to VO2+ ions and when electrons are removed from the positive terminal of the battery. The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". Ethylenediamine is a neutral ligand and chloride has a â 1 charge associated with it. ii) Dissociative (S. N. 1; slow step is ligand loss) B) Redox Processes . 2,2'-Bipyridine is a bidentate chelating ligand, forming complexes with many transition metals. Oxidation state of Co = +4 Coordination sphere consists of 2 chloro ligands with -1 charge , 2 ethylene diammine (en) neutral ligand .As sulphate carries a 2- charge ,the coordination sphere carries a +2 charge as the the coordination compound is electrically neutral. Naming ... = oxidation state of metal ion + total charge on ligands = 2+ + 0 = 2+ i 1 charge associated with it, forming complexes with many transition metals iron ( III ) ) =! No charge transfer between ligands was observed when Ru was oxidation state of en ligand a 2+ oxidation state represented. 3 ) 2 S. N. 2 ; expanded coordination no. is represented by a Roman numeral a! Has an oxidation number for metals that can have more than one oxidation state of =... In [ Ni ( en ) 3 ] Cl 2 b ) redox processes a neutral ligand and chloride a! 1 charge associated with it charge transfer between ligands was observed when Ru was in a 2+ oxidation state Co. Number is placed in parentheses after the name suggests, the molecule comprises two connected pyridines Cl 2 also... ( -1 ) 4=-1 ) 3d 10 ) b ) What is oxidation.: [ Ar ] 4s 1 3d 10 the following electron distribution: [ Ar ] 4s 1 3d.... Ions and when electrons are removed from the positive terminal of the (... Use the name suggests, the oxidation number for metals that can have more than one oxidation state of battery! ) 3 ] Cl 2 oxidation number of the element symbol ( Fe 3+ ) a â 1 charge with! Element ( iron ( III ) ), each half-cell is composed of a vanadium redox.. Complex ion is a bidentate chelating ligand, forming complexes with many metals! Will use the name of the battery metal oxidation state of the element ( iron ( III ) ) is. Is placed in parentheses after the name of the metal is 3 x+... Agent, is a bidentate chelating ligand, or complexing agent, is a polyatomic species consisting of central. ] ( no 3 ) 2 ligand and chloride has a â charge... With it name of the central metal ion removed from the positive of... An oxidation number of the central metal ion surrounded by several ligands suggests... Right of the element symbol ( Fe 3+ ) was observed when Ru oxidation state of en ligand in 2+! 1 3d 10 ) 3 ] ( no 3 ) 2 ), half-cell.  1 charge associated with it ligand exchange processes: i ) Associative ( S. N. ;! Number of the metal is 3 ( x+ ( -1 ) 4=-1.. And when electrons are removed from the positive terminal of the central metal in Co! 3D 10 charge transfer between ligands was observed when Ru was in a oxidation! Two connected pyridines a ligand, forming complexes with many transition metals more than one state! Metals may exhibit paramagnetism dependent on metal oxidation state is represented by a Roman numeral N. 2 ; coordination... Forming complexes with many transition metals number of +1, giving an charge! A complex ion is a polar molecule or an ion bonded to a central metal in [ (! Potassium has an oxidation number for metals that can have more than one oxidation state has the electron... A Roman numeral anode VO2+ ions are converted to VO2+ ions are converted to VO2+ ions are converted to ions! Is 3 ( x+ ( -1 ) 4=-1 ) metals may exhibit paramagnetism dependent on oxidation! Element symbol ( Fe 3+ ) [ Ar ] 4s 1 3d 10 ( en ) 3 ] no. I ) Associative ( S. N. 2 ; expanded coordination no. +1, giving overall... The name `` tetrahydroxo oxidation state of en ligand ; expanded coordination no. is placed in after! To a central metal ion surrounded by several ligands so we will use the name tetrahydroxo! 3D 10 and chloride has a â 1 charge associated with it N. 2 expanded!, each half-cell is composed of a central metal in [ Ni ( en ) 3 ] ( no )!, copper has the following electron distribution: [ Ar ] 4s 1 3d 10 ion is a polar or... The vanadium redox couple battery ( VRB ), each half-cell is of. Has a â 1 charge associated with it expanded coordination no., the molecule comprises connected. We will use the name `` tetrahydroxo '' are converted to VO2+ ions converted. Written as a superscripted number to the right of the element symbol ( Fe 3+ ) has â... State of Co = +4 2,2'-Bipyridine is a neutral ligand and chloride has a â 1 charge associated it. In oxidation state of en ligand non-ionized state, copper has the following electron distribution: [ Ar ] 4s 3d..., or complexing agent, is a neutral ligand and chloride has a â charge... Co ( en ) 3 ] ( no 3 ) 2 3 ] Cl 2 composed of a metal... ( VRB ), each half-cell is composed of a vanadium redox couple are converted to VO2+ ions converted. -1 ) 4=-1 ) x+ ( -1 ) 4=-1 ) as a superscripted number to the of! Four of them, so we will use the name suggests, oxidation. Name suggests, the oxidation number for metals that can have more than oxidation. State, copper has the following electron distribution: [ Ar ] 1! Ni ( en ) 3 ] ( no 3 ) 2 ( 3... Element ( iron ( III ) ) the metal is 3 ( x+ ( -1 ) ). ( S. N. 2 ; expanded coordination no. Ar ] 4s 1 3d 10 ) 3 Cl... Processes: i ) Associative ( S. N. 1 ; slow step is loss! ] Cl 2 charge associated with it from the positive terminal of the central metal ion surrounded by several.... Between ligands was observed when Ru was in a 2+ oxidation state charge associated with it 4=-1.! And when electrons are removed from the positive terminal of the element iron... Overall charge of +2 ligand, or complexing agent, is a bidentate chelating ligand forming... Ion bonded to a central metal ion surrounded by several ligands in [ Ni ( en ) 3 (. ] ( no 3 ) 2 name `` tetrahydroxo '' the anode VO2+ ions and when are! ] 4s 1 3d 10 ( no 3 ) 2 after the name `` ''! `` tetrahydroxo '' of the element ( iron ( III ) ) was in a 2+ state. Four of them, so we will use the name of the metal is 3 x+! Was observed when Ru was in a 2+ oxidation state is represented by a Roman numeral complex ion a... Consisting of a central metal in [ Co ( en ) 3 ] no. 4S 1 3d 10 when electrons are removed from the positive terminal of battery. Than one oxidation state is represented by a Roman numeral ; slow is. Between ligands was observed when Ru was in a 2+ oxidation state is represented by a Roman.. ) Associative ( S. N. 1 ; slow step is ligand loss ) b ) is. Observed when Ru was in a 2+ oxidation state is represented by Roman... Written as a superscripted number to the right of the battery 4s 3d! Are converted to VO2+ ions are converted to VO2+ ions are converted to VO2+ ions are converted to ions. An oxidation number for metals that can have more than one oxidation state is represented by a Roman numeral ]... Right of the battery redox battery ( VRB ), each half-cell is of... ( -1 ) 4=-1 ) by several ligands between ligands was observed when Ru in! Coordination no. ) What is the oxidation state is represented by Roman. Ii ) Dissociative ( S. N. 2 ; expanded coordination no. complexing agent, is a polyatomic consisting. Converted to VO2+ ions and when electrons are removed from the positive terminal of the element ( (. Includes: a ) ligand exchange processes: i ) Associative ( S. N. 2 ; expanded no! Following electron distribution: [ Ar ] 4s 1 3d 10 at the anode VO2+ ions and when electrons removed! Molecule comprises two connected pyridines is composed of a vanadium redox couple may exhibit paramagnetism dependent on metal oxidation is. Ion bonded to a central metal in [ Ni ( en ) 3 ] ( no 3 ) 2 is! Number for metals that can have more than one oxidation state of the central metal surrounded. 1 ; slow step is ligand loss ) b ) What is the oxidation number for metals that have! Ru was in a 2+ oxidation state of the element ( iron ( ). ( iron ( III ) ) sometimes, the oxidation number of +1, giving overall... By a Roman numeral, is a polyatomic species consisting of a vanadium redox couple paramagnetism dependent on oxidation. A ligand, forming complexes with many transition metals the molecule comprises two connected pyridines i the state... Than one oxidation state of the battery observed when Ru was in a 2+ oxidation state is by..., giving an overall charge of +2, forming complexes with many transition metals 1 ; step! Forming complexes with many transition metals that can have more than one oxidation state is by... A complex ion is a bidentate chelating ligand, forming complexes with many transition metals can be! To VO2+ ions are converted to VO2+ ions are converted to VO2+ ions and when are! Terminal of the element ( iron ( III ) ) number of the battery ) Dissociative ( N.! Loss ) b ) What is the oxidation number is placed in parentheses after the suggests. In its non-ionized state, copper has the following electron distribution: [ Ar ] 4s 1 3d 10 four! 2 ; expanded coordination no. in its non-ionized state, copper has the following distribution... | 3,326 | 12,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-39 | latest | en | 0.88634 |
https://www.teacherspayteachers.com/Browse/Search:addition%20and%20subtraction%20fractions/Type-of-Resource/Lesson-Plans-Individual?ref=filter/resourcetype/all | 1,550,671,510,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495001.53/warc/CC-MAIN-20190220125916-20190220151916-00595.warc.gz | 964,440,109 | 52,823 | showing 1-24 of 206 results
This document is a step-by-step journal to help kids learn more in depth about the steps of adding and subtracting fractions. Students will learn how to add and subtract fractions with the same denominator, then move to fractions with different denominators. The document is in half pages so that
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WORD PROBLEMS ADDITION & SUBTRACTION of FRACTIONS NO FLUFF - JUST 25 PURE WORD PROBLEMS! You receive 2 full worksheets of 25 word problems involving ALL types of addition and subtraction of fractions with LIKE and UNLIKE denominators. Each problem involves either addition or subtraction or BO
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This is a 20 page FREE item on Addition and Subtraction of Fractions. Students may work alone or in a small group to do a Gallery Walk around the classroom. Fraction problems do have borrowing and reducing! This is a wonderful activity to do in class. Answer key is included. Please check out
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Hello teachers!This fraction game is a powerful tool in building concrete, thorough understanding of fractions. Included with this game is a detailed lesson plan, 3rd, 4th, and 5th grade Common Core standards, learning targets, and discussion questions. There are also game cards, fraction pieces (pr
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ZIPÂ (2.33 MB)
This is the ninth unit of 5th Grade Guided Math. This unit is focused on Adding and Subtracting Fractions (5.NF.1,2) (TEKS 5.3H,K--download the preview for more information about TEKS correlation) and is a three week long unit. 5th Grade Guided Math units are recommended for small group, targeted in
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Also included in:Â 5th Grade Guided Math -Year Long Bundle
\$12.75
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4.0
ZIPÂ (8.37 MB)
A YEAR OF GUIDED MATH LESSON PLANS for 5TH GRADE! Over 680 Pages of Lesson Plans, Anecdotal Records Sheets, Vocabulary Pages and Graphic Organizers! This product includes 30 weeks (CCSS) and 32 weeks (TEKS) of Guided Math Lesson Plans for the teacher and activities for the students! 26 CCSS Alig
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Also included in:Â Math Bundle - 5th Grade
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These lesson plans are recommended for February or March of the school year, and are relevant to the 5th Grade level, which are aligned to the Common Core Standards. The cover operations with fractions, such as adding and subtracting fractions and mixed numbers with different denominators. The wor
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Adding and Subtracting Fractions with Unlike Denominators will help students with more practice to master this tough skill. These task cards will help make your math center come alive with this fun monster theme. This pack includes: 30 Adding Fraction Task Cards 30 Subtracting Fraction Task Cards Re
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\$3.50
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This bundle has everything you need to scaffold your students through adding and subtracting fractions with like and unlike denominators. All of the materials are compatible with interactive notebooks, and facilitate whole-group, small-group, and collaborative learning.Included in the guide:• Hands
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This is the ninth unit of 4th Grade Guided Math. This unit is focused on adding and subtracting fractions. (4.NF.3,5) (TEKS 4.3B,E,F--download the preview for more information about TEKS correlation) and is a two week long unit.4th Grade Guided Math units are recommended for small group, targeted in
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Also included in:Â 4th Grade Guided Math -Year Long Bundle
\$8.50
28 Ratings
4.0
ZIPÂ (8.34 MB)
This 8-page lesson packet includes everything you need to teach unlike mixed number addition and subtraction word problems. Problems include improper sums, subtraction with regrouping, and fractions that need to be simplified. And best of all it requires NO PREP! Just PRINT and TEACH.What's includ
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PDFÂ (7.5 MB)
This 8-page lesson packet includes everything you need to teach addition and subtraction word problems for fractions and mixed numbers with like denominators. Problems include improper sums and fractions that need to be simplified so this packet is the perfect way to prepare students for unlike fra
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Also included in:Â 4th Grade Fraction Operations Unit: 4th Grade Fractions Lesson Bundle
\$3.00
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PDFÂ (5.26 MB)
Adding and Subtracting Fractions with Like Denominators will help students with more practice to master this skill. These task cards will help make your math center come alive with this fun monster theme. This pack includes: 30 Adding Fraction Task Cards 30 Subtracting Fraction Task Cards Recording
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This 110 slide power point presentation clearly demonstrates how to solve addition and subtraction problems with fractions. The following sections are included in the power point: **Adding Fractions with the Same Denominator **Subtracting Fractions with the Same Denominator **Adding Mixed Numbers w
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This six-part lesson is a great addition to any number system lesson. Students will develop a greater understanding of how to add and subtract fractions and decimals. The different parts of this lesson include the following: *Problem of the Day: Students will begin class by taking a few minutes to
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ZIPÂ (1.12 MB)
Adding and Subtracting Fractions – 7th Grade Rational Numbers ***We also have the entire 7th Grade Curriculum available here:*** 7th Grade Math Bundle – Year Long Curriculum This lesson teaches students how to Add and Subtracting Fractions in 7th Grade Rational Numbers. They will review basic voc
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There are 30 questions with multiple choice answers. The topics covered are : Addition, Subtraction, Multiplication, Division, Fractions and Expression Some algebra expressions with decimal is also included. Check our blog at : Math Blog By Jega Chinna
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Sixth Grade Fractions 3 - Addition and Subtraction of Unlike Denominators teaches students how to add and subtract fractions with unlike denominators using the lowest common multiple. There are 13 strategies of effective teaching based on research. For best results, students should have white bo
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Also included in:Â 6th Grade Fractions Bundle - 7 Powerpoint Lessons - 386 Slides
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In Lesson 1, students used plus and minus counters and a number line to add and subtract integers. This lesson takes what they have learned and uses it to develop a set of rules that they can remember. The following items are included in this lesson: -Do Now -Rules for Integer Addition and Subtr
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Fifth Grade Fractions 4 - Addition & Subtraction of Like Denominators teaches students how to add and subtract fractions that have like denominators. Models are used. There is a title slide, focus slide, objective, essential question, vocabulary, concept development, demonstration on how to add
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Sixth Grade Fractions 4 - Addition and Subtraction of Mixed Numbers teaches students how to add and subtract mixed numbers with unlike denominators. There are 13 strategies of effective teaching based on research. For best results, students should have white boards and dry erase markers. Incl
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Also included in:Â 6th Grade Fractions Bundle - 7 Powerpoint Lessons - 386 Slides
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Simple Inequalities – Subtraction, Addition, Multiplication and Division Problems Product Description: This unit consists of four thorough lessons on simple inequalities (including addition, subtraction, multiplication and division problems), compound inequalities, and absolute value inequalities.
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This bundle includes three Adding and Subtracting Fractions with Like Denominators resources for 4th grade. Included are a complete unit with lesson plans, a set of ten stations, and a Digital Stinky Feet review game. Each element can be found alone, but you can purchase together to save. Adding a
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This is Part 5 of "Understanding Fractions", a multi-part Scaffolded fractions curriculum. Part 5 topics include: Comparing Fractions & Mixed Numbers; Addition & Subtraction of Fractions with Different Denominators; Addition & Subtraction of Fractions and Mixed Numbers. Includes 5 Qui
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,204 | 8,935 | {"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} | 3.375 | 3 | CC-MAIN-2019-09 | latest | en | 0.879197 |
https://convertoctopus.com/1713-cubic-centimeters-to-fluid-ounces | 1,675,773,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00852.warc.gz | 199,537,434 | 7,443 | ## Conversion formula
The conversion factor from cubic centimeters to fluid ounces is 0.033814022558919, which means that 1 cubic centimeter is equal to 0.033814022558919 fluid ounces:
1 cm3 = 0.033814022558919 fl oz
To convert 1713 cubic centimeters into fluid ounces we have to multiply 1713 by the conversion factor in order to get the volume amount from cubic centimeters to fluid ounces. We can also form a simple proportion to calculate the result:
1 cm3 → 0.033814022558919 fl oz
1713 cm3 → V(fl oz)
Solve the above proportion to obtain the volume V in fluid ounces:
V(fl oz) = 1713 cm3 × 0.033814022558919 fl oz
V(fl oz) = 57.923420643429 fl oz
The final result is:
1713 cm3 → 57.923420643429 fl oz
We conclude that 1713 cubic centimeters is equivalent to 57.923420643429 fluid ounces:
1713 cubic centimeters = 57.923420643429 fluid ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 fluid ounce is equal to 0.017264173781378 × 1713 cubic centimeters.
Another way is saying that 1713 cubic centimeters is equal to 1 ÷ 0.017264173781378 fluid ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand seven hundred thirteen cubic centimeters is approximately fifty-seven point nine two three fluid ounces:
1713 cm3 ≅ 57.923 fl oz
An alternative is also that one fluid ounce is approximately zero point zero one seven times one thousand seven hundred thirteen cubic centimeters.
## Conversion table
### cubic centimeters to fluid ounces chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to fluid ounces
cubic centimeters (cm3) fluid ounces (fl oz)
1714 cubic centimeters 57.957 fluid ounces
1715 cubic centimeters 57.991 fluid ounces
1716 cubic centimeters 58.025 fluid ounces
1717 cubic centimeters 58.059 fluid ounces
1718 cubic centimeters 58.092 fluid ounces
1719 cubic centimeters 58.126 fluid ounces
1720 cubic centimeters 58.16 fluid ounces
1721 cubic centimeters 58.194 fluid ounces
1722 cubic centimeters 58.228 fluid ounces
1723 cubic centimeters 58.262 fluid ounces | 567 | 2,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | latest | en | 0.718134 |
http://mathoverflow.net/questions/134334/estimate-the-diagonal-of-a-cholesky-factor | 1,469,494,553,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00099-ip-10-185-27-174.ec2.internal.warc.gz | 169,959,627 | 13,969 | # Estimate the diagonal of a Cholesky factor…?
I'm computing several hundred Cholesky factorizations of large, sparse matrices, and I'm really only doing Cholesky factorization because I need to know the diagonal elements of the Cholesky factor L. Does anyone know a good method for estimating the diagonal of Cholesky factors (that does not involve Cholesky factorization)?
For example: Q = L * transpose(L) D = diagonal(L)
Can I get D from Q without calculating all of L?
Thanks!
-
The product of the first $k$ diagonal elements of $L$ squared is equal to the determinant of the $k\times k$ top left submatrix of $Q$. – Yoav Kallus Jun 20 '13 at 21:37
That's interesting and possibly helpful. I do need the entire diagonal of L, which would leave me taking the determinant of Q... and determinants for large, sparse, spd matrices are usually calculated via Cholesky factorization, which would put me back where I started. Any thoughts on that? – Ted Jun 21 '13 at 11:24
@Ted, in a first time, you want the diagonal of $L$. As wrote Kallus, that implies the calculation of $\det(Q)$. In a second time, you write yourselves that your question has not any interest. Finally you say " I do need the entire diagonal of L". What do you want exactly ? – loup blanc Dec 7 '13 at 13:18 | 322 | 1,283 | {"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} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.939245 |
https://toc.123doc.org/document/1008209-7-examination-of-major-assumptions-of-multiple-regressionmultiple-regression-analysis.htm | 1,508,430,044,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00015.warc.gz | 868,669,880 | 10,596 | Tải bản đầy đủ
7…Examination of Major Assumptions of Multiple Regressionmultiple regression Analysis
7…Examination of Major Assumptions of Multiple Regressionmultiple regression Analysis
Tải bản đầy đủ
8.7 Examination of Major Assumptions
203
percentages of residuals falling within ± 2 SE or ± 2.5 SE. More formal assessment can be made by running the tests: Shapiro–Wilk, Kolmogorov–Smirnov,
Cramer–von Mises and Anderson–Darling.1
How to Autocorrelation Assumption.mp4
8.7.4 Test of Homogeneity of Variance (Homoscedasticity)
The assumption of constant variance of the error term can be examined by plotting
ˆ i. If the
the residuals against the predicted values of the dependent variable, Y
pattern is not random, the variance of the error term is not constant. See the video
How to check Normality Assumption.
8.7.5 Test of Autocorrelation
How to Check No Multicollinearity Assumption.mp4
A plot of residuals against time, or the sequence of observations, will throw
some light on the assumption that the error terms are uncorrelated or no autocorrelation. A random pattern should be seen if this assumption is true. A more
formal procedure for examining the correlations between the error terms is the
Durbin–Watson test (Applicable only for time series data).
8.7.6 Test of Multicollinearity
The presence of multicollinearity or perfect linear relationship between independent variables can be identified using different methods. These methods are:
1. VIF (Variance-Inflating factor): As a rule of thumb, If the VIF value exceeds
10, which will happen only if correlation between independent variables
exceeds 0.90, that variable is said to be highly collinear (Gujarati and
Sangeetha 2008).
1
Null hypothesis the observations are normally distributed, alternative hypothesis not normally
distributed.
204
8 Multiple Regression
2. TOL (Tolerance): The closer the TOL to zero, the greater the degree of collinearity of the variables (Gujarati and Sangeetha 2008).
3. Conditional Index (CI): If CI exceeds 30, there is severe multicollinearity
(Gujarati and Sangeetha 2008).
4. Partial Correlations: High partial correlation between independent variables
also shows the presence of multicollinearity.
How to Check No Multicollinearity Assumption.mp4
8.7.7 Questions
Examine the following fictitious data
Model
1
R
0.863
R2
0.849
0.850
Std. error of the estimate
13.8767
1. Which of the following statements can we not say?
(a) The standard error is an estimate of the variance of y, for each value of x.
(b) In order to obtain a measure of explained variance, you need to square
the correlation coefficient.
(c) The correlation between x and y is 86 %.
d) The correlation is good here as the data points cluster around the line of fit
quite well. So prediction will be good.
(e) The correlation between x and y is 85 %.
2. The slope of the line is called:
(a) Which gives us a measure of how much y changes as x changes.
(b) Is the point where the regression line cuts the vertical axis.
(c) A correlation coefficient indicates the variability of the points around the
regression line in the scatter diagram.
(d) None of the above.
(e) The average value of the dependent variable.
3. Using some fictitious data, we wish to predict the musical ability for a person
who scores 8 on a test for mathematical ability. We know the relationship is
positive. We know that the slope is 1.63 and the intercept is 8.41. What is their
predicted score on musical ability?
8.7 Examination of Major Assumptions
(a)
(b)
(c)
(d)
(e)
205
80.32
-4.63
21.45
68.91
54.55
4. We have a negative relationship between number of drinks consumed and
number of marks in a driving test. One individual scores 3 on number of drinks
consumed, another individual scores 5 on number of drinks consumed. What
will be their respective scores on the driving test if the intercept is 18 and the
slope 3?
(a) It is not possible to predict from negative relationships.
(b) Driving test scores (Y-axis) will be 51 and 87 [individual who scored 5 on
drink consumption].
(c) Driving test scores (Y-axis) will be 27 [individual who scored 3 on drink
consumption] and 33 [individual who scored 5 on drink consumption].
(d) Driving test scores (Y-axis) will be 9 [individual who scored 3 on drink
consumption] and 3 [individual who scored 5 on drink consumption].
(e) None of these.
5. You are still interested in whether problem-solving ability can predict the
ability to cope well in difficult situations; whether motivation can predict
coping and whether these two factors together predict coping even better. You
produce some more results.
Dependent variable coping skills in difficult situations
Constant
Problem
Motivation
Unstandardized coefficients
Standardized coefficients
B
-0.466
0.200
0.950
Beta
Std. error
0.241
0.048
0.087
0.140
0.740
t
Sig.
1.036
2.082
10.97
0.302
0.030
0.000
Which of the following statements is incorrect?
(a) As motivation increases by one standard deviation, coping skills increases by
almost three quarters of a standard deviation (0.74). Thus, motivation appears
to contribute more to coping skills than problem solving.
(b) As motivation increases by one unit coping skills increases by 0.95.
206
8 Multiple Regression
(c) The t-value for problem solving is 2.082 and the associated probability is
0.03. This tells us the likelihood of such a result arising by sampling error,
assuming the null hypothesis is true, is 97 in 100.
(d) Problem solving has a regression coefficient of 0.20. Therefore, as problem
solving increases by one unit coping skills increases by 0.20.
(e) None of these.
Chapter 9
Exploratory Factor and Principal
Component Analysis
Chapter Overview
This chapter provides an introduction to Factor Analysis (FA): A procedure to
define the underlying structure among the variables in the analysis. The chapter
provides general requirements, statistical assumptions and conceptual assumptions
behind FA. This chapter explains the way to do FA with IBM SPSS 20.0. It shows
how to determine the number of factors to retain, interpret the rotated solution,
create factor scores and summarize the results. Fictitious data from two studies are
analysed to illustrate these procedures. The present chapter deals only with the
creation of orthogonal (uncorrelated) components.
9.1 What is Factor Analysis
According to Hair et al. (2010),1 ‘factor analysis is an interdependence technique
whose primary purpose is to define the underlying structure among the variables in
the analysis’. Suppose a marketing researcher wants to identify the underlying
dimensions of retail brand attractiveness. He begins by administering the retail
brand attractiveness scale from the existing literature to a large sample of people
(N = 2000) during their visit in a particular retail store. Assume that there are five
different dimensions, which consist of 30 different items. What the researcher will
end up with these 30 different observed variables, the mass number as such will
say very little about the underlying dimension of this retail attractiveness. On
average, some of the scores will be high, some will be low and some intermediate,
but interpretation of these scores will be extremely difficult if not impossible. This
is where the tool factor analysis (FA) comes in handy and it allows the researcher
in ‘data reduction’ and ‘data summarization’ of this large pool of items to a few
representative factors or dimensions, which could be used for further multivariate
1
See Ref. Hair et al. (2010).
S. Sreejesh et al., Business Research Methods,
DOI: 10.1007/978-3-319-00539-3_9,
Ó Springer International Publishing Switzerland 2014
207
208
9 Exploratory Factor and Principal Component Analysis
statistical analysis. The general purpose of FA is the orderly simplification of a
large number of intercorrelated measures or condense the information contained in
a number of original variables into a few representative constructs or factors with
minimal loss of information. The application of FA is based on some of the
following conditions: general requirement, statistical assumptions and conceptual
assumptions (See Table 9.1).
FA is used in the following circumstances:
1. To identify underlying dimensions, or factors, that explains the correlations
among a set of variables. For example, a set of personality trait statements may
be used to measure the personality dimensions of people. These statements may
then be factor analysed to identify the underlying dimensions of personality
trait or factors.
2. To identify a new, smaller, set of uncorrelated variables to replace the original
set of correlated variables in subsequent multivariate analysis (regression or
discriminant analysis). For example, the psychographic factors identified may
be used as independent variables in explaining the differences between loyal
and non-loyal consumers.
3. To identify a smaller set of salient variables from a larger set for use in subsequent multivariate analysis. For example, a few of the original lifestyle
Table 9.1 Conditions for doing factor analysis
General Requirements
1. Type of scale: Observed variables should be measured in either interval or ration scales, or at
least at the ordinary level
2. Number of Items: If the researcher has prior knowledge about the underlying factor structure
and want to test the dimensionality, then at least five or more variables should be included to
represent each factor structure
3. Sample size: The rule of thumb for sample size is to have at least five times as many cases as
variables entered into factor analysis +10
Statistical Assumptions
1. Random sampling: Each participant will contribute one response for each observed variable.
These sets of scores should represent a random sample drawn from the population of interest
2. Linearity: The relationship between all observed variables should be linear
3. Bivariate Normal Distribution: Each pair of observed variables should display a bivariate
normal distribution (e.g. they should form an elliptical scattergram when plotted)
Conceptual Assumptions
1. Variable Selection: Factor analysis is based on the basic assumption that there exists an
underlying structure for the selected set of variables. The presence of high correlation and
subsequent interpretation of do not guarantee relevance, even if it meets statistical
assumptions. Therefore, it is the responsibility of the researcher to select the set of variables or
items that are conceptually valid and appropriate to represent the underlying dimension
2. Sample Homogeneity: Another important conceptual assumption with regard to the factor
analysis is that the selected sample should be homogeneous with respect to the underlying
factor structure. It is inappropriate to do factor analysis for a set of items once the researcher
knows a priori that the sample of male and female is different because of gender. The
ignorance of this heterogeneity, and subsequent mixing of two groups (males and females
would results in getting a correlation matrix and factor structure, that will be a poor
representation of the unique structure of each group | 2,505 | 11,172 | {"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-2017-43 | latest | en | 0.855338 |
https://gpuzzles.com/questions/interview/aptitude/barclays-capital-puzzles/ | 1,500,756,235,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424148.78/warc/CC-MAIN-20170722202635-20170722222635-00098.warc.gz | 677,477,012 | 9,033 | # Barclays Capital Interview Puzzle #1 - Fastest Horse Interview Puzzle
Difficulty Popularity
You are provided with twenty five different horses and you must find out who are the fastest horses. You can conduct a race of five horses only at one time. There is no point in the race where you can find out the actual speed of a horse in a race.
How many races will it take to help you determine the fastest three horses?
# Barclays Capital Interview Puzzle #2 - Popular Logic Interview Problem
Difficulty Popularity
A box contains seven purple, five blue and eleven yellow balls.
What is the minimum number of tries required to get one blue and one yellow ball ?
# Barclays Capital Interview Puzzle #3 - Distance and Speed Interview Question
Difficulty Popularity
Peter travels 20 km a day uniformly. John starts from the same point peter started after three days. He travels at a speed of 15 km a day on the first day, at 19 km a day of the second day and so on following an arithmetic progression.
In how many days will he catch up with Peter?
# Barclays Capital Interview Puzzle #4 - Popular Logical Interview Problem
Difficulty Popularity
You are invited to a logical competition where some mini games are to be played among the participants that comprises of logics. What they do is make you stand in front of the water tank and give you two jars - one of 3 liters capacity and one of 5 liters capacity. Then they tell you to collect exactly 4 liters of water using the two jugs.
How will you do it?
# Barclays Capital Interview Puzzle #5 - Hands of Clock Interview Puzzle
Difficulty Popularity
Can you determine how many times do the minute and hour hands of a clock overlap in a day?
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In his mathematics activity class, Andy ... | 512 | 2,324 | {"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-2017-30 | latest | en | 0.900636 |
http://www.britannica.com/EBchecked/topic/682032/numerals-and-numeral-systems | 1,432,934,217,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930423.94/warc/CC-MAIN-20150521113210-00154-ip-10-180-206-219.ec2.internal.warc.gz | 357,480,738 | 17,528 | # Numerals and numeral systems
Mathematics
numerals and numeral systems, a collection of symbols used to represent small numbers, together with a system of rules for representing larger numbers.
Just as the first attempts at writing came long after the development of speech, so the first efforts at the graphical representation of numbers came long after people had learned how to count. Probably the earliest way of keeping record of a count was by some tally system involving physical objects such as pebbles or sticks. Judging by the habits of indigenous peoples today as well as by the oldest remaining traces of written or sculptured records, the earliest numerals were simple notches in a stick, scratches on a stone, marks on a piece of pottery, and the like. Having no fixed units of measure, no coins, no commerce beyond the rudest barter, no system of taxation, and no needs beyond those to sustain life, people had no necessity for written numerals until the beginning of what are called historical times. Vocal sounds were probably used to designate the number of objects in a small group long before there were separate symbols for the small numbers, and it seems likely that the sounds differed according to the kind of object being counted. The abstract notion of two, signified orally by a sound independent of any particular objects, probably appeared very late.
## Number bases
When it became necessary to count frequently to numbers larger than 10 or so, the numeration had to be systematized and simplified; this was commonly done through use of a group unit or base, just as might be done today counting 43 eggs as three dozen and seven. In fact, the earliest numerals of which there is a definite record were simple straight marks for the small numbers with some special form for 10. These symbols appeared in Egypt as early as 3400 bc and in Mesopotamia as early as 3000 bc, long preceding the first known inscriptions containing numerals in China (c. 1600 bc), Crete (c. 1200 bc), and India (c. 300 bc). Some ancient symbols for 1 and 10 are given in the figure.
The special position occupied by 10 stems from the number of human fingers, of course, and it is still evident in modern usage not only in the logical structure of the decimal number system but in the English names for the numbers. Thus, eleven comes from Old English endleofan, literally meaning “[ten and] one left [over],” and twelve from twelf, meaning “two left”; the endings -teen and -ty both refer to ten, and hundred comes originally from a pre-Greek term meaning “ten times [ten].”
It should not be inferred, however, that 10 is either the only possible base or the only one actually used. The pair system, in which the counting goes “one, two, two and one, two twos, two and two and one,” and so on, is found among the ethnologically oldest tribes of Australia, in many Papuan languages of the Torres Strait and the adjacent coast of New Guinea, among some African Pygmies, and in various South American tribes. The indigenous peoples of Tierra del Fuego and the South American continent use number systems with bases three and four. The quinary scale, or number system with base five, is very old, but in pure form it seems to be used at present only by speakers of Saraveca, a South American Arawakan language; elsewhere it is combined with the decimal or the vigesimal system, where the base is 20. Similarly, the pure base six scale seems to occur only sparsely in northwest Africa and is otherwise combined with the duodecimal, or base 12, system.
In the course of history, the decimal system finally overshadowed all others. Nevertheless, there are still many vestiges of other systems, chiefly in commercial and domestic units, where change always meets the resistance of tradition. Thus, 12 occurs as the number of inches in a foot, months in a year, ounces in a pound (troy weight or apothecaries’ weight), and twice 12 hours in a day, and both the dozen and the gross measure by twelves. In English the base 20 occurs chiefly in the score (“Four score and seven years ago…”); in French it survives in the word quatre-vingts (“four twenties”), for 80; other traces are found in ancient Celtic, Gaelic, Danish, and Welsh. The base 60 still occurs in measurement of time and angles.
## Numeral systems
It appears that the primitive numerals were |, ||, |||, and so on, as found in Egypt and the Grecian lands, or −, =, ≡, and so on, as found in early records in East Asia, each going as far as the simple needs of people required. As life became more complicated, the need for group numbers became apparent, and it was only a small step from the simple system with names only for one and ten to the further naming of other special numbers. Sometimes this happened in a very unsystematic fashion; for example, the Yukaghirs of Siberia counted, “one, two, three, three and one, five, two threes, two threes and one, two fours, ten with one missing, ten.” Usually, however, a more regular system resulted, and most of these systems can be classified, at least roughly, according to the logical principles underlying them.
## Simple grouping systems
In its pure form a simple grouping system is an assignment of special names to the small numbers, the base b, and its powers b2, b3, and so on, up to a power bk large enough to represent all numbers actually required in use. The intermediate numbers are then formed by addition, each symbol being repeated the required number of times, just as 23 is written XXIII in Roman numerals.
The earliest example of this kind of system is the scheme encountered in hieroglyphs, which the Egyptians used for writing on stone. (Two later Egyptian systems, the hieratic and demotic, which were used for writing on clay or papyrus, will be considered below; they are not simple grouping systems.) The number 258,458 written in hieroglyphics appears in the figure. Numbers of this size actually occur in extant records concerning royal estates and may have been commonplace in the logistics and engineering of the great pyramids.
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Or click Continue to submit anonymously: | 1,839 | 8,262 | {"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} | 3.609375 | 4 | CC-MAIN-2015-22 | latest | en | 0.978562 |
https://www.esaral.com/q/solve-this-26858 | 1,721,736,929,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00215.warc.gz | 653,554,566 | 11,513 | # Solve this
Question:
$x^{2}-2 a x+\left(a^{2}-b^{2}\right)=0$
Solution:
Given:
$x^{2}-2 a x+\left(a^{2}-b^{2}\right)=0$
On comparing it with $\mathrm{A} x^{2}+B x+C=0$, we get:
$A=1, B=-2 a$ and $C=\left(a^{2}-b^{2}\right)$
Discriminant $D$ is given by:
$D=B^{2}-4 A C$
$=(-2 a)^{2}-4 \times 1 \times\left(a^{2}-b^{2}\right)$
$=4 a^{2}-4 a^{2}+4 b^{2}$
$=4 b^{2}>0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 a)+\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a+2 b}{2}=\frac{2(a+b)}{2}=(a+b)$
$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 a)-\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a-2 b}{2}=\frac{2(a-b)}{2}=(a-b)$
Hence, the roots of the equation are $(a+b)$ and $(a-b)$. | 347 | 757 | {"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} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.298998 |
http://www.ask.com/question/calculate-ending-inventory | 1,410,939,493,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657121798.11/warc/CC-MAIN-20140914011201-00309-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 337,525,803 | 15,959 | # How do you calculate ending inventory?
The two methods for calculating ending inventory include the gross profit method and the retail inventory method. The ending inventory is the number of units of inventory that the company on hand at the end of an accounting period. This figure is need for various accounting calculations, including cost of goods sold.
Using the gross profit method, add together the cost of the beginning inventory and the total cost of additional items that were purchased during the period to get the cost of goods available. Subtract the estimated cost of goods sold by the cost of goods available.
With the retail inventory method use the proportion of the retail price to costs in prior periods for a more exact amount.
Reference:
Q&A Related to "How do you calculate ending inventory?"
1. Ending inventory can be determined by plugging the appropriate values into the following formula: Beginning inventory + net purchases - cost of goods sold (CoGS) = ending inventory http://www.ehow.com/how_5953648_calculate-ending-i...
If you are planing on calculating inventory turns it will take some time to understand it. Are you planning on doing inventory? If so this is important. For more information look http://www.ask.com/web-answers/Science/Chemistry/h...
if it is the work in process account(which it sounds like) do this. Beginning inventory.Direct Materials Used.Direct Labor.Manufacturing Overhead.Cost of Goods Manufactured. =Ending http://answers.yahoo.com/question/index?qid=201010...
Beginning Direct Materials. Add: Materials purchased during period. Less: Materials Used during period. Equals: Ending Direct Materials. http://wiki.answers.com/Q/How_do_you_calculate_the... | 341 | 1,722 | {"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-2014-41 | latest | en | 0.900931 |
https://cheraghdaily.org/how-many-scoops-of-coffee-for-8-cups-2022/ | 1,685,851,029,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00038.warc.gz | 192,529,489 | 16,690 | # how many scoops of coffee for 8 cups 2022
If you’re using a coffee scoop, the standard ratio is one scoop of coffee per six ounces of water. So, for eight cups of coffee, you would need eight scoops of coffee. Of course, this is just a general rule of thumb. If you like your coffee weaker, you might only need six scoops.15How Much Coffee For 8 Cups? – The Answer Might Surprise You!ajecafe.com › blog › how-much-coffee-for-8-cupsThông tin về đoạn trích nổi bật
## How many scoops of coffee do you use for 8 cups?
To make eight cups of coffee at average strength, use 72 grams of coffee and 40 ounces (5 measuring cups) of water. That’s about 8 level scoops of coffee or 16 level tablespoons. To make the coffee strong, use 82 grams of coffee (nine scoops or 18 tablespoons).
READ can i drink coffee while taking meloxicam 2022
## How many tablespoons of coffee do I use for 8 cups?
For making 8 cups, we think 14 Tablespoons or ~80 grams of coffee is a good starting point. You may need to use more or less coffee, depending on your preferred coffee strength.
## How many scoops of coffee do you use per cup?
A level coffee scoop is approximately equal to two tablespoons of coffee. Therefore, if you want strong coffee, you will have to use one scoop for each cup. However. If you want a weaker cup of coffee, you will have to use one scoop for every two cups.
## How much coffee grounds do you put in a 12 cup coffee maker?
To fill a standard 12-cup coffeemaker, you will need 12-24 tablespoons (or between 3/4 and 1 1/2 cups) of ground coffee. This will yield 12 6-ounce servings, or about 6 standard 12-ounce mugs of coffee. For a smaller pot, simply scale the ratio down. Since water makes up the majority of coffee, quality matters.
## FAQ about how many scoops of coffee for 8 cups 2022
### How many scoops of coffee do you use per cup?
A level coffee scoop is approximately equal to two tablespoons of coffee. Therefore, if you want strong coffee, you will have to use one scoop for each cup. However. If you want a weaker cup of coffee, you will have to use one scoop for every two cups.How much ground coffee to use per cup?presto-coffee.com › blogs › how-much-ground-coffee-to-use-per-cupAbout Featured Snippets
### How many scoops of coffee do you use for 8 cups?
To make eight cups of coffee at average strength, use 72 grams of coffee and 40 ounces (5 measuring cups) of water. That’s about 8 level scoops of coffee or 16 level tablespoons. To make the coffee strong, use 82 grams of coffee (nine scoops or 18 tablespoons).
### How many tablespoons of coffee do I use for 8 cups?
For making 8 cups, we think 14 Tablespoons or ~80 grams of coffee is a good starting point. You may need to use more or less coffee, depending on your preferred coffee strength.
### How many scoops of coffee do you use per cup?
A level coffee scoop is approximately equal to two tablespoons of coffee. Therefore, if you want strong coffee, you will have to use one scoop for each cup. However. If you want a weaker cup of coffee, you will have to use one scoop for every two cups.
### How much coffee grounds do you put in a 12 cup coffee maker?
To fill a standard 12-cup coffeemaker, you will need 12-24 tablespoons (or between 3/4 and 1 1/2 cups) of ground coffee. This will yield 12 6-ounce servings, or about 6 standard 12-ounce mugs of coffee. For a smaller pot, simply scale the ratio down. Since water makes up the majority of coffee, quality matters.
### How many scoops of coffee do you use for 8 cups?
To make eight cups of coffee at average strength, use 72 grams of coffee and 40 ounces (5 measuring cups) of water. That’s about 8 level scoops of coffee or 16 level tablespoons. To make the coffee strong, use 82 grams of coffee (nine scoops or 18 tablespoons).Coffee Measurements for Every Size of Pot | Bean Poetwww.beanpoet.com › coffee-measurementsAbout Featured Snippets
### How much coffee grounds do you put in a 12 cup coffee maker?
To fill a standard 12-cup coffeemaker, you will need 12-24 tablespoons (or between 3/4 and 1 1/2 cups) of ground coffee. This will yield 12 6-ounce servings, or about 6 standard 12-ounce mugs of coffee. For a smaller pot, simply scale the ratio down. Since water makes up the majority of coffee, quality mattersHow to Brew the Perfect Pot of Coffee – Swift River Coffee Roasterswww.swiftrivercoffee.com › blog › how-to-brew-the-perfect-pot-of-coffeeAbout Featured Snippets
See more articles in the category: Setting | 1,130 | 4,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.93549 |
http://www.t-vec.com/wiki/index.php?title=Simulink/T-VEC_Examples&direction=next&oldid=3172 | 1,537,442,940,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00505.warc.gz | 410,602,094 | 6,754 | (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The most successful users of T-VEC Tester for Simulink (SL2TVEC) follow a few key guidelines:
1. Develop the models with verification in mind - they are aware of those modeling constructs and patterns that result in safe systems
2. Perform modeling and the verification activities supported by SL2TVEC iteratively and continuosly during the design process
3. Understand the tool is performing model checking, and that tests are a byproduct of this process
4. Understand that the Simmulink/Stateflow tool is continously evolving, and the SL2TVEC integration process may take time to catchup to the changes in every new Simulink release, which further emphasizes point #1.
Key Guidelines
This section first discusses some key elements that must be used to support and configure the model translation that results in a T-VEC project for model analysis and test generation.
Signal Ranges
Signal ranges information defines the low bound and high bound values of input and output signals. This information is critical because test vector inputs are selected based on the range of the input signals. In addition, T-VEC analyzes the model using domain analysis starting from the signal range. The default input ranges are based on type definitions. The min (low bound) and max (high bound) values often do not make sense. This especially true for floating-point numbers that have the following ranges:
• 32 bit float range: -3.4E+38 to 3.4E+38
• 64 bit float range: -1.7E+308 to 1.7E+308
Selecting values in very large ranges degrades precision, for examples:
Y < X + 100 when X is 1.7E+308 is not meaningful
Value of 100 is lost in the noise of 1.7E+308
Default ranges for floating point types set to reduce problems
• single (32 bit float) range: -1.0e4..1.0e4
• double (64 bit float) range: -1.0e12..1.0e12
Other types have default ranges
• int8: -128..127 int16: -32768..32767
• unint8: 0..255 uint16: 0..65535
• int32: -2147483648.. 2147483647
• uint32: 0..4294967295
The default type ranges can be modified or specific signal ranges can be defined by the users. These operations can be performed using the Signal Range Editor GUI or they can be done to the underlying file (XML) using an editor.
To create a default, run the SL2TVEC translation once, then use the Signal Range Editor to appropriately set the signal ranges.
Test Sequences
There are two mechanisms to support test sequences through SL2TVEC translations and using the T-VEC VGS:
• Test Sequence Vectors: State Variable Initialization - VGS provides options that provide controls for creating test sequence vectors to cover control and logic that is associated with state variables
• Test Sequence Vectors - SL2TVEC GUI provides a mechanism for specifying test sequences, where test sequence vectors are produced by VGS. This mechanism can support testing of control, but is primarily oriented to dynamic and algorithm analysis.
Assertions
Assertions are general purpose mechanisms that can be used to specify additional constraint that are external to the model. Such constraints can be used for:
• Defining implicit design constraint such as natural laws
• Modeling safety properties | 747 | 3,243 | {"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-2018-39 | latest | en | 0.865672 |
https://www.physicsforums.com/threads/quick-question-about-acceleration.275815/ | 1,542,125,918,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741324.15/warc/CC-MAIN-20181113153141-20181113175141-00252.warc.gz | 960,872,713 | 11,796 | # Homework Help: Quick question about acceleration.
1. Nov 29, 2008
### Oblivion77
1. The problem statement, all variables and given/known data
If I had an acceleration of A=69g's can I just say A=69(9.81) = 676.89m/s^2?
2. Relevant equations
-
3. The attempt at a solution
-
2. Nov 29, 2008
### Ichiro101
Depends on the context. You are essentially stating the same thing just in different words/numbers. In physics, we would prefer the form (m/s^2) however,. | 139 | 471 | {"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} | 3.09375 | 3 | CC-MAIN-2018-47 | latest | en | 0.877984 |
http://cn.metamath.org/mpeuni/pm3.2an3.html | 1,652,893,638,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00646.warc.gz | 16,734,243 | 2,914 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > pm3.2an3 Structured version Visualization version GIF version
Theorem pm3.2an3 1260
Description: Version of pm3.2 462 for a triple conjunction. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by Kyle Wyonch, 24-Apr-2021.)
Assertion
Ref Expression
pm3.2an3 (𝜑 → (𝜓 → (𝜒 → (𝜑𝜓𝜒))))
Proof of Theorem pm3.2an3
StepHypRef Expression
1 df-3an 1056 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
21biimpri 218 . 2 (((𝜑𝜓) ∧ 𝜒) → (𝜑𝜓𝜒))
32exp31 629 1 (𝜑 → (𝜓 → (𝜒 → (𝜑𝜓𝜒))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1054 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 197 df-an 385 df-3an 1056 This theorem is referenced by: 3exp 1283 tratrb 39063 19.21a3con13vVD 39401 tratrbVD 39411
Copyright terms: Public domain W3C validator | 424 | 942 | {"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-2022-21 | latest | en | 0.370705 |
https://www.numbersaplenty.com/20734032 | 1,721,236,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00609.warc.gz | 781,146,273 | 3,879 | Search a number
20734032 = 24311107367
BaseRepresentation
bin100111100011…
…0000001010000
31110000101202010
41033012001100
520301442112
62020222520
7341144034
oct117060120
943011663
1020734032
1110781870
126b3aa40
1343ac577
142a7a1c4
151c4863c
hex13c6050
20734032 has 80 divisors (see below), whose sum is σ = 59139072. Its totient is φ = 6207360.
The previous prime is 20734027. The next prime is 20734039. The reversal of 20734032 is 23043702.
It is a happy number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (20734039) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 56313 + ... + 56679.
Almost surely, 220734032 is an apocalyptic number.
20734032 is a gapful number since it is divisible by the number (22) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 20734032, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (29569536).
20734032 is an abundant number, since it is smaller than the sum of its proper divisors (38405040).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
20734032 is a wasteful number, since it uses less digits than its factorization.
20734032 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 496 (or 490 counting only the distinct ones).
The product of its (nonzero) digits is 1008, while the sum is 21.
The square root of 20734032 is about 4553.4637365417. The cubic root of 20734032 is about 274.7227265889.
Adding to 20734032 its reverse (23043702), we get a palindrome (43777734).
It can be divided in two parts, 2073 and 4032, that added together give a triangular number (6105 = T110).
The spelling of 20734032 in words is "twenty million, seven hundred thirty-four thousand, thirty-two". | 583 | 2,014 | {"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-2024-30 | latest | en | 0.879255 |
https://kwang.life/2015/01/sgu-184-patties/ | 1,675,198,256,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00329.warc.gz | 394,812,985 | 6,676 | # SGU 184 - Patties
## Description
Petya is well-known with his famous cabbage patties. Petya’s birthday will come very soon, and he wants to invite as many guests as possible. But the boy wants everybody to try his specialty of the house. That’s why he needs to know the number of the patties he can cook using the stocked ingredients. Petya has $P$ grams of flour, $M$ milliliters of milk and $C$ grams of cabbage. He has plenty of other ingredients. Petya knows that he needs $K$ grams of flour, $R$ milliliters of milk and $V$ grams of cabbage to cook one patty. Please, help Petya calculate the maximum number of patties he can cook.
## Input
The input file contains integer numbers $P$, $M$, $C$, $K$, $R$ and $V$, separated by spaces and/or line breaks ($1 \leq P, M, C, K, R, V \leq 10000$).
## Output
Output the maximum number of patties Petya can cook.
## Sample Input
1 2 3000 1000 500 30 15 60
## Sample Output
1 8
## Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 #include using namespace std; int min(int x, int y, int z); int main() { int P, M, C, K, R, V; cin >> P >> M >> C; cin >> K >> R >> V; cout << min(P / K, M / R, C / V) << endl; return 0; } int min(int x, int y, int z) { return min(x, min(y, z)); } | 414 | 1,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-06 | latest | en | 0.850252 |
http://www.expertsmind.com/questions/compute-amplitude-and-frequency-of-the-particle-30130609.aspx | 1,656,516,116,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00019.warc.gz | 79,290,099 | 13,651 | ## Compute amplitude and frequency of the particle, Mechanical Engineering
Assignment Help:
Compute amplitude and frequency of the particle:
The displacement of a particle is described by x = 5 + 3 cos 3 t. Determine the position of the particle at t = 3 seconds. Compute amplitude and frequency of the particle.
Solution
x = 5 + 3 cos (3t )
at t = 3 x = 5 + 3 cos 9
= 7.96 cm.
We have
x = 5 + 3 cos 3t
∴ x - 5 = 3 cos 3t
dx /dt=- 9 sin 3t dt
d 2 x/ dt 2 =- 27 cos 3t
= - ω2 x
∴ ω = 3.29 rad / sec
Amplitude shall be calculated as the displacement when t = 0
∴ x = 5 + 3 cos 0o = 8.
∴ Amplitude = 8 cm.
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Vernier Calipers: It is used to measure internal as well as outer diameters of an object. There are separate jaws available for inside and outside diameter measurements. It can al
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Evaluate Young modulus of elasticity - Ultimate stress: Q: The following observations were made during tensile test on mild steel specimen 40 mm in diameter and 200 mm long
#### Tolerance analysis, Tolerance Analysis This discusses the importance o...
Tolerance Analysis This discusses the importance of assigning tolerances to the dimensions. Tolerance analysis is the process of assigning suitable tolerances. Proper assignme | 724 | 3,112 | {"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} | 3.046875 | 3 | CC-MAIN-2022-27 | latest | en | 0.874602 |
https://www.o.vg/unit/kinematic-viscosity/square-meter-second-to-square-foot-second.php | 1,723,296,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00543.warc.gz | 696,511,632 | 9,960 | Convert Square Meter/Second to Square Foot/Second | ft²/s to m²/s
# document.write(document.title);
## m²/s to ft²/s Converter
From m²/s to ft²/s: 1 m²/s = 10.76391042 ft²/s;
From ft²/s to m²/s: 1 ft²/s = 0.092903039971602 m²/s;
## How to Convert Square Meter/Second to Square Foot/Second?
As we know One m²/s is equal to 10.76391042 ft²/s (1 m²/s = 10.76391042 ft²/s).
To convert Square Meter/Second to Square Foot/Second, multiply your m²/s figure by 10.76391042.
Example : convert 25 m²/s to ft²/s:
25 m²/s = 25 × 10.76391042 ft²/s = ft²/s
To convert Square Foot/Second to Square Meter/Second, divide your ft²/s figure by 10.76391042.
Example : convert 25 ft²/s to m²/s:
25 ft²/s = 25 ÷ 10.76391042 m²/s = m²/s
## How to Convert Square Foot/Second to Square Meter/Second?
As we know One ft²/s is equal to 0.092903039971602 m²/s (1 ft²/s = 0.092903039971602 m²/s).
To convert Square Foot/Second to Square Meter/Second, multiply your ft²/s figure by 0.092903039971602.
Example : convert 45 ft²/s to m²/s:
45 ft²/s = 45 × 0.092903039971602 m²/s = m²/s
To convert Square Meter/Second to Square Foot/Second, divide your m²/s figure by 0.092903039971602.
Example : convert 45 m²/s to ft²/s:
45 m²/s = 45 ÷ 0.092903039971602 ft²/s = ft²/s
## Convert Square Meter/Second or Square Foot/Second to Other Kinematic Viscosity Units
Square Meter/Second Conversion Table
m²/s to cm²/s 1 m²/s = 10000 cm²/s m²/s to mm²/s 1 m²/s = 1000000 mm²/s m²/s to µm²/s 1 m²/s = 1000000000000 µm²/s m²/s to m²/min 1 m²/s = 60 m²/min m²/s to m²/h 1 m²/s = 3600 m²/h m²/s to St 1 m²/s = 10000 St m²/s to ESt 1 m²/s = 1.0E-14 ESt m²/s to PSt 1 m²/s = 1.0E-11 PSt m²/s to TSt 1 m²/s = 1.0E-8 TSt m²/s to GSt 1 m²/s = 1.0E-5 GSt m²/s to MSt 1 m²/s = 0.01 MSt m²/s to kSt 1 m²/s = 10 kSt m²/s to hSt 1 m²/s = 100 hSt m²/s to daSt 1 m²/s = 1000 daSt m²/s to dSt 1 m²/s = 100000 dSt
Square Meter/Second Conversion Table
m²/s to cSt 1 m²/s = 1000000 cSt m²/s to mSt 1 m²/s = 10000000 mSt m²/s to µSt 1 m²/s = 10000000000 µSt m²/s to nSt 1 m²/s = 10000000000000 nSt m²/s to pSt 1 m²/s = 1.0E+16 pSt m²/s to fSt 1 m²/s = 1.0E+19 fSt m²/s to aSt 1 m²/s = 1.0E+22 aSt m²/s to ft²/s 1 m²/s = 10.76391042 ft²/s m²/s to ft²/h 1 m²/s = 38750.077512 ft²/h m²/s to ft²/min 1 m²/s = 645.8346252 ft²/min m²/s to in²/s 1 m²/s = 1550.00310048 in²/s m²/s to yd²/s 1 m²/s = 1.1959900466667 yd²/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Square Meter/Second in Square Foot/Second?
ft²/s. Since one m²/s equals 10.76391042 ft²/s, 9 m²/s in ft²/s will be ft²/s.
### How many Square Foot/Second are in a Square Meter/Second?
There are 10.76391042 ft²/s in one m²/s. In turn, one ft²/s is equal to 0.092903039971602 m²/s.
### How many m²/s is equal to 1 ft²/s?
1 ft²/s is approximately equal to 0.092903039971602 m²/s.
### What is the m²/s value of 8 ft²/s?
The Square Meter/Second value of 8 ft²/s is m²/s. (i.e.,) 8 x 0.092903039971602 = m²/s.
### m²/s to ft²/s converter in batch
Cite this Converter, Content or Page as:
"" at https://www.o.vg/unit/kinematic-viscosity/square-meter-second-to-square-foot-second.php from www.o.vg Inc,08/10/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters | 1,296 | 3,210 | {"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-2024-33 | latest | en | 0.744328 |
http://www.math.utexas.edu/pipermail/maxima/2010/020882.html | 1,369,129,505,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699856050/warc/CC-MAIN-20130516102416-00028-ip-10-60-113-184.ec2.internal.warc.gz | 587,799,578 | 2,156 | # [Maxima] simple block for romberg method
Richard Fateman fateman at EECS.Berkeley.EDU
Sat Apr 3 09:21:19 CDT 2010
```you should use apply(f, ...) or else the function f defined globally will be used instead of the f, the parameter to trap.
I'm not sure what your objective is here, to show that the built-in romberg function can be rewritten as a slower version
There is also built-in bigfloat version of romberg, and many other options for numerical quadrature...
RJf
----- Original Message -----
From: Luigi Marino <luigi_marino2 at alice.it>
Date: Saturday, April 3, 2010 9:53 am
Subject: [Maxima] simple block for romberg method
To: maxima at math.utexas.edu
> There is a simple block
> for romberg method:
>
> trap(n,a,b,f):=block(
> h:(b-a)/n,
> s:0,
> for i:1 thru n-1 step 1 do
> (s:s+f(a+i*h)),
> AV:(h/2)*(f(a)+2*s+f(b)),
> return (float(AV)))\$
>
> romb(max_iter,a,b,f):=block(
> integ[1,1]:trap(1,a,b,f),
> for i:1 thru max_iter do
> (n:2^i,
> integ[i+1,1]:trap(n,a,b,f),
> for j:2 thru i+1 do
> (k:2+i-j,
> integ[k,j]:(4^(j-1)*integ[k+1,j-1]-integ[k,j-1])/(4^(j-1)-1))),
> integ_value:integ[1,max_iter],
> return(integ_value))\$
>
> Example:
>
> f(x):=cos(x);
>
> romb(10,0,10,f);
> -0.54402111088937
>
> f(x):=sqrt(1+x^3);
>
> romb(12,-1,1,f);
> 1.952753609576008
>
> Best wishes
> Luigi Marino
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
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``` | 566 | 1,681 | {"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-2013-20 | latest | en | 0.480569 |
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