url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://blog.cluster-text.com/tag/acceptance-test/ | 1,669,972,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00748.warc.gz | 173,433,658 | 14,894 | Substantial Reduction in Review Effort Required to Demonstrate Adequate Recall
Measuring the recall achieved to within +/- 5% to demonstrate that a production is defensible can require reviewing a substantial number of random documents. For a case of modest size, the amount of review required to measure recall can be larger than the amount of review required to actually find the responsive documents with predictive coding. This article describes a new method requiring much less document review to demonstrate that adequate recall has been achieved. This is a brief overview of a more detailed paper I’ll be presenting at the DESI VII Workshop on June 12th (slides available here).
The proportion of a population having some property can be estimated to within +/- 5% by measuring the proportion on a random sample of 400 documents (you’ll also see the number 385 being used, but using 400 will make it easier to follow the examples). To measure recall we need to know what proportion of responsive documents are produced, so we need a sample of 400 random responsive documents. Since we don’t know which documents in the population are responsive, we have to select documents randomly and review them until 400 responsive ones are found. If prevalence is 10% (10% of the population is responsive), that means reviewing roughly 4,000 documents to find 400 that are relevant so that recall can be estimated. If prevalence is 1%, it means reviewing roughly 40,000 random documents to measure recall. This can be quite a burden.
Once recall is measured, a decision must be made about whether it is high enough. Suppose you decide that if at least 300 of the 400 random responsive documents were produced (75%) the production is acceptable. For any actual level of recall, the probability of accepting the production can be computed (see figure to right). The probability of accepting a production where the actual recall is less than 70% will be very low, and the probability of rejecting a production where the actual recall is greater than 80% will also be low — this comes from the fact that a sample of 400 responsive documents is sufficient to measure recall to within +/- 5%.
The idea behind the new method is to achieve the same probability profile for accepting/rejecting a production using a multi-stage acceptance test. The multi-stage test gives the possibility of stopping the process and declaring the production accepted/rejected long before reviewing 400 random responsive documents. The procedure is shown in the flowchart to the right (click to enlarge). A decision may be reached after reviewing enough documents to find just 25 random documents that are responsive. If a decision isn’t made after reviewing 25 responsive documents, review continues until 50 responsive documents are found and another test is applied. At worst, documents will be reviewed until 400 responsive documents are found (the same as the traditional direct recall estimation method).
The figure to the right shows six examples of the multi-stage acceptance test being applied when the actual recall is 85%. Since 85% is well above the 80% upper bound of the 75% +/- 5% range, we expect this production to virtually always be accepted. The figure shows that acceptance can occur long before reviewing a full 400 random responsive documents. The number of random responsive documents reviewed is shown on the vertical axis. Toward the bottom of the graph the sample is very small and the percentage of the sample that has been produced may deviate greatly from the right answer of 85%. As you go up the sample gets larger and the proportion of the sample that is produced is expected to get closer to 85%. When a green decision boundary is touched, causing the production to be accepted as having sufficiently high recall, the color of the remainder of the path is changed to yellow — the yellow part represents the document review that is avoided by using the multi-stage acceptance method (since the traditional direct recall measurement would involve going all the way to 400 responsive documents). As you can see, when the actual recall is 85% the number of random responsive documents that must be reviewed is often 50 or 100, not 400.
The figure to the right shows the average number of documents that must be reviewed using the multi-stage acceptance procedure from the earlier flowchart. The amount of review required can be much less than 400 random responsive documents. In fact, the further above/below the 75% target (called the “splitting recall” in the paper) the actual recall is, the less document review is required (on average) to come to a conclusion about whether the production’s recall is high enough. This creates an incentive for the producing party to aim for recall that is well above the minimum acceptable level since it will be rewarded with a reduced amount of document review to confirm the result is adequate.
It is important to note that the multi-stage procedure provides an accept/reject result, not a recall estimate. If you follow the procedure until an accept/reject boundary is hit and then use the proportion of the sample that was produced as a recall estimate, that estimate will be biased (the use of “unbiased” in the paper title refers to the sampling being done on the full population, not on a subset [such as the discard set] that would cause a bias due to inconsistency in review of different subsets).
You may want to use a splitting recall other than 75% for the accept/reject decision — the full paper provides tables of values necessary for doing that. | 1,134 | 5,633 | {"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-2022-49 | longest | en | 0.908479 |
http://mathforum.org/mathtools/cell.html?co=a&sortBy=C!desc&offset=0&limit=25&resort=1 | 1,500,656,716,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423787.24/warc/CC-MAIN-20170721162430-20170721182430-00086.warc.gz | 215,654,564 | 8,561 | You are not logged in.
Resource Name Topic (Course) Technology Type $? Rating Browsing: All Content in Algebra Browse discussions Login to Subscribe / Save Results Exploring Algebra 1 with The G... General (Algebra) Sketchpad Activity$ [1] This collection of 56 activities compiled specifically for Algebra 1—including updated activities from Exploring Algebra with The Geometer's Sketchpad and many new ones—covers topics such as: TI InterActive! Data Collectio... General (Algebra) TI Calculator Activity $[0] TI InterActive! is a new product that enables high school and college teachers and students to easily investigate ideas in mathematics and science. The purpose of this workbook is to introduce algebra... More: lessons, discussions, ratings, reviews,... Algebra Nspirations: Linear Fu... Linear Equations (Algebra)+ TI-Nspire Activity$ [0] A video that focuses on the TI-Nspire graphing calculator in the context of teaching algebra. In this program the TI-Nspire is used to explore the nature of linear functions. Examples ranging from ... More: lessons, discussions, ratings, reviews,... Apples and Oranges 1 Decimal Linear Equations (Algebra)+ Android Tool $[0] This application is a basic tool for learning algebra. Users try to solve a system of linear equations with two equations and two unknown variables. It is based on the same problem used in the free ap... More: lessons, discussions, ratings, reviews,... Minutes Text Notes Pro General (Algebra) Android Tool$ [0] A note taking Android app that syncs to Dropbox and can render Markdown along with mathematical equations (MathJax support requires Android 4.4). It includes dictionary support, Tasker extension supp... More: lessons, discussions, ratings, reviews,... MathPoint General (Algebra) Calculator Tool $[0] MathPoint is a suite of math tools for students in grades 6 through 12 and college including color graphing, graphing calculator and interactive solving, and an open library for lessons and activit... More: lessons, discussions, ratings, reviews,... Abacus MathWriter General (Algebra) Computer Tool$ [0] MathWriter is a stand-alone Windows program that allows for the production of mathematical, scientific and engineering equations and formulae. More: lessons, discussions, ratings, reviews,... Dimenxian Linear Equations (Algebra)+ Computer Tool $[4] Students experience, explore, and apply math concepts in an engaging three-dimensional virtual world. The topics covered in Dimenxian include: coordinate system and scatter p... More: lessons, discussions, ratings, reviews,... Green Globs and Graphing Equat... Linear Equations (Algebra)+ Computer Tool$ [1] Students explore the relationship between equations and their graphs in this hands-on learning environment where they investigate, manipulate, and understand linear, quadratic and other graphs. They ... More: lessons, discussions, ratings, reviews,... Infinite Algebra 1 General (Algebra) Computer Tool $[1] Infinite Algebra 1 is a test and worksheet generator for algebra teachers. Example problems, assignments, and assessments are designed by the teacher by selecting the topics to include and the exac... More: lessons, discussions, ratings, reviews,... MATLAB General (Algebra) Computer Tool$ [0] MATLAB® is a high-level technical computing language and interactive environment for algorithm development, data visualization, data analysis, and numerical computation. Using MATLAB, you can solve te... More: lessons, discussions, ratings, reviews,... Matamicus General (Algebra) Computer Tool $[0] A user may enter math problems into the program and the output is a step-by-step solution. It's used primarily for solving expressions, relations, factoring, systems of relations, and other step-by-s... More: lessons, discussions, ratings, reviews,... Math911 Properties of Reals (Algebra)+ Computer Tool$ [4] Tutorial fee-based software for PCs that must be downloaded to the user's computer. It covers topics from pre-algebra through pre-calculus, including trigonometry and some statistics. The software pos... More: lessons, discussions, ratings, reviews,... Mathpad General (Algebra) Computer Tool $[0] Mathpad is a stand alone math text editor. For math teachers, it can be used to create math quizzes, tests, and handouts. Also you can save any math expression or text as an image for inclusion i... More: lessons, discussions, ratings, reviews,... Maple General (Algebra) Computer Algebra System Tool$ [0] Maple is an environment for scientific and engineering problem-solving, mathematical exploration, data visualization and technical authoring. It provides both numeric and symbolic capabilities. More: lessons, discussions, ratings, reviews,... Math Foundation General (Algebra) Flash Tool $[0] Math Foundation's interactive online math course includes animated lessons with step-by-step examples covering basic math and pre-algebra with introductions to algebra and geometry. Features include s... More: lessons, discussions, ratings, reviews,... Mathematica General (Algebra) Mathematica Tool$ [0] Comprehensive symbolic and numerical software. Advanced programming language, graphics and sound, programmable interface, notebook documents, and applications library. More: lessons, discussions, ratings, reviews,... Cram 1.0 General (Algebra) PDA Tool $[0] Cram is test preparation software to use on a mobile device. It allows you to create, import, share, and study for tests. Cram is suited for studying for job training, certifications, homework help, t... More: lessons, discussions, ratings, reviews,... ImagiGraph Functions / Relations (Algebra)+ Palm OS Tool$ [1] One of three software applications included in the ImagiMath suite that turns a Palm OS handheld computer into a mathematics learning environment. The suite of three applications can be purchased for ... More: lessons, discussions, ratings, reviews,... ImagiSolve General (Algebra) Palm OS Tool $[0] One of three software applications included in the ImagiMath suite that turns a Palm OS handheld computer into a mathematics learning environment. The suite of three applications can be purchased for ... More: lessons, discussions, ratings, reviews,... Math Trick Trainer General (Algebra) Palm OS Tool$ [3] MathTrick Trainer is a math/mental training program for all age groups. It is a flash card with unlimited randomly generated math quizes, correct answer is provided as you advances through the quiz... More: lessons, discussions, ratings, reviews,... powerOneTM Graph v4 General (Algebra) Palm OS Tool $[1] powerOne(r) Graph supports computation & graphing in the mathematics classroom. The following tasks can be done with this software on a handheld computer including: Data plotting More: lessons, discussions, ratings, reviews,... Linear Inequalities in Two Va... Graphing (Algebra) Shockwave Tool$ [0] This site allows one access per day per computer free of charge. This applet allows students to graph up to six different linear inequalities in six different colors with solid or broken lines. The s... More: lessons, discussions, ratings, reviews,... Dividing Exponential Expressio... Basic Rules (Algebra) Shockwave Tool $[0] This is a pay for use site. You are permitted one free access per computer per day. In this activity, students divide exponential expressions by dragging and dropping the next step in the process into... More: lessons, discussions, ratings, reviews,... Exponents and Power Rules Gizm... Basic Rules (Algebra) Shockwave Tool$ [0] This is a pay for use site, but you may access it once a day per computer for free. In this exercise, students are required to raise a product to a power by dragging and dropping the correct next ste... More: lessons, discussions, ratings, reviews,... Page 1 2 3 4 5 6 ... 10 ... 20 next » | 1,645 | 7,828 | {"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} | 2.53125 | 3 | CC-MAIN-2017-30 | longest | en | 0.832191 |
https://e-learnteach.com/calculate-percentage-in-c/ | 1,685,278,364,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00059.warc.gz | 267,044,732 | 13,792 | # Calculate percentage in c
Percentage = ( sum * 100 ) / tmp printf(“nPercentage Of Student : %dn”, Percentage) here tmp. Logic to find total, average and percentage · Input marks of five subjects. · Calculate sum of all subjects and store in total = eng + phy + chem. 4.Calculate percentage using percentage = (average / 500) * 100. 5.Finally, print resultant values total, average and percentage.Algorithm to calculate Average and Percentage · First Received marks of each subject. · Calculate sum of all subjects marks. · Divide total marks by number of. You divide by 0. num is the variable you use for input. For unknown reason you divide with it in the end. You have a counter for how many.
View this answer now! It’s completely free.
## how to calculate percentage in c++
You divide by 0. num is the variable you use for input. For unknown reason you divide with it in the end. You have a counter for how many. 35 : C Program to Calculate total marks and Percentage · Variables : We have six integer variables for subjects English, French, Algebra, Geometry, History and. Calculate percentage using percentage = (total / 500) * 100. Finally, print resultant values total , average and percentage. C Program to. To find percentage, divide the total scores with the marks obtained and then multiply the result with 100. Formula to Find Percentage of Marks: Percentage of. C program to calculate total average and percentage of five subjects · 1.Input marks of five subjects. · 2.Calculate sum of all subjects and store in total = eng.
## c program to calculate sum of 5 subjects and find percentage and grade
Divide sum of all subjects by total number of subject to find average i.e. average = total / 5. Calculate percentage using percentage = (total. Example. Here we will create a program in which user enter the marks of five subjects and calculate percentage. And you will learn how to find percentage of. OUTPUT: Enter Marks of 5 Subjects: 80 70 85 83 90 Sum of 5 Subjects is: 408 Percentage is: 81.000000. Author: sk.shahida. C Program to calculate sum of 5 subjects and find percentage · main() { · int s1, s2, s3, s4, s5, sum, total = 500 · float per · printf(“nEnter. Write a Program in c to calculate sum of 5 subjects and find percentage float avg printf(” Enter the marks of five subjects : “) scanf(“%d %d %d %d %d ” sum. | 558 | 2,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-23 | latest | en | 0.830595 |
http://www.studymode.com/essays/Ski-Jacket-Production-724220.html | 1,529,763,156,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00208.warc.gz | 498,820,719 | 26,054 | # Ski Jacket Production
Pages: 5 (1420 words) Published: June 21, 2011
Ski Jacket Production
Executive Summary
The problem is to determine the optimal production level of the Egress new designed jacket given the uncertainty in the forecasted demand. As oppose to determining a single profit value in the deterministic approach, the probabilistic method will incorporate the uncertainty in estimated demand and provide insights of the range of profit outcomes and its associated risk (deviation from mean). The key issue is to understand impact of demand uncertainty and production level to the profit range and its distribution. In this case, we will determine the optimal decision variables (jacket production quantity) that will maximize our objective (average profit).
The given information from management is the variable production cost, the selling price, and the salvage value per unit. There is also the fixed production cost regardless of production level. Production can all be sold if it is not greater than the demand level. However, if the production quantity is greater than the demand, it will be sold to discounters at the salvage value price.
When dealing with uncertainty, the simulation model is useful to explicitly incorporate uncertainty into the input variables. This random input variable and the resulting output variables of interest are keep recorded. Hence, we are able to analyze how the outputs vary as of function of the varying inputs. In this case, the demand distribution was assumed normally distributed with the mean and standard deviation from 12 discrete forecasts from Egress employees. The optimal production level of Egress’ new designed is around 10,174 jackets with the average profit of \$61,849 and standard deviation of \$94,136.
Model
* Decision:Production quantity of ski jacket
* Objective:Maximize the profit
* Assumption:
* The decision variable (production quantity)) are non-negative * Variable production cost = \$80/jacket
* Selling price = \$100/jacket
* Salvage value = \$50/jacket
* Fixed production cost = \$100,000
* Demand is non-negative and normally distributed.
* Excel’s Add-in: Data Analysis (histogram and descriptive statistics) are used to obtain and analyze statistical data and uncertainty. * RAND function is used to randomly determine the probability of the demand distribution. RAND will generate values between 0 – 1. Number of samples is set at 1,000 to well represent the random range. * NORMINV function is used to convert the random probability to the normal cumulative distribution for the specified mean and standard deviation.
Structure
Demand is the normal inverse of the random probability given the mean and standard deviation Normal sales quantity = minimum btween demand and production quantity Salvage sales quantity = extra quantity when demand > production Revenue from normal sales = Selling Price/unit * Normal sales quantity Revenue from salvage value = Salvage Price/unit * Salvage sales quantity Variable costs = Variable Cost/unit * Production quantity
Profit = Revenue from normal sales + Revenue from salvage - Variable costs – Fixed cost Average profit = Average profit from the demand distribution at each production level
Analysis
* Inputs are all numerical data in the problem
* The changing cells (decision variables) in blue box are the jacket production level. * The target cell (objective) in green box is average profit to be maximized * No constraint
Conclusion
Answer (1): If the normal distribution is a reasonable model for the demand of their ski jacket design in the upcoming season, based on the estimated demand forecast from 12 Egress employees, the mean of the demand distribution would be 12,000 jackets and standard deviation of 3,497. Figure 1 shows the Demand Distribution and Statistical Summary. This demand distribution will be used in calculating the profit at each production level. Figure 1: Demand Distribution and... | 764 | 3,954 | {"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-2018-26 | latest | en | 0.890755 |
https://www.boatdesign.net/threads/some-questions-on-polars-and-upwind-downwind-performance.66050/ | 1,713,792,979,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00795.warc.gz | 591,712,002 | 17,565 | # Some questions on polars and upwind/downwind performance
Discussion in 'Sailboats' started by Tedd McHenry, Sep 7, 2021.
1. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
I made the polar below using ORC Sailboat Data. I picked the Nordic Folkboat only because it's a boat I like and is of the size and configuration that most interests me. (And the data was readily available.) I've marked the points of maximum upwind and downwind VMG, by eye. I have two questions about this (below) and I'm also interested in other thoughts from people with experience using polars.
1. I've fitted a radial line through the upwind VMGmax points, by regression. They all lie close to 53 degrees. What strikes me about this is how far maximum VMG is from the minimum beat angle--nearly 10 degrees in some cases. Is that typical for a boat like this, or is my methodology somehow exaggerating the difference?
2. I'm also intrigued by the downwind VMGmax points. For 10 kt and below, they lie well away from a dead run. It's not obvious to me why that would be, but one explanation that occurred to me is that a spinnaker might have been used for the 10-kt and below data points. I would have thought that an asymmetrical spinnaker, especially, would tend to give VMGmax at some angle away from a dead run. Does that make sense, or is there a better explanation?
I originally attempted to do this analysis by fitting various nth order polynomials through the data. That allows the VMGmax points to be accurately determined by finding the zero points on the derivative of the polynomial. But I found that it wasn't possible to find polynomials that gave a good fit. Lower order polynomials rounded off the curves too much in the beating area while higher order polynomials introduced wavy bits in the downwind and broad reach areas. So I settled on spline curves so that I at least know the curve matches the data at the actual data points. Unfortunately, I've yet to work out how to find the equation used at each point in the spline, so I have to eyeball the VMGmax points instead of using calculus. I've got a reference for working out the actual spline equations but I haven't digested it, yet.
2. Joined: Aug 2017
Posts: 1,456
Likes: 418, Points: 83
### BlueknarrSenior Member
1. Very typical results.
2. Your explanation is probable. Could also be result of further reduction in sail area from reefing the main. Surfing dead down wave could also change vector of vmg
3. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
Thanks for your thoughts, @Blueknarr . I hadn't even thought about waves and following seas but I agree, that could be quite significant.
I still remember the first time I found myself going downwind at above hull speed in a keel boat. At that time I'd never even heard of "surfing" in a sailboat.
4. Joined: Aug 2017
Posts: 1,523
Likes: 667, Points: 113
Location: France
### DolfimanSenior Member
Your results upwind are intriguing, not usual I would say, the usual optimum VMG with a modern sailboat is obtained with twa from ~ 44 deg with 6 Knots of wind to ~39 deg with 20 Knots. Example of the Sydney 38 :
Downwind, it also depends of the capacity to go fast when closer to the wind, same example : twa ~ 140 deg with 6 Knots to ~ 170 deg with 20 Knots (0r less, see also the targets which assume to go planing ?) .
5. Joined: Jul 2021
Posts: 56
Likes: 8, Points: 8
Location: 31 42S 152 04 E
### AJBJunior Member
Tedd,
You need to get out more!
But seriously...
Upwind
Your regression around TWA is a source of confusion, not enlightenment. A useful approach is to calculate the AWA for the optimum VMG numbers.
So, for example in 10 knots TWS, BS is 5.0 and AWA is ~ 35 degrees, at 52 degrees TWA from the graph.
This AWA kind of summarises the lift/drag characteristics of the whole apparatus; and what you will probably find is that the AWA for optimum VMG across the range 6 - 20 TWS does not vary by more than about 2 degrees.
This AWA tells us that the foil and underwater surfaces are not very efficient; from a modern standpoint, the wetted surface of the Folkboat is very high. For reference, a 'modern' upwind AWA would be around 25 degrees. Ballpark estimate for your wetted surface drag is around 2/3 of the total hydro drag.
Downwind VMG
These are flat water polars, so no surfing is involved. The ORC downwind aero coefficients are a fair way away from reality, particularly for the VMG run part of the spectrum.
But two major points:
1. Because attached flow generates aero force that is a multiple of stalled flow, some degree of gybing downwind will always be better than DDW, i.e. sailing with TWA at 180, where the proportion of unstalled flow over the sails becomes very low. So for most boats (with spinnakers, asymmetric or symmetric) the optimum AWA for VMG running in less than 8 knots TWS is around and even less than 90 degrees. The AWS might be more than double that of the boat trying to sail DDW. Aero force is proportional to V^2.... so the aero force might be 4 times as large...
2. Because the hydro drag of the relatively heavy design rises rapidly as the speed exceeds about 5 knots, the downwind gybing angles in good breeze (say 14 knots and up) will be quite small. But there is plenty of evidence to suggest that around 165 degrees TWA is a good rule of thumb maximum ; i.e. minimum gybe angle of 30 degrees.
For some larger modern designs with large masthead spinnakers, the ORC VPP continues to predict zero gybe angles for VMG run, when if fact, the aero force of the spinnaker in that situation will lead to sufficient windward heel and force misalignment as to propel the yacht into an involuntary gybe! The more accurate designer VPPs have the maximum TWAs at around 165 and AWAs not greater than about 155, for moderately heavy designs.
6. Joined: Aug 2002
Posts: 16,809
Likes: 1,721, Points: 123, Legacy Rep: 2031
Location: Milwaukee, WI
### gonzoSenior Member
Downwind, the boats speed gets subtracted from the windspeed. As the boat accelerates, the apparent wind decelerates slowing the boat until it reaches a balance. At lesser angles, the boat speed adds to the windspeed, so the apparent wind is higher than the actual windspeed. As the boat reaches "hull speed", the power needed to go faster increases exponentially, so the difference in apparent wind does not make as much of a difference.
7. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
Thanks for the Sydney 38 data, @Dolfiman . I plugged it into my spreadsheet so I could see all the curves. It certainly looks more like what I expected to see, with max. VMG upwind at the minimum beat angle and max. VMG downwind at 180 degrees (for most wind speeds).
8. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
I do. But I'm not currently working and I'm between boats, so what the heck else am I going to do?
That is a great idea, I will do that. I started out only interested in where maximum VMG would lie relative to TWA, but I think you're right that considering AWA will provide more insight. I'm also going to apply the same analysis to a few different boats. Since I can't currently sail anything that will have to substitute for experience!
Are the ORC numbers not actual measurements? I assumed that they were.
I'm going to have to think about your last two points for a while. But thanks very much for the input!
9. Joined: Jul 2021
Posts: 56
Likes: 8, Points: 8
Location: 31 42S 152 04 E
### AJBJunior Member
All good Tedd
ORC is just reheated IMS, not really very sophisticated - just a bunch of spreadsheets, no actual performance feedback.
Which is why, with good resistance, flow and dynamic performance models the leading designers can game it very significantly.
10. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
Hi @AJB ,
Is this the kind of thing you expected to see? This is from the Sydney 38 data that @Dolfiman linked to. It seemed to be good quality data so I used it. The polar below is based on AWA, and the scalar quantity is the boat-speed/wind-speed ratio. As you predicted, the maximum point for each wind speed is found in a pretty narrow range of angles.
View attachment 171421
[Attachment updated with better version. This version uses 4th order polynomial fit to the original data, rather than a spline curve. I think that reflects reality more closely and also allows the maximum speed ratio points to be calculated from the polynomial.]
#### Attached Files:
• ###### Sydney_38_AWA_Polar.png
File size:
45 KB
Views:
238
Last edited: Sep 11, 2021
11. Joined: Jul 2021
Posts: 56
Likes: 8, Points: 8
Location: 31 42S 152 04 E
### AJBJunior Member
Tedd,
Nice!
If only the real world was quite so smooth!
Now you should consider the VMG vs VMC speed downwind for offshore stuff...
You can find the Farr 40 polars on the class website; and they are well closer to reality than ORC or the Sydney 38 numbers.
12. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
Thanks again, @AJB . Now that I have the spreadsheet as a template I can easily plug in the Farr 40 data. And I'll have a look at downwind, as well.
Yes, fitting polynomials creates unrealistically smooth data, but I think it's fine for conceptual purposes. I'm not trying to learn anything about any particular boat, but more just get a sense of how all the parameters work together.
13. Joined: Jun 2013
Posts: 282
Likes: 100, Points: 43, Legacy Rep: 10
Location: Cambridge, UK
### tlouth7Senior Member
Polars should always be created with respect to TWS/TWA not AWA. Otherwise if you aren't sailing to the target speed you can't use the polar at all.
Folkboats have long keels so a wider optimum close-hauled angle than modern boats doesn't surprise me much. The downwind angles don't surprise me either - at low windspeed the boat is sailing an appreciable proportion of the windspeed, so AWS is very low and there is a comparatively big increase in AWS from heading up a few degrees. For higher windspeed the subtraction of STW has less effect, so the advantage of heading up is reduced. It is a general rule that whether under white sails or spinnaker, boats benefit more from sailing angles (and those angles are wider) in light than strong winds.
14. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
Thanks for your comment. The purpose of looking at the effect of AWA, as @AJB mentioned above, is to gain an understanding of how the L/D of the entire system (sails and hull) perform under different conditions. It's not so much something a sailor would use, as you noted. But it is of interest from the perspective of boat design.
15. Joined: Feb 2020
Posts: 54
Likes: 8, Points: 8
### Tedd McHenryJunior Member
@AJB ,
Here's what I got from the Farr 40 data.
• "SRmax" is the point where the ratio of boat speed to wind speed is maximim, for the given wind speed.
• "VMGmax" is the point where VMG is maximum, for the given wind speed.
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different. | 2,868 | 11,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.962098 |
https://www.physicsforums.com/threads/suvat-question-seemingly-too-little-info.139375/ | 1,723,000,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00854.warc.gz | 727,799,534 | 18,740 | # Suvat Question - seemingly too little info
• TicTac2
In summary, the conversation is about a question regarding throwing a ball vertically into the air and finding its initial velocity, maximum height, and speed at impact. The correct formula to use is \Delta x = v_{i}t + \frac{1}{2}at^{2} and it is important to consider the sign of displacement, which in this case is negative as the ball finishes lower than its initial position.
TicTac2
Hi all,
I am new here with probably what is a very simple question for most. I hope I am posting this in the right place. The question is:
A ball is thrown vertically into the air, and leaves the thrower's hand when it is 1.6m above the ground. It hits the ground 3.1s later.
Assume acceleration as 9.8, and assume no air resistance.
You have to find out everything really - initial velociy, maximum hieght, and speed at impact.
It seems quite simple but I can't get anyway.
I tried assuming displacement as 1.6m and a as 9.8, and did 1.6m=3.1t+4.9t^2. I got the answer 14.7m/s but you can't do that can you as a changes from + to -. Our teacher reckons the answer is closer to 12 but isn't really sure.
I also tried some simultaneous questions but ended up with stuff like 4.9t^2=4.9t^2
So what is the correct method?
Thanks very much
Welcome to the Forums,
Your almost there with your first method, note that the ball finishes lower than it's initial position. What does this tell you about the sign of the displacement? Also note that the correct formula is;
$$\Delta x = v_{i}t + \frac{1}{2}at^{2}$$
It may have been a typo in your above post but your equation seemed incorrect.
Hootenanny said:
Welcome to the Forums,
Your almost there with your first method, note that the ball finishes lower than it's initial position. What does this tell you about the sign of the displacement? Also note that the correct formula is;
$$\Delta x = v_{i}t + \frac{1}{2}at^{2}$$
It may have been a typo in your above post but your equation seemed incorrect.
I think I did have the equation correct I just shortened 1/2xa to 4.9.
The negative displacement is the mistake I'm making.
Thanks very much for the info!
TicTac2 said:
I tried assuming displacement as 1.6m and a as 9.8, and did 1.6m=3.1t+4.9t^2.
This should be vit or 3.1vi as 3.1 is the time t. A minor typing error though as it looks like you worked it out correctly.
My pleasure
## What is a "Suvat Question"?
A "Suvat Question" is a type of physics problem that involves using the Suvat equations to solve for the unknown variables of motion (displacement, initial velocity, final velocity, acceleration, and time).
## Why does a "Suvat Question" seem to have too little information?
Suvat questions often seem to have too little information because they typically only provide a few of the five variables needed to solve the problem. However, with the use of the Suvat equations and some algebraic manipulation, the missing variables can be calculated.
## What are the Suvat equations?
The Suvat equations are five equations that relate the five variables of motion (displacement, initial velocity, final velocity, acceleration, and time). They are:
• s = ut + 1/2at^2
• v = u + at
• v^2 = u^2 + 2as
• s = (u + v)/2 * t
• u = (v + u)/2 * t
## How do I solve a "Suvat Question"?
To solve a "Suvat Question", you must first identify which variables are given and which are unknown. Then, you can use the appropriate Suvat equation(s) to manipulate and solve for the unknown variable(s). It is important to remember to use consistent units and pay attention to the direction of motion.
## Are there any tips for solving "Suvat Questions"?
Yes, some tips for solving "Suvat Questions" include:
• Draw a diagram to visualize the problem and label the known and unknown variables.
• Use the most appropriate Suvat equation(s) based on the given information.
• Be consistent with units and pay attention to the direction of motion.
• Check your answer by plugging it back into the original equation or using a different equation.
• Practice and familiarize yourself with the Suvat equations and their applications.
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
18
Views
377
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
7K | 1,217 | 4,702 | {"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-2024-33 | latest | en | 0.965383 |
http://www.jiskha.com/display.cgi?id=1364152201 | 1,493,543,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124478.77/warc/CC-MAIN-20170423031204-00592-ip-10-145-167-34.ec2.internal.warc.gz | 567,848,027 | 4,411 | # College Algerbra
posted by on .
THE TEMPERATURE ON A SUMMER DAY IS // DEGREES FAHRENHEIT.the formula to convert the temperature to degrees Celsius is C = 5/9(F-32) what is the corresponding temperature in degrees Celsius?
• College Algerbra - ,
C = 5/9(F-32)
C = (5/9) (11 - 32)
C = 5/9 (-21)
C = -105/9
C = 11.67
• College Algerbra - ,
THE TEMPERATURE ON A SUMMER DAY IS 77 DEGREES FAHRENHEIT.the formula to convert the temperature to degrees Celsius is C = 5/9(F-32) what is the corresponding temperature in degrees Celsius?
Sorry Ms. Sue I cut of the top of the screenshot, it is 77 degrees Fahrenheit.
• College Algerbra - ,
Use the same method I used, except substitute 77 for 11.
• College Algerbra - ,
So I did what you said but I am not sure how you got to C= 5/9(-21)
to C=105/9
So on mine C=(5/9)(77-32)
I got C=5/9(-45)
I am unsure how you did the math to come to your C=-105/9 answer and then the C=11.67 answer. Please show me those last steps if you do not mind, Thanks for all the help.
• College Algerbra - ,
C = 5/9 (-21) I subtracted 32 from 11 (the degrees F).
C = -105/9 I multiplied 5/9 times -21
C = -11.67 I divided the numerator by the denominator
When the temperature is 77 degrees F, it is 25 degrees C.
Work with the formula until you get 25 degrees C.
• College Algerbra - ,
Thank You so much for your invaluable help and patience with walking me through these problems Ms. Sue, I hope you have a great day!
• College Algerbra - ,
You're very welcome, Eric.
• College Algerbra - ,
The temperature on a summer day is 77 degrees fahreheit. The formula to convert temperature to degrees Celsuis is C =5/9(F-32). What is the corresponding in degrees Celsius? | 492 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-17 | latest | en | 0.831144 |
https://winnerscience.com/einstein-coefficient-relations/ | 1,695,418,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00583.warc.gz | 705,513,905 | 20,380 | # Einstein Coefficient Relation
Einstein Coefficient Relation derivation and discussion:
Einstein showed the interaction of radiation with the matter with the help of three processes called stimulated absorption, spontaneous emission, and stimulated emission. He showed in 1917 that for a proper description of radiation with matter, the process of stimulated emission is essential. Let us first derive the Einstein coefficient relation on the basis of the above theory:
Let N1 be the number of atoms per unit volume in the ground state E1 and these atoms exist in the radiation field of photons of energy E2-E1 =h v such that the energy density of the field is E.
## Einstein Coefficient for Stimulated Absorption:
Let R1 be the rate of absorption of light by E1 -> E2 transitions by the process called stimulated absorption (Refer below figure):
This rate of absorption R1 is proportional to the number of atoms N1 per unit volume in the ground state and proportional to the energy density E of radiations.
That is R1∞ N1 E
Or R1 = B12N1 E (1)
Where B12 is known as the Einstein’s coefficient of stimulated absorption and it represents the probability of absorption of radiation. Energy density e is defined as the incident energy on an atom as per unit volume in a state.
## Einstein Coefficient for Spontaneous Emission:
Now atoms in the higher energy level E2 can fall to the ground state E1 automatically after 10-8 sec by the process called spontaneous emission (Refer below figure).
The rate R2 of spontaneous emission E2-> E1 is independent of energy density E of the radiation field.
R2 is proportional to number of atoms N2 in the excited state E2 thus
R2∞ N2
R2=A21 N2 (2)
Where A21 is known as Einstein’s coefficient for spontaneous emission and it represents the probability of spontaneous emission.
## Einstein Coefficient for Stimulated Emission:
Atoms can also fall back to the ground state E1 under the influence of the electromagnetic field of an incident photon of energy E2-E1 =hv by the process called stimulated emission (Refer below Figure):
Rate R3 for stimulated emission E2-> E is proportional to energy density E of the radiation field and proportional to the number of atoms N2 in the excited state,thus
R3α N2 E
Or R3=B21N2 E (3)
Where B21 is known as the Einstein coefficient for stimulated emission and it represents the probability of stimulated emission.
## Einstein Coefficient Relation Derivation:
In steady-state (at thermal equilibrium), the two emission rates (spontaneous and stimulated) must balance the rate of absorption.
Thus R1=R2+R3
Using equations (1,2, and 3) ,we get
N1B12E=N2A21+N2B21E
Or N1B12E –N2B21E=N2A21
Or (N1B12-N2B21) E =N2A21
Or E= N2A21/N1B12-N2B21
= N2A21/N2B21[N1B12/N2B21 -1]
[by taking out common N2B21from the denominator]
Or E=A21/B21 {1/N1/N2(B12/B21-1)) (4)
Einstein proved thermodynamically, that the probability of stimulated absorption is equal to the probability of stimulated emission. thus
B12=B21
Then equation(4) becomes
E=A21/B21(1/N1/N2-1) (5)
From Boltzman’s distribution law, the ratio of populations of two levels at temperature T is expressed as
N1/N2=e(E2E1)/KT
N1/N2=ehv/KT
Where K is the Boltzman’s constant and h is Planck’s constant.
Substituting value of N1/N2in equation (5) we get
E= A21/B21(1/ehv/KT-1) (6)
Now according to Planck’s radiation law, the energy density of the black body radiation of frequency v at temperature T is given as
E = 8πhv3/c3(1/ehv/KT) (7)
By comparing equations (6 and 7),we get
A21/B21=8πhv3/c3
This is the relation between Einstein’s coefficients in laser.
Significance of Einstein coefficient relation: This shows that the ratio of Einstein’s coefficient of spontaneous emission to the Einstein’s coefficient of stimulated absorption is proportional to the cube of frequency v. It means that at thermal equilibrium, the probability of spontaneous emission increases rapidly with the energy difference between two states. If you want to learn derivation of Einstein Coefficient relations, then following is our youtube video also: | 1,037 | 4,619 | {"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-2023-40 | latest | en | 0.898737 |
https://inchestometers.com/32-93-inches-to-meters | 1,611,205,918,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00699.warc.gz | 393,492,422 | 9,834 | # 32.93 Inches to Meters
Welcome to 32.93 inches to meters. The United States customary unit inch is abbreviated as in or ″, whereas the result in meters is often written using the unit symbol m.
Here you can find all about 32.93″ in m, including a converter as well as the formula. If you have been looking for what is 32.93 inches in meters, then you are right here,too.
You may overwrite the length in inches in the calculator right below, 32.93; our application then conducts the math automatically.
In the next section you can find the formula and the math explained.
## 32.93″ in Meters
How many meters is 32.93″? With the formula explained on our home page: [m] = 32.93 in x 0.0254, we get the following result, rounded to 5 decimal places:
32.93″ = 0.83642 m
To convert the units you have to multiply the imperial and US customary unit of length by 0.0254.
The result of the multiplication, 0.83642 meters, is the height, length or width in the base unit of length in the International System of Units (SI).
Similar conversions on inchestometers.com include, for example:
## Convert 32.93 Inches to Meters
By reading so far, you know everything about the 32.93 inch to m conversion. It’s a simple multiplication.
Visitors who have come here in search for, for example, 32.93 in to meters, have definitely found all their answers, too.
This also applies to 32.93 ″ in m, 32.93″ to meters and lots of similar terms searched terms such as, for instance, 32.93 inches to m.
Note that you can find many inches to meters conversions including 32.93 inch to meters by means of the search form we have placed in the sidebar.
Give it a try right now! Enter, for example, 32.93 ″ in meters or 32.93 in meters, or something alike. The result page contains all relevant posts.
BTW: 32.93 inches in related units is:
• 836.422 millimeters
• 83.6422 centimeters
• 8.36422 decimeters
The following paragraph wraps our content up.
## Summary
We summarize our content with this image: | 512 | 1,996 | {"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-2021-04 | latest | en | 0.904959 |
https://www.shaalaa.com/question-bank-solutions/draw-velocity-time-graph-show-following-motion-car-accelerates-uniformly-rest-5-s-then-it-travels-steady-velocity-5-s-velocity-time-graphs_29732 | 1,660,261,777,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571536.89/warc/CC-MAIN-20220811224716-20220812014716-00361.warc.gz | 866,076,133 | 9,843 | # Draw a Velocity-time Graph to Show the Following Motion : a Car Accelerates Uniformly from Rest for 5 S ; Then It Travels at a Steady’ Velocity for 5 S. - Science
Short Note
Draw a velocity-time graph to show the following motion :
A car accelerates uniformly from rest for 5 s ; then it travels at a steady’ velocity for 5 s.
#### Solution
we have to draw a velocity verses time curve for a moving body .
Concept: Velocity - Time Graphs
Is there an error in this question or solution?
#### APPEARS IN
Lakhmir Singh Science Class 9 Physics
Chapter 1 Motion
Short Answers | Q 31 | Page 41 | 149 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.760119 |
http://people.sc.fsu.edu/~jburkardt/m_src/sudoku/sudoku_check.m | 1,369,177,955,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00007-ip-10-60-113-184.ec2.internal.warc.gz | 192,817,029 | 1,307 | function check = sudoku_check ( s ) %*****************************************************************************80 % %% SUDOKU_CHECK checks a partial or filled-in Sudoku puzzle. % % Discussion: % % The routine ensures that % 1) each entry of S is either 0 (unfilled) or between 1 and 9. % 2) each row contains no more than one occurrence of a digit. % 3) each column contains no more than one occurrent of a digit. % 4) each box contains no more than one occurrence of a digit. % % Licensing: % % This code is distributed under the GNU LGPL license. % % Modified: % % 22 October 2007 % % Author: % % John Burkardt % % Parameters: % % Input, integer S(9,9) contains the Sudoku puzzle. Unfilled % entries should be set to 0. % . % Output, integer CHECK, is % 0, if no errors were detected. % 10*I+J, if A(I,J) contains an illegal digit, % 100+10*I+J, if A(I,J) violates the row condition. % 200+10*I+J, if A(I,J) violates the column condition. % 300+10*I+J, if A(I,J) violates the box condition. % % % Digit check. % i = 0; for i3 = 1 : 3 for i2 = 1 : 3 i = i + 1; j = 0; for j3 = 1 : 3 for j2 = 1 : 3 j = j + 1; if ( s(i,j) < 0 | 9 < s(i,j) ) check = 10 * i + j; return end end end end end % % Row check. % i = 0; for i3 = 1 : 3 for i2 = 1 : 3 i = i + 1; digit(1:9) = 0; j = 0; for j3 = 1 : 3 for j2 = 1 : 3 j = j + 1; k = s(i,j); if ( 1 <= k & k <= 9 ) digit(k) = digit(k) + 1; if ( 1 < digit(k) ) check = 100 + 10 * i + j; return end end end end end end % % Column check. % j = 0; for j3 = 1 : 3 for j2 = 1 : 3 j = j + 1; digit(1:9) = 0; i = 0; for i3 = 1 : 3 for i2 = 1 : 3 i = i + 1; k = s(i,j); if ( 1 <= k & k <= 9 ) digit(k) = digit(k) + 1; if ( 1 < digit(k) ) check = 200 + 10 * i + j; return end end end end end end % % Box check % for i3 = 1 : 3 for j3 = 1 : 3 digit(1:9) = 0; i = ( i3 - 1 ) * 3; for i2 = 1 : 3 j = ( j3 - 1 ) * 3; i = i + 1; for j2 = 1 : 3 j = j + 1; k = s(i,j); if ( 1 <= k & k <= 9 ) digit(k) = digit(k) + 1; if ( 1 < digit(k) ) check = 300 + 10 * i + j; return end end end end end end % % No errors discovered. % check = 0; return end | 802 | 2,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2013-20 | latest | en | 0.476643 |
https://www.vedantu.com/question-answer/expand-each-of-the-following-using-suitable-class-9-maths-cbse-5ee34eee6067f248d1692cde | 1,701,526,230,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00619.warc.gz | 1,163,309,667 | 29,397 | Courses
Courses for Kids
Free study material
Offline Centres
More
Last updated date: 27th Nov 2023
Total views: 381.6k
Views today: 5.81k
# Expand each of the following, using suitable identities:$(i) {\left( {x + 2y + 4z} \right)^2}$ $(ii) {\left( {2x - y + z} \right)^2}$ $(iii) {\left( { - 2x + 3y + 2z} \right)^2}$ $(iv) {\left( {3a - 7b - c} \right)^2}$ $(v) {\left( { - 2x + 5y - 3z} \right)^2}$ $(iv) {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2}$
Verified
381.6k+ views
Hint: Let’s substitute the values of a, b and c in the formula given below for determining the square of sum of three numbers and reach the answer by simplifying.
$\Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
All the above parts of the question contain the form of ${\left( {a + b + c} \right)^2}$ , and as we know that:
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right),$ we will use this formula for expanding every expression:
$(i)$ If we compare ${\left( {x + 2y + 4z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = x,b = 2y$ and $c = 4z$. Thus, using the formula, we’ll get:
$\Rightarrow {\left( {x + 2y + 4z} \right)^2} = {x^2} + {\left( {2y} \right)^2} + {\left( {4z} \right)^2} + 2\left( {x.2y + 2y.4z + x.4z} \right), \\ \Rightarrow {\left( {x + 2y + 4z} \right)^2} = {x^2} + 4{y^2} + 16{z^2} + 4xy + 16yz + 8xz \\$
$(ii)$ If we compare ${\left( {2x - y + z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = 2x,b = - y$ and $c = z$. Thus, using the formula, we’ll get:
$\Rightarrow {\left( {2x - y + z} \right)^2} = {\left( {2x} \right)^2} + {\left( { - y} \right)^2} + {z^2} + 2\left[ {2x.\left( { - y} \right) + \left( { - y} \right).z + 2x.z} \right], \\ \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + {y^2} + {z^2} - 4xy - 2yz + 4xz \\$
$(iii)$ If we compare ${\left( { - 2x + 3y + 2z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = - 2x,b = 3y$ and $c = 2z$. Thus, using the formula, we’ll get:
$\Rightarrow {\left( { - 2x + 3y + 2z} \right)^2} = {\left( { - 2x} \right)^2} + {\left( {3y} \right)^2} + {\left( {2z} \right)^2} + 2\left[ {\left( { - 2x} \right).3y + 3y.2z + \left( { - 2x} \right).2z} \right], \\ \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + 9{y^2} + 4{z^2} - 12xy + 12yz - 8xz \\$
$(iv)$ If we compare ${\left( {3a - 7b - c} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, in place of a, b and c here we have 3a, -7b and –c respectively. Thus, using the formula, we’ll get:
$\Rightarrow {\left( {3a - 7b - c} \right)^2} = {\left( {3a} \right)^2} + {\left( { - 7b} \right)^2} + {\left( { - c} \right)^2} + 2\left[ {3a.\left( { - 7b} \right) + \left( { - 7b} \right).\left( { - c} \right) + 3a.\left( { - c} \right)} \right],$
$\Rightarrow {\left( {3a - 7b - c} \right)^2} = 9{a^2} + 49{b^2} + {c^2} - 42ab + 14bc - 6ac$
$(v)$ If we compare ${\left( { - 2x + 5y - 3z} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, we will get $a = - 2x,b = 5y$ and $c = - 3z$. Thus, using the formula, we’ll get:
$\Rightarrow {\left( { - 2x + 5y - 3z} \right)^2} = {\left( { - 2x} \right)^2} + {\left( {5y} \right)^2} + {\left( { - 3z} \right)^2} + 2\left[ {\left( { - 2x} \right).5y + 5y.\left( { - 3z} \right) + \left( { - 2x} \right).\left( { - 3z} \right)} \right], \\ \Rightarrow {\left( {x + 2y + 4z} \right)^2} = 4{x^2} + 25{y^2} + 9{z^2} - 20xy - 30yz + 12xz \\$
$(vi)$ If we compare ${\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2}$ with ${\left( {a + b + c} \right)^2}$, in place of a, b and c here we have $\dfrac{1}{4}a, - \dfrac{1}{2}b$ and 1 respectively. Thus, using the formula, we’ll get:
$\Rightarrow {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2} = {\left( {\dfrac{1}{4}a} \right)^2} + {\left( { - \dfrac{1}{2}b} \right)^2} + {1^2} + 2\left[ {\left( {\dfrac{1}{4}a} \right).\left( { - \dfrac{1}{2}b} \right) + \left( { - \dfrac{1}{2}b} \right).\left( 1 \right) + \left( {\dfrac{1}{4}a} \right).\left( 1 \right)} \right], \\ \Rightarrow {\left( {\dfrac{1}{4}a - \dfrac{1}{2}b + 1} \right)^2} = \dfrac{1}{{16}}{a^2} + \dfrac{1}{4}{b^2} + 1 - \dfrac{1}{4}ab - b + \dfrac{1}{2}a. \\$
Note: If we miss the formula for ${\left( {a + b + c} \right)^2}$, we can apply general multiplication method for expanding the above expressions:
$\Rightarrow {\left( {a + b + c} \right)^2} = \left( {a + b + c} \right).\left( {a + b + c} \right), \\ \Rightarrow {\left( {a + b + c} \right)^2} = a\left( {a + b + c} \right) + b\left( {a + b + c} \right) + c\left( {a + b + c} \right) \\$
On expansion, we’ll get the same result. | 2,118 | 4,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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} | 4.71875 | 5 | CC-MAIN-2023-50 | longest | en | 0.544465 |
https://gzipwtf.com/what-is-the-difference-between-binomial-distribution-and-hyper-geometric-distribution/ | 1,718,884,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00835.warc.gz | 250,840,576 | 42,876 | # What is the difference between binomial distribution and hyper geometric distribution?
## What is the difference between binomial distribution and hyper geometric distribution?
The difference between the hypergeometric and the binomial distributions. For the binomial distribution, the probability is the same for every trial. For the hypergeometric distribution, each trial changes the probability for each subsequent trial because there is no replacement.
Is hyper geometric distribution continuous?
The hypergeometric distribution is an example of a discrete probability distribution because there is no possibility of partial success, that is, there can be no poker hands with 2 1/2 aces. Said another way, a discrete random variable has to be a whole, or counting, number only.
### Which distribution maximizes entropy?
normal distribution
distributed about the unit circle, the Von Mises distribution maximizes the entropy when the real and imaginary parts of the first circular moment are specified or, equivalently, the circular mean and circular variance are specified. are specified, the wrapped normal distribution maximizes the entropy.
What is the relationship between binomial and geometric distribution?
Geometric distribution is a special case of negative binomial distribution, where the experiment is stopped at first failure (r=1). So while it is not exactly related to binomial distribution, it is related to negative binomial distribution.
#### What is the difference between negative binomial distribution and geometric distribution?
In the binomial distribution, the number of trials is fixed, and we count the number of “successes”. Whereas, in the geometric and negative binomial distributions, the number of “successes” is fixed, and we count the number of trials needed to obtain the desired number of “successes”.
What are various types of distributions?
Gallery of Distributions
Normal Distribution Uniform Distribution Cauchy Distribution
Power Normal Distribution Power Lognormal Distribution Tukey-Lambda Distribution
Extreme Value Type I Distribution Beta Distribution
Binomial Distribution Poisson Distribution
## What is the variance of geometric distribution?
The mean of the geometric distribution is mean = 1 − p p , and the variance of the geometric distribution is var = 1 − p p 2 , where p is the probability of success.
Is geometric distribution discrete?
The geometric distribution is the only discrete memoryless random distribution. It is a discrete analog of the exponential distribution.
### Which distribution has highest variance?
1 Answer. The normal distribution has the maximum entropy among all continuous distributions with fixed mean and variance on real support. The variance for a given entropy can be made arbitrarily large using a mixture of two Gaussians that are spread farther and farther apart.
Why is entropy maximum at uniform distribution?
The reason why entropy is maximized for a uniform distribution is because it was designed so! Yes, we’re constructing a measure for the lack of information so we want to assign its highest value to the least informative distribution.
#### What is hypergeometric distribution in statistics?
Statistics – Hypergeometric Distribution. A hypergeometric random variable is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Hypergeometric distribution is defined and given by the following probability function:
What is the hypergeometric distribution of marbles?
marbles are drawn without replacement and colored red. Then, the number of marbles with both colors on them (that is, the number of marbles that have been drawn twice) has the hypergeometric distribution. The symmetry in balls and colouring them red first. The classical application of the hypergeometric distribution is sampling without replacement.
## How to produce the hypergeometric cumulative distribution function in R?
Figure 1: Hypergeometric Density. The second example shows how to produce the hypergeometric cumulative distribution function (CDF) in R. Similar to Example 1, we first need to create an input vector of quantiles… …then we can apply the phyper function to this vector… …and finally we can produce a plot representing the hypergeometric CDF:
What is a Hypergeometric random variable?
A hypergeometric random variable is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Hypergeometric distribution is defined and given by the following probability function:
Begin typing your search term above and press enter to search. Press ESC to cancel. | 895 | 4,815 | {"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-2024-26 | latest | en | 0.861106 |
https://www.coursehero.com/file/6644917/MIT6-045JS11-lec13/ | 1,513,226,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00450.warc.gz | 730,803,207 | 23,927 | MIT6_045JS11_lec13
# MIT6_045JS11_lec13 - 6.080/6.089 GITCS April 4th 2008...
This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: 6.080/6.089 GITCS April 4th, 2008 Lecture 16 Lecturer: Scott Aaronson Scribe: Jason Furtado Private-Key Cryptography 1 Recap 1.1 Derandomization In the last six years, there have been some spectacular discoveries of deterministic algorithms, for problems for which the only similarly-efficient solutions that were known previously required randomness. The two most famous examples are the Agrawal-Kayal-Saxena (AKS) algorithm for determining if a number is prime or composite • in deterministic polynomial time, and the algorithm of Reingold for getting out of a maze (that is, solving the undirected s-t con- • nectivity problem) in deterministic LOGSPACE. Beyond these specific examples, mounting evidence has convinced almost all theoretical com- puter scientists of the following Conjecture: Every randomized algorithm can be simulated by a deterministic algorithm with at most polynomial slowdown. Formally, P = BP P . 1.2 Cryptographic Codes 1.2.1 Caesar Cipher In this method, a plaintext message is converted to a ciphertext by simply adding 3 to each letter, wrapping around to A after you reach Z. This method is breakable by hand. 1.2.2 One-Time Pad The “one-time pad” uses a random key that must be as long as the message we want to en- crypt. The exclusive-or operation is performed on each bit of the message and key ( Msg ⊕ Key = EncryptedMsg ) to end up with an encrypted message. The encrypted message can be decrypted by performing the same operation on the encrypted message and the key to retrieve the message ( EncryptedMsg ⊕ Key = Msg ). An adversary that intercepts the encrypted message will be unable to decrypt it as long as the key is truly random. The one-time pad was the first example of a cryptographic code that can proven to be secure, even if the adversary has all the computation time in the universe. The main drawback of this method is that keys can never be reused, and the key must be the same size as the message to encrypt. If you were to use the same key twice, an eavesdropper could compute ( Enc ⊕ Msg 1) ⊕ ( Enc ⊕ Msg 2) = Msg 1 ⊕ Msg 2. This would leak information about Msg 1 and Msg 2....
View Full Document
{[ snackBarMessage ]}
### Page1 / 5
MIT6_045JS11_lec13 - 6.080/6.089 GITCS April 4th 2008...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 672 | 2,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-51 | latest | en | 0.911485 |
https://encyclopedia2.thefreedictionary.com/rectangular+hyperbola | 1,555,693,338,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527865.32/warc/CC-MAIN-20190419161226-20190419183226-00028.warc.gz | 412,708,343 | 11,622 | # rectangular hyperbola
Also found in: Dictionary, Wikipedia.
## rectangular hyperbola
[rek′taŋ·gyə·lər hī′pər·bə·lə]
(mathematics)
A hyperbola whose major and minor axes are equal.
Mentioned in ?
References in periodicals archive ?
Figure 11 demonstrates the application of the proposed rectangular hyperbola model on unfilled LCP samples as a function of shear rate.
Hence, the variation of shear stress with normal stress is small and the rectangular hyperbola model is not applicable to the [N.
Rectangular hyperbola function fitted to TCL and wheat yield indicates the increase of weeds density based on TCL caused the decrease in yield (Figure 1).
8] showed that the increase in TCL followed by rectangular hyperbola function reduced wheat yield.
Thus rhodium provides an independent verification of the method of rectangular hyperbolas for the Periodic Table of elements of D.
0006756 a) columns 4 and 5 contain coordinates of peaks of rectangular hyperbolas of elements; b) in a column 6 are presented abscissas the secants which are starting with the peak center (0,1) up to crossings with line Y = 0:5; at prolongation they cross the valid axis in points peaks; c) in a column 7 are resulted abscissa points of crossing of a direct and adjacent hyperbola each element presented here; d) the column 8 contains a difference between sizes of 6 and 7 columns; e) in a column 9 tangents of a corner of an inclination of secants are resulted; at an element "rhodium" this line crosses an axis X in a point with abscissa, equal 411.
With only one exception, the parameters of the inverse and rectangular hyperbola equations fitted to the data of these experiments did not differ (P > 0.
Inverse or rectangular hyperbola models described accurately the relationship between KNP and PGR, and the effect of reducing N availability was similar to the effect of reducing water availability or both N and water availability (Fig.
IDENTICAL ISO-HIRING FUNCTION RECTANGULAR HYPERBOLA
If it is assumed that the hiring functions of each sector are rectangular hyperbolas such as H, [([U.
Taking into consideration that the semiaxis of the rectangular hyperbola a=b= [square root of 2 [absolute value of K]], the coordinates of the point [X.
In the case of two rectangular hyperbolas with the same centre and rotated 45[degrees] with respect to each other, there will be just two points of intersection and so the quadratic [z.
Site: Follow: Share:
Open / Close | 572 | 2,457 | {"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-2019-18 | longest | en | 0.872099 |
https://chhapdesign.com/computer-laguages/automata-theory/ | 1,701,433,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00075.warc.gz | 195,355,056 | 23,667 | # Theory-of-automata
The Automata theory is the study of automata and abstract machines and also the computational problems which solved using them. Automata mean self-acting.
• ## Chomsky Normal Form (CNF)
Chomsky Normal Form (CNF) It cannot contain the null, that CFG in which non-terminals belongs to two non-terminal. Or non-terminal…
• ## CFG Properties and Applications
CFG Properties and Applications CFG- Context-free grammar is closure under: Union Concatenation Kleene Star Operation Union: Let suppose L1 and…
• ## Ambiguity in CFG
Ambiguity in CFG Something when having more meaning and confusion occurs when select appropriate mean for the selected condition called…
• ## CFG- Context Free Grammar
CFG- Context-Free Grammar CFG- Context-free grammar is semantic free grammar. CFG has the following terms. Terminals Non Terminals Production ǀ…
• ## Pumping Lemma
Pumping Lemma (PL) Without any proof the string or language is accepted as true, this process is known as Pumping…
• ## FA to Regular Expression construction
FA to Regular Expressions construction This topic designed for finite automata FA to regular expression (RE) construction. Case 1: Finite…
• ## Regular language
Definition of Regular Language A Regular language whose regular expressions can be drawn is known as regular language. Property of…
• ## Regular expression
Regular expression A regular expression can be defined as the following properties. Kleene Star Closure: Kleene star closure can be…
• ## Moore machine and Mealy machine
Moore machine A Moore machine that consists of the following The finite number of states such as q0, q1, q2…qn…
• ## Non-deterministic finite automata
Non-deterministic finite automata Non-deterministic finite automata have the same DFA characteristic but a slight change. The total number of the alphabet…
• ## Finite automata
Finite automata Finite automata can be described into two types. 1: Deterministic Finite Automata (DFA). 2: Non Deterministic Finite Automata…
• ## Automata theory
Automata theory Automata theory is the branch of computer science that deals with the self-making languages and follows the predetermined… | 464 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.822856 |
http://singaporemathguru.com/question/primary-5-problem-sums-word-problems-whole-numbers-and-fractions-time-for-a-little-challenge-whole-numbers-amp-fractions-exercise-11-508 | 1,537,626,958,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158450.41/warc/CC-MAIN-20180922142831-20180922163231-00378.warc.gz | 227,209,585 | 11,378 | ### Primary 5 Problem Sums/Word Problems - Try FREE
Score :
(Single Attempt)
#### Question
There are 3 rows of pupils.
Each row had 48 pupils.
A teacher arranged the pupils into a marching contingent.
The marching contingent was shaped in a form of a non-hollow square.
Find the number of pupils along each side of the square.
The correct answer is : 12
pupils | 87 | 368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-39 | longest | en | 0.942675 |
https://foreach.id/EN/fluids/flowratemolar/attomol%7Csecond-to-millimol%7Csecond.html | 1,624,071,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00545.warc.gz | 242,115,637 | 8,465 | # Convert attomol/second to millimol/second (amol/s to mmol/s)
Batch Convert
• millimol/second [mmol/s]
• attomol/second [amol/s]
Copy
_
Copy
• millimol/second [mmol/s]
• attomol/second [amol/s]
## Attomol/second to Millimol/second (amol/s to mmol/s)
### Attomol/second (Symbol or Abbreviation: amol/s)
Attomol/second is one of molar flow rate units. Attomol/second abbreviated or symbolized by amol/s. The value of 1 attomol/second is equal to 1e-18 mol/second. In its relation with millimol/second, 1 attomol/second is equal to 1e-15 millimol/second.
#### Relation with other units
1 attomol/second equals to 1e-18 mol/second
1 attomol/second equals to 1e-36 examol/second
1 attomol/second equals to 1e-33 petamol/second
1 attomol/second equals to 1e-30 teramol/second
1 attomol/second equals to 1e-27 gigamol/second
1 attomol/second equals to 1e-24 megamol/second
1 attomol/second equals to 1e-21 kilomol/second
1 attomol/second equals to 1e-20 hectomol/second
1 attomol/second equals to 1e-19 dekamol/second
1 attomol/second equals to 1e-17 decimol/second
1 attomol/second equals to 1e-16 centimol/second
1 attomol/second equals to 1e-15 millimol/second
1 attomol/second equals to 1e-12 micromol/second
1 attomol/second equals to 1e-9 nanomol/second
1 attomol/second equals to 0.000001 picomol/second
1 attomol/second equals to 0.001 femtomol/second
1 attomol/second equals to 6e-17 mol/minute
1 attomol/second equals to 3.6e-15 mol/hour
1 attomol/second equals to 8.64e-14 mol/day
1 attomol/second equals to 6e-14 millimol/minute
1 attomol/second equals to 3.6e-12 millimol/hour
1 attomol/second equals to 8.64e-11 millimol/day
1 attomol/second equals to 6e-20 kilomol/minute
1 attomol/second equals to 3.6e-18 kilomol/hour
1 attomol/second equals to 8.64e-17 kilomol/day
### Millimol/second (Symbol or Abbreviation: mmol/s)
Millimol/second is one of molar flow rate units. Millimol/second abbreviated or symbolized by mmol/s. The value of 1 millimol/second is equal to 0.001 mol/second. In its relation with attomol/second, 1 millimol/second is equal to 1000000000000000 attomol/second.
#### Relation with other units
1 millimol/second equals to 0.001 mol/second
1 millimol/second equals to 1e-21 examol/second
1 millimol/second equals to 1e-18 petamol/second
1 millimol/second equals to 1e-15 teramol/second
1 millimol/second equals to 1e-12 gigamol/second
1 millimol/second equals to 1e-9 megamol/second
1 millimol/second equals to 0.000001 kilomol/second
1 millimol/second equals to 0.00001 hectomol/second
1 millimol/second equals to 0.0001 dekamol/second
1 millimol/second equals to 0.01 decimol/second
1 millimol/second equals to 0.1 centimol/second
1 millimol/second equals to 1,000 micromol/second
1 millimol/second equals to 1,000,000 nanomol/second
1 millimol/second equals to 1,000,000,000 picomol/second
1 millimol/second equals to 1,000,000,000,000 femtomol/second
1 millimol/second equals to 1,000,000,000,000,000 attomol/second
1 millimol/second equals to 0.06 mol/minute
1 millimol/second equals to 3.6 mol/hour
1 millimol/second equals to 86.4 mol/day
1 millimol/second equals to 60 millimol/minute
1 millimol/second equals to 3,600 millimol/hour
1 millimol/second equals to 86,400 millimol/day
1 millimol/second equals to 0.00006 kilomol/minute
1 millimol/second equals to 0.0036 kilomol/hour
1 millimol/second equals to 0.0864 kilomol/day
### How to convert Attomol/second to Millimol/second (amol/s to mmol/s):
#### Conversion Table for Attomol/second to Millimol/second (amol/s to mmol/s)
attomol/second (amol/s) millimol/second (mmol/s)
0.01 amol/s 1e-17 mmol/s
0.1 amol/s 1e-16 mmol/s
1 amol/s 1e-15 mmol/s
2 amol/s 2e-15 mmol/s
3 amol/s 3e-15 mmol/s
4 amol/s 4e-15 mmol/s
5 amol/s 5e-15 mmol/s
6 amol/s 6e-15 mmol/s
7 amol/s 7e-15 mmol/s
8 amol/s 8e-15 mmol/s
9 amol/s 9e-15 mmol/s
10 amol/s 1e-14 mmol/s
20 amol/s 2e-14 mmol/s
25 amol/s 2.5e-14 mmol/s
50 amol/s 5e-14 mmol/s
75 amol/s 7.5e-14 mmol/s
100 amol/s 1e-13 mmol/s
250 amol/s 2.5e-13 mmol/s
500 amol/s 5e-13 mmol/s
750 amol/s 7.5e-13 mmol/s
1,000 amol/s 1e-12 mmol/s
100,000 amol/s 1e-10 mmol/s
1,000,000,000 amol/s 0.000001 mmol/s
1,000,000,000,000 amol/s 0.001 mmol/s
#### Conversion Table for Millimol/second to Attomol/second (mmol/s to amol/s)
millimol/second (mmol/s) attomol/second (amol/s)
0.01 mmol/s 10,000,000,000,000 amol/s
0.1 mmol/s 100,000,000,000,000 amol/s
1 mmol/s 1,000,000,000,000,000 amol/s
2 mmol/s 2,000,000,000,000,000 amol/s
3 mmol/s 3,000,000,000,000,000 amol/s
4 mmol/s 4,000,000,000,000,000 amol/s
5 mmol/s 5,000,000,000,000,000 amol/s
6 mmol/s 6,000,000,000,000,000 amol/s
7 mmol/s 7,000,000,000,000,000 amol/s
8 mmol/s 8,000,000,000,000,000 amol/s
9 mmol/s 9,000,000,000,000,000 amol/s
10 mmol/s 10,000,000,000,000,000 amol/s
20 mmol/s 20,000,000,000,000,000 amol/s
25 mmol/s 25,000,000,000,000,000 amol/s
50 mmol/s 50,000,000,000,000,000 amol/s
75 mmol/s 75,000,000,000,000,000 amol/s
100 mmol/s 100,000,000,000,000,000 amol/s
250 mmol/s 250,000,000,000,000,000 amol/s
500 mmol/s 500,000,000,000,000,000 amol/s
750 mmol/s 750,000,000,000,000,000 amol/s
1,000 mmol/s 1,000,000,000,000,000,000 amol/s
100,000 mmol/s 100,000,000,000,000,000,000 amol/s
1,000,000,000 mmol/s 1e+24 amol/s
1,000,000,000,000 mmol/s 1e+27 amol/s
#### Steps to Convert Attomol/second to Millimol/second (amol/s to mmol/s)
1. Example: Convert 414 attomol/second to millimol/second (414 amol/s to mmol/s).
2. 1 attomol/second is equivalent to 1e-15 millimol/second (1 amol/s is equivalent to 1e-15 mmol/s).
3. 414 attomol/second (amol/s) is equivalent to 414 times 1e-15 millimol/second (mmol/s).
4. Retrieved 414 attomol/second is equivalent to 4.14e-13 millimol/second (414 amol/s is equivalent to 4.14e-13 mmol/s).
▸▸ | 2,361 | 5,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.82065 |
https://cameramath.com/es/expert-q&a/Algebra/Tish-and-Emma-are-knitting-scarves-Tish-s-scarf-is17-75-inches | 1,642,721,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00096.warc.gz | 221,597,939 | 6,779 | ### ¿Todavía tienes preguntas de matemáticas?
Pregunte a nuestros tutores expertos
Algebra
Question
Tish and Emma are knitting scarves. Tish's scarf is17.75 inches long, and she knits 2$$\frac{3}{8}$$ inches per minute.Emma's scarf is 4 inches long, and she knits 3$$\frac{3}{4}$$ inchd per minute.After how many minutes will Emma's scarf be longer than Tish's scarf? | 115 | 369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.501846 |
https://www.svpwiki.com/equilibrium | 1,716,343,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00402.warc.gz | 896,339,091 | 31,417 | equilibrium
Balance or balanced; scalar.
a stable situation in which forces equate one another.
A system in equilibrium is held to be neutral.
See List of Synonyms for Scalar
Jules Verne "We cannot prevent equilibrium from producing its effects. We may brave human laws, but we cannot resist natural ones." [Jules Verne]
Keely
"The aerial propeller has a sympathetic polar accumulator and disperser in one instrument, which is entirely distinct from any of the devices intended for terrestrial use; also other mechanical adjuncts not needed on land nor water.
All forms of non-sympathetic machinery have, associated with them, conditions of centrifugal force on the ratio of the velocity induced; the diverging power from the center of induction being governed specifically by its gravital weight according to the diameter it occupies in its circle of rotation.
In a sympathetic negative circuit this order of conditions is reversed; for the power of neutral attraction draws the molecules of any mass, no matter what the weight, toward the center of rotation (instead of toward its periphery) according to the intensity of the negative vibration that is induced upon that particular circle.
Our earth, in its routine of revolution, is governed by the same law in every particular; its mass tending toward its centre of neutrality with a force that is equivalent to the character and velocity of its rotation.(3) If its rotation were increased, the tendency of everything associated with it would be increased toward its centre of neutrality on the same ratio. That is, a pound in weight would, under certain conditions of increased velocity, become two pounds in weight. The laws governing the sympathetic rotation of vibratory machinery are the same laws that govern planetary suspension. To those who have not witnessed the operation of my devices, my theories must indeed seem wild; but the laws of nature are the same yesterday today and forever. They know no change; and sympathetic physics, demonstrated mechanically, must triumph over all ridicule and opposition in the end. To contradict the laws governing sympathetic rotation is to contradict the laws governing planetary suspension, as I am prepared to demonstrate.
If the earth were rotated on a shaft by mechanical force, the present condition of its rotations would be reversed; everything on its surface would fly off at a tangent, on the ratio of the velocity induced. The equilibrium of all things would also be changed.
The gyroscope reveals astounding facts in relation to this philosophy, even when operated mechanically. No other known device is so nearly associated with sympathetic vibratory physics.
The vitalization of the disks for the polar and depolar field is established on the ratio of thirds, sixths, and ninths; the ninths being the circuit occupied by the polar field, must represent, in the scale of vitalized focalized intensity, 100 in my system: sixths in the depolar field, or 66 2/3; and in the neutral field, or thirds, 33 1/3. The triplets must represent one true chord of equation. The sympathetic transmitter transfers any degree of intensity desired from zero up to disintegration; all the transfers being made above the line of the first inaudible, as associated with my resonating system of transfer. On the sixths and ninths, in the progressive triple subdivision of the elements (4) of water, the nearest sympathetic approach is made to the high luminous, which is the main sympathetic link to the earth's polar negative envelope, and the one whereby coordination is effected for commercial work. In short, this progressive condition establishes the necessary association between celestial radiation and terrestrial outreach, in regard to controlling the polar negative attractive force in mechanics; whether for aerial navigation or for terrestrial commercial work, in all its multiplied forms.
The atmospheric envelope of our earth owes its activity and its volume entirely to celestial radiating forces.(5) Reception and dispersion are kept up by atomic and interatomic conflict between the dominant and enharmonic currents of the triune polar stream. The harmonic and enharmonic current with the dominant (in the electric stream) by their sympathetic association evolve the energy of matter.
The mechanical proof of the correctness of my theories, in sympathetic or spiritual physics, is so overwhelming in its simplicity that it needs but to be witnessed to convince the most learned or the most simple mind that this system will place both science and commerce on a platform which will elevate each to a level far higher than those they now occupy." [The Operation of the Vibratory Circuit]"The earth's rotation is caused and continued by the action of the POSITIVE AND NEGATIVE SYMPATHETIC CELESTIAL STREAMS." He proved this by the action of the metal sphere mentioned elsewhere. "It's motion is governed to the utmost mathematical nicety, free from intermittent impulses, by the mobility of the earth's oceans and the oceans' "anastomosis". Herein is the elephant mentioned in the fable, on which the tortoise stands."
"The fixed neutral center of the earth, being the concentration of all the molecular neutral centers in its mass, is the unit of the Universe held in suspension through its own properties. We cannot say "This is what the elephant stands on" but we can say "This is the power by which the elephant is suspended."
"The earth's rotation excites and disturbs the equilibrium of the three sympathetic streams, of which the material manifestations are the three forms of electricity. The alternate light and darkness assists in preserving their activity through disturbance of equilibrium, with consequent "dissimilation" (radiation) and "assimilation" (absorption). The dark zone, following the light zone, holds the fluctuations of the polar power constant. This is the tortoise mentioned in the fable, upon which the world rests." [Snell Manuscript - the book, FORCES CONTROLLING THE EARTH]
"The power of attractive vibration of the solar forces is the great coincident towards which the terrestrial magnetic sympathetic flow is diverted. This force is the celestial current that makes up the prime third of the triple association. It also induces aqueous disintegration and thermal concentration, the two prime conductors towards this coincident chord of sympathy with itself. Without this aqueous disintegration there would be no connective link between the celestial and terrestrial. There would exist nothing but a condition of luminous radiation on the order of the aurora - a reaching out for the concordant without any sympathetic diversion to create unstable equilibrium of terrestrial magnetism. In fact under such a condition, the absence of the sun on one side, or the absence of water on the other, the magnetic or electric force would remain in a stable state of equilibrium, or the highest order of the chaotic." [Vibratory Physics - The Connecting Link between Mind and Matter]
"In analyzing this triple union in its vibratory philosophy, I find the highest order of perfection in this assimilative action of Nature. The whole condition is atomic, and is the introductory one which has an affinity for terrestrial centres, uniting magnetically with the Polar stream, in other words, uniting with the Polar stream by neutral affinity. The magnetic or electric forces of the earth are thus kept in stable equilibrium by this triune force, and the chords of this force may be expressed as 1st, the dominant, 2nd, the harmonic, and 3rd, the enharmonic. The value of each is, one to the other, in the rates of figures, true thirds. E?, - transmissive chord or dominant; A? - harmonic; A?? - enharmonic. The unition of the two prime thirds is so rapid, when the negative and the positive conditions reach a certain range of vibratory motion, as to be compared to an explosion. During this action the positive electric stream is liberated, and immediately seeks its neutral terrestrial centre, or centre of highest attraction." [True Science]
Law of Corporeal Oscillations
"All coherent aggregates not isolated from like bodies, oscillate at a period-frequency varying with the tensions that augment and diminish the state of equilibrium." [Keely, 1894]
Schauberger
This explains why, for example, the resistance to motion increases by the square of the unnatural (centrifugal) rotational velocity in all technical, hydraulic and dynamic pressure- and heat-increasing machines. This, however, does not occur if medial masses (earth, water & air) are accelerated 'originally' in the manner of the natural motion of the Earth's mass (see fig. 9), namely centripetally along the longitudinal axis. Such motion, which builds up everything indirectly and maintains the whole Earth in delicate equilibrium, can be copied in a manner true to Nature with the aid of very specially constructed and alloyed devices. [The Energy Evolution - Harnessing Free Energy from Nature, Cadaverine Poison in Ray-Form - Ptomaine Radiation]
In conclusion it is necessary to address in more detail the question: Whence do these gigantic levitational forces actually come? Taking a very large-scale example from Nature, these are the forces that maintain the whole Earth in a fragile state of equilibrium, because they also move in cycloid-spiral-space-curves like every falling filament of water. [The Energy Evolution - Harnessing Free Energy from Nature, Cadaverine Poison in Ray-Form - Ptomaine Radiation]
What is here involved are the colossal carrying and tractive forces that maintain the whole Earth in an unstable floating state and cause it to rotate in peculiar spiral space-curves (cycloid-space-curve-motion). They are the same forces that maintain the delicate equilibrium of a trout in fast-flowing water and enable everything that crawls and flies on this dung-heap - Earth - to overcome their own physical weight. [The Energy Evolution - Harnessing Free Energy from Nature, The Life-Current in Air and Water]
Curiously enough, since then apparently no scientifically trained individual has bothered about the causes of the mysterious motion of this dung-heap Earth, and the purpose of its exceptional form of motion. This disinterest, however, was only feigned. In reality, since Galilei's discovery, people were very much on their guard in order to avert the immeasurable repercussions resulting from the revelation of: 'Why' does the Earth ceaselessly rotate about her own peculiar system of axes in an unstable state of equilibrium in so-called free space in very particular looping movements in defiance of the laws of gravity? [The Energy Evolution - Harnessing Free Energy from Nature, The Life-Current in Air and Water]
When the fruit juices that have flourished in the Sun begin to ferment, it is a sign that reactive temperature differences have triggered the fermenting movement, which in the initial stages has to separate the inferior from the superior. The former will be separated and centrifugated out through cycloid-space-curve-motion. The latter interconnects with itself, as it were, and the product of this remarkable marriage process is that something, which also brings us out of the delicately balanced equilibrium that the Earth ceaselessly enjoys. Not being in a state of stable equilibrium, she has constantly to rotate about her own axis in a form of movement that declinates and oscillates in two directions. This is how she generates the inversely symmetrically- and inversely proportionally-acting force, which enables everything that crawls and flies on this dung-heap Earth to move 'originally' (autonomously) in order to seek out its food, which also contains allotropic energy-producing substances. These are then broken down through the vibration of the organs, leading to the familiar development of physical strength and mental vigour, whose ur-causes have not been perceived by physico-technical scientists, who calculate with common calories and measure them with decimal scales and rulers. Moreover, they also believe they have captured everything with their tweezers that is proper to Nature. Unfortunately, however, this cannot be accomplished with such an instrument, because the producer of 'measure' and 'weight' is an imperceptible extraordinary energy-form, which as the 'essential' is the 'reactivated' itself, which actively moves the ordinary. [The Energy Evolution - Harnessing Free Energy from Nature, New Forms of Motion and Energy]
Ramsay
Different writers have put forth different views of what constitute a musical vibration, but their various views do not make any difference in the ratios which the notes of this sound-host bear to each other. Whether the vibrations be counted as single or double vibrations, the ratios of their relative motions are the same. Nevertheless, a musical vibration is an interesting thing in itself, and ought to be correctly defined. A string when vibrating musically is passing and re-passing the central line of its rest or equilibrium with a certain range of excursion. Some writers have defined a vibration to be the passage of the string from one extreme of its excursion to the other, while some have preferred to define it as the passage of the string from the one extreme of its excursion to the other and back again. D. C. Ramsay has been led in his researches to define a vibration as the movement of the string from its central line of rest to the extreme of its excursion on one side, and back to the central line of rest; and from the central line of rest to the extreme of its excursion on the other side, and back again to the "right line," as he calls it, as a second vibration. His reasoning on this will be seen in what follows. (See Fig. 3, Plate IV.) [Scientific Basis and Build of Music, page 21]
"If the particles of the air be slightly disturbed from a position of stable equilibrium, every particle makes an effort to return to that position; and it can be shown that the force of restitution varies as the distance from the position of equilibrium ... a musical note is the consequence of the vibrations excited in the disturbed system, and the permanency of its musical pitch is the consequence of these vibrations being all made in the same time ... If the vibrations gradually slackened in their times, as they do in the extent of these excursions from the effect of the resistances they meet, there would be no note of sustained pitch; the sound would then be a sliding descent in pitch, and music would be impossible." - Penny Cyclopedia, Art. "Vibration." [Scientific Basis and Build of Music, page 24]
"By affinity the notes group in chords. The tonic is the center chord, the key of harmony; the dominant is the fifth above it and the subdominant the fifth below it, and these two are balanced on the center chord as the scales on a balance beam. The dominant chord is vigorous and active, tending to soar; the subdominant is solemn, soft, and grave, tending to sink; the tonic is melodious and restful, and in it the harmony of equilibrium. This far AFFINITY." [Scientific Basis and Build of Music, page 91]
Russell
"The master-tone of each octave is the inheritance of the original motion of the thinking process of Mind. These master-tones are the "inert gases" which are classified in the zero group of the Mendeleef table.
The state of motion of these inert gases is that of motion-in-inertia."
"Motion-in-inertia is that state of pressure equilibrium which lies between any two masses.
"The inertial line, or plane is that dividing line, or plane, toward which all masses discharge their potential.
"It is the line, or plane, of lowest potential of two opposing areas of potential, where opposing pressures neutralize. This is the plane of minimum pressure of two opposing areas.
"The master-tones which represent a state of motion-in-inertia and are the inert gases, bear the same relationship to the elements that white bears to the colors. They are a registration of them all. White is not included in the spectrum, it has no place there. The inert gases should not be included in the elements. They have no place there. Of this more shall be written later in its proper place." [Russell, The Universal One]
The attraction of gravitation and the repulsion of radiation is nature's simple method of distribution and redistribution of all masses, so that each mass will find its proper position. [See Universal Heart Beat]
It is not proper to conceive either of these apparently opposite forces as two forces.
The south wind and the north wind are not two winds. They are the same wind blowing in opposite directions.
It is more correct to say that gravitation and radiation are processes.
The one motive force which directs these processes is equally divided into opposite effects, but these opposite effects are unequally balanced. [See 14.35.1 - Keely 3 6 and 9]
The unequal divisions of the two opposites totaled together constitute an equilibrium.
The One force never subdivides into any minus expression of force without counterbalancing that minus with an equal and opposite plus. [See Reciprocating Proportionality]
Gravitation is a synthetic process of putting things together, and analytic one of taking them apart.
The chemist uses these processes in every action and reaction.
Consider for example the reduction of iron oxide at high temperature by passing a jet of hydrogen over it.
What happens? The hydrogen falls toward the higher potential of the oxygen of the iron which is sufficiently expanded by the high temperature to absorb the hydrogen, and the hot iron is sufficiently expanded to release the oxygen.
This is an effect of gravitation in respect to the oxygen and the hydrogen.
They unite, they mutually integrate, and freeze into amorphous crystals of such extended orbits that they assume the liquid state known as "water."
On the contrary, it is an effect of radiation in respect to the iron and the oxygen. [The Universal One, Book 02 - Chapter 12 - Gravitation and Radiation, Gravitation and Radiation - page 141]
DIMENSION CHART No. 7. STABILITY IS THE INACTIVITY OF A STATE OF EQUILIBRIUM. MOTION IS ACTIVE AND OPPOSED TO EQUILIBRIUM. ALL MOTION IS RELATIVELY UNSTABLE. STABILITY OF MATTER IN MOTION MAY ONLY BE SIMULATED BY THE UNION OF EXACTLY EQUAL AND OPPOSITE PLUS AND MINUS DIMENSIONS. [Energy Displacement]
Hughes
Helmholtz's experiments on developing colours shown to agree with the scheme
—The sounds of the Falls of Niagara are in triplets or trinities
—The Arabian system divides tones into thirds
—Two trinities springing from unity apparently the germ of never-ending developments in tones and colours
—Inequality of the equinoctial points; is the want of equilibrium the motive power of the entire universe?
—The double tones of keyed instruments, the meetings by fifths, the major and minor keys, so agree with the development of colours, that a correct eye would detect errors in a piece of coloured music
Numbers not entered upon, but develope by the same laws
Bass notes omitted in order to simplify the scheme, 18 [Harmonies of Tones and Colours, Table of Contents2 - Harmonies]
The inequality of the equinoctial points is a well-known fact. It will be seen how apparent this is in the developments of harmonies. From the moment that trinities depart from unity, the balance is unequal, and the repeated endeavours after closer union cause a perpetual restlessness. May not this want of equilibrium be the life or motive power of the entire universe, with its continuous struggle after concord, even to oneness? "Closer and closer union is the soul of perfect harmony." In tracing harmonies of tones and colours, the double tones of keyed instruments will be seen to correspond with the intermediate tints and shades of colours. The twelve notes, scales, and chords in the major and minor series, the meetings by fifths, &c., all agree so exactly in their mode of development, that if a piece of music is written correctly in colours with the intermediate tints and shades, the experienced musician can, as a rule, detect errors more quickly and surely with the eye than the ear, and the correct eye, even of a non-musical person, may detect technical errors. Although the arithmetical relation has been most useful in gaining the laws, it is not here entered upon; but numbers equally meet all the intricacies both of tones and colours. The bass notes have been omitted, in order to simplify the scheme. [Harmonies of Tones and Colours, The Arabian System of Music, page 21]
Bearden
"The 3-space symmetry of energy flow equilibrium is actually between the energy we input (from its external environment) to the system, and the energy that escapes from the system back into its external environment, either in its losses or in its loads. As we stated, equilibrium condition is a balance between ongoing entropic and negentropic operations." [Bearden, Energy from the Vacuum] [See Chemical Equilibrium]
Figure 13.14 - Equilibrium as Reciprocal Forces
Christ Returns - Speaks His Truth
"The DIVINE INTENTION of UNIVERSAL CONSCIOUSNESS was to express ITs own wholeness through creation by individualising, in physical form, each of the two equally balanced aspects of ITSELF and then bringing them together again in physical form, to experience the unity and wholeness of DIVINE CONSCIOUSNESS from which they originally took their individuality.
As they come together in love and unity of spirit and body, they discover the joy and ecstasy of UNIVERSAL CONSCIOUSNESS in equilibrium. (This entire process is clearly set out in Letters 5 and 7.) Therefore, the combination of the masculinity of the man and the femininity of the woman is essential to make a 'whole' drawn from the SOURCE of BEING. Out of this combination is formed a whole child." [Christ Returns - Speaks His Truth, Letter 4, page 17]
"Because 'FATHER-MOTHER' tools: electro-magnetism is in
equilibrium within
UNIVERSAL CONSCIOUSNESS
IT will never be detected within SPACE by scientists no matter how they may probe
space."
[Christ Returns - Speaks His Truth, Letter 5, page 19]
"I want you to notice that the equilibrium is impossible the moment that thought is introduced." [Christ Returns - Speaks His Truth, Letter 5, page 19]
"The equilibrium - the restraint between opposing IMPULSES - 'to move about' and 'remain bonded' creates an infinite spiral of self-contained energy. The SELF-CONTAINED ENERGY of MUTUAL RESTRAINT is beyond the power of individuality to even imagine." [Christ Returns - Speaks His Truth, Letter 5, page 20]
"You now know that the Father-Mother Creative Process and the tools of physical creation are all in a state of equilibrium within the Universal Dimension, but now that equilibrium is to be exploded to bring about
INDIVIDUAL FORM.
You know, too, that since the Infinite Eternal IMPULSES are contained in a state of mutual restraint, these IMPULSES are of an unimaginable intensity of energy - against which your atomic energy contained within the splitting of an atom is a mere 'pouff', an infinitesimal twitch of no importance.
I want you to fully realise all the foregoing, since your realisation of what happened at the time of the Big Bang will give you a glimpse of what happened at the time of the sundering of UNIVERSAL CONSCIOUSNESS to permit the creation of individual form to take place." [Christ Returns - Speaks His Truth, Letter 5, page 20]
"'Father' and 'Mother' Consciousness energies are IMPULSES both restrained within the UNIVERSAL DIMENSION and when they have been released from equilibrium they powerfully perform the work of creation." [Christ Returns - Speaks His Truth, Letter 5, page 23]
"DIVINE CONSCIOUSNESS is the re-union of the original IMPULSES within UNIVERSAL CONSCIOUSNESS which were released to become both the activity and the substance of creation at the moment of the Big Bang.
These IMPULSES were explosively divided and then came together in a state of mutual restraint. They were also destined to work forever in the created realm either separately, manifested as energies, or together restrained in equilibrium." [Christ Returns - Speaks His Truth, Letter 6, page 134]
"Bear in mind, at all times, that your SOURCE of BEING is in two states -
in equilibrium and activity.
The ACTIVE state of your SOURCE of BEING is the dimension in which you have been conceived and given individualization.
The EQUILIBRIUM of UNIVERSAL CONSCIOUSNESS is the dimension of perfect silence and stillness, in which the underlying IMPULSES OF CREATIVITY are locked together in an embrace of mutual restraint." [Christ Returns - Speaks His Truth, Letter 9, page 8]
Buckminster Fuller
“The vector equilibrium is the true zero reference of the energetic mathematics… the zerophase of conceptual integrity inherent in the positive and negative asymmetries that propagate the differentials of consciousness.” [Buckminster Fuller, Synergetics]
10.05 - The Neutral Zone or Undifferentiated Mind
10.10 - Stabilities
13.12 - Compound Equilibrium
13.12.1 - Disturbance of Equilibrium
2.11 - Beginning as Undifferentiated One
Balance
Book 02 - Chapter 07 - Instability, and the Illusion of Stability in Motion
Center of Gravity
Center of Gyration
Center of Moments
Center of Motion
Center of Oscillation
Center of Percussion
Center of Tension
Center of Velocity
center point balance equilibrium fulcrum
central line of rest
central line
Centralization
Chemical Equilibrium
Depolar
differential equilibrium
disequilibrium
disturbance of atomic equilibrium
DISTURBANCE OF EQUILIBRIUM - Snell
Disturbance of Equilibrium
disturbance of negative equilibrium
disturbance of sympathetic equilibrium
electrically divided equilibrium
Entropy
equative centre of equilibrium
equilibrium exchange
equilibrium level
equilibrium zone of pressure
Equilibrium
equilibrium-harmony
equivalence
equivalency
Figure 13.14 - Equilibrium as Reciprocal Forces
Figure 13.15 - Equilibrium as Musical Tonal Equivalents
Figure 13.16 - Compound Equilibrium States
Figure 18.13 - Scalar or Undifferentiated Mind Force
forces in equilibrium
Fulcrum
Harmonic
harmony of equilibrium
Hydrostatic equilibrium
Instability and Stability 102
Instability
Latent
Law of Assimilation
Law of Compensation and Equilibrium
law of etheric compensation and restoration
Le Chateliers Principle of Equilibrium Shift
List of Synonyms for Scalar
metastable
Mill of God
multiplication of resistant pressure
Neutral Equilibrium
Neutral
Potential
resistant equilibrium
Rest
restoration of equilibrium
Restorative Force
retrogression of harmonies
Rhythmic Balanced Interchange
Scalar
stable
stability
sympathetic disturbers of its equilibrium
sympathetic equilibrium
Syntropy
Theremin
thermodynamic equilibrium
transformation of equilibrium
Triple Point
Undifferentiated Mind
Undifferentiated
universal equilibrium level
universal equilibrium
Universal Heart Beat
unstable equilibrium
Vector Equilibrium
zero of equilibrium
Created by Dale Pond. Last Modification: Tuesday January 30, 2024 10:20:50 MST by Dale Pond. | 5,705 | 27,138 | {"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-2024-22 | latest | en | 0.953789 |
https://community.home-assistant.io/t/templeting-average-temp/304162 | 1,709,361,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00358.warc.gz | 180,649,941 | 6,664 | # Templeting Average temp
I’m having issues getting the average temp. I have 3 thermostats in my house and was using this temple to test to make sure it would give to correct number
{{ state_attr(‘climate.basement_heat’, ‘current_temperature’)
• state_attr(‘climate.basement_heat’, ‘current_temperature’)
• state_attr(‘climate.basement_heat’, ‘current_temperature’) /3 }}
Im getting back 154 as a number if my math is right and if and if all of these are at 66 then 66+66+66/3 = 66 . I dont know how it got 154 for a number?
It’s hard to know why your calculations are off without you posting the entire code where you are trying to calculate it.
This is the temples code. It changed it when I submitted. The dots are supposed to be +
Because the formula you created is doing this:
``````{{ 66 + 66 + 66/3 }}
``````
It’s not dividing the sum by 3, it’s only dividing the last number by 3.
Try this:
``````{{ ( state_attr('climate.basement_heat', 'current_temperature') +
state_attr('climate.basement_heat', 'current_temperature') +
state_attr('climate.basement_heat', 'current_temperature') ) / 3 }}
``````
Obviously your final template should include the names of each one of three thermostats (and not the same thermostat three times).
Have you considered the Min/Max integration?
@rysm83 The Average Sensor works pretty well for me. There are some features of the Min/Max sensor which I find undesireable, as also mentioned in the Average Sensor readme. Although I haven’t looked into the Min/Max Sensor in a while so it might have changed.
I use the Min/Max sensor for my temperature averages and it works great. I know some folks don’t like Min/Max but for this purpose it suited my needs perfectly.
1 Like | 424 | 1,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.879138 |
https://qualifiednursingtutors.com/1-1-suppose-that-a-car-rental-agency-offers-insurance-for-week-that-will-cost-10-per-day-a-minor-fender-bender-will-cost-1500-while-a-major-accident-might-cost-15000-in-repairs-witho-2/ | 1,627,657,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00070.warc.gz | 486,217,639 | 15,776 | # (1) (1) Suppose that a car rental agency offers insurance for week that will cost \$10 per day. A minor fender bender will cost \$ 1,500, while a major accident might cost \$ 15,000 in repairs. Without the insurance, you would be personally liable for an
(1) (1) Suppose that a car rental agency offers insurance for week that will cost \$10 per day. A minor fender bender will cost \$ 1,500, while a major accident might cost \$ 15,000 in repairs. Without the insurance, you would be personally liable for any damages. What should you do? Clearly, there are two decision alternatives: take the insurance or do not take the insurance. The uncertain consequences, or events that might occur, are that you would not be involved in an accident, that you will involved in a fender bender, or that you would be involved in a major accident. Assume that researched insurance industry statistics and found out that probability of a major accident is 0.05% and that the probability of a fender bender is 0.16%. What is the expected value decision? Would you choose this? Why or why not? What would be some alternate ways to evaluate risk?
(2) Suppose that the service rate to a waiting line system in 10 customers per hour (exponentially distributed). Analyze how the average waiting time is expected to change as the arrival rate varies from two to ten customers per hour ( exponentially distributed)
Please be sure your work is organized, legible, and your responses are substantive. You need to submit all details of your work including excel sheets used to arrive to the solution. It is not enough to attach your excel sheet. You MUST provide interpretation of results and describe conclusions
### Save your time - order a paper!
Get your paper written from scratch within the tight deadline. Our service is a reliable solution to all your troubles. Place an order on any task and we will take care of it. You won’t have to worry about the quality and deadlines
Order Paper Now | 433 | 1,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-31 | latest | en | 0.951628 |
http://infosciphi.info/poker/how-to-throw-dice-in-craps.php | 1,540,295,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00273.warc.gz | 163,116,231 | 12,685 | 4692
# How to throw dice in craps
How to play craps in such a way as to 'flip the game' in The technique of dice control So we want the initial conditions of a dice throw to be as uniform. Dice control in casino craps is a controversial theory where proponents claim that individuals can learn to carefully toss the dice so as to influence the outcome. A small but dedicated community of dice shooters claim proof of dice influencing in casino conditions. How to roll dice in craps, shooting craps like a master, and how to throw the dice at craps table with precision. Roll a die, with dice control to change the odds.
## Dice control
By taking forever to line up your dice in your lucky orientation and applying some sort of mojo to them, you delay the game, which frustrates the other players. Don't odds are capped on the maximum allowed win some casino allow the odds bet itself to be larger than the maximum bet allowed as long as the win is capped at maximum odds. Regardless of what else is occurring at the table e. Players must wait until next roll as long as a pass line point has been established players cannot bet don't come on come out rolls before they can make a new don't come bet. Each individual bet has the same payout as a single bet on the specific numbers, A lay bet is the opposite of a buy bet, where a player bets on a 7 to roll before the number that is laid.
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The annual profit from the gaming industry in the US is 18 billion dollars.
## CrapsPit.org
Casinos want all their games to have an edge for the house, and, for the most part, all their games do have house edges, some high, some low. Blackjack is a singular exception because it is possible for some players to get the mathematical edge at that game by playing perfect strategy and keeping track of the cards that have been played. Not many people can do that in the real world and blackjack makes a boatload of money for the casinos, mostly from people who think they can beat the game if they play this or that homegrown system long enough.
The mathematical underpinnings of all the other casino games rely on randomness; as in the spin of a roulette ball that then bounces randomly from pocket to pocket and where it stops nobody knows; the random shuffling of cards, usually by machine, at Let It Ride and Caribbean Stud before dealing out only a single round; the throwing of dice against a back wall that has foam-rubber pyramids to deflect and make them randomly bounce hither and yon.
Randomness is the key to these games and the math and money the casinos haul in is based on a random distribution of results. Given such a random distribution of results, in the long run the casinos must win, as they rarely pay off bets at the true odds. They must win and indeed they do. However, in craps if the game is "derandomized," the mathematical underpinnings of it must change.
Take a simple example. The 6 comes up five times for every six times a 7 comes up. But what if a player can learn how to play craps in a way that changes the game so that the 6 and 7 come up the same number of times, say, six times each?
Желаем всем мирного неба над головой и поменьше работы нашим военным в России и в мире. Спортсмены натягивают черлидершу эта сексуальная девушка справляется сразу с двумя половыми членами. I felt this hand grab my cock and I saw her dive her hands into my underwear and pull them down. I look forward to your bells. [end] Source: Human Events, p. Просто красавице очень нравиться заниматься онанизмом прямо у них на глазах с обязательным условием, что они снимут ее на камеру и выставят в интернет.
He saw young Jake, as a sheep, ready to be slaughtered.
### Details
Dice control in casino craps is a controversial theory where proponents claim that individuals can learn to carefully toss the dice so as to influence the outcome. A small but dedicated community of dice shooters claim proof of dice influencing in casino conditions.
The concept of such precision shooting claims to elevate craps from a random game of chance to a sport, not unlike bowling , darts , or pool. The concept of "controlled shooting" goes beyond simply " setting the dice " prior to shooting. It purports to involve limiting the rotational characteristics of the dice.
The theory is that if the dice are properly gripped and tossed at the correct angle they will land just before the back wall of the craps table, then gently touch the wall, greatly increasing the probability of their remaining on the same axis.
If executed properly and consistently this technique would be able to change the game's long-term odds from the house's favor to the player's favor. How to Control the Dice explains the math and science behind dice control. Stanford Wong , well-known advantage player and gaming author, also discusses dice control in his book Wong on Dice. Pawlicki and Jerry L. Jim Klimesh, director of casino operations for Indiana's Empress Casino Hammond believes it is sometimes possible to control the dice with certain throws that do not hit the back wall of the craps table.
In the army blanket roll, a player sets the dice on an axis and gently rolls or slides them down the table. If the shooter is successful, the dice will not leave the axis they are rolled on and will come to rest before hitting the back wall. A successful shooter would affect the odds significantly. But most casinos require that the dice touch the wall in order for a throw to be valid. The chances of altering the odds when the dice bounce off a surface of rubber pyramids are much slimmer, no matter what axis the dice were on before they hit.
## Craps natural crossword puzzle clue
We cannot vouch for accuracy as this is a joint project through the help and submissions of our datagrabber fans and awesome volunteers! List is Updated Frequently!!! According to its commercial jingle, consumers are advised to use what amount of Daisy sour cream? What punny prize is doled out each year by real Nobel laureates to highlight unusual research? The catastrophic BP oil spill resulted from the explosion of an offshore drilling rig with what name?
Established in , the James Beard Foundation Awards recognizes excellency in which of these fields? By definition, an analgesic is a type of medication intended primarily to do what? Introduced in ,Viking is a popular high-end brand of which of these household items?
## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,443 | 6,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.965692 |
https://media4math.com/TEKS-6-2B | 1,601,372,776,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00487.warc.gz | 534,352,080 | 8,537 | ## TEKS-6-2B
There are 40 resources.
Title Description Thumbnail Image Curriculum Topics
## Related Resources
### To see the complete collection of Videos on this topic, click on this link: https://bit.ly/377WM5K
Numerical Expressions
## Related Resources
### To see the complete collection of Videos on this topic, click on this link: https://bit.ly/3eZIU07
Rational Expressions
## MATH EXAMPLES--Absolute Value Functions in Tabular and Graph Form
### This set of tutorials provides 40 examples of absolute value functions in tabular and graph form. NOTE: The download is a PPT file.
Special Functions
## MATH EXAMPLES--Absolute Value
### The complete set of 21 examples that make up this set of tutorials. NOTE: The download is a PPT file.
Special Functions
## Related Resources
### To see the complete collection of glossary terms in the Visual Glossary, click on this link: https://media4math.com/Visual-Glossary
Numerical Expressions
## Related Resources
### To see the complete collection of glossary terms in the Visual Glossary, click on this link: https://media4math.com/Visual-Glossary
Special Functions
## Video Transcript: Integers: Integers and Absolute Value
### This is the transcript that goes with the video segment entitled Video: Integers: Integers and Absolute Value. To access the corresponding Video Tutorial, click here: https://media4math.com/library/video-tutorial-integers-integers-and-absolute-value
Numerical Expressions
## Video Transcript: Rational Numbers: Rational Numbers and Absolute Value
### This is the transcript that goes with the video segment entitled Video Tutorial: Rational Numbers: Rational Numbers and Absolute Value. To access the corresponding Video Tutorial, click here: https://media4math.com/library/video-tutorial-rational-numbers-rational-numbers-and-absolute-value
Rational Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/3gpwGj0
Numerical Expressions
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money
## Related Resources
### To see the complete collection of Math Examples on this topic, click on this link: https://bit.ly/2TfKgvl
Working with Money | 1,068 | 4,498 | {"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-2020-40 | latest | en | 0.66829 |
https://brilliant.org/problems/very-odd/ | 1,532,298,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594018.55/warc/CC-MAIN-20180722213610-20180722233610-00622.warc.gz | 611,787,331 | 10,998 | very odd!
Level pending
(1+3+5+7+.....+2p-1)+(1+3+5+7...+2q-1)=(1+3+5+....+2r-1)
What is the smallest possible value of $$p \times q \times r$$
× | 68 | 149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.50855 |
http://www.wearcam.org/cvpr2001mann180/node6.html | 1,669,914,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00372.warc.gz | 99,420,483 | 3,114 | Next: Smoothness and monotonicity constraints Up: . Introduction: Variable gain image Previous: Estimation in the presence
## Estimating camera response function
From (4), we may find by minimizing the sum of squared errors: =_x _y ( F(f_1(x)) - F(f_2( x )) + K ) ^2
Since can only be found up to a single unknown scalar constant, is fixed at zero3, where is the number of greyvalues (typically ).
Equation 4 gives a set of equalities of the form: , where , and is the number of pixels in one of the images . The first rows of are constructed by inserting in the column index corresponding to the pixel value of and inserting into the column index corresponding to the pixel value of : A(x+wy,f_1(x,y))&=&1
A(x+wy,f_2(x,y))&=&-1 where is the width of one of the images, and the last row of is all zeros except its last entry which is 1: A(L+1,N)=1 All unspecified entries of matrix are zero. Vector is constructed by placing the value in the first entries and in the last entry. This is an overdetermined system of equations.
The solution that minimizes the error in (9) is the maximum likelihood solution according to the noise model of (5), assuming [7], and is given by: ddF = 2A^TAF + 2A^TK = 0 giving F=(A^TA)^-1A^T (-K), assuming additive white Gaussian noise. This solution gives us a way of estimating the camera response function from two or more differently exposed pictures of overlapping subject matter.
Although this system is massively overdetermined, the constraints follow a comparametric form that admits solutions having sinusoidal components (e.g. solutions tend to be wavy'') [7]. (Indeed, comparametric forms are very similar to difference equations as can be seen from the form of (4) which constrains the function over an interval step size of allowing ripples in the solution.)
Subsections
Next: Smoothness and monotonicity constraints Up: . Introduction: Variable gain image Previous: Estimation in the presence
Steve Mann 2002-05-25 | 480 | 1,969 | {"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} | 3.21875 | 3 | CC-MAIN-2022-49 | latest | en | 0.876097 |
https://www.lotterypost.com/thread/250437 | 1,529,466,759,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00150.warc.gz | 869,390,960 | 15,110 | Welcome Guest
You last visited June 19, 2018, 11:30 pm
All times shown are
Eastern Time (GMT-5:00)
# what is it called ?
Topic closed. 3 replies. Last post 6 years ago by mathhead.
Page 1 of 1
New Member
Las Vegas
United States
Member #132336
September 1, 2012
8 Posts
Offline
Posted: September 4, 2012, 5:00 pm - IP Logged
lets say you have the number 93 . what's it called when you find all the additions that equal 93 ?
thank you
New Member
Las Vegas
United States
Member #132336
September 1, 2012
8 Posts
Offline
Posted: September 4, 2012, 5:43 pm - IP Logged
example 2+5+13+18+22+33 = 93
how do you find all the additions that equal 93 ?
Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
Posted: September 6, 2012, 1:58 pm - IP Logged
lets say you have the number 93 . what's it called when you find all the additions that equal 93 ?
thank you
It is called wasted time.
United States
Member #130795
July 25, 2012
80 Posts
Offline
Posted: September 6, 2012, 8:42 pm - IP Logged
It is called wasted time.
-----
winrockywin wrote: ``lets say you have the number 93 . what's it called when you find all the additions that equal 93 ?``
I don't know of any particular name. It is simply as you described: all combinations of n numbers that sum to x (93).
Perhaps some lottery "system" (aka "snake oil") has a special name for this kind of "filter" process. But in that case, SergeM has the "right" answer, IMHO.
-----
winrokcywin wrote: ``example 2+5+13+18+22+33 = 93[.] how do you find all the additions that equal 93 ?``
If you're adept at Excel VBA programming, you can write a macro to generate all possible selections and count the ones that sum to x (93). You might even write all qualifying sums to the worksheet.
But you still have not properly explained the problem. Are you trying to test the sum of all possible subsets of n numbers, that is sums of 2, 3, 4 numbers etc up to n? Or are you trying to test the sum of, say, any 6 of n numbers?
And the number of such subsets can be daunting, depending the size of n and the subsets to be summed.
For example, there are 13,983,816 sums of 6 out of 49 numbers. And there are 169,339 combinations that sum to 93.
It took less than 2 sec on my (ancient) computer to generate all of those sums and simply count the qualifying ones. YMMV.
Page 1 of 1 | 721 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-26 | latest | en | 0.874047 |
http://www.dmulholl.com/docs/pymatrix/guide.html | 1,590,471,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00259.warc.gz | 145,553,229 | 3,463 | # Pymatrix
A simple matrix library built in Python.
# Guide
Pymatrix exports a lightweight, general purpose matrix class, `Matrix`. A matrix element can be any arbitrary object that supports the required arithmethic and comparison operators. All of Python's native numeric types — integers, floats, complex numbers, and rational numbers — are supported.
Note that Pymatrix has been built for comfort, not for speed. If you have heavy-duty computational needs you should turn to a C-based alternative like NumPy instead.
## Instantiation
You can instantiate a matrix object directly, optionally specifying a fill value:
```m = Matrix(rows, cols, fill=0)
```
You can instantiate a matrix object from a list of lists using the `from_list()` static method:
```m = Matrix.from_list([
[1, 2, 3],
[4, 5, 6]
])
```
You can instantiate a matrix object from a string using the `from_string()` static method:
```string = '''
1 2 3/7
4/7 5 6
'''
m = Matrix.from_string(
string,
rowsep=None,
colsep=None,
parser=fractions.Fraction
)
```
Row separators default to newlines, column separators default to spaces. Leading and trailing whitespace is stripped from the string. Elements are parsed as fractions (rational numbers) by default.
You can instantiate an n x n identity matrix using the `identity()` static method:
```m = Matrix.identity(n)
```
The shortcut `matrix()` function supports the syntax of all three static methods:
```m = matrix([[1, 2, 3]])
m = matrix('1 2 3')
m = matrix(3)
```
## Iteration
Matrix objects are iterable. Iteration proceeds left-to-right by column, then top-to-bottom by row; i.e. the top-left element will be returned first, the bottom-right element will be returned last.
The iterator returns a tuple containing the row number, the column number, and the element:
```for row, col, element in matrix:
...
```
Alternatively, the `elements()` method returns an iterator over just the matrix elements:
```for element in matrix.elements():
...
```
## Indexing
Matrices are indexed as two-dimensional arrays:
```matrix[row][col] = element
element = matrix[row][col]
```
Note that indices are zero-based in accordance with programming convention rather than one-based in typical math style, i.e. the matrix's top-left element is `matrix[0][0]` rather than `matrix[1][1]`.
## Matrix Methods
Matrix objects support the following methods:
`.adjoint()`
Returns the adjoint matrix as a new object.
`.cofactor(row, col)`
Returns the specified cofactor.
`.cofactors()`
Returns the matrix of cofactors as a new object.
`.col(n)`
Returns an iterator over the specified column.
`.cols()`
Iterator returning a column iterator for each column in the matrix.
`.colvec(n)`
Returns the specified column as a new column vector.
`.copy()`
Returns a copy of the matrix.
`.cross(other)`
Returns the cross/vector product of the matrix with `other` as a new matrix. The cross product is only defined for pairs of 3-dimensional column vectors.
`.del_col(col)`
Returns a new matrix with the specified column deleted.
`.del_row(row)`
Returns a new matrix with the specified row deleted.
`.det()`
Returns the determinant of the matrix.
`.dir()`
Vectors only. Returns the unit vector in the direction of the vector.
`.dot(other)`
Returns the dot/scalar product of the matrix with `other`. The dot product is only defined for pairs of vectors.
`.elements()`
Returns an iterator over the matrix's elements.
`.equals(other, delta=None)`
If `delta` is `None`, two matrices are equal if they are the same size and their corresponding elements are equal, i.e. `e1 == e2`.
If `delta` is not `None`, two matrices are equal if they are the same size and their corresponding elements agree to within `delta`, i.e. `abs(e1 - e2) <= delta`.
`.inv()`
Returns the inverse matrix if it exists, otherwise raises `MatrixError`.
`.is_invertible()`
True if the matrix is invertible. Note that determining whether a matrix is invertible is as computationally expensive as actually calculating the inverse.
`.is_square()`
True if the matrix is square.
`.len()`
Vectors only. Returns the length of the vector.
`.map(func)`
Returns a new matrix formed by mapping `func` to each element.
`.minor(row, col)`
Returns the specified minor.
`.rank()`
Returns the rank of the matrix.
`.ref()`
Returns the row echelon form of the matrix.
`.row(n)`
Returns an iterator over the specified row.
`.rowop_add(r1, m, r2)`
In-place row operation. Adds `m` times row `r2` to row `r1`.
`.rowop_multiply(row, m)`
In-place row operation. Multiplies the specified row by the scalar `m`.
`.rowop_swap(r1, r2)`
In-place row operation. Interchanges the two specified rows.
`.rows()`
Iterator returning a row iterator for each row in the matrix.
`.rowvec(n)`
Returns the specified row as a new row vector.
`.rref()`
Returns the reduced row echelon form of the matrix.
`.trans()`
Returns the transpose of the matrix as a new object.
## Module Functions
The `pymatrix` module exports the following functions:
`dot(u, v)`
Returns `u . v` - the inner/scalar/dot product of the vectors `u` and `v`.
`cross(u, v)`
Returns `u x v` - the vector/cross product of the 3D column vectors `u` and `v`.
`matrix()`
Shortcut function for instantiating `Matrix` objects; supports the syntax of the various static instantiation methods.
## Exceptions
An invalid operation on a matrix object will raise a `MatrixError` exception. | 1,299 | 5,454 | {"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-2020-24 | latest | en | 0.722925 |
https://education.casio.co.uk/blog/visualising-probability-distributions-on-the-fx-cg50/ | 1,726,346,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00099.warc.gz | 201,316,063 | 14,089 | Visualising Probability Distributions on the fx-CG50 - Casio Calculators
# Visualising Probability Distributions on the fx-CG50
Feb 2023 Medium Read: 5 Min
Ask any teacher, and they’ll likely agree: displaying probability distributions has typically been a challenge on many graphing calculators. Fortunately, that’s all changing. With the new Distribution App on the fx-CG50, it’s now possible to visualise both discrete and continuous distributions with a graph. We spoke with experienced Maths Education Consultant Gerard Dummett to learn more about this big leap forward.
## How the Distribution App enhances probability visualisation
It’s well known that many graphic calculators have struggled to display probability distributions, particularly discrete probability distributions. But with the new Distribution App, the fx-CG50 overcomes this challenge and provides students with a clear visual representation, which they can explore and interrogate to improve their understanding.
The second big step forward are changes to how information is inputted. Gerard explains that an input screen was used to enter the parameters. Then, once inputted, students pressed the button and either saw a probability or, if in a continuous distribution, they also saw a graph. But if they wanted to change it, they had to go back to the input screen and start over again.
Now, with the updated app, students are on the input screen; you’ve added the inputs and then want to change these inputs. You stay on the screen, and the input is immediately above where you are, immediately above the graph.
## Exploring the teaching possibilities with the Distribution App
These two big improvements lead to many teaching possibilities. First, when dealing with a Normal distribution the visuals in the distribution app provide an ideal way for students to explore that symmetry.
Gerard explains with the example of standardised Normal distribution. In that case, if we go one standard deviation on either side of the mean, we can see an important property of the normal distribution, which is 68.3% of the area covered.
They’ll also see it’s symmetrical. That’s quite important to an understanding of the normal distribution.
If you are dealing with the binomial distribution, by contrast, it is not generally symmetrical. It can be, but it typically is not. And so you want to see that values on either side of a given value might not be equal in the way they are in a normal distribution. And just getting the student to visualise that is quite important, and you can use it for teaching, which is one important advantage of the new distribution app.
### Ease of use gives students the confidence to explore
Another teaching point is the ease with which students can do things. For example, as Gerard explains, a question might ask the probability that the height of people in school falls within given bounds. If we’re looking at the distribution of that height, what is the probability between 1.48 and 1.53 metres, given that they are distributed in a particular way? Students can visualise the answer to that on the app.
Students could then explore what would happen if the value was 1.47 instead of 1.48. They can see what happens with that change, and the ease of use means students quickly build confidence when exploring new concepts.
They can then take it further, with more complicated questions. For instance, if the height is distributed equally on either side of the mean, what are the limits for us to cover 80% of students, which is 40% on either side of the mean? That is incredibly easy to do on the calculator now. Students type in your 80% in the probability; it returns the limits for you. The calculator draws the graph to show you how it looks, which is great for continuous distribution, especially for continuous symmetrical distributions.
But, with discrete distribution, where it might not be able to take the exact value you want to get that probability, the fact that it’s not continuous makes it harder because you’re not going to get the exact probability. The calculator helps here by finding the best probability between limits for students, and students can decide how to interpret that.
In this case, students will want to know the value of the variable that will cause the hypothesis to be either accepted or rejected. That value is typically 5% or 2.5%, but the problem is 5% or 2.5% may only correspond to a specific value if the distribution is continuous.
With discrete distributions, asking the calculator to provide the value of 5% means students can see in the visual representation if it’s too much or not. For example, if it comes out at 5.3% and they only wanted 5%, they’ll know they need to select a smaller value left tail) or larger value (right tail). In the case illustrated it has to be 12 instead of X being 11.
On the diagram, this distinction is very obvious. It’s an area of maths that students often need help with; the ability to switch values, see the graph and interpret it is so powerful.
Are you looking to introduce the fx-CG50 in the classroom? We provide free skills training sessions to help you build confidence using graphic calculators in the classroom, which, in turn, will increase your students’ engagement. Sign up for your free session here! | 1,092 | 5,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.917524 |
https://www.experts-exchange.com/questions/21111143/Running-Debugged-vs-Non-Debugged.html | 1,506,411,601,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695113.88/warc/CC-MAIN-20170926070351-20170926090351-00513.warc.gz | 774,214,820 | 27,662 | Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
x
Solved
# Running Debugged vs. Non-Debugged
Posted on 2004-08-28
Medium Priority
217 Views
Hi,
I have a program that simulates a game many times and returns the percent of the time that each team won (8 teams). I believe I have all the logics correct, but I'm finding some odd results. When I run the program stepping through the lines (in debug mode obviously), it distributes the wins correctly. However, when I simply run the program through for 100 trials, one team will win all 100 times. And then if I run it again, another team will win 100 times. Then again, and team 1 will win 84 times and team 2 will win 16 times and the other 6 teams won't win at all. Now if I run the program through a million trials, it seems to distribute nicely. I'm really confused as to why this would ever happen. My logic is correct since as I said before, running the program debugged works fine. This is C# dot net console program btw.
Here's my code:
using System;
namespace Project1
{
class Class1
{
public static void Main(string[] args)
{
int gameWinner;
double percentWon;
int[] teamGames = new int[8];
for (int xx = 1; xx <= 100; xx++)
{
gameWinner = Class1.RunGame();
teamGames[gameWinner]++;
}
for (int yy = 1; yy <= 8; yy++)
{
percentWon = (teamGames[yy - 1]/100.000);
Console.WriteLine("Team " + yy + " = " + percentWon);
}
}
public static int RunGame()
{
int team1, team2, teamondeck, pointmod, winner;
int[] teamWins = new int[8];
team1 = 1;
team2 = 2;
teamondeck = 3;
pointmod = 1;
Random r = new Random();
while (true)
{
if (r.NextDouble() >= .5)
{
teamWins[team1 - 1] = teamWins[team1 - 1] + pointmod;
if (teamWins[team1 - 1] >= 7)
{
winner = team1 - 1;
break;
}
team2 = teamondeck;
teamondeck++;
}
else
{
teamWins[team2 - 1] = teamWins[team2 - 1] + pointmod;
if (teamWins[team2 - 1] >= 7)
{
winner = team2 - 1;
break;
}
team1 = team2;
team2 = teamondeck;
teamondeck++;
}
if (teamondeck > 8)
{
teamondeck = 1;
pointmod = 2;
}
}
return winner;
}
}
}
I'd appreciate any help. Thanks!!
0
Question by:schwarz2
[X]
###### Welcome to Experts Exchange
Add your voice to the tech community where 5M+ people just like you are talking about what matters.
• Help others & share knowledge
• Earn cash & points
• Learn & ask questions
LVL 3
Accepted Solution
GrumbleBot earned 500 total points
ID: 11925787
I found by making the variable r a global variable your program works correctly. Not sure if I can explain why it doesn't work the way that you have it. But if i basically set up a random numbe rin a fuction and fill a listbox continually I get the same number for many consecutive values then it changes and repeats those many times , on and on.
Well here is your code modified and it seems to work now.
using System;
namespace Project1
{
class Class1
{
static Random r;
public static void Main(string[] args)
{
int gameWinner;
double percentWon;
int[] teamGames = new int[8];
r = new Random(); //((int)DateTime.Now.Ticks)
for (int xx = 1; xx <= 100; xx++)
{
gameWinner = Class1.RunGame();
teamGames[gameWinner]++;
}
for (int yy = 1; yy <= 8; yy++)
{
percentWon = (teamGames[yy - 1]/100.000);
Console.WriteLine("Team " + yy + " = " + percentWon);
}
}
public static int RunGame()
{
int team1, team2, teamondeck, pointmod, winner;
int[] teamWins = new int[8];
team1 = 1;
team2 = 2;
teamondeck = 3;
pointmod = 1;
while (true)
{
if (r.NextDouble() >= .5)
{
teamWins[team1 - 1] = teamWins[team1 - 1] + pointmod;
if (teamWins[team1 - 1] >= 7)
{
winner = team1 - 1;
break;
}
team2 = teamondeck;
teamondeck++;
}
else
{
teamWins[team2 - 1] = teamWins[team2 - 1] + pointmod;
if (teamWins[team2 - 1] >= 7)
{
winner = team2 - 1;
break;
}
team1 = team2;
team2 = teamondeck;
teamondeck++;
}
if (teamondeck > 8)
{
teamondeck = 1;
pointmod = 2;
}
}
return winner;
}
}
}
Hope this helps.
Good luck
0
Author Comment
ID: 11926547
Excellent, GrumbleBot!! This was driving me crazy .. I guess C# doesn't re-seed the random variable when it's re-created. Very strange, but I'm glad you caught it. Thanks millions.
0
## Featured Post
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question
C# And Nullable Types Since 2.0 C# has Nullable(T) Generic Structure. The idea behind is to allow value type objects to have null values just like reference types have. This concerns scenarios where not all data sources have values (like a databa…
Having new technologies does not mean they will completely replace old components. Recently I had to create WCF that will be called by VB6 component. Here I will describe what steps one should follow while doing so, please feel free to post any qu…
This course is ideal for IT System Administrators working with VMware vSphere and its associated products in their company infrastructure. This course teaches you how to install and maintain this virtualization technology to store data, prevent vuln…
Do you want to know how to make a graph with Microsoft Access? First, create a query with the data for the chart. Then make a blank form and add a chart control. This video also shows how to change what data is displayed on the graph as well as form…
###### Suggested Courses
Course of the Month4 days, 23 hours left to enroll
#### 670 members asked questions and received personalized solutions in the past 7 days.
Join the community of 500,000 technology professionals and ask your questions. | 1,552 | 5,517 | {"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-2017-39 | longest | en | 0.865664 |
https://www.teacherspayteachers.com/Product/Multiplication-Comparison-Models-Ratio-Tape-Diagram-Match-Games-4157757 | 1,553,484,326,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203548.81/warc/CC-MAIN-20190325031213-20190325053213-00478.warc.gz | 911,943,352 | 24,525 | # Multiplication Comparison Models/ Ratio/ Tape Diagram Match Games
Subject
Resource Type
Common Core Standards
Product Rating
4.0
4 Ratings
File Type
PDF (Acrobat) Document File
23 MB|29 pages
Share
Product Description
Multiplicative Comparison begins in 4th grade, returns in 5th grade and evolves into Ratios in 6th grade. This match game offers a word problem or an algebraic expression on one card, and a model (often called a tape diagram in grade 6) on its matching card. The 4 Sets were designed around the 4. OA standards:
• 4.OA.A1/5.OA.A1 Match Game with Simple Expressions
• 4.OA.A2/6.RP.A1; 6.RP.A3 Comparison Word Problems and Matching Models
• 4.OA.A2/6.RP.A1; 6.RP.A3 More Word Problems (Additive)
• 6.EE.A.2A; 6.EE.B6 The algebraic expressions and the modeling of the word problems fit these standards as well.
• 2.OA.A1 & 3.OAA2 Simple Bar Models: Students must distinguish multiplicative comparison from simple comparison (subtraction).
INCLUDES:
4 Different Match Games of 12-18 matching card pairs.
Teacher page for each game
Visual Answer keys so students can learn from them if needed... only three to a page so they can check an answer and self-correct before seeing ALL of the answers.
These cards offer SO MANY OPTIONS!
• Print as a packet or single worksheets. (In fourth grade, I do this first.)
• Print desired quantity of model matching card sets; Cut apart model cards, separating model or visual from equations; Mix up cards for students to match; Answer Keys are provided for self-checking.
• EXTENSION: Mix up multiple sets. Blank cards are provided for students to create their own! (They can create their own from scratch, or write a problem for a model card, or a model for a problem card!
• MODIFICATION: Give students fewer cards to match. I intentionally repeated many numbers so they couldn't just match solely by the numbers.. in most cases.
Thank you for EXPLORING MATH with me!! If you like this resource PLEASE Review and Pin it! Remember, TPT rewards us for reviewing!!
-Math Viking
Fraction Division Visual Match Game Task Cards and Scoot
Sort & Solve NUMBERLESS Fraction Word Problems
Sort & Solve NUMBERLESS Word Problems Grades 3-5
Grade 4-6 Number of the Day REASONING templates for whole numbers, fractions and decimals
Place Value Chart BOARD TOPPER poster
Fractions in the Pie Hands-on Division Game FREEBIE
Division Area Model Level I/II 4.NBT.B6
Division Area Model Level III 5.NBT.B6
FORTNITE Superhero Race Car MASTERMIND Smartboard/hands on Reasoning Game
FREEBIE Multiplication Fact Sorts: Simple and Sophisticated
Total Pages
29 pages
Included
Teaching Duration
N/A
Report this Resource
\$5.50 | 661 | 2,678 | {"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-2019-13 | latest | en | 0.887832 |
https://starlink.eao.hawaii.edu/devdocs/sun210.htx/sun210ss135.html | 1,660,811,580,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00129.warc.gz | 497,702,877 | 2,891 | ### AST_MATRIXMAP
Create a MatrixMap
#### Description:
This function creates a new MatrixMap and optionally initialises its attributes.
A MatrixMap is a form of Mapping which performs a general linear transformation. Each set of input coordinates, regarded as a column-vector, are pre-multiplied by a matrix (whose elements are specified when the MatrixMap is created) to give a new column-vector containing the output coordinates. If appropriate, the inverse transformation may also be performed.
#### Invocation
RESULT = AST_MATRIXMAP( NIN, NOUT, FORM, MATRIX, OPTIONS, STATUS )
#### Arguments
##### NIN = INTEGER (Given)
The number of input coordinates, which determines the number of columns in the matrix.
##### NOUT = INTEGER (Given)
The number of output coordinates, which determines the number of rows in the matrix.
##### FORM = INTEGER (Given)
An integer which indicates the form in which the matrix elements will be supplied.
A value of zero indicates that a full NOUT x NIN matrix of values will be supplied via the MATRIX argument (below). In this case, the elements should be given in row order (the elements of the first row, followed by the elements of the second row, etc.).
A value of 1 indicates that only the diagonal elements of the matrix will be supplied, and that all others should be zero. In this case, the elements of MATRIX should contain only the diagonal elements, stored consecutively.
A value of 2 indicates that a " unit" matrix is required, whose diagonal elements are set to unity (with all other elements zero). In this case, the MATRIX argument is not used.
##### MATRIX( $\ast$ ) = DOUBLE PRECISION (Given)
The array of matrix elements to be used, stored according to the value of FORM.
##### OPTIONS = CHARACTER $\ast$ ( $\ast$ ) (Given)
A character string containing an optional comma-separated list of attribute assignments to be used for initialising the new MatrixMap. The syntax used is identical to that for the AST_SET routine.
##### STATUS = INTEGER (Given and Returned)
The global status.
#### Returned Value
##### AST_MATRIXMAP = INTEGER
A pointer to the new MatrixMap.
#### Notes:
• In general, a MatrixMap s forward transformation will always be available (as indicated by its TranForward attribute), but its inverse transformation (TranInverse attribute) will only be available if the associated matrix is square and non-singular.
• As an exception to this, the inverse transformation is always available if a unit or diagonal matrix is specified. In this case, if the matrix is not square, one or more of the input coordinate values may not be recoverable from a set of output coordinates. Any coordinates affected in this way will simply be set to the value zero.
• A null Object pointer (AST__NULL) will be returned if this function is invoked with STATUS set to an error value, or if it should fail for any reason.
#### Status Handling
The protected interface to this function includes an extra parameter at the end of the parameter list descirbed above. This parameter is a pointer to the integer inherited status variable: " int $\ast$status" . | 648 | 3,122 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "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-2022-33 | latest | en | 0.760761 |
http://www.jiskha.com/display.cgi?id=1195173055 | 1,498,438,437,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320595.24/warc/CC-MAIN-20170625235624-20170626015624-00215.warc.gz | 580,003,070 | 4,151 | # physics
posted by .
The speed of propagation of a sound wave in air at 27 degrees Celsius is about 350 m/s. Calculate, for comparison, v_rms for nitrogen (N2) molecules at this temperature. The molar mass of nitrogen is 28.0 g/mol.
• physics -
You will find the two velocites to be roughly the same.
The mean kinetic energy of the molecules is
Eav = (3/2) kT = 6.21*10^-21 J
That equals (1/2) M V^2 (where V is the rms average).
M = (28.0 g/mole)/[(6.02*10^23 molecules/mole)* 1 kg/1000 g]
= 4.65*10^-26 kg/molecule
V^2 = 2 Eav/M = 2.67*10^5 m^2/s^2
Vrms = sqrt V^2 = 517 m/s
Check my reasoning and numbers.
• physics - | 217 | 627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-26 | latest | en | 0.842386 |
http://www.cteonline.org/portal/default/Curriculum/Viewer/Curriculum?action=2&cmobjid=195837&view=viewer&refcmobjid=132950 | 1,369,506,548,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706082529/warc/CC-MAIN-20130516120802-00089-ip-10-60-113-184.ec2.internal.warc.gz | 411,281,695 | 16,946 | Lesson Plan Industry Sector
Engineering & Design
## Mousetrap Powered Racer 3/4 Modeling 2
### Lesson Plan Overview / Details
Wheels and Axles (circles) In this lesson we will be working on making the wheels for the mousetrap vehicle in AutoCAD. This will require the students to begin thinking about the specific dimensions of their wheels and axle.
One Class Period
55 Minutes
### Objectives and Goals
Students will model the wheels and axles of their mousetrap vehicle in AutoCAD.
### Activities in this Lesson
• When Wheels FAIL! - Hooks / Set
• Wheels and Axles - Demo / Modeling
With theLCD projector broadcasting and the student machines running AutoCAD, you will demonstrate to the students how to make a circle in AutoCAD. The students should be following along with the teacher on their own computers to form the wheel and axle solids.
-In AutoCAD click on the circle icon in the menu bar at the top of the screen.
-Orient your curser to the workspace and (left) click the mouse button to begin drawing the circle. At this point, you can see a dialog box appear for you to enter the specific radius or diameter that you would like to use as the value for your circle.
-The Command: line at the bottom of the screen wil indicate wether you are entering a R (radius) or D (diameter). Make sure that you are entering the correct value for your particular wheel.
-Click ENTER to accept the measurement.
-Click ESC to deselect your shape.
-At this point, you have a 2 dimensional circle on your workspace.
-On the Command Line at the bottom of your screen, type extrude (ext) then enter.
-Now you may either type in the thickness of the wheel or manually drag the vehicle's wheel to the thickness that you desire.
REMEMBER: IF YOU ARE GOING TO BE USING A CD, RECORD ALBUM, PLASTIC LID, ETC., YOU WILL NEED TO ACCURATELY MEASURE THE EXISTING MATERIAL TO DETERMINE ITS THICKNESS.
-Press Esc to deselect
### Assessment
Assessment Types:
Demonstrations, Portfolios, Observations,
Students will need to use these files when they begin to 'assemble' their mousetrap cars in a later lesson. Make sure that the students are cognisant of the diameter/radius of their wheels because they will need to be able to use these measurements to calculate their projected distance their car will cover. | 525 | 2,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-20 | latest | en | 0.851648 |
https://community.esri.com/t5/arcgis-network-analyst-questions/travel-time-of-vehicle-routing-problems-gives/m-p/655261/highlight/true | 1,675,760,272,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00847.warc.gz | 184,082,532 | 43,713 | # TRAVEL TIME OF VEHICLE ROUTING PROBLEMS GIVES WRONG RESULTS
2322
6
01-27-2014 06:10 AM
New Contributor
Hello,
I have a network dataset with historical traffic.
Every time that I run a vehicle routing problem, the solver gives me results with the time-neutral travel time.
This is not happening when I do Routing. The travel times of routes depend on the time of the day.
What could be a reason for the fact that historical traffic data are only read with the Route option but are not read in the VRP option?
Best,
Fabio
Tags (2)
6 Replies
by
Occasional Contributor
Hi,
VRP solver currently does not support optimization with historical traffic information. So the solutions returned in NA classes are all based on time-neutral time. After VRP solve finishes, you can click on Directions button. It will call Route solver to generate the travel time with the historical traffic information based on the route start time and sequence returned in VRP solver.
Thanks.
Anna
New Contributor
THANK YOU very much Anna
New Contributor
But, even though the solver uses neutral travel time, the best path is still depending on traffic condition right? I think the solver knows the streets that have more traffic
Thanks
Fabio
by
Occasional Contributor
I'm not sure where you get the network dataset. Usually, when we prepare the network dataset, we take the weekly average travel speed on a street and divide it by the distance to get the time neutral travel time. If the street has a lot of traffic, the weekly average travel speed will be slower. So even with time neutral solve, VRP solver still knows where the high traffic streets are.
Thanks.
Anna
New Contributor
Hello. In this problem. can solve?
I have a network dataset with historical traffic.
Every time that I run a vehicle routing problem, the solver gives me results with the time-neutral travel time.
This is not happening when I do Routing. The travel times of routes depend on the time of the day.
What could be a reason for the fact that historical traffic data are only read with the Route option but are not read in the VRP option?
Esri Contributor
Hello,
What is described by Anna is still accurate for the VRP Solver. It uses a time-neutral travel time for solving and that is what is returned in the NA classes. The directions will reflect the traffic if included in the network dataset.
Thanks,
Heather | 516 | 2,393 | {"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-2023-06 | latest | en | 0.92809 |
http://www.education.com/activity/article/Chalk_Outdoor_Classification_third/ | 1,477,384,292,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720000.45/warc/CC-MAIN-20161020183840-00074-ip-10-171-6-4.ec2.internal.warc.gz | 420,518,913 | 29,050 | Activity:
# Chalk it Up! An Outdoor Classification Activity
4.0 based on 3 ratings
### What You Need:
• Sidewalk chalk
### What You Do:
1. Bring your chalk along with you as you visit a neighborhood park or your own front yard. Ask your child to name what they see in the park. Then encourage her to group what she sees into categories. Ask your child what title she would give each category (for instance, “Things that are green”, “Things that are bumpy,” etc).
2. Using the chalk, write the category on the ground in a wide open space, whether it’s your driveway or the sidewalk at the park.
3. Now do the same with two or three more categories your child comes up with. If your child is not a reader yet, feel free to make an illustration of the category on the ground alongside the word.
4. Now it's time to sort and find objects that fit the categories listed. You can either place the actual object under the category title (for example, a leaf under the category of “Things that are green) or write the object’s name or draw its picture (for example, a stop sign under the category of “things made of metal”).
5. Once the categories are substantially filled in, look at what objects you found for each group. Which category had the most “things”? Which had the least? Which “things” fit into more than one group?
6. Rewrite the list of objects that appeared on multiple lists and make a Venn diagram (explained in the next step) to take this classification activity to another level.
7. To make a Venn diagram, draw two large circles on the ground that overlaps in the middle. In one circle, avoiding the overlapping part, write the first list of objects, and on the second side of the circle, write the other. Objects that can fit in both categories are written in the overlapping part shared by the two circles.
8. On the walk home, talk about the discoveries that your child made. What outdoor “things” overlapped in two categories? Where these “things” found in nature or man-made? Or both? Encouraging your child to share her observations is a great way to reinforce the exercise and ensure that your child will be a master classifier in no time! | 477 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-44 | latest | en | 0.95299 |
https://www.reference.com/web?q=number+model+in+math&qo=contentPageRelatedSearch&o=600605&l=dir | 1,591,272,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00250.warc.gz | 875,612,486 | 28,462 | ARTICLES
A number model in math is a sentence that illustrates how the parts of a number story are related. The equation may include addition, subtraction, division and multiplication and may be expressed as words or in number fo...
www.reference.com/world-view/number-model-math-4919107bd2f566de
A verbal model is a representation of a math problem in words. A verbal model is the step between translating a verbal phrase in math to a math model.
www.reference.com/article/verbal-model-math-52d2d85b376e8d37
The bar model is a visual representation of a number using unit cubes, as stated on SplashMath.com. One cube is equal to one. One line is equal to 10 cubes. One sheet is equal to 100 cubes or 10 lines.
www.reference.com/article/bar-model-694e5a35444d50de
Related Search
SIMILAR ARTICLES
In math, expanded form can refer to any type of expression, equation or notation that is completely broken down into its individual parts. Expanded form is commonly used in teaching students place value and factoring.
www.reference.com/world-view/expanded-form-math-6e5dacfc41994817
"Digits" are the symbols or characters used to represent a number visually. A number like five contains one digit, whereas a number like 555 contains three digits. This is easily seen when they are written as numerals: 5...
www.reference.com/world-view/digit-math-4616644a88f876d9
In math, the multiples of a number include all the numbers that result from multiplying that number by any whole number. A number's multiples include the number itself plus the numbers that are divisible by it without le...
www.reference.com/world-view/multiple-math-ba9bfbd8282ce9a7
A term in mathematics is defined as a number, variable or number-variable combination in an algebraic expression or equation. Terms are separated from each other by a plus, minus or equal sign.
www.reference.com/world-view/term-math-a75e4e8b2755b61c
Related Search | 449 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.881031 |
http://www.docstoc.com/docs/166475762/Overview-of-Alaska%e2%80%99s-Proposed-Mathematics-Standards | 1,432,446,935,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927843.59/warc/CC-MAIN-20150521113207-00210-ip-10-180-206-219.ec2.internal.warc.gz | 411,808,983 | 43,823 | # Overview of Alaska’s Proposed Mathematics Standards.pptx by pptfiles
VIEWS: 6 PAGES: 38
• pg 1
``` Overview of Alaska’s Proposed
Mathematics Standards
Alaska Department of Education & Early Development
Topics for the Webinar
Encourage Changes to Practice
Understand Alaska’s Mathematics Standards Organization
Knowledge of the Standards for Mathematical Practice
Examine the Standards for Mathematical Content
Next Steps http://www.eed.state.ak.us/standfaqs.html
2
Encourage Changes to Practice
The proposed mathematics standards support
improved curriculum and instruction due to increased:
FOCUS, via critical areas at each grade level
COHERENCE, through carefully developed
progression of topics within and across grades
RIGOR, including a focus on College and Career Readiness
and Standards for Mathematical Practice throughout K-12
3
Mathematics Standards Organization
Introduction/Description
Standards for Mathematical Practice
Taught in all grade levels
Standards for Mathematical Content
High school standards are in domains
Glossary for the Alaska Mathematics Standards
4
Standards for Mathematical Practice
Both a goal and a vehicle -
The goal is to move students along a continuum –
always deepening their use of the mathematical practices.
They are a vehicle for learning the content standards.
5
Background Information
The National Research
National Council of
Council’s Report
Teachers of Mathematics
Principal and (2001)
(2000)
Problem Solving
Reasoning and Proof
Connections
Communication
Representation
6
Standards for Mathematical Practice
1. Make sense of problems and persevere in solving them
2. Reason abstractly and quantitatively
3. Construct viable arguments and critique the
reasoning of others
4. Model with mathematics
5. Use appropriate tools strategically
6. Attend to precision
7. Look for and make use of structure
8. Look for and express regularity in repeated reasoning
Turn to
Page 4
7
Mathematical Practice Descriptions
Turn to
Page 11
Read the paragraph description for Practice #5.
Notice the verbs used to describe actions students will take.
8
5. Use appropriate tools strategically
Page 11
Notice the continuum of development.
9
In October, John’s PFD check for \$1100
arrives. His parents give him 2 choices so the
rest can be saved for post-secondary options.
• Choice 1: Spend 1/5 of the PFD
• Choice 2: Spent 15% of the PFD
Which would give him the most spending
10
Standards for Mathematical Practice
Graphic
A
B
C
11
K-8 Standards for Mathematical Content
Turn to
Instructional Domains
Focus Counting, Cardinality and Ordinality
Standards for Operations and Algebraic Thinking
Mathematical Number and Operations in Base Ten
Content Measurement and Data
Number and Operations – Fractions
Ratios and Proportional Relationships
Domains The Number System
Clusters Expressions and Equations
Functions
Standards Statistics and Probability
12
Focus – Critical Areas
Turn to
Page 16
For each grade level, there are typically two
to four critical areas for instructional focus.
Let’s look at Kindergarten Instructional Focus
13
Kindergarten Instructional Focus
Instructional time should focus on two critical areas:
(1) representing, relating, and operating on
whole numbers, initially with sets of objects;
(2) describing shapes and space.
More learning time in Kindergarten should be
devoted to number than to other topics.
Page 16
14
Examining a critical area further
For each of the critical areas
more detailed information
Kindergarten students should
equations and writing is
encouraged but not required.
Remember the Glossary
Page 16
15
K–2
measurement using whole number quantities
Multiplication and division of whole numbers
3–5
and fractions
Ratios and proportional reasoning;
6
early expressions and equations
Ratios and proportional reasoning;
7
arithmetic of rational numbers
8 Linear algebra
16
Coherence - Domains
Mathematics - Domain Progression Overview by Grade
K 1 2 3 4 5 6 7 8 HS
Counting, Cardinality and
Ordinality
Ratios and Proportional Number &
Relationships Quantity
Number and Operations in Base Ten
Number and Operations - Fractions Number System
Expressions and Equations Algebra
Operations and Algebraic Thinking Functions Functions
Geometry Geometry
Statistics
and
Measurement and Data Statistics and Probability
Probability
Domains – large groups of related standards.
They may begin and end in different grades.
17
Coherence - Clusters
Clusters – groups of closely related standards
inside domains, subsets of domains.
Let’s look at the grades 3-5 clusters. Turn to
Page 30
18
Operations and Algebraic Thinking (3-5)
• Represent and solve • Use the four operations • Write and interpret
problems involving with whole numbers to numerical expressions.
multiplication and division. solve problems.
• Understand properties of • Analyze patterns and
multiplication and the • Gain familiarity with relationships.
relationship between factors and multiples.
multiplication and division.
• Multiply and divide within • Generate and analyze
100. patterns.
• Solve problems involving
the four operations, and
identify and explain
patterns in arithmetic.
Pages 30-31
19
Number and Operations in Base Ten (3-5)
• Use place value • Generalize place value • Understand the place
understanding and understanding for multi- value system.
properties of operations digit whole numbers.
to perform multi-digit • Perform operations with
arithmetic. • Use place value multi-digit whole numbers
understanding and and with decimals to
properties of operations hundredths.
to perform multi-digit
arithmetic.
Pages 33-34
20
Number and Operations—Fractions (3-5)
• Develop understanding • Extend understanding of • Use equivalent fractions
of fractions as numbers. fraction equivalence and as a strategy to add and
ordering. subtract fractions.
• Build fractions from unit • Apply and extend
fractions by applying and previous understandings
extending previous of multiplication and
understandings of division to multiply and
operations on whole divide fractions.
numbers.
• Understand decimal
notation for fractions, and
compare decimal fractions.
Pages 35-37
21
Measurement and Data (3-5)
• Solve problems involving • Solve problems • Convert like
measurement and estimation of involving measurement measurement units
intervals of time, liquid volumes, and conversion of within a given
and masses of objects. measurements from a measurement system.
• Represent and interpret data. larger unit to a smaller • Represent and
• Geometric measurement: unit. interpret data.
•understand concepts of • Represent and • Geometric
area and relate area to interpret data. measurement:
multiplication and to • Geometric understand concepts of
•recognize perimeter as understand concepts of volume to multiplication
an attribute of plane angle and measure and to addition.
figures and distinguish angles.
between linear and area
measures.
Pages 39-43
22
Geometry (3-5)
• Reason with shapes and • Draw and identify lines • Graph points on the
their attributes. and angles, and classify coordinate plane to solve
shapes by properties of real-world and
their lines and angles. mathematical problems.
• Classify two-dimensional
figures into categories
based on their properties.
Pages 43-44
23
Examining a cluster
Domain – Number and
Operations in Base Ten
Generalize place value . . .
Use place value. . .
Standards listed under
more closely related.
Page 33
24
Mathematics K-8 Standards Guide
K-8 Math Domains:
• Counting, Cardinality, and Ordinality – CC
• Operations and Algebraic Thinking – OA
• Number and Operations in Base Ten – NBT
• Measurement and Data - MD
• Number and Operations—Fractions - NF
• Geometry - G
• Ratios and Proportional Relationships – RP
• The Number System - NS
• Expressions and Equations - EE
• Functions - F
• Statistics and Probability - SP
Page 48
25
Closer Look at Standards (6-8)
Domain
Example
Clusters
Subparts
Page 54
26
A Strong Foundation for Algebra
Through the K-8 Standards Progression
Focus on number, operations, and fractions in early
Increased attention to proportionality, probability and
In depth study of linear algebra and introductions of
Provide good preparation for high school mathematics
27
High School Standards for Math Content
Turn to
Page 66
Organized in six Domains
Crosses course boundaries Domains
Span high school years 9-12 Number &
Highlights key points in the narratives Quantity
Algebra
Standards Functions
Majority are core for all students Modeling
prepare students to take courses such as calculus, Geometry
discrete mathematics, or advanced statistics. Statistics &
* Standards indicate connection to Modeling Probability
28
Focus - Modeling
Narrative is provided for each domain
Highlights key points of focus Turn to
Page 84
29
9-12 Domains and Subdomains
Number and Algebra Functions
Quantity
The Real Number Seeing Structure in Interpreting Functions
System Expressions
Building Functions
Quantities Arithmetic with
The Complex Number Rational Expressions Exponential Models*
System
Creating Equations* Trigonometric Functions
Vector and Matrix
Quantities Reasoning with
Equations and
Inequalities
Pages 67-71 Pages 72-77 Pages 78-83
30
9-12 Domains and Subdomains
Geometry Statistics and Probability*
Congruence Interpreting Categorical and
Quantitative Data
tin the other five domains
Modeling (*) – integrated
Similarity, Right Triangles, and
Trigonometry Making Inferences and Justifying
Conclusions
Circles
Conditional Probability and the
Expressing Geometric Properties with Rules of Probability
Equations
Using Probability to Make
Geometric Measurement and Decisions
Dimension
Modeling with Geometry
Pages 86-92 Pages 93-97
31
Mathematics 9-12 Standards Guide
9-12 Math Domains:
• Number and Quantity - N
• Algebra - A
• Functions - F
• Modeling - M
• Geometry - G
• Statistics and Probability - P
32
Closer Look at Standards (9-12)
Domain - Geometry Turn to
Page 92
Subdomains
Clusters
Standards
33
Rigor and Relevance – Modeling
Links classroom mathematics and statistics to
everyday life, work and decision-making.
In Grades K-8, modeling skills are developed as
one of the Standards for Mathematical Practice.
Modeling is both a High School Domain
and a Standards for Mathematical Practice
Domains, Strands, Clusters or Individual Standards
have been identified by *
34
Encourage Changes to Practice
The proposed mathematics standards support
improved curriculum and instruction due to increased:
FOCUS, via critical areas at each grade level
COHERENCE, through carefully developed progression of
topics (domains, strands and clusters) within and across grades
RIGOR, including a focus on College and Career Readiness
and Standards for Mathematical Practice throughout K-12
35
Timeline
January – May 12, 2012 Public comment period
February– April 2012 Webinars and presentations about the new
standards scheduled
June 2012 New standards potentially adopted
Fall 2012 – Spring 2015 Transition from GLEs to new standards
Spring 2016 New assessments potentially administered
36
Support for Districts
EED Website provides information and updated weekly
http://www.eed.state.ak.us/tls/assessment/GLEHome.html
Frequently asked questions have been created for
distribution http://www.eed.state.ak.us/standfaqs.html
Webinars have been scheduled for an in-depth review of
the proposed standards (dates are posted on the
website)
Publications for parents, teachers, and the community will
be available.
Curriculum and Alignment Institute will support districts
with the transition of curriculum.
37
Thank You
Your time is much appreciated!
Public comment can be made at
http://eed.state.ak.us/regs
Questions about the math standards
Cecilia Miller, Math Content Specialist | 3,000 | 13,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2015-22 | latest | en | 0.839518 |
https://books.google.gr/books?id=QJsBAAAAYAAJ&qtid=f87e7872&dq=editions:UOMDLPabq7928_0001_001&lr=&hl=el&output=html_text&source=gbs_quotes_r&cad=7 | 1,619,062,419,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039560245.87/warc/CC-MAIN-20210422013104-20210422043104-00309.warc.gz | 261,051,330 | 6,288 | Αναζήτηση Εικόνες Χάρτες Play YouTube Ειδήσεις Gmail Drive Περισσότερα »
Είσοδος
Βιβλία Βιβλία
Parallelograms upon the same base and between the same parallels, are equal to one another.
Geometry Without Axioms; Or the First Book of Euclid's Elements. With ... - Σελίδα 111
των Thomas Perronet Thompson - 1833 - 150 σελίδες
Πλήρης προβολή - Σχετικά με αυτό το βιβλίο
## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books
Euclid, Edmund Stone - 1765 - 464 σελίδες
...demqnftrated. PROP. XXXVII. THEO R. Triangles cm-Jiituted upon the fame bafe, and hetwtett the fame parallels, are equal to one another. Let the triangles ABc, DEC be conftituted upon the fame bafe B c, and between the fame parallels AD, B c : I fay, the triangle AB...
## The Popular Educator, Τόμοι 1-2;Τόμος 12
1867
...to show. The reader will remember that in Problem XXIV (page 308) it was shown that triangles on the same base and between the same parallels are equal to one another, and that triangles on equal bases and between the same parallels are also equal to one another. Now...
## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1781 - 466 σελίδες
...parallelograms, &c. Q-_E. D. PROP. XXXVII. THEO R. HPRiANGLES upon the fame bafe, and between the fame parallels, are equal to one another. Let the triangles ABC, DEC be upon the fame bafe BC and » -n A Jl -n J? between the fame parallels-i AD, BC. the mangle ABC isj equal to...
## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1806 - 518 σελίδες
...divides the paralkio c 4. 1. gram ACDB into two equal parts. QED PROP. XXXV. THEOR. See N. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. Sue OH- Let the parallelograms ABCD, EBCF be upon the same base 2dand3d BC, and between the same parallels...
## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1806 - 311 σελίδες
...the parallelogram ACDB intq two equal parts. Therefore, &c. QED j / PROP. XXXV. THEOR. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. gures. Seethe 2d Let the parallelograms ABCD, EBCF be upon the same and 3d fi- base BC, and between...
## Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory ..., Τόμος 5
John Mason Good - 1813
...diameter bisects them, that is, divides them in two equal parts. Prop. XXXV. Theor. Parallelograms upon the same base and between the same parallels, are equal to one another. Prop. XXXVI. Theor. Parallelograms upon equal basis, and between the same parallels, are equal to one...
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclides - 1816 - 528 σελίδες
...i divides the parallelogram ACDB into two equal" parts. PROP. XXXV. THEOR. , s«eN. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. See the ad Let the parallelograms ABCD, EBCF be upon the same and 3d fi. base BC, and between the same...
## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1819 - 333 σελίδες
...the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c* Q,. ED PROP. XXXVII. THEOR. Triangles upon the same base, and between the same...parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, ^_ BC : The triangle ABC, is e- •&...
## Elements of Geometry: Containing the First Six Books of Euclid: With a ...
John Playfair - 1819 - 317 σελίδες
...the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q, ED PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. AD Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC : The...
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid, Robert Simson - 1821 - 516 σελίδες
...diameter BC divides the parallelogram ACDB into two equal parts. QED PROP. XXXV. THEOR. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another.* "Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram... | 1,269 | 4,380 | {"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-2021-17 | latest | en | 0.553842 |
https://lindasm.com/the-two-blocks-shown-are-connected-by-a-string-of-negligible-mass-passing-over-a-pulley-of-radius-0-254-m-and-moment-of-inertia-l-the-block-on-the-frictionless-incline-is-moving-up-with-a-constant-acc/ | 1,718,899,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00422.warc.gz | 332,052,639 | 20,243 | # The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string
The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string
The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string
x A cos tot 2 TT f Laney College Physics MA (o points) The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string (b) Find the moment of inertia ofthe pulley (4) s cast Share this:TwitterFacebook
The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string
x A cos tot 2 TT f Laney College Physics MA (o points) The two blocks shown are connected by a string of negligible mass passing over a pulley of radius 0.254 m and moment of inertia L The block on the frictionless incline is moving up with a constant acceleration of 2.50 m/s 7 kg 10.0 (6) (a) Determine TI and T2, the tensions in the two parts of the string (b) Find the moment of inertia ofthe pulley (4) s cast Share this:TwitterFacebook | 561 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.824333 |
https://www.doubtnut.com/question-answer/in-fig-1599-a-square-o-a-b-c-is-inscribed-in-a-quadrant-o-p-b-q-of-a-circle-if-o-a21-c-m-find-the-ar-642571642 | 1,638,276,011,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00384.warc.gz | 815,268,603 | 75,542 | Login
# In Fig. 15.99, a square O A B C is inscribed in a quadrant O P B Q of a circle. If O A=21\ c m , find the area of the shaded region. (FIGURE)
Answer
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 20-3-2021
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
CLICK HERE
Loading DoubtNut Solution for you
Watch 1000+ concepts & tricky questions explained!
8.4 K+
400+
Transcript
TimeTranscript
00:00 - 00:59hello friends Mera question is in the given figure a square oabc is inscribed in a quadrant of a circle if a is equal to 21 CM then find the area of the shaded region so according to our question we have the figure op is the quadrant and a square oabc is inscribed in this quadrant so area of shaded season is equal to area of quadrant minus area of care and now we have to find the area of square
01:00 - 01:59as you know that the formula for area of square is side X side and going to have a question is equal to 21 CM that is 21 multiply 21 which is equal to 44128 if we draw the diagonal if we draw the diagonal of B then diagonal of square is equal to under root 2 X side means a under to multiply 21 that is equal to 21 under root 2 cm and diagonal of square
02:00 - 02:59is equal to radius of circle and hence the area of quadrant is equal to 1 / 45 are here and now we calculate area of water that is 1 / 4 multiply 22/7 multiply and this is also equal to radius is 21 under root 2 cm this is 21 under root 2 X 21 under root 2 that is equal to 1 / 4 multiply 22 /
03:00 - 03:59multiply 441 multiplied to that is world to 693 CM square so area of shaded region is area of quadrant - area of sphere sunao calculate the area of shaded region area of shaded region is equal to area of quadrant is 693 CM square minus and area of square is 441 CM here so here it is equal to 252 care so this is our
04:00 - 04:59area of this shaded portion answer thank you | 521 | 1,954 | {"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} | 3.796875 | 4 | CC-MAIN-2021-49 | latest | en | 0.919152 |
https://stephengalvan.medium.com/sorting-algorithms-radix-sort-5be2620ab2c7?responsesOpen=true&source=user_profile---------0---------------------------- | 1,632,262,872,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00546.warc.gz | 588,879,016 | 32,540 | This week’s sorting algorithm is going to be a little different compared to the other sorting algorithms. With Radix Sort, we are not comparing two values based on: “is one greater than the other?”, instead we are sorting the data by: binary numbers and not by comparing elements.
Radix sort works on lists of numbers, or binary data, NOT by comparing elements. The more digits the bigger a number. And the way we sort with radix, is by sorting the numbers through 10 distinct buckets. These buckets are reused until we’ve gone through sorting the digits, and each bucket represents a number from 0–9 or base 10.
We organize the numbers in order, based on the value of the number in the 1s position first and then we iterate through the numbers again in the 10s, 100s and so on and so forth.
So if we were to sort the following list of numbers:
3, 92, 11, 214, 3160, 26, 817, 2000, 33
we first sort by the numbers in the one digits position first. So the numbers:
3, 92, 11, 214, 3160, 26, 817, 2000, 33 would be sorted into:
3160, 2000, 11, 92, 3, 33, 214, 26, 817
We are only looking at the single digit value, and not at the rest of the digits in the numbers. In the next run of our sort we will look at the values in the 10s place. And our next sorted list will be:
2000, 3, 11, 214, 817, 26, 33, 3160, 92
We’ll repeat this process for the hundreds and thousands until we arrive at our sorted list. If you notice in the image above, 3 gets sorted into the 0’s bucket. Because we’ve run out of digits for this number (3 does not have a digit in the 10s place) anything to the left of a number will be given an imaginary 0 value. This allows the number to hold its space in the sorted list. The number of times we go through this sorting process depends on the number of digits in the largest number. In this case, the largest number has 4 digits so we go through this process 4 times.
Hundreds:
2000, 3, 11, 26, 33, 92, 3160, 214, 817
Thousands:
3, 11, 26, 33, 92, 214, 817, 2000, 3160
# Helper Methods (from Stack Overflow)
## getDigit(number, place)
— returns the digit inside of the number at the given place value. So if our number is 365 and the place is 0 the digit is 5, if the place is 1 the digit is 6, and if the place is 2 the digit is 3.
`function getDigit(number, place) { return Math.floor(Math.abs(number) / Math.pow(10, place)) % 10;}`
For example let’s say our number is 365, and place is 2. Math.abs(number) ensures that even if we were dealing with negative values, the number wouldn’t be returned as negative, so we are just dealing with absolute numbers. Math.pow(10, place) returns 10 to the place power or in this case 10 to the 2nd power. And we get 100. Then we Math.floor(365/100), which is 3. And lastly we find 3 % 10 which returns 3.
Looks like:
getDigit(365, 2)
Math.abs(365) => 365
Math.pow(10, 2) => 100
Math.floor(365/100) => 3
3%10 => 3
## digitCount(number)
— this will return how many digits there are in the number. So if the number is 365, digitCount returns 3. We must include the edge case below `if (num === 0) return 1` because if we run Math.log10(Math.abs(0)) the output would be`-Infinity.`
`function digitCount(num) { if (num === 0) return 1; return Math.floor(Math.log10(Math.abs(num))) + 1;}`
For example if our number is again 365, Math.log10(365) is 2.5622928644564746 and if we Math.floor(2.5622928644564746) the value it is 2. Lastly we add one, 2+1 and we get the length of our number, 3.
## mostDigits(numbers)
taking in a list of numbers, this method tells us what is the size of the largest number in the list. If these were our numbers, [1000, 1786300, 365], most Digits would return 7.
`function mostDigits(nums) { let maxDigits = 0; for (let num of nums) { maxDigits = Math.max(maxDigits, digitCount(num)); } return maxDigits;}`
For example, let’s say mostDigits takes in this number list: [1000, 1786300, 365]. First we declare a variable maxDigits equal to 0. We are going through each number of the list of numbers or num of nums to compare the digit length and assign the max value to the maxDigits variable. Under the hood this looks like:
1. maxDigits = Math.max(0, digitCount(1000))
which is greater, 0 or 4? => maxDigit = 4
2. maxDigits = Math.max(4, digitCount(1786300))
which is greater, 4 or 7? => maxDigit = 7
3. maxDigits = Math.max(7, digitCount(365))
which is greater, 7 or 3? => maxDigit = 7
Return 7
• Our function will take in a list of numbers
• Find the max digits of the largest number in the list of numbers
• Create a loop with a variable k starting at 0 and going up to the value of max digits from the largest number.
• where inside each loop we create buckets for each digit (0–9)
• And place each number in the correct bucket based on the index of k for that number (ie. 0 is the ones, 1 is the 10s, 2 is the 100s, etc.).
• Replace our existing array with values from the buckets from 0 to 9
• Lastly, return the sorted number list.
`function radixSort(nums){ let maxDigitCount = mostDigits(nums); for(let k = 0; k < maxDigitCount; k++){ let digitBuckets = Array.from({length: 10}, () => []); for(let i = 0; i < nums.length; i++){ let digit = getDigit(nums[i],k); digitBuckets[digit].push(nums[i]); } nums = [].concat(...digitBuckets); } return nums;}`
## Let’s break this down (full code snippet below):
First: We take in a list of numbers, and declare a variable maxDigitCount which is the length of the largest number in the list.
Example numbers: [90, 881, 16, 320722, 531, 5, 91, 4760]
maxDigitCount = 6, because 320722 is the largest number and there are 6 digits within it.
Second: Then inside a loop, we declare variable k which represents the place of the number(1s, 10s, 100s, 1000s)
Inside the k loop we declare a variable called digitBuckets, and inside this variable, we create a hash of 10 empty arrays (`[]`).
Third: We create a nested loop with variable i which represents the current number in the list that we are iterating through in this case we start at 0 out of 8 numbers. While i is less than the length of the list of numbers(8), we will increment i.
Fourth: Inside the nested i loop, we declare variable digit = getDigit(nums[i], k) and then based on the digit, we push the number into the digitBucket, with digitBuckets[digit].push(nums[i]).
Example: First iteration of i:
digit = getDigit(nums[0], k=0)
nums[0] = 90
digit = 0
digitBuckets[0].push(90)
Example: Second iteration of i:
digit = getDigit(nums[1], k=0)
nums[1] = 881
digit = 1
digitBuckets[1].push(881)
Example: Third iteration of i:
digit = getDigit(nums[2], k=0)
nums[2] = 16
digit = 6
digitBuckets[6].push(16)
Fifth: Once we complete the nested loop, we reassign the value of nums into a new array, concatenating the numbers in order from digitBuckets.
Example: First iteration of k:
nums = [90, 881, 16, 320722, 531, 5, 91, 4760]
digitBuckets = {
0: [90, 4760],
1: [881,531, 91],
2: [320722],
3: [],
4: [],
5: [5],
6: [16],
7: [],
8: [],
9: []
}
nums = [].concat(…digitBuckets)
=> [90, 4760, 881, 531, 91, 320722, 5, 16]
Final Step: we repeat all previous steps, until k is no longer less than maxDigitCount, and lastly we return the sorted list of numbers.
`function getDigit(number, place) { return Math.floor(Math.abs(number) / Math.pow(10, place)) % 10;}function digitCount(num) { if (num === 0) return 1; return Math.floor(Math.log10(Math.abs(num))) + 1;}function mostDigits(nums) { let maxDigits = 0; for (let num of nums) { maxDigits = Math.max(maxDigits, digitCount(num)); } return maxDigits;}function radixSort(nums){ let maxDigitCount = mostDigits(nums); for(let k = 0; k < maxDigitCount; k++){ let digitBuckets = Array.from({length: 10}, () => []); for(let i = 0; i < nums.length; i++){ let digit = getDigit(nums[i],k); digitBuckets[digit].push(nums[i]); } nums = [].concat(...digitBuckets); } return nums;}`
A Software Engineer with a background in Education Technology and Dance. Recent grad form FlatIron Bootcamp, and passion for the arts and working with databases
## More from Stephen Galvan
A Software Engineer with a background in Education Technology and Dance. Recent grad form FlatIron Bootcamp, and passion for the arts and working with databases | 2,394 | 8,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-39 | latest | en | 0.872751 |
https://socratic.org/questions/what-is-21-75-m-in-centimeters | 1,642,468,823,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00113.warc.gz | 590,732,887 | 5,964 | # What is 21.75 m in centimeters?
Sep 20, 2016
color(green)(21.75" metres "=2175" centimetres"
#### Explanation:
It takes 100 cm to make 1 metre.
Using ratio:
$\frac{\text{centimetres")/("metres") ->100/1 -=("centimetres}}{21.75}$
Write as:
$\left(\frac{100}{1} \textcolor{m a \ge n t a}{\times 1}\right) \equiv \frac{\text{centimetres}}{21.75}$
$\left(\frac{100}{1} \textcolor{m a \ge n t a}{\times \frac{21.75}{21.75}}\right) \equiv \frac{\text{centimetres}}{21.75} = \frac{2175}{21.75}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So $21.75 \text{ metres"=2175" centimetres}$ | 222 | 593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.3125 | 4 | CC-MAIN-2022-05 | latest | en | 0.401542 |
https://www.fasps.org/academics/grade-8-subjects/mathematics | 1,718,225,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00321.warc.gz | 716,768,486 | 14,784 | });
Mathematics in Grade 8 | 4ème
Mathematics Class in Grade 8 Meets:
• Three Times / Week (Instruction in French)
• Three Times / Week (Instruction in English)
Structure of Program:
• Algebra 1 Second Year (Standard Level) — Students Study Second Half of Algebra 1 Curriculum
• Introductory Algebra 2 (Extended Level) — Students Study Much of the Algebra 2 Curriculum
• We Follow the French Ministry of National Education Program in French Mathematics
Contents:
• Data Management (French)
• Algebra: Linear Functions, Exponents and Exponential Functions, Polynomials and Quadratic Functions and Equations (U.S., Standard)
• Algebra: Transformations of Functions, Quadratics, Complex Numbers, Exponents and Logarithmic Functions, Sequences and Series, Trigonometry (U.S., Extended)
• Proofs (French)
• Geometry: Thales, Pythagorean Properties, Trigonometry (French)
• Area and Volume (French, U.S.)
• Algorithms (French)
• Solids: Pyramids and Cones (French)
• Proportionality, Percent (French)
Objectives:
• Complete the Skills and Concepts Outlined in the Washington State Standards for Algebra 1 (U.S.) and Geometry (French)
• Use of Appropriate Tools Strategically in Mathematics
• Develop Inductive and Deductive Reasoning Skills
• Strategic Problem-solving
Assessment:
• Formative and Summative Tests and Quizzes
• Long-term Projects
• Organizational Skills
• Open Problems
Resources:
• Mathématiques Transmath (French)
• KWYK (French) Exercises Website
• Algebra 1, Holt McDougal (U.S., Standard)
• Algebra 2, Big Ideas Learning (U.S., Extended)
• Geogebra Geometry Software (French)
• Excel Software (French)
• Desmos.com Graphing Calculator and Interactive Activities (U.S.)
Teachers
Mme Anne Sophie Cune
French Mathematics | Geometry
Ms. Sarah Blick Vandivort
U.S. Mathematics Co-teacher | Mathematics Coordinator | 430 | 1,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.766008 |
https://www.physicsforums.com/threads/calculating-the-dimensions-of-an-arc.508374/ | 1,542,473,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743717.31/warc/CC-MAIN-20181117164722-20181117190722-00284.warc.gz | 990,158,085 | 13,842 | # Calculating the dimensions of an arc
1. Jun 20, 2011
### kennethgilpin
Dear all,
I feel this should be a simple problem but I can't solve it. Could you give me a hand?
Imagine if an arc is bounded by a rectangle of dimensions width and height. The arc starts in the bottom left corner of the rectangle, and ends in the bottom right corner. The apex of the arc is the center of the top side of the rectangle. A picture of this is shown half way down this page:
http://www.mathopenref.com/chordsintersecting.html" [Broken]
Assuming that the arc is less than a semi circle, if I know the length of the arc and the height of the rectangle, how can I calculate it's width?
kenneth
Last edited by a moderator: May 5, 2017
2. Jun 20, 2011
### micromass
Hi kennethgilpin!
So you know some calculus? In my opinion that's the easiest way to solve this.
First we will have to find the equation of the circle given the coordinates of the rectangle.
Say that the vertices of the rectangle have coordinates (0,L),(0,-L),(H,L),(H,-L).
The general equation of a circle is
$$(x-x_0)^2+(y-y_0)^2=R^2$$
with (x0,y0) the center of the circle and R the radius. We see easily that the center of the circle must lie on the y-axis, and thus x0=0.
Now, what you have to do is to find y0 and R such that (0,L) and (H,0) lie on the circle
$$x^2+(y-y_0)^2=R^2$$
3. Jun 20, 2011
### I like Serena
@MM: I'm afraid this won't work.
It's the arc length that is given, and the width that is asked.
I believe that the resulting equations can not be solved algebraically.
Of course it can be approximated numerically.
4. Jun 20, 2011
### micromass
Yes, I know. What I was attempting is to find an equation which calculated the arc length if you have the length and width given. Then I would use this equation to find an equation calculating the width... I think this ought to work, given that the equations are not too difficult...
5. Jun 20, 2011
### I like Serena
I created the set of equations:
$$\begin{eqnarray} S &=& 2 R \alpha \\ W &=& 2 R \sin \alpha \\ R &=& H + R \cos \alpha \end{eqnarray}$$
where S is the arc length, and $\alpha$ is half of the angle of the arc.
If you try to solve it for W, with given S and H, you'll quickly see that you're left with an equation that is afaik not solvable algebraically.
6. Jun 21, 2011
### kennethgilpin
thank you for your replies. @likeSerena - i got similar equations to you which I could not solve.
This is for use within an iterative model so I will try to solve it iteratively.
kenneth | 703 | 2,543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-47 | latest | en | 0.919928 |
https://www.physicsforums.com/threads/plasma-physics-question-making-a-protecting-shield.755000/ | 1,531,905,821,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590074.12/warc/CC-MAIN-20180718080513-20180718100513-00611.warc.gz | 971,785,927 | 13,099 | # Plasma Physics Question - making a protecting shield
Tags:
1. May 23, 2014
### thetao
I need to find the correct voltage to apply to a wire grid in order to repel a plasma of which I know the density (free electrons per volume) and the particle energy in eV. I see a Debye sheath as being analogous to what I will create with such a charge.
The way I see it is that I will have two wire layers at opposite potentials, creating a capacitance in between them, and this will create an outside charge. This charge will repel the plasma on the outside so that it does not enter the system.
Would someone tell me what I need to consider in order to solve this problem? I understand that different voltages would achieve a larger or smaller distance between the outside wire grid and the plasma, more or less debye lengths, but is there a saturation point, a voltage beyond which the distance would not be noticeably increased? How do I go about calculating this voltage?
Thank you!
2. May 23, 2014
### UltrafastPED
The grid voltage must be greater than the voltage part of the electron energy (eV).
It's not guaranteed that every electron will be repelled ... there is a statistical variation of energies inside the plasma.
Not clear on what you are trying to accomplish - diagrams are preferred over verbal descriptions - but here is a patent that discusses some similar issues, and has diagrams:
http://www.google.com/patents/US20030102402
I've used the voltage on a grid to repel electrons that were _below_ a given energy; in this application they would only have contributed to the noise, and so had to be rejected.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 373 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-30 | latest | en | 0.947079 |
https://web.math.princeton.edu/generals/humphries_peter | 1,632,361,273,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-39/segments/1631780057416.67/warc/CC-MAIN-20210923013955-20210923043955-00703.warc.gz | 660,201,818 | 7,016 | Peter Humphries' generals. 10 May 2013, 1pm. Special topics: analytic number theory and representation theory of compact Lie groups. Committee: Peter Sarnak (chair), Charlie Fefferman, Nicolas Templier. Algebra (all Sarnak): Say you have a finite field. What's its cardinality? (Prime power - must be prime characteristic to be a field, and must be a vector space over F_p for some p.) Prove that the unit group of a finite field is cyclic. (I of course used the structure theorem for finite abelian groups. They made me explain why x^m - 1 can only have at most m roots, which to me felt like a tautology. A few hints about factorising out roots eventually did the trick.) You mentioned the structure theorem for finite abelian groups. What is the module and what is the PID in this case? (The group, and Z.) How do you prove the theorem? (Structure theorem for finitely generated modules over a PID.) How do you prove THAT? (Can't remember.) Seriously?! (Yep, forgot to revise this. Oops!) When are two matrices conjugate over a given field? (Is the field algebraically closed?) Okay, let's say the field is algebraically closed at first. (Jordan canonical form.) How do you prove it? What is the module and what is the PID? (I fluffed around with this a bit at first before getting it right.) What about over a field that isn't algebraically closed? (Rational canonical form. I again had to state the module and the PID.) Why is Q[x] a PID? (Euclidean algorithm.) Explain it. Give an example of a ring that is not a PID? (Z[sqrt{-5].) Why not? (It's not even a UFD, as 6 factorises in two ways. We talked about ideals and norms of rings of integers, in order to calculate the unit group of Z[sqrt{-5].) What are Sylow's theorems? Give an application. (I falsely claimed that if G was a group of order n = mp with m < p, then G was abelian, before correcting myself and stating rather that it was not a simple group.) Complex: T: What is a Laurent series? (I started talking about where a Laurent series converges, with some difficulty remembering the radii of convergence.) How do you calculate the coefficients? (I differentiated Cauchy's integral formula.) T: Say f is holomorphic in the unit disc. How do you know its power series converges inside the disc? (I looked at this for ages before remembering to expand the denominator as a power series.) F: Why does this work? (Estimate the tail, or use the dominated convergence theorem.) S: Say you have an entire function that is bounded. What can you say about it? (Constant, by Liouville's theorem.) What if it's bounded by a polynomial, say |z|^100? (Cauchy's integral formula shows that it must be equal to c z^100 for some |c| = 1.) S: Suppose now that you have a bounded function on a punctured disc. What can you say? (Riemann's theorem on removable singularities.) Proof? (Use a keyhole contour.) What if instead it may not be bounded, but blows up like 1/sqrt{|z|}? (Can't happen - the same method for removable singularities.) So what's the weakest condition for a singularity at the origin to be removable? (Only need o(1/|z|).) S: When are two doubly-connected domains conformally equivalent? (I stated that any doubly-connected domain is conformally equivalent to an annulus.) Proof? (Don't know - it's not in Stein & Shakarchi!) Really? (Yep, a notable omission. But I can prove that two annuli are conformally equivalent iff the ratios of the radii are equal!) Please do. (I gave a proof from Rudin that he hadn't seen before, by looking at g(z) = 2(log(|f(z)|/r_2) - a log(|z|/r_1)) for a = log(R_2/r_2)/log(R_1/r_1). Sarnak hadn't seen this before and thought it was quite neat.) S: Do you know a proof via Schwarz reflection? (Yes, and we went through it.) Real: F: Define the Fourier transform. What is it useful for? (I couldn't think of anything, so I said if you integrate by parts you can turn differential operators in phase space into polynomials on frequency space.) F: What is the Poisson summation formula? Prove it. (I did.) F: Why does the Fourier series of a smooth function convergs to the function itself? (I started working through the proof but was quickly stopped when he saw I knew how to do it.) Sarnak decided he wanted to generalise this (uh-oh). How do you define a (tempered) distribution? (Linear functional on Schwartz space.) What's the topology? (No idea. Oops. They spent a while giving hints as to what this should be, without success. Along the way I falsely claimed that the space of smooth functions in the unit circle is a Banach space, before digging myself out of that whole by quoting the Stone-Weierstrass theorem.) S: How do you define the Fourier transform of a tempered distribution? What about in the case of the sum of Dirac delta functions, each giving mass one at an integer? Is it a distribution? (Yes.) Now what does Plancherel's theorem state about this sum applied to a Schwartz function? (Gives back the Poisson summation formula.) Sarnak then told me some rigidity results about this kind of thing. Analytic number theory: T: Define a Dirichlet character. What is a primitive character? (I struggled to define it correctly. I probably should've mentioned the word conductor at some point!) Why are Dirichlet characters useful? (Orthogonality relations allow you to pick out elements of an arithmetic progression.) What can you say about primes in arithmetic progressions? (Stated Dirichlet's theorem.) Anything stronger? (Dirichlet density is 1/phi(q).) S: No, something else. (PNT in APs.) Which is stronger, that or the Dirichlet density result? (The PNT version, of course.) S: How do you show L(1,chi) doesn't vanish? (I explained why this is true for complex characters, by multiplying together all the L-functions of a fixed modulus.) What is this product of L-functions, and why are the coefficients of the Dirichlet series nonnegative? (Dedekind zeta function of a cyclotomic field.) Is this immediate? (No, but I didn't know much about how to factorise Dedekind zeta functions of abelian extensions into products of zeta functions and Dirichlet L-functions. I said I could still prove that the coefficients were nonnegative via their Euler products, but Sarnak was unimpressed - he pointed out that the only recipe to cook up such products of L-functions for which the resulting Dirichlet series has nonnegative coefficients is via Dedekind zeta functions and their like.) S: What about when chi is real? (Multiply by zeta(s).) What is the zeta function in this case? (Quadratic field. He made me write out what it was.) How do you proceed from here? (Show that the Dirichlet series has coefficients bounded below by 1 when n is a square, so the series diverges at s = 1/2. Then Landau's lemma for the rest.) Sarnak pointed out that this lemma was actually due to Pringsheim earlier, but I countered with the fact that Pringsheim's version was only for power series, not Dirichlet series, which Sarnak found amusing (and in fact Vivanti published this result a year before Pringsheim). S: What lower bound on L(1,chi) can you get from this (>> 1/sqrt{q}, though he didn't press for details.) T: Do you know another way to show this? (I mentioned the analytic class number formula, in full generality, though I didn't prove it.) S: Can you get a better bound? (Landau-Siegel.) What's bad about this? (Ineffective.) Why? (I went through the proof and discussed where the ineffectivity arises. Once again, I had to mention that the product of L-functions in this case is the Dedekind zeta function of a biquadratic field.) S: Let's talk about the circle method. (I started talking about three primes, but then he changed tack.) Show that for every sufficiently large N can be written as the sum of 20 cubes. (I talked about Waring's problem.) How do you bound the minor arcs? (Hua's lemma and Weyl's inequality.) What bounds do they give? How do you prove them? (I mentioned Parseval for Hua's lemma and he was content. For Weyl, I couldn't remember how the differencing method worked, so they talked me through how to bound sum_{n < N} e(alpha n^2).) Why can you give a good bound for an exponential sum where the polynomial inside is linear? (He wanted me to say that it was a geometric series!) They didn't ask about the major arcs at all. Representation theory: F: What are the irreducible representations of SO(n)? (I mentioned harmonic polynomials.) Okay, let's assume that this representation is irreducible. Does this give all representations? (I floundered for a bit before mentioning the Weyl character formula, and that the theorem of highest weights gives a way of indexing irreducible representations.) S: Let's step back for a second. Suppose that I just have some compact topological group. How do I prove the existence of an irreducible representation? (I started talking about the Peter-Weyl theorem, and decomposing the left regular representation. This took a while, and they were very quick to point out when a claim of mine was false because I hadn't stated a necessary condition - e.g. why the convolution operator I defined was actually compact and self-adjoint.) S: Okay, all you've shown is that finite-dimensional representations exist. How do you know that there are any finite-dimensional irreducible ones? (Maschke's theorem - as soon as I said average over the group, he was happy.) S: You mentioned the Weyl character formula earlier on. What is it? (I wrote it down. Sarnak didn't like that everything was in terms of Lie algebras, so he made me define everything.) What IS a Lie algebra? (This was embarrassing. I had a complete mental blank on how to define it. After a few painful minutes, I mumbled something incomprehensible about invariant vector fields at the identity, but did not stating the crucial words ``tangent space''.) S: How about we take a five minute break? (Fine with me! When Sarnak and Fefferman left the room, Templier whispered the answer to me, though thankfully by that point I had remembered the missing words!) S: Back to Lie algebras. What are maximal tori? Cartan subalgebras? (He never asked me prove that maximal tori are conjugate! I was very surprised.) What are roots? Weights? S: Let's get back to the original problem. Maybe make it a bit easier - what are the irreducible representations of SO(3)? (I knew the dimensions, the maximal torus, and the roots, but was stuggling to work out the weights.) S: How about some other group, of rank at least 2? (I talked about SU(3): the maximal torus, Cartan subalgebra, simple and positive roots.) T: Can you draw the diagram associated to the roots? (I could not - I forgot to practise this in my study for generals!) They were unimpressed. S: How about a simpler example still? (I looked at the representations of SU(2) on the space of homogeneous polynomials, in particular the action of the maximal torus.) T: Is there a connection between SO(3) and SU(2)? (The latter is a double cover of the former - I mentioned the adjoint representation giving the cover.) They decided to end here, perhaps to put me out of my misery. I left the room and waited for what seemed like an eternity (but was more likely three or four minutes) before Sarnak welcomed me back in and told me that I had passed (somehow) but that there were some serious things that I needed to improve on (which was painfully clear to me, especially in the last 45 minutes). The exam lasted 2h45m. Some general advice: there is a lot to study, but you will learn a lot along the way. Enjoy it; you'll come out the other end much more knowledgeable. Make sure you look at past exams, and try to actually solve plenty of these questions yourself (I found this to be the most productive way to study). Some examiners ask the same questions very often, so make sure you know how to answer those ones in some detail. In the weeks leading up to the exam, it's a good idea to test your knowledge by doing mock generals (i.e. getting your friends to ask you questions in front of a blackboard). This is especially useful for more lengthy problems that you might be asked about, such as the Riemann mapping theorem or Vinogradov's three primes theorem. For representation theory of compact Lie groups in particular, it would be worthwhile to completely memorise everything about SU(n), SO(2n), Sp(n), and SO(2n+1) (i.e. their Lie algebras, maximal tori, Cartan subalgebras, simple roots, positive roots, roots, Weyl chambers, Weyl groups, Dynkin diagrams, and so on). I had most of this covered but made the mistake of not knowing how to calculate the weights of these groups, which my committee were not happy about. Finally, don't worry about how screwed you might be. Everyone freezes up and forgets things (in my case, some extremely obvious and simple things). Your committee expect that and will give you some leeway, and will try to help you dig yourself out any hole you might find yourself in. For what it's worth, here are the books that I used to study: For complex analysis, Sarnak recommended Ahlfors, though I couldn't get a hold of a copy so instead I studied Stein & Shakarchi (volume II of their Princeton Lectures in Analysis). Its only faults are missing details about conformal mappings for multiply-connected domains (which Ahlfors does cover). Both don't really have anything about the uniformisation theorem, which is something worth knowing (especially if Sarnak is on your committee). For real, I used the third volume of Stein & Shakarchi. It has pretty much everything you need except some functional analysis (in particular L^p spaces), which is covered at the beginning of the fourth volume. Don't bother with the first volume - although it covers Fourier analysis, it's all in terms of Riemann integration, which is pretty silly. For algebra, I used both Dummit and Foote's ``Abstract Algebra'' and Michael Artin's ``Algebra'', though I can't say I was a huge fan of either. For number theory, I used Montgomery and Vaughan's ``Multiplicative Number Theory I. Classical Theory'' and Davenport's ``Multiplicative Number Theory'' for multiplicative number theory - both books are written similarly (Montgomery revised the later editions of the latter), with the latter being briefer and less comprehensive, though the proofs were often neater. For additive number theory, I liked Nathanson's ``Additive Number Theory'', though it covers a lot of extra stuff and can be a bit lengthy. For the proof of the three primes theorem, I actually preferred Montgomery's lecture notes (see http://www-personal.umich.edu/~hlm/math775/handouts.html). For representation theory, Sarnak recommended Adams' ``Lectures on Lie Groups'', which I found pretty heavy going - I much preferred the Lie algebra approach. Hall's ``Lie Groups, Lie Algebras, and Representations'' is a good introduction but has too many proofs ommitted. I mostly ended up using Sepanski's ``Compact Lie Groups'' and Fegan's ``Introduction to Compact Groups''. Note that if you want to read about noncompact groups (e.g. anything about SL_n(R)) then you're going to have to look elsewhere. | 3,554 | 15,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-39 | latest | en | 0.94862 |
https://blogs.msdn.microsoft.com/lucabol/2008/09/26/downloading-stock-prices-in-f-part-v-adjusting-historical-data/ | 1,526,950,330,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864572.13/warc/CC-MAIN-20180521235548-20180522015548-00415.warc.gz | 542,770,525 | 15,739 | Other parts:
Here is the problem. When you download prices/divs/splits from Yahoo you get a strange mix of historical numbers and adjusted numbers. To be more precise, the dividends are historically adjusted. The prices are not adjusted, but there is one last column in the data for "Adjusted close". If you don't know what 'adjusted' means in this context read here.
The problem with using the 'adjusted close' column is that, for a particular date in the past, 'adjusted close' changes whenever the company pays a dividend or splits its stock. So if I retrieve the value on two different days I might get different numbers because, in the meantime, the company paid a dividend. This prevents me from storing a subset of the data locally and then retrieving other subsets later on. It also has the limitation that just the closing price is present while I might need adjusted opening price, adjusted high price or even adjusted volume depending on the operations I want to perform on the data (i.e. calculating oscillators or volume-adjusted moving averages).
The solution I came up with is to download the data and transform it to an 'asHappened' state. This state is simply an unadjusted version of what happened in the past. Data in this state is not going to change in the future, which means that I can safely store it locally. I can then on demand produce 'historically adjusted' data whenever I need to.
Ok, to the code. As it often happens, I need some auxiliary functions before I get to the core of the algorithms. The first one is a way to compare two observations, I will use it later on to sort a list of observations.
```let compareObservations obs1 obs2 =
if obs1.Date <> obs2.Date then obs2.Date.CompareTo(obs1.Date)
else
match obs1.Event, obs2.Event with
| Price _, Price _ | Div _, Div _ | Split _, Split _ -> failwith "Two same date/ same kind observations"
| Price _, _ -> -1
| _, Price _ -> 1
| _ -> 0```
This is rather simple. If the dates of these observations are different, just compare them. If they are the same then the two observations cannot be of the same type (i.e. I cannot have two prices for a particular date). Given that they are not of the same, then &(&^%!#\$!4. Crap, that teaches me to put comments in my code! I think I'm putting the price information first, but I'm not sure. Anyhow my universal excuse not to figuring it out is that the testcases succeed so I must be doing it right (how lame, testcase-addiction I guess ...).
The next auxiliary function is just a wrapper over fold. I always tend to wrap fold calls in a method with a better name because I remember the old times when I didn't know what fold was. I want a reader of my code to be able to understand it even if they are not familiar with fold (the 'universal functional Swiss-Army-Knife). This function is a map that needs to know the value of an accumulator to correctly perform its mapping over each element.
```let mapAcc acc newAccF newItemF inl =
let foldF (acc, l) x = newAccF acc x, (newItemF acc x)::l
let _, out = inl |> List.fold_left foldF (acc, [])
out```
Apart from the implementation details, this function takes an accumulator, an accumulator function, an item function and an input list. For each element in the list it calculates two things:
1. a new value for the accumulator: newAccumulatorValue = newAccF oldAccValue itemValue
2. a new value for the item: new ItemValue = newItemF accValue oldItemValue
Maybe there is a standard functional way to do such a thing with a specific name that I'm not aware of. Luke might know. He is my resident fold expert.
All right, now to he main algorithm.
```let asHappened splitFactor observations =
let newSplitFactor splitFactor obs =
match obs.Event with
| Split(factor) -> splitFactor * factor
| _ -> splitFactor
let newObs splitFactor obs =
let date = obs.Date
let event = match obs.Event with
| Price(p) -> Price(p)
| Div(amount) -> Div(amount * splitFactor)
| Split(factor) -> Split(factor)
{Date = date; Event = event}
observations
|> List.sort compareObservations
|> mapAcc splitFactor newSplitFactor newObs```
To understand what's going on start from the bottom. I'm taking the observation list downloaded from Yahoo and sorting it using my compareObservations function. I then take the resulting list and apply the previously described mapAcc to it. For this function splitFactor is the accumulator, newSplitFactor is the accumulator function and newObs is the function that generate a new value for each item in the list.
NewSplitFactor is trivial: every time it sees a Split observation it updates the value of the split factor. That's it. NewObs is rather simple as well. Every time it sees a dividend, it 'unadjust' it by multiplying its amount by the split factor. The end result is to transform the dividends downloaded from Yahoo (which are adjusted) to an unadjusted state. I could have filtered out the price observations before doing all of this and add them back afterward, but didn't. It'd probably be slower ...
Now that I can recreate the state of the world as it was at a particular point in time, what if I want to adjust the data? I can call adjusted ...
```let adjusted (splitFactor, lastDiv, oFact, hFact, lFact, cFact, vFact) asHappenedObs =
let newFactor (splitFactor, lastDiv, oFact, hFact, lFact, cFact, vFact) obs =
match obs.Event with
| Split(split) -> splitFactor * split, lastDiv, oFact, hFact, lFact, cFact, vFact
| Div(div) -> splitFactor, div, oFact, hFact, lFact, cFact, vFact
| Price(p) -> splitFactor, 0.<money>, oFact / (1. - lastDiv / p.Open), hFact / (1. - lastDiv / p.High), lFact / (1. - lastDiv / p.Low), cFact / (1. - lastDiv / p.Close), vFact / (1. - lastDiv / p.Close)
let newObs (splitFactor, lastDiv, oFact, hFact, lFact, cFact, vFact) obs =
let date = obs.Date
let event = match obs.Event with
| Price(p) -> Price({Open = p.Open / splitFactor / oFact; High = p.High / splitFactor / hFact; Low = p.Low / splitFactor / lFact; Close = p.Close / splitFactor / cFact; Volume = p.Volume / splitFactor / vFact })
| Div(amount) -> Div (amount / splitFactor)
| Split(split) -> Split(split)
{Date = date; Event = event}
asHappenedObs
|> List.sort compareObservations
|> mapAcc (splitFactor, lastDiv, oFact, hFact, lFact, cFact, vFact) newFactor newObs
|> List.filter (fun x -> match x.Event with Split(_) -> false | _ -> true) ```
Wow, ok, this looks messy. Let's go through it. Starting from the bottom: sort the observations, perform the right algorithm and filter away all the splits. It doesn't make sense to have splits in adjusted data.
The interesting piece is the mappAcc function. It take a tuple of factors as accumulator and the usual two functions to update such tuple and create new observations. The newObs function creates a new Observation using the factors in the accumulator tuple. Notice how the dividends are divided by the splitFactor (which is the opposite of our asHappened algorithm where we were multiplying them). Also notice how the prices are divided by both the splitFactor and the pertinent price factor. This is needed because the prices need to be adjusted by the dividends paid out and the adjustment factor is different for each kind of price (i.e. open, close, etc...). The newFactor function simply updates all the factors depending on the current observation.
Notice how asHappened and adjusted are structurally similar. This is an artifact of having a functional approach to writing code: it kind of forces you to identify these commonality in the way an algorithm behave and abstract them out (in this case in the mapAcc function). You often discover that such abstracted-out pieces are more generally useful than the case at hand.
Tags | 1,865 | 8,163 | {"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-2018-22 | latest | en | 0.913993 |
https://oercommons.org/browse?f.keyword=counting | 1,714,010,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00295.warc.gz | 383,691,539 | 21,813 | Updating search results...
# 49 Results
View
Selected filters:
• counting
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
In this lesson, students will develop their fluency with addition and subtraction through a "real-world" application of collecting data about insects in the garden.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Conditional Remix & Share Permitted
CC BY-SA
Rating
0.0 stars
In this activity, kids will work on two fundamental early math skills – sorting/classifying, and graphing. There will also be some great fine motor skill practice! Includes place-based discussion questions, activity instructions, extension activities, songs, and student graph worksheets.
NGSS: K-LS1-1, 1-LS1-1, partially meets K-ESS3-1 (book and discussion)
Common Core: MP.4
Time: 45 minutes
Matierals: bag of dried beans ("16 bean soup"), paper bowls, glue, chart paper, the book "One Bean" or similar book about growing food plants, especially beans.
Subject:
Agriculture
Career and Technical Education
Early Childhood Development
Education
Elementary Education
Mathematics
Measurement and Data
Material Type:
Activity/Lab
Lesson Plan
Author:
Columbia Gorge STEM Hub
08/07/2020
Unrestricted Use
CC BY
Rating
0.0 stars
In this card game students play in pairs to practice recognizing the biggest number.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
In this lesson, students use the real-world application of insect monitoring to practice their fluency with addition and subtraction.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This template is meant to be a guide for Nebraska Physical Education Teachers when creating digital online lessons. Headings and/or topics not included in the lesson plan should be marked N/A.
Subject:
Education
Material Type:
Activity/Lab
Lesson
Lesson Plan
07/10/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
In this lesson, students will collect data about an observation that can be made in the outdoor classroom, record the data on a table, and compare the numbers collected by writing greater-than, less-than statements.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
Students will understand that we use numbers to represent a quantity and to compare them.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Only Sharing Permitted
CC BY-NC-ND
Rating
0.0 stars
A suite of counting games for the whiteboard, tablet or computer. Games provide a rich interactive learning setting.
Subject:
Mathematics
Material Type:
Activity/Lab
Game
Interactive
Author:
ComputerMice
03/10/2019
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
Students will practice counting up to 20 in a very fun and exciting way. The lesson includes a hands-on component where students will be asked to collect leaves and use them as manipulatives. This lesson is great for kindergarten through first grade.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Unrestricted Use
CC BY
Rating
0.0 stars
I have attatched the board Game overview in the resource library with a turtorial on how to play the game and the actual lesson plan. This is a fun board game to help Kindergarten students count 1-6 while also learning about the do's and don't's of digital citizenship.
Subject:
Elementary Education
Material Type:
Lesson Plan
Author:
Melanie Szklarek
12/05/2017
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
This intensive micro-subject provides the necessary skills in Microsoft® Excel spreadsheet modeling for ESD.71 Engineering Systems Analysis for Design. Its purpose is to bring entering students up to speed on some of the advanced techniques that we routinely use in analysis. It is motivated by our experience that many students only have an introductory knowledge of Excel, and thus waste a lot of time thrashing about unproductively. Many people think they know Excel, but overlook many efficient tools, such as Data Table and Goal Seek. It is also useful for a variety of other subjects.
Subject:
Applied Science
Engineering
Mathematics
Material Type:
Full Course
Provider:
MIT
Provider Set:
MIT OpenCourseWare
Author:
Cardin, Michel-Alexandre
de Neufville, Richard
09/01/2009
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
Students estimate the numbers of objects in situations in which counting is not feasible or necessary.
Subject:
Mathematics
Material Type:
Lesson Plan
Author:
Out Teach
07/22/2021
Educational Use
Rating
0.0 stars
In this video segment from Cyberchase, the Teleporter breaks and the CyberSquad must try to fix it using clues given to them by Ms. Fileshare.
Subject:
Mathematics
Material Type:
Lecture
Provider:
PBS LearningMedia
Provider Set:
PBS Learning Media: Multimedia Resources for the Classroom and Professional Development
Author:
U.S. Department of Education
WNET
09/10/2008
Educational Use
Rating
0.0 stars
In this Cyberchase video segment, the CyberSquad finds the right combination switch, lever and button that will deactivate the force field protecting the stolen Black Crystal.
Subject:
Mathematics
Material Type:
Lecture
Provider:
PBS LearningMedia
Provider Set:
PBS Learning Media: Multimedia Resources for the Classroom and Professional Development
Author:
U.S. Department of Education
WNET
09/10/2008
Unrestricted Use
CC BY
Rating
0.0 stars
This task aims to give students practice counting and recording the given number on quantities up to 20.
Subject:
Mathematics
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
08/27/2012
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
In this first module of Grade 1, students make significant progress towards fluency with addition and subtraction of numbers to 10 as they are presented with opportunities intended to advance them from counting all to counting on which leads many students then to decomposing and composing addends and total amounts.
Find the rest of the EngageNY Mathematics resources at https://archive.org/details/engageny-mathematics.
Subject:
Mathematics
Numbers and Operations
Material Type:
Module
Provider:
New York State Education Department
Provider Set:
EngageNY
05/11/2013
Conditional Remix & Share Permitted
CC BY-NC-SA
Rating
0.0 stars
Module 4 marks the next exciting step in math for kindergartners, addition and subtraction! They begin to harness their practiced counting abilities, knowledge of the value of numbers, and work with embedded numbers to reason about and solve addition and subtraction expressions and equations. In Topics A and B, decomposition and composition are taught simultaneously using the number bond model so that students begin to understand the relationship between parts and wholes before moving into formal work with addition and subtraction in the rest of the module.
Find the rest of the EngageNY Mathematics resources at https://archive.org/details/engageny-mathematics.
Subject:
Mathematics
Numbers and Operations
Material Type:
Module
Provider:
New York State Education Department
Provider Set:
EngageNY | 1,733 | 7,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.85369 |
https://leetcode.ca/2019-07-26-1334-Find-the-City-With-the-Smallest-Number-of-Neighbors-at-a-Threshold-Distance/ | 1,695,981,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00797.warc.gz | 389,945,862 | 10,432 | ##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1334.html
# 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance (Medium)
There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.
Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.
Notice that the distance of a path connecting cities i and j is equal to the sum of the edges' weights along that path.
Example 1:
Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2]
City 1 -> [City 0, City 2, City 3]
City 2 -> [City 0, City 1, City 3]
City 3 -> [City 1, City 2]
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
Example 2:
Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1]
City 1 -> [City 0, City 4]
City 2 -> [City 3, City 4]
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3]
The city 0 has 1 neighboring city at a distanceThreshold = 2.
Constraints:
• 2 <= n <= 100
• 1 <= edges.length <= n * (n - 1) / 2
• edges[i].length == 3
• 0 <= fromi < toi < n
• 1 <= weighti, distanceThreshold <= 10^4
• All pairs (fromi, toi) are distinct.
Related Topics:
Graph
## Solution 1.
• class Solution {
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
Map<Integer, List<int[]>> distancesMap = new HashMap<Integer, List<int[]>>();
for (int[] edge : edges) {
int city1 = edge[0], city2 = edge[1], distance = edge[2];
List<int[]> list1 = distancesMap.getOrDefault(city1, new ArrayList<int[]>());
List<int[]> list2 = distancesMap.getOrDefault(city2, new ArrayList<int[]>());
distancesMap.put(city1, list1);
distancesMap.put(city2, list2);
}
int[] reachableCounts = new int[n];
for (int i = 0; i < n; i++) {
int[] distances = getDistance(n, i, distancesMap);
for (int distance : distances) {
if (distance > 0 && distance <= distanceThreshold)
reachableCounts[i]++;
}
}
int minReachable = n - 1;
for (int i = 0; i < n; i++)
minReachable = Math.min(minReachable, reachableCounts[i]);
int cityIndex = 0;
for (int i = n - 1; i >= 0; i--) {
if (reachableCounts[i] == minReachable) {
cityIndex = i;
break;
}
}
return cityIndex;
}
public int[] getDistance(int n, int source, Map<Integer, List<int[]>> distancesMap) {
int[] distances = new int[n];
for (int i = 0; i < n; i++)
distances[i] = Integer.MAX_VALUE;
distances[source] = 0;
PriorityQueue<CityDistance> priorityQueue = new PriorityQueue<CityDistance>();
priorityQueue.offer(new CityDistance(source, 0));
while (!priorityQueue.isEmpty()) {
CityDistance cityDistance = priorityQueue.poll();
int city = cityDistance.city, distance = cityDistance.distance;
if (distance > distances[city])
continue;
List<int[]> adjacentCities = distancesMap.getOrDefault(city, new ArrayList<int[]>());
int newDistance = distance + nextDistance;
if (newDistance < distances[nextCity]) {
distances[nextCity] = newDistance;
priorityQueue.offer(new CityDistance(nextCity, newDistance));
}
}
}
return distances;
}
}
class CityDistance implements Comparable<CityDistance> {
public int city;
public int distance;
public CityDistance(int city, int distance) {
this.city = city;
this.distance = distance;
}
public int compareTo(CityDistance cityDistance2) {
return this.distance - cityDistance2.distance;
}
}
############
class Solution {
private int n;
private int[][] g;
private int[] dist;
private boolean[] vis;
private int inf = 1 << 30;
private int distanceThreshold;
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
this.n = n;
this.distanceThreshold = distanceThreshold;
g = new int[n][n];
dist = new int[n];
vis = new boolean[n];
for (var e : g) {
Arrays.fill(e, inf);
}
for (var e : edges) {
int f = e[0], t = e[1], w = e[2];
g[f][t] = w;
g[t][f] = w;
}
int ans = n, t = inf;
for (int i = n - 1; i >= 0; --i) {
int cnt = dijkstra(i);
if (t > cnt) {
t = cnt;
ans = i;
}
}
return ans;
}
private int dijkstra(int u) {
Arrays.fill(dist, inf);
Arrays.fill(vis, false);
dist[u] = 0;
for (int i = 0; i < n; ++i) {
int k = -1;
for (int j = 0; j < n; ++j) {
if (!vis[j] && (k == -1 || dist[k] > dist[j])) {
k = j;
}
}
vis[k] = true;
for (int j = 0; j < n; ++j) {
dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
}
}
int cnt = 0;
for (int d : dist) {
if (d <= distanceThreshold) {
++cnt;
}
}
return cnt;
}
}
• // OJ: https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/
// Time: O(V * (E + VlogV))
// Space: O(E)
class Solution {
typedef pair<int, int> iPair;
unordered_map<int, vector<iPair>> graph;
int th, N, minNum = INT_MAX;
int dijkstra(int i) {
priority_queue<iPair, vector<iPair>, greater<iPair>> q;
vector<int> dist(N, INT_MAX);
q.push(make_pair(0, i));
dist[i] = 0;
while (q.size()) {
int u = q.top().second;
q.pop();
for (auto ne : graph[u]) {
int v = ne.first;
int w = ne.second;
if (dist[v] > dist[u] + w) {
dist[v] = dist[u] + w;
q.push(make_pair(dist[v], v));
}
}
}
int cnt = 0;
for (int j = 0; j < N; ++j) {
if (j != i && dist[j] <= th) ++cnt;
}
return cnt;
}
public:
int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
th = distanceThreshold;
N = n;
for (auto e : edges) {
m[e[0]].push_back(make_pair(e[1], e[2]));
m[e[1]].push_back(make_pair(e[0], e[2]));
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt = dijkstra(i);
if (cnt <= minNum) {
minNum = cnt;
ans = i;
}
}
return ans;
}
};
• class Solution:
def findTheCity(
self, n: int, edges: List[List[int]], distanceThreshold: int
) -> int:
def dijkstra(u):
dist = [inf] * n
dist[u] = 0
vis = [False] * n
for _ in range(n):
k = -1
for j in range(n):
if not vis[j] and (k == -1 or dist[k] > dist[j]):
k = j
vis[k] = True
for j in range(n):
dist[j] = min(dist[j], dist[k] + g[k][j])
return sum(d <= distanceThreshold for d in dist)
g = [[inf] * n for _ in range(n)]
for f, t, w in edges:
g[f][t] = g[t][f] = w
ans = n
t = inf
for i in range(n - 1, -1, -1):
if (cnt := dijkstra(i)) < t:
t = cnt
ans = i
return ans
• func findTheCity(n int, edges [][]int, distanceThreshold int) int {
g := make([][]int, n)
dist := make([]int, n)
vis := make([]bool, n)
const inf int = 1e7
for i := range g {
g[i] = make([]int, n)
for j := range g[i] {
g[i][j] = inf
}
}
for _, e := range edges {
f, t, w := e[0], e[1], e[2]
g[f][t], g[t][f] = w, w
}
ans, t := n, inf
dijkstra := func(u int) (cnt int) {
for i := range vis {
vis[i] = false
dist[i] = inf
}
dist[u] = 0
for i := 0; i < n; i++ {
k := -1
for j := 0; j < n; j++ {
if !vis[j] && (k == -1 || dist[j] < dist[k]) {
k = j
}
}
vis[k] = true
for j := 0; j < n; j++ {
dist[j] = min(dist[j], dist[k]+g[k][j])
}
}
for _, d := range dist {
if d <= distanceThreshold {
cnt++
}
}
return
}
for i := n - 1; i >= 0; i-- {
cnt := dijkstra(i)
if t > cnt {
t = cnt
ans = i
}
}
return ans
}
func min(a, b int) int {
if a < b {
return a
}
return b
} | 2,427 | 7,515 | {"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} | 3.65625 | 4 | CC-MAIN-2023-40 | longest | en | 0.830985 |
http://www.education.com/reference/article/math-1-words-words-words/?page=2 | 1,371,685,436,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709805610/warc/CC-MAIN-20130516131005-00001-ip-10-60-113-184.ec2.internal.warc.gz | 421,949,018 | 29,675 | Education.com
# Math Terms to Know for CBEST Exam Study Guide (page 2)
By LearningExpress Editors
LearningExpress, LLC
Updated on Mar 29, 2011
#### Zero
Zero is an integer that is neither positive nor negative.
#### Whole Numbers
Whole numbers include all positive integers, as well as zero {0, 1, 2, 3…}. Like integers, whole numbers do not include numbers with fractions or decimals.
#### Digit
A digit is a single number symbol. In the number 1,246, each of the four numerals is a digit. Six is the ones digit, 4 is the tens digit, 2 is the hundreds digit, and 1 is the thousands digit. Knowing place names for digits is important when you're asked to round to a certain digit. Rounding will be covered in Math 3: Rounding, Estimation, and Decimal Equivalents.
#### Real Numbers
Real numbers include all numbers: negative, positive, zero, fractions, decimals, most square roots, and so on. Usually, the numbers used on the CBEST will be real numbers, unless otherwise stated.
#### Variables
Variables are symbols, such as x and y, that are used to replace numbers. The symbol is usually a letter of the alphabet, although occasionally, other symbols are used. When a math problem asks you to "solve for y," that means to figure out what number the letter is replacing. At other times, the problem requires you to work with the letters as if they were numbers. Examples of both will be covered in the lesson on algebra on page 120.
#### Reciprocal
The reciprocal of a fraction is the fraction turned upside down. For example, the reciprocal of is , and vice versa. The reciprocal of an integer is 1 over the integer. For example, the reciprocal of 2 (or ) is , and vice versa. To get the reciprocal of a mixed number such as , first change the number to an improper fraction and then turn it over .
#### Numerator and Denominator
The numerator of a fraction is the number on top, and the denominator is the number on the bottom. The numerator of is 6 and the denominator is 7.
#### = and ≠
The symbol = is called an equal sign. It indicates that the values on both sides of the sign are equal to each other. For example, 7 = 2 + 5. A line drawn through an equal sign (≠) indicates that the values on either side are not equal: 8 ≠ 4 + 5.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
### Today on Education.com
#### SUMMER LEARNING
June Workbooks Are Here!
#### EXERCISE
Get Active! 9 Games to Keep Kids Moving
#### TECHNOLOGY
Are Cell Phones Dangerous for Kids? | 593 | 2,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2013-20 | longest | en | 0.904618 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/4/lesson/4.2.3/problem/4-75 | 1,725,952,479,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00507.warc.gz | 275,944,572 | 16,167 | ### Home > APCALC > Chapter 4 > Lesson 4.2.3 > Problem4-75
4-75.
Given $f\left(x\right) = 2x$, write the equation of a vertical line that will divide $\int_0^{10}f\left(x\right)dx$ in half.
Sketch a graph of
$\int_{0}^{10}2xdx$
Find the area under the graph from $[0, 10]$ through integration.
$\int_{0}^{10}2xdx=x^{2}\left|\begin{matrix} 10\\ 0 \end{matrix}\right.=100$
Because you want to find the equation that would divide the area of the graph in half, you must solve the following equation:
$x^{2}\left|\begin{matrix} x\\ 0 \end{matrix}\right.=50$
$x^{2}= 50$
$x=\sqrt{50}=5\sqrt{2}$
Use the eTool below to examine the graph.
Click the link at right for the full version of the eTool: Calc 4-75 HW eTool | 242 | 720 | {"found_math": true, "script_math_tex": 8, "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-2024-38 | latest | en | 0.732439 |
http://sage-doc.sis.uta.fi/reference/modabvar/sage/modular/abvar/homspace.html | 1,555,627,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00086.warc.gz | 149,469,356 | 8,174 | # Spaces of homomorphisms between modular abelian varieties¶
EXAMPLES:
First, we consider J0(37). This Jacobian has two simple factors, corresponding to distinct newforms. These two intersect nontrivially in J0(37).
sage: J = J0(37)
sage: D = J.decomposition() ; D
[
Simple abelian subvariety 37a(1,37) of dimension 1 of J0(37),
Simple abelian subvariety 37b(1,37) of dimension 1 of J0(37)
]
sage: D[0].intersection(D[1])
(Finite subgroup with invariants [2, 2] over QQ of Simple abelian subvariety 37a(1,37) of dimension 1 of J0(37),
Simple abelian subvariety of dimension 0 of J0(37))
As an abstract product, since these newforms are distinct, the corresponding simple abelian varieties are not isogenous, and so there are no maps between them. The endomorphism ring of the corresponding product is thus isomorphic to the direct sum of the endomorphism rings for each factor. Since the factors correspond to abelian varieties of dimension 1, these endomorphism rings are each isomorphic to ZZ.
sage: Hom(D[0],D[1]).gens()
()
sage: A = D[0] * D[1] ; A
Abelian subvariety of dimension 2 of J0(37) x J0(37)
sage: A.endomorphism_ring().gens()
(Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(37) x J0(37),
Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(37) x J0(37))
sage: [ x.matrix() for x in A.endomorphism_ring().gens() ]
[
[1 0 0 0] [0 0 0 0]
[0 1 0 0] [0 0 0 0]
[0 0 0 0] [0 0 1 0]
[0 0 0 0], [0 0 0 1]
]
However, these two newforms have a congruence between them modulo 2, which gives rise to interesting endomorphisms of J0(37).
sage: E = J.endomorphism_ring()
sage: E.gens()
(Abelian variety endomorphism of Abelian variety J0(37) of dimension 2,
Abelian variety endomorphism of Abelian variety J0(37) of dimension 2)
sage: [ x.matrix() for x in E.gens() ]
[
[1 0 0 0] [ 0 1 1 -1]
[0 1 0 0] [ 1 0 1 0]
[0 0 1 0] [ 0 0 -1 1]
[0 0 0 1], [ 0 0 0 1]
]
sage: (-1*E.gens()[0] + E.gens()[1]).matrix()
[-1 1 1 -1]
[ 1 -1 1 0]
[ 0 0 -2 1]
[ 0 0 0 0]
Of course, these endomorphisms will be reflected in the Hecke algebra, which is in fact the full endomorphism ring of J0(37) in this case:
sage: J.hecke_operator(2).matrix()
[-1 1 1 -1]
[ 1 -1 1 0]
[ 0 0 -2 1]
[ 0 0 0 0]
sage: T = E.image_of_hecke_algebra()
sage: T.gens()
(Abelian variety endomorphism of Abelian variety J0(37) of dimension 2,
Abelian variety endomorphism of Abelian variety J0(37) of dimension 2)
sage: [ x.matrix() for x in T.gens() ]
[
[1 0 0 0] [ 0 1 1 -1]
[0 1 0 0] [ 1 0 1 0]
[0 0 1 0] [ 0 0 -1 1]
[0 0 0 1], [ 0 0 0 1]
]
sage: T.index_in(E)
1
Next, we consider J0(33). In this case, we have both oldforms and newforms. There are two copies of J0(11), one for each degeneracy map from J0(11) to J0(33). There is also one newform at level 33. The images of the two degeneracy maps are, of course, isogenous.
sage: J = J0(33)
sage: D = J.decomposition()
sage: D
[
Simple abelian subvariety 11a(1,33) of dimension 1 of J0(33),
Simple abelian subvariety 11a(3,33) of dimension 1 of J0(33),
Simple abelian subvariety 33a(1,33) of dimension 1 of J0(33)
]
sage: Hom(D[0],D[1]).gens()
(Abelian variety morphism:
From: Simple abelian subvariety 11a(1,33) of dimension 1 of J0(33)
To: Simple abelian subvariety 11a(3,33) of dimension 1 of J0(33),)
sage: Hom(D[0],D[1]).gens()[0].matrix()
[ 0 1]
[-1 0]
Then this gives that the component corresponding to the sum of the oldforms will have a rank 4 endomorphism ring. We also have a rank one endomorphism ring for the newform 33a (since it is again 1-dimensional), which gives a rank 5 endomorphism ring for J0(33).
sage: DD = J.decomposition(simple=False) ; DD
[
Abelian subvariety of dimension 2 of J0(33),
Abelian subvariety of dimension 1 of J0(33)
]
sage: A, B = DD
sage: A == D[0] + D[1]
True
sage: A.endomorphism_ring().gens()
(Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(33),
Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(33),
Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(33),
Abelian variety endomorphism of Abelian subvariety of dimension 2 of J0(33))
sage: B.endomorphism_ring().gens()
(Abelian variety endomorphism of Abelian subvariety of dimension 1 of J0(33),)
sage: E = J.endomorphism_ring() ; E.gens() # long time (3s on sage.math, 2011)
(Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3)
In this case, the image of the Hecke algebra will only have rank 3, so that it is of infinite index in the full endomorphism ring. However, if we call this image T, we can still ask about the index of T in its saturation, which is 1 in this case.
sage: T = E.image_of_hecke_algebra() # long time
sage: T.gens() # long time
(Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3)
sage: T.index_in(E) # long time
+Infinity
sage: T.index_in_saturation() # long time
1
AUTHORS:
• William Stein (2007-03)
• Craig Citro, Robert Bradshaw (2008-03): Rewrote with modabvar overhaul
class sage.modular.abvar.homspace.EndomorphismSubring(A, gens=None, category=None)
A subring of the endomorphism ring.
INPUT:
• A - an abelian variety
• gens - (default: None); optional; if given should be a tuple of the generators as matrices
EXAMPLES:
sage: J0(23).endomorphism_ring()
Endomorphism ring of Abelian variety J0(23) of dimension 2
sage: sage.modular.abvar.homspace.EndomorphismSubring(J0(25))
Endomorphism ring of Abelian variety J0(25) of dimension 0
sage: E = J0(11).endomorphism_ring()
sage: type(E)
<class 'sage.modular.abvar.homspace.EndomorphismSubring_with_category'>
sage: E.homset_category()
Category of modular abelian varieties over Rational Field
sage: E.category()
Category of endsets of modular abelian varieties over Rational Field
sage: E in Rings()
True
sage: TestSuite(E).run(skip=["_test_prod"])
abelian_variety()
Return the abelian variety that this endomorphism ring is attached to.
EXAMPLES:
sage: J0(11).endomorphism_ring().abelian_variety()
Abelian variety J0(11) of dimension 1
discriminant()
Return the discriminant of this ring, which is the discriminant of the trace pairing.
Note
One knows that for modular abelian varieties, the endomorphism ring should be isomorphic to an order in a number field. However, the discriminant returned by this function will be $$2^n$$ ( $$n =$$ self.dimension()) times the discriminant of that order, since the elements are represented as 2d x 2d matrices. Notice, for example, that the case of a one dimensional abelian variety, whose endomorphism ring must be ZZ, has discriminant 2, as in the example below.
EXAMPLES:
sage: J0(33).endomorphism_ring().discriminant()
-64800
sage: J0(46).endomorphism_ring().discriminant() # long time (6s on sage.math, 2011)
24200000000
sage: J0(11).endomorphism_ring().discriminant()
2
image_of_hecke_algebra(check_every=1)
Compute the image of the Hecke algebra inside this endomorphism subring.
We simply calculate Hecke operators up to the Sturm bound, and look at the submodule spanned by them. While computing, we can check to see if the submodule spanned so far is saturated and of maximal dimension, in which case we may be done. The optional argument check_every determines how many Hecke operators we add in before checking to see if this condition is met.
INPUT:
• check_every – integer (default: 1) If this integer is positive, this integer determines how many Hecke operators we add in before checking to see if the submodule spanned so far is maximal and saturated.
OUTPUT:
• The image of the Hecke algebra as an subring of self.
EXAMPLES:
sage: E = J0(33).endomorphism_ring()
sage: E.image_of_hecke_algebra()
Subring of endomorphism ring of Abelian variety J0(33) of dimension 3
sage: E.image_of_hecke_algebra().gens()
(Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3,
Abelian variety endomorphism of Abelian variety J0(33) of dimension 3)
sage: [ x.matrix() for x in E.image_of_hecke_algebra().gens() ]
[
[1 0 0 0 0 0] [ 0 2 0 -1 1 -1] [ 0 0 1 -1 1 -1]
[0 1 0 0 0 0] [-1 -2 2 -1 2 -1] [ 0 -1 1 0 1 -1]
[0 0 1 0 0 0] [ 0 0 1 -1 3 -1] [ 0 0 1 0 2 -2]
[0 0 0 1 0 0] [-2 2 0 1 1 -1] [-2 0 1 1 1 -1]
[0 0 0 0 1 0] [-1 1 0 2 0 -3] [-1 0 1 1 0 -1]
[0 0 0 0 0 1], [-1 1 -1 1 1 -2], [-1 0 0 1 0 -1]
]
sage: J0(33).hecke_operator(2).matrix()
[-1 0 1 -1 1 -1]
[ 0 -2 1 0 1 -1]
[ 0 0 0 0 2 -2]
[-2 0 1 0 1 -1]
[-1 0 1 1 -1 -1]
[-1 0 0 1 0 -2]
index_in(other, check=True)
Return the index of self in other.
INPUT:
• other - another endomorphism subring of the same abelian variety
• check - bool (default: True); whether to do some type and other consistency checks
EXAMPLES:
sage: R = J0(33).endomorphism_ring()
sage: R.index_in(R)
1
sage: J = J0(37) ; E = J.endomorphism_ring() ; T = E.image_of_hecke_algebra()
sage: T.index_in(E)
1
sage: J = J0(22) ; E = J.endomorphism_ring() ; T = E.image_of_hecke_algebra()
sage: T.index_in(E)
+Infinity
index_in_saturation()
Given a Hecke algebra T, compute its index in its saturation.
EXAMPLES:
sage: End(J0(23)).image_of_hecke_algebra().index_in_saturation()
1
sage: End(J0(44)).image_of_hecke_algebra().index_in_saturation()
2
class sage.modular.abvar.homspace.Homspace(domain, codomain, cat)
A space of homomorphisms between two modular abelian varieties.
Element
calculate_generators()
If generators haven’t already been computed, calculate generators for this homspace. If they have been computed, do nothing.
EXAMPLES:
sage: E = End(J0(11))
sage: E.calculate_generators()
free_module()
Return this endomorphism ring as a free submodule of a big $$\ZZ^{4nm}$$, where $$n$$ is the dimension of the domain abelian variety and $$m$$ the dimension of the codomain.
OUTPUT: free module
EXAMPLES:
sage: E = Hom(J0(11), J0(22))
sage: E.free_module()
Free module of degree 8 and rank 2 over Integer Ring
Echelon basis matrix:
[ 1 0 -3 1 1 1 -1 -1]
[ 0 1 -3 1 1 1 -1 0]
gen(i=0)
Return i-th generator of self.
INPUT:
• i - an integer
OUTPUT: a morphism
EXAMPLES:
sage: E = End(J0(22))
sage: E.gen(0).matrix()
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
gens()
Return tuple of generators for this endomorphism ring.
EXAMPLES:
sage: E = End(J0(22))
sage: E.gens()
(Abelian variety endomorphism of Abelian variety J0(22) of dimension 2,
Abelian variety endomorphism of Abelian variety J0(22) of dimension 2,
Abelian variety endomorphism of Abelian variety J0(22) of dimension 2,
Abelian variety endomorphism of Abelian variety J0(22) of dimension 2)
identity()
Return the identity endomorphism.
EXAMPLES:
sage: E = End(J0(11))
sage: E.identity()
Abelian variety endomorphism of Abelian variety J0(11) of dimension 1
sage: E.one()
Abelian variety endomorphism of Abelian variety J0(11) of dimension 1
sage: H = Hom(J0(11), J0(22))
sage: H.identity()
Traceback (most recent call last):
...
TypeError: the identity map is only defined for endomorphisms
matrix_space()
Return the underlying matrix space that we view this endomorphism ring as being embedded into.
EXAMPLES:
sage: E = End(J0(22))
sage: E.matrix_space()
Full MatrixSpace of 4 by 4 dense matrices over Integer Ring
ngens()
Return number of generators of self.
OUTPUT: integer
EXAMPLES:
sage: E = End(J0(22))
sage: E.ngens()
4 | 3,980 | 11,838 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-18 | latest | en | 0.761666 |
https://www.univerkov.com/the-abc-triangle-is-isosceles-the-angle-abc-is-equal-to-20-degrees-through-the-vertex-b-of-this-triangle/ | 1,726,560,537,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00436.warc.gz | 963,209,441 | 6,283 | # The ABC triangle is isosceles. the angle ABC is equal to 20 degrees through the vertex B of this triangle
The ABC triangle is isosceles. the angle ABC is equal to 20 degrees through the vertex B of this triangle a straight line EF is drawn parallel to the straight line AC, find 1) angle ABE 2) angle CBF 3) angle BAC and CB
Since the triangle ABC is isosceles, its angles at the base of the AC are equal.
Angle BAC = BCA = (180 – ABC) / 2 = (180 – 20) / 2 = 80.
According to the condition, the straight line EF is parallel to AC, then the angle ABE = BAC = 80, as the angles lying crosswise at the intersection of parallel straight lines EF and AC secant AC.
Similarly, the angle CBF = BCA = 80.
Answer: 1) Angle ABC = 80, 2) angle CBF = 800, 3) angle BAC = BCA = 80.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 286 | 1,087 | {"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} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.91007 |
https://justaaa.com/statistics-and-probability/609115-bill-alther-is-a-zoologist-who-studies-annas | 1,701,378,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00797.warc.gz | 382,498,596 | 10,104 | Question
# Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna).† Suppose that in a remote...
Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna).† Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were as follows.
3.7 2.9 3.8 4.2 4.8 3.1
The sample mean is x = 3.75 grams. Let x be a random variable representing weights of hummingbirds in this part of the Grand Canyon. We assume that x has a normal distribution and σ = 0.72gram. Suppose it is known that for the population of all Anna's hummingbirds, the mean weight is μ = 4.45 grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than 4.45 grams? Use α = 0.10.
(a) What is the level of significance?
State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
H0: μ < 4.45 g; H1: μ = 4.45 g; left-tailedH0: μ = 4.45 g; H1: μ > 4.45 g; right-tailed H0: μ = 4.45 g; H1: μ < 4.45 g; left-tailedH0: μ = 4.45 g; H1: μ ≠ 4.45 g; two-tailed
(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The Student's t, since n is large with unknown σ.The Student's t, since we assume that x has a normal distribution with known σ. The standard normal, since we assume that x has a normal distribution with unknown σ.The standard normal, since we assume that x has a normal distribution with known σ.
What is the value of the sample test statistic? (Round your answer to two decimal places.)
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Sketch the sampling distribution and show the area corresponding to the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?
At the α = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
(e) State your conclusion in the context of the application.
There is sufficient evidence at the 0.10 level to conclude that humming birds in the Grand Canyon weigh less than 4.45 grams.There is insufficient evidence at the 0.10 level to conclude that humming birds in the Grand Canyon weigh less than 4.45 grams.
The statistical software output for this problem is :
(a)
Level of significance = 0.10
Option C is correct.
(b)
Option D is correct.
Test statistics = -2.38
(c)
P-value = 0.0086
Graph is correct .
(d)
Option A is correct.
(e)
Option A is correct.
#### Earn Coins
Coins can be redeemed for fabulous gifts.
##### Need Online Homework Help?
Most questions answered within 1 hours.
##### Active Questions
• Case 12.1 Determine what variables are categorical (either nominal or ordinal scales), perform the appropriate descriptive...
• GDebi Enterprises is thinking of building a chemical processing plant to produce 4-hydroxy-3-methoxybenzaldehyde. The firm estimates...
• Suppose you are picking one number from the set of natural numbers from 97 to 4....
• What is the wavelength (in meters) of photons with the following energies? Part A.) 136 kJ/mol...
• In your own words, define nonverbal communication and explain three differences between verbal and nonverbal communication...
• A new fuel injection system has been engineered for pickup trucks. The new system and the... | 952 | 3,823 | {"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-2023-50 | latest | en | 0.922571 |
https://forum.allaboutcircuits.com/threads/unbalanced-three-phase-how-to-calculate-total-power.113527/ | 1,611,232,457,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00376.warc.gz | 332,854,986 | 20,592 | Unbalanced Three Phase : How to calculate total power ?
kbarb
Joined Dec 29, 2008
6
In our building, we have a four wire 208v/120v Wye service.
I'm trying to measure the watts used by a subpanel, but the loads are unbalanced.
For example, for the 208v lines (black, red, blue) I get :
-- 42A
-- 84A
-- 80A
I know about P = √3 Vl × Il × cos φ where cos φ = power factor, but that's for balanced scenarios, right ?
I'm going to have to guess at the power factor - most of the loads are lighting & computers/monitors.
But anyone know how to calculate the total power draw ?
If you look on this page, Three Phase Current - Simple Calculation and look at the "Balanced Voltages" section it implies that for unbalanced systems you can still calc as such :
• the line to neutral (phase) voltage VLN = 208/√3 = 120 V
• phase 1 apparent power = 42 x 120 = 5040 VA = 5.04 kVA
• phase 2 apparent power = 84 x 120 = 10080 VA = 10.08 kVA
• phase 3 apparent power = 80 x 120 = 9600 VA = 9.6 kVA
• Total three phase power = 5.04 + 10.08 + 9.6 = 24.72 kVA
Anyone know if this is correct ?
WBahn
Joined Mar 31, 2012
26,398
Yes, that looks correct.
As you say, you would have to guess at the power factor and that could have a significant effect on your real power.
As a quick sanity check, you can take the average of the three currents (which would be 68.7 A) and throw that at the balanced three-phase formula and see if you get something that is in the ball park of when you do the per-leg unbalanced computation. In this case that comes out to 24.74 kVA, which is a pretty small ballpark.
Looking at the formulas, these work out exactly.
kbarb
Joined Dec 29, 2008
6
And I like your sanity check exercise - I can always use a bit of that.
Thanks again,
Kent
KL7AJ
Joined Nov 4, 2008
2,229
Yikes....this is not as simple as it appears. But yes....it is probably unsolvable without knowing the power factor. | 550 | 1,909 | {"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} | 3.796875 | 4 | CC-MAIN-2021-04 | longest | en | 0.945461 |
https://www.jiskha.com/questions/1183203/p-5-and-q-1-2-are-points-on-a-straight-line-find-the-equation-of-the-perpendicular | 1,590,815,704,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00530.warc.gz | 777,211,723 | 6,138 | # math
P (5,) and Q (-1, 2) are points on a straight line. Find the equation of the perpendicular bisector of PQ: y = mx+c
1. 👍 0
2. 👎 0
3. 👁 141
1. Perpendicular bisector is the negative reciprocal of the slope so if the slope is 2 the negative reciprocal is -1/2.
1. 👍 0
2. 👎 0
2. how do you get?
1. 👍 0
2. 👎 0
3. what is the coordinates of P
1. 👍 0
2. 👎 0
## Similar Questions
1. ### graphing
Q: How could the function y=3t^2 +4 be plotted on a cartesian graph to produce a straight line? What would be the numerical values of the slope and the intercept of the line? For this it's explained in the lab book for physics as
asked by ~christina~ on September 2, 2007
2. ### MATH HELP
Which of the following statements best describes the graph of 4x − y = 1? It is a straight line joining the points (0, −1), (1, 3), and (−1, −5). It is a straight line joining the points (4, −3), (−1, 2), and (−4,
asked by Oscar on December 20, 2015
3. ### Maths
Hi, need a bit of help on this one. A straight line passes through the points (4,3) and (10,0), I need to write down the equation of the line in the form y=mx+c, I have tried but I cant seem to get the second coordinates to
asked by Del on July 9, 2010
4. ### Chemistry
Given data collected for a liquid. Equilibrium vapor pressure in torr: 4.6, 17.5, 149.4, 525.8 T in *C 0., 20., 60., 90. *Find T in K, Find 1/T in K-1 and 1n VP. *Then make a graph of 1n vapor pressure (1n VP) versus 1/T (where
asked by Anonymous on July 29, 2008
5. ### algebra
how would i find an equation of a line that goes through points(1,6) and (3,10)?? thanks A straight line is y=mx+b Substitute the points to make two equations. 6=m(1)+b 10=m(3)+b Two equations; two unknowns, m and b. Solve for m
asked by jasmine on December 22, 2006
1. ### math help & correction
Problem #1 Is this correct or wrong? Find the slope of the line passing through the points(-1, -1)and(-1, 2). Write the equation of the line. For this one I KEEP GETTING Y= - (3)/(2)x-2.5 Problem #2 Find the y-intecept and slope
asked by jacky on March 31, 2007
2. ### calculus
hi we have been given a graph and we need to fing its equation. the graph is logistic. we are asked to take any six points on it and then apply appropriate log to get a straight line. i don't know how to get that straight line? i
asked by thoung on June 11, 2009
3. ### maths
What would the equation be of a straight line that passes through the points (1,1) and (3,3) ?? write the equation for the line: y= mx+ b put the first set of points in.. 1=1m + b then the second equation 3=3m + b subtract the
asked by kat on June 9, 2007
4. ### Math
The graph of the equation y=4-x consists of all the points in the coordinate plane that satisfy the equation. List 5 points that satisfy y= 4-x. Also, what do you think is the minimum number of points you need to plot in order to
asked by victoria on November 21, 2006
5. ### math
Q2.Plot three points with second coordinate equal yo (1.3) and join them . Q3.What is the slope of the line y= -1/3x +3 (B)Does it slope upwards? (C) It is steeper than the line y=-1/2x -5? Give reason for your answer. Q4.The
asked by athanasia on March 16, 2015
6. ### help!
Can you please give me one of them so I cansolve for the other one?PLease! The general formula for a straight line is y = mx + b. Just plug the points into the equation to arrive at two equations. 6=2m+b 10=4m+b you have two
asked by margie on November 1, 2006 | 1,101 | 3,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-24 | latest | en | 0.902827 |
http://nrich.maths.org/public/leg.php?code=149&cl=2&cldcmpid=70 | 1,506,314,135,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690318.83/warc/CC-MAIN-20170925040422-20170925060422-00110.warc.gz | 253,454,719 | 9,530 | # Search by Topic
#### Resources tagged with Area similar to The Great Tiling Count:
Filter by: Content type:
Stage:
Challenge level:
### There are 93 results
Broad Topics > Measures and Mensuration > Area
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
### Ribbon Squares
##### Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
### Lawn Border
##### Stage: 1 and 2 Challenge Level:
If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time?
### Numerically Equal
##### Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
### It Must Be 2000
##### Stage: 2 Challenge Level:
Here are many ideas for you to investigate - all linked with the number 2000.
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### Uncanny Triangles
##### Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
### Through the Window
##### Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
### Tiles in the Garden
##### Stage: 2 Challenge Level:
How many tiles do we need to tile these patios?
### Shaping It
##### Stage: 1 and 2 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
### Geoboards
##### Stage: 2 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Rope Mat
##### Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
### Making Boxes
##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### The Big Cheese
##### Stage: 2 Challenge Level:
Investigate the area of 'slices' cut off this cube of cheese. What would happen if you had different-sized block of cheese to start with?
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### Making Squares
##### Stage: 2
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### Fencing Lambs
##### Stage: 2 Challenge Level:
A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales.
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### A Square in a Circle
##### Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Wrapping Presents
##### Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
### Cutting it Out
##### Stage: 1 and 2 Challenge Level:
I cut this square into two different shapes. What can you say about the relationship between them?
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Transformations on a Pegboard
##### Stage: 2 Challenge Level:
How would you move the bands on the pegboard to alter these shapes?
### Inside Seven Squares
##### Stage: 2 Challenge Level:
What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?
### A Day with Grandpa
##### Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
### From One Shape to Another
##### Stage: 2
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### How Random!
##### Stage: 2 Challenge Level:
Explore this interactivity and see if you can work out what it does. Could you use it to estimate the area of a shape?
### Shape Draw
##### Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
### Overlapping Squares
##### Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
### Fitted
##### Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Triangle Island
##### Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
### Different Sizes
##### Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted? | 1,967 | 8,726 | {"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} | 3.78125 | 4 | CC-MAIN-2017-39 | latest | en | 0.880257 |
http://portraitofacreative.com/books/on-riemanns-theory-of-algebraic-functions-and-their-integrals | 1,542,170,112,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741628.8/warc/CC-MAIN-20181114041344-20181114063344-00081.warc.gz | 288,722,863 | 8,928 | # Download On Riemann's Theory of Algebraic Functions and their by Felix Klein PDF
By Felix Klein
This quantity introduces Riemann's method of multiple-value services and the geometrical illustration of those features via Riemann surfaces. It concentrates at the kind of services that may be outlined upon those surfaces, limiting the remedy to rational features and their integrals demonstrating how Riemann's mathematical rules approximately Abelian integrals should be derived by means of reflecting upon the move of electrical present on surfaces. This paintings is without doubt one of the top introductions to the origins of topological difficulties.
Similar algebra books
Topics in Computational Algebra
The most objective of those lectures is first to in short survey the basic con nection among the illustration idea of the symmetric team Sn and the idea of symmetric services and moment to teach how combinatorial tools that come up certainly within the idea of symmetric features bring about effective algorithms to specific a variety of prod ucts of representations of Sn when it comes to sums of irreducible representations.
Additional resources for On Riemann's Theory of Algebraic Functions and their Integrals
Example text
1867, pp. 169 et seq. ] introductory remarks. 22 be particularly well observed in Plateau’s experiments. We shall attempt to define such states of motion also by a potential and we shall especially enquire what is the case in steady motion. The proper extension of our conception of a potential presents itself at once. Let u be a function of position on the surface and let the curves u = const. be drawn; moreover let the direction of fluid-motion on the surface at every point be perpendicular to the curve u = const.
It is more difficult to prove the converse, that the equality of the p’s is a sufficient condition for the possibility of a uniform correspondence between the two surfaces. For proof of this ∗ Deformations by means of continuous functions only are considered here. Moreover in the arbitrary surfaces of the text certain particular occurrences are for the present excluded. It is best to imagine them without singular points; branch-points and hence the penetration of one sheet by another will be considered later on (§ 13).
Riemann’s theory. 45 In the conjugate streaming, the curves of latitude play the part of the meridians in the first example; this is shown in the following drawing: The direction of motion in this case is the same on the upper and lower sides. Let us now deform the anchor-ring, p = 1, by causing two excrescences to the right of the figure, roughly speaking, to grow from it, which gradually bend towards each other and finally coalesce. We then have a surface p = 2 and on it a pair of conjugate streamings as illustrated by Figures 23 and 24. | 572 | 2,842 | {"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-2018-47 | longest | en | 0.929674 |
https://bloga8.com/pregnancy/are-you-6-months-pregnant-at-24-weeks.html | 1,680,065,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00052.warc.gz | 175,666,680 | 24,693 | # Question: Are You 6 Months Pregnant At 24 Weeks?
Contents
If you’re 24 weeks pregnant, you’re in month 6 of your pregnancy.
## What month is 24 weeks of pregnancy?
Now that you’re 24 weeks pregnant, you’re five months and about one week pregnant. (If you do simple multiplication, it seems like 24 weeks would be six months pregnant, but remember: all months except February are longer than just four weeks, and those extra days add up!)
## How many weeks do you have to be to be 6 months pregnant?
1-4 weeks= 1 month 5-8 weeks= 2 months 9-12 weeks= 3 months 13-17 weeks= 4 months 18-22 weeks= 5 months 23-27 weeks= 6 months 28-32 weeks= 7 months 33-36 weeks = 8 months 36-40 weeks= 9 months The confusion here is that while at 9 weeks you are fully 2 months pregnant and in the first week of your third month of
## What position should baby be in at 24 weeks?
The size of your belly at 24 weeks pregnant might make it difficult to find a comfortable sleeping position. Some well-placed pillows can help! Try sleeping on your side with your knees bent and with one pillow between your legs and another one under your belly for support.
## How big should I be at 24 weeks?
24 weeks pregnant ultrasound
The baby is the size of a foot, his body averaging 12.6 inches (32 cm), weight is around 1.5 pounds (670 g). This week, your baby’s normal heart rate to will range anywhere from 120 to 160 beats per minute. At this stage, babies have developed most of their systems and organs.
## How often should you feel your baby move at 24 weeks?
How often should I feel my baby moving? Once you easily feel your baby’s movements (around 20–24 weeks), you’ll notice them more often. However, you may not always feel your baby’s movements, especially if you are busy and are not paying attention.
## Is 25 weeks the 3rd trimester?
The third trimester begins in week 28 of pregnancy and lasts until you give birth, which may be around week 40 of pregnancy. In other words, your third trimester lasts from month 7 through month 9 of pregnancy.
## What to expect when expecting 6 months?
You might be energized and excited about your baby’s arrival in a few months, even if you’re dealing with common pregnancy symptoms like heartburn, hot flashes, and backaches.
Common Pregnancy Symptoms at 6 Months Pregnant
• Heartburn.
• Backaches.
• Hot flashes.
• Dizziness.
• Leg cramps.
• Fast heartbeat.
## What are signs of having a boy?
20 Ways to Tell if You Are Pregnant with a Boy
1. Baby’s heart rate is slower than 140 beats per minute.
2. Morning sickness that’s not too bad.
3. Lustrous hair and skin.
4. If it’s all out front.
5. A hankering for chips, not ice cream.
6. Big appetite.
7. Keeping it on the downlow.
8. The wedding ring spin.
## Do babies kick at 6 months?
In the 6th month of pregnancy, baby movements tend to grow in intensity as the pregnancy progresses. “Around the 24th week, my baby moved in response to my voice!” Towards the end of the 6th month of pregnancy, these movements turn into noticeable kicks.
## Is 24 weeks second or third trimester?
Conception to about the 12th week of pregnancy marks the first trimester. The second trimester is weeks 13 to 27, and the third trimester starts about 28 weeks and lasts until birth. This slide show will discuss what occurs to both the mother and baby during each trimester.
## Do boy babies kick more?
One study, published in 2001 in the journal Human Fetal and Neonatal Movement Patterns, found that boys may move around more in the womb than girls. The average number of leg movements was much higher in the boys compared to the girls at 20, 34 and 37 weeks, that study found.
## What happens at 24 weeks in pregnancy?
At 24 weeks into your pregnancy, your baby would be considered viable and with a chance of survival if they were born prematurely. Even though your baby is mainly just laying down flat now, and growing a bit longer, there are some subtle changes also taking place within their body.
## What should my belly measure at 24 weeks?
The measurement in centimeters from the top of your pubic bone to the top of your uterus (the fundus) should be about the same as the number of weeks you are pregnant, with an allowance of up to 2 cm either way. For example, if you are 26 weeks’ pregnant, you should measure between 24 and 28 cm.
## How much weight should I have gained at 24 weeks pregnant?
On a trimester basis in a woman with a normal pre-pregnancy weight: First trimester: 1-4.5 pounds. Second trimester: 1-2 pounds per week. Third trimester: 1-2 pounds per week.
## How do you feel at 24 weeks pregnant?
During week 24, your symptoms may include:
• stretch marks.
• itchy skin.
• dry or itchy eyes.
• slight breast colostrum production.
• occasional Braxton-Hicks contractions.
• backaches.
• constipation.
## How can I get baby to kick?
8 Tricks for Getting Your Baby to Move in Utero
1. Have a snack.
2. Do some jumping jacks, then sit down.
3. LEARN MORE: Fetal Movement During Pregnancy and How to Do a Kick Count.
4. Gently poke or jiggle your baby bump.
5. Shine a flashlight on your tummy.
6. Lie down.
7. Talk to baby.
8. Do something that makes you nervous (within reason).
## What are the signs of a dead baby in the womb?
Symptoms of miscarriage and stillbirth
• Spotting or bleeding from your vagina.
• Mild to severe abdominal cramps.
• Dizziness.
• High fever.
• No foetal movements.
• No heartbeat can be detected.
## When can dads feel kicks?
You’ll probably begin to feel your baby move, yourself, between 16 and 22 weeks. It usually takes a little longer for your baby’s activity to be noticeable outside your belly. That said, each pregnancy is unique, so your partner may feel your baby kick before week 20.
## What trimester do you gain the most weight?
How much weight to gain and when
Recommended Weight Gain According to ACOG
Pre-pregnancy Weight Category Body Mass Index Recommended Range of Total Weight (pounds)
Underweight Less than 18.5 28–40
Normal Weight 18.5–24.9 25–35
Overweight 25–29.9 15–25
1 more row
## What should I eat to increase fetal weight?
Daily recommendations: Include 2 to 3 servings of vegetables, 2 servings of fruits, at least 3 servings of whole grain bread, cereals, pasta, 2 to 3 servings of lean protein (e.g., meat, fish, and poultry). Vitamin D: Vitamin D works with calcium to help the baby’s bones and teeth develop.
## What a fetus looks like at 25 weeks?
Baby at 25 weeks is as big as a head of cauliflower, measuring 13.6 inches in length and weighing nearly 1.5 pounds.
## What does a 6 month fetus look like?
By the end of the sixth month, your baby is about 12 inches long and weighs about 2 pounds. His or her skin is reddish in color, wrinkled, and veins are visible through the baby’s translucent skin. Baby’s finger and toe prints are visible. The eyelids begin to part and the eyes open.
## How does it feel when baby moves in womb?
Pregnant women describe their baby’s movements as butterflies, nervous twitches, or a tumbling motion. At first, it may be hard to tell whether your baby has moved. Second- and third-time moms are more adept at distinguishing those first baby movements fromgas,hungerpangs, and other internal motions.
## Why does baby move after I eat?
Babies often kick in reaction to foods you’ve eaten. Spicy foods get many babies up and kicking. Most fetuses will move around thirty times every hour, although you won’t necessarily feel all of those movements due to the baby’s positioning within the womb.
## Can the baby feel when you rub your belly?
By now, you have probably already experienced the sensation of your baby kicking and turning inside you. He or she can feel you as well. As you rub your belly, you may be able to identify parts of your baby. If you gently massage the lump, the baby may pull back.
## Are you more hungry when pregnant with a boy?
Boys make expectant mothers feel hungry. But the idea that the more hungry the expectant mother is, the more likely she is to have a boy, has been backed in the United States. A study of 244 pregnant women found that, on average, those carrying boys consumed 10 per cent more calories than the women carrying girls.
## Are you more tired when pregnant with a boy?
Fatigue During Pregnancy First Trimester
Your body is producing more blood to carry nutrients to your growing baby. Your blood sugar levels and blood pressure are also lower. Hormones, especially increased progesterone levels, are responsible for making you sleepy.
Photo in the article by “Flickr” `https://www.flickr.com/photos/littlelovemonster/10433739754` | 2,080 | 8,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-14 | latest | en | 0.94482 |
http://www.cplusplus.com/forum/beginner/115583/ | 1,503,182,724,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00306.warc.gz | 537,353,854 | 4,187 | ### Long value print
Hi there.
I have a simple problem. I know You solve it easily.. :p
How can I print this type of floating number?
I dont need straight cout. I need this value as a result of some calculation.
5.48816e+23
Such a number exceeds FP precision. Why do you need such a huge number with so many significant digits?
This is actually an online judge problem.
I'm trying to write a function which returns exponent value.
The above number is a result of this-
95.123^12
Last edited on
Think of it like you would a bit shift register and just parse it inside of a custom struct.
You'll need to use a bignum implementation.
95123**12 =
548815620517731830194541899025343415715973535967221869852721
Remember how you handle decimal places in multiplication problems, so 3 places * 12 = 36 places, giving you
548815620517731830194541.899025343415715973535967221869852721
So to print it, print all but the last n-places of the digits, then a decimal point, then the remaining digits.
Remember, this number is so big, we are talking a bignum -- an integer -- that we are treating as if it were a floating point value.
Going the other way (playing with differences) doesn't help much with precision, so I can't suggest that. Unless I'm missing something really obvious.
Hope this helps.
Topic archived. No new replies allowed. | 325 | 1,337 | {"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-2017-34 | latest | en | 0.897397 |
https://au.mathworks.com/matlabcentral/cody/problems/2516-element-by-element-multiplication-of-two-vectors/solutions/1916273 | 1,579,349,510,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592565.2/warc/CC-MAIN-20200118110141-20200118134141-00412.warc.gz | 341,764,907 | 15,517 | Cody
# Problem 2516. Element by element multiplication of two vectors
Solution 1916273
Submitted on 30 Aug 2019 by Abbie Lund
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A=[1 2 3]; B=[1 1 1]; y_correct=[1 2 3]; assert(isequal(ele_wise(A,B),y_correct));
2 Pass
A=[1 2 3]; B=[1 2 3]; y_correct=[1 4 9]; assert(isequal(ele_wise(A,B),y_correct));
3 Pass
A=[1 8]; B=[8 1]; y_correct=[8 8]; assert(isequal(ele_wise(A,B),y_correct)); | 191 | 559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.574199 |
https://testbook.com/question-answer/what-is-the-average-number-of-girls-enrolled-in-th--5ff4191e45d43d57e8a7977b | 1,627,445,237,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00382.warc.gz | 573,760,413 | 18,661 | ## Question:
### Comprehension:
Direction: Consider the following table that shows the number of girls enrolled in three different Universities A, B and C over the years 2015 to 2019. Based on the data in the table, answer the question.
Year-wise Number of Girls in Different Universities Year Number of girls in a University A B C 2015 15,000 10,000 5,000 2016 10,000 17,500 20,000 2017 22,500 12,000 15,000 2018 20,000 25,000 15,000 2019 25,000 20,000 22,500
# What is the average number of girls enrolled in the year 2017 in all the three Universities together?
Free Practice With Testbook Mock Tests
## Options:
1. 20,000
2. 16,500
3. 10,000
4. 18,500
### Correct Answer: Option 2 (Solution Below)
This question was previously asked in
Official Paper 14: Held on 4th Nov 2020 Shift 1
## Solution:
The number of girls enrolled in University A in the year 2017 = 22500
The number of girls enrolled in University B in the year 2017 = 12000
The number of girls enrolled in University C in the year 2017 = 15000
The total number of girls enrolled in all three Universities A, B and C together = 22500 + 12000 + 15000
⇒ 49500
Average = sum of enrolled girls/ number of Universities
⇒ 49500/3 = 16500
∴ The Average number of Enrolled girls in the year 2017 in three Universities is 16500 | 390 | 1,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-31 | latest | en | 0.852556 |
https://wisdombase.cf/free-energy/what-is-free-energy-with-magnet-spontaneous-gibbs-free-energy-equation-with-entropy-journal/ | 1,620,481,970,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.7/warc/CC-MAIN-20210508121446-20210508151446-00346.warc.gz | 644,613,217 | 31,653 | Categories
# What is free energy with magnet? – Spontaneous Gibbs Free Energy Equation With Entropy Journal
What kind of energy does magnet have?
Free energy with magnetic field can be defined as the amount of energy (potential energy) the magnetic field has to transfer between the magnet (the body) and whatever the material is (whether it is mass or energy). That is, if a magnet with a strong magnetic field and a field of force is placed on a mass, free energy can be said to be the product of the net force and the magnitude of the magnetic field.
What is magnet’s magnetic force?
How strong do magnets have magnetic torque?
Which kind of energy (electromagnetic or non-magnetic) can be created by magnets?
Which kind of energy (electromagnetic or non-magnetic) can be de-energized by magnets?
A magnet always needs another magnet to be turned to work, for the force of action comes from that of the other magnet (an electromagnet). We use the term ‘magnet’ to describe a magnetic field and the term ‘field’ to describe the direction of the magnetic field. If we have two magnets and point towards each other, the two magnets will pull each other towards the other, the magnetic field (as a whole) can be called a field (like an electric field) and the direction of its direction will be called a magnetic field. The two magnet’s fields will, when combined into a single vector (a vector that points in three directions), cancel each other out. What we mean by that is that they will not interfere in any way, and it can be said that they are mutually repelling magnets.
An electric conductor (as opposed to a magnetic conductor, which is an electromagnetic conductor) is one type of magnetic field (an electric field), and a non-magnetized conductor is another type. So, for example, a conductive belt, made up of a set of magnets, is called an electric conductor and a non-magnetized belt is called a non-magnetic conductor. In some cases (e.g. when magnetized conductors are used in an electric belt system), this is also called an electromagnet. A magnetic field produces a force (a force that makes one magnet push another and the two magnets will cancel each other out). The strength of the force depends on the strength of the magnetic field. If the magnetic field is strong enough, the force between the magnets is strong enough to turn a magnet to work.
The
free energy generator motor kit, nikola tesla free energy secret work affairs relationship, free energy definition for kids, gibbs free energy cell biology, free energy generator homemade with magnet | 546 | 2,583 | {"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} | 3.71875 | 4 | CC-MAIN-2021-21 | latest | en | 0.952954 |
https://quomodocumque.wordpress.com/tag/kowalski/ | 1,679,662,587,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00612.warc.gz | 543,755,406 | 18,474 | ## Expander graphs, gonality, and variation of Galois representations
Suppose you have a 1-dimensional family of polarized abelian varieties — or, just to make things concrete, an abelian variety A over Q(t) with no isotrivial factor.
You might have some intuition that abelian varieties over Q don’t usually have rational p-torsion points — to make this precise you might ask that A_t[p](Q) be empty for “most” t.
In fact, we prove (among other results of a similar flavor) the following strong version of this statement. Let d be an integer, K a number field, and A/K(t) an abelian variety. Then there is a constant p(A,d) such that, for each prime p > p(A,d), there are only finitely many t such that A_t[p] has a point over a degree-d extension of K.
The idea is to study the geometry of the curve U_p parametrizing pairs (t,S) where S is a p-torsion point of A_t. This curve is a finite cover of the projective line; if you can show it has genus bigger than 1, then you know U_p has only finitely many K-rational points, by Faltings’ theorem.
But we want more — we want to know that U_p has only finitely many points over degree-d extensions of K. This can fail even for high-genus curves: for instance, the curve
C: y^2 = x^100000 + x + 1
has really massive genus, but choosing any rational value of x yields a point on C defined over a quadratic extension of Q. The problem is that C is hyperelliptic — it has a degree-2 map to the projective line. More generally, if U_p has a degree-d map to P^1, then U_p has lots of points over degree-d extensions of K. In fact, Faltings’ theorem can be leveraged to show that a kind of converse is true.
So the relevant task is to show that U_p admits no map to P^1 of degree less than d; in other words, its gonality is at least d.
Now how do you show a curve has large gonality? Unlike genus, gonality isn’t a topological invariant; somehow you really have to use the geometry of the curve. The technique that works here is one we learned from an paper of Abramovich; via a theorem of Li and Yau, you can show that the gonality of U_p is big if you can show that the Laplacian operator on the Riemann surface U_p(C) has a spectral gap. (Abramovich uses this technique to prove the g=1 version of our theorem: the gonality of classical modular curves increases with the level.)
We get a grip on this Laplacian by approximating it with something discrete. Namely: if U is the open subvariety of P^1 over which A has good reduction, then U_p(C) is an unramified cover of U(C), and can be identified with a finite-index subgroup H_p of the fundamental group G = pi_1(U(C)), which is just a free group on finitely many generators g_1, … g_n. From this data you can cook up a Cayley-Schreier graph, whose vertices are cosets of H_p in G, and whose edges connect g H with g_i g H. Thanks to work of Burger, we know that this graph is a good “combinatorial model” of U_p(C); in particular, the Laplacian of U_p(C) has a spectral gap if and only if the adjacency matrix of this Cayley-Schreier graph does.
At this point, we have reduced to a spectral problem having to do with special subgroups of free groups. And if it were 2009, we would be completely stuck. But it’s 2010! And we have at hand a whole spray of brand-new results thanks to Helfgott, Gill, Pyber, Szabo, Breuillard, Green, Tao, and others, which guarantee precisely that Cayley-Schreier graphs of this kind, (corresponding to finite covers of U(C) whose Galois closure has Galois group a perfect linear group over a finite field) have spectral gap; that is, they are expander graphs. (Actually, a slightly weaker condition than spectral gap, which we call esperantism, is all we need.)
Sometimes you think about a problem at just the right time. We would never have guessed that the burst of progress in sum-product estimates in linear groups would make this the right time to think about Galois representations in 1-dimensional families of abelian varieties, but so it turned out to be. Our good luck.
## Speaking of Emmanuel Kowalski
How can I have forgotten to put his blog on my blogroll? Well, it’s up there now — a great place for thoughtful posts on number theory both contemporary and historical, not to mention engaging diversions on mysterious symbols on slide rules and the important question of whether Grothendieck appears in the movie of Zazie dans le métro.
In Emmanuel’s most recent post, he reports on something I too should have mentioned; that the beautiful result of Bilu and Parent about rational points of X^split(p), whose original version fell prey to a subtle error, has apparently been corrected, and the original result is now once again independent of GRH. | 1,177 | 4,731 | {"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-2023-14 | latest | en | 0.934508 |
https://physics.stackexchange.com/questions/81700/how-can-a-block-which-is-not-receiving-the-direct-force-have-a-greater-accelerat/81737 | 1,717,017,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00248.warc.gz | 385,938,883 | 43,579 | # How can a block which is not receiving the direct force have a greater acceleration?
I solved it like this:
$$F(\text{st max})=5\text{ N}$$
For the top block,
\begin{align} 6\text{ N} - 5\text{ N} &= 1a \\ a &= 1\ \mathrm{m/s^2} \end{align}
For the lower block, the driving force will be the frictional force, so
\begin{align} 2a &= 5\text{ N} \\ a &= \frac{5}{2} = 2.5\ \mathrm{m/s^2} \end{align}
I am confused as to how the lower block could have acceleration greater than the upper block, since the force is acting on the top block.
• I thought lower block never get more acc than above in this case because if it does then friction force would act on it towards left. Oct 21, 2013 at 20:35
• This question is related, and possibly even a duplicate. Oct 21, 2013 at 21:06
• I got the answer 5N is not critical or max force because it is also depend upon mass of lower block. Oct 21, 2013 at 21:30
• Solution: F-f=m1a and f=ma2 after solving it we get F=((m1+m2)/m2)*f so your F(max) will be 7.5N not 5N Oct 21, 2013 at 21:30
• No, that's still a comment. Look further down, to the box just above the "Post Your Answer" button. Oct 21, 2013 at 22:25
The basic assumption that friction = u x N u = coefficient of friction N = Normal force (in this case the weight of the block) Above assumption is valid only if there is relative motion between the two blocks i.e a case of sliding motion, but before we consider that sliding occurs we should verify whether the block are moving relative to each other of not i.e. checking for static friction.
Now the maximum value static friction can reach is uN i.e. Sliding/kinetic friction but can also be lesser than that. Taking that into account and assuming friction to be f (a variable) and no relative motion between the blocks. No relative motion means that both blocks will have same acceleration.
Calculations :
6N− f =a m/s2 (for small block) f = 2a m/s2 (for big block)
substituting f=2a for small block
6N - 2a = a m/s2
6N = 3a m/s2
2 m/s2 = a
both blocks having same acceleration, hence no relative motion.
Value of friction in this condition is 2 x 2 = 4N which is less than uxN = 5N
You have got the right answers, just are interpreting it a little bit wrong;
The net force on upper block is $6N - 5N = 1N$ and it's accelerations is $1 ms^{-2}$ while the force on bottom block is $5N$ and it's acceleration is $2.5ms^{-2}$ but you are missing the fact that the acceleration of the bottom block is with respect to ground while that of upper block is with respect to lower block. Therefore, if you see from ground bottom block accelerates with $2.5ms^{-2}$ while upper block moves with $3.5ms^{-2}$.
I apologise for misunderstanding and posting the wrong answer earlier
• This is incorrect, accelerations are always measured and calculated relative to an inertial frame (ie one undergoing zero acceleration (in Newtonian mechanics)) Thus assuming the acceleration of the upper block is relative to the lower block is incorrect.
– Rick
Aug 18, 2015 at 12:58
• @Rick: With utmost respect please try and understand what I have done properly before saying that the answer is incorrect. In this particular solution the force "1N" is calculated with respect to the lower block and hence the acceleration is also with respect to the lower block. If you understand now, kindly withdraw your downvote! Aug 18, 2015 at 15:42
• The 1 N is force acting on the upper block. It doesn't matter where the force came from, it could be 1N of gravity from the moon, in calculating the acceleration via F=ma, the sources of the forces that add to the net force do not matter. The acceleration is always calculated with respect to an inertial frame. The lower block is not an inertial frame as it's accelerating. The ground is an inertial frame and thus it a valid reference frame for the acceleration. Please read potato's answer for an explanation to the question. I'd gladly upvote your answer after you correct it.
– Rick
Aug 18, 2015 at 15:53
• Looking back at your edit history, your first and second versions were actually much better. Your third version however, is incorrect.
– Rick
Aug 18, 2015 at 16:03
• @Rick: LOL, did you not read the question properly or what? the actual applied force is 6N and with respect to lower block it becomes 1N, get it? Aug 18, 2015 at 16:13
This the result of not keeping track of what your calculation or measurement is relative to. The small block is not accelerating very much relative to the large block, this is correct:
5N-6N=-1(kg)*1(m/s^2)
The large block is accelerating much relative to the surface, this is correct:
2(kg)*(5/2)(m/s^2)=5N
But how is the small block moving relative to the surface? It is accelerating much faster than the large block:
(5/2)(m/s^2)+1(m/s^2)=(7/2)(m/s^2) or as you say (7/2)a
This is the acceleration of the small block you are missing out on.
Yes relative to the large block the small block is only accelerating at "a" or 1(m/s^2). But to find this relative to the surface you must add the acceleration of the large block.
• All the numbers imrran gave looked correct to me assuming a gravitational field of 10(m/s^2). did David make it so? Oct 22, 2013 at 4:57
• This is incorrect, accelerations are always measured and calculated relative to an inertial frame (ie one undergoing zero acceleration (in Newtonian mechanics)) Thus assuming the acceleration of the upper block is relative to the lower block is incorrect.
– Rick
Aug 18, 2015 at 12:57
• "assuming the acceleration of the upper block is relative to the lower block is incorrect." ~ 5N was subtracted from the constantly provided 6N, why? How is it that an object in acceleration of the same direction can provide a force of 5N opposite to its acceleration, would not an equal and opposite force be present? What you are trying to say is that the 5N provided by the lower block is not relative to the lower block but relative to the friction-less surface, and I must disagree. The lower block is being accelerated not standing still. Jan 30, 2016 at 16:06
• The 5N is incorrect. That is the maximum friction. However, in this case there is less friction as the blocks are not moving relative to each other. In this case the friction is 4N, giving each block an acceleration of 2m/s/s (relative to an inertial frame). Forces are on objects and are not relative quantities. Changing from one frame to another does not change the forces between objects. So you don't need to specify a reference frame when referring to forces between objects. So specifying what the 5N of force is relative to doesn't make sense.
– Rick
Jan 31, 2016 at 16:39
• "you don't need to specify a reference frame when referring to forces between objects" -> this is why I stated "I must disagree." But now I am to understand that you are trying to say the "u=0.5" is a coefficient of static friction. While I am inclined to believe this is true it is not made clear in the question. Feb 4, 2016 at 3:22 | 1,814 | 6,986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-22 | latest | en | 0.897926 |
https://music.stackexchange.com/questions/135891/how-do-you-find-roman-numerals-out-of-voicings/135897 | 1,718,422,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00348.warc.gz | 367,076,689 | 40,698 | # how do you find Roman Numerals out of voicings?
the author says "In measure 2, V7/IV has b9 in the lead, ...".
I guess then 1st chord of measure 2 is G-7, 2nd chord is 2nd inversion of I7 (V7/IV).
But this is a lot of guess work (to me),
1. contextually dominant is sought for,
2. 1st chord of measure 2 has G in the bass so it MAY be G something chord(if it is in root position)
3. target chord has F in the bass so it MAY be F something chord (again, if it is in root position)
4. b2 is good to be emphasized(in the lead) for more forward motion (but not a MUST, only recommended).
Based on above contextual information (which could easily be wrong), from voicings of G Bb C E Db, I7 could be assumed.
I wonder such guess work really becomes the basis of the RN analysis that is to be done further on, a common practice of RN analysis for all classical pieces also?
• Voicing isn't particularly relevant to Roman numeral analysis. In fact, the primary point of Roman numeral analysis is to relegate the voicing of chords to secondary status -- to call attention to the functional equivalence of differently voiced chords. It's not clear to me from the question why your uncertainty concerns voicing. For example, would you approach it differently if the second chord's D flat were in the left hand and the C natural in the right, or if its bass note were E or C instead of G? Can you clarify? Commented May 3 at 9:43
• What is the key signature in the example? Commented May 3 at 9:56
• @user1079505 we can infer that the piece is in C major from the fact that the second chord is identified as V7/IV and the B flat is given as an accidental. Commented May 3 at 10:00
RNA requires interpretation at times, but basic chord progressions tend to be well understood.
In the example presented, and with the understanding the larger passage is in C major, then the analysis is `ii7/IV V7/IV IV`. This is based on a couple of standard conventions: 1) chords are understood as "stacks of thirds", and 2) the basics of functional harmony.
`V-7`, for example, has no special functional meaning, but `ii7` is a core predominant. Since the following chord is C7 — the dominant of F — and G-7 is the `ii` chord in F (major), there really can't be any other interpretation of these three chords, unless there is some larger context that could change things.
how do you find Roman Numerals out of voicings?
To answer this question literally, you first determine the root of the chord. This involves rearranging the chord as a stack of thirds and identifying the lowest note in the stack. This can involve some guesswork but usually is pretty straightforward.
Then you identify the function of the chord from its form and context. This can more likely require inferences that might be considered "guesswork," but for the most part it involves identifying where the chord falls in any of a small number of common harmonic progressions. The dominant seventh chord is easy, though, because it is always a dominant (that which is known as tritone substitution is, in classical music, an augmented sixth chord, so it has the same shape as a dominant seventh chord enharmonically but is spelled differently).
If the chord does not have the harmonic function that is expected given its root, then it is generally "secondary harmony" so it is necessary to identify the scale degree that is the center of the secondary progression. That is also easy for the dominant seventh chord because it is always a perfect fifth below the root. It can be straightforward in general, but in harmonically complex pieces it can indeed require a good deal of insight, if not actual guesswork. But this is independent of the inversion.
Your thought process seems to skip the part about identifying the root, because your description begins by identifying the bass note and then wondering whether the chord is in root position. The first chord is plainly in root position; there's no need to consider "if" it is. Conversely, the second chord is plainly not in root position, so your first step should not be to look at the bass note but to find the root. If you stack in thirds, you have C, E, G, B♭, D♭ so the root is C.
• "Your thought process seems to skip the part about identifying the root" I agree to this. you said first chord is plainly in root position while second chord is not, do you discern by the second degree apart tones? what if there are tensions then? If a chord is in root position with tenstions, second degree apart tones will also be present.
– Sean
Commented May 10 at 2:33 | 1,051 | 4,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.963558 |
https://www.physicsforums.com/threads/newtons-second-law-in-non-inertial-frame-of-reference.159656/ | 1,695,599,369,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00638.warc.gz | 1,020,196,744 | 14,652 | # Newton's Second Law in NON-inertial frame of reference
• Spiewgels
## Homework Statement
The steel ball is suspended from the accelerating frame by the two cords A and B. The angles (they are on the inside) are both 60 degrees.
Determine the acceleration of the frame which will cause the tension in A to be twice that in B. The acceleration is going to the right and the cord A is to the right of the moving frame.
## Homework Equations
I want to know how to relate the forces on the inside of the accelerating frame to the accelerating frame itself.
## The Attempt at a Solution
Thus far, I have drawn free body diagrams to the inside cords and steel ball. I broke down the components of cord A and B and found the x-coordinate of cord A to be Acos60 and y-coordinate of Asin60. I got these same results for cord B. Combining knowns I've determined both cords tension to be .87w where w equals the weight of the ball. I'm now stuck and don't know how this relates to the moving frame where I think the force is F=ma(of x) and a(of x)=F/m...Have I screwed up this entire problem?
You've apparently solved the problem of a ball hanging in a gravitational field using F=mg. In an frame with acceleration vector a, the equivalent force is given by F=ma (obviously) in the direction opposite to the acceleration. So redraw your force diagram, but this time instead of drawing the 'external' force as pointing straight down, let it point at some angle. Your job is to determine that angle so you get the right tension relation. | 348 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-40 | latest | en | 0.933719 |
https://www.physicsforums.com/threads/motion-on-a-curve.179472/ | 1,544,879,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00289.warc.gz | 993,495,677 | 13,600 | # Motion on a Curve
1. Aug 6, 2007
### Swapnil
Why is the velocity vector always tangent to the curve?
P.S.: I know it makes sense!
But I can't prove it and its driving me crazy!!
2. Aug 6, 2007
### matness
just compare the definitions
if your curve is given by r(t)
T=r'(t)/|r'(t)|
v=r'(t)
3. Aug 6, 2007
### AiRAVATA
In the case your curve is parametrized by $\vec{M}(t)=(x(t),y(t))$, then the tangent vector is given by $\vec{T}(t)=(\dot{x}(t),\dot{y}(t))$. Now, you can do two things to prove that $\vec{M}$ and $\vec{T}$ are parallel. One is to calculate the normal vector to the curve and then the dot product, proving that they are orthogonal. The second one is to calculate the cross product between the curve and the tangent.
This is really simple in $\mathbb{R}^2$, so i'll recommend you to prove it for $\mathbb{R}^n$ as well.
(If you are working with functions in $\mathbb{R}$, then $\vec{M}(t)=(t,f(t))$)
Last edited: Aug 6, 2007
4. Aug 6, 2007
### Swapnil
Why would $\vec{M}(t)$ and $\vec{T}(t)$ be parallel to each other? $\vec{M}(t)$ is the position vector correct?
5. Aug 6, 2007
### quasar987
It is my guess that you're trying to prove that the velocity vector is always tangent to the curve, but you don't have a clear idea (i.e. a definition!) of what it means for a vector to be tangent to a curve.
Like you noted, it makes sense that the velocity is tangent to the curve. So that is how we decide to define "tangency to the curve" (see matness). We'll say that some vector is tangent to the curve at some point if that vector is parallel to the derivative at that point.
With that definition, your problem is more than trivial.
6. Aug 6, 2007
### Swapnil
OH Yes! I didn't even think about how tangent vectors were defined! How foolish of me... :rofl: | 526 | 1,795 | {"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 | 0.947334 |
https://questions.llc/questions/649/factor-completely-or-state-that-this-is-prime-cannot-be-factored-2y-4-28y-3-98y-2 | 1,660,319,109,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00045.warc.gz | 450,259,027 | 4,361 | # Factor completely or state that this is prime (cannot be factored)
2y^4 + 28y^3 + 98y^2
Note: x^2 is x-squared (that is, x with the superscript 2), etc.
2y^4 + 28y^3 + 98y^2 = 2 y^2 (y^2 + 14 y + 49)
The last expression (in parentheses) is a perfect square, and can be factored again. See if you can finish it.
2y^2[(y = 7)]^2
Is that right?
If you change the = sign to a + sign (an obvious typo), yes, that is right.
Thank you
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩 | 183 | 463 | {"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} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.89888 |
http://www.chegg.com/homework-help/questions-and-answers/you-start-making-deposits-of-5000-into-an-account-3-years-from-now-you-make-your-last-depo-q3455648 | 1,368,944,952,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384181/warc/CC-MAIN-20130516092624-00097-ip-10-60-113-184.ec2.internal.warc.gz | 362,621,328 | 8,092 | ## you start making deposits of $5,000 into an account 3 years from now. You make your last deposit at t=8. What is the present value today of this stream of cash flows if the discount rate is 12%? you start making deposits of$5,000 into an account 3 years from now. You make your last deposit at t=8. What is the present value today of this stream of cash flows if the discount rate is 12%?
• Anonymous commented
• Anonymous commented | 107 | 436 | {"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} | 3.234375 | 3 | CC-MAIN-2013-20 | latest | en | 0.942972 |
http://stackoverflow.com/questions/14688235/standardize-all-numerical-predictors-in-a-regression-formula | 1,464,181,622,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274985.2/warc/CC-MAIN-20160524002114-00091-ip-10-185-217-139.ec2.internal.warc.gz | 275,526,002 | 18,655 | # Standardize all numerical predictors in a regression formula
How do you standardize only the numerical predictors in a linear model?
I know that I can simply scale the original numerical data. However, I want to write a function that takes an `lm` object as an argument and returns the standardized beta coefficients for the numerical predictors only.
Here is an example:
``````data(iris)
mod1 <- lm(Sepal.Length ~ Petal.Width, data = iris)
summary(mod1)
mod1.b <- update(mod1, scale(.) ~ scale(.))
summary(mod1.b)
``````
This works without problems. But when I include a factor, it gives an error message.
``````mod2 <- lm(Sepal.Length ~ Petal.Width + Species, data = iris)
summary(mod2)
mod2.b <- update(mod2, scale(.) ~ scale(.)) #Gives an error
``````
So, how can I scale only the numerical predictors in the second example?
-
## migrated from stats.stackexchange.comFeb 4 '13 at 13:54
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
Hi @johannes this question seems to be just about how to do something in R, so it is a better fit on StackOverflow. – Peter Flom Feb 4 '13 at 11:34
@PeterFlom Seems like I missed the obvious - you are of course right. Could someone of the moderators move or close this question? – Johannes Feb 4 '13 at 11:58
Try to change the design matrix of the lm object. For example, we could do the following:
``````design.matrix <- mod2\$model
numeric.columns <- design.matrix[,unlist(lapply(design.matrix,is.numeric))]
scaled.numeric.columns <- scale(numeric.columns)
``````
Now we replace the numeric columns in the data.frame with the scales ones:
``````design.matrix[,unlist(lapply(design.matrix,is.numeric))] <- scaled.numeric.columns
``````
Finally, update the lm object:
``````mod2.b <- update(mod2, data = design.matrix)
``````
- | 471 | 1,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-22 | latest | en | 0.733783 |
https://www.physicsforums.com/threads/mosfet-differential-amplifier-gain.758284/ | 1,532,238,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593010.88/warc/CC-MAIN-20180722041752-20180722061752-00193.warc.gz | 976,665,917 | 22,095 | # Homework Help: MOSFET differential amplifier gain
1. Jun 16, 2014
### etf
Hi!
Here is my task:
For differential amplifier on scheme calculate voltage gains Av1=vout1/(vin1-vin2) and Av2=vout2/(vin1-vin2). MOSFET's M1 and M2 have same characteristics. MOFSTE's M3 and M4 have same characteristics. All MOSFET's are in saturation. Assume that gm*rds>>1.
How to use fact that gm*rds>>1?
I did AC Analysis of this amplifier and found vout1 in function of vin1 and vin2 and vout2 in function of vin1 and vin2. For the sake of simplicity, I used variables a,b,... . I got these results:
$vout1=\frac{bgn-cfn+dfm-dgl}{afm-agl-bem+bgk+cel-cfk}\, \, \,\, vin1+\frac{chl-bhm+bgo-cfo}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin2$
$vout2=\frac{cen-agn-dem+dgk}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin1+\frac{ahm-chk-ago+ceo}{afm-agl-bem+bgk+cel-cfk}\, \, \, \, vin2$
My question is, how to get rid of vin1 and vin2 in Av1 and Av2? V1 and v2 shouldn't figure in gains Av1 and Av2.
Last edited: Jun 16, 2014
2. Jun 16, 2014
### rude man
I don't think there is enough info to determine the gains.
But if you feel you know those four fractions which I will call F1, F2, F3 and F4, then
Av1 = Vout1/(Vin1 - Vin2)
1/Av1 = Vin1/Vout1 - Vin2/Vout2
and Vin1/Vout1 = 1/F1 etc.
It should be apparent that F1, F2, F3 and F4 all have the same magnitude.
3. Jun 17, 2014
### The Electrician
You could replace those complicated expressions you have with single variables x and y, and make use of the extreme symmetry of the circuit. You get this:
$vout1=x*vin1 - y*vin2$
$vout2=-y*vin1+x*vin2$
Also make use of the fact that for purely differential input, Vin1 = -Vin2
I think that then you can find a solution.
4. Jun 18, 2014
### The Electrician
Why do you have about 14 single letter variables for this problem? Have you assigned a variable to each branch current? If that's what you've done, I would recommend using some other approach. Your circuit only has 6 nodes; wouldn't it be simpler to use nodal analysis rather than branch current analysis?
5. Jun 18, 2014
### etf
I used nodal analysis. I will post complete work very soon.
6. Jun 18, 2014
### etf
7. Jun 18, 2014
### The Electrician
So is your final answer what you have shown in the last page, expressions involving a11, a12, a21 and a22?
I would have thought you would need expressions in terms of the MOSFET parameters, such as gm1, gm2, gm3, rds1, rds2, rds3, etc.
Your problem stated that gm*rds>>1, and you asked how to use that fact. Wouldn't you expect that expressions involving gm*rds would appear in your algebra, and that's where you could make use of the strong inequality? But I don't see any expressions like in your work. I wonder why?
I find it hard to follow your work. For example, on page 2 you have a subexpression you've labeled q. It looks like the subexpression is:
(||rds1 + ||(2 R) + gm3/2 + ||rds3) "phi"d1
What are those 2 vertical bars in front of rds1 (why are they there?), and what is the character I've labeled "phi"? Are the vertical bars || really 1/ (1 divided by)?
Also on page 2 is what looks like a signal flow graph. Is it supposed to be a signal flow graph, or is it just a representation of the circuit topology to be used as an aid in setting up the nodal equations?
Between nodes d1 and s1 I see a branch with an arrow in series with rds1; beside the arrow I see what looks like qds1. What is qds1?
8. Jun 18, 2014
### etf
I have bad handwriting :) I used "phi" to represent node potentials...There are no vertical bars, "(||rds1 + ||(2 R) + gm3/2 + ||rds3) "phi"d1" is (1/rds1 + 1/(2 R) + gm3/2 + 1/rds3) "phi"d1 :) I will rewrite solution and upload it again :)
9. Jun 18, 2014
### The Electrician
What about my question regarding the form of your final solution? Shouldn't your answer involve things like gm1, rds1 and not just a11, a12, etc.
10. Jun 18, 2014
### etf
Firstly, we draw original circuit with shorted Voltage sources (VB, VSS, VDD):
Then we draw 5 small-signal MOSFET models and connect them according to first circuit:
Here is result of "simplification" of second circuit:
Now we form system of 3 equations with 3 uknowns:
Here is solution (I didn't write potential for node S1 since it isn't required):
Now we can find gains:
11. Jun 18, 2014
### etf
Now I should replace variables a11, a12, a21 and a22 with firstly introduced a,b,c,... and then replace them with gm1, rds1,... and use fact that gmrds>>1 ? :)
Last edited: Jun 18, 2014
12. Jun 18, 2014
### The Electrician
I'm off to lunch with some of my former EE professors at the local University.
I'll have a look at your work when I get back.
In the meantime, answer me this. Do you have access to Matlab, Mathcad, or some similar modern program to do the tedious algebra involved in this problem?
If you don't have any of those, consider downloading the free package Maxima and learning how to use it:
http://maxima.sourceforge.net/
13. Jun 18, 2014
### etf
Ok :)
I use Matlab. It would be very exhausting to do all these calculations by hand :)
14. Jun 18, 2014
### etf
Ok :)
I use Matlab. It would be very exhausting to do all these calculations by hand :)
15. Jun 18, 2014
### The Electrician
Don't bother replacing your subexpressions with a, b, c, ...
Just plug the complicated expressions into the Matlab solver and let it do the work. Look at your results and see if you see any thing like this: (1+gm*rds). If you do, then you can replace that with (gm*rds). That may help simplify your results.
Post your results here.
16. Jun 19, 2014
### The Electrician
By the way, do you have the correct answer for this problem, from the text or from your instructor?
17. Jun 19, 2014
### etf
I don't have solution for this problem in my book...
Below is code and final result from Matlab. I did DC analysis and got gm1=gm2, gm3=gm4, rds1=rds2, rds3=rds4.
>> syms gm1 gm2 gm3 gm4 gm5 rds1 rds2 rds3 rds4 rds5 R
>> rds1=rds2;
>> rds3=rds4;
>> gm1=gm2;
>> gm3=gm4;
>> a=1/rds1+1/(2*R)+gm3/2+1/rds3;
>> b=-(1/(2*R)-gm3/2);
>> c=-(gm1+1/rds1);
>> d=-gm1;
>> e=-(1/(2*R)-gm4/2);
>> f=1/rds2+1/(2*R)+gm4/2+1/rds4;
>> g=-(1/rds2+gm2);
>> h=-gm2;
>> k=-1/rds1;
>> l=-1/rds2;
>> m=gm1+1/rds1+gm2+1/rds2+1/rds5;
>> n=gm1;
>> o=gm2;
>> a11=(b*g*n-c*f*n+d*f*m-d*g*l)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a12=(c*h*l-b*h*m+b*g*o-c*f*o)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a21=(c*e*n-a*g*n-d*e*m+d*g*k)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> a22=(a*h*m-c*h*k-a*g*o+c*e*o)/(a*f*m-a*g*l-b*e*m+b*g*k+c*e*l-c*f*k);
>> Av1=simplify((1/2)*(a11-a12))
Av1 =
-(R*gm2*rds2*rds4)/(2*(R*rds2 + R*rds4 + rds2*rds4))
>> Av2=simplify((1/2)*(a21-a22))
Av2 =
(R*gm2*rds2*rds4)/(2*(R*rds2 + R*rds4 + rds2*rds4))
Last edited: Jun 19, 2014
18. Jun 19, 2014
### The Electrician
That's the result I get:
You have made your solution more cumbersome than it needs to be by introducing all those single variables a, b, c, ... m, n, o.
As you can see, the most compact way to do it is with the matrix form of your equations. I have more equations than you do because I set up the general problem. I provided equations for the g1, g2, and g3g4 nodes, and I allowed for the possibility of a gate resistor Rg, which turns out not matter in the end. Also we see that rds5 doesn't matter because of the symmetry.
You can see 2 matrices; that's because I kept the matrix for M1 and M2 separate from the matrix for M3 and M4. They could be combined into one matrix; in fact as you can see, I added them, but didn't bother displaying the single matrix which is the sum of the two.
You can use your same technique to solve the CMRR problem thread. You will need to label the MOSFETs corresponding to M1, M2 and M3, M4 of this problem. You will need more equations (at least one more), but the equations you used in this problem will apply in the CMRR problem, possibly with different designators.
The reason I asked you in the CMRR problem if you need to keep the various parameters for each MOSFET separate, is because if you do, the solutions become very complicated. For example, in this problem, if you keep gm1 and gm2 separate, and rds1 and rds2 separate, the answers become (now we see that rds5 appears, because the symmetry is lost):
If you let all of the MOSFETs have only a single gm and rds, the result becomes very simple.
#### Attached Files:
File size:
7.9 KB
Views:
296
• ###### DiffAmpX.png
File size:
69.4 KB
Views:
315
19. Jun 19, 2014
### The Electrician
The results shown in red above were obtained in another Matlab session. If you would just let Matlab do the work of solving with your original equations, you should get the same answer.
What I mean is this. You set up some equations for Matlab to solve, using variables a, b, c...m, n, o. Just leave the original subexpressions in place. Everywhere you have the variable a, just put 1/rds1+1/(2*R)+gm3/2+1/rds3 instead. Do the same for all the other single variables. Put the subexpressions for each in the equations Matlab is solving.
If you were doing this by hand, it would be very tedious, but Matlab can do it without mistakes.
20. Jun 19, 2014
### etf
Thanks a lot! One more question for thread about CMRR: Can I analyse diff. amp. with Iss source from b), calculate CMRR for that circuit and use that CMRR to calculate CMRR for diff. amp. with Iss source from a) (i.e. seting for M2 gm2=0, rds2=∞) ? More precisely, for CMRR of diff. amp. with ISS from b) I will got CMRR in function of gm1, rds1, gm2, rds2 and gm's and rds's of other MOSFETS so I will just put there gm2=0 and rds2=∞ and that would be my CMRR for diff. amp. with source Iss from a)?
Last edited: Jun 19, 2014
21. Jun 19, 2014
### The Electrician
What you must do is set rds2=0, not rds2=∞. If you do that, then your proposed method should work.
What I did is use a symbol to represent the impedance of the (not so good) current source; I called it "zs". Then you can see the effect of that source on CMRR and differential gain, if you wanted to examine the effect.
I then calculated the impedance looking down into the a) and b) circuits, and substituted those values for zs in the calculations for CMRR and AV.
I should warn you that the circuit in your CMRR problem doesn't have the extreme symmetry that this one does and if you use different values for the parameters of the 4 MOSFETs making up the diff amp, plus the two MOSFETs of the current source, you will get very complicated answers! You could use just gm and rds for the 4 diff amp MOSFETs.
I did a solution with just gm and rds for all the MOSFETs, including M1 and M2, and even then the solutions begin to expand.
Your description of the CMRR problem seems to indicate that you should keep gm1, rds1 and gm2, rds2 as separate parameters, but I see no mention of what to do about the parameters of the 4 diff amp parameters. Maybe you should just go with gm and rds.
You should move over to the CMRR thread for further discussion. Give your calculations a try and post your results. | 3,340 | 11,024 | {"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} | 3.203125 | 3 | CC-MAIN-2018-30 | latest | en | 0.838703 |
https://math.eretrandre.org/tetrationforum/archive/index.php?thread-40.html | 1,576,298,587,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540584491.89/warc/CC-MAIN-20191214042241-20191214070241-00447.warc.gz | 443,170,603 | 2,120 | # Tetration Forum
Full Version: Bell formula for iterated exponentiation
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Gottfried Wrote:m=arbitrary, integer>0:
$\hspace{24} T^{\tiny(m)}_s(x)=
\sum_{ k_m...k_1 =0..\infty \\ k_1+k_2+...+k_m=n }
x^{k_m}
* \left( k_{m-1}^{k_m} \dots k_2^{k_3} k_1^{k_2} \right)
* \begin{pmatrix} n \\ k_2,k_3,...,k_m \end{pmatrix}
* \frac{ log(s)^n}{n!}
$
(hope I didn't make an index-error).
This formula can also be found in
E.T.BELL ,"The iterated Exponential Integers", Annals of Mathematics (193 p539-557
Are you sure Gottfried? I didnt find it in the mentioned article.
bo198214 Wrote:Are you sure Gottfried? I didnt find it in the mentioned article.
Too bad. I don't know, where my brain was... I had just skimmed through some articles and meant to have caught the source correctly. I'll seek, where I've actually have seen it, maybe in Abramowitsch/Stegun...
Anyway, it is easy to derive it with pen & paper.
I'll be back in the evening, going for a walk today -
Gottfried | 334 | 1,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-51 | longest | en | 0.879896 |
https://math.stackexchange.com/questions/4306147/rank-of-a-pair-of-coprime-integers | 1,709,413,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00020.warc.gz | 370,867,022 | 35,720 | # Rank of a pair of coprime integers
Let's say two pairs of coprime integers $$(a, b)$$ and $$(c, d)$$ are connected if $$ac + bd = 1$$.
A connected chain, or just a chain, is a sequence of coprime pairs in which every two consecutive pairs are connected.
Clearly, if $$(a, b)$$ is connected to $$(c, d)$$, then $$(a, b)$$ is connected to $$(c + nb, d - na)$$ for an integer $$n$$.
Using the property, we may find a connected pair $$(x, y)$$ such that $$max(|x|,|y|) < max(|a|,|b|)$$.
Continuing the process we may build a chain that ends on $$(0, 1)$$ or $$(1, 0)$$ in absolute values.
Examples (in absolute values):
• $$(41, 53)$$, $$(22, 17)$$, $$(7, 9)$$, $$(4, 3)$$, $$(1, 1)$$, $$(0, 1)$$.
• $$(41, 53)$$, $$(31, 24)$$, $$(7, 9)$$, $$(4, 3)$$, $$(1, 1)$$, $$(0, 1)$$.
The described algorithm may give chains of different length:
• $$(41, 61)$$, $$(58, 39)$$, $$(2, 3)$$, $$(1, 1)$$, $$(0, 1)$$.
• $$(41, 61)$$, $$(3, 2)$$, $$(1, 1)$$, $$(0, 1)$$.
Let's define the rank of a pair of coprime integers $$(a, b)$$ as the minimal length within all possible chains connecting $$(a, b)$$ with $$(0, 1)$$ or $$(1, 0)$$ in absolute values.
Let's call a chain with the minimal length a minimal chain for $$(a, b)$$.
Questions:
1. Does the rank exist for any pair of coprime or prime integers?
2. Is there a maximal rank within all pairs of coprime or prime integers with rank?
3. Is there an algorithm of constructing a minimal chain or calculating the rank of a given coprime pair?
• not sure what you want. In continued fractions, consecutive convergents are connected. Anyway, I wrote a gcd thing that replaces the usual "back-substitution" with the continued fraction for the ratio of the given positive integers Nov 14, 2021 at 21:51
• @WillJagy Could you explain the connection with continued fractions, please? I am struggling to prove that we can always obtain $(x, y)$ such that $max(|x|, |y|)$ < $min(|a|, |b|)$ or $max(|x|, |y|)$ < $max(|a|, |b|)$. Nov 14, 2021 at 23:18
• Alex, your first chain for (41,53) are just the convergents for $\frac{53}{41},$ in reverse order. I put that in an answer. Your second chain for (41,61) are the convergents for $\frac{61}{41}.$ I suggest that the string of convergents always gives your "rank" If you have some more examples you have worked, let me know and I will put in the c.f., you may compare Nov 15, 2021 at 0:02
• $$\begin{array}{cccccccccccc} & & 1 & & 3 & & 2 & & 2 & & 2 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 4 }{ 3 } & & \frac{ 9 }{ 7 } & & \frac{ 22 }{ 17 } & & \frac{ 53 }{ 41 } \end{array}$$ Nov 15, 2021 at 0:34
• your $\frac{22}{17}$ is the penultimate convergent for $\frac{53}{41}.$ Perhaps you are looking at the initial string of divisions: the fractions in the left column are of no importance, those being $\frac{41}{12}, \frac{12}{5}, \frac{5}{2}, \frac{2}{1}.$ Ignore them. Just look at $$\begin{array}{cccccccccccc} & & 1 & & 3 & & 2 & & 2 & & 2 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 4 }{ 3 } & & \frac{ 9 }{ 7 } & & \frac{ 22 }{ 17 } & & \frac{ 53 }{ 41 } \end{array}$$ Nov 15, 2021 at 1:32
Your question (3), begin with your coprime pair $$(a,b)$$ with $$a > b > 0,$$ the minimal chain comes from using the Extended Euclidean Algorithm and writing the (simple) continued fraction for $$\frac{a}{b},$$ then let each convergent $$\frac{p}{q}$$ be part of the chain, as pair $$(p,q)$$
$$\gcd( 53, 41 ) = ???$$
$$\frac{ 53 }{ 41 } = 1 + \frac{ 12 }{ 41 }$$ $$\frac{ 41 }{ 12 } = 3 + \frac{ 5 }{ 12 }$$ $$\frac{ 12 }{ 5 } = 2 + \frac{ 2 }{ 5 }$$ $$\frac{ 5 }{ 2 } = 2 + \frac{ 1 }{ 2 }$$ $$\frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 }$$ Simple continued fraction tableau:
$$\begin{array}{cccccccccccc} & & 1 & & 3 & & 2 & & 2 & & 2 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 4 }{ 3 } & & \frac{ 9 }{ 7 } & & \frac{ 22 }{ 17 } & & \frac{ 53 }{ 41 } \end{array}$$ $$53 \cdot 17 - 41 \cdot 22 = -1$$
$$\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$$
$$\gcd( 61, 41 ) = ???$$
$$\frac{ 61 }{ 41 } = 1 + \frac{ 20 }{ 41 }$$ $$\frac{ 41 }{ 20 } = 2 + \frac{ 1 }{ 20 }$$ $$\frac{ 20 }{ 1 } = 20 + \frac{ 0 }{ 1 }$$ Simple continued fraction tableau:
$$\begin{array}{cccccccc} & & 1 & & 2 & & 20 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 3 }{ 2 } & & \frac{ 61 }{ 41 } \end{array}$$ $$61 \cdot 2 - 41 \cdot 3 = -1$$
• Your sequence does not agree with his, so you should explain how they are related. I see no answers above to any of his questions. Nov 14, 2021 at 23:06
• $\frac{53}{41} = [1,3,2,2,2]$, $\frac{22}{17} = [1,3,2,2]$, $\frac{9}{7} = [1,3,2]$, etc. Nov 15, 2021 at 1:48
• The remaining piece is the proof of minimality of the chain obtained this way. Nov 15, 2021 at 2:09
• @AlexC induction on the length of the continued fraction. Nov 15, 2021 at 2:14 | 1,977 | 5,030 | {"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": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.782612 |
https://www.mersenneforum.org/search.php?s=cb5b70e2c1ff94ec07f3b95d0e886c5d&searchid=3343928 | 1,601,542,761,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00521.warc.gz | 861,011,553 | 9,364 | mersenneforum.org Search Results
Register FAQ Search Today's Posts Mark Forums Read
Showing results 1 to 25 of 1000 Search took 0.53 seconds. Search: Posts Made By: axn
Forum: Aliquot Sequences 2020-09-30, 13:01 Replies: 606 Views: 26,922 Posted By axn All numbers with Nov 4 date is a lie. It just... All numbers with Nov 4 date is a lie. It just means the database was rebuilt on that date and they lost all history of any existing numbers. Hence the "before"
Forum: enzocreti 2020-09-29, 11:01 Replies: 3 Views: 65 Posted By axn 69660 is a pg prime thingie. Sorry to disappoint... 69660 is a pg prime thingie. Sorry to disappoint you.
Forum: GpuOwl 2020-09-28, 05:44 Replies: 39 Views: 1,011 Posted By axn Do we have some hard numbers comparing these two... Do we have some hard numbers comparing these two schemes (i.e. traditional P-1-before-PRP, and P-1-during-PRP) and the expected savings and optimal bounds for a current wavefront PRP? Especially with...
Forum: Lounge 2020-09-24, 11:45 Replies: 1,496 Views: 98,680 Posted By axn Dean Jones (former Aussie cricketer) Dean Jones (former Aussie cricketer)
Forum: Hardware 2020-09-24, 09:11 Replies: 124 Views: 7,294 Posted By axn Hmmm... Haven't run into these. Mostly seen their... Hmmm... Haven't run into these. Mostly seen their hardware contents only. Will check these out.
Forum: Hardware 2020-09-24, 08:25 Replies: 124 Views: 7,294 Posted By axn What Linux content? What Linux content?
Forum: Wagstaff PRP Search 2020-09-23, 17:11 Replies: 391 Views: 36,814 Posted By axn 181k-182k 2^181031+1 = 4539282868521690809873 2^181087+1 = 583220928363898193 2^181201+1 = 378599683202077370507 2^181211+1 = 78063361679692049 2^181219+1 = 1568762826538016763403 2^181303+1 =...
Forum: Wagstaff PRP Search 2020-09-15, 02:02 Replies: 391 Views: 36,814 Posted By axn 180k-181k 2^180073+1 = 529177529574833417081 2^180073+1 = 57294368098941995181747710561 2^180097+1 = 13481683846649636744156659 2^180161+1 = 3338880716108152547 2^180179+1 = 955625595136915380089...
Forum: Software 2020-09-14, 15:14 Replies: 377 Sticky: Prime95 v30.3 Views: 19,323 Posted By axn :tu: :tu:
Forum: Software 2020-09-13, 15:08 Replies: 377 Sticky: Prime95 v30.3 Views: 19,323 Posted By axn I'm guessing mprime doesn't have "execute"... I'm guessing mprime doesn't have "execute" permission. I'm also guessing mprime's owner is not your login account. If you do ls -l and paste it here, we can confirm. Use sudo chown to change...
Forum: Marin's Mersenne-aries 2020-09-13, 14:57 Replies: 8 Views: 456 Posted By axn Does the p-1 probability calculation account for... Does the p-1 probability calculation account for the fact that k from 2kp+1 is more likely to have smaller factors compared to random number of the same size (because 2kp+1 is prime and does not have...
Forum: Wagstaff PRP Search 2020-09-07, 14:41 Replies: 391 Views: 36,814 Posted By axn 179k-180k 2^179029+1 = 41310431965823413656683449 2^179083+1 = 1792200143196195457427 2^179083+1 = 49812943670932589645593 2^179089+1 = 16737373736913886523 2^179089+1 = 957393083474638075481 2^179111+1 =...
Forum: GPU Computing 2020-09-06, 12:02 Replies: 3,337 Views: 285,239 Posted By axn Out of curiosity, what is your power consumption... Out of curiosity, what is your power consumption and thruput if you lock the graphics clock to base clock (nvidia-smi -lgc 1530)? You can reset graphics clock by nvidia-smi -rgc afterwards.
2020-09-06, 02:34 Replies: 4 Views: 420 Posted By axn This is wrong. y does not have to divide k. ... This is wrong. y does not have to divide k. 2*k*p = y(y + 2) = 2*k1*p*(y+2) Therefore, k = k1*(y+2). i.e. Only k1 needs to divide k p*(k/y - k1) = 1 ==> For this we need that p*k/y be an...
Forum: PrimeNet 2020-09-05, 01:59 Replies: 59 Views: 2,473 Posted By axn Current wavefront is > 5M (5.5? 6?). 5.5M FFT =... Current wavefront is > 5M (5.5? 6?). 5.5M FFT = 44MB + overhead. So your statement will work if it was 64MB per chiplet. Still, that ought to give non-trivial IPC improvement over Zen 2
Forum: mersenne.ca 2020-09-03, 02:34 Replies: 22 Views: 4,372 Posted By axn And then you click "Top factors for all users",... And then you click "Top factors for all users", et voila!
Forum: Puzzles 2020-08-30, 15:09 Replies: 20 Views: 1,038 Posted By axn No need to brute force. R.Gerbicz already showed... No need to brute force. R.Gerbicz already showed big numbers, but basically, using the periodicity of residues, we can iteratively improve a record. For example: ...
Forum: Puzzles 2020-08-30, 14:17 Replies: 20 Views: 1,038 Posted By axn This approach is not going to work. If we... This approach is not going to work. If we consider n lower order digits, they cycle with a periodicity 5^(n-1). So you need to find an n such that all 5^(n-1) residues turn up bad -- then that...
Forum: GPU to 72 2020-08-28, 15:12 Replies: 5,048 Sticky: GPU to 72 status... Views: 230,592 Posted By axn Who dat? Who dat?
Forum: Wagstaff PRP Search 2020-08-28, 01:33 Replies: 391 Views: 36,814 Posted By axn 178k-179k 2^178021+1 = 53291761850599835531 2^178067+1 = 1320625435316852603 2^178067+1 = 7198050638355970137286058849 2^178117+1 = 28267026588553964047770067 2^178187+1 = 1666952508666434561953...
Forum: Factoring 2020-08-24, 11:00 Replies: 9 Views: 615 Posted By axn ecm -resume ecm_input.txt 44e6-44e6 >... ecm -resume ecm_input.txt 44e6-44e6 > ecm_output.txt Ask gmp-ecm to do stage 1 from 44e6 to 44e6 (or whatever the B1 limit was for P95 stage1 run), and it will correctly figure out the B2 and do...
Forum: Lone Mersenne Hunters 2020-08-20, 17:01 Replies: 1,675 Views: 139,488 Posted By axn P-1 found a factor in stage #2, B1=30000000,... P-1 found a factor in stage #2, B1=30000000, B2=600000000, E=12. UID: axn/R5, M3640333 has a factor: 75280105019644427503116372295759257424213063 (P-1, B1=30000000, B2=600000000, E=12) 44 digits...
Forum: Cloud Computing 2020-08-20, 06:37 Replies: 1,022 Views: 50,535 Posted By axn Yes, I can confirm this behavior too. Has been... Yes, I can confirm this behavior too. Has been happening for the past couple of hours.
Forum: GPU Computing 2020-08-19, 01:11 Replies: 23 Views: 1,852 Posted By axn I see 467+430 = 897 ? I see 467+430 = 897 ?
Forum: Wagstaff PRP Search 2020-08-18, 02:55 Replies: 391 Views: 36,814 Posted By axn 177k-178k 2^177011+1 = 106247300131063066483 2^177013+1 = 465958696873229803 2^177019+1 = 2315470461171687702587 2^177043+1 = 164772305082767428151107 2^177101+1 = 2389566866428137169 2^177109+1 =...
Showing results 1 to 25 of 1000
All times are UTC. The time now is 08:59.
Thu Oct 1 08:59:21 UTC 2020 up 21 days, 6:10, 0 users, load averages: 1.65, 1.64, 1.62 | 2,344 | 6,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.769288 |
http://www.drps.ed.ac.uk/18-19/dpt/cxecnm11072.htm | 1,723,066,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00501.warc.gz | 31,698,864 | 7,876 | # DEGREE REGULATIONS & PROGRAMMES OF STUDY 2018/2019
University Homepage DRPS Homepage DRPS Search DRPS Contact
DRPS : Course Catalogue : School of Economics : Economics
# Postgraduate Course: Advanced Mathematical Economics (ECNM11072)
School School of Economics College College of Humanities and Social Science Credit level (Normal year taken) SCQF Level 11 (Postgraduate) Availability Available to all students SCQF Credits 20 ECTS Credits 10 Summary This course is about the advanced mathematical tools that are used in economics research. Each mathematical topic is explored in the context of an important economic problem. Course description The topics covered vary from year to year. An example curriculum would be the following mathematics concepts illustrated in the context of general equilibrium theory: * Naive Set Theory. This is the language of mathematics, and is widely used by economists. This is important for making precise hypotheses, such as "in every equilibrium, real wages increase over time", and for verifying these hypotheses with logically sound proofs. The main concepts are: sets, functions, logical connectives, quantifiers, countability, induction, proof by contradiction. * Real Analysis and Metric Spaces. This branch of mathematics focuses on continuity and nearness (topology) while putting geometric concepts like distance and angles into the background. These ideas are useful for determining whether an optimal decision is possible, whether an equilibrium of an economy exists, and determining when optimal decisions change drastically when circumstances change. The main concepts are: open sets, continuity, limits, interior, boundary, closure, function spaces, sup metric, Cauchy sequences, connected spaces, complete spaces, compact spaces, Bolzano-Weierstrass theorem, Banach fixed point theorem, Brouwer fixed point theorem. * Convex Analysis. This branch of geometry focuses on comparing extreme points and intermediate points that lie between extremes. These tools are useful for determining whether there is one or several optimal decisions in a particular situation, and determining in which direction optimal choices move when circumstances change. Convex analysis is related to the economic notions of increasing marginal cost and decreasing marginal benefit. The main concepts are: convex sets, convex and concave functions, quasi-convex and quasi-concave functions, supporting hyperplane theorem, separating hyperplane theorem. * Dynamic Programming. This branch of mathematics is about breaking up a complicated optimisation problem involving many decisions into many simple optimisation problems involving few decisions. For example, a lifetime of choices can be broken up into simple choices made day-by-day. The main concepts are: value functions, Bellman equations, Bellman operators. * Envelope Theorem. This is a calculus formula for calculating marginal values, such marginal benefit of saving money. The main concepts are: differentiable support functions, the Benveniste-Scheinkman theorem.
Pre-requisites Co-requisites Prohibited Combinations Other requirements None
Pre-requisites None
Academic year 2018/19, Available to all students (SV1) Quota: None Course Start Semester 1 Timetable Timetable Learning and Teaching activities (Further Info) Total Hours: 200 ( Lecture Hours 20, Seminar/Tutorial Hours 18, Summative Assessment Hours 6, Programme Level Learning and Teaching Hours 4, Directed Learning and Independent Learning Hours 152 ) Assessment (Further Info) Written Exam 80 %, Coursework 20 %, Practical Exam 0 % Additional Information (Assessment) Coursework (20%), Exam (80%).«br /» «br /» - Mathematical Economics Project 20%«br /» - 3 Hour Examinations in December and May 80% (using the best mark)«br /» «br /» While we recommend that most Continuing Professional Development students take this as a full-year course, this course is also available in a one-semester format (without the May exam). For our internal record-keeping purposes, we call this option 'part-year visiting student' (because we offer the same format to exchange students), even though this is a Continuing Professional Development course.«br /» Feedback All tutorials will involve problem solving, and opportunities for formative feedback. Exam Information Exam Diet Paper Name Hours & Minutes Main Exam Diet S1 (December) Advanced Mathematical Economics Class Exam 3:00 Main Exam Diet S2 (April/May) Advanced Mathematical Economics Degree Exam 3:00
Mathematical maturity, i.e. the ability to: distinguish between definitions, conjectures, theorems, and proofs, generalise and specialise theorems and proofs, devise counter-examples, and determine whether objects conform to definitions and conditions of theorems. Experience in applying mathematical tools to derive economic conclusions.
Indicative readings: * Boyd and Vandenburghe (2004), "Convex Optimization", Cambridge University Press. * Luenberger (1968), "Optimization by Vector Space Methods", Wiley. * de la Fuente (2000), "Mathematical Methods and Models for Economists", Cambridge University Press
Graduate Attributes and Skills Not entered Keywords ADVMath
Course organiser Dr Andrew Clausen Tel: (0131 6)51 5131 Email: Andrew.Clausen@ed.ac.uk Course secretary Miss Sophie Bryan Tel: (0131 6)50 9905 Email: Sophie.Bryan@ed.ac.uk
Navigation Help & Information Home Introduction Glossary Search DPTs and Courses Regulations Regulations Degree Programmes Introduction Browse DPTs Courses Introduction Humanities and Social Science Science and Engineering Medicine and Veterinary Medicine Other Information Combined Course Timetable Prospectuses Important Information | 1,150 | 5,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.85257 |
https://studylib.net/doc/5615468/trendline | 1,623,884,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626122.27/warc/CC-MAIN-20210616220531-20210617010531-00200.warc.gz | 482,129,686 | 12,794 | # Trendline
advertisement
``` Studies of the metabolism of alcohol consistently show
that blood alcohol content (BAC), after rising rapidly
after ingesting alcohol, declines linearly.
For example, in one study, BAC in a fasting person rose
to about 0.018 % after a single drink. After an hour
the level had dropped to 0.010 %. Assuming that BAC
continues to decline linearly (meaning at a constant
rate of change), approximately when will BAC drop to
0.002%?
Need to make an equation and then use the equation
to answer the question (make a prediction)
Define the variables and assign each a letter to
represent it
We can make it a table for fun
Make the equation (y intercept)
y = -.008x + .018
Make predictions!
The question is "when will the BAC reach .002%?“
Plug in .002 for Y and solve for X.
.002 = -.008x + .018
-.016 = -.008x
x = 2 hours
Therefore the BAC will reach .002% after 2 hours.
What if the data isn’t perfectly linear?
Can use trendlines – make predictions
Interpolations / extrapolations
Women’s world records
Trendline - estimate of a linear function that fits the
data
Legend – identifies what the dots on the chart
represent. For two variable graphs we don’t really need
this
Title + axes – always have these labeled
Source
What is the projected record in 1999?
By equation
Trendline estimate
Data points – 7 data points is good, lower than this is
not as good.
2. R squared (r2)
1.
1.
2.
3.
4.
This is a measure of strength of the linear relationship
between x and y.
.7 to 1 typically indicates a strong linear relationship
.4 - .7 is more moderate linear relationship
< .4 is weak and probably shouldn’t be used to make a
prediction
3.
Practical / Physical / Social / Political Sense ??
- Think about your prediction. What outside factors
could affect it?
- How far away from the data is your prediction?
What will be the record in 2100?
What was the record in 500?
``` | 534 | 1,951 | {"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-2021-25 | latest | en | 0.894739 |
http://www.eltronicschool.com/2015/02/36v-battery-level-led-indicator-circuit.html | 1,506,105,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689102.37/warc/CC-MAIN-20170922183303-20170922203303-00373.warc.gz | 441,848,975 | 32,749 | ## Tuesday, February 3, 2015
### 36V Battery Level LED Indicator Circuit Schematic
Eltronicschool. - One way to know battery level is using LED indicator. With this way we will know how is the level of our battery and we can anticipate to do more. In this time we will show you one circuit schematic for battery level indicator using LED specially for 36 volt dc as like in figure 1 below.
Circuit Schematic
Figure 1. 36V Battery Level LED Indicator Circuit (Source: Electroschematic)
Component List
1. Zener Diodes 6.2 Volt
2. Resistors
3. Transistors
4. IC LM329
5. Variable resistor
6. LED Red, Orange, Yellow, Green, and Blue
Description
Circuit schematic like in figure 1 above is 36V Battery Level LED Indicator Circuit that will indicate some battery level indication like in table 1 below.
Table 1. Note of LED indicator of battery level condition
Main component of this circuit is voltage comparator IC LM329. So, the output of this IC will give output on and off depending the two input condition. As the input of this IC are come from voltage as reference and adjustable from the resistors that combine.
So, from the main article of 36V Battery Level LED Indicator Circuit describe of circuit operation as follow:
D1 is the voltage reference zener. Tied to this is a string of divider resistors (R2-6) that set the various fixed voltage levels. R7 & 8 form a voltage divider to that divides the battery voltage by a factor of 9. The quad comparator compares the various voltages from the two dividers.
For calibration, connect to a voltage source that can be set to the highest LED threshold (41V in this case). Then adjust the calibration pot until D2 flickers. The remaining LEDs will switch on close to the indicated voltage –accuracy of those voltages may suffer slightly, but should be close.
The LEDs are biased to operate at 2.3mA which is reasonably bright for high efficiency LEDs. This current can be adjusted simply by varying the emitter driver resistors (R9 through R13). To reduce standby power, a push-to-test pushbutton may be used.
All of the 36V Battery Level LED Indicator Circuit schematic can you read more from the original article below.
Thank you for your coming here in www.eltronicschool.com site, we hope the article above will help you to know more about your an electronic circuit design and software in this time, etc. Please comment here when you want to share and other. Thank you. | 551 | 2,440 | {"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-2017-39 | longest | en | 0.879323 |
http://www.explorelearning.com/index.cfm?method=cResource.dspBookCorrelation&id=131 | 1,436,189,452,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375098464.55/warc/CC-MAIN-20150627031818-00144-ip-10-179-60-89.ec2.internal.warc.gz | 454,474,519 | 8,128 | ## Mathematics: Course 2
• Author(s): Charles et. al. Publisher: Pearson/Prentice Hall 2004
This correlation lists the recommended Gizmos for this textbook. Click any Gizmo title below to go to the Gizmo Details page.
### Chapter 1. Decimals and Integers
#### 1-3. Multiplying and Dividing Decimals
Multiplying with Decimals
#### 1-6. Comparing and Ordering Integers
Comparing and Ordering Integers
#### 1-10. Mean, Median, and Mode
Mean, Median and Mode
Describing Data Using Statistics
### Chapter 2. Equations and Inequalities
#### 2-1. Evaluating and Writing Algebraic Expressions
Using Algebraic Expressions
#### 2-3. Solving Equations by Adding or Subtracting
Modeling One-Step Equations - Activity A
#### 2-4. Solving Equations by Multiplying or Dividing
Modeling and Solving Two-Step Equations
Solving Two-Step Equations
#### 2-6. Solving Two-Step Equations
Modeling and Solving Two-Step Equations
Solving Two-Step Equations
#### 2-7. Write an Equation
Using Algebraic Expressions
#### 2-8. Graphing and Writing Inequalities
Solving Linear Inequalities using Addition and Subtraction
#### 2-9. Solving Inequalities by Adding or Subtracting
Solving Linear Inequalities using Addition and Subtraction
#### 2-10. Solving Inequalities by Multiplying or Dividing
Solving Linear Inequalities using Multiplication and Division
### Chapter 3. Exponents, Factors, and Fractions
#### 3-1. Exponents and Order of Operations
Order of Operations
#### 3-4. Prime Factorization
Finding Factors with Area Models
#### 3-6. Comparing and Ordering Fractions
Comparing and Ordering Fractions
#### 3-8. Mixed Numbers and Improper Fractions
Improper Fractions and Mixed Numbers
#### 3-9. Fractions and Decimals
Ordering Percents, Fractions and Decimals
#### 3-10. Rational Numbers
Comparing and Ordering Rational Numbers
### Chapter 4. Operations With Fractions
#### 4-1. Estimating With Fractions and Mixed Numbers
Estimating Sums and Differences
#### 4-2. Adding and Subtracting Fractions
Fractions with Unlike Denominators
#### 4-3. Adding and Subtracting Mixed Numbers
Improper Fractions and Mixed Numbers
#### 4-4. Multiplying Fractions and Mixed Numbers
Multiplying Mixed Numbers
#### 4-5. Dividing Fractions and Mixed Numbers
Dividing Mixed Numbers
### Chapter 5. Ratios, Rates, and Proportions
#### 5-1. Ratios
Part:Part and Part:Whole Ratios
#### 5-4. Proportions
Percents and Proportions
Proportions and Common Multipliers
#### 5-5. Using Proportional Reasoning
Similar Figures - Activity A
#### 5-6. Using Similar Figures
Similar Figures - Activity A
Similar Polygons
### Chapter 6. Percents
#### 6-1. Understanding Percents
Percents, Fractions and Decimals
#### 6-2. Percents, Fractions, and Decimals
Percents, Fractions and Decimals
#### 6-4. Finding a Percent of a Number
Percents and Proportions
#### 6-5. Solving Percent Problems Using Proportions
Percents and Proportions
#### 6-8. Finding Percent of Change
Percent of Change
### Chapter 7. Geometry
#### 7-1. Lines and Planes
Investigating Parallel Lines and Planes
#### 7-3. Constructing Bisectors
Construct Parallel and Perpendicular Lines
Constructing Congruent Segments and Angles
#### 7-4. Triangles
Classifying Triangles
Finding Patterns
Chords and Arcs
### Chapter 8. Geometry and Measurement
#### 8-2. Areas of Parallelograms and Triangles
Area of Parallelograms - Activity A
#### 8-3. Areas of Other Figures
Perimeter, Circumference, and Area - Activity A
#### 8-4. Circumferences and Areas of Circles
Circle: Perimeter, Circumference and Area
#### 8-5. Square Roots and Irrational Numbers
Ordering and Approximating Square Roots
Square Roots
#### 8-6. The Pythagorean Theorem
Geoboard: The Pythagorean Theorem
Pythagorean Theorem - Activity A
#### 8-7. Three-Dimensional Figures
Prisms and Cylinders
Pyramids and Cones - Activity A
#### 8-8. Surface Areas of Prisms and Cylinders
Surface and Lateral Area of Prisms and Cylinders
#### 8-9. Volumes of Rectangular Prisms and Cylinders
Prisms and Cylinders
### Chapter 9. Patterns and Rules
#### 9-2. Number Sequences
Finding Patterns
Arithmetic and Geometric Series
#### 9-3. Patterns and Tables
Using Tables, Rules and Graphs
#### 9-4. Function Rules
Introduction to Functions
Linear Functions
Using Tables, Rules and Graphs
#### 9-5. Using Tables, Rules, and Graphs
Using Tables, Rules and Graphs
#### 9-7. Simple and Compound Interest
Simple and Compound Interest
#### 9-8. Write an Equation
Using Algebraic Equations
#### 9-9. Transforming Formulas
Solving Formulas for any Variable
### Chapter 10. Graphing in the Coordinate Plane
#### 10-1. Graphing Points in Four Quadrants
Points in the Coordinate Plane - Activity A
#### 10-3. Finding the Slope of a Line
Slope - Activity A
#### 10-5. Make a Table and Make a Graph
Using Tables, Rules and Graphs
#### 10-6. Translations
Rotations, Reflections and Translations
#### 10-7. Symmetry and Reflections
Rotations, Reflections and Translations
#### 10-8. Rotations
Rotations, Reflections and Translations
### Chapter 11. Displaying and Analyzing Data
Histograms
Line Plots
#### 11-3. Other Displays
Constructing Box-and-Whisker Plots
#### 11-6. Estimating Population Size
Estimating Population Size
#### 11-8. Exploring Scatter Plots
Scatter Plots - Activity A
### Chapter 12. Using Probability
#### 12-1. Probability
Probability Simulations
Geometric Probability - Activity A
#### 12-2. Experimental Probability
Theoretical and Experimental Probability
#### 12-5. Compound Events
Compound Independent Events
Permutations
#### 12-7. Combinations
Permutations and Combinations
Content correlation last revised: 4/22/2009 | 1,555 | 5,743 | {"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} | 3.203125 | 3 | CC-MAIN-2015-27 | latest | en | 0.703768 |
https://electronics.stackexchange.com/questions/322244/why-capacitor-accumulate-less-charge-in-higher-frequency | 1,575,725,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499439.6/warc/CC-MAIN-20191207132817-20191207160817-00444.warc.gz | 358,166,571 | 31,660 | # why capacitor accumulate less charge in higher frequency
it always said that the higher the frequency, the less charge will accumulate because when in higher frequency, there is less time for capacitor to accumulate electrons. and in lower frequency, there will be more time for capacitor to accumulate electrons.
but in higher frequency, although the time for accumulating is less but the current is larger so how to conclude that in higher frequency, the total charge accumulated in capacitor is small ?
• Please provide a reference for "it always said" claim. It does sound wrong as stated in your question. – Dmitry Grigoryev Aug 4 '17 at 9:31
• @DmitryGrigoryev learnabout-electronics.org/ac_theory/reactance62.php The lower the frequency of the applied voltage, the more time the capacitor has to reach the fully charged, zero current state before the voltage reverses its polarity and begins to discharge the capacitor again. The capacitor therefore spends more time fully charged and passing much less current, the average value of current flow is therefore less at low frequencies. When a higher frequency is applied, the capacitor changes from charging to discharging sooner in its charge curve – John Lu Aug 4 '17 at 9:43
• This writeup has a lot of inaccuracies. I stopped reading after the line where it says the capacitor never reaches zero current on AC. – Dmitry Grigoryev Aug 4 '17 at 10:09
• Why do you say the current is larger? – user253751 Aug 4 '17 at 10:35
• In every circuit you have real resistance which will limit the maximum current to some value independent of the frequency. At higher frequencies this fixed maximum current has less time to charge the capacitor, so the cap cannot accumulate as much charge before starting to discharge again. – JimmyB Aug 4 '17 at 12:02 | 395 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-51 | latest | en | 0.941245 |
https://app.seesaw.me/activities/5xyqzr/monday-math-piggy-bank-one-more-one-less-equal | 1,642,842,958,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00098.warc.gz | 153,966,824 | 38,047 | Student Instructions
### Monday Math: Piggy Bank One More, One Less, Equal
Tap to begin Page 1: Click on the black triangle and watch as I model this activity. Pages 2-9: Tap to hear the directions. Count how many pennies are in the first pig. Use to write the number in the box under the pig. Click next to the gold coin. If the gold coin says "one more" put an amount of pennies that is one more than the amount of pennies in the first pig. If the gold coin says "one less" put an amount of pennies that is one less than the amount of pennies in the first pig. If the gold coin says "equal" put an amount of pennies that is equal to the amount of pennies in the first pig. Use to write the number in the box under the second pig. Use to record telling me how many pennies are in the second pig and if the amount is "one more" "one less' or "equal" to the amount of pennies in the first pig. Tap to submit completed activity.
50 teachers like this
Compatible with: Chromebooks, computers, iPads, iPhones, Android tablets, Android phones, Kindle Fire
Students will edit this template: | 263 | 1,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-05 | latest | en | 0.917913 |
https://mycartablog.com/2014/11/03/what-your-brain-does-with-colours-when-you-are-not-looking-part-2/comment-page-1/ | 1,702,015,858,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00406.warc.gz | 475,085,036 | 22,909 | # What your brain does with colours when you are not “looking” – part 2
In What your brain does with colours when you are not “looking”, part 1, I displayed some audio spectrogram data (courtesy of Giuliano Bernardi at the University of Leuven) using 5 different colormaps to render the amplitude values: Jet (until recently Matlab’s standard colormap), grayscale, linear lightness rainbow, modified heated body, and cube lightness rainbow. I then asked readers to cast a vote for what they thought was the best colormap to visualize this dataset.
I was curious to see how all these colormaps fared, but my expectation was that Jet would sink to the bottom. I was really surprised to see it came on top, one vote ahead of the linear lightness rainbow (21 and 20 votes out of 62, respectively). The modified heated body followed with 11 votes.
My surprise comes from the fact that Jet carries perceptual artifacts within the progression of colours (see for example this post). One way to demonstrate these artifacts is to convert the 2D map into a 3D surface where again we use Jet to colour amplitude values, but we use the intensities from the 2D map for the elevation. This can be done for example using the Interactive 3D Surface Plot plugin for ImageJ (as in my previous post ). The resulting surface is shown in Figure 1. This is almost exactly what your brain would do when you look at the 2D map colored with Jet in the previous post.
Figure 1
In Figure 2 the same data is now displayed as a surface where amplitude values were used for the elevation, with a very light sun shading to help a bit with the perception of relief, but no colormap at all. to When comparing Figure 1 with Figure 2 one of the artifacts is immediately recognized: the highest values in Figure 2, which honours the data, become a relative low in Figure 1. This is because red has lower intensity than yellow and therefore data colored in red in 2D are plotted at a lower elevation than data colored in yellow, even though the amplitudes of the latter were lowest.
Figure 2
For these reasons, I did not expect Jet to be the top pick. On the other hand, I think Jet is perhaps favoured because with consistent use, our brain, learns in part to accommodate for these non-perceptual artifacts in 2D maps, and because it has at least two regions of higher contrast (higher magnitude gradient) than other colormaps. Unfortunately, as I wrote in a recently published tutorial, these regions are randomly placed in the colormap, and the gradients are variable, so we gain on contrast but lose on faithfulness in representing the data structure.
Matt Hall wrote a great comment following the previous post, really making an argument for switching between multiple colormaps in the interpretation stage to explore and highlight features in both the signal and the noise in the data, and that perhaps no single colormap is best overall. I agree 100% on almost everything Matt said, except perhaps on the best overall: looking at the 2D maps, at least with this dataset, I feel the heated body could be the best overall colormap, even if marginally. In Figure 3, Figure 4, Figure 5, and Figure 6 I show the 3D displays obtained by converting the 2D grayscale, linear lightness rainbow, modified heated body, and cube llightness rainbow, respectively. Looking at the 3D displays altogether gives me a confirmation of that feeling.
What do you think?
Figure 3
Figure 4
Figure 5
Figure 6
## 7 thoughts on “What your brain does with colours when you are not “looking” – part 2”
1. Thanks for the post Matteo.
I agree with you that “modified heated body” is probably the most appealing in this case, although “linear lightness rainbow” has become my default colormap and, maybe, I’m getting a bit biased on the choice 🙂
Among the last 4, I’d say that the “cube llightness rainbow” is the one that, to me, conveys the most difficult information to interpret.
• Ciao Giuliano
It took me a while to get this one finished, but I think it came out interesting.
Have you tried Parula, the new default Matlab colormap?
The cube lightness rainbow is less succesfull because it has the least total lightness contrast (60% of the full 1-100 range). This is cvclose to the minimum contrast that should be used in any colormap (according to Rogowitz and Kalvin in The” Which Blair Project”: a quick visual method for evaluating perceptual color maps).
— Sent from Mailbox
• Thanks Evan
I’m not done yet, there’s one more post in the blitz.
🙂
About the link: sometimes a post will get published with the date of the first draft instead of the date it was posted. If I change it, then the email link is lost. I have to live with this. Would there be a workaround with wordpress.org?
3. Matteo – two guesses about the votes for jet: familiarity, and a perceived “greater contrast.” I have heard the contrast comment from a MATLAB user who preferred jet to parula for that reason. jet goes dark to bright in the lower third and bright to dark in the upper third (roughly) of the colormap, so within certain data ranges there is a greater contrast. (For this discussion, I’m defining contrast as rate of change of lightness.) Parula, as well as the other colormaps shown above, go dark to bright over the whole data range, so the rate of change of lightness is lower.
• Hi Steve | 1,262 | 5,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.889415 |
https://alex.state.al.us/learningasset_view.php?asset_id=1544&res_id=1544&res_type=LA | 1,660,502,215,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00051.warc.gz | 126,553,153 | 7,616 | # ALEX Learning Activity
## Kahooting Equations
A Learning Activity is a strategy a teacher chooses to actively engage students in learning a concept or skill using a digital tool/resource.
You may save this Learning Activity to your hard drive as an .html file by selecting “File”,then “Save As” from your browser’s pull down menu. The file name extension must be .html.
This learning activity provided by:
Author: Pamela West System: Montgomery County School: Montgomery County Board Of Education
General Activity Information
Activity ID: 1544 Title: Kahooting Equations Digital Tool/Resource: Kahoot Web Address – URL: https://create.kahoot.it/details/9a0e8bc8-b88b-44cc-97f0-042a1d515b1b Overview: Students will use the Kahoot! response system to solve one-step equations and inequalities with one variable.The teacher will be able to monitor the use of Kahoots as a type of formative assessment - through quizzing, collaboration, and presentation of content. Kahoot! initiates peer-led discussions, with students left on the edge of their seats.This activity was created as a result of the ALEX Resource Development Summit.
Associated Standards and Objectives
Content Standard(s):
Mathematics MA2015 (2016) Grade: 9-12 Algebra I 17 ) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. [A-REI3] Alabama Alternate Achievement Standards AAS Standard: M.AAS.A.HS.17- Solve an equation of the form ax + b = c where a, b, and c are positive whole numbers and the solution, x, is a positive whole number to represent a real-world problem.
Learning Objectives:
The student will be able to solve equations and inequalities with addition, subtraction, multiplication, and division.
Strategies, Preparations and Variations | 395 | 1,792 | {"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} | 3.6875 | 4 | CC-MAIN-2022-33 | latest | en | 0.894639 |
https://bytes.com/topic/python/answers/36972-proposition-syntax-initialisation-multidimensional-lists | 1,713,857,585,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818468.34/warc/CC-MAIN-20240423064231-20240423094231-00771.warc.gz | 137,505,792 | 8,577 | 473,394 Members | 1,769 Online
# proposition for syntax for initialisation of multidimensional lists
i always have trouble explaining why creation of multidimensional lists
is not as straight forward as it could be in python. list comprehensions
are ugly if you are new to the language. i really would like to see it
made easy. i propose using tuples in the same way as integers are now.
example:
[0] * (2,3,4)
[[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]], [[0, 0, 0, 0], [0, 0, 0,
0], [0, 0, 0, 0]]]
this is even less "mathematically bad" syntax than multiplication with
an integer, as multiplication of vectors is only well defined if they
are of the same length. [1] * 7 could be interpreted as a vector of
length one multiplied with a scalar 7 resulting in the vector [7]. (this
was not meant as a proposition to remove the current syntax.)
i would be easy to implement:
class mylist(list):
def __mul__(self, dims):
if dims.__class__ == (()).__class__:
if len(dims) > 1:
return [self.__mul__(dims[1:]) for i in range(dims[0])]
else:
return list.__mul__(self, dims[0])
else:
return list.__mul__(self, dims)
li = mylist()
li.append(0)
print li * (2,3,4)
what do you think?
klem fra nils
Jul 18 '05 #1
1 1324
Nils Grimsmo wrote:
i always have trouble explaining why creation of multidimensional lists
is not as straight forward as it could be in python.
Maybe because that is not really a sensible thing to do with the list
data structure, which can change in length after instantiation?
>>> [0] * (2,3,4) [[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]], [[0, 0, 0, 0], [0, 0, 0,
0], [0, 0, 0, 0]]]
[SNIP]
what do you think?
I think I will continue doing this:
import numarray
numarray.zeros((2,3,4))
array([[[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]],
[[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]])
--
Michael Hoffman
Jul 18 '05 #2
This thread has been closed and replies have been disabled. Please start a new discussion.
### Similar topics
7 by: 2mc | last post by: I am finding out all kinds of ways to do things in NumPy through the many suggestions I have received. It's exciting. Thanks to all who have replied on my other threads. I'm still having... 1 by: Gabriel Birke | last post by: Given the multidimensional list l: l = ] ] I want to access specific items the indices of which are stored in another list. For now, I created a function to do this: def getNestedValue(l,... 106 by: A | last post by: Hi, I have always been taught to use an inialization list for initialising data members of a class. I realize that initialsizing primitives and pointers use an inialization list is exactly the... 2 by: Tim | last post by: Please advise if you can. Presumably initialisation of members in member initialisation lists is perfomed by 'C' run-time startup. If the CRT was never started-up would those members be garbage?... 13 by: Kevin | last post by: Help! Why are none of these valid? var arrayName = new Array(); arrayName = new Array('alpha_val', 1); arrayName = ; I'm creating/writing the array on the server side from Perl, but I 2 by: d[ - - ]b | last post by: Hi there, Just wondering, if there is any way to have a dynamic / unspecified size of an array? It obviously needs to be updated, deleted etc. The only way I thought is by making the array... 2 by: BB | last post by: Hello, I have a HTML form containing multidimensional selects listing equipments and their quantitites. This allow the users to select the kind of equipment and quantitites they would like to... 16 by: Rehceb Rotkiv | last post by: Hello everyone, can I sort a multidimensional array in Python by multiple sort keys? A litte code sample would be nice! Thx, Rehceb 7 by: krishna | last post by: What is the need of this syntax for initializing values, isn't this ambiguous to function call? e.g., int func_name(20); this looks like function call (of course not totally as there is no... 0 by: Charles Arthur | last post by: How do i turn on java script on a villaon, callus and itel keypad mobile phone 0 by: ryjfgjl | last post by: If we have dozens or hundreds of excel to import into the database, if we use the excel import function provided by database editors such as navicat, it will be extremely tedious and time-consuming... 0 by: ryjfgjl | last post by: In our work, we often receive Excel tables with data in the same format. If we want to analyze these data, it can be difficult to analyze them because the data is spread across multiple Excel files... 0 by: emmanuelkatto | last post by: Hi All, I am Emmanuel katto from Uganda. I want to ask what challenges you've faced while migrating a website to cloud. Please let me know. Thanks! Emmanuel 1 by: Sonnysonu | last post by: This is the data of csv file 1 2 3 1 2 3 1 2 3 1 2 3 2 3 2 3 3 the lengths should be different i have to store the data by column-wise with in the specific length. suppose the i have to... 0 by: Hystou | last post by: There are some requirements for setting up RAID: 1. The motherboard and BIOS support RAID configuration. 2. The motherboard has 2 or more available SATA protocol SSD/HDD slots (including MSATA, M.2... 0 by: Hystou | last post by: Most computers default to English, but sometimes we require a different language, especially when relocating. Forgot to request a specific language before your computer shipped? No problem! You can... 0 by: Hystou | last post by: Overview: Windows 11 and 10 have less user interface control over operating system update behaviour than previous versions of Windows. In Windows 11 and 10, there is no way to turn off the Windows... 0 by: tracyyun | last post by: Dear forum friends, With the development of smart home technology, a variety of wireless communication protocols have appeared on the market, such as Zigbee, Z-Wave, Wi-Fi, Bluetooth, etc. Each... | 1,619 | 5,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.87482 |
https://lists.defectivebydesign.org/archive/html/help-octave/2019-08/msg00095.html | 1,686,374,674,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00138.warc.gz | 411,212,618 | 3,289 | help-octave
[Top][All Lists]
Re: Order of Evaluation
From: fi Subject: Re: Order of Evaluation Date: Tue, 20 Aug 2019 11:27:37 +0200 User-agent: Mutt/1.5.24 (2015-08-30)
```On Mon, Aug 19, 2019 at 03:17:00PM -0700, Mike Miller wrote:
> On Mon, Aug 19, 2019 at 20:56:00 +0200, address@hidden wrote:
> > Dear List,
> >
> > consider the following functions (assumed free of side effects). The
> > functions themself do not matter. Here are some simple examples for
> > clearness:
> >
> >
> > f = @(x) x .^ pi + log(x); % ... some expression ...
> > g = @(x) sin(x) * exp(-x/5); % ... another expression ...
> > h = @(x) sqrt(x .^ cos(x)); % ... normally a costly (in time) expression
> > ...
> >
> > % and last:
> >
> > y = @(x) f(h(x)) + g(h(x));
> >
> > % here h(x) will be evaluated twice. This is inefficient, especially
> > % if y() is evaluated frequently (e.g. in finding roots or
> > % integrating).
> > %
> > % Unfortunately Octave (like Matlab) does not have a sequence operator like
> > C.
> > % So constructs like the following lead to syntax errors:
> >
> > y = @(x) H = h(x), f(H) + g(H); % also wrong if in brackets
> >
> > % The only idea I found was:
> >
> > y = @(x) f(H = h(x)) + g(H);
>
> You could try using eval or evalin to evaluate a compound expression
> inside an anonymous function, for example
>
> y = @(x) evalin ("caller", "H = h(x); f(H) + g(H)");
that is an interesting application of the somewhat exotiv
"evalin()". But I am afraid, that the evaluation could have a
performance impact. My goal was to increase the performance as much as
possible.
> Couldn't you simply define another auxiliary y1 function, for example
>
> y1 = @(x) f(x) + g(x);
> y = @(x) y1(h(x));
This trick is great and it is robust even if the order of evaluation
is not guaranteed.
> Lastly, you could define y as a true function rather than an anonymous
> function
>
> function res = y(x); H = h(x); res = f(H) + g(H); endfunction
>
> Is there a reason to constrain yourself to use only anonymous functions?
I have noticed, that anonymous functions are handled faster than true
functions (but maybe this is wrong for recent versions of Octave). My
expressions are basically simple and my goal was maximum performance
(because evaluation takes place frequently).
best regards
Torsten Finke
--
------------------------------------------------------------------------
Dr.-Ing. Torsten Finke | 691 | 2,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.833898 |
https://www.chase2learn.com/word-order-in-python-hackerrank-solution/ | 1,675,416,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00619.warc.gz | 709,256,014 | 54,981 | # Word Order in python HackerRank Solution
Hello coders, In this post, you will learn how to solve Word Order in python HackerRank Solution. This problem is a part of the Python Hacker Rank series.
We also provide Hackerrank solutions in CC++Java programming, and Python Programming languages so whatever your domain we will give you an answer in your field.
You can practice and submit all HackerRank problem solutions in one place. Find a solution for other domains and Sub-domain. I.e. Hacker Rank solution for HackerRank C ProgrammingHackerRank C++ ProgrammingHackerRank Java Programming, HackerRank Python ProgrammingHackerRank Linux ShellHackerRank SQL Programming, and HackerRank 10 days of Javascript.
As you already know that this site does not contain only the Hacker Rank solutions here, you can also find the solution for other problems. I.e. Web Technology, Data StructuresRDBMS ProgramsJava Programs Solutions, Fiverr Skills Test answersGoogle Course AnswersLinkedin Assessment, and Coursera Quiz Answers.
## Word Order in python HackerRank Solution
### problem
You are given n words. Some words may repeat. For each word, output its number of occurrences. The output order should correspond with the input order of appearance of the word. See the sample input/output for clarification.
Note: Each input line ends with a “\n” character.
#### Constraints :
• 1 <= n <= 10^5
The sum of the lengths of all the words do not exceed 10^6
All the words are composed of lowercase English letters only.
#### Input Format :
The first line contains the integer, n.
The next n lines each contain a word.
#### Output Format :
Output 2 lines.
On the first line, output the number of distinct words from the input.
On the second line, output the number of occurrences for each distinct word according to their appearance in the input.
```4
bcdef
abcdefg
bcde
bcdef
```
```3
2 1 1
```
#### Explanation :
There are 3 distinct words. Here, “bcdef” appears twice in the input at the first and last positions. The other words appear once each. The order of the first appearances are “bcdef”, “abcdefg” and “bcde” which corresponds to the output.
### Word Order in python HackerRank Solution
```# Word Order in python - Hacker Rank Solution
# Python 3
# Enter your code here. Read input from STDIN. Print output to STDOUT
# Word Order in python - Hacker Rank Solution START
import collections;
N = int(input())
d = collections.OrderedDict()
for i in range(N):
word = input()
if word in d:
d[word] +=1
else:
d[word] = 1
print(len(d));
for k,v in d.items():
print(v,end = " ");```
Disclaimer: The above Problem (Word Order in python ) is generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes. Authority if any of the queries regarding this post or website fill the following contact form thank you.
### FAQ:
#### Does HackerRank have Python?
HackerRank’s programming challenges can be solved in a variety of programming languages (including Java, C++, PHP, Python, SQL, JavaScript) and span multiple computer science domains.
#### Where can I practice Python coding?
• HackerRank is a great site for practice that’s also interactive.
#### Where can I find HackerRank solutions in Python?
in this post, you will get all the solutions to HackerRank Python Problems.
#### Is possible HackerRank Solution in Python?
HackerRank’s programming challenges can be solved in a variety of programming languages (including Java, C++, PHP, Python, SQL, and JavaScript) and span multiple computer science domains. When a programmer submits a solution to a programming challenge, their submission is scored on the accuracy of their output.
Finally, we are now, in the end, I just want to conclude some important message for you
Note:- I compile all programs, if there is any case program is not working and showing an error please let me know in the comment section. If you are using adblocker, please disable adblocker because some functions of the site may not work correctly. | 921 | 4,063 | {"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-2023-06 | latest | en | 0.796605 |
https://math.stackexchange.com/questions/3323998/rigorously-proving-lim-x-to-2n1-tan-left-frac-pi-x-2-right-i/3324007 | 1,623,540,615,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586465.3/warc/CC-MAIN-20210612222407-20210613012407-00510.warc.gz | 351,345,334 | 37,613 | # Rigorously proving $\lim_{x \to (2n+1)^+} \tan\left(\frac{\pi x} 2\right) = - \infty$
How to rigorously prove this limit? $$\lim_{x \to (2n+1)^+} \tan\left(\frac{\pi x} 2\right) = - \infty$$
My studying book used function composition and Heine definition for limits (Using sequences) to do so. But I didn't really understand the proof.
How can I prove this in a rigorous way?
Since:
• $$\displaystyle\tan\left(\frac{\pi x}2\right)=\frac{\sin\left(\frac{\pi x}2\right)}{\cos\left(\frac{\pi x}2\right)}$$;
• $$\displaystyle\lim_{x\to(2n+1)^+}\sin\left(\frac{\pi x}2\right)=\begin{cases}1&\text{ if n is even}\\-1&\text{ if n is odd;}\end{cases}$$
• if $$x$$ is close to and greater than $$2n+1$$, then $$\displaystyle\cos\left(\frac{\pi x}2\right)\begin{cases}>0&\text{ if n is even}\\<0&\text{ if n is odd}\end{cases}$$
you have$$\lim_{x\to(2n+1)^+}\tan\left(\frac{\pi x}2\right)=-\infty.$$
As $$f(x)= \tan\left(\frac{\pi x} 2\right)$$ is a periodic function of period $$2$$, the limit is equal to
$$\lim_{x \to 1^+} \tan\left(\frac{\pi x} 2\right).$$
And
$$\lim_{x \to 1^+} \sin\left(\frac{\pi x} 2\right)=1$$ while $$\lim_{x \to 1^+} \cos\left(\frac{\pi x} 2\right)=0$$ by taking only negative values in the interval $$(1,2)$$.
This provides the expected limit. | 496 | 1,275 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-25 | latest | en | 0.52369 |
https://www.homeworkmarket.com/content/age-onset-dementia-was-determined-sample-adults-between-ages | 1,571,079,679,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00115.warc.gz | 907,011,218 | 38,127 | # Age at onset of dementia was determined for a sample of adults between the ages
Age at onset of dementia was determined for a sample of adults between the ages of 60 and 75. For 15 subjects, the results were ΣX = 1008, and Σ(X-M)2 = 140.4. Use this information to answer the following:
1 of 2: Based on the data you have and the Normal Curve Tables, what percentage of people might start to show signs of dementia at or before age 62?
2 of 2: A neuropsychologist is interested only in studying the most deviant portion of this population, that is, those individuals who fall within the top 10% and the bottom 10% of the distribution. She must determine the ages that mark these boundaries. What are these ages?
• Posted: 5 years ago
Age at onset of dementia was determined for a sample of adults between the ages
Purchase the answer to view it
Save time and money!
Our teachers already did such homework, use it as a reference!
• Not rated
### Age at onset of dementia was determined for a sample of adults between the ages of 60
Age at onset of dementia was determined for a sample of adults between the ages of 60 and 75. For 15 subjects, the results were ΣX = 1008, and Σ(X-M)2 = 140.4. Use this information to answer the … | 300 | 1,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-43 | latest | en | 0.969704 |
palmoilmill.co.nz | 1,563,907,379,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529481.73/warc/CC-MAIN-20190723172209-20190723194209-00371.warc.gz | 126,882,228 | 5,791 | #### Continuous Clarification Tank Working principle in a palm oil mill
A= Skimmer height
B= Height of oil layer
C = Height of Sludge
D = Height of some oil + Sludge
Density of Palm Oil is 0.9
Density of sludge is 1.0 (Same as water)
Forces are always equal and opposite if there is no movement. (Newton’s Law)
Weight of SLUDGE = Volume of sludge x density of sludge
Weight of SLUDGE = surface area x height x density of sludge
Weight of SLUDGE = π d2/4 x D x 1.0
Weight of SLUDGE + Oil = surface area x height x density of sludge + surface area (oil) x height x density of oil
=π d2/4 x B x 0.9+ π d2/4 x C x 1.0
Therefore
π d2/4 x D x 1.0 = π d2/4 x B x 0.9+ π d2/4 x C x 1.0
To solve the above equation.
Divide both sides by π d2/
D = 0.9B + C …………………..Equation 1
A+D = B+C
Assume A = 10
Therefore
10+D=B+C
C = 10 + D – B ……………………2
Substitution equation 2 into 1
D = 0.9B + ( 10 + D – B)
D= 0.9B + 10 + D – B
Subtract D from both sides
0 = 0.9B + 10 – B
0 = 0.1B + 10
0.1B = 10 + 0
B = 10/0.1
B= 100
If A = 10 Then B= 100
Therefore the height of the oil in a balanced tank is 10 times the skimmer height.
During mill operations the skimmer heignt can be adjusted to achieve different thickness of oil layer. | 454 | 1,255 | {"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-2019-30 | latest | en | 0.824529 |
http://mathhelpforum.com/discrete-math/15997-tricky-induction.html | 1,481,201,666,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00445-ip-10-31-129-80.ec2.internal.warc.gz | 179,213,850 | 12,839 | Thread: Tricky Induction
1. Tricky Induction
Suppose that $n$ points lie on a circle are all joined in pairs. The points are positioned so that no three joining lines are concurrent in the interior of the circle. Let $a_n$ be the number of regions into which the interior of the circle is divided. Draw diagrams to find $a_n$ is given by the following formula $a_n = n + \binom{n-1}{2} + \binom{n-1}{3} + \binom{n-1}{4} = 1+ \frac{n(n-1)(n^{2}-5n+18)}{24}$.
I am pretty sure that I can do the induction part. I tried plugging in $n = 2,3$ and the number of regions were 2 and 4. The diagrams also had 2 and 4 regions respectively. But when I plugged in $n = 4$ $a_n = 8$. However, when I drew it I got $a_n = 9$.
Here are the circles:
Am I drawing the points wrong in the third circle?
2. actually I think I know what I am doing wrong. I was overconnecting points.
3. Also, I think depending on where you put the points $a_n$ will be different?
4. Point locations and connections
I believe that the points in your diagrams would lie along the perimeter of the circle, not inside of it. Also, your formula works when each point is connected to every other point, so in your n=4 diagram, there would be a total of 6 lines. Doing that, my n=2,3,4 worked out to the desired number of regions. Hope that helps a little!
5. To prove the inductive step is it as simple as just plugging in $k+1$ for $a_k$?
6. counterexample?
I'm actually having difficulty proving this, since I can't seem to figure out how it's true. Have you tried n=6? When I plug it into the formula, I get 31, but when I actually draw it out, I get 30 areas. (I encountered this while exploring the powers of two pattern, did you notice that?) Maybe I'm not clear on the definition of concurrent lines, but I just took that part of the problem to mean that no two lines connected the same two points. Any new thoughts?
7. Here are the drawings I got:
I think the concurrent lines mean that no three lines intersect. I got 31 for $n = 6$
Yeah the pattern was:1,2,4,8,16,31,57,99,163. We might have to use strong induction. But I cant see how $a_{k-1}$ and $a_k$ relate to $a_{k+1}$. Seeing that there is no pattern, I was wondering if this is valid.
Our induction hypothesis is: $a_{k} = k + \binom{k-1}{2} + \binom{k-1}{3} + \binom{k-1}{4} = 1 + \frac{k(k-1)(k^{2}-5k+18)}{24}$ for $n \leq k$ points joined in pairs. It really doesnt matter if we use strong induction or not, since regular induction is just a special case of strong induction. So our goal is to show that $a_{k} = k + \binom{k-1}{2} + \binom{k-1}{3} + \binom{k-1}{4} = 1 + \frac{k(k-1)(k^{2}-5k+18)}{24}$ $\Rightarrow a_{k+1} = k+1 + \binom{k}{2} + \binom{k}{3} + \binom{k}{4} = 1 + \frac{(k+1)k((k+1)^{2}-5(k+1)+18)}{24}$
So then would it be valid to substitute $k+1$ for $k$? I dont think we can do this since we have to use the induction hypothesis.
Is this valid: $a_{k+1} = a_{k} + \frac{1}{6}k^{3} - \frac{1}{2}k^{2}+\frac{4}{3}k$?
8. To get $P(k) \Rightarrow P(k+1)$ we need to use the inductive hypothesis. Maybe we have to use the definition of the binomial coefficient: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$?
9. anybody know how to prove the inductive step?
10. I have an inkling that strong induction is needed, but cant see the relationships.
11. The correct formula is given here.
12. Isnt the formula that I gave also correct?
13. I know why nobody can answer this. This is a very difficult problem I believe. | 1,066 | 3,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-50 | longest | en | 0.964727 |
https://www.unitconverters.net/flow-mass/pound-day-to-gram-hour.htm | 1,718,981,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00599.warc.gz | 909,063,619 | 3,067 | Home / Flow - Mass Conversion / Convert Pound/day to Gram/hour
# Convert Pound/day to Gram/hour
Please provide values below to convert pound/day [lb/d] to gram/hour [g/h], or vice versa.
From: pound/day To: gram/hour
### Pound/day to Gram/hour Conversion Table
Pound/day [lb/d]Gram/hour [g/h]
0.01 lb/d0.1889968208 g/h
0.1 lb/d1.8899682083 g/h
1 lb/d18.8996820833 g/h
2 lb/d37.7993641667 g/h
3 lb/d56.69904625 g/h
5 lb/d94.4984104167 g/h
10 lb/d188.9968208333 g/h
20 lb/d377.9936416667 g/h
50 lb/d944.9841041667 g/h
100 lb/d1889.9682083333 g/h
1000 lb/d18899.682083333 g/h
### How to Convert Pound/day to Gram/hour
1 lb/d = 18.8996820833 g/h
1 g/h = 0.0529109429 lb/d
Example: convert 15 lb/d to g/h:
15 lb/d = 15 × 18.8996820833 g/h = 283.49523125 g/h | 295 | 761 | {"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-2024-26 | latest | en | 0.446025 |
http://www.physicsforums.com/showthread.php?t=451172 | 1,369,448,872,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705318091/warc/CC-MAIN-20130516115518-00052-ip-10-60-113-184.ec2.internal.warc.gz | 663,253,357 | 7,456 | Heat exchanger problem in Thermodynamics.
Hi!
I get the basics now, but this problem has got me stumped. I am trying to figure out where to start.
1. The problem statement, all variables and given/known data
In a heat exchanger, cold water flows through the tubes of the exchanger. It enters at 5.5°C, and leaves at 65°C. In the shell of the exchanger steam is being condensed. The steam enters at atmospheric pressure and 100°C, and the condensate that is formed leaves the exchanger before it can be cooled. The flow rate of the steam is 15.0 kg/s. What is the cooling flow rate?
2. Relevant equations
Oil flows through a heat exchanger at a flow rate of 125 L/min. It enters at 20°C and leaves at 75°C. The oil has a specific heat of 2.84 kJ/kgK and a relative density of 0.8. The oil is being heated by steam. The steam enters the heat exchanger at 100°C and the condensate formed leaves the heat exchanger at 85°C. Calculate the mass flow rate of the steam.
3. The attempt at a solution
First thing we know is that cool water flows through the tube, and exits as warm water. In the shell, steam is being condensed. So basically water turns into steam.
Q (cool water absorbing heat) becomes hot water.
Q = mcΔt + mLv + mcΔt (water becomes steam)
The idea is solid, but I am not sure how I take the above, and find the flow rate. For all I know, those two are different categories.
Thank you for your help. I am this close to finishing up the course, so stumbling at this point does hurt, but the help which I receive will mean a lot to me.
PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
Tags heat exchanger, mass flow rate, specific heat, steam, thermodynamics | 449 | 1,831 | {"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} | 3.453125 | 3 | CC-MAIN-2013-20 | longest | en | 0.953409 |
https://homeworksonline.wordpress.com/category/class-vii-s-ii/ | 1,586,205,863,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00195.warc.gz | 491,145,531 | 18,294 | Home » Class VII (S-II)
# Category Archives: Class VII (S-II)
## English VII A, B, C
1. Make a diary entry of your Summer Vacation. (any 20days)
2. Read a short story by any Indian English Writer (R K Narayan or Ruskin Bond) and write a short summary.
NOTE:
• Use A4 size paper
• Do not use file.
## English, VII
Read the lesson The Story of Cricket and prepare a short project on the history of Cricket and its changes through the time.
## Social Science, VII
1. Describe the art of miniature painting.
2. Describe how kathak dance evolved into the present form?
3. What was called Rajputana by the British?
4. What were the activities that people engaged in after migrating to forested and marshal areas ?
5. Collect pictures of six forms of classical Indian Dances and make an album.
## Sanskrit, VII
चर् धातोः लट् ,लङ् , लृट् रूपाणि लिखत ।
## Maths, VII
1.Write five monomials with the variable x, five binomials with the variable y and five trinomials with the variable z
2.Verify that area of a parallelogram is base x altitude with help of paper cutting and paste the same in your notebook
## Science, VII A, B, C
1. Draw a labeled diagram of stomata. Write its functions.
2. Explain photosynthesis with the help of a diagram. Write the relevant chemical equations.
## Maths VII A,B,C
1. Draw 4 rectangle and 4 squares. Find the length of their size and find their perimeter and area.
2. Take a rectangle of size 8cm and 5cm. Cut the rectangle along its diagonal to get two traingle. Supercose one triangle on the other and find the areas of 4 triangle and compare their areas.
3. Draw a circle shade one half of the circle. Fold it into 8 parts and cut along fold. Arrange the seperate pieces which is roughly a parallelogram. Compare the area of circle and the parallelogram.
## English VII A,B,C
1. Prepare family tree in a court paper with members name, occupation and relation.
2. Read supplementary reader (lesson 7,8,9 )and prepare pocket dictionary of new words.
## Sanskrit, VII-X
Sanskrit VII-X (Shift-II)
## Social Science, VII
VII – Social Science Holiday homework for Winter Break
1.Define forest
2.Define grassland
3.Write shortly on prairies
4.Who are Red Indians
5. What is Chinook? How is Chinook useful to animals?
6.What are the following:
Ranches, cowboys, granaries of the world, veld
7.How is climate in the veld?
8.What are the activities of people in veld?
## Social Science, VIII
TOPIC: HUMAN RESOURCE DEVELOPMENT( MEANING, IMPORTANCE OF HUMAN RESOURCE , DIFFERENCE BETWEEN HUMAN RESOURCE AND PHYSICAL RESOURCE)
## Social Science VII A,B
Shift-II
1. Name three major types of rocks and explain each.
2. What is rock cycle? Explain with picture.
3. Why can’t we go to the centre of the Earth?
4. What is universal adult franchise?
5. What are the common forms of inequality?
6. What is known as civil rights movement?
7. What were the problems Omprakash Valmiki faced?
8. Who is a cartographer?
9. What are the sources to learn about the past?
10. What did Amir Khusrau write about Sanskrit?
11. Draw the outline map of India in a chart paper, mark the important rivers and name them.
## Sanskrit VII
एतत् शब्दः पुंल्लिंङ्गे ।(द्विवारं लेखनीयम् )
Shift-II
## Hindi VII
कक्षा : 7
(१)स्वरचित रचना-
२) निबन्ध –
समाचार पत्र
गणतंत्र दिवस
मेरा भारत महान
पंछी
३) बीस पर्यायवाचि
४) बीस विलोम शब्द | 929 | 3,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-16 | latest | en | 0.875572 |
https://doubtnut.com/question-answer/a-ray-of-light-comes-light-comes-along-the-line-l-0-and-strikes-the-plane-mirror-kept-along-the-plan-34611518 | 1,582,205,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00400.warc.gz | 354,142,216 | 41,024 | or
# A ray of light comes light comes along the line L = 0 and strikes the plane mirror kept along the plane P = 0 at B. A(2, 1, 6) is a point on the line L = 0 whose image about P = 0 is A'. It is given that L = 0 is (x-2)/(3)= (y-1)/(4)= (z-6)/(5) and P =0 is x+y-2z=3. <br> If L_(1) =0 is the reflected ray, then its equation is
Question from Class 12 Chapter Three-Dimensional Geometry
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
200+ views | 7.7 K+ people like this
Share
Share | 187 | 555 | {"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-2020-10 | latest | en | 0.919458 |
http://docs.factorcode.org/content/vocab-math.blas.matrices.html | 1,513,511,029,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00736.warc.gz | 78,580,086 | 2,989 | Documentation
BLAS interface matrix operations
Meta-data
Tags: math, bindings Authors: Joe Groff
Words
Tuple classes
Class Superclass Slots blas-matrix-base tuple underlying ld rows cols transpose blas-matrix-rowcol-sequence tuple parent inc rowcol-length rowcol-jump length complex-double-blas-matrix blas-matrix-base complex-float-blas-matrix blas-matrix-base double-blas-matrix blas-matrix-base float-blas-matrix blas-matrix-base
Parsing words
Word Syntax cmatrix{ cmatrix{ { 1.0 0.0 0.0 1.0 } { 0.0 C{ 0.0 1.0 } 0.0 2.0 } { 0.0 0.0 -1.0 3.0 } { 0.0 0.0 0.0 C{ 0.0 -1.0 } } } dmatrix{ dmatrix{ { 1.0 0.0 0.0 1.0 } { 0.0 1.0 0.0 2.0 } { 0.0 0.0 1.0 3.0 } { 0.0 0.0 0.0 1.0 } } smatrix{ smatrix{ { 1.0 0.0 0.0 1.0 } { 0.0 1.0 0.0 2.0 } { 0.0 0.0 1.0 3.0 } { 0.0 0.0 0.0 1.0 } } zmatrix{ zmatrix{ { 1.0 0.0 0.0 1.0 } { 0.0 C{ 0.0 1.0 } 0.0 2.0 } { 0.0 0.0 -1.0 3.0 } { 0.0 0.0 0.0 C{ 0.0 -1.0 } } }
Generic words
Word Stack effect n*M.M+n*M! ( alpha A B beta C -- C=alpha*A.B+beta*C ) n*M.V+n*V! ( alpha A x beta y -- y=alpha*A.x+b*y ) n*V(*)V+M! ( alpha x y A -- A=alpha*x(*)y+A ) n*V(*)Vconj+M! ( alpha x y A -- A=alpha*x(*)yconj+A )
Ordinary words
Word Stack effect (Mcols) ( A -- columns ) (Mrows) ( A -- rows ) (Msub) ( matrix row col height width -- data ld rows cols ) (define-blas-matrix) ( TYPE T U C -- ) ( parent inc rowcol-length rowcol-jump length -- blas-matrix-rowcol-sequence ) ( underlying ld rows cols transpose -- matrix ) ( underlying ld rows cols transpose -- matrix ) ( underlying ld rows cols transpose -- matrix ) ( rows cols exemplar -- matrix ) ( underlying ld rows cols transpose -- matrix ) >complex-double-blas-matrix ( arrays -- matrix ) >complex-float-blas-matrix ( arrays -- matrix ) >double-blas-matrix ( arrays -- matrix ) >float-blas-matrix ( arrays -- matrix ) M*n ( A n -- A*n ) M. ( A B -- A.B ) M.V ( A x -- A.x ) M/n ( A n -- A/n ) Mcols ( A -- cols ) Mheight ( matrix -- height ) Mrows ( A -- rows ) Msub ( matrix row col height width -- sub ) Mtranspose ( matrix -- matrix^T ) Mtransposed? ( matrix -- ? ) Mwidth ( matrix -- width ) V(*) ( x y -- x(*)y ) V(*)conj ( x y -- x(*)yconj ) define-complex-blas-matrix ( TYPE T -- ) define-real-blas-matrix ( TYPE T -- ) n*M ( n A -- n*A ) n*M! ( n A -- A=n*A ) n*M.M ( alpha A B -- alpha*A.B ) n*M.M+n*M ( alpha A B beta C -- alpha*A.B+beta*C ) n*M.V ( alpha A x -- alpha*A.x ) n*M.V+n*V ( alpha A x beta y -- alpha*A.x+b*y ) n*V(*)V ( alpha x y -- alpha*x(*)y ) n*V(*)V+M ( alpha x y A -- alpha*x(*)y+A ) n*V(*)Vconj ( alpha x y -- alpha*x(*)yconj ) n*V(*)Vconj+M ( alpha x y A -- alpha*x(*)yconj+A )
Class predicate words
Word Stack effect blas-matrix-base? ( object -- ? ) blas-matrix-rowcol-sequence? ( object -- ? ) complex-double-blas-matrix? ( object -- ? ) complex-float-blas-matrix? ( object -- ? ) double-blas-matrix? ( object -- ? ) float-blas-matrix? ( object -- ? )
Files
Children
Vocabulary Summary math.blas.matrices.private math.blas.matrices.private vocabulary | 1,099 | 2,979 | {"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-2017-51 | longest | en | 0.329624 |
https://gmatclub.com/forum/what-is-the-value-of-z-1-y-2-3z-2-6z-2y-1858.html | 1,496,050,749,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612069.19/warc/CC-MAIN-20170529091944-20170529111944-00072.warc.gz | 930,994,563 | 45,116 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 29 May 2017, 02:39
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# What is the value of z ? 1) y=2-3z 2) 6z+2y=4
Author Message
Manager
Joined: 24 Jun 2003
Posts: 91
Location: Moscow
Followers: 1
Kudos [?]: 4 [0], given: 0
What is the value of z ? 1) y=2-3z 2) 6z+2y=4 [#permalink]
### Show Tags
05 Aug 2003, 05:05
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
What is the value of z ?
1) y=2-3z
2) 6z+2y=4
_________________
Respect,
KL
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2
Kudos [?]: 2 [0], given: 0
### Show Tags
05 Aug 2003, 05:15
Konstantin Lynov wrote:
What is the value of z ?
1) y=2-3z
2) 6z+2y=4
Solving the two equations we get Z=2/3 and y=0
Manager
Joined: 24 Jun 2003
Posts: 91
Location: Moscow
Followers: 1
Kudos [?]: 4 [0], given: 0
### Show Tags
05 Aug 2003, 05:18
Here we go, one has eaten the bate
_________________
Respect,
KL
Manager
Joined: 24 Jun 2003
Posts: 91
Location: Moscow
Followers: 1
Kudos [?]: 4 [0], given: 0
### Show Tags
05 Aug 2003, 05:22
Other ideas, please. Careful with DS practice question I have posted!
They are from Caplan's GMAT 800 book and, to my opinion, one has to be rather smart to have a bunch of questions alike done in the allotted time...
Have fun
_________________
Respect,
KL
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2
Kudos [?]: 2 [0], given: 0
### Show Tags
05 Aug 2003, 05:39
Konstantin Lynov wrote:
Here we go, one has eaten the bate
Agree Lynov....Answer is E... I was too quick to bite... the bait looked delicious ...!!!
Manager
Joined: 22 Jul 2003
Posts: 62
Location: CA
Followers: 1
Kudos [?]: 0 [0], given: 0
### Show Tags
05 Aug 2003, 07:46
good one. Isn't it called inconsistant equation or something?
Intern
Joined: 01 Aug 2003
Posts: 48
Location: Mumbai
Followers: 1
Kudos [?]: 1 [0], given: 0
### Show Tags
05 Aug 2003, 22:12
I think its called infinite solutions!
05 Aug 2003, 22:12
Display posts from previous: Sort by | 910 | 2,813 | {"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-2017-22 | latest | en | 0.860576 |
http://mathfour.com/general/we-use-math-and-grammar-rules-they-dont-use-us | 1,726,236,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00629.warc.gz | 17,180,803 | 13,259 | # We Use Math and Grammar Rules – They Don’t Use Us!
I got engaged in a twitter fight about grammar with Chiew from @aClilToClimb. Yes, I’m the math mom, but my college minor is English. And I tend to be a sharpie carrying, sign correcting, grammar vigilante.
I complained that Twitter has the link “Who To Follow” when it should be “Whom To Follow.” Here’s an excerpt from the fight:
### The fight raged on.
This guy was so adamant that you could use “who” as the object in a sentence (clearly wrong), and just wouldn’t let it go. After quite a few tweets I got curious. “What’s this guy’s deal? All he has to do is pull out the Little Brown Handbook and read it in plain black and white.”
So I went and looked at his site. Holy cow! He’s a grammar blogger!
I couldn’t find his “About” page, but from what I could gather in his fervor in our twitter fight, he’s trying to do for grammar what I’m trying to do for math. Demystify, take away the “have to” rules, and make it accessible, acceptable and appreciated by everyone.
### We make the rules!
The rules of grammar, like the rules of math, are created by humans and used by humans. They are changeable.
Of course the difference is that, in grammar, if you deviate slightly from the rules that others follow, you’ll most likely be understood. In math, you really have to define how you’re using things before you begin to work.
For instance, if I wanted to have a conversation with someone about a new way of adding fractions I was inventing, I could totally do it. As long as I started the conversation with, “Here’s how we are going to talk about adding fractions for the next hour…”
Teaching math and teaching grammar are two of the fundamental things we do for our children. And neither should be hard, creepy or frustrating. They should be a normal, natural flow of who we are as people.
Another great math mom says, “Make math your own, to make your own math.” I’d bet Chiew would say something similar for grammar.
And for this, I respect him. </fight>
This post may contain affiliate links. When you use them, you support us so we can continue to provide free content!
### 6 Responses to We Use Math and Grammar Rules – They Don’t Use Us!
1. I’m glad, mom, that it’s just a Twitter fight! I’d hate to be in the same ring with you! 😉 The thing is that there is a lot of difference, sometimes, between spoken and written English. There is also a clear boundary between what is acceptable and what isn’t. For example, ‘To who shall I write’ is not acceptable, just as ‘She don’t love me’ isn’t. Both can be heard, however, more often than they ought to be.
I quote from Cambridge Grammar of English (Carter/McCarthy) 205: “Who can be used in both subject and object forms. Whom is used in object forms and following prepositions in more formal contexts”. An example given: Who did the prime minister promote to the cabinet?
• Bon says:
Thanks so much for the engaging interaction. It was a joy!
And thanks for the information. I might continue to say “whom” but I will know in my heart that it’s okay for others to say “who.”
2. I’ve always had difficulties with “who” and “whom” usage, but I do have to say your way sounds right to me. I think it’s great, though, that the man is a grammar blogger! And you unknowingly got into a grammar fight with him. Sounds like somethign I’d get myself into!
• Bon says:
After I realized that he was a grammar blogger, I really felt like I knew the other side of the coin. I often wondered what people thought when I went nuts rallying against a math topic that is typically known as absolute fact. Now I know.
3. This is basically an argument about whether grammar should be descriptive or prescriptive. You will hear people say “It was me what done it,” in the street in the UK… you’d probably lose street-cred points, or get beaten up badly, if you said “It was I who did it.” Spoken language has a different grammar/usage from written language too, aside from any ethnic/class use. “To boldly split the infinitive” and not to use no double negatives, etc., were impositions from the Latinists three or four centuries ago (“neither… nor” is a double negative, after all!) = prescriptive. I was schooled during the days of prescriptive English grammar, however, insisting on prescribing some types of language use can give the impression of being ‘overly stuffy’ or pedantic, rather than treating it as a means of communication. It all comes down to the impression you want to create, I suppose, and whom one would wish to impress!
• Bon says:
Indeed. And it’s easier to go from “proper” to “improper” when you know proper.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,105 | 4,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.95907 |
https://www.brainbashers.com/showcalcudoku.asp?date=0514&diff=2&size=5 | 1,624,007,011,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00358.warc.gz | 596,780,193 | 4,535 | Daily CalcuDoku
May 14 - 5 x 5 Medium
Show mistakes when checking
Warn when leaving page
Auto-pencil marks
Today's CalcuDoku Puzzles
4 x 4 - Easy - Medium - Hard
5 x 5 - Easy - Medium - Hard
6 x 6 - Easy - Medium - Hard
7 x 7 - Easy - Medium - Hard
8 x 8 - Easy - Medium - Hard
9 x 9 - Weekly Special
All daily items change at midnight GMT.
Notes
• Objective / Rules
- Complete the grid such that every row and column contains the digits 1 to the size of the grid.
- Each row and column contains each digit only once.
- A cage clue tells you the answer after the cage values have undergone the specified mathematical operation.
- The clue doesn't tell you which way around the digits occur, just the answer to the calculation.
- A digit can appear more than once in a cage.
• Help
Read the help/walkthrough page on CalcuDoku for a more detailed explanation.
• Keyboard Usage [only when your cursor is in the grid]
Note: not all options work on all browsers.
A = auto-pencil marks (or click the check box).
CTRL + arrows = move around the grid.
SHIFT + number = highlight all squares with that number [SHIFT+0 to clear].
• Checking
If you click 'Check' the system will check for incorrect squares. If 'Show mistakes when checking' is checked they will be marked in red. There are two types of error checked for:
1. A single number that isn't the same as the solution.
2. Multiple numbers that don't contain the solution number.
• Uniqueness
Each puzzle has exactly one solution, which can be found using logic alone and no guesses are ever required. If you think you've found another solution, then please double check the rules.
• Note
BrainBashers CalcuDoku are not approved, nor sponsored, nor endorsed by any other website, creator or company.
• JavaScript
This game requires JavaScript to be enabled.
[Puzzle Code = CalcuDoku-0514-5x5-Medium-078225] | 458 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-25 | latest | en | 0.817469 |
https://course.ccs.neu.edu/cs3500/lec_commands_notes.html | 1,591,199,691,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00044.warc.gz | 296,339,090 | 7,899 | 6.3
## Lecture 12: The Command Design Pattern
### Objectives of the lecture
This lecture starts by working out the design of a model and controller of a new small application in detail. It streamlines code in the controller in several iterations to introduce the command design pattern.
Link to Pre-lecture-survey (take survey after reading notes, but before lecture)
### 1Context of the example program
This lecture uses the example of turtle graphics. Turtle graphics uses the notion of a turtle moving on a 2D plane. At any point in time, the turtle occupies a fixed position on the plane, and points in some direction. The turtle has limited mobility: it is able to turn in its place, and move only in the direction that it is pointing.
Turtle graphics is used to measure and draw in various applications. Many such applications find it convenient to specify movement relative to current position and direction (as the turtle) instead of absolute Cartesian coordinates. For example, driving directions provide navigation relative to the current position and direction (“Drive 0.3 miles”, “Take left onto...”).
#### 1.1Basic Design
Based on the basic operations identified above for turtle model, we begin by designing an interface for the model of our program:
/**
* This interface specifies the operations on a 2D turtle
* <p>
* A 2D turtle is characterized by a position (x,y) and a
* heading (where it is looking).
* <p>
* It can be asked to draw the path it has moved using one of
* the commands below.
*/
public interface TurtleModel {
/**
* Move the turtle by the specified distance along its
*
* @param distance
*/
void move(double distance);
/**
* Turn the turtle's heading by the given angle.
* A positive angle means counter-clockwise
* turning. The turtle turns in place, i.e.
* it does not change position.
*
* @param angleDegrees
*/
void turn(double angleDegrees);
/**
* Save the current turtle state (position + heading)
*/
void save();
/**
* Retrieve the last saved turtle state (position + heading)
*/
void retrieve();
/**
* Get the current position of the turtle
*
* @return
*/
Position2D getPosition();
/**
* Get the current heading of the turtle
*
* @return
*/
}
The Position2D class represents a single, immutable 2D position.
/**
* This class represents a 2D position
*/
public final class Position2D {
private final double x;
private final double y;
/**
* Initialize this object to the specified position
*/
public Position2D(double x, double y) {
this.x = x;
this.y = y;
}
/**
* Copy constructor
*/
public Position2D(Position2D v) {
this(v.x, v.y);
}
public double getX() {
return x;
}
public double getY() {
return y;
}
@Override
public String toString() {
return String.format("(%f, %f)", this.x, this.y);
}
@Override
public boolean equals(Object a) {
if (this == a) {
return true;
}
if (!(a instanceof Position2D)) {
return false;
}
Position2D that = (Position2D) a;
return ((Math.abs(this.x - that.x) < 0.01) && (Math.abs(this.y - that.y) < 0.01));
}
@Override
public int hashCode() {
return Objects.hash(this.x, this.y);
}
}
We implement the TurtleModel interface as a SimpleTurtle class.
/**
* This class manages a 2D turtle and implements all
* its associated operations
*/
public class SimpleTurtle implements TurtleModel {
// the position of the turtle
private Position2D position;
// the heading of the turtle in degrees
// stacks to save and retrieve turtle states
Stack<Position2D> stackPositions;
/**
* Initializes the turtle to the default state.
* Default state = position (0,0) and heading (0) meaning
* looking in the +X direction.
*/
public SimpleTurtle() {
this(new Position2D(0, 0), 0);
}
/**
* Initializes the turtle to the given position and heading.
*/
public SimpleTurtle(Position2D startPos, double startHeading) {
position = Objects.requireNonNull(startPos);
stackPositions = new Stack<>();
stackHeadings = new Stack<>();
}
@Override
public void move(double distance) {
//trigonometry to move by distance along angle
position = new Position2D(position.getX() + x, position.getY() + y);
}
@Override
public void turn(double angleDegrees) {
}
@Override
public void save() {
stackPositions.push(position);
}
@Override
public void retrieve() {
if ((stackPositions.isEmpty()) || (stackHeadings.isEmpty())) {
throw new IllegalArgumentException("no state to retrieve");
}
position = stackPositions.pop();
}
@Override
public Position2D getPosition() {
return position;
}
@Override
}
#### 1.2Enhancement
The above turtle model is able to move and turn, but is not able to draw anything. Although drawing is within the scope of the view, the model must provide it with the data to draw. In this specific example, our turtle must trace its path, and provide a way for the client to retrieve its traces (in the form of lines).
Specifically we need the following operations:
void trace(double distance);
List<Line> getLines();
It may be tempting to simply add these methods to the TurtleModel interface and implement them in the SimpleTurtle class. However doing so has several problems:
How do we add these operations? We extend the existing interface and then implement it by reusing the existing implementation (with inheritance).
public interface TracingTurtleModel extends TurtleModel {
/**
* Move the turtle by the specified distance along its
* Draw a line from its initial position to its
* final position.
*
* @param distance
*/
void trace(double distance);
/**
* Get the lines traced by this turtle, caused by the
* trace method above.
*
* @return a list of {@code Line} objects, in the order they were drawn.
*/
List<Line> getLines();
}
public class SmarterTurtle extends SimpleTurtle implements TracingTurtleModel {
public SmarterTurtle() {
super();
lines = new ArrayList<Line>();
}
@Override
public void trace(double distance) {
Position2D cur = this.getPosition();
move(distance);
}
@Override
public List<Line> getLines() {
return new ArrayList<>(lines);
}
//list of lines traced since this object was created
List<Line> lines;
}
### 2Controller
We offer a text-based synchronous controller for our application. The application (through the controller) has the following loop:
1. Take a one-word command from the user. This command is one of “move”, “turn”, “trace”, “show” and “quit”.
2. Depending on the command, take additional input (e.g. “move” requires a distance to move).
3. Call the appropriate operation on the model, or quit (if the command is “quit”).
This results in the following code:
public class SimpleController {
public void go() {
Scanner s = new Scanner(System.in);
TracingTurtleModel m = new SmarterTurtle();
while (s.hasNext()) {
String in = s.next();
switch(in) {
case "q":
case "quit":
return;
case "show":
for (Line l : m.getLines()) {
System.out.println(l);
}
break;
case "move":
try {
double d = s.nextDouble();
m.move(d);
} catch (InputMismatchException ime) {
...
}
break;
case "trace":
try {
double d = s.nextDouble();
m.trace(d);
} catch (InputMismatchException ime) {
...
}
break;
case "turn":
try {
double d = s.nextDouble();
m.turn(d);
} catch (InputMismatchException ime) {
...
}
break;
default:
System.out.println(String.format("Unknown command %s", in));
break;
}
}
}
}
### 3Scaling up the controller
Imagine if we support additional text commands in the controller. Although the current set of supported commands use all available operations in the model (except save and retrieve), we can support new drawing commands at the controller level. For example, drawing a square is a sequence of 4 move and 4 turn operations on the model. Instead of letting the user draw it as a sequence of 8 text commands, we could offer it as a new text command.
switch(in) {
...
case "square":
try {
double d = s.nextDouble();
m.trace(d);
m.turn(90);
m.trace(d);
m.turn(90);
m.trace(d);
m.turn(90);
m.trace(d);
m.turn(90);
} catch (InputMismatchException ime) {
...
}
break;
}
The possibilities are now endless! We can support similar higher-level drawing text commands. Every such text commands adds a new case to our switch statement. The number of lines of code in each case statement depends on the complexity of the text command (e.g. the square text-command added 13 lines of code). As a result, the switch statement quickly grows in size. Moreover the go method is increasingly incohesive.
### 4The Command Design Pattern
In order to make each case statement shorter, we can put all its code into a separate helper method. Since all the helper methods operate on the model, we pass the model object to them. Also since some of the text command require additional input, we pass the additional input to each of them (in an attempt to make their signatures the same). The switch statement would now become:
switch(in) {
case "move":
double d = s.nextDouble();
moveHelper(d,m);
break;
case "turn":
double d = s.nextDouble();
traceHelper(d,m);
break;
..
}
We can now characterize each such helper method as follows: Take the model object and an additional piece of data, and execute a set of operations on the model. Note that although the operations that each helper method executes are different, all of the helper methods can be characterized this way. Since the methods differ only by name we can unify them under a single interface that has a method of the same signature: void go(TracingTurtleModel model);.
Design-wise, how can we justify the purpose of such an interface? It represents a high-level command: a set of operations that must be executed. This is an example of the command design pattern. This pattern unifies different sets of operations under one umbrella, so that they can be treated uniformly.
public interface TracingTurtleCommand {
void go(TracingTurtleModel m);
}
We then implement this interface, once for each text command. Each implementation would take additional data during instantiation.
public class Move implements TracingTurtleCommand {
double d;
public Move(Double d) {
this.d = d;
}
@Override
public void go(TracingTurtleModel m) {
m.move(this.d);
}
}
public class Trace implements TracingTurtleCommand {
double d;
public Trace(Double d) {
this.d = d;
}
@Override
public void go(TracingTurtleModel m) {
m.trace(this.d);
}
...
}
Now we can change the logic of our controller to:
1. Take a one-word command from the user.
2. Create the corresponding TracingTurtleCommand object..
3. Execute the command object.
String in = s.next();
try {
switch (in) {
case "q":
case "quit":
return;
case "show":
for (Line l : m.getLines()) {
System.out.println(l);
}
break;
case "move":
cmd = new Move(s.nextDouble());
break;
case "trace":
cmd = new Trace(s.nextDouble());
break;
case "turn":
cmd = new Turn(s.nextDouble());
break;
case "square":
cmd = new Square(s.nextDouble());
break;
default:
System.out.println(String.format("Unknown command %s", in));
cmd = null;
break;
}
if (cmd != null) {
cmd.go(m); //execute the command
cmd = null;
}
} catch (InputMismatchException ime) {
System.out.println("Bad length to " + in);
}
#### 4.1Advantages of the command design pattern
As the above example shows, the command design pattern offers several advantages:
• The command design pattern has a unifying effect, making unrelated lines of code appear as if working towards the same purpose. This increases cohesion: the controller is no longer doing 1 of 10 unrelated things, but executing commands.
• The command design pattern promotes delegation. Details of each command are now kept in separate classes, instead of all appearing within the controller. This allows us to support new, more complicated commands such as drawing fractal objects without cluttering the controller (see the attached code for the Koch snowflake).
• The command design pattern extends its unifying effect by allowing common operations across commands.
For example we can think of supporting a new text command: undo the last operation. Although our model supports this in a primitive way, the controller should be able to undo entire sequences of model operations (such as square). Although the details of undoing differ with the operation being undone, we can think of the operation itself at the “command” level.
We may implement this in one of two ways:
• Extend the TracingTurtleCommand interface to offer an undo() operation. Each command would determine how to undo itself.
• Store a list of command objects in the controller and undo them.
Again, note that both of these ways use the abstraction created by the command design pattern.
### 5Improving the controller
Although we made our controller code more cohesive, the switch statement will still grow as more text commands are supported.
Consider what virtually every case is doing: it takes additional input from a Scanner object, and creates a command object. We can conceptualize this operation as a method:
TracingTurtleCommand create(Scanner sc)
.
If we have such methods for each case block, then the switch statement is simply executing the correct method depending on the entered text command. We can store all such text-command -> method in a map of function objects.
Map<String, Function<Scanner, TracingTurtleCommand>> knownCommands;
knownCommands = new HashMap<>();
knownCommands.put("move",s->new Move(s.nextDouble()));
knownCommands.put("turn",s->new Turn(s.nextDouble()));
knownCommands.put("trace", s->new Trace(s.nextDouble()));
knownCommands.put("square", s -> new Square(s.nextDouble()));
Then our controller logic becomes:
1. Take a one-word command from the user.
2. Find if the command exists in the map. If so, execute the corresponding function object to get the command object.
3. Execute the command object.
The second step above becomes a map lookup, instead of a switch statement!
while(scan.hasNext()) {
TracingTurtleCommand c;
String in = scan.next();
if (in.equalsIgnoreCase("q") || in.equalsIgnoreCase("quit"))
return;
Function<Scanner, TracingTurtleCommand> cmd =
knownCommands.getOrDefault(in, null);
if (cmd == null) {
throw new IllegalArgumentException();
} else {
c = cmd.apply(scan);
c.go(m);
}
}
Adding support for a new text command is now as easy as adding a new entry to the map! This allows us to quickly assemble a list of supported text commands for a controller. The controller’s logic does not change depend on the number and complexity of supported text commands. | 3,204 | 14,354 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | longest | en | 0.855583 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.