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[ "# SDK数据库 Command·聚合操作符·日期操作符\n\n2020-07-27 14:31 更新\n\n# AggregateCommand.dateFromParts(value: any): Object\n\n## API 说明\n\n``````db.command.aggregate.dateFromParts({\nyear: <year>,\nmonth: <month>,\nday: <day>,\nhour: <hour>,\nminute: <minute>,\nsecond: <second>,\nmillisecond: <ms>,\ntimezone: <tzExpression>\n})\n``````\n\n``````db.command.aggregate.dateFromParts({\nisoWeekYear: <year>,\nisoWeek: <week>,\nisoDayOfWeek: <day>,\nhour: <hour>,\nminute: <minute>,\nsecond: <second>,\nmillisecond: <ms>,\ntimezone: <tzExpression>\n})\n``````\n\n## 示例代码\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\ndate: \\$.dateFromParts({\nyear: 2017,\nmonth: 2,\nday: 8,\nhour: 12,\ntimezone: 'America/New_York'\n}),\n})\n.end()\n``````\n\n``````{\n\"date\": ISODate(\"2017-02-08T17:00:00.000Z\")\n}``````\n\n# AggregateCommand.dateFromString(value: any): Object\n\n## API 说明\n\n``````db.command.aggregate.dateFromString({\ndateString: <dateStringExpression>,\ntimezone: <tzExpression>\n})\n``````\n\n## 示例代码\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\ndate: \\$.dateFromString({\ndateString: \"2019-05-14T09:38:51.686Z\"\n})\n})\n.end()\n``````\n\n``````{\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}``````\n\n# AggregateCommand.dateToString(value: any): Object\n\n## API 说明\n\ndateToString 的调用形式如下：\n\n``````db.command.aggregate.dateToString({\ndate: <日期表达式>,\nformat: <格式化表达式>,\ntimezone: <时区表达式>,\nonNull: <空值表达式>\n})\n``````\n\n%d 月份的日期（2位数，0填充） 01 - 31\n%G ISO 8601 格式的年份 0000 - 9999\n%H 小时（2位数，0填充，24小时制） 00 - 23\n%j 一年中的一天（3位数，0填充） 001 - 366\n%L 毫秒（3位数，0填充） 000 - 999\n%m 月份（2位数，0填充） 01 - 12\n%M 分钟（2位数，0填充） 00 - 59\n%S 秒（2位数，0填充） 00 - 60\n%w 星期几 1 - 7\n%u ISO 8601 格式的星期几 1 - 7\n%U 一年中的一周（2位数，0填充） 00 - 53\n%V ISO 8601 格式的一年中的一周 1 - 53\n%Y 年份（4位数，0填充） 0000 - 9999\n%z 与 UTC 的时区偏移量 `+/-[hh][mm]`\n%Z 以分钟为单位，与 UTC 的时区偏移量 `+/-mmm`\n%% 百分号作为字符 `%`\n\n## 示例代码\n\n``````{ \"date\": \"1999-12-11T16:00:00.000Z\", \"firstName\": \"Yuanxin\", \"lastName\": \"Dong\" }\n{ \"date\": \"1998-11-10T16:00:00.000Z\", \"firstName\": \"Weijia\", \"lastName\": \"Wang\" }\n{ \"date\": \"1997-10-09T16:00:00.000Z\", \"firstName\": \"Chengxi\", \"lastName\": \"Li\" }\n``````\n\n#### 格式化日期\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('students')\n.aggregate()\n.project({\n_id: 0,\nformatDate: \\$.dateToString({\ndate: '\\$date',\nformat: '%Y-%m-%d'\n})\n})\n.end()\n``````\n\n``````{ \"formatDate\": \"1999-12-11\" }\n{ \"formatDate\": \"1998-11-10\" }\n{ \"formatDate\": \"1997-10-09\" }\n``````\n\n#### 时区时间\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('students')\n.aggregate()\n.project({\n_id: 0,\nformatDate: \\$.dateToString({\ndate: '\\$date',\nformat: '%H:%M:%S',\ntimezone: 'Asia/Shanghai'\n})\n})\n.end()\n``````\n\n``````{ \"formatDate\": \"00:00:00\" }\n{ \"formatDate\": \"00:00:00\" }\n{ \"formatDate\": \"00:00:00\" }\n``````\n\n#### 缺失情况的默认值\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('students')\n.aggregate()\n.project({\n_id: 0,\nformatDate: \\$.dateToString({\ndate: '\\$empty',\nonNull: 'null'\n})\n})\n.end()\n``````\n\n``````{ \"formatDate\": \"null\" }\n{ \"formatDate\": \"null\" }\n{ \"formatDate\": \"null\" }``````\n\n# AggregateCommand.dayOfMonth(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.dayOfMonth(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\ndayOfMonth: \\$.dayOfMonth('\\$date')\n})\n.end()\n``````\n\n``````{\n\"dayOfMonth\": 14\n}``````\n\n# AggregateCommand.dayOfWeek(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.dayOfWeek(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\ndayOfWeek: \\$.dayOfWeek('\\$date')\n})\n.end()\n``````\n\n``````{\n\"dayOfWeek\": 3\n}``````\n\n# AggregateCommand.dayOfYear(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.dayOfYear(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\ndayOfYear: \\$.dayOfYear('\\$date')\n})\n.end()\n``````\n\n``````{\n\"dayOfYear\": 134\n}``````\n\n# AggregateCommand.hour(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.hour(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nhour: \\$.hour('\\$date')\n})\n.end()\n``````\n\n``````{\n\"hour\": 9\n}``````\n\n# AggregateCommand.isoDayOfWeek(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.month(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nisoDayOfWeek: \\$.isoDayOfWeek('\\$date')\n})\n.end()\n``````\n\n``````{\n\"isoDayOfWeek\": 2\n}``````\n\n# AggregateCommand.isoWeek(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.isoWeek(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nisoWeek: \\$.isoWeek('\\$date')\n})\n.end()\n``````\n\n``````{\n\"isoWeek\": 20\n}``````\n\n# AggregateCommand.isoWeekYear(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.isoWeekYear(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nisoWeekYear: \\$.isoWeekYear('\\$date')\n})\n.end()\n``````\n\n``````{\n\"isoWeekYear\": 2019\n}``````\n\n# AggregateCommand.millisecond(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.millisecond(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nmillisecond: \\$.millisecond('\\$date'),\n})\n.end()\n``````\n\n``````{\n\"millisecond\": 686\n}``````\n\n# AggregateCommand.minute(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.minute(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nminute: \\$.minute('\\$date')\n})\n.end()\n``````\n\n``````{\n\"minute\": 38\n}``````\n\n# AggregateCommand.month(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.month(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nmonth: \\$.month('\\$date')\n})\n.end()\n``````\n\n``````{\n\"month\": 5\n}``````\n\n# AggregateCommand.second(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.second(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nsecond: \\$.second('\\$date')\n})\n.end()\n``````\n\n``````{\n\"second\": 51\n}``````\n\n# AggregateCommand.week(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.week(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nweek: \\$.week('\\$date')\n})\n.end()\n``````\n\n``````{\n\"week\": 19\n}``````\n\n# AggregateCommand.year(value: Expression<string>): Object\n\n## API 说明\n\n``````db.command.aggregate.year(<日期字段>)\n``````\n\n## 示例代码\n\n``````{\n\"_id\": 1,\n\"date\": ISODate(\"2019-05-14T09:38:51.686Z\")\n}\n``````\n\n``````const \\$ = db.command.aggregate\ndb\n.collection('dates')\n.aggregate()\n.project({\n_id: 0,\nyear: \\$.year('\\$date')\n})\n.end()\n``````\n\n``````{\n\"year\": 2019\n}``````\n\n# AggregateCommand.subtract(value: Expression[]): Object\n\n## 参数\n\n### value: Expression[]\n\n[<expression1>, <expression2>]\n\n## API 说明\n\n``````db.command.aggregate.subtract([<expression1>, <expression2>])\n``````\n\n## 示例代码\n\n``````{ \"_id\": 1, \"max\": 10, \"min\": 1 }\n{ \"_id\": 2, \"max\": 7, \"min\": 5 }\n{ \"_id\": 3, \"max\": 6, \"min\": 6 }\n``````\n\n``````const \\$ = db.command.aggregate\ndb.collection('scores').aggregate()\n.project({\ndiff: \\$.subtract(['\\$max', '\\$min'])\n})\n.end()\n``````\n\n``````{ \"_id\": 1, \"diff\": 9 }\n{ \"_id\": 2, \"diff\": 2 }\n{ \"_id\": 3, \"diff\": 0 }``````\n\n## 您可能还喜欢：\n\nApp下载", null, "", null, "" ]

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[ "# Pre & post increment operator behavior in C, C++, Java, and C#\n\nC++Server Side ProgrammingProgramming\n\n#### C in Depth: The Complete C Programming Guide for Beginners\n\n45 Lectures 4.5 hours\n\n#### Practical C++: Learn C++ Basics Step by Step\n\nMost Popular\n\n50 Lectures 4.5 hours\n\n#### Master C and Embedded C Programming- Learn as you go\n\n66 Lectures 5.5 hours\n\nThe Pre increment and post increment both operators are used as increment operations. The pre increment operator is used to increment the value of some variable before using it in an expression. In the pre increment the value is incremented at first, then used inside the expression.\n\nif the expression is a = ++b; and b is holding 5 at first, then a will hold 6. Because increase b by 1, then set the value of a with it.\n\n## Example Code\n\n#include <iostream>\nusing namespace std;\nmain () {\nint a, b = 15;\na = ++b;\ncout << a;\n}\n\n## Output\n\n16\n\n## Example Code\n\n#include <stdio.h>\nmain () {\nint a, b = 15;\na = ++b;\nprintf(“%d”, a);\n}\n\n## Output\n\n16\n\n## Example Code\n\npublic class IncDec {\npublic static void main(String[] args) {\nint a, b = 15;\na = ++b;\nSystem.out.println(“” + a);\n}\n}\n\n## Output\n\n16\n\n## Example Code\n\nusing System;\nnamespace IncDec {\nclass Inc {\nstatic void Main() {\nint a, b = 15;\na = ++b;\nConsole.WriteLine(\"\"+a);\n}\n}\n}\n\n## Output\n\n16\n\nThe post increment operator is used to increment the value of some variable after using it in an expression. In the post increment the value is used inside the expression, then incremented by one.\n\nif the expression is a = b++; and b is holding 5 at first, then a will also hold 5. Because increase b by 1 after assigning it into a.\n\n## Example Code\n\n#include <iostream>\nusing namespace std;\nmain () {\nint a, b = 15;\na = b++;\ncout << a;\ncout << b;\n}\n\n## Output\n\n15\n16\n\n## Example Code\n\n#include <stdio.h>\nmain () {\nint a, b = 15;\na = ++b;\nprintf(“%d”, a);\nprintf(“%d”, b);\n}\n\n## Output\n\n15\n16\n\n## Example Code\n\npublic class IncDec {\npublic static void main(String[] args) {\nint a, b = 15;\na = ++b;\nSystem.out.println(“” + a);\nSystem.out.println(“” + b);\n}\n}\n\n## Output\n\n15\n16\n\n## Example Code\n\nusing System;\nnamespace IncDec {\nclass Inc {\nstatic void Main() {\nint a, b = 15;\na = ++b;\nConsole.WriteLine(\"\"+a);\nConsole.WriteLine(\"\"+b);\n}\n}\n}\n\n## Output\n\n15\n16" ]

[ null ]

[ "# Correlation matrix\n\nCorrelation matrix shows correlation coefficients for column pairs of a given dataset. TableTorch supports both the Pearson and the Spearman’s rank coefficients.\n\nIn the following sections, some variants of correlation matrices are computed for the vehicle dataset. The video presented below demonstrates how to use TableTorch to produce a colorful correlation matrix right in a spreadsheet.\n\nWatch on YouTube: Compute a correlation matrix in Google Sheets with TableTorch 4:33\n\n## Start TableTorch\n\n1. Install TableTorch to Google Sheets via Google Workspace Marketplace. More details on initial setup.\n2. Click on the TableTorch icon", null, "on right-side panel of Google Sheets.", null, "## Pearson coefficients for every column pair\n\nSelect the whole range of the sheet and click Correlation matrix menu item inside TableTorch.", null, "Non-numeric columns will be automatically filtered out.", null, "Click on the Compute button to produce a new sheet with the correlation matrix. The produced matrix will look like the following one:", null, "The default settings produce an already useful matrix. For example, if the aim is to do a regression in order to be able to predict the value of selling_price column, it is possible to draw the following conclusions from the produced matrix:\n\n1. `selling_price` is negligibly correlated with `from trustmark dealer` column suggesting that it wouldn’t make sense to use this feature in a regression. It is same with the `more than 5 seats` column.\n2. `petrol` is highly correlated with `max torque min RPM` indicating that it would make sense to use only one of those features in a regression of `selling_price`.\n3. The same could be true for the `engine cc` and `max power bhp` columns.\n\n## Configuration\n\nTableTorch can be configured to produce different matrices via editing the following settings:", null, "• Coefficient: either Pearson or Spearman’s rank. If you would like to rank-scale only certain columns before computing the coefficients, use the standalone Scaling function of the TableTorch before creating a correlation matrix.\n• Color scheme: whether and how to highlight the cells of the produced matrix.\n• Choose columns: by default, TableTorch will calculate the coefficients for every single column pair of the table. However, it is possible to select only the needed columns using this section.\n\n## Spearman’s rank coefficients for specified column pairs\n\nSimply put, Spearman’s rank coefficient is a Pearson coefficient applied to rank-scaled column data. TableTorch uses fractional ranking which is equivalent to `RANK.AVG(value, data, TRUE)` Google Sheets function, i.e. two identical values will be assigned their average rank. The produced correlation matrix will thus identify non-linear correlations as well as the linear ones. However, it shall be noted that TableTorch’s Linear regression implementation does not support automatic rank-scaling. Therefore, if some strong non-linear correlations are identified, it is advisable to use standalone Scaling and run the regression against scaled data.\n\nTableTorch produces following matrix with Spearman’s rank coefficients for select columns:", null, "## Conclusion\n\nHaving reviewed the correlation matrix for the given data, it is now possible to do a linear regression in order to fit a model predicting desired variable." ]

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[ "# Events and Probabilites", null, "The relevance of randomness in divination as been debated at length. One side of the camp claims that the divination process, for its very nature, reveals a deeper connection with the structure of reality and, for this very reason, the response is not random but determined by the moment in which the divination process occurs.\nThe other side of the camp responds that nothing can escape the law of physics and that reality itself seems to be governed, at quantum level, by randomness. Hence the statistics related to the divination process play a huge role.\n\nI tend to agree with the latter view: the divination process, to me, is fully governed by the law of physics; the randomness of each method is key in interpreting the results. The probability to get a certain moving line, especially when multiple changing lines are present, is a factor to be taken into account. Getting six moving yin line is near to impossible with the the yarrow stalk method (i.e. possible but highly unlikely to happen), while using the dutch sticks method is as probable as any other outcome. You need to be aware of these differences to correctly interpret the response.\n\nAny method for casting I Ching hexagrams can be characterized, at the most abstract level, by the  probability it assigns to the each possible response.\n\n#### Line probabilities\n\nThe most used methods build the hexagram one line at the time (usually from bottom to top); it is, hence, important to consider the probability assigned to each line.\nWe will use Probi (e) to signify the \"probability that the event e will happen at line i\". In the majority of cases the line i is not significant and will be omitted.\nFor example for the three coins method, for which the probabilities do not depend on the line, the indication Prob(7) = 3/8 means that a non movin yang line (", null, ") has 3 out of 8 chances to occur in any position.\n\n#### Response probabilites", null, "Given the line probabilities, the probability to get a specific response (out of the 4096 which are possible) is given by the formula on the left (i.e. by the multiplication of the probabilities of each line) where ei is the type of line i  (6,7,8,9).", null, "For eaxample, the probability to get the hexagram on the right as response is (counting from bottom to top): Prob1(7) * Prob2(6) * Prob3(8) * Prob4(7) * Prob5(9) * Prob6(8)\n\nFor the three coins method it would be:\n3/8 * 1/8 * 3/8 * 3/8 * 1/8 * 3/881/262144 = 0.031%\n\nFor the yarrow stalks method it would be:\n5/16 * 1/16 * 7/16 * 5/16 * 3/16 * 7/163675/16777216 = 0.022%\n\nIf we want to calculate the probability of getting one of the 64 hexagrams (regardless of moving lines) as response, the formula in the previous paragraph is still applicable. The difference is that ei represent the fact that the line is either yin or a yang line:\nProbi(yin) = Probi(6) + Probi(8)\nProbi(yang) = Probi(9) + Probi(7)\n\nNote that if  the probability is independent of the line and Prob(yin) = Prob(yang) = 1/2  (as it is often the case) the formula above returns 1/26 = 1/64 which means that each hexagram is equiprobable.\n\nSome simple formula for the most common cases (and with the most common assumptions):\n\nProbability of getting a response with n yin lines:         [Prob(6) + Prob(8)]n\nProbability of getting a response with n yang lines:      [Prob(9) + Prob(7)]n\nProbability of getting a response with n moving lines:  [Prob(6) + Prob(9)]n\n\n#### Secondary Hexagram\n\nTo determine the probability to get a given hexagram as the secondary one we can consider that a secondary hexagram contains a yang line in position i in two cases: if the primary hexagram contains a yang line in that position or if the primary hexgram has a moving yin line in that position; and similarly for a yin line.\nProbi(secondary_yin) = Probi(9) + Probi(8)\nProbi(secondary_yang) = Probi(6) + Probi(7)\n\nOnly if the method is symmetric (i.e. [Prob(6) = Prob(9)] and [Prob(7) = Prob(8)]) each hexagram will have the same probability of being the secondary hexagram." ]

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[ "# Shutdown Point\n\nAn operating level where a business does not benefit in continuing production operations in the short run\n\n## What is a Shutdown Point?\n\nA shutdown point is an operating level where a business does not benefit in continuing production operations in the short run when revenue from selling their product is unable to cover variable costs of production. The shutdown point represents a point where a firm will incur higher and increasing losses if it continues production, as opposed to reduced losses if production is ceased. The shutdown point occurs at a point where marginal profit reaches a negative scale.", null, "### Understanding Shutdown Points\n\nA shutdown arises when price or average revenue (AR) falls below average variable cost (AVC) at the profit-maximizing output level. Continued production will incur additional variable costs but will not generate enough revenue to cover them. At the same time, the firm will still have fixed costs to pay, further increasing the losses.\n\nA shutdown point is typically a short-run position; however, in the long run, the firm should shut down and leave the industry if its product price is less than its average total cost.  Therefore, there are two shutdown points for a firm – in the short run and the long run. The decision to shut down is dependent on which costs the firm can avoid by shutting down production. The short run is a period where at least one of the firm’s inputs is fixed, resulting in fixed costs incurred despite the decision to shut down.\n\nIn summary, the shutdown point has the following characteristics:\n\n• It is the output and price point where a firm is able to just cover its total variable cost.\n• The average variable cost (AVC) is at its minimum point.\n• It is where the marginal cost (MC) curve intercepts the average variable cost (AVC) curve.\n• The firm is indifferent between shutting down and continuing production where losses equal to the total fixed costs are incurred regardless of either decision.\n\n### Shutdown Point Diagram", null, "Where:\n\n• MC – Marginal Cost\n• ATC – Average Total Cost\n• AVC – Average Variable Cost\n• SP – Shutdown Price\n• BEP – Break-even Price\n\n### Short-Run Shutdown Decision\n\nThe cost of production is divided into two parts – fixed costs and variable costs. The break-even point is a point where revenue generated from sales of a product is equal to the production cost (fixed cost plus variable cost). Zero profit is generated at the break-even point. On the graph above, it is the point where the average total cost (ATC) is equal to marginal cost (MC) (i.e., MC = ATC). Marginal cost equals a change in total costs for each additional unit produced. Fixed costs do not change in the short run; hence, the change in total costs refers to variable cost only.\n\nThe shutdown zone represents an area between the break-even point and the shutdown point. it is an area where production can continue, as average revenue (AR) will still be able to cover average variable cost (AVC). However, in the shutdown zone, the firm will be making losses as the price is below average total cost (ATC). The firm operates at any level above the AVC curve as long as it is where MC = MR (price). The MC curve above the AVC is also the short-run supply curve of the firm.\n\nThe shutdown rule states that a firm should continue operations as long as the price (average revenue) is able to cover average variable costs. The firm can continue operating, as it will be producing where marginal revenue (price, average revenue) is equal to marginal cost, a condition that ensures profit maximization or loss minimization.\n\nA continuation of the shutdown rule states that in the short run, fixed costs are considered as sunk costs. Hence, it should not be considered in the decision of whether to shut down or continue with operations. In addition, in the short run, if the firm’s total revenue is less than variable costs, the firm should shut down.\n\nA short-run decision to shut down is not the same as exiting the industry. Several firms in seasonal industries – such as agriculture, fishing, etc. – shut down their firms during the offseason to avoid unnecessary operating costs. They will not be generating any revenue during the off-season; hence they are unable to cover variable costs arising. It makes sense to temporarily shut down until the upcoming season commences.\n\n### Shutdown Point Illustration\n\nEnderby Manufacturing’s production details are as follows:", null, "Enderby Manufacturing is operating at a loss of \\$2,800. The firm cannot avoid paying fixed costs, whether they operate or not. If they choose to shut down and cease operations, they will generate zero revenue, zero variable costs, and incur fixed costs of \\$10,000, which means the total loss will increase to \\$10,000.\n\nHowever, if the firm continues to operate, it will still generate revenue of \\$16,000, where \\$8,800 will be expended to cover variable costs, and the balance of \\$7,200 will meet part of the fixed costs. Therefore, by continuing operations, the firm will only make a loss of \\$2,800 instead of \\$10,000 if they decide to shut down in the short run.\n\nHowever, if the selling price falls below \\$11 per unit and costs remain the same, the firm will have reached the shutdown point (AR < AVC). Such a condition satisfies the shutdown rule where shutdown is recommended.\n\n### Calculation of the Short-Run Shutdown Point\n\nAs illustrated above, the shutdown point is the output level at the minimum of the average variable cost curve (AVC).\n\nThe shutdown point can be calculated using the total cost (TC) function. Suppose the total cost function is as follows:", null, "### Long-Run Shutdown (Industry Exit)\n\nAs a rule of thumb, a decision to shut down in the long run – i.e., exiting the industry – should only be undertaken if revenues are unable to cover total costs. It means in the long run, a firm making losses should shut down permanently and exit the industry. The short run is defined as a period where at least one fixed input or cost is present in the business. Fixed expenses such as rentals are incurred whether the firm undertakes production or not. In the long run, all inputs and costs are variable.\n\nHowever, in the long run, if the firm is unable to raise the selling price per unit (to increase overall revenue) to cover total costs, the losses will continue ballooning until average revenue (AR) is exceeded by the average total cost (ATC). The firm will have reached the shutdown point where the only viable option is to shut down.\n\nShutdown, as indicated above, is a short-run decision to minimize losses. It is because a firm can shut down in the interim, but if market conditions permit, it can still resume production. Even if the firm shuts down, it will still have incurred sunk costs in terms of investment in plant and equipment. Hence, it is possible for a firm to shut down in the short run and resume production in the long run.\n\nHowever, if conditions do not improve, firms will resultantly decide to exit the industry, which is a long-run decision. The long-run exit decision is guided by the relationship between the price (P) and the long-run average cost (LRAC). Firms will exit the industry if P < LRAC. In the long run, if the firm decides to operate, it will still operate where the long-run marginal cost (LRMC) is equal to marginal revenue (MR).\n\nThe long-run shutdown point is defined by the output corresponding to the minimum average total cost (ATC). The long-run shutdown point can be calculated much the same way we did for the short-run shutdown point. We take the derivative of the ATC and solve for Q by setting it to zero. We plug it into the ATC function to get the price.\n\n### Monopoly Market Structure Shutdown Point\n\nIn the short run, a monopolist market structure shutdown point is reached when average revenue (price) is below average variable cost (AVC) at every output level. In such a case, it means that the demand curve is completely below the average variable cost curve.\n\nEven though a firm may be producing where marginal revenue is equal to marginal cost (MR = MC: the profit-maximizing level of output), average revenue would be less than average variable cost. The monopolist would be wise to shut down at such a point.\n\n### Real-World Application of the Shutdown Point\n\nThere are circumstances where firms can reach a shutdown point where the price is below AVC, but they decide not to shut down and keep operating because of any of the following reasons:\n\n1. To retain long-term customers of the business. When a firm thinks that they are in a passing period of falling demand, they can opt to keep producing.\n2. Some financially-strong firms are able to ride out a period of loss-making due to readily available credit support or a healthy standby reserve fund.\n3. Firms can also decide to cut costs or increase product prices if they reach the shutdown point.\n4. Firms can also shut down and leave the industry if they perceive a gloomy forecast of the long-term performance of the industry. It can happen well before the firm reaches the shutdown point – i.e., where AR < AVC.\n5. Firms can also continue operating past the shutdown point, as it may take time to realize that they are operating at a loss. They may find out through management accounts once they are released." ]

[ null, "https://corporatefinanceinstitute.com/resources/management/shutdown-point/", null, "https://corporatefinanceinstitute.com/resources/management/shutdown-point/", null, "https://corporatefinanceinstitute.com/resources/management/shutdown-point/", null, "https://corporatefinanceinstitute.com/resources/management/shutdown-point/", null ]

[ "# Checking Flags in C# Enums\n\n#### March 05, 2008c-sharp code\n\nUpdate 2021/09/22: The Enum class has a built-in `HasFlag()` method now.\n\nI like C# enums and I also like using them as bitfields, even though apparently not everyone does. I realize they aren’t perfectly typesafe, but then I don’t think that’s the problem Abrams and company were trying to solve anyway.\n\nHere’s an example one:\n\n``````[Flags]\npublic enum Fruits\n{\nApple = 1,\nBanana = 2,\nCherry = 4,\nDate = 8,\nEggplant = 16\n}\n``````\n\nNice, clean syntax. The way they solved C++’s name collision issues with enum values is genius: `Fruits.Apple`. Clearly these guys are using the old noggin.\n\n## The annoying bit (argh, a pun)\n\nThe one thing that does annoy me about flag enums is the syntax to see if a given flag (or set of flags) is set:\n\n``````if ((myFruit & Fruits.Date) == Fruits.Date)\n``````\n\nI’m not afraid of bitwise operators, but there’s some serious lameness in here. Needing to specify the explicit `==` for type safety and having to use the parenthesis because the operator precedence puts `==` before `&` first? Gross.\n\n## For every nail there is a hammer\n\nBehold the solution:\n\n``````public static class FruitsExtensions\n{\npublic static bool IsSet(this Fruits fruits, Fruits flags)\n{\nreturn (fruits & flags) == flags;\n}\n}\n``````\n\nWith that, you can just do:\n\n``````if (myFruit.IsSet(Fruits.Date))\n``````\n\nMuch nicer. For kicks, here’s some other useful methods:\n\n``````public static class FruitsExtensions\n{\npublic static bool IsSet(this Fruits fruits, Fruits flags)\n{\nreturn (fruits & flags) == flags;\n}\n\npublic static bool IsNotSet(this Fruits fruits, Fruits flags)\n{\nreturn (fruits & (~flags)) == 0;\n}\n\npublic static Fruits Set(this Fruits fruits, Fruits flags)\n{\nreturn fruits | flags;\n}\n\npublic static Fruits Clear(this Fruits fruits, Fruits flags)\n{\nreturn fruits & (~flags);\n}\n}\n``````\n\nUseful, no?\n\n## Why solve one when you can solve n?\n\nSo, if you’re like me and this guy, right now you’re thinking, “This just fixes one enum. Can I solve it for all enums?” Ideally, you’d do something like:\n\n``````public static class EnumExtensions\n{\npublic static bool IsSet<T>(this T value, T flags) where T : Enum\n{\nreturn (value & flags) == flags;\n}\n}\n``````\n\nUnfortunately, that doesn’t fly. You can’t use `Enum` as a constraint. Likewise, there’s no way to require a typeparam to implement an operator (like `&` above). You can implement a generic solution for this:\n\n``````public static class EnumExtensions\n{\npublic static bool IsSet<T>(this T value, T flags)\nwhere T : struct\n{\nType type = typeof(T);\n\n// only works with enums\nif (!type.IsEnum) throw new ArgumentException(\n\"The type parameter T must be an enum type.\");\n\n// handle each underlying type\nType numberType = Enum.GetUnderlyingType(type);\n\nif (numberType.Equals(typeof(int)))\n{\nreturn Box<int>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(sbyte)))\n{\nreturn Box<sbyte>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(byte)))\n{\nreturn Box<byte>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(short)))\n{\nreturn Box<short>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(ushort)))\n{\nreturn Box<ushort>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(uint)))\n{\nreturn Box<uint>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(long)))\n{\nreturn Box<long>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(ulong)))\n{\nreturn Box<ulong>(value, flags, (a, b) => (a & b) == b);\n}\nelse if (numberType.Equals(typeof(char)))\n{\nreturn Box<char>(value, flags, (a, b) => (a & b) == b);\n}\nelse\n{\nthrow new ArgumentException(\n\"Unknown enum underlying type \" +\nnumberType.Name + \".\");\n}\n}\n\n/// Helper function for handling the value types. Boxes the\n/// params to object so that the cast can be called on them.\nprivate static bool Box<T>(object value, object flags,\nFunc<T, T, bool> op)\n{\nreturn op((T)value, (T)flags);\n}\n}\n``````\n\n…but, yeah, it’s not exactly fun using reflection for this." ]

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[ "What is the Electromagnetic Spectrum?\n\nElectromagnetic Spectrum\n\nThe electromagnetic spectrum is defined by frequency and wavelength. Frequency is abbreviated as a lower-case f and is frequently measured in Hertz, and wavelength is abbreviated as a symbol that looks kind of like a lower-case h, with more squiggles, and wavelength is measured in meters.\n\nNow because light travels at a fairly constant speed, frequency is inversely proportional to wavelength. This concept is illustrated in the formula: frequency equals c, divided by wavelength. C stands for the speed of light. The speed of light is about 300 million meters per second.\n\nNow frequency multiplied by wavelength equals the speed of the wave, and for electromagnetic waves this is going to be the speed of light with some variance for the medium in which it is traveling. Now electromagnetic waves include many different types of waves, as you can see here.\n\nWhat I’ve done here is I’ve arranged these different types of waves from largest to smallest in terms of wavelength: we have radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, x-rays, and gamma rays. Now, the energy of electromagnetic waves is carried in packets that have a magnitude inversely proportional to the wavelength.\n\n771761\n\nby Mometrix Test Preparation | Last Updated: August 15, 2019" ]

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[ "Feet To Meters\n\n# 73.8 ft to m73.8 Foot to Meters\n\nft\n=\nm\n\n## How to convert 73.8 foot to meters?\n\n 73.8 ft * 0.3048 m = 22.49424 m 1 ft\nA common question is How many foot in 73.8 meter? And the answer is 242.125984252 ft in 73.8 m. Likewise the question how many meter in 73.8 foot has the answer of 22.49424 m in 73.8 ft.\n\n## How much are 73.8 feet in meters?\n\n73.8 feet equal 22.49424 meters (73.8ft = 22.49424m). Converting 73.8 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 73.8 ft to m.\n\n## Convert 73.8 ft to common lengths\n\nUnitUnit of length\nNanometer22494240000.0 nm\nMicrometer22494240.0 µm\nMillimeter22494.24 mm\nCentimeter2249.424 cm\nInch885.6 in\nFoot73.8 ft\nYard24.6 yd\nMeter22.49424 m\nKilometer0.02249424 km\nMile0.0139772727 mi\nNautical mile0.0121459179 nmi\n\n## What is 73.8 feet in m?\n\nTo convert 73.8 ft to m multiply the length in feet by 0.3048. The 73.8 ft in m formula is [m] = 73.8 * 0.3048. Thus, for 73.8 feet in meter we get 22.49424 m.\n\n## 73.8 Foot Conversion Table", null, "## Alternative spelling\n\n73.8 ft in Meter, 73.8 Foot to m, 73.8 Feet in Meter, 73.8 ft in m, 73.8 Foot to Meters, 73.8 Foot in Meters, 73.8 Foot in Meter, 73.8 Feet in Meters," ]

[ null, "https://feet-to-meters.appspot.com/image/73.8.png", null ]

[ "# Vector calculus\n\ncalculus of vector-valued functions\n\nVector calculus is a branch of mathematics that investigates vector fields and how they change over time. Vector calculus usually studies two- or three-dimensional vector fields, but can be used in higher dimensions, too. It is a part of the study of multivariable calculus. Vector calculus is useful in physics and engineering because of how it can look at electromagnetic and gravitational fields." ]

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[ "# frequency\n\nAlso found in: Dictionary, Thesaurus, Medical, Legal, Financial, Acronyms, Wikipedia.\nRelated to frequency: frequency modulation, Sound frequency\n\n## frequency:\n\nsee harmonic motionharmonic motion,\nregular vibration in which the acceleration of the vibrating object is directly proportional to the displacement of the object from its equilibrium position but oppositely directed.\n; wavewave,\nin physics, the transfer of energy by the regular vibration, or oscillatory motion, either of some material medium or by the variation in magnitude of the field vectors of an electromagnetic field (see electromagnetic radiation).\n.\n\n## Frequency (wave motion)\n\nThe number of times which sound pressure, electrical intensity, or other quantities specifying a wave vary from their equilibrium value through a complete cycle in unit time. The most common unit of frequency is the hertz (Hz), which is equal to 1 cycle per second. In one cycle there is a positive variation from equilibrium, a return to equilibrium, then a negative variation, and return to equilibrium. This relationship is often described in terms of the sine wave, and the frequency referred to is that of an equivalent sine-wave variation in the parameter under discussion. See Frequency measurement, Sine wave, Wave motion\n\n## frequency\n\nSymbol: f , ν. The number of oscillations per unit time of a vibrating system. Frequency is measured in hertz. The frequency of a wave is the number of wave crests passing a point per unit time. For light and other electromagnetic radiation, it is related to wavelength λ by ν = c /λ, where c is the speed of light.\n\n## Frequency\n\nThe frequency of an event A is the ratio m/n of the number m of occurrences of A in a given series of trials to the total number n of trials. If the trials are independent and there is a definite probability p of the occurrence of A in an individual trial, then, for arbitrarily small ∊ > 0, at sufficiently large m it is practically certain that the frequency m/n satisfies the inequality", null, "(seeLARGE NUMBERS, LAW OF and PROBABILITY).\n\nThe term “frequency” is used in mathematical statistics to designate the number of elements of a set that have a specified attribute.\n\n## frequency\n\n[′frē·kwən·sē]\n(physics)\nThe number of cycles completed by a periodic quantity in a unit time.\n(statistics)\nThe number of times an event or item falls into or is expected to fall into a certain class or category.\n\n## frequency\n\nThe number of oscillations per second (a) of the current or voltage in an alternating-current electric circuit, or (b) of a sound wave, or (c) of a vibrating solid object; expressed in hertz (abbr. Hz) or in cycles per second (abbr. cps).\n\n## frequency\n\ni. The number of recurrences of a periodic phenomenon in a unit of time.\nii. The number of cycles completed in one second. One cycle per second is the basic unit of measurement of frequency and is called a hertz.\niii. The number of services operated by an airline per day or per week over a particular route.\n\n## frequency\n\n1. Physics the number of times that a periodic function or vibration repeats itself in a specified time, often 1 second. It is usually measured in hertz.\n2. Ecology\na. the number of individuals of a species within a given area\nb. the percentage of quadrats that contains individuals of a species\n\n## frequency\n\nThe number of oscillations (vibrations) in one second. Frequency is measured in Hertz (Hz), which is the same as \"oscillations per second\" or \"cycles per second.\" For example, the alternating current in a wall outlet in the U.S. and Canada is 60Hz. Electromagnetic radiation is measured in kiloHertz (kHz), megahertz (MHz) and gigahertz (GHz). See wavelength, frequency response, audio, carrier and space/time.\n\nFrequency", null, "The frequency is the number of oscillations per second. The higher the frequency (the closer the ripples would be in this diagram) and the shorter the wavelength.\nReferences in periodicals archive ?\nRunning with automatic frequency steps, the duration, deflection, starting and final frequency, as well as the amount of steps, can be set.\nAs a set, the defective components have a lower average frequency and a broader distribution because they represent an out-of-control process state.\nIt is the square root of the power spectral density of magnetic field fluctuations at the magnetometer frequency.\nAAF is a frequency-boosting scheme using a first-order zero (s) up to the maximum frequency of the digital information, attenuating higher frequencies that do not carry information to reduce noise.\nDistortion-product otoacoustic emissions in humans with high frequency sensorineural hearing loss.\nIf each device needs a different frequency, there may not be enough frequencies to go around.\nTypical a-c frequency drives are limited to point-to-point movement and are not capable of complex motions, says AEC's Schmitz.\nThe measurement system has been optimized so that parasitic instrument resonances occur above the maximum measurement frequency of 1,000 Hz.\nThe value of a standard capacitor may vary slightly with frequency because the imperfect medium between its electrodes has varying degrees of dielectric relaxation over the frequency range and the leads and electrodes of the capacitor have residual inductance.\nIn 1994, the company was the first to commercialize the SIEGET 25 family of ultra-high-frequency transistors with a cut-off frequency (fT) of 25 GHz.\nFrequency Method - With the introduction of the solid state power supply, an inductor could be designed to operate at a frequency other than 60 Hertz (Hz).\n\nSite: Follow: Share:\nOpen / Close" ]

[ null, "http://img.tfd.com/ggse/8b/gsed_0001_0029_0_img9079.png", null, "http://img.tfd.com/cde/WAVELEN.GIF", null ]

[ "Calculate Kramers-Moyal coefficients for stochastic process of any dimension, up to any order.\n\n# KramersMoyal\n\n`kramersmoyal` is a python package designed to obtain the Kramers–Moyal coefficients, or conditional moments, from stochastic data of any dimension. It employs kernel density estimations, instead of a histogram approach, to ensure better results for low number of points as well as allowing better fitting of the results\n\n# Installation\n\nFor the moment the library is available from TestPyPI, so you can use\n\n``````pip install -i https://test.pypi.org/simple/ kramersmoyal\n``````\n\nThen on your favourite editor just use\n\n```from kramersmoyal import km, kernels\n```\n\n## Dependencies\n\nThe library depends on `numpy` and `scipy`.\n\n# A one-dimensional stochastic process\n\nA Jupyter notebook with this example can be found here\n\n## The theory\n\nTake, for example, the well-documented one-dimension Ornstein–Uhlenbeck process, also known as Vašíček process, see here. This process is governed by two main parameters: the mean-reverting parameter θ and the diffusion parameter σ", null, "which can be solved in various ways. For our purposes, recall that the drift coefficient, i.e., the first-order Kramers–Moyal coefficient, is given by", null, "and the second-order Kramers–Moyal coefficient is", null, ", i.e., the diffusion.\n\nGenerate an exemplary Ornstein–Uhlenbeck process with your favourite integrator, e.g., the Euler–Maruyama or with a more powerful tool from `JiTCSDE` found on GitHub. For this example let's take θ=.3 and σ=.1, over a total time of 500 units, with a sampling of 1000 Hertz, and from the generated data series retrieve the two parameters, the drift -θy(t) and diffusion σ.\n\n## Integrating an Ornstein–Uhlenbeck process\n\nHere is a short code on generating a Ornstein–Uhlenbeck stochastic trajectory with a simple Euler–Maruyama integration method\n\n```# integration time and time sampling\nt_final = 500\ndelta_t = 0.001\n\n# The parameters theta and sigma\ntheta = 0.3\nsigma = 0.1\n\n# The time array of the trajectory\ntime = np.arange(0, t_final, delta_t)\n\n# Initialise the array y\ny = np.zeros(time.size)\n\n# Generate a Wiener process\ndw = np.random.normal(loc = 0, scale = np.sqrt(delta_t), size = time.size)\n\n# Integrate the process\nfor i in range(1,time.size):\ny[i] = y[i-1] - theta*y[i-1]*delta_t + sigma*dw[i]\n```\n\nFrom here we have a plain example of an Ornstein–Uhlenbeck process, always drifting back to zero, due to the mean-reverting drift θ. The effect of the noise can be seen across the whole trajectory.", null, "## Using `kramersmoyal`\n\nTake the timeseries `y` and let's study the Kramers–Moyal coefficients. For this let's look at the drift and diffusion coefficients of the process, i.e., the first and second Kramers–Moyal coefficients, with an `epanechnikov` kernel\n\n```# Choose number of points of you target space\nbins = np.array()\n\n# Choose powers to calculate\npowers = np.array([, ])\n\nbw = 0.15\n\n# The kmc holds the results, where edges holds the binning space\nkmc, edges = km(y, kernel = kernels.epanechnikov, bw = bw, bins = bins, powers = powers)\n```\n\nThis results in", null, "Notice here that to obtain the Kramers–Moyal coefficients you need to multiply `kmc` by the timestep `delta_t`. This normalisation stems from the Taylor-like approximation, i.e., the Kramers–Moyal expansion (`delta t` → 0).\n\n# A two-dimensional diffusion process\n\nA Jupyter notebook with this example can be found here\n\n## Theory\n\nA two-dimensional diffusion process is a stochastic process that comprises two", null, "and allows for a mixing of these noise terms across its two dimensions.", null, "where we will select a set of state-dependent parameters obeying", null, "with", null, "and", null, ".\n\n## Choice of parameters\n\nAs an example, let's take the following set of parameters for the drift vector and diffusion matrix\n\n```# integration time and time sampling\nt_final = 2000\ndelta_t = 0.001\n\n# Define the drift vector N\nN = np.array([2.0, 1.0])\n\n# Define the diffusion matrix g\ng = np.array([[0.5, 0.0], [0.0, 0.5]])\n\n# The time array of the trajectory\ntime = np.arange(0, t_final, delta_t)\n```\n\n## Integrating a 2-dimensional process\n\nIntegrating the previous stochastic trajectory with a simple Euler–Maruyama integration method\n\n```# Initialise the array y\ny = np.zeros([time.size, 2])\n\n# Generate two Wiener processes with a scale of np.sqrt(delta_t)\ndW = np.random.normal(loc = 0, scale = np.sqrt(delta_t), size = [time.size, 2])\n\n# Integrate the process (takes about 20 secs)\nfor i in range(1, time.size):\ny[i,0] = y[i-1,0] - N * y[i-1,0] * delta_t + g[0,0]/(1 + np.exp(y[i-1,0]**2)) * dW[i,0] + g[0,1] * dW[i,1]\ny[i,1] = y[i-1,1] - N * y[i-1,1] * delta_t + g[1,0] * dW[i,0] + g[1,1]/(1 + np.exp(y[i-1,1]**2)) * dW[i,1]\n```\n\nThe stochastic trajectory in 2 dimensions for 10 time units (10000 data points)", null, "## Back to `kramersmoyal` and the Kramers–Moyal coefficients\n\nFirst notice that all the results now will be two-dimensional surfaces, so we will need to plot them as such\n\n```# Choose the size of your target space in two dimensions\nbins = np.array([300, 300])\n\n# Introduce the desired orders to calculate, but in 2 dimensions\npowers = np.array([[0,0], [1,0], [0,1], [1,1], [2,0], [0,2], [2,2]])\n# insert into kmc: 0 1 2 3 4 5 6\n\n# Notice that the first entry in [,] is for the first dimension, the\n# second for the second dimension...\n\n# Choose a desired bandwidth bw\nbw = 0.1\n\n# Calculate the Kramers−Moyal coefficients\nkmc, edges = km(y, bw = bw, bins = bins, powers = powers)\n\n# The K−M coefficients are stacked along the first dim of the\n# kmc array, so kmc[1,...] is the first K−M coefficient, kmc[2,...]\n# is the second. These will be 2-dimensional matrices\n```\n\nNow one can visualise the Kramers–Moyal coefficients (surfaces) in green and the respective theoretical surfaces in black. (Don't forget to normalise: `kmc * delta_t`).", null, "# Contributions\n\nWe welcome reviews and ideas from everyone. If you want to share your ideas or report a bug, open an issue here on GitHub, or contact us directly. If you need help with the code, the theory, or the implementation, do not hesitate to contact us, we are here to help. We abide to a Conduct of Fairness.\n\n# TODOs\n\nNext on the list is\n\n• Include more kernels\n• Work through the documentation carefully\n• Create a sub-routine to calculate the Kramers–Moyal coefficients without a convolution\n\n# Changelog\n\n• Version 0.4 - Added the documentation, first testers, and the Conduct of Fairness\n• Version 0.32 - Adding 2 kernels: `triagular` and `quartic` and extenting the documentation and examples.\n• Version 0.31 - Corrections to the fft triming after convolution.\n• Version 0.3 - The major breakthrough: Calculates the Kramers–Moyal coefficients for data of any dimension.\n• Version 0.2 - Introducing convolutions and `gaussian` and `uniform` kernels. Major speed up in the calculations.\n• Version 0.1 - One and two dimensional Kramers–Moyal coefficients with an `epanechnikov` kernel.\n\n# Literature and Support\n\n### Literature\n\nThe study of stochastic processes from a data-driven approach is grounded in extensive mathematical work. From the applied perspective there are several references to understand stochastic processes, the Fokker–Planck equations, and the Kramers–Moyal expansion\n\n• Tabar, M. R. R. (2019). Analysis and Data-Based Reconstruction of Complex Nonlinear Dynamical Systems. Springer, International Publishing\n• Risken, H. (1989). The Fokker–Planck equation. Springer, Berlin, Heidelberg.\n• Gardiner, C.W. (1985). Handbook of Stochastic Methods. Springer, Berlin.\n\nYou can find and extensive review on the subject here1\n\n### History\n\nThis project was started in 2017 at the neurophysik by Leonardo Rydin Gorjão, Jan Heysel, Klaus Lehnertz, and M. Reza Rahimi Tabar. Francisco Meirinhos later devised the hard coding to python. The project is now supported by Dirk Witthaut and the Institute of Energy and Climate Research Systems Analysis and Technology Evaluation.\n\n### Funding\n\nHelmholtz Association Initiative Energy System 2050 - A Contribution of the Research Field Energy and the grant No. VH-NG-1025 and STORM - Stochastics for Time-Space Risk Models project of the Research Council of Norway (RCN) No. 274410.\n\n1 Friedrich, R., Peinke, J., Sahimi, M., Tabar, M. R. R. Approaching complexity by stochastic methods: From biological systems to turbulence, Phys. Rep. 506, 87–162 (2011).\n\n## Project details\n\nThis version", null, "0.4", null, "0.3.2" ]

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[ "Cody\n\n# Problem 2307. length of string on cylinder\n\nSolution 1741277\n\nSubmitted on 3 Mar 2019 by Athi\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nH=90 C=20 N=6 y_correct =150; assert(isequal(your_fcn_name(H,C,N),y_correct))\n\nH = 90 C = 20 N = 6\n\n2   Pass\nH=72 C=16 N=6 y_correct =120; assert(isequal(your_fcn_name(H,C,N),y_correct))\n\nH = 72 C = 16 N = 6\n\n3   Pass\nH=105 C=20 N=7 y_correct =175; assert(isequal(your_fcn_name(H,C,N),y_correct))\n\nH = 105 C = 20 N = 7" ]

[ null ]

[ "Home | | Compiler Design | | Compilers Principles, Techniques, & Tools | | Compiler Design | Syntax-Directed Translation Schemes\n\n# Syntax-Directed Translation Schemes\n\n1 Postfix Translation Schemes 2 Parser-Stack Implementation of Postfix SDT's 3 SDT's With Actions Inside Productions 4 Eliminating Left Recursion From SDT's 5 SDT's for L-Attributed Definitions 6 Exercises for Section 5.4\n\nSyntax-Directed Translation Schemes\n\n1 Postfix Translation Schemes\n\n2 Parser-Stack Implementation of Postfix SDT's\n\n3 SDT's With Actions Inside Productions\n\n4 Eliminating Left Recursion From SDT's\n\n5 SDT's for L-Attributed Definitions\n\n6 Exercises for Section 5.4\n\nSyntax-directed translation schemes are a complementary notation to syntax-directed definitions. All of the applications of syntax-directed definitions in Section 5.3 can be implemented using syntax-directed translation schemes.\n\nFrom Section 2.3.5, a syntax-directed translation scheme (SDT) is a context-free grammar with program fragments embedded within production bodies. The program fragments are called semantic actions and can appear at any position within a production body. By convention, we place curly braces around actions; if braces are needed as grammar symbols, then we quote them.\n\nAny SDT can be implemented by first building a parse tree and then per-forming the actions in a left-to-right depth-first order; that is, during a preorder traversal. An example appears in Section 5.4.3.\n\nTypically, SDT's are implemented during parsing, without building a parse tree. In this section, we focus on the use of SDT's to implement two important classes of SDD's:\n\nThe underlying grammar is LR-parsable, and the SDD is S-attributed.\n\nThe underlying grammar is LL-parsable, and the SDD is L-attributed.\n\nWe shall see how, in both these cases, the semantic rules in an SDD can be converted into an SDT with actions that are executed at the right time. During parsing, an action in a production body is executed as soon as all the grammar symbols to the left of the action have been matched.\n\nSDT's that can be implemented during parsing can be characterized by in-troducing distinct marker nonterminals in place of each embedded action; each marker M has only one production, M -» e. If the grammar with marker non-terminals can be parsed by a given method, then the SDT can be implemented during parsing.\n\n1. Postfix Translation Schemes\n\nBy far the simplest SDD implementation occurs when we can parse the grammar bottom-up and the SDD is S-attributed. In that case, we can construct an SDT in which each action is placed at the end of the production and is executed along with the reduction of the body to the head of that production. SDT's with all actions at the right ends of the production bodies are called postfix SDT's.\n\nExample 5.14 : The postfix SDT in Fig. 5.18 implements the desk calculator SDD of Fig. 5.1, with one change: the action for the first production prints a value. The remaining actions are exact counterparts of the semantic rules. Since the underlying grammar is LR, and the SDD is S-attributed, these actions can be correctly performed along with the reduction steps of the parser.", null, "2. Parser-Stack Implementation of Postfix SDT's\n\nPostfix SDT's can be implemented during LR parsing by executing the actions when reductions occur. The attribute(s) of each grammar symbol can be put on the stack in a place where they can be found during the reduction. The best plan is to place the attributes along with the grammar symbols (or the LR states that represent these symbols) in records on the stack itself.\n\nIn Fig. 5.19, the parser stack contains records with a field for a grammar symbol (or parser state) and, below it, a field for an attribute. The three grammar symbols X YZ are on top of the stack; perhaps they are about to be reduced according to a production like A —> X YZ. Here, we show X.x as the one attribute of X, and so on. In general, we can allow for more attributes, either by making the records large enough or by putting pointers to records on the stack. With small attributes, it may be simpler to make the records large enough, even if some fields go unused some of the time. However, if one or more attributes are of unbounded size — say, they are character strings — then it would be better to put a pointer to the attribute's value in the stack record and store the actual value in some larger, shared storage area that is not part of the stack.", null, "If the attributes are all synthesized, and the actions occur at the ends of the productions, then we can compute the attributes for the head when we reduce the body to the head. If we reduce by a production such as A X Y Z , then we have all the attributes of X, Y, and Z available, at known positions on the stack, as in Fig. 5.19. After the action, A and its attributes are at the top of the stack, in the position of the record for X.\n\nExample 5.15 :  Let us rewrite the actions of the desk-calculator SDT of Ex ample 5.14 so that they manipulate the parser stack explicitly. Such stack manipulation is usually done automatically by the parser.", null, "Suppose that the stack is kept in an array of records called stack, with top a cursor to the top of the stack. Thus, stack[top] refers to the top record on the stack, stack[top — 1] to the record below that, and so on. Also, we assume that each record has a field called val, which holds the attribute of whatever grammar symbol is represented in that record. Thus, we may refer to the attribute E.val that appears at the third position on the stack as stack[top2].val. The entire SDT is shown in Fig. 5.20.\n\nFor instance, in the second production, E -» E1 + T, we go two positions below the top to get the value of E±, and we find the value of T at the top. The resulting sum is placed where the head E will appear after the reduction, that is, two positions below the current top. The reason is that after the reduction, the three topmost stack symbols are replaced by one. After computing E.val, we pop two symbols off the top of the stack, so the record where we placed E.val will now be at the top of the stack.\n\nIn the third production, E —> T, no action is necessary, because the length of the stack does not change, and the value of T.val at the stack top will simply become the value of E.val.  The same observation applies to the productions T ->• F and F digit .  Production F ->• ( E) is slightly different.  Although the value does not change, two positions are removed from the stack during the reduction, so the value has to move to the position after the reduction.\n\nNote that we have omitted the steps that manipulate the first field of the stack records — the field that gives the LR state or otherwise represents the grammar symbol. If we are performing an LR parse, the parsing table tells us what the new state is every time we reduce; see Algorithm 4.44. Thus, we may simply place that state in the record for the new top of stack.\n\n3. SDT's With Actions Inside Productions\n\nAn action may be placed at any position within the body of a production.\n\nIt is performed immediately after all symbols to its left are processed. Thus,\n\nif we have a production B -» X {a} Y, the action a is done after we have recognized X (if X is a terminal) or all the terminals derived from X (if X is a nonterminal). More precisely,\n\n• If the parse is bottom-up, then we perform action a as soon as this oc-currence of X appears on the top of the parsing stack.\n\n• If the parse is top-down, we perform a just before we attempt to expand this occurrence of Y (if Y a nonterminal) or check for Y on the input (if Y is a terminal).\n\nSDT's that can be implemented during parsing include postfix SDT's and a class of SDT's considered in Section 5.5 that implements L-attributed defini-tions. Not all SDT's can be implemented during parsing, as we shall see in the next example.\n\nExample 5.16 : As an extreme example of a problematic SDT, suppose that we turn our desk-calculator running example into an SDT that prints the prefix form of an expression, rather than evaluating the expression. The productions and actions are shown in Fig. 5.21.", null, "Unfortunately, it is impossible to implement this SDT during either top-down or bottom-up parsing, because the parser would have to perform critical actions, like printing instances of * or +, long before it knows whether these symbols will appear in its input.\n\nUsing marker nonterminals M2 and M4 for the actions in productions 2 and 4, respectively, on input 3, a shift-reduce parser (see Section 4.5.3) has conflicts between reducing by M2 -> e, reducing by M4 —> e, and shifting the digit.\n\nAny SDT can be implemented as follows:\n\nIgnoring the actions, parse the input and produce a parse tree as a result.\n\nThen, examine each interior node N, say one for production A -± a.  Add additional children to N for the actions in a,  so the children of N from left to right have exactly the symbols and actions of a.\n\n3. Perform a preorder traversal (see Section 2.3.4)  of the tree, and as soon as a node labeled by an action is visited, perform that action.\n\nFor instance,  Fig.  5.22 shows the parse tree for expression 3 * 5 + 4 with actions inserted. If we visit the nodes in preorder, we get the prefix form of the expression: + * 3 5 4.", null, "4. Eliminating Left Recursion From SDT's\n\nSince no grammar with left recursion can be parsed deterministically top-down, we examined left-recursion elimination in Section 4.3.3. When the grammar is part of an SDT, we also need to worry about how the actions are handled.\n\nFirst, consider the simple case, in which the only thing we care about is the order in which the actions in an SDT are performed. For example, if each action simply prints a string, we care only about the order in which the strings are printed. In this case, the following principle can guide us:\n\nWhen transforming the grammar, treat the actions as if they were termi-nal symbols.\n\nThis principle is based on the idea that the grammar transformation preserves the order of the terminals in the generated string. The actions are therefore executed in the same order in any left-to-right parse, top-down or bottom-up.\n\nThe  \"trick\" for eliminating left recursion is to take two productions\n\nA -> Aaa | b\n\nthat generate strings consisting of a j3 and any number of en's, and replace them by productions that generate the same strings using a new nonterminal R (for \"remainder\") of the first production:\n\nA->bR\n\nR —»• aR | e\n\nIf (3 does not begin with A, then A no longer has a left-recursive production. In regular-definition terms, with both sets of productions, A is defined by 0(a)*. See Section 4.3.3 for the handling of situations where A has more recursive or nonrecursive productions.\n\nExampl e 5 . 1 7 : Consider the following E-productions from an SDT for trans-lating infix expressions into postfix notation:\n\nE       ->      E i + T   { print('+'); }\n\nE       ->      T\n\nIf we apply the standard transformation to E, the remainder of the left-recursive production is\n\na        =       +  T { print('-r'); }\n\nand    the body of the other production is T.  If we introduce R for the remain-\n\nder of E, we get the set of productions:\n\nE        -->         T  R\n\nR       -->      + T { printC-h'); } R\n\nR       ->     e\n\nWhen the actions of an SDD compute attributes rather than merely printing output, we must be more careful about how we eliminate left recursion from a grammar. However, if the SDD is S-attributed, then we can always construct an SDT by placing attribute-computing actions at appropriate positions in the new productions.\n\nWe shall give a general schema for the case of a single recursive production, a single nonrecursive production, and a single attribute of the left-recursive nonterminal; the generalization to many productions of each type is not hard, but is notationally cumbersome. Suppose that the two productions are A", null, "Here, X.a is the synthesized attribute of left-recursive nonterminal A, and X and Y are single grammar symbols with synthesized attributes X.x and Y.y, respectively. These could represent a string of several grammar symbols, each with its own attribute(s), since the schema has an arbitrary function g computing A.a in the recursive production and an arbitrary function / computing A.a in the second production. In each case, / and g take as arguments whatever attributes they are allowed to access if the SDD is S-attributed.\n\nWe want to turn the underlying grammar into", null, "Figure 5.23 suggests what the SDT on the new grammar must do. In (a) we see the effect of the postfix SDT on the original grammar. We apply / once, corresponding to the use of production A -> X, and then apply g as many times as we use the production A AY. Since R generates a \"remainder\" of Y's, its translation depends on the string to its left, a string of the form XYY • • Y. Each use of the production R -> YR results in an application of g. For R, we use an inherited attribute R.i to accumulate the result of successively applying g, starting with the value of A.a.", null, "In addition, R has a synthesized attribute R.s, not shown in Fig. 5.23.\n\nThis attribute is first computed when R ends its generation of Y symbols, as signaled by the use of production R —>• e. R.s is then copied up the tree, so it can become the value of A.a for the entire expression XYY • • • Y. The case where A generates XYY is shown in Fig. 5.23, and we see that the value of A.a at the root of (a) has two uses of g. So does R.i at the bottom of tree (b), and it is this value of R.s that gets copied up that tree.\n\nTo accomplish this translation, we use the following SDT:", null, "Notice that the inherited attribute R.i is evaluated immediately before a use of R in the body, while the synthesized attributes A.a and -R.s are evaluated at the ends of the productions. Thus, whatever values are needed to compute these attributes will be available from what has been computed to the left.\n\n3. SDT's for L-Attributed Definitions\n\nIn Section 5.4.1, we converted S-attributed SDD's into postfix SDT's, with actions at the right ends of productions. As long as the underlying grammar is LR, postfix SDT's can be parsed and translated bottom-up.\n\nNow, we consider the more general case of an L-attributed SDD. We shall assume that the underlying grammar can be parsed top-down, for if not it is frequently impossible to perform the translation in connection with either an LL or an LR parser. With any grammar, the technique below can be imple-mented by attaching actions to a parse tree and executing them during preorder traversal of the tree.\n\nThe rules for turning an L-attributed SDD into an SDT are as follows:\n\n1. Embed the action that computes the inherited attributes for a nonterminal A immediately before that occurrence of A in the body of the production. If several inherited attributes for A depend on one another in an acyclic fashion, order the evaluation of attributes so that those needed first are computed first.\n\nPlace the actions that compute a synthesized attribute for the head of a production at the end of the body of that production.\n\nWe shall illustrate these principles with two extended examples. The first involves typesetting. It illustrates how the techniques of compiling can be used in language processing for applications other than what we normally think of as programming languages. The second example is about the generation of intermediate code for a typical programming-language construct: a form of while-statement.\n\nE x a m p l e 5 . 1 8 : This example is motivated by languages for typesetting math - ematical formulas. Eqn is an early example of such a language; ideas from Eqn are still found in the TEX typesetting system, which was used to produce this book.\n\nWe shall concentrate on only the capability to define subscripts, subscripts of subscripts, and so on, ignoring superscripts, built-up fractions, and all other mathematical features. In the Eqn language, one writes a sub i sub j to set the expression aij. A simple grammar for boxes (elements of text bounded by a rectangle) is", null, "Corresponding to these four productions, a box can be either\n\n1.       Two boxes, juxtaposed, with  the     first,  B1, to the    left of the other, B2.\n\n2. A box and a subscript box. The second box appears in a smaller size, lower, and to the right of the first box.\n\n3. A parenthesized box, for grouping of boxes and subscripts. Eqn and I g X both use curly braces for grouping, but we shall use ordinary, round paren-theses to avoid confusion with the braces that surround actions in SDT's.\n\n4. A text string, that is, any string of characters.\n\nThis grammar is ambiguous, but we can still use it to parse bottom-up if we make subscripting and juxtaposition right associative, with s u b taking precedence over juxtaposition.\n\nExpressions will be typeset by constructing larger boxes out of smaller ones.\n\nIn Fig. 5.24, the boxes for E1 and .height are about to be juxtaposed to form the box for Ei.height. The left box for E1 is itself constructed from the box for E and the subscript 1. The subscript 1 is handled by shrinking its box by about 30%, lowering it, and placing it after the box for E. Although we shall treat .height as a text string, the rectangles within its box show how it can be constructed from boxes for the individual letters.", null, "In this example, we concentrate on the vertical geometry of boxes only. The horizontal geometry — the widths of boxes — is also interesting, especially when different characters have different widths. It may not be readily apparent, but each of the distinct characters in Fig. 5.24 has a different width.\n\nThe values associated with the vertical geometry of boxes are as follows:\n\na)                 The point size is used to set text within a box. We shall assume that characters not in subscripts are set in 10 point type, the size of type in this book. Further, we assume that if a box has point size p, then its subscript box has the smaller point size 0.7p. Inherited attribute B.ps will represent the point size of block B. This attribute must be inherited, because the context determines by how much a given box needs to be shrunk, due to the number of levels of subscripting.\n\nb)        Each box has a baseline, which is a vertical position that corresponds to the bottoms of lines of text, not counting any letters, like \"g\" that extend below the normal baseline. In Fig. 5.24, the dotted line represents the baseline for the boxes E, .height, and the entire expression. The baseline\n\nfor the box containing the subscript 1 is adjusted to lower the subscript.\n\nc)          A box has a height, which is the distance from the top of the box to the baseline. Synthesized attribute B.ht gives the height of box B.\n\nd)         A box has a depth, which is the distance from the baseline to the bottom of the box. Synthesized attribute B.dp gives the depth of box B.\n\nThe SDD in Fig. 5.25 gives rules for computing point sizes, heights, and depths. Production 1 is used to assign B.ps the initial value 10.", null, "Production 2 handles juxtaposition. Point sizes are copied down the parse tree; that is, two sub-boxes of a box inherit the same point size from the larger box. Heights and depths are computed up the tree by taking the maximum. That is, the height of the larger box is the maximum of the heights of its two components, and similarly for the depth.\n\nProduction 3 handles subscripting and is the most subtle. In this greatly simplified example, we assume that the point size of a subscripted box is 70% of the point size of its parent. Reality is much more complex, since subscripts cannot shrink indefinitely; in practice, after a few levels, the sizes of subscripts shrink hardly at all. Further, we assume that the baseline of a subscript box drops by 25% of the parent's point size; again, reality is more complex.\n\nProduction 4 copies attributes appropriately when parentheses are used. Fi-nally, production 5 handles the leaves that represent text boxes. In this matter too, the true situation is complicated, so we merely show two unspecified func-tions getHt and getDp that examine tables created with each font to determine the maximum height and maximum depth of any characters in the text string. The string itself is presumed to be provided as the attribute lexval of terminal t e x t .\n\nOur last task is to turn this SDD into an SDT, following the rules for an L-attributed SDD, which Fig. 5.25 is. The appropriate SDT is shown in Fig. 5.26. For readability, since production bodies become long, we split them across lines and line up the actions. Production bodies therefore consist of the contents of all lines up to the head of the next production.", null, "Our next example concentrates on a simple while-statement and the gener-ation of intermediate code for this type of statement. Intermediate code will be treated as a string-valued attribute. Later, we shall explore techniques that involve the writing of pieces of a string-valued attribute as we parse, thus avoid-ing the copying of long strings to build even longer strings. The technique was introduced in Example 5.17, where we generated the postfix form of an infix expression \"on-the-fly,\" rather than computing it as an attribute. However, in our first formulation, we create a string-valued attribute by concatenation.\n\nE x a m p l e 5 . 1 9 :  For this example, we only need one production:\n\nS -> while ( C ) Si\n\nHere, S is the nonterminal that generates all kinds of statements, presumably including if-statements, assignment statements, and others. In this example, C stands for a conditional expression — a boolean expression that evaluates to true or false.\n\nIn this flow-of-control example, the only things we ever generate are labels. All the other intermediate-code instructions are assumed to be generated by parts of the SDT that are not shown. Specifically, we generate explicit instruc-tions of the form l a b e l L, where L is an identifier, to indicate that L is the label of the instruction that follows. We assume that the intermediate code is like that introduced in Section 2.8.4.\n\nThe meaning of our while-statement is that the conditional C is evaluated. If it is true, control goes to the beginning of the code for Si. If false, then control goes to the code that follows the while-statement's code. The code for Si must be designed to jump to the beginning of the code for the while-statement when finished; the jump to the beginning of the code that evaluates C is not shown in Fig. 5.27.\n\nWe use the following attributes to generate the proper intermediate code:\n\n1. The inherited attribute S.next labels the beginning of the code that must be executed after S is finished.\n\n2. The synthesized attribute S.code steps s the sequence of intermediate-code S and that implements a statement ends with a jump to S.next.\n\n3. The inherited attribute C.true labels the beginning of the code that must be executed if C is true.\n\n4. The inherited attribute C.false labels the beginning of the code that must be executed if C is false.\n\nThe synthesized attribute C.code is the sequence of intermediate-code steps that implements the condition C and jumps either to C.true or to C.false, depending on whether C is true or false.\n\nThe SDD that computes these attributes for the while-statement is shown in Fig. 5.27. A number of points merit explanation:\n\n• The function new generates new labels.\n\nThe variables LI and L2 hold labels that we need in the code. LI is the beginning of the code for the while-statement, and we need to arrange", null, "Figure 5.27: SDD for while-statements that Si jumps there after it finishes. That is why we set Si.next to LI.\n\nL2 is the beginning of the code for Si, and it becomes the value of C.true, because we branch there when C is true.\n\n• Notice that C.false is set to S.next, because when the condition is false, we execute whatever code must follow the code for S.\n\nWe use || as the symbol for concatenation of intermediate-code fragments. The value of S.code thus begins with the label LI, then the code for condition C, another label L2, and the code for Si.\n\nThis SDD is L-attributed. When we convert it into an SDT, the only re-maining issue is how to handle the labels LI and L2, which are variables, and not attributes. If we treat actions as dummy nonterminals, then such variables can be treated as the synthesized attributes of dummy nonterminals. Since LI and L2 do not depend on any other attributes, they can be assigned to the first action in the production. The resulting SDT with embedded actions that implements this L-attributed definition is shown in Fig. 5.28.", null, "6. Exercises for Section 5.4\n\nExercise 5 . 4 . 1 : We mentioned in Section 5.4.2 that it is possible to deduce, from the LR state on the parsing stack, what grammar symbol is represented by the state. How would we discover this information?\n\nExercise 5 . 4 . 2:  Rewrite the following SDT:\n\nA       A {a} B | A B {b}          |         0\n\nB ->   B {c} A | B A       {d}    |         1\n\nso that the underlying grammar becomes non-left-recursive. Here, a, 6, c, and d are actions, and 0 and 1 are terminals.\n\n! Exercise 5 . 4 . 3 : The following SDT computes the value of a string of O's and l's interpreted as a positive, binary integer.\n\nB       -»      Br      0  {B.val     = 2 x Bx.val]\n\n|         Bx     1  {B.val     = 2 x Bx.val + 1}\n\nj         1        {B.val = 1}\n\nRewrite this SDT so the underlying grammar is not left recursive, and yet the same value of B.val is computed for the entire input string.\n\n! Exercise 5 . 4 . 4: Write L-attributed SDD's analogous to that of Example 5.19 for the following productions, each of which represents a familiar flow-of-control construct, as in the programming language C. You may need to generate a three-address statement to jump to a particular label L, in which case you should generate goto L.\n\na)       S       -)• if ( C )     Si else S2\n\nb)      S       do Si while ( C )\n\nc)       S       -4'{' L         '}';     L       L S    1  e\n\nNote that any statement in the list can have a jump from its middle to the next Statement, so it is not sufficient simply to generate code for each statement in order.\n\nExercise 5 . 4 . 5 : Convert each of your SDD's from Exercise 5.4.4 to an SDT in the manner of Example 5.19.\n\nExercise 5 . 4 . 6 : Modify the SDD of Fig. 5.25 to include a synthesized attribute B.le, the length of a box. The length of the concatenation of two boxes is the sum of the lengths of each. Then add your new rules to the proper positions in the SDT of Fig. 5.26\n\nExercise 5 . 4 . 7: Modify the SDD of Fig. 5.25 to include superscripts denoted by operator sup between boxes. If box B2 is a superscript of box Bi, then position the baseline of B2 0.6 times the point size of Bi above the baseline of Bi. Add the new production and rules to the SDT of Fig. 5.26.\n\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n\nRelated Topics" ]

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The article spinner route is certainly much less expensive especially if you want around 50 to 100 pages of content for your website. The cost can run into the thousands and the less than \\$100 per year cost of Spin Rewriter 6.0 begins to seem extremely reasonable.\nIf you are someone who is trying to have a proper understanding of the functionality of an article rewriter or a paraphrasing tool then you are at the right place. An article rewriter tool is a simple tool that can work both online or offline, depending on how it functions. An instant article spinner or a paraphrasing tool is an efficient yet very easy to use tool that understand and identifies the text you give it, then rewriting it for you. It enables you to get your article rewritten with no trouble. Any quality article spinner saves your time and the trouble of sitting in front of your computer for hours in an effort to write unique and quality content. 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[ "# Diffusion coefficients\n\n## Top PDF Diffusion coefficients:", null, "### A Cumulative migration method for computing rigorous transport cross sections and diffusion coefficients for LWR lattices with Monte Carlo\n\nThe generation of multi-group cross section data for LWR analysis usually starts by identifying some characteristic “lattice” — be it a pin-cell, a fuel assembly, or a collection of fuel assemblies. For each such lattice, a very-fine-group transport calculation (e.g., 50 – 10,000 groups) is performed to obtain the neutron flux and reaction rate distributions within the lattice. Unless this transport calculation explicitly models anisotropic scattering, an approximation for transport-corrected-P 0 cross sections for each nuclide must be introduced before the multi-group lattice transport calculation can be performed. In addition, lattice reaction rates and fluxes are used to compute energy-condensed and/or spatially- homogenized transport cross sections (or diffusion coefficients) for use in downstream multi-group (e.g., 2 – 100 groups) core calculations. Here, additional approximations are required to compute the appropriate transport cross section that preserves some selected characteristic of the lattice calculation. All production lattice physics codes [ 3 – 6 ] make such approximations, often without substantial justifi- cation. Moreover, the most useful of these approximations are often considered to be proprietary, and the literature remains largely silent on useful methods. One example might be that of the transport- corrected-P 0 methods that have been employed in CASMO for more than 40 years. Only recently has Herman [ 7 ] published details of the method used in CASMO to generate transport-corrected-P 0 cross sections for 1 H in LWR lattices. Herman was able to compute CASMO’s “exact” transport cross section that matched continuous-energy Monte Carlo (MC) neutron leakages (integrated into 70 fine energy groups) from a slab of pure hydrogen. This transport correction is markedly different from that computed using the “micro-balance” argument [ 8 ] which produces the classic “out-scatter” approxi- mation — with its transport-to-total ratio of 1/3 for purely isotropic center-of-mass neutron scattering with free gas 1 H model. CASMO developers recognized long ago that this definition of transport cross section produced excellent eigenvalues for small LWR critical assemblies with large neutron leakages, while the classic out-scatter approximation produced errors in eigenvalue as large as 1000 pcm. In addition, SIMULATE-3 nodal code developers observed (more than 30 years ago) that the CASMO transport cross section also produced two-group diffusion coefficients that eliminated radial power tilts observed in large 4-loop PWR cores when using the out-scatter approximation.", null, "### Diffusion coefficients of solids in supercritical carbon dioxide: Modelling of near critical behaviour\n\nMost of the models proposed in literature for binary diffusion coefficients of solids in supercritical fluids are restricted to infinite dilution; this can be explained by the fact that most of experimental data are performed in the dilute range. However some industrial processes, such as supercritical fluid separation, operate at finite concentration for complex mixtures. In this case, the concentration dependence of diffusion coefficients must be considered, especially near the upper critical endpoint (UCEP) where a strong decrease of diffusion coefficients was experimentally observed. In order to represent this slowing down, a modified version of the Darken equation was proposed in literature for naphthalene in supercritical carbon dioxide. In this paper, the conditions of application of such a modelling are investigated. In particular, we focus on the order of magnitude of the solubility of the solid and on the vicinity of the critical endpoint. Various equations proposed in literature for the modelling of the infinite dilution diffusion coefficients of the solutes are also compared. Ten binary mixtures of solids with supercritical carbon dioxide were considered for this purpose.", null, "### On the relationship between the formation factor and diffusion coefficients of Portland cement mortars\n\nFigure 5: Comparison between formation factor of lithium, and HTO obtained experimentally Differences obtained between these factors must normally tend to zero since it is the same formula- tions which were tested by the two tracers. Because of some disturbances that can take place during tests, such an error exists. Among these disturbing factors, may be mentioned two possible reasons. From one perspective, it was noticed during the tests, a slight drop of temperature in the room where were placed the cells, this fall of temperature can cause a decrease in effective diffusion coefficients and thus increase the formation factor F. From another perspective, the increase in the formation factor may be due to the presence of chlorides ions in the upstream solution (solution of LiCl). These ions diffusing into mortars can involve some possible modifications in the mi- crostructure of materials, in particular in the porosity of the hydrates. Otherwise, since hydrates porosity in CEM I based materials represents a smaller fraction of total porosity (compared to a CEM V material for example), the effect produced by chloride ions on the", null, "### Note: Determination of effective gas diffusion coefficients of stainless steel films with differently shaped holes using a Loschmidt diffusion cell", null, "### Cumulative migration method for computing multi-group transport cross sections and diffusion coefficients with Monte Carlo calculations\n\nand improve the errors caused by spatial homogenization and energy condensation with MGXS generated using Monte Carlo codes [1,23–27], but the accuracy improvement for angular approximations remains a challenging task and open question. Most Monte Carlo-based MGXS generation applications aim to improve the accuracy of full core diffusion calculations and are focused on generating few-group cross sections and diffusion coefficients. The recent study by Boyd has investigated the accuracy by directly using full core continuous energy Monte Carlo simulations to compute MGXS for full core deterministic multi-group transport calculations. But due to the difficulty caused by the anisotropy in scattering reactions, the study in adopted the isotropic- in-LAB (isotropic in the laboratory system) approximation in both the Monte Carlo and deterministic simulations.", null, "### Experimental study of oxygen diffusion coefficients in clean water containing salt, glucose or surfactant: Consequences on the liquid-side mass transfer coefficients\n\nFor sugar solutions, the reduction of oxygen diffusion coef- ficients reaches 30% for the highest concentrations (100 g L −1 ), leading to a minimal value of 1.39 × 10 −9 m 2 s −1 . This latter value is fully relevant when compared to the results of Van Stroe-Biezen et al. who measured, using an electrochemical method, the diffusion coefficient of oxygen in glucose solutions; indeed, these authors found a reduction in diffusivity of 26% when the glucose concentration varied from 0 to 100 g L −1 in their fermentation media. The decrease in viscosity with increasing concentrations of glucose ( Table 1 ) is mainly responsible for such change in diffusion coefficients. This is illustrated in Fig. 4 b where the usual depen- dence of D with the inverse of viscosity is verified, as predicted by the Stokes–Einstein equation and the correlation of Wilke and Chang (1954) . The rate of change of D/D water with concentra-", null, "### Measurement of effective gas diffusion coefficients of catalyst layers of PEM fuel cells with a Loschmidt diffusion cell\n\nExperimental measurement techniques, e.g., gas chromatogra- phy, nuclear magnetic resonance (e.g., PFG NMR), and diffusion cell methods, have been developed to determine the EGDCs of porous materials [9–13] . Kramer and co-workers measured the EGDC in carbon paper using electrochemical impedance spectroscopy [14,15] . Zhang et al. used a Wicke–Kallenbach diffusion cell to measure the EGDC of a catalyst monolith washcoat. A closed- tube method with a Loschmidt diffusion cell is considered as one of the most reliable methods to determine binary diffusion coeffi- cients of gases [16–19] . Using a Loschmidt cell, bulk binary diffusion coefficients of O 2 –N 2 were precisely measured under the experi-", null, "### Methods to calculate gas diffusion coefficients of cellular plastic insulation from experimental data on gas absorption\n\nhttps://doi.org/10.1177/109719639101400408 Access and use of this website and the material on it are subject to the Terms and Conditions set forth at Methods to calculate gas diffusion coefficients of cellular plastic insulation from experimental data on gas absorption", null, "### Acquisition of anion profiles and diffusion coefficients in the Opalinus Clay at the Mont Terri rock laboratory (Switzerland)\n\nAix Marseille Université UMR 6635 CEREGE, BP80 F13545 Aix-en-Provence Cedex 4, France Abstract Chloride, bromide and sulphate concentration profiles have been analysed through the Opalinus Clay of Mont Terri in the framework of the Deep Borehole Experiment. Aqueous leaching and out diffusion experiments were carried out to acquire anion concentrations and estimate pore diffusion coefficients. Out diffusion technique gave consistent values of chloride and bromide compared to the concentration profiles acquired so far at the tunnel level of the Mont Terri rock laboratory. Concentrations acquired by leaching experiments show a maximum chloride concentration of 16.1 ± 1.7 g/l at the basal part of Opalinus Clay which is higher than the value of 14.4 ± 1.4 g/l obtained by out diffusion at the same level. This excess of chloride is likely due to dissolution of Cl - bearing minerals or release of Cl initially contained in unaccessible porosity. Bromide to chloride ratios are virtually the same as that of seawater, whereas sulphate to chloride ratios are significantly higher. Those latter are probably due to pyrite oxidation and dissolution of sulphate-bearing minerals occurring during sample collection and preparation. An anisotropy ratio of 2.4 was estimated for pore diffusion coefficient in the Opalinus Clay sandy facies.", null, "### Analysis of volatile organic compounds (VOCs) in soil via passive sampling : measuring partition and diffusion coefficients\n\nAnalysis of Volatile Organic Compounds (VOCs) in Soil via Passive Sampling: Measuring Partition and Diffusion Coefficients.. by Hanqing Liu.[r]", null, "### Coherent quantum transport in disordered systems: II. Temperature dependence of carrier diffusion coefficients from the time-dependent wavepacket diffusion method\n\nIn the present work, we use the time-dependent wavepacket diffusion (TDWPD) method [ 41 ] to investigate the diffusion coef ficient of the one-dimensional molecular-crystal model when both the static and dynamic disorders are present. Formally, TDWPD can be understood as an approximation to the exact stochastic Schrödinger equation (SSE) [ 42 – 46 ] or stochastic wavefunction methods [ 47 ]. The TDWPD has a similar structure to the HSR; except we start from the coherent-state representation of the phonon, and the dynamic disorders are then incorporated by stochastic complex-valued forces, which are quantitatively generated from their quantum correlation functions. The TDWPD method essentially overcomes the de ficiency of HSR, i.e. it incorporates the quantum tunneling effect and approximately satis fies the detailed balance principle. Moreover, its demand on computational time is similar to that of HSR, and it can thus be applied to nanoscale systems. To compare with existing theories, we also present the results from Red field theory [ 38 , 48 – 50 ], which is a standard approach to describe the coherent motions of carriers. The diffusion coef ficients from the (FGR) and Marcus formula are further illustrated to investigate the validity of the TDWPD in the strong dynamic disorder limit.", null, "### Extension of the bifurcation method for diffusion coefficients to porous medium transport\n\nExtension of the bifurcation method for diffusion coefficients to porous medium transport.. Cédric Descamps, Gerard LG[r]", null, "### Calculation of combined diffusion coefficients from the simplified theory of transport properties\n\nL’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.", null, "### Theoretical and experimental determination of effective diffusion and thermodiffusion coefficients in porous media\n\n1.4.2 The Thermogravitational Column Another method for measuring thermal diffusion coefficients is the thermogravitational column which consists of two vertical plates separated by a narrow space under a horizontal or vertical thermal gradient. The principle is to use a thermal gradient to simultaneously produce a mass flux by thermal diffusion and a convection flux. Starting from a mixture of homogeneous composition, the coupling of the two transport mechanisms leads to a separation of the components. In most experimental devices, the applied thermal gradient is horizontal and the final composition gradient is globally vertical. The separation rate in this system defined as the concentration difference between the top and the bottom cell. Thermogravitational column was devised by Clusius and Dickel (1938). The phenomenology of thermogravitational transport was exposed by Furry et al. (1939), and was validated by many experiments. The optimal coupling between thermal diffusion and convection ratio (maximum separation) correspond to an optimal thickness of the cell in free fluid (less than one millimetre for usual liquids) and an optimal permeability in porous medium [56, 57]. The so called packed thermal diffusion cell (PTC) was described and intensively used to perform experiments on varieties of ionic and organic mixtures [54, 21, 66]. The separation in a thermogravitational column can be substantially increased by inclining the column . Recently, Mojtabi et al., 2003, showed that the vibrations can lead whether to increase or to decrease heat and mass transfers or delay or accelerate the onset of convection .", null, "### Bounds between contraction coefficients\n\nOur discussion regarding model selection using contraction coefficients can be naturally extended to include proba- bility density functions (pdfs). For instance, the local approximations introduced in Subsection I-A were used to study AWGN channels in a network information theory context in [ 18 ]. We now consider the relationship between the local and global contraction coefficients in the Gaussian regime. To this end, we introduce the classical AWGN channel [ 7 ]. Definition 4 (AWGN Channel). The single letter AWGN channel has jointly distributed input random variable X and output random variable Y , where X and Y are related by the equation:", null, "### Design heat transmission coefficients\n\nNote 3: When heat transfer is by conduction alone, the average thermal conductivity is the product of the thermal conductance per unit area and the thickness.. The aver[r]", null, "### Reductions for branching coefficients\n\nrecently proved independently by Derksen and Weyman in [DW11, Theorem 7.4] and King, Tollu and Toumazet in [KTT09, Theorem 1.4] if G = GL n.. and for any reductive group by Roth in [Rot1[r]", null, "### Évolution historique des coefficients aérodynamiques\n\nF X = ρV 2 SC X (α) F Y = ρV 2 SC Y (α) où ρ est la masse volumique du fluide en mouvement autour de l’aile, C X (α) et C Y (α) les coefficients d’influence de l’angle d’incidence α, pour la force de traînée et pour la force de portance. De cette façon, les coefficients C X (α) et C Y (α) devenaient indépendants du fluide considéré. Des mesures", null, "### Synthèse indépendante des coefficients de réflexions de la matrice de diffusion d'une méta-surface\n\nd’une cellule réflectrice afin d’approcher une matrice de diffusion voulue. Cette étude est motivée par le besoin grandissant de contrôler le coefficient de réflexion en co- polarisation des méta-surfaces ainsi que le couplage entre polarisations incidente et croisée, en amplitude et en phase. L’utilité de la synthèse est montrée avec une application de conversion de polarisation linéaire en polarisation circulaire, la bande sous 3dB du rapport elliptique obtenue est de 17% autour de 16GHz. Toutes les simulations sont effectuées sous CST.", null, "" ]

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[ "## Algebra and Trigonometry 10th Edition\n\n$x=4.535$\n$8(3^{6-x})=40$ $3^{6-x}=5$ $\\log_33^{6-x}=\\log_35~~$ (Using the Inverse Property $\\log_aa^x=x$): $6-x=\\log_35=\\frac{\\ln5}{\\ln3}$ $6-\\frac{\\ln5}{\\ln3}=x$ $x=4.535$" ]

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[ "Discrete & Computational Geometry\n\n, Volume 37, Issue 1, pp 43–58\n\n# Improved Approximation Algorithms for Geometric Set Cover\n\nArticle\n\n## Abstract\n\nGiven a collection S of subsets of some set $${\\Bbb U},$$ and $${\\Bbb M}\\subset{\\Bbb U},$$ the set cover problem is to find the smallest subcollection $$C\\subset S$$ that covers $${\\Bbb M},$$ that is, $${\\Bbb M} \\subseteq \\bigcup (C),$$ where $$\\bigcup(C)$$ denotes $$\\bigcup_{Y \\in C} Y.$$ We assume of course that S covers $${\\Bbb M}.$$ While the general problem is NP-hard to solve, even approximately, here we consider some geometric special cases, where usually $${\\Bbb U} = {\\Bbb R}^d.$$ Combining previously known techniques , , we show that polynomial-time approximation algorithms with provable performance exist, under a certain general condition: that for a random subset $$R\\subset S$$ and nondecreasing function f(·), there is a decomposition of the complement $${\\Bbb U}\\backslash\\bigcup (R)$$ into an expected at most f(|R|) regions, each region of a particular simple form. Under this condition, a cover of size O(f(|C|)) can be found in polynomial time. Using this result, and combinatorial geometry results implying bounding functions f(c) that are nearly linear, we obtain o(log c) approximation algorithms for covering by fat triangles, by pseudo-disks, by a family of fat objects, and others. Similarly, constant-factor approximations follow for similar-sized fat triangles and fat objects, and for fat wedges. With more work, we obtain constant-factor approximation algorithms for covering by unit cubes in $${\\Bbb R}^3,$$ and for guarding an x-monotone polygonal chain.\n\n## Keywords\n\nApproximation Algorithm Lower Envelope Improve Approximation Algorithm Visibility Polygon Jordan Region\nThese keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves." ]

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[ "# Solving Black-Scholes PDE using Laplace transform\n\nI'm trying to obtain the Laplace transform of Call option price with repect to time to maturity under the CEV process.\n\nThe well known Black scholes PDE is given by $$\\frac{1}{2}\\sigma(x)^2x^2\\frac{\\partial^2}{\\partial x^2}C(x,\\tau)+\\mu x\\frac{\\partial}{\\partial x}C(x,\\tau)-rC(x,\\tau)-\\frac{\\partial}{\\partial \\tau}C(x,\\tau)=0.$$ where the initial condition $C(x,0)=max(x-K,0)$ and $\\sigma(x)=\\delta x^\\beta$.\n\nTaking the Laplace transform with respect to $\\tau$, we obtain the following ODE: $$\\frac{1}{2}\\delta x^{2\\beta+2}\\frac{\\partial^2}{\\partial x^2}\\hat{C}(x,\\lambda)+\\mu x\\frac{\\partial}{\\partial x}\\hat{C}(x,\\lambda)-(\\lambda+r)\\hat{C}(x,\\lambda)=-max(x-K,0).$$ where $\\hat{C}(x,\\lambda)=\\int_0^\\infty e^{-\\lambda \\tau}C(x,\\tau)d\\tau$\n\nand the initial condition is transformed to $$\\hat{C}(x,\\lambda)=\\int_0^\\infty e^{-\\lambda \\tau}C(x,0) d\\tau=max(x-K,0)/\\lambda$$(is this right??? it seems wrong..)\n\nThen, $\\hat{C}(x,\\lambda)$ can be analytically formulated by the case $x>K$ and $x\\leq K$.\n\nHow to get explicit formula for $\\hat{C}(x,\\tau)$? I can't proceed from this stage.\n\nI know one paper, \"(2001 Dmitry) Pricing and Hedging Path-Dependent Options under the CEV\", related to this question. However, there's big jumps for me to understand readily. Could you explain it step by step?\n\n• It's perfectly legitimate to use the Laplace transform (and vonjd's linked paper does a fine job), but I've personally always preferred to solve the PDE by changing variables until the PDE turns into the standard diffusion equation. – Brian B Sep 18 '14 at 15:39\n• Welcome to Quant Stackexchange :-) Thank you for your interesting first question. If the answer was helpful you could upvote and accept it :-) – vonjd Sep 21 '14 at 6:38" ]

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[ "We included HMH Into Math Grade 5 Answer Key PDF Module 16 Multiply Decimals to make students experts in learning maths.\n\n## HMH Into Math Grade 5 Module 16 Answer Key Multiply Decimals\n\nFour friends want to see a show. Each friend brings a bag full of coins to pay for a ticket. A ticket costs $20. (Hint:$20 = 2,000 cents)\n\nWhich of the friends has enough money to buy a ticket?", null, "Penny, Dmitri and Carter has enough cents to purchase the ticket.\n\nExplanation:\nCost of the ticket = \\$20 = 2000 cents.\nNumber of pennies Penny gets = 2316.\nNumber of nickels Nicki gets = 382.\nNumber of dimes Dmitri gets = 217.\nNumber of quarters Carter gets = 83.\nConversion:\n1 penny =   cent.\n=> Number of pennies Penny gets = 2316 × 1 = 2316 cents.\n1 nickel =  5 cents.\n=> Number of nickels Nicki gets = 382 × 5 = 1910 cents.\n1 dime =  10 cents.\n=> Number of dimes Dmitri gets = 217 × 10 = 2170 cents.\n1 quarter = 25 cents.\n=>  Number of quarters Carter gets = 83 × 25 = 2075 cents.\n\nTurn and Talk\nHow do you find the number of cents each person has?\nThe number of cents for each person has been calculated by converting nickels, dimes, quarters into cents.\n\nExplanation:\nI found the number of cents for each person has by converting nickels, dimes, quarters into cents.\n\nHow can the four friends share their money so that everyone can buy a ticket to the show?\nThey can share the rest amount after the ticket amount and can buy four tickets to watch the show.\n\nExplanation:\nThe four friends share their money so that everyone can buy a ticket to the show by sharing their extra amount after having 2000 cents with them n rest to buy the tickets.\n\nComplete these problems to review prior concepts and skills you will need for this module.\nMultiplication Facts\n\nFind the product.\nQuestion 1.\n4 × 8 = __________\n4 × 8 = 32.\n\nExplanation:\nMultiplication of 4 and 8:\n4 × 8 = 32.\n\nQuestion 2.\n_________ = 8 × 8\n64 = 8 × 8.\n\nExplanation:\nMultiplication of 8 and 8:\n8 × 8 = 64.\n\nQuestion 3.\n9 × 3 = ____________\n9 × 3 = 27.\n\nExplanation:\nMultiplication of 9 and 3:\n9 × 3 = 27.\n\nQuestion 4.\n__________ = 9 × 10\n90 = 9 × 10.\n\nExplanation:\nMultiplication of 9 and 10:\n9 × 10 = 90.\n\nQuestion 5.\n7 × 2 = _________\n7 × 2 = 14.\n\nExplanation:\nMultiplication of 7 and 2:\n7 × 2 = 14.\n\nQuestion 6.\n_________ = 7 × 6\n42 = 7 × 6.\n\nExplanation:\nMultiplication of 7 and 6:\n7 × 6 = 42.\n\nEstimate. Then find the sum.\nQuestion 7.\n386 + 246 = _________\nEstimate: ___________\n386 + 246 = 632.\nEstimate: 650.\n\nExplanation:\n386 + 246 = 632.\nEstimate: 650.\n\nQuestion 8.\n35 + 346 = ___________\nEstimate: ___________\n35 + 346 = 381.\nEstimate: 400.\n\nExplanation:\n35 + 346 = 381.\nEstimate: 400.\n\nQuestion 9.\n82 + 579 = ___________\nEstimate: ___________\n82 + 579 = 661.\nEstimate: 690.\n\nExplanation:\n82 + 579 = 661.\nEstimate: 690.\n\nQuestion 10.\n463 + 96 = ___________\nEstimate: ____________\n463 + 96 = 559.\nEstimate: 600.\n\nExplanation:\n463 + 96 = 559.\nEstimate: 600.\n\nMultiply with 3-Digit Numbers\nFind the product.\nQuestion 11.", null, "", null, "Explanation:\nMultiplication of 582 and 34:\n582 × 34 = 19,788.\n\nQuestion 12.", null, "", null, "Explanation:\nMultiplication of 435 and 73:\n435 × 73 = 31,755.\n\nQuestion 13.", null, "", null, "Explanation:\nMultiplication of 391 and 42:\n391 × 42 = 16,422.\n\nQuestion 14.", null, "", null, "Explanation:\nMultiplication of 843 and 58:\n843 × 58 = 48,894.\n\nQuestion 15.", null, "", null, "Explanation:\nMultiplication of 496 and 95:\n496 × 95 = 47,120.\n\nQuestion 16.", null, "", null, "" ]

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[ "QQ网名设计网请您输入需设计的网名\n\nQQ内涵网名\n\n• 骚年歌声眠于旧巷╮\n• ゛年少怎能不多情╮。\n• 一刻暧昧。\n• So/要识趣\n• 假戏-/\n• 枯萎的玫瑰ゝ\n• 可劲矫情可劲造。\n• 陪你到世纪末日、\n• 　 心动则痛,\n• 陪你看日出、说不离╮\n• 空中飞人#\n• 凉凉凉”凉不过人心。\n• 优雅式、妖媚\n• 唯㈠飠尓\n• 末日不孤单 ▲\n• 丿寂寥丶夜丨\n• 小鸟爱天空丶\n• 他春了春天\n• 所有的痛都不算痛\n• 你走了、心空了▍\n• 是梦终空 -\n• 只有爱情、何来生活╰\n• ◇所谓的你\n• _____旧情劫′\n• 没 企图\n• moment°誓言如狗屁\n• 在等一个感动,\n• 本女子ァ你们玩不起的女人\n• 目断魂销°\n• 有你、已滿足。\n• 度娘°\n• 念得寝食难安\n• 眼眸有珎旳温柔㏄\n• 阳光腐朽了心 、\n• 不够你ヽ愛\n• 丶失控的情绪\n• 无所谓的记忆\n• 那些甜言蜜語\n• ∞华丽的谎言\n• 碎了一地的阳光 Sun\n• 代替╮沉默\n• 逞不完的强\n• 喜新恋旧。\n• 空白答卷\n• 怀念的钟表。\n• 。冷情\n• 其实狠陌生°\n• 想你难捱的夜\n• 幸福眩晕而至-\n• 寂寞在唱歌丶\n• 幸福的距離\n• 直到最后一句。\n• - 如果有一天\n• 我们都不帅\n• 爱到忘记流月经ノ\n• 交横的铁轨。\n• //似有若无\n• 刺痛我的心。\n• 西瓜熟了夏走了\n• *在文字里诉说的悲伤*\n• 心最暖的怀抱\n• 你心丢了那个我″\n• 你是我的独家回忆-\n• - 纠缠的神经线\n• 忽而令夏 ゝ\n• 无言 ゛\n• 华丽蜕变优雅丶\n• ′`背影﹖ノ\n• 你没有看我\n• 沒有未來以後\n• 撕裂了温暖的阳光丶\n• 为欲而生ノ\n• 年少怎么不叛逆╮\n• 丑却有无限可能\n• 一纸荒凉。\n• 洋葱浓汤。\n• ﹋ 私奔、\n• 從開始到現在\n• 小雏菊、\n• 抚筝、游移" ]

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[ "# Questions Asked onDecember 9, 2012\n\n1. ## Physics\n\nAs a box is pushed 30 meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the box increases by 300 joules. How much total internal energy is produced during this process?\n\n2. ## Physics (PLEASE CHECK WORK)\n\nFor safety in climbing, a mountaineer uses a nylon rope that is 50m long and 1.0cm in diameter. When supporting a 90kg climber, the rope elongates 1.6m. Find its Young's modulus This is what I have so far, I am getting the wrong answer. Can you please fix\n\n3. ## Physics\n\nTwo objects of different mass start from rest, are pulled by the same magnitude net force, and are moved through the same distance. The work done on object A is 500J. After the force pulled each object, object A moves twice as fast as object B. Answer the\n\n4. ## Finance\n\nkollo enterprise has beta 0.82 real risk freerate 2.00% investors expected 3.00% future inflation rate, market risk premium is 4.70% what kollo's required rate of return?\n\n5. ## physics\n\nA CONVEX LENS PRODUCES O REAL AND INVERTED IMAGE 2.5 TIMES MAGNIFIED AT A DISTANCE OF 25CM FROM THE LENS.CALCULATE THE FOCAL LENGTH OF THE LENS,\n\n6. ## math\n\nMary and Peter had a total of Rs 480.After Peter spent Rs 20, he had 3 times as much as Mary.How much did Mary have?\n\n7. ## physics- TEST TMRW PLS HELP!\n\nWhy is the following situation impossible? A librarian lifts a book from the ground to a high shelf, doing 20.0 J of work in the lifting process. As he turns his back, the book falls off the shelf back to the ground. The gravitational force from the Earth\n\n8. ## Physics\n\nAn object of mass .550kg is lifted from the floor to a height of 3.5m at a constant speed. How much work is done by the lifting force (include units)? b.) How much work is done by the Earth on the object? c.) What is the net work done on the object? d.)\n\n9. ## Chemistry\n\nThe figure below is a heat curve for an unknown outerspace substance discovered by the Mars rover. By understanding the heat curve, scientists can make inferences about the composition of the planet. What is the temperature of fusion, Tfusion, and\n\n10. ## math/algebra\n\nA kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second. Write an equation that gives the height (in feet) of the football as a function of the time (in seconds) since it was punted. Find the height (in feet)\n\n11. ## Psychology 220\n\n1.65 (1 pts.) David is a junior at the local high school. The school's administration believes his recent emotional outbursts during class may be due to problems in his home life. Who might be best suited to deal with this situation? A) the principal B) a\n\n12. ## Chemistry\n\nHow many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al(s) + 3FeO(s)-- 3Fe(s) + Al2O3(s)\n\n13. ## calculus\n\na norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. Find the dimensions of a Norman window of maximum area it the total perimiter is 52 feet.\n\n14. ## science\n\nMost of the eastern United States and the West Coast have a climate that is- F)mountain H)desert G)polar J)temperate My answer is temperate, am i right if not please tell me why?\n\n15. ## Maths\n\nAn ice cream store sells 20 different flavors of ice cream and 3 different types of cones. How many different combinations of cone and ice cream flavor are possible, if customers are permitted to choose only one flavor of ice cream to be served in one of\n\n16. ## psychology 220\n\n# 1.49 (1 pts.) Donna's clinician has requested that she undergo electroconvulsive therapy in an attempt to control her current, drastic suicidal impulses. What stage of treatment is she in? A) immediate management B) assessment of objectives C) assessment\n\n17. ## physics\n\nA particle starts from the origin with a velocity of 6.0 i m/s and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s^2 . at he instant the particle achieves maximum positive x-coordinate, how far is it from the origin?\n\n18. ## calculus\n\ndetermine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a,b) such that f(c) =f(b) - f(a) / b - a\n\na circle has the equation x^2+y^2+6x-6y-46=0 find the center and the radius of the euation. If there are any intercepts list them.\n\n20. ## psychology 220\n\n1.67 (1 pts.) Shelly meets twice a week with a number of individuals who also have similar problems. They openly share their problems with others in a trusting, receptive environment which also facilitates improvement of their interpersonal skills. Shelly\n\n21. ## Chemistry\n\nA certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound? I had some ideas as to\n\n22. ## college biology\n\nif you have a father who is type A heterozygote blood and married to a woman who is type B heterozygote blood, find all the children's possible blood types and ratios.\n\n23. ## Math\n\nthe ferris wheel at a carnival has a diameter of 18 m and descends to 2m above the ground at its lowest point. Assume that a rider enters a car at this point and rides the wheel for two revolutions\n\n24. ## Chemistry\n\nMagnetite, an iron ore with formula Fe3O4, can be reduced by treatment with hydrogen to yield iron metal and water vapor. (a) Write the balanced equation. (Use the lowest possible coefficients.) 1Fe3O4(s) + 4H2(g)-> 3Fe(s) + 4H2O(g) (b) This process\n\n25. ## math/algebra\n\nthe length of a rectangular frame is 5 cm more than the width. the area inside the frame is 66 square cm. find the width of the frame.\n\n26. ## physics\n\nThe figure below shows the circular wave fronts emitted by two sources. Make a table with rows labeled P, Q, and R and columns labeled r1, r2, Δr, and C/D. Fill in the table for points P, Q and R, giving distances as multiples of λ and indicating, with a\n\n27. ## calc\n\nWhich of the following statements would always be true? I. If f is differentiable at x = c, then f is continuous at x = c. II. If f is continuous at x = c, then f is differentiable at x = c. III. If f is not continuous at x = c, then f is not\n\n28. ## PSYchology 220\n\n# 1.59 (1 pts.) Ralph has been sad and listless for 3 years. He can't sleep and has lost his appetite. Using a diagnostic system, one clinician arrives at a diagnosis of depression. Using the exact same diagnostic system, another clinician arrives at a\n\n29. ## hi\n\nWhat is the sixth term of the arithmetic sequence if the 10th term is 33 and the 15th term is 53.\n\n30. ## college algebra-plz help\n\na circle has the equation 3(x-5)^2_3y^2=12. Find the center and radius and graph the circle. find the intercepts, if any. plz show work\n\n31. ## Psychology 220\n\n# 1.36 (1 pts.) Based on the incidence of Fetal Alcohol Syndrome in various ethnic groups, which of the following individuals would be considered high risk? A) a woman of French Canadian descent B) a woman of Irish descent C) a woman of Italian descent D)\n\n32. ## math\n\n3 boys and 3 girls,how many boy - girl pairs can be formed ?\n\n33. ## psychology 220\n\n# 1.56 (1 pts.) Mr. Kang, a recent immigrant from China, is extremely overwhelmed and distressed by the untimely death of his wife. His co-workers are concerned about his mental health because he believes he is in constant contact with her and can be seen\n\n34. ## algebra/math\n\nthe length of a rectangular flower bed is 7 feet less than 3 times its width. the area of the bed is 20 feet squared. find the dimensions of the flower bed.\n\n35. ## Math\n\nThe line L has equation of 4x + 3y = 7. Find the equation of the line perpendicular to L which passes through the point of intersection of L and the y-axis. Give answer in ax + by + c = 0.\n\n36. ## Intermediate Algebra\n\nA painter can paint a fence in 12 hours working alone. His apprentice who works slower can paint the same fence in 24 hours. How many hours would it take to do the job if they worked together?\n\n37. ## Statistics\n\nA number of grade school students were asked about their perceived math ability and the results are in the table below. Below Average = BA Average = A Above Average = AV BA A AV Girls 61 54 21 Boys 43 61 42 Test at the 0.01 level if there is a difference\n\n38. ## Intermediate Algebra\n\nYou are 6 feet in height. If you are standing away from the base of the lamppost, you can see where your shadow stops. If you are 10 feet from the base of the lamppost and your shadow stops 5 feet from where you stand, what is the height of the lamppost?\n\n39. ## Intermediate Algebra\n\nThe amount of current flowing into an electrical circuit varies inversely as the resistance in the circuit. When the resistance of a particular circuit is 5 ohms, the current is 42 amperes. what is the current when the resistance is 4 ohms?\n\n40. ## math\n\nI need to find the sum of 2/5 and 3/4. The LCD is 20 so I did 2x4=8 and 3x5=15. so I got 8/20 + 15/20 = 23/20, but this does not match any of the answers I have to pick from. What am I doing wrong?\n\n41. ## Chemistry\n\nIn the \"Methode Champenoise,\" grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is given below. C6H12O6(aq)--> 2 C2H5OH(aq) + 2 CO2(g) Fermentation of 746 mL grape juice (density = 1.0 g/cm3) is allowed to take place in a\n\n42. ## Physics\n\nThe (nonconservative) force propelling a 1.50E3 kg car up a mountain road does 4.70E6 j of work on the car. The car starts from rest at sea level and has a speed of 27.0 m/s at an altitude of 2.00E2 m above sea level. Obtain the work done on the car by the\n\n43. ## Chemistry\n\nA compound of carbon, hydrogen, and oxygen was burned in oxygen, and 2.00g of the compound produced 2.868 g CO2 and 1.567 g H2O. In another experiment, 0.1107 g of the compound was dissolved in 25.0 g of water. This solution had a freezing point of -0.0894\n\n44. ## Q Research\n\n2. The average commute time via train from the Chicago O'Hare Airport to downtown is 60 minutes with a standard deviation of 15 minutes. Assume that the commute times are normally distributed. What proportion of commutes would be: a. longer than 80\n\n45. ## calculus\n\nfind the points of inflection and discuss the concavity of the graph of the function f(x) = -sin x + cos X 0\n\n46. ## Physics\n\nAn elevator's cable is cut causing the elevator to slide down thhe elevator shaft on the emergency brakes and wheels. Suppose that the mass of the elevator and passengers were 1500 kg, and the frictional force (wheels on tracks)was 9000N Draw a force\n\nUsing the simplex method, solve the following linear programming problem: Maximize: P= 5x+2y Subject to: 4x+3y < or equal to 30 2x-3y < or equal to 6 x > or equal to 0, y > or equal to 0.\n\n48. ## calculus\n\nuse differentials to approximate the value of the expression: square root of 64.07\n\n49. ## physics\n\nHow much work is needed to lift a 400kg engine upward 1.5 meters out of a car?\n\n50. ## physics\n\nHow much work is needed to lift a 400kg engine upward 1.5 meters out of a car?\n\n51. ## physics\n\nHow much work is needed to lift a 400kg engine upward 1.5 meters out of a car?\n\n52. ## chemistry\n\n1)A solution has a [OH-] of 5.2 x 10-4. What is the [H3O+] in the solution? 2) A 0.25 M solution of a monoprotic acid, HA, has a pH of 2.54. What is Ka for this acid? The dissociation of HA is: HA + H2O H3O+ + A- (Hint: write the expression for pH) 3 A\n\n53. ## college algebra--need help please!!\n\nwrite the standard form of the equation and the general form of the equation of the circle with radius r and center (h,k) r=10; (h,k)=(8,-6)\n\n54. ## math\n\nfind the maximum and minimum values of f(x,y)=y-3x for feasible regions\n\n55. ## torque physics\n\nA bowling ball with a radius of 12 cm is given a backspin (i.e. if you picture the ball moving to the right it is spinning counterclockwise at the same time) of 8.3 rad/ s and a forward velocity of 8.5 m/s.As a result, the ball slides along the floor. What\n\n56. ## math\n\nif a point is equidistant from the endpoints of a segment,then the point lies on the perpendicular bisector of the segment.\n\n57. ## Psychology 220\n\n1.68 (1 pts.) In order to evaluate the effectiveness of a new psychotherapeutic technique, Dr. Schwartz is testing the therapy on clients that have sought treatment from him at his private practice. According to Martin Seligman, Dr. Schwartz is conducting\n\n58. ## Chemis\n\nSteve, in an attempt to impress his girlfriend, plops 18.2 grams of sodium metal into 500 mL of concentrated phosphoric acid (density = 1.88 g/cm3) in a beaker. The reactants were both initially at 25 oC. The reaction gently produced bubbles of hydrogen\n\n59. ## chemistry\n\nSteve, in an attempt to impress his girlfriend, plops 18.2 grams of sodium metal into 500 mL of concentrated phosphoric acid (density = 1.88 g/cm3) in a beaker. The reactants were both initially at 25 oC. The reaction gently produced bubbles of hydrogen\n\n60. ## Chemistry\n\nHow many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al(s) + 3FeO(s)-- 3Fe(s) + Al2O3(s)\n\n61. ## chemistery\n\nA student dissolves 196 g of H3PO4 (Molar mass = 98 g) in enough water to make one liter of solution. The molarity of the solution is ?\n\n62. ## chemistery\n\nA gas occupied 224 L at 25 c . At what temperature would this gas occupy 164 liters pressure constants\n\n63. ## psychology 220\n\n1.72 (1 pts.) Which of the following types of disorders would not meet the criteria for insanity according to the Insanity Defense Reform Act of 1984? A) schizophrenia B) major depression C) bipolar disorder D) antisocial personality disorder (my answer is\n\n64. ## analytic geometry\n\nWhat can be said of points with equal ordinates? with equal abscissa?\n\n65. ## combusion -science\n\nthe reaction order with respect to C4H10 =.15 and the reaction order with respect to O=1.6 , the pre exponential factor =4.16.10^9 . the activation energy =125000 kl/kaul. write out an expression for the rate of butane destruction\n\n66. ## CJS/250\n\nWill someone please help me find a website to research the following questions? Write a 750- to 1,050-word paper discussing the relationship between private and public police. Is this relationship positive or negative? Provide specific examples to support\n\n67. ## calculus\n\nfind the distanced(p_1&P_2) between points P_1 and P_2. P_1=(4,-5) and P_2=(3,4) plz show work\n\n68. ## Math\n\nWhat is the equation in point-slope form of the line that passes through the given point(1,3); M=5?\n\n69. ## calculus\n\nfind the midpoint of the line segment joining the points P_1 and P_2. P_1=(-2,6); P_2=(5,0) m=(x,y) please show work\n\n70. ## calculus\n\nfind the distance d(P_1 & P_1) between points P_1 and P_2. P_1=(4,-5)and P_2=(3,4) please show work\n\n71. ## pre-cal\n\nfind the intercepts. y=2x-2 show work\n\n72. ## English\n\n1. He spent a lot of money on dancing. 2. He spent a lot of money in dancing. 3. He spent a lot of money on buying things at the shop. 4. He spent a lot of money in buying things at the shop. 5. He spent a lot of money dancing. 6. He spent a lot of money\n\nA usefful personal and business investment site with in-depth detail on personal financial planning. After reading a March 19,2009 , article\"Preparing a Portfolio for Retirement,\" Arlene Supple, 47 years old, is evaluating her retirement portifolio. She\n\n74. ## English narrative\n\nHello and thanks for starting reading this post :) I have written a english narrative that has very very bad grammar and the rest... :D I would be very thankful if you could help me fix that narrative :) Here it is (BTW it is not a real story, I made it up\n\n75. ## Algebra 2..\n\nSolve. 1. (-16)^1/4 2. -16^1/4 *Please post some links so I can have extra practice.\n\n76. ## Algebra\n\nSolve the inequality and graph. 7-12 Is that correct?\n\n77. ## Social Studies\n\nWhat were Boomtowns and ghost towns? My history book dosn't describe it very well! Thanks for your help!\n\n78. ## math\n\npercent of change 20 is decreased to 11. What is the percent of change?\n\n79. ## Chemistry\n\nhow to prepare 100cm3 of 1.0moldm3 copper sulphate and from this prepare 100cm3 of 0.1 moldm3 solution.similarly use the 0.1moldm3 to prepare a 0.05 moldm3 ,and the 0.05moldm3 solution to prepare 0.01 moldm3 and 0.001moldm3solution.\n\n80. ## math steve\n\nsteve can you explain step by step please 8x=5y and 3y=14z then 12x= how many z's\n\n81. ## Chemistry\n\nHow much energy, in joules, must be absorbed to convert to Na+ all the atoms present in 1.00mg of gaseous Na? The first ionization energy of Na is 495.8 kJ/mol\n\n82. ## physics\n\nAn object is placed 5 cm in front of a concave lens of focal length 7 cm. Find the image location by drawing a ray tracing diagram to scale.\n\n83. ## precalc h\n\nsin(theta)/1-cos(theta) + 1-cos(theta)/sin(theta) = 2csc(theta) That question makes absolutely no sense.. could someone help me? or lead me in the direction to figuring it out? and.. (Beside the trig functions is theta) 1+1/cos = tan^2/sec-1\n\n84. ## math\n\nthere are 4 students sitting to receive their prizes. Alan and Julia wanted to sit together. how many different arrangements of students would there be\n\n85. ## math\n\n25 is increase to 100 What is the percent of change?\n\n86. ## maths\n\nLet a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .\n\n87. ## Math(help)\n\nwhen 120 guests take seat in an auditorium,only 34 of the seats occupied.what is the total number of seats in the auditorium?\n\n88. ## Ms.Sue\n\nThe Secular Student Alliance will resume their meetings on January 13th, 2013. Again at the Bernhard Center in the Martin Luther King room 214 at 4:00 PM. Erdman says big their primary wish is to bring more people into the group, even those who are not\n\n89. ## Statistics of Psychology\n\nCreate a numerical and verbal hypothesis statement: People who are cyber bullied have, for the most part, been bullied in more traditional manners as well.\" Their research showed that the incidence increased over time, with 23 cases (56%) taking place\n\n90. ## algerbra\n\nKaty has a rectangle field that is 700 feet wide one portion of the field is 2x long the other portion is 14y. Use the distributive property to find an expression for the area of the field.\n\n92. ## English\n\nPosted by rfvv on Sunday, December 9, 2012 at 8:06pm. 1. He spent a lot of money on dancing. 2. He spent a lot of money in dancing. 3. He spent a lot of money on buying things at the shop. 4. He spent a lot of money in buying things at the shop. 5. He\n\n93. ## math\n\nHow much stock solution of hydrogen peroxide 8% will you need to make 320ml of hydrogen peroxide 2% solution?\n\n94. ## math\n\nwe decided on a new logo for our brand of shirts.It was to be 3 stripes. How many different logos we make, if we used 3 colors\n\n95. ## Algebra\n\nFind the real solutions of the equation by graphing. 6x^3-2x=-9x^2\n\n96. ## college physics\n\na bullet of mass 8g is fired into a block of mass 260g that is initially at rest at the edge of a table of height 1m the bullet remains in the block and after the impact the block land d=2.20m for the bottom of the table. determine the speed of the bullet\n\n97. ## physics\n\nhow much energy would it take to melt 250kg of lead assuming it starts at 30 degrees centigrade.\n\n98. ## psychology 220\n\n# 1.55 (1 pts.) Which of the following information led Dr. Tobin to conclude that Peter was suffering from bipolar disorder? A) the fact that he reported having intense bouts of anxiety B) the fact that he had recently been married and soon after was\n\n99. ## english\n\ndoes this sound and look grammatically correct? The ruined city of Troy looked to be a catastrophic place. There was blood everywhere. Men’s bodies were cut open and infants were crying. The whole city looked as Poseidon, the god of earthquakes, gave\n\n100. ## Chemistry\n\nAssume that 57.0 cal of heat is applied to a 28g sample of sulfur at 27 degrees C. What is the final temperature of the sample if the specific heat of sulfur is 0.175 cal/g degrees C?\n\n101. ## Chemistry\n\nThe Ksp value for FeS is 6 x 10^-18 M. FeS(s) --> Fe^(2+)(aq) + S^(2-)(aq) What is DG at 25 degrees C for the reaction?\n\n102. ## Chemistry\n\nFluorine reacts with oxygen to yield oxygen difluoride. 2 F2(g) + O2(g) 2 OF2(g) What is the value of K if the following concentrations are found at equilibrium: [O2] = 0.200 mol/L, [F2] = 0.0100 mol/L, and [OF2] = 0.0633 mol/L\n\n103. ## Chemistry\n\nFluorine reacts with oxygen to yield oxygen difluoride. 2 F2(g) + O2(g) 2 OF2(g)K = 2.00x 10^2 (a) Calculate [O2] at equilibrium when [OF2] = 0.25 mol/L and [F2] = 0.0208 mol/L. mol/L (b) Calculate [OF2] at equilibrium when [F2] = 0.088 mol/L and [O2] =\n\n104. ## SOCIALS HELP...THANX\n\nThis question is based on: the Reign of Terror which government replaced Robespierre and what did that signify for France? Pleese help... Thanx\n\nSome critics of corporate social responsibility view spending money on CSR as:\n\n106. ## algebra\n\nA family refers to a group of two or more people related by birth, marriage, or adoption who reside together. In 2000, in Country A, the average family net worth was \\$460,000, and there were about 7.6*10 to the seventh power families. Calculate the total\n\n107. ## algebra\n\ny - 2 = -2(x - 3)\n\n108. ## Statistics of Psychology\n\nCreate a numerical and verbal hypothesis statement: People who are cyber bullied have, for the most part, been bullied in more traditional manners as well.\" Their research showed that the incidence increased over time, with 23 cases (56%) taking place\n\n109. ## Social Studies\n\nWhat did railroads contribute to the development of the west? Pls Help! Soical studies is my worst subject ever! Thanks if you can help!\n\n110. ## Life science\n\nHow long could it take an ecosystem to reach climax community in a primary succession? And a secondary succession?\n\n111. ## british lit\n\nwhat are the ALLUSIONs in the book Frankenstein? I need help with a list of seven of them.\n\n112. ## chemistry\n\nHow many gallons of C6H12 , measured at 20 degrees Celsius, must be burned to provide enough heat to warm 26.4m3 of water from 17.4 degrees Celsius to 32.8 degrees Celsious , assuming that all the heat of combustion is transferred to the water, which has a\n\n113. ## Statistics\n\nA tax specialist knows that tax refunds for people who use a firm to do their tax returns are normally distributed with a mean of \\$150 and a standard deviation of \\$43. The specialist believes that the company, Taxco, uses under trained staff and gets a\n\n114. ## Statistics\n\n90% of the world’s large corporations are actively involved in data warehousing. In a random sample of 10 large corporations, what is the probability that at least 8 of them are actively involved in data warehousing.\n\n115. ## Statistics\n\nA provincial politician would like to determine the average amount earned during summer employment by the provinces teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within \\$25 of the actual\n\n116. ## Calculus\n\nLet f be a continuous function on the interval [0,2] which satisfies integral of f(x)dx=5 when b is 2 and a is 0. Give this information compute the definite integral f(2y)dy when b is 1 and a is 0. Solve this\n\n117. ## Statistics\n\nBottles of a popular cola contain a mean amount of 300 mL of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is normal with a standard deviation of 3 mL. a) What\n\n118. ## Economics\n\nn auto-service establishment has estimated its monthly cost function as follows: TC = 6000 + 10 Q where Q is the number of cars it services each months and TC represents its total cost. The firm is targeting 35,000 net monthly profit servicing 2000 cars.\n\n119. ## Statistics\n\n18 workers at a large assembly plant were asked the time it took to commute to work. The data was analyzed in Minitab and the Descriptive Statistics for the data appears below. Descriptive Statistics: Minutes Variable N Mean Median TrMean StDev SE Mean\n\n120. ## physics\n\nA U-shaped tube contains a fluid of density 1.5x10^3 kg/m^3. the right side of the tube is closed, with vacuum above the fluid. The left side of the tube is open to the atmosphere. Assume the atmospheric pressure outside the tube is exactly 10^5 Pa. a) How\n\n121. ## Statistics\n\nCigarette companies advertise that the mean amount of nicotine in each cigarette is 1.4 milligrams. A health advocacy group believes that the amount of nicotine is greater than the advertised amount. They hire a consultant who collects a sample of size 31\n\n122. ## physics\n\nA 20.0kg solid gold statue is raised from the sea bottom. What is the tension in the hoisting cable (assumed massless) when the statue is accelerating upward at 2m/s^2? (assume the statue is still completely underwater) The density of gold is\n\n123. ## Social Studies\n\ntrue or false: many cowhands on the cattle drives were chinese imigrants, african americans,or veterans of the confederate Army. I guessed false and instead of chinese imigrants, i think it's Vaqueros? but i don't know?\n\n124. ## stem cells\n\nStem cells may open the door to discoveries about many diseases and may allow treatments for diseases to be developed. However many ethical issues surround stem cell research which need to be addressed by scientists, governments, etc. a. Explain why\n\n125. ## Statistics\n\nThe data below represents six years of the harvest of oranges in millions of boxes and the price per 75 pound box. (a) Plot the data and calculate the regression line. Plot that line. (b)Calculate the coefficient of determination (c)What percent of the\n\n126. ## poli sci\n\nWhich of the fundamental American political values does the \"Tea Party\" most fundamentally support? Do you believe the \"Tea Party\" is a legitimate heir of the Founding Father's protest against the British? Do you believe the Tea Party will transform\n\n127. ## english\n\ndoes this sound right with grammatical errors too Odysseus sent three men to see what the land of the Lotus Eaters, who live on the flower, is about. The three men had gotten fed the Lotus and now they had no desire to go back to the ship or go back home.\n\n128. ## Intermediate Algebra\n\nin still water, a paddle boat averages 10 miles per hour. It takes the boat the same amount of time to travel 36 mile downstream, with the current as 24 miles upstream, against the current. What is the rate of the water's current?\n\n129. ## chemistry\n\nIf you add 2 liters of .45% NaCl to 36 liters of blood that has a pH balance of 7.44, what will the new pH be?\n\n130. ## Intermediate Algebra\n\nA car travels 30 miles in the same time that a car traveling 5 miles per hours faster travels 45 miles. What is the rate of each car?\n\n131. ## Math\n\n\\$5000 for 4yrs at 3.2% compounding monthly Find the amount of interest earned?\n\n132. ## Physics\n\nA(n) 80.0-kg student eats a 160-Calorie doughnut. To burn-it-off, he decides to climb the steps of a tall building. How high (in m) would he have to climb to expend an equivalent amount of work?\n\n133. ## Economics\n\nAssuming a constant marginal cost, a lower price elasticity of demand would call for a relatively lower mark-up ration.\n\n134. ## chemistry\n\nIf you add 2 liters of .45% NaCl to 36 liters of blood that has a pH balance of 7.44, what will the new pH be?\n\n135. ## Intermediate Algebra\n\nYour gas bill varies directly as the amount of gas used. The bill for 2800 cubic feet of gas comes to \\$196. If you used 1400 cubic feet of gas, what will the bill be?\n\n136. ## Social Studies\n\nThese are multipal choice: According to the (Dawes Act,or Fort Laramie Treaty), Indians would receive money,domestic animals,tools,and land that would be theirs forever.\n\n137. ## biology\n\nStem cells may open the door to discoveries about many diseases and may allow treatments for diseases to be developed. However many ethical issues surround stem cell research which need to be addressed by scientists, governments, etc. a. Explain why\n\n138. ## Social Studies\n\nThese are mulitpal choice: Reforms failed because Indians did not want to become (farmers, or miners) and reservation life changed their culture.\n\n139. ## Physics\n\nA force compresses a bone by 0.10mm. A second bone has the same length but twice the cross-sectional area as the first. By how much would the same force compress the bone?\n\n140. ## psychology\n\nRecent neuroimaging research indicates that people with OCD show increased levels of grey matter in the _____ and the _____ relative to those who do not have OCD. 1)splenium; corpus collosum 2)locus ceruleus; hypothalamus 3)thalamus; left frontal cortex\n\n141. ## psychology\n\nMarla is talking to a friend and spreading a rumor that she heard about a co-worker. When the friend asks where she heard the rumor, Marla cannot remember. Marla is experiencing what psychologists call 1)retrograde amnesia. 2)anterograde amnesia. 3)source\n\n142. ## physics\n\nA child and sled with a combined mass of 58.9 kg slide down a frictionless hill that is 8.09 m high at an angle of 23◦ from horizontal. The acceleration of gravity is 9.81 m/s2 . If the sled starts from rest, what is its speed at the bottom of the hill?\n\n143. ## geometry math\n\nthree teenagers can eat 6 pizzas in half an hour. Assuming that their friends eat at the same rate, how many pizzas can 5 teenagers eat in 90 minutes?\n\n144. ## Math\n\nA curve has equation y= 2 + x - x^2. Find coordinates of 3 unknown points on the line. 1st point is at negative x, y value. 2nd point is at x=0 and y is positive. 3rd value is at y=o and x is positive.\n\n145. ## geometry math\n\nhow many different triangles can be traces along the line segments in the diagram below?\n\n146. ## Dynamics\n\nA 20-lb disk has a radius of 1ft and can roll without slip on a rough plane that is inclined by an angle è to the horizontal. The disk is attached to the lower end of a linear spring of stiffness 10lb/in. The upper end of the spring is attached to a wall.\n\n147. ## engines\n\n1. Look at the automotive engine shown in the figure below. Auto Technician A says that the each of the camshafts in this engine makes two revolutions for every one revolution of the crankshaft. Auto Technician B says that the camshafts in this engine\n\n148. ## psychology\n\nhow can i explain racism in relations to cognitive perspective?\n\n149. ## Math\n\nLet f(x) = x - 5 and g(x) = 1/2x^3. Find g^-1(108). Write an expression for (f•g)(x). Solve equation (f•g)(x)= 27.\n\n150. ## psychology 220\n\n1.73 (1 pts.) Dr. Strickland has neglected to notify Gary, one of his newest clients, about the potential risks of therapy and the limits of confidentiality during therapy sessions. Dr. Strickland has not provided him with the information needed for Gary\n\n151. ## Math\n\na. Express f(x)=x^2 - 6x + 14 in the form f(x)=(x-h)^2 + k, where h and k are to be determined. b.Hence, or otherwise, write down the coordinates of the vertex of the parabola equation y=x^2 - 6x + 14.\n\n152. ## physics\n\n)A block with mass m = 11.3 kg slides down an inclined plane of slope angle 43.4 ° with a constant velocity. It is then projected up the same plane with an initial speed 1.45 m/s. How far up the incline will the block move before coming to rest?\n\n153. ## physics\n\n)A block with mass m = 11.3 kg slides down an inclined plane of slope angle 43.4 ° with a constant velocity. It is then projected up the same plane with an initial speed 1.45 m/s. How far up the incline will the block move before coming to rest?\n\n154. ## Social Studies\n\nWhat was the purpose of the Homstead Act? Did congress achieve this goal? I got what the homstead act was, but i'm confused on the second part. I put that congress did not achieve there goal because only about 20% of the homstead land went directly to\n\n155. ## calc-tangent line\n\nfind the equation to the line tangent to y=(x+3)^(1/3) at x=-3.\n\n156. ## english\n\ndoes this sound write and look grammatically correct? Please and thank you Odysseus the Curious took us inside a cave to eat. After we ate some cheeses the ugly Cyclops came along with his flock of sheep and rams. He put a gigantic boulder to block the\n\n157. ## math\n\nA kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second. Write an equation that gives the height (in feet) of the football as a function of the time (in seconds) since it was punted. Find the height (in feet)\n\n158. ## algebra\n\n90 degress= ? radians How is the answer pie over 2? 330 degrees = 11pie over 6 -105 degrees= -7pie over 12 Converting the angle in radians to degrees 3pie over 4 =135 degrees pie over 2 = 90 degrees pie over 30 = 6 degrees -7pie over 5 =-135 degrees I have\n\n159. ## algebra\n\n90 degress= ? radians How is the answer pie over 2? 330 degrees = 11pie over 6 -105 degrees= -7pie over 12 Converting the angle in radians to degrees 3pie over 4 =135 degrees pie over 2 = 90 degrees pie over 30 = 6 degrees -7pie over 5 =-135 degrees I have\n\n160. ## trigonometry\n\nA 100-foot vertical antenna is on the roof of a building. From a point on the ground, the angle of elevation to the top and the bottom of the antenna are 51 degrees and 37 degrees, respectively. Find the height of the building to the nearest foot. How\n\n161. ## math\n\na computer store sold a total of 300 items last month. the store sold six times as amny hard drives as they did CD-ROM drives, andhalf as many floppy drives as hard drives. based on this information, how many of each items were sold\n\n162. ## psychology\n\nFred was diagnosed as schizophrenic at age 27. According to the information presented in your textbook, what is a probable description of his life at age 68? a)He has been institutionalized for 41 years. b)He may have shown improvement or even recovery.\n\n163. ## algebra\n\nof radius 54 meters subtended by the central angle 1/9 radian. s(arc length = meters How is the answer 6? s denotes the lenght of the arc of a circle of radius r subtended by the central angle 0. Find the missing quantity. The radius r of the circle is ?\n\nasked by Find the length s of the arc of a circle\n164. ## calc\n\nFind the slope of the tangent line to the graph y = x-3 at the point (0.5, 8)\n\n165. ## GENERAL\n\nWhich of the following would you use to sort through slides for a slide presentation? A. Camera B. Opaque Projector C. Copy machine D. Light box\n\n167. ## calc\n\nfind derivative d/dt of sec * square root of t\n\n168. ## psychology\n\nBy dissociating key parts of personality, people are able to keep ___________ from reaching conscious awareness, thereby reducing their anxiety. a)symptoms b)compulsions c)confused thoughts d)disturbing memories\n\n169. ## Calc\n\nIf f (3) = 9 and f ´(3) = 4, find the tangent line to the graph of f when x = 3\n\n170. ## math\n\nmultiply the fraction 9/15 by the fraction 30/72 and reduce your answer to the lowest terms\n\n171. ## general\n\nWhich of the following would you use to sort through slides for a slide presentation? A. Camera B. Opaque Projector C. Copy machine D. Light box\n\n172. ## english\n\ndoes this sound write and look grammatically correct? Please and thank you(nohbdy is suppose to be spelled that way) Odysseus was brave from the rest of the crew so he led us into fooling Cyclops. He gave some wine to Cyclops to get him drunk. Then\n\n173. ## PSYchology 220\n\n1.60 (1 pts.) \"Coughing, sniffling, sneezing, runny nose, achiness, scratchy throat, and inability to sleep.\" This collection of symptoms represents a A) constellation. B) diagnosis. C) prognosis. D) syndrome. (MY answer is D) AM I RIGHT?\n\n174. ## calculus\n\nfind the equation of line L. L is parellel to y=4x show work pleaase..\n\n175. ## Algr 1\n\nWhen your graphing linear equations the rate of change can be the same as?\n\n176. ## Chemistry\n\nA 0.75–L bottle is cleaned, dried, and closed in a room where the air is 22oC and 44% relative humidity (that is, the water vapour in the air is 0.44 of the equilibrium vapour pressure at 22oC). The bottle is brought outside and stored at 0.0oC. a) What\n\n177. ## Science\n\nHow do you prepare 500mL of a 2.5M solution of NaOH from the solid compound?\n\n178. ## Ms.Sue 1 more time\n\nThe Secular Student Alliance will resume their meetings on January 13th, 2013. The meetings will take place at the Bernhard Center in the Martin Luther King room, room 214 at 4:00 PM. Erdman says that he hopes the group will continue to grow, even if you\n\n179. ## Chemistry\n\nAt a local convenience store, you purchase a cup of coffee, but, at 98.4oC, it is too hot to drink. You add 23.0 g of ice that is –2.2oC to 248 mL of the coffee. What is the final temperature of the coffee? (Assume the heat capacity and density of the\n\n180. ## Math Definition Algr\n\nWhen your graphing linear equations the rate of change can be the same as?\n\n181. ## PSYchology 220\n\n# 1.61 (1 pts.) Dr. Johnson believes that the frequency and intensity of compulsive behavior exists on a continuum from mild to severe. Dr. Johnson prefers to use which approach when viewing this behavior? A) categorical B) dimensional C) medical D)\n\n182. ## Intermiate Algebra\n\nReduce the Rational expression to lowest terms: 3x2-9x+6/10-5x\n\n183. ## maths --plse help me..\n\nLet a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .\n\n184. ## chemistry\n\nA 0.200g sample containing mn02 was dissolved and analyzed by addiction of 50ml of 0.100M fe to drive the reation. The excess fe required 15ml of 0.0200 m kmn04. Find % mn304 fw 228.8mg/mmol in the sample? who helps me\n\n185. ## psychology 220\n\n# 1.62 (1 pts.) A psychologist suspects that Dan's psychotic episode may be related to the recent death of his wife. On which DSM-IV Axis would this stressful event be recorded? A) Axis I B) Axis II C) Axis III D) Axis IV (My answer is D) Am i right?\n\n186. ## Keyboarding and word processing\n\nIf you're compling a list of sources you used for your report in MLA style, the list is called a A. references list B. bigliography C. work-cited list. D. source list. I think the answer is \"C\"\n\n187. ## math\n\n320ml of active ingredient is added to 680ml of an alcohol base solution. what is the final strength (v/v)?\n\n188. ## Q. Research\n\n3. Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15, and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob?\n\n189. ## math\n\nSue was in excellent shape in high school, she could ride her 10-speed bicycle 4.8miles to her grandmother's house in 11 minutes. Her return trip home took 14 minutes. The times are different due to there being a slight downhill slop on part of the trip to\n\n190. ## Calculus\n\nFind the critical numbers of f, find the open intervals on which the function is increasing or decreasing and use the first derivative test to identify all relative extremes f(x) = 3 + cos^2 x 0\n\n191. ## calculus\n\noptimization find the point on the graph of the function that is closest to the given point f(X)= square root of x point:(8,0)\n\n192. ## CALCULUS\n\na bacteria culture grows with a constant per capita growth rate. After 2 hours there are 500 bacteria and after 6 hours the count is 312500, find the initial population and the population after 8 hours\n\n193. ## English-Grammar\n\nWhich of the following is an example of an adverbial clause fragment? A.wherever there is a chance for a fabulous meal B.who described the book as a compelling masterpiece C.that the poet choose words carefully to convey a particular tone I think the\n\n194. ## Psychology 220\n\n1.63 (1 pts.) Robert has a serious drinking problem, but his therapist feels that the drinking is a result of Robert's constant bouts with depression. Based on this information, what would Robert's principle diagnosis be? A) alcoholism B) bipolar disorder\n\n195. ## Physics\n\nA truck is moving 40 m/s when it encounters a ramp inclined by 30% above the horizontal. Gravel provides a frictional force to slow the truck and has a coefficient of friction(.5). How far along the ramp would the truck travel before coming to a stop?\n\n196. ## calculus\n\nfind the differential dy of the given function y=3x-4 sec(2x-1)\n\n197. ## Calculus\n\nlim as x->infinity [(x^3+x^2)^1/3] - [(x^3-x^2)^1/3]\n\n198. ## Psychology 220\n\n1.60 (1 pts.) \"Coughing, sniffling, sneezing, runny nose, achiness, scratchy throat, and inability to sleep.\" This collection of symptoms represents a A) constellation. B) diagnosis. C) prognosis. D) syndrome. (MY answer is B) AM I RIGHT?\n\n199. ## calculus\n\nI have two questions please. 1. find an equation for the line with the given properties. Express in slope-inntercept form. 2. find the equation of a line that is perpendicular to the line y=1/2x+4 and contains the point (-3,0) please show work\n\n200. ## calculus\n\nI have two questions please. 1. find an equation for the line with the given properties. Express in slope-inntercept form. 2. find the equation of a line that is perpendicular to the line y=1/2x+4 and contains the point (-3,0) please show work\n\n201. ## physics\n\npls explain this to me ? \"momentum is inertia....but inertia is NOT momentum\" :S confused\n\n202. ## Algebra!!\n\n(x-1)/(x-3)/1/x2-x-6-4/(x+2)\n\n203. ## Psychology 220\n\n# 1.55 (1 pts.) Which of the following information led Dr. Tobin to conclude that Peter was suffering from bipolar disorder? A) the fact that he reported having intense bouts of anxiety B) the fact that he had recently been married and soon after was\n\n204. ## SCIENCE\n\nAT WHAT HEIGHT ABOVE THE EARTH'S SURFACE DOES THE ACCELERATION DUE TO GRAVITY FALL TO 1% OF ITS VALUE AT THE EARTH SURFACE\n\n206. ## MATH!!!!!!!!!!!! I need help please\n\nSimplify Complex fraction: (x-1) ----- (x-3) --------- 1 4 ---- - ---- x2-x-6 (x+2)\n\n207. ## math pre-calc h\n\nIt says which expressiion is equivalent to sec T + csc T / 1+tan T a. sin t b. cos t c. tan t d. csc t I'm not sure what to do first to find it out.\n\n208. ## physics\n\nA ferry boat is 4.9 m wide and 6.9 m long. When a truck pulls onto it, the boat sinks 3.34 cm in the water. What is the weight of the truck? The acceleration of gravity is 9.81 m/s2 .\n\n209. ## Psychology 220\n\n1.64 (1 pts.) Sangue dormido, or \"sleeping blood\" is a condition observed among individuals from the Cape Verde Islands (located off the west coast of Africa) and involves paralysis, convulsions, blindness, and tremors. This condition is referred to in\n\n210. ## Physics\n\nIf a sphere and paper are dropped from 10 meters and the sphere hit the ground in 1.4 seconds and the paper hit in 2 seconds, why did the impact time in case of the piece of paper change in comparison to the sphere? A. a=(Wt-R)/m;a=g B. a=(Wt-R)/m;a>g C.\n\n211. ## math\n\nthe ph of a solution is 3.6 what is it hydrogen ion concentration\n\n212. ## Chemistry\n\nI need help with this problem A 0.75–L bottle is cleaned, dried, and closed in a room where the air is 22oC and 44% relative humidity (that is, the water vapour in the air is 0.44 of the equilibrium vapour pressure at 22oC). The bottle is brought outside\n\n213. ## physics\n\nAn elevator weigting 400 kg/s is to be lifted up at a constant velovity of 3cm/s.what would be the minimum power of the motor to be used ?\n\n214. ## trigonometry\n\nProve cotx-1/cotx+1=1-sin2x/cos2x\n\n215. ## phicicks\n\nA 0.300- kg brass block is attached to a spring of negligible mass. What is the value of the spring constant if the vibrational frequency is 0.855Hz?\n\nI can get part of the answer then I get stumped. find the center and radius of the circle, Write the standard form of the equation. (4,3)(1,3) to get center: (1+4/2, 3+3/2)=5/2,3 answer:5/2,3 for center radius: then i am stuck\n\n217. ## psychology 220\n\n# 1.66 (1 pts.) James is a person with schizophrenia and has delusions that the CIA is after him. He believes that everyone on the street is trying to kill him and, as a matter of fact, he has been brought to a psychologist by the police after being\n\n218. ## calculus\n\nfind the center (h,k) and radius r of the circle and then use these to (a)graph circle, (b) find intercepts if any\n\n219. ## math\n\nsolve by completing the square x^2-3x=4 please show work\n\n220. ## Biology\n\nThe half-life of Carbon-14 is 5730 years. What is the age of a fossil containing 1/16 the amount of Carbon-14 of living organisms? Explain your calculation.\n\n221. ## English\n\nIdentify the nouns and verb/verb phrases in the sentence. Employers value honesty and hard work. Nouns: Employers, honesty, and work Verb: Value\n\n222. ## algebra\n\n-4r^2+21r=r+13 3s^2-26s+2=5s^2+1 Please, please could you be so nice and help me with step by step, I can't find right solution, Thank you so much!!!!\n\n223. ## chemistry\n\nWhat volume in mL of 0.0985 M sodium hydroxide solution is required to reach the equivalence point in the complete titration of a 15 mL sample of 0.124 M phosphoric acid?\n\n224. ## math\n\nAccording to the statistics of a famous high school in Canada, all their graduates went to either A or B for university study. Following are the proportions of graduates who went to the two universities. University Proportion A 1-k B k Suppose k = 0.4 (a)\n\n225. ## Chemistry\n\nWhat is true for water? A) In a phase diagram the solid-liquid coexistance line has a negative slope. B) In a phase diagram the solid-liquid coexistance line has a positive slope. C) The solid is more dense than the liquid. D) The solid is less dense than\n\n226. ## Calculus\n\nfind the derivative (df/dx) of the squareroot of sin4x\n\n227. ## math\n\nIf a point is equidistantf rom th€ €ndpointso f a segm€ntt,h en thc poinl lies on thc perpendicular bisector of the segment\n\n228. ## Music\n\ni have a test tommorrow and i need help! What is that one note called that has 4 beats?\n\n229. ## Chemistry\n\nExplain in step by step and provide calculations to show how you can prepare 1.50 L of 0.10 M H2SO4 solution starting with 16 M H2SO4\n\n230. ## physics\n\nA truck travels up a hill with a 10 ◦ incline. The truck has a constant speed of 21 m/s. What is the horizontal component of the truck’s velocity? Answer in units of m/s What is the vertical component of the truck’s velocity? Answer in units of m/s\n\n231. ## Calculus\n\nSuppose a reflector is attached to a 26 inch diameter bike wheel, 8 inches from the center. If the bike wheel rolls without slipping, find parametric equations for the position of the reflector. I thing i use parametrics????\n\n232. ## math\n\nA boat travels down the river in 2.8 hours. The return trip up river takes 3.8 hours. A one way trip down or up river is a distance of 24 miles. If the boat is traveling at maximum speed, what is the boats speed measured in waters with no current? You need\n\n233. ## Chemistry\n\nWhen 100 mL of Ba(NO3)2 solution at 25 degrees Celsius is mixed with 100 mL solution CaSO4 solution at 25 degrees Celsius in calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 degrees Celsius. Assuming that the\n\n234. ## Math\n\nA ladder leans against a vertical wall and the top of the ladder is sliding down the wall at a constant rate of 1/2 ft/sec. At the moment when the top of the ladder is 16 feet above the ground, the bottom of the ladder is sliding away from the wall\n\n235. ## Science\n\nConvert 150g Hydrogen Fluoride (Molecular Weight 20) to moles\n\n236. ## Calculus\n\nEvaluate the limit as h approaches zero of (the cubed root of 8-h)-2/h. Can also be written ((8-h)^1/3)-2)/h\n\na Geiger ionization chamber contains 60 cubic cm of air and records 500 mR/min. calculate (a). the current produce by the chamber (b). the total number of ion pairs produced per hour of exposure. please answer, thank you!,\n\n238. ## Astronomy\n\nEnter the expression for kinetic energy ( ) below. Use m_1 for mass and v for velocity.\n\n239. ## Math\n\nSue was in excellent shape in high school, she could ride her 10-speed bicycle 4.8miles to her grandmother's house in 11 minutes. Her return trip home took 14 minutes. The times are different due to there being a slight downhill slop on part of the trip to\n\n240. ## Physics\n\nAn object of mass .550kg is lifted from the floor to a height of 3.5m at a constant speed. If the object is released from rest after it is lifted, what is its kinetic energy just before it hits the floor? What is its velocity? Show your work and units.\n\n241. ## math\n\nBernie put in a garden in her back yard. The garden has a perimeter of 272 feet. She wants to know the total surface area of the garden and all she knows is the perimeter and that the width is 20% or (1/5) of the length. (Remember 20% and the fraction 1/5\n\n242. ## Physics\n\nAn ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of 2m/s. Assuming that the only horizontal force that acts on her is a slight frictional force between the blades of the skates and the ice: a.) Identify the\n\n243. ## Math -urgent :)\n\nHello A straight line passes through the point (1,27) and intersects the positive x-axis at the point A and the positive y-axis at the point B.Find the shortest possible distance between A and B.. Pleaase could you help me? Thank you in advance\n\n244. ## Physics\n\nA plane is flying horizontally with speed 152 m/s at a height 2110 m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8 m/s 2 . Neglecting air resistance,A second package is thrown downward from the plane with a\n\n245. ## chemistry\n\nCalculate the number of moles of Al2O3 that are produced when 0.6 mol of Fe is produced in the following reaction. 2Al(s) + 3Fe(s)-- 3Fe(s) + Al2O3(s)\n\n246. ## Anthropology\n\nThe difference between a state and a cheifdom is?\n\n247. ## Chemistry\n\nHow many moles of glucose , C6H12O6, can be \"burned\" biologically when 10.0 mol of oxygen is available. C6H12O6(s) + 602(g)---6CO2(g) + 6H2O(l)\n\n248. ## General chemistry\n\nThe density of the vapor of a compound at 95 degrees C and 827 mmhg is 7.41 grams/liter. Calculate the molecular weight.\n\nfind the line with the given properties. Parellel to the line 7x-y= -7; containing rhe point (0,0) plz show work\n\n250. ## math\n\nShow that -1/2+ 1/2(i3 root) is a cube root of 1\n\n251. ## geography\n\nCan you check my answers please? 1. Describe North Dakota's geographic borders using cardinal directions? To the North is Saskatchewan and Manitoba, Canada. To the East, it is Minnesota. To the South, it is South Dakota. To the West, it is Montana 2.Which\n\n252. ## accounting\n\nBy Saturday, January 5, 2013, submit the following assignment: As a financial consultant, you have contracted with Wheel Industries to evaluate their procedures involving the evaluation of long term investment opportunities. You have agreed to provide a\n\n253. ## chemistry\n\nplease explain How % errors would affected the mass/moles of substances in lab?\n\n254. ## chemistery\n\nwhat pressure is exerted by 0.30 mole of CL2 in a 4.00L container at 20 degree C ?\n\n255. ## psychology 220\n\n1.69 (1 pts.) Dr. Ramirez is a clinical psychologist who treats individuals with somatoform disorders, yet he is still troubled by his own somatic complaints. Which of the following roles and responsibilities would be in question in this case? A) therapist\n\n256. ## maths\n\nLet a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .\n\n257. ## Chemisrty\n\nA 10.0 g sample of ice at -14.0°C is mixed with 124.0 g of water at 81.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J/g°C, respectively, and the\n\n258. ## physics\n\nhow much heat enrgy must be taken in 24 gram of water to reduce the temperature of 7 degrres celcius\n\n259. ## Calc-Derivative\n\nFind the derivative of x^(1/x). Can someone show me steps for this problem...I go round in circles\n\n260. ## kendriya vidyalaya\n\ncalculate the mass of the following 1)3.11* 10 to the power 23 atoms of o.\n\n261. ## chemistery\n\nin a solution of pH of 3.26, the molar concentration of the hydronium ion is ?\n\n262. ## chemistery\n\nthe mass percent of a solution prepared by dissolving 0.500 moll KCL ( Molar mass = 74.6 g) in 850 g of water is ?\n\n263. ## Physical Education\n\nWhat is the average weight for 8th grade girls?\n\n264. ## physics- TEST TMRW PLS HELP!\n\nw w w . m e d i a . c h e g g c d n . c o m / m e d i a / a c 3 / a c 3 b b 1 e 1 - 5 b a 8 - 4 1 9 e - 9 a 8 4 -a d 5 0 e 4 6 a 2 9 4 e / p h p s U O o L 3 .p n g (delete all spaces for link to work) Determine the magnitude of the resultant moment\n\n265. ## algebra\n\nX^2 * x+15 X^2+7 x\n\n266. ## psychology 220\n\nA) Alzheimer's clients suffer from sudden cognitive losses; clients with vascular dementia suffer from gradual losses. B) Alzheimer's clients suffer from global cognitive losses; clients with vascular dementia suffer from selective losses. C) Alzheimer's\n\n267. ## chemistery\n\nHow many milliliters of 4.00M NaOH wll react with 100 mL of 0.312 M H2SO4?\n\n268. ## algebra\n\nX^2 x+15 ----- * ------ = X^2+7 x\n\n269. ## psychology 220\n\n1.71 (1 pts.) Wendy has been involuntarily committed to a psychiatric hospital. When she tries to use the ward pay phone, the staff members stand nearby and listen to her conversation. Occasionally they have actually interrupted her calls and have hung the\n\n270. ## psychology 220\n\n1.72 (1 pts.) Which of the following types of disorders would not meet the criteria for insanity according to the Insanity Defense Reform Act of 1984? A) schizophrenia B) major depression C) bipolar disorder D) antisocial personality disorder (my answer is\n\n271. ## PSYchology 220\n\n1.18 (1 pts.) Harry has recently been getting lost in the factory where he works. He is forgetting some of the names of his coworkers. He is having difficulty concentrating and his reading comprehension is poor. He denies having any memory problems but\n\n272. ## history\n\ncompare the status of Europe and the United States after World War I. How were they different financially, politically, and psychologically?\n\n273. ## Math\n\nIf a person throws a water balloon, the height (in feet) if the water balloon is modeled by the equation: h=-16t^2-5t+100, where t is in seconds. When will the water balloon hit the ground?\n\n274. ## 5th grade math (dividing fractions)\n\nHi.. I need help figuring out this problem, sovling for x: 3 divided by x/9 = 5 and 2/5 I was able to solve: x divided by 3/4 = 6 by multiplying 6 x 3/4 = 4 1/2. Just not sure how to solve the first problem. If I change 5 2/5 into mixed number,I still\n\n275. ## physics\n\nA grinding wheel of radius 0.370 m rotating on a frictionless axle is brought to rest by applying a constant friction force tangential to its rim. The constant torque produced by this force is 79.2 N • m. Find the magnitude of the friction force.\n\n276. ## psychology 220\n\n# 1.29 (1 pts.) Mrs. Johansen is pregnant and has accidentally been exposed to a person with German measles. She should be most concerned about the possibility of producing a child with mental retardation if she is in the A) fetal period. B) 35th week. C)\n\n277. ## Statistics\n\n3. A professor is interested in whether there is a difference in counseling students’ statistics competency scores among 1) those who have never taken any statistics course, 2) those who have only taken an undergraduate statistics course, 3) those who\n\n278. ## statics\n\nA 10kg block rest on top of a 50kg block. A force of 120n is being pushed on the 10kg block at an angle of 45 deg. What is the magnitude of frictional force between the blocks? The cofficient is force is .4\n\n279. ## physics- TEST TMRW PLS HELP!\n\nhow would i do this? -----------------A-----------B-------- Illl a carpenter carries a 6kg board . the downward force he feels on his shoulder (A) is .... ? (B is where his hand is) pls help!! i have so much difficulty with these questions!!!\n\n280. ## Trigonometry\n\nIM STUCK ON THESE :( 1. What is the equation for shifting the standard sine curve +2 units horizontally? A. y = sin (x + 2) B. y = sin x + 2 C. y = sin x − 2 D. y = sin (x − 2) 3. What is tan-¹ √3/3 ? A. π/4 B. -π/3 C. π/6 D. -π/4 4. cot–1\n\n281. ## calculus\n\nHow many gallons of gas is in a cylindrical tank (that is laying on it’s side, the circular bases are vertical, the diameter of the bases are 5 feet, the length of the tank is 8 feet, the gas is at a 2 ft level)?\n\nwhat are the most difficult aspects of simultaneously balancing the four perspectives in the balanced scorecard?\n\n283. ## calculus\n\nThe base of a cell phone tower is an equilateral triangle with sides 10 ft long. How far is one corner of the triangle from the center of the of the triangle?\n\n284. ## Calculus\n\nFind the equation of the tangent line to the graph of 4y^2− xy − 3 = 0, at the point P=(1,1)\n\n285. ## math\n\nA number has 2 digits. The digit in the tens place is 4 times bigger than the digit in the ones place. If the product of the digits is 16, what is the number ?\n\n286. ## calculus\n\nIf a bone is found to have 20% of its normal amount of carbon 14, how old is the bone?\n\n287. ## calculus\n\nIf a bone is found to have 20% of its normal amount of carbon 14, how old is the bone?\n\n288. ## algebra II\n\nexamples of sum and product of two numbers.\n\n289. ## trig\n\nAn airplane flies with a speed of 425 mph and a heading of 63°. If the heading of the wind is 24°and the speed of the wind is 31 mph, what is the heading of the plane and the ground speed? I've done a couple questions like this but still a little fuzzy\n\n290. ## Physics\n\nIf a sphere and paper are dropped from 10 meters and the sphere hit the ground in 1.4 seconds and the paper hit in 2 seconds, why did the impact time in case of the piece of paper change in comparison to the sphere? A. a=(Wt-R)/m;a=g B. a=(Wt-R)/m;a>g C.\n\n291. ## Physics\n\nPretend the hot brick extracts heat from the cold brick and becomes hotter. Would this violate the Law of Conservation of Energy?\n\n292. ## Math\n\nFind the value of each expression using the pemdas method. 5(to the second power)+8 divided by 2\n\n293. ## Calc-extrema\n\nFind all relative extrema in the function h(x)=cos(2pix/3) on the interval [0,2]\n\n294. ## English/reporting\n\nI need this paragraph proofread To test Erdman’s statement, Deja Guinn, a Christian and senior at WMU decided to join the SSA group. Guinn says unlike most of the group members, who are atheist agnostic, she is a Christian and believes in how Jesus\n\n295. ## Trigonometry\n\nStuck :( 1. Simplify (5-√-36)(2+√-1) A. 16 − 7i B. 16 + 7i C. 4 + 7i 2. Choose the point that lies on the curve r = 2 – 3 sin θ. A. (-5, 3π/3) B. (-2, π) C. (1, π/2) 3. Which of the following is not an approximate solution of x5 – 1 = 0? A.\n\n296. ## Statistics\n\nA tax specialist knows that tax refunds for people who use a firm to do their tax returns are normally distributed with a mean of \\$150 and a standard deviation of \\$43. The specialist believes that the company, Taxco, uses under trained staff and gets a\n\n297. ## statistics\n\nA provincial politician would like to determine the average amount earned during summer employment by the provinces teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within \\$25 of the actual\n\n298. ## Chemistry\n\nIf the heat from burning 6.000 g of C6H6 2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l) + 6542 kJ is added to 5691 g of water at 21 °C, what is the final temperature of the water? Please tell me what I am doing wrong: 293,551= 5691*4.2*(Tf-21C) 293551=\n\n299. ## statistics\n\n90% of the world’s large corporations are actively involved in data warehousing. In a random sample of 10 large corporations, what is the probability that at least 8 of them are actively involved in data warehousing\n\n300. ## Statistics\n\nBottles of a popular cola contain a mean amount of 300 mL of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is normal with a standard deviation of 3 mL. a) What\n\n301. ## calculus\n\nfind an equation for the line with the given properties. express the equation in slope-intercept form containing the points. P=(-5,4) and Q=(-4,2) plz show work\n\n302. ## statistics\n\n18 workers at a large assembly plant were asked the time it took to commute to work. The data was analyzed in Minitab and the Descriptive Statistics for the data appears below. Descriptive Statistics: Minutes Variable N Mean Median TrMean StDev SE Mean\n\n303. ## Chemistry\n\nConsider the molecular level representation of a gas There are: (a picture) 6 blue dots 5 orange dots 3 green dots (i think they are the diatomic one, because it is pairs) If the partial pressure of the diatomic gas is .750, what is the total temperature?\n\n304. ## English - ms. sue\n\nms. sue i writing an essay on : is it possible to live a peaceful/happy life in times of today? I going to write yes it be for this essay, but teacher say that write essay by saying WE, so what be other forms of that I use, I think she said don't use they,\n\n305. ## Chemistry\n\nAt constant volume, the heat combustion of a particular compound is -3447.0 KJ/mol. When 1.205 g of this compound (molar mass = 119.94 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 5.007\n\n306. ## Honors Physics\n\nIf the potential energy is 1962000 what is the kinetic energy of a boulder weighing 200 kg" ]

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[ "# Regularity of euler equations for a class of three-dimensional initial data\n\nAlex Mahalov, B. Nicolaenko, C. Bardos, F. Golse\n\nResearch output: Chapter in Book/Report/Conference proceedingChapter\n\n3 Scopus citations\n\n### Abstract\n\nThe 3D incompressible Euler equations with initial data characterized by uniformly large vorticity are investigated. We prove existence on long time intervals of regular solutions to the 3D incompressible Euler equations for a class of large initial data in bounded cylindrical domains. There are no conditional assumptions on the properties of solutions at later times, nor are the global solutions close to some 2D manifold. The approach is based on fast singular oscillating limits, nonlinear averaging and cancellation of oscillations in the nonlinear interactions for the vorticity field. With nonlinear averaging methods in the context of almost periodic functions, resonance conditions and a nonstandard small divisor problem, we obtain fully 3D limit resonant Euler equations. We establish the global regularity of the latter without any restriction on the size of 3D initial data and bootstrap this into the regularity on arbitrary large time intervals of the solutions of 3D Euler equations with weakly aligned uniformly large vorticity at t = 0.\n\nOriginal language English (US) Progress in Nonlinear Differential Equations and Their Application Springer US 161-185 25 https://doi.org/10.1007/3-7643-7317-2_13 Published - Jan 1 2005\n\n### Publication series\n\nName Progress in Nonlinear Differential Equations and Their Application 61 1421-1750 2374-0280\n\n### Keywords\n\n• Conservation laws\n• Fast singular oscillating limits\n• Three-dimensional Euler equations\n• Vorticity\n\n### ASJC Scopus subject areas\n\n• Analysis\n• Computational Mechanics\n• Mathematical Physics\n• Control and Optimization\n• Applied Mathematics\n\n### Cite this\n\nMahalov, A., Nicolaenko, B., Bardos, C., & Golse, F. (2005). Regularity of euler equations for a class of three-dimensional initial data. In Progress in Nonlinear Differential Equations and Their Application (pp. 161-185). (Progress in Nonlinear Differential Equations and Their Application; Vol. 61). Springer US. https://doi.org/10.1007/3-7643-7317-2_13" ]

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[ "# #0c4109 Color Information\n\nIn a RGB color space, hex #0c4109 is composed of 4.7% red, 25.5% green and 3.5% blue. Whereas in a CMYK color space, it is composed of 81.5% cyan, 0% magenta, 86.2% yellow and 74.5% black. It has a hue angle of 116.8 degrees, a saturation of 75.7% and a lightness of 14.5%. #0c4109 color hex could be obtained by blending #188212 with #000000. Closest websafe color is: #003300.\n\n• R 5\n• G 25\n• B 4\nRGB color chart\n• C 82\n• M 0\n• Y 86\n• K 75\nCMYK color chart\n\n#0c4109 color description : Very dark lime green.\n\n# #0c4109 Color Conversion\n\nThe hexadecimal color #0c4109 has RGB values of R:12, G:65, B:9 and CMYK values of C:0.82, M:0, Y:0.86, K:0.75. Its decimal value is 803081.\n\nHex triplet RGB Decimal 0c4109 `#0c4109` 12, 65, 9 `rgb(12,65,9)` 4.7, 25.5, 3.5 `rgb(4.7%,25.5%,3.5%)` 82, 0, 86, 75 116.8°, 75.7, 14.5 `hsl(116.8,75.7%,14.5%)` 116.8°, 86.2, 25.5 003300 `#003300`\nCIE-LAB 23.265, -29.141, 27.284 2.091, 3.878, 0.897 0.305, 0.565, 3.878 23.265, 39.921, 136.885 23.265, -19.652, 26.038 19.693, -15.51, 11.085 00001100, 01000001, 00001001\n\n# Color Schemes with #0c4109\n\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #3e0941\n``#3e0941` `rgb(62,9,65)``\nComplementary Color\n• #284109\n``#284109` `rgb(40,65,9)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #094122\n``#094122` `rgb(9,65,34)``\nAnalogous Color\n• #410928\n``#410928` `rgb(65,9,40)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #220941\n``#220941` `rgb(34,9,65)``\nSplit Complementary Color\n• #41090c\n``#41090c` `rgb(65,9,12)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #090c41\n``#090c41` `rgb(9,12,65)``\n• #413e09\n``#413e09` `rgb(65,62,9)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #090c41\n``#090c41` `rgb(9,12,65)``\n• #3e0941\n``#3e0941` `rgb(62,9,65)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #041403\n``#041403` `rgb(4,20,3)``\n• #082b06\n``#082b06` `rgb(8,43,6)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #10570c\n``#10570c` `rgb(16,87,12)``\n• #146e0f\n``#146e0f` `rgb(20,110,15)``\n• #188412\n``#188412` `rgb(24,132,18)``\nMonochromatic Color\n\n# Alternatives to #0c4109\n\nBelow, you can see some colors close to #0c4109. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #1a4109\n``#1a4109` `rgb(26,65,9)``\n• #154109\n``#154109` `rgb(21,65,9)``\n• #114109\n``#114109` `rgb(17,65,9)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #09410b\n``#09410b` `rgb(9,65,11)``\n• #09410f\n``#09410f` `rgb(9,65,15)``\n• #094114\n``#094114` `rgb(9,65,20)``\nSimilar Colors\n\n# #0c4109 Preview\n\nThis text has a font color of #0c4109.\n\n``<span style=\"color:#0c4109;\">Text here</span>``\n#0c4109 background color\n\nThis paragraph has a background color of #0c4109.\n\n``<p style=\"background-color:#0c4109;\">Content here</p>``\n#0c4109 border color\n\nThis element has a border color of #0c4109.\n\n``<div style=\"border:1px solid #0c4109;\">Content here</div>``\nCSS codes\n``.text {color:#0c4109;}``\n``.background {background-color:#0c4109;}``\n``.border {border:1px solid #0c4109;}``\n\n# Shades and Tints of #0c4109\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #020d02 is the darkest color, while #fbfefb is the lightest one.\n\n• #020d02\n``#020d02` `rgb(2,13,2)``\n• #061f04\n``#061f04` `rgb(6,31,4)``\n• #093007\n``#093007` `rgb(9,48,7)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #0f520b\n``#0f520b` `rgb(15,82,11)``\n• #12630e\n``#12630e` `rgb(18,99,14)``\n• #167510\n``#167510` `rgb(22,117,16)``\n• #198613\n``#198613` `rgb(25,134,19)``\n• #1c9715\n``#1c9715` `rgb(28,151,21)``\n• #1fa817\n``#1fa817` `rgb(31,168,23)``\n• #22ba1a\n``#22ba1a` `rgb(34,186,26)``\n• #25cb1c\n``#25cb1c` `rgb(37,203,28)``\n• #29dc1e\n``#29dc1e` `rgb(41,220,30)``\n• #36e22c\n``#36e22c` `rgb(54,226,44)``\n• #46e43e\n``#46e43e` `rgb(70,228,62)``\n• #57e74f\n``#57e74f` `rgb(87,231,79)``\n• #67e960\n``#67e960` `rgb(103,233,96)``\n• #78eb71\n``#78eb71` `rgb(120,235,113)``\n• #88ee82\n``#88ee82` `rgb(136,238,130)``\n• #99f094\n``#99f094` `rgb(153,240,148)``\n• #a9f3a5\n``#a9f3a5` `rgb(169,243,165)``\n• #baf5b6\n``#baf5b6` `rgb(186,245,182)``\n• #caf7c7\n``#caf7c7` `rgb(202,247,199)``\n``#dafad9` `rgb(218,250,217)``\n• #ebfcea\n``#ebfcea` `rgb(235,252,234)``\n• #fbfefb\n``#fbfefb` `rgb(251,254,251)``\nTint Color Variation\n\n# Tones of #0c4109\n\nA tone is produced by adding gray to any pure hue. In this case, #232723 is the less saturated color, while #044a00 is the most saturated one.\n\n• #232723\n``#232723` `rgb(35,39,35)``\n• #202a20\n``#202a20` `rgb(32,42,32)``\n• #1e2d1d\n``#1e2d1d` `rgb(30,45,29)``\n• #1b301a\n``#1b301a` `rgb(27,48,26)``\n• #193317\n``#193317` `rgb(25,51,23)``\n• #163614\n``#163614` `rgb(22,54,20)``\n• #143812\n``#143812` `rgb(20,56,18)``\n• #113b0f\n``#113b0f` `rgb(17,59,15)``\n• #0f3e0c\n``#0f3e0c` `rgb(15,62,12)``\n• #0c4109\n``#0c4109` `rgb(12,65,9)``\n• #094406\n``#094406` `rgb(9,68,6)``\n• #074703\n``#074703` `rgb(7,71,3)``\n• #044a00\n``#044a00` `rgb(4,74,0)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c4109 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "Addition and Subtraction Worksheets, Free Simple Printable Addition and Subtraction Worksheets - BYJU'S FutureSchool\n\nAddition and Subtraction are the basic operations for elementary school arithmetic. You can give your kid a chance to plunge into practice with a whole range of addition and subtraction worksheets, which have plenty of exercises to practice adding and subtracting single-digit, 2-digit, 3-digit, 4-digit, and 5-digit numbers. These worksheets come according to your child's grade level and have varying levels of difficulty. There are pdf practice sets as well as online sheets designed for elementary school children. Children not only get exposure to a wide variety of addition and subtraction sums  but build solid foundational concepts....Read MoreRead Less\n\n## Benefits of Addition and Subtraction Worksheets:\n\nUnderstand drilling exercises\n\nThese online addition and subtraction worksheets provide your kid with an ample amount of practice questions and drilling exercises under various categories. Your kid learns to work on small to big numbers covering the basic operations on numbers from 1 to 5 digits.\n\nHave fun while learning\n\nThese worksheets for kids contain appealing graphics and visuals, making them more interesting for the young ones. Moreover, they get a chance to get an ample amount of practice with fun facts and activities included in the worksheets.\n\nUnderstand simplified concepts\n\nAddition and subtraction worksheets for kids are available for both pre-schoolers and higher grade kids. The worksheets present the basic concepts of addition and subtraction in the easiest way possible. The kid can easily understand grouping and regrouping and how to use these skills to add or subtract higher numbers.\n\nLearn with puzzles and riddles\n\nMath can be more fun with games, puzzles, and riddles. These worksheets do just that. After a concept is introduced and exercises are given to the children to practice, the worksheets provide several topic-based games and riddles to test or recap what the kids have learned." ]

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[ "# Re: [isabelle] partial functions: is f_rel too weak too?\n\nThis looks like another case of a missing congruence rule. This congreunce rule would express that your set comprehension only makes calls to f where B has become smaller. Grep for fundef_cong in the sources to see many examples. You may have to introduce an auxiliary constant to hide the complex set comprehension.\n```\nTobias\n\nOn 29/02/2016 07:33, Peter Gammie wrote:\n```\n```Hi,\n\nConsider this function and proof attempt:\n\nfunction f :: \"'a::finite set \\<Rightarrow> 'a set \\<Rightarrow> 'a list set\" where\n\"f C B =\n(if B = {} then {[]}\nelse if B \\<subseteq> C\nthen {c # cs |c cs. c \\<in> B \\<and> cs \\<in> f {} (B - {c})}\nelse {})\"\nby pat_completeness auto\n\nlemma f_termination:\nshows \"f_dom (C, B)\"\nproof(induct t \\<equiv> \"card B\" arbitrary: B C)\ncase (Suc i B C) then show ?case\napply -\napply (rule accpI)\napply (erule f_rel.cases)\n\nIntuitively this function terminates because we keep removing elements of B from B. However f_rel does not seem to note that c is drawn from B in the recursion. (domintros is too weak, as is apparently well-known.)\n\nAny tips?\n\nIn my use case f is actually only partially defined.\n\ncheers,\npeter\n\n```\n```\n```\n\nAttachment: smime.p7s\nDescription: S/MIME Cryptographic Signature\n\nThis archive was generated by a fusion of Pipermail (Mailman edition) and MHonArc." ]

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[ "In :\nimport spot\nspot.setup()\n\n\nThis notebook demonstrates the implementation of Zielonka Tree and ACD (Alternating Cycle Decomposition) in Spot.\n\nThese two structures are used to decompose an acceptance condition (or automaton) into trees that alternate accepting and rejecting elements in order to help converting an automaton to parity acceptance. Spot implements those structures, includes some display code to better explore them iteractively, and finally use them to implement a transformation to parity acceptance.\n\nFor a formal tratement of these, in a slightly different formalism, see Optimal Transformations of Games and Automata Using Muller Conditions by Casares, Colcombet, and Fijalkow. In Spot those definitions have been adapted to use Emerson-Lei acceptance, and support transitions labeled by multiple colors (the main differences are for the Zielonka Tree, the ACD is almost identical).\n\n# Zielonka Tree¶\n\nThe Zielonka tree is built from an acceptance formula and is labeled by sets of colors. The root contains all colors used in the formula. If seing infinitely all colors of one node would satisfy the acceptance condition, we say that the node is accepting and draw it with an ellipse, otherwise is is rejecting and drawn with a rectangle. The children of an accepting (resp. rejecting) node, are the largest subsets of colors that are rejecting (resp. accepting).\n\nHere is an example:\n\nIn :\nc = spot.acc_cond('Fin(0)&Inf(1)&(Inf(2)|Fin(3)) | (Inf(0)|Fin(1))&Fin(2)&Inf(3)')\nt = spot.zielonka_tree(c)\nt\n\nOut:\n\nAbove we can see that {0,1,2,3} is a rejecting root, but it has two accepting maximal subsets: {1,2,3} and {0,1,3}. Similarly, {1,2,3} has two rejecting maximal subsets: {2,3} and {1,3}.\n\nThe leaves of this tree have some additional numbers in angle brackets: those are the node numbers. Each node has a number (it can be seen by hovering over it with the mouse), but those leaf numbers play a particular role when the tree is used to paritize an automaton.\n\nThis tree is also layered: all nodes in each layers are alternatively rejecting and accepting. Layers are numbered incrementally from 0 at the root. In this example, leaves are in layer 3. Since it is conventional to put the root at the top, we will say that a node is high in the tree when it has a small level.\n\nIn this example, odd levels are accepting: we say the tree is odd. On another example, it could be the other way arround. The is_even() method tells us which way it is.\n\nIn :\nt.is_even()\n\nOut:\nFalse\n\nThe number of leaves, and the number of the left-most leaf can be found with the following functions:\n\nIn :\nt.num_branches(), t.first_branch()\n\nOut:\n(6, 7)\n\nThe step() methods takes one leaf L, one set of colors S, and returns a pair: (leaf, level). It works as follows: starting from node L, it walks up the tree until it has seen all colors in S, and then return the left-most leaf of the next branch of the node it has seen (assuming branches are ordered circularly from left to right). The level returned is that of the highest node seen.\n\nWe refer to the process performed by step(L,S) as reading S from L.\n\nFor instance reading from 7 does not change the leaf, since that leaf is already labeled by {3}. The level emitted is 3, since that's the level of 7.\n\nIn :\nt.step(7, )\n\nOut:\n(7, 3)\n\nReading from 8, would however move to leave 9 and emit level 2. Indeed, color 3 is not recognized by 8, so we move up to node 4 (the one labelled by {1,3}), emit its level (2), and go down to the left-most leaf of the next branch.\n\nIn :\nt.step(8, )\n\nOut:\n(9, 2)\n\nIf we read [0,2] from 9, we will have to go all the way up to the root (emitting 0) and then down to leaf 10:\n\nIn :\nt.step(9, [0,2])\n\nOut:\n(10, 0)\n\nNow, let's take this a step further, and chain multiple steps, by reading a sequence of Sᵢ and producing a sequence of levels.\n\nIn :\ndef steps(tree, leaf, seq_of_s):\nseq_of_lvl = []\nfor s in seq_of_s:\nleaf, lvl = tree.step(leaf, s)\nseq_of_lvl.append(lvl)\nreturn leaf, seq_of_lvl\n\nIn :\nprint(steps(t, 7, [,,,,,,,,,,,]))\nprint(steps(t, 7, [,,,,,,,]))\n\n(10, [1, 0, 2, 2, 1, 1, 3, 1, 1, 3, 1, 1])\n(9, [1, 2, 2, 2, 2, 2, 2, 2])\n\n\nIf we imagine an run looping on tree transitions labeled by , , , we know (from the original acceptance condition) that it should be accepting. Infinite repetition of the step() procedure will emit many levels, but the smallest level we see infinitely often is 1. It corresponds to node 2, labeled by {0,1,3}: this is the highest node we visit infinitely often while steping through this tree in a loop.\n\nSimilarly, a loop of two transitions labeled by and should be rejecting. Stepping through the tree will emit infinitely many 2, a rejecting level.\n\nAll of this can be used to transform an automaton with arbitrary acceptance into a parity automaton, where the emitted levels will correspond to the priorities of the parity automaton.\n\nThe states of the new automaton will be pairs of states of the form (original state, leaf). If the input edge (src,dst) has colors S, then the output edge ((src,leaf),(dst,nextleaf)) will have priority L where L and nextleaf are computed as (nextleaf,L)=t.step(leaf, S). The leaf used for the initial state does not really matter, we use the left-most one.\n\nLet's create a random automaton with this example acceptance condition, and paritize it:\n\nIn :\na1 = spot.automaton(\"randaut -Q4 --colored -e.7 -A '{}' 2 |\".format(c.get_acceptance()))\na1\n\nOut:\nIn :\np1 = spot.zielonka_tree_transform(a1)\np1.copy_state_names_from(a1)\np1\n\nOut:\n\nHere the parity automaton output has as many proprities as there are levels in the Zielonka tree.\n\nThe call to copy_state_names_from() above causes the states to be labeled by strings of the form orig#leaf when orig is the original state number, and leaf is a leaf of the Zielonka tree.\n\nSince this notebook is part of our test-suite, let us also make sure those automata are equivalent.\n\nIn :\nspot.are_equivalent(p1, a1)\n\nOut:\nTrue\n\nNote that a1 above was generated as a so-called \"colored\" automaton, i.e., each edge has exactly one color. But the transformation also works when edges have arbitrary and possibly empty subsets of colors. In that case, reading a transition without color may emit a level that is not in the Zielonka tree, as if an additional imaginary layer labeled with the empty set was below the tree: we do not store this level for simplicity, but the step() function is aware of that.\n\nIn :\na2 = spot.automaton(\"randaut -Q3 -e.8 --seed=4 -A '{}' 2 |\".format(c.get_acceptance()))\na2\n\nOut:\nIn :\np2 = spot.zielonka_tree_transform(a2)\np2.copy_state_names_from(a2)\np2\n\nOut:\n\nThe shape of the Zielonka tree (and later of the ACD), can also tell us if the acceptance condition can be converted into Rabin/Streett/parity without changing the automaton structure. This follows from the following properties:\n\n• an automaton is Rabin-type iff the union of any two accepting cycle is necessarily accepting\n• an automaton is Streett-type iff the union of two rejecting cycle is necessarily rejecting\n• an automaton is parity-type iff it is Rabin-type and Streett-type\n\nHere, X-type means that the acceptance condition of the automaton can be changed into X without changing the transition structure (just by coloring it differently). The Zielonka tree does not really look at the automaton, however its shape can still implies some typeness:\n\n• an automaton is Rabin-type if (not iff) accepting nodes of the Zielonka tree have at most one child\n• an automaton is Street-type if (not iff) rejecting nodes of the Zielonka tree have at most one child\n\nThe following methods provide this information:\n\nIn :\nt.has_rabin_shape(), t.has_streett_shape(), t.has_parity_shape()\n\nOut:\n(False, False, False)\n\nLet's look at some examples:\n\nIn :\nfor ct in ('Rabin 3', 'Streett 3', 'parity min odd 5',\n'Inf(0)&Fin(1) | (Inf(2)&Fin(3)&Fin(4))', 'f', 't'):\ncond = spot.acc_cond(ct)\ntcond = spot.zielonka_tree(cond)\nprint(cond.get_acceptance(), \":\", tcond.has_rabin_shape(),\ntcond.has_streett_shape(), tcond.has_parity_shape());\ndisplay(tcond)\n\n(Fin(0) & Inf(1)) | (Fin(2) & Inf(3)) | (Fin(4) & Inf(5)) : True False False\n\n(Fin(0) | Inf(1)) & (Fin(2) | Inf(3)) & (Fin(4) | Inf(5)) : False True False\n\nFin(0) & (Inf(1) | (Fin(2) & (Inf(3) | Fin(4)))) : True True True\n\n(Inf(0) & Fin(1)) | (Inf(2) & Fin(3) & Fin(4)) : True False False\n\nf : True True True\n\nt : True True True\n\n\n# Alternating Cycle Decomposition¶\n\nWe now turn to the ACD, which extends the Zielonka tree to take the automaton structure into account. Instead of storing subsets of colors, it stores SCCs (not necessarily maximal). In Spot, those SCCs are stored as bitvectors of edges. The principle is similar: instead of one tree, we are building a forest, with one root per non-trivial maximal SCC. The root of each tree is labeled by all the transitions in this maximal SCC: if such a huge cycle is accepting the node is said accepting and drawn as an ellipse, otherwise it is rejecting and drawn as a rectangle. For children, we look for maximal subsets of cycles that can be rejecting (if the parent was accepting), or accepting (if the parent was rejecting).\n\nIn other words all nodes correspond to SCCs that may not be maximal from a graph point of view, but that are maximal with respect to their accepting or rejecting status: adding more cycles to it would change that status.\n\nIn :\na3 = spot.automaton(\"\"\"HOA: v1 name: \"(FGp0 & ((XFp0 & F!p1) | F(Gp1 &\nXG!p0))) | G(F!p0 &\\n(XFp0 | F!p1) & F(Gp1 | G!p0))\" States: 10 Start:\n0 AP: 2 \"p0\" \"p1\" Acceptance: 6 (Fin(0) & Fin(1)) | ((Fin(4)|Fin(5)) &\n(Inf(2)&Inf(3))) properties: trans-labels explicit-labels trans-acc\ncomplete properties: deterministic --BODY-- State: 0 [!0&!1] 0 {0 1 2\n3 5} [!0&1] 1 [0&!1] 2 [0&1] 3 State: 1 [!0&1] 1 {1 3} [0&!1] 2\n[!0&!1] 4 {0 1 2 3 5} [0&1] 5 State: 2 [!0&!1] 2 {1 2 3 5} [0&!1] 2 {2\n4 5} [!0&1] 3 {1 3} [0&1] 3 {4} State: 3 [0&!1] 2 {2 4 5} [!0&1] 3 {1\n3} [0&1] 5 {2 4} [!0&!1] 6 {1 2 3 5} State: 4 [!0&1] 1 {1 3} [0&!1] 2\n[0&1] 3 [!0&!1] 7 State: 5 [0&!1] 2 {2 4 5} [0&1] 3 {4} [!0&1] 5 {1 3}\n[!0&!1] 8 {1 3 5} State: 6 [0&!1] 2 {2 4 5} [!0&1] 3 {1 3} [0&1] 3 {4}\n[!0&!1] 8 {1 3 5} State: 7 [0&!1] 2 [0&1] 5 [!0&!1] 7 {0 1 2 3 5}\n[!0&1] 9 {1 2 3} State: 8 [0&!1] 2 {2 4 5} [!0&1] 5 {1 2 3} [0&1] 5 {2\n4} [!0&!1] 8 {1 2 3 5} State: 9 [0&!1] 2 [0&1] 3 [!0&!1] 7 {0 1 3 5}\n[!0&1] 9 {1 3} --END--\"\"\")\n\n\nHere is the ACD for automaton a3, displayed below a copy of the automaton. Note that the automaton and ACD are linked interactively: you may click on nodes of the ACD to highlight the cycles it stores, and you may click on a state or an edge in the automaton to highlight to relevant ACD nodes (it is easier to click on edge labels than on edges).\n\nIn :\ntheacd = spot.acd(a3); theacd\n\nOut:\n\nThe nodes of the ACD represent various bits of informations. First, the root of each tree shows the number of the maximal SCC (as computed by spot.scc_info). Trivial maximal SCCs (without cycles) are omitted from the construction. Children are sorted by decreasing size of the SCCs they represent (i.e., decreasing count of edges).\n\nThe numbers after T: list the edges that belong to the strongly connected component. The numbers are simply the indices of those edges in the automaton. Between braces are the colors covered by all those edges, to decide if a cycle formed by all these edges would be accepting. The numbers after Q: are the states touched by those edges. lvl: indicates the level of each node. Since some trees can have accepting roots while others have rejecting roots, the levels of some tree have been shifted down (i.e., the root start at levl 1 instead of 0) to ensure that all trees use the same parity. The parity of the decomposition (i.e., whether even levels are accepting) is indicated by the is_even() method.\n\nIn :\ntheacd.is_even()\n\nOut:\nFalse\n\nFinally, the leaves have their node number indicated between angle brackets, but again, node numbers can be obtained by hovering over the node with the mouse. This construction may step in more that just those indicated leaves because it will consider subtrees of the ACD.\n\nFor each state of the original automaton, there exists a subtree in the ACD formed by all the node whose Q: contain that state. In the display of theacd above, you can click on any state of the input automaton to highlight the corresponding subtree of the ACD.\n\nThe step() method works similarly to Zielonka trees, except it uses edge number instead of colors. It is given a node of the ACD and the number of an edge whose source state touched belong to Q: in the current node. Then it walks up the tree until it finds a node that contains the given edge, emit the level of that node, and returns the left-most node of the next-branch in the sub-tree associated to the destination state.\n\nFor instance, let's consider the edges 2->3 and 3->2 in a1. There are actually two edges going from 2 to 3. Hovering above those arrows tells us the number of these edges:\n\n• 2->3 with {1,3} is edge 11\n• 2->3 with {4} is edge 12\n• 3->2 with {2,4,5} is edge 13\n\nLet us assume we are in node 4 of the ACD (a node that contains state 2) and want to step through edges 12 and 13:\n\nIn :\ntheacd.step(4, 12)\n\nOut:\n(4, 1)\nIn :\ntheacd.step(4, 13)\n\nOut:\n(4, 1)\n\nIn both case, we stay in node 4 and emit level 1, this is the level of node 4, which contains both edges.\n\nIf we were to iterate through edges 11 and 13 instead, the story would be different:\n\nIn :\ntheacd.step(4, 11)\n\nOut:\n(12, 0)\nIn :\ntheacd.step(12, 13)\n\nOut:\n(8, 0)\nIn :\ntheacd.step(8, 11)\n\nOut:\n(4, 0)\nIn :\ntheacd.step(4, 13)\n\nOut:\n(4, 1)\n\nAnd then it would loop again.\n\nLet's consider the first step. Node 4 does not contains edge 11, so we move up the tree until we find node 0 that contains it, and emit level 0. Then the destination of edge 11 is 3, so we consider the subtree associated to 3 (clicking on state 3 shows us this is the tree that has nodes 4,7,13 as leaves) and return the left-most node of the next branch: 12.\n\nWhen we read edge 13 from node 12, go all the way up to node 0 again, and consider the left-most node of the next branch in the tree for the destination 2: that's node 8.\n\nReading edges 11 and 13 will get us back to node 4. If we repeat this infinitely often, the smallest level we will see infinitely often is therefore 0, meaning that is a rejecting cycle.\n\nThe initial node we use to start this process can be any node that contains this state. The first_branch(s) method returns the left-most node containing state s.\n\nIn :\ntheacd.first_branch(2)\n\nOut:\n4\n\nThis first_branch() method is actually being used when step() is used to take an edge between two SCCs. In that case the edge does not appear in the ACD, and step() will return the first_branch() for its destination, and level 0.\n\nAn automaton can be paritized using ACD following a procedure similar to Zielonka tree, with this more complex stepping process.\n\nIn :\np3 = spot.acd_transform(a3, True)\np3.copy_state_names_from(a3)\np3\n\nOut:\n\nNote how the cycles 2#4 → 3#4 → 2#4 and 2#4 → 3#12 → 2#8 → 3#4 → 2#4 corresponds to the two cycles stepped though before.\n\nIn :\na3.equivalent_to(p3)\n\nOut:\nTrue\n\nBy default, the construction will try to save colors by not emitting colors between SCCs, and not emitting colors for the largest level of an SCC when the parity allows it. So this automaton can actually be paritized with two colors:\n\nIn :\np3 = spot.acd_transform(a3) # second argument defaults to False\np3.copy_state_names_from(a3)\nassert a3.equivalent_to(p3)\np3\n\nOut:\n\nThis transformation can have substantiually fewer states than the one based on Zielonka tree, because the branches are actually restricted to only those that matter for a given state.\n\nIn :\np3.num_states()\n\nOut:\n15\nIn :\nspot.zielonka_tree_transform(a3).num_states()\n\nOut:\n27\n\n## Typeness checks¶\n\nThe ACD can be used to decide Rabin, Streett, and parity-typeness. Testing Rabin-typeness requires making sure that round nodes have no children sharing a common state, and testing Streett-typeness needs a similar test for square node. These additional tests have a cost, so they need to be enabled by passing spot.acd_options_CHECK_RABIN, spot.acd_options_CHECK_STREETT, or spot.acd_options_CHECK_PARTIY to the ACD constructor. The latter option implies the former two.\n\nAs an example, automaton a3 is Rabin-type because nodes 12 and 13, which are the children of the only round node with more than one child, do not share any state. It it not Street-type, because for instance state 2 appears in both nodes 4 and 8, two children of a square node.\n\nIn :\ntheacd = spot.acd(a3, spot.acd_options_CHECK_PARITY)\n\nIn :\nprint(theacd.has_rabin_shape(), theacd.has_streett_shape(), theacd.has_parity_shape())\n\nTrue False False\n\n\nCalling the above methods withtout passing the relevant option will raise an exception.\n\nAdditonally, when the goal is only to check some typeness, the construction of the ACD can be aborted as soon as the typeness is found to be wrong. This can be enabled by passing the additional option spot.acd_options_ABORT_WRONG_SHAPE. In case the construction is aborted the ACD forest will be erased (to make sure it is not used), and node_count() will return 0.\n\nIn :\nprint(theacd.node_count())\n\n15\n\nIn :\ntheacd = spot.acd(a3, spot.acd_options_CHECK_STREETT | spot.acd_options_ABORT_WRONG_SHAPE)\n\nIn :\nprint(theacd.has_streett_shape(), theacd.node_count())\n\nFalse 0\n\n\n## State-based transformation¶\n\nThe ACD usage can be modified slightly in order to produce a state-based automaton. The rules for stepping through the ACD are similar, except that when we detect that a cycle through all children of a node has been done, we return the current node without going to the leftmost leave of the next children. When stepping a transition from a node a child, we should act as if we were in the leftmost child of that node containing the source of tha transition. Stepping through a transition this way do not emit any color, instead the color of a state will be the level of its associated node. (This modified transformation do not claim to be optimal, unlike the transition-based version.)\n\nThe ORDER_HEURISTIC used below will be explained in the next section, it justs alters ordering of children of the ACD in a way that the state-based transformation prefers.\n\nIn :\ntheacd = spot.acd(a3, spot.acd_options_ORDER_HEURISTIC)\n\nIn :\ntheacd\n\nOut:\n\nLet set what happens when we step through the two self-loops of state 2 (edges 9 and 10).\n\nIn :\ntheacd.first_branch(2)\n\nOut:\n4\nIn :\ntheacd.state_step(4, 9)\n\nOut:\n8\nIn :\ntheacd.state_step(8, 10)\n\nOut:\n0\nIn :\n# since 0 is not a leaf of the tree associated to state 2, the following acts like state_step(first_branch(2), 9)\ntheacd.state_step(0, 9)\n\nOut:\n8\n\nThe state-based version of the ACD transformation is acd_transform_sbacc(). In the output, the cycle between 2#8 and 2#0 corresponds to the repeatitions of edges 9 and 10 we stepped through above.\n\nIn :\ns3 = spot.acd_transform_sbacc(a3);\nassert s3.equivalent_to(a3)\ns3.copy_state_names_from(a3)\ns3\n\nOut:\n\nIn this case, the number of states we obtain with acd_transform_sbacc() is smaller than what we would obtain by calling sbacc() on the result of acd_transform().\n\nIn :\nprint(s3.num_states(),\nspot.sbacc(spot.acd_transform(a3)).num_states())\n\n22 28\n\n\n## An ordering heuristic for state-based transformation¶\n\nWe now explain the ORDER_HEURISTIC option, by first looking at an example without it.\n\nIn :\na4 = spot.automaton(\"\"\"HOA: v1 States: 3 Start:\n0 AP: 1 \"p\" acc-name: generalized-Buchi 2 Acceptance:\n2 Inf(1)&Inf(0) properties: trans-labels implicit-labels\ntrans-acc complete deterministic --BODY-- State: 0 0 {0}\n1 State: 1 1 {1} 2 State: 2 2 {1} 0 --END--\"\"\")\nspot.acd(a4)\n\nOut:\nIn :\ns4 = spot.acd_transform_sbacc(a4, False, False) # The third argument disables the heuristic\ns4.copy_state_names_from(a4)\ns4\n\nOut:\n\nOne this example, we can see that whenever an edge leaves not 2, it creates a copy of the target state.\n\nThe ORDER_HEURISTIC option of the ACD construction, attemps to order the children of a node by decreasing number of number of successors that are out of the node. It is activated by default inside spot.acd_transform_sbacc().\n\nIn :\nspot.acd(a4, spot.acd_options_ORDER_HEURISTIC)\n\nOut:\nIn :\ns4b = spot.acd_transform_sbacc(a4)\ns4b.copy_state_names_from(a4)\ns4b\n\nOut:\n\n# More examples¶\n\nThese additional examples also contribute to our test suite.\n\nIn :\nc = spot.automaton(\"\"\"\nHOA: v1\nStates: 2\nStart: 0\nAP: 2 \"p1\" \"p0\"\nAcceptance: 5 (Fin(0) & (Fin(3)|Fin(4)) & (Inf(1)&Inf(2))) | Inf(3)\nproperties: trans-labels explicit-labels trans-acc complete\nproperties: deterministic stutter-invariant\n--BODY--\nState: 0\n[0&!1] 0 {2 3}\n[!0&!1] 0 {2 3 4}\n[!0&1] 1\n[0&1] 1 {2 4}\nState: 1\n[!0&!1] 0 {0 2 3 4}\n[!0&1] 1 {1}\n[0&!1] 1 {2 3}\n[0&1] 1 {1 2 4}\n--END--\n\"\"\"); c\n\nOut:\nIn :\nspot.zielonka_tree(c.acc())\n\nOut:\nIn :\nd = spot.zielonka_tree_transform(c); d\n\nOut:\nIn :\nassert c.equivalent_to(d)\n\nIn :\ncacd = spot.acd(c); cacd\n\nOut:\nIn :\ncacd.state_step(1, 7)\n\nOut:\n0\nIn :\nd = spot.acd_transform(c); d\n\nOut:\nIn :\ne = spot.acd_transform_sbacc(c); e\n\nOut:\nIn :\nassert c.equivalent_to(d)\nassert c.equivalent_to(e)\n\n\nParitizing a generalized-Büchi automaton with acd_transform() produces a Büchi automaton. The Zielonka-tree transformation produces a larger automaton because it ignores the transition structure, and also it uses an extra and unnecessary color (because it was not taught how to omit it).\n\nIn :\ng = spot.automaton(\"\"\"HOA: v1 States: 4 Start: 0 AP: 2 \"p0\" \"p1\"\nacc-name: generalized-Buchi 4 Acceptance: 4\nInf(0)&Inf(1)&Inf(2)&Inf(3) properties: trans-labels explicit-labels\ntrans-acc --BODY-- State: 0 [!0&1] 2 {0 1 2} [!0&!1] 3 {0 1 2 3}\nState: 1 [!0&1] 0 {0 1 2} [!0&!1] 2 {0} State: 2 [!0&1] 1 {0} State: 3\n[!0&!1] 2 {0 1 2 3} [0&!1] 3 {0} --END--\"\"\")\n\nspot.acd(g)\n\nOut:\nIn :\nspot.acd_transform(g)\n\nOut:\nIn :\nspot.acd_transform_sbacc(g)\n\nOut:\nIn :\nspot.zielonka_tree_transform(g)\n\nOut:\n\nAn issue in Spot is to always ensure that property bits of automata (cleaming that an automaton is weak, inherently weak, deterministic, etc.) are properly preserved or reset.\n\nHere if the input is inherently weak, the output should be weak.\n\nIn :\nw = spot.automaton(\"\"\"HOA: v1\nStates: 2\nStart: 1\nAP: 1 \"a\"\nAcceptance: 2 Inf(0) | Inf(1)\nproperties: trans-labels explicit-labels trans-acc colored complete\nproperties: deterministic inherently-weak\n--BODY--\nState: 0\n[t] 1 {0}\nState: 1\n 0 {1}\n[!0] 1 {0}\n--END--\"\"\"); w\n\nOut:\nIn :\nw.prop_weak(), w.prop_inherently_weak()\n\nOut:\n(spot.trival_maybe(), spot.trival(True))\nIn :\nw2 = spot.acd_transform(w); w2\n\nOut:\nIn :\nw2.prop_weak(), w2.prop_inherently_weak()\n\nOut:\n(spot.trival(True), spot.trival(True))\nIn [ ]:" ]

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[ "# #5e8a69 Color Information\n\nIn a RGB color space, hex #5e8a69 is composed of 36.9% red, 54.1% green and 41.2% blue. Whereas in a CMYK color space, it is composed of 31.9% cyan, 0% magenta, 23.9% yellow and 45.9% black. It has a hue angle of 135 degrees, a saturation of 19% and a lightness of 45.5%. #5e8a69 color hex could be obtained by blending #bcffd2 with #001500. Closest websafe color is: #669966.\n\n• R 37\n• G 54\n• B 41\nRGB color chart\n• C 32\n• M 0\n• Y 24\n• K 46\nCMYK color chart\n\n#5e8a69 color description : Mostly desaturated dark cyan - lime green.\n\n# #5e8a69 Color Conversion\n\nThe hexadecimal color #5e8a69 has RGB values of R:94, G:138, B:105 and CMYK values of C:0.32, M:0, Y:0.24, K:0.46. Its decimal value is 6195817.\n\nHex triplet RGB Decimal 5e8a69 `#5e8a69` 94, 138, 105 `rgb(94,138,105)` 36.9, 54.1, 41.2 `rgb(36.9%,54.1%,41.2%)` 32, 0, 24, 46 135°, 19, 45.5 `hsl(135,19%,45.5%)` 135°, 31.9, 54.1 669966 `#669966`\nCIE-LAB 53.574, -22.36, 12.958 16.254, 21.576, 16.672 0.298, 0.396, 21.576 53.574, 25.844, 149.907 53.574, -21.658, 20.676 46.45, -18.827, 11.234 01011110, 10001010, 01101001\n\n# Color Schemes with #5e8a69\n\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #8a5e7f\n``#8a5e7f` `rgb(138,94,127)``\nComplementary Color\n• #698a5e\n``#698a5e` `rgb(105,138,94)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #5e8a7f\n``#5e8a7f` `rgb(94,138,127)``\nAnalogous Color\n• #8a5e69\n``#8a5e69` `rgb(138,94,105)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #7f5e8a\n``#7f5e8a` `rgb(127,94,138)``\nSplit Complementary Color\n• #8a695e\n``#8a695e` `rgb(138,105,94)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #695e8a\n``#695e8a` `rgb(105,94,138)``\n• #7f8a5e\n``#7f8a5e` `rgb(127,138,94)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #695e8a\n``#695e8a` `rgb(105,94,138)``\n• #8a5e7f\n``#8a5e7f` `rgb(138,94,127)``\n• #3f5c46\n``#3f5c46` `rgb(63,92,70)``\n• #496c52\n``#496c52` `rgb(73,108,82)``\n• #547b5d\n``#547b5d` `rgb(84,123,93)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #699975\n``#699975` `rgb(105,153,117)``\n• #78a383\n``#78a383` `rgb(120,163,131)``\n``#87ad91` `rgb(135,173,145)``\nMonochromatic Color\n\n# Alternatives to #5e8a69\n\nBelow, you can see some colors close to #5e8a69. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #5e8a5e\n``#5e8a5e` `rgb(94,138,94)``\n• #5e8a62\n``#5e8a62` `rgb(94,138,98)``\n• #5e8a65\n``#5e8a65` `rgb(94,138,101)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #5e8a6d\n``#5e8a6d` `rgb(94,138,109)``\n• #5e8a70\n``#5e8a70` `rgb(94,138,112)``\n• #5e8a74\n``#5e8a74` `rgb(94,138,116)``\nSimilar Colors\n\n# #5e8a69 Preview\n\nThis text has a font color of #5e8a69.\n\n``<span style=\"color:#5e8a69;\">Text here</span>``\n#5e8a69 background color\n\nThis paragraph has a background color of #5e8a69.\n\n``<p style=\"background-color:#5e8a69;\">Content here</p>``\n#5e8a69 border color\n\nThis element has a border color of #5e8a69.\n\n``<div style=\"border:1px solid #5e8a69;\">Content here</div>``\nCSS codes\n``.text {color:#5e8a69;}``\n``.background {background-color:#5e8a69;}``\n``.border {border:1px solid #5e8a69;}``\n\n# Shades and Tints of #5e8a69\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #070a07 is the darkest color, while #fdfefd is the lightest one.\n\n• #070a07\n``#070a07` `rgb(7,10,7)``\n• #0f1510\n``#0f1510` `rgb(15,21,16)``\n• #162119\n``#162119` `rgb(22,33,25)``\n• #1e2d22\n``#1e2d22` `rgb(30,45,34)``\n• #26382b\n``#26382b` `rgb(38,56,43)``\n• #2e4434\n``#2e4434` `rgb(46,68,52)``\n• #36503d\n``#36503d` `rgb(54,80,61)``\n• #3e5b45\n``#3e5b45` `rgb(62,91,69)``\n• #46674e\n``#46674e` `rgb(70,103,78)``\n• #4e7357\n``#4e7357` `rgb(78,115,87)``\n• #567e60\n``#567e60` `rgb(86,126,96)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #669672\n``#669672` `rgb(102,150,114)``\n• #719e7c\n``#719e7c` `rgb(113,158,124)``\n• #7da687\n``#7da687` `rgb(125,166,135)``\n• #88ae92\n``#88ae92` `rgb(136,174,146)``\n• #94b69d\n``#94b69d` `rgb(148,182,157)``\n• #a0bea7\n``#a0bea7` `rgb(160,190,167)``\n• #abc6b2\n``#abc6b2` `rgb(171,198,178)``\n• #b7cebd\n``#b7cebd` `rgb(183,206,189)``\n• #c3d6c7\n``#c3d6c7` `rgb(195,214,199)``\n• #ceded2\n``#ceded2` `rgb(206,222,210)``\n• #dae6dd\n``#dae6dd` `rgb(218,230,221)``\n• #e6eee8\n``#e6eee8` `rgb(230,238,232)``\n• #f1f6f2\n``#f1f6f2` `rgb(241,246,242)``\n• #fdfefd\n``#fdfefd` `rgb(253,254,253)``\nTint Color Variation\n\n# Tones of #5e8a69\n\nA tone is produced by adding gray to any pure hue. In this case, #707872 is the less saturated color, while #05e33c is the most saturated one.\n\n• #707872\n``#707872` `rgb(112,120,114)``\n• #67816d\n``#67816d` `rgb(103,129,109)``\n• #5e8a69\n``#5e8a69` `rgb(94,138,105)``\n• #559365\n``#559365` `rgb(85,147,101)``\n• #4c9c60\n``#4c9c60` `rgb(76,156,96)``\n• #43a55c\n``#43a55c` `rgb(67,165,92)``\n• #3aae57\n``#3aae57` `rgb(58,174,87)``\n• #31b753\n``#31b753` `rgb(49,183,83)``\n• #28c04e\n``#28c04e` `rgb(40,192,78)``\n• #20c84a\n``#20c84a` `rgb(32,200,74)``\n• #17d145\n``#17d145` `rgb(23,209,69)``\n• #0eda41\n``#0eda41` `rgb(14,218,65)``\n• #05e33c\n``#05e33c` `rgb(5,227,60)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #5e8a69 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "", null, "Download", null, "Download Presentation", null, "AQA GCSE Physics 3-3 Electromagnetism\n\n# AQA GCSE Physics 3-3 Electromagnetism\n\nDownload Presentation", null, "## AQA GCSE Physics 3-3 Electromagnetism\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. AQA GCSE Physics 3-3Electromagnetism GCSE Physics pages 254 to 265 April 10th 2010\n\n2. THE MOTOR EFFECT 13.7 How can electricity be used to make things move? Using skills, knowledge and understanding of how science works: • to explain how the motor effect is used in simple devices. Skills, knowledge and understanding of how science works set in the context of: • When a conductor carrying an electric current is placed in a magnetic field, it may experience a force. • The size of the force can be increased by: – increasing the strength of the magnetic field – increasing the size of the current. • The conductor will not experience a force if it is parallel to the magnetic field. • The direction of the force is reversed if either the direction of the current or the direction of the magnetic field is reversed. ELECTRICAL GENERATORS 13.8 How do generators work? Using skills, knowledge and understanding of how science works: • to explain from a diagram how an a.c. generator works, including the purpose of the slip rings and brushes. Skills, knowledge and understanding of how science works set in the context of: • If an electrical conductor .cuts. through magnetic field lines, an electrical potential difference is induced across the ends of the conductor. • If a magnet is moved into a coil of wire, an electrical potential difference is induced across the ends of the coil. • If the wire is part of a complete circuit, a current is induced in the wire. • If the direction of motion, or the polarity of the magnet, is reversed, the direction of the induced potential difference and the induced current is reversed. • The generator effect also occurs if the magnetic field is stationary and the coil is moved. • The size of the induced potential difference increases when: – the speed of the movement increases – the strength of the magnetic field increases – the number of turns on the coil increases – the area of the coil is greater. TRANSFORMERS 13.9 How do transformers work? Using skills, knowledge and understanding of how science works: • to determine which type of transformer should be used for a particular application. Skills, knowledge and understanding of how science works set in the context of: • The basic structure of the transformer. • An alternating current in the primary coil produces a changing magnetic field in the iron core and hence in the secondary coil. This induces an alternating potential difference across the ends of the secondary coil. • The potential difference (p.d.) across the primary and secondary coils of a transformer are related by the equation: p.d. across primary / p.d. across secondary = number of turns on primary / number of turns on secondary • In a step-up transformer the potential difference across the secondary coil is greater than the potential difference across the primary coil. • In a step-down transformer the potential difference across the secondary coil is less than the potential difference across the primary coil. • The uses of step-up and step-down transformers in the National Grid. AQA GCSE Specification\n\n3. S N - + - - + + - + The motor effect When a conductor carrying an electric current is placed in a magnetic field, it may experience a force. This is called the motor effect. Motor effect - Fendt\n\n4. The force increases if: – the strength of the magnetic field is increased – the current is increased The direction of the force is reversed if either the direction of the current or the direction of the magnetic field is reversed. The conductor will not experience a force if it is parallel to the magnetic field. Motor effect - Fendt\n\n5. The left-hand motor rule Note: Magnetic field direction is from NORTH to SOUTH Current direction is from PLUS to MINUS Motor effect - Fendt\n\n6. N N N N S S S S Note: means current out of the page means current into the page Insert the missing information Q1. Force direction ? Q2 Current direction ? Q3 N and S poles ? Q4 Force directions ? Motor effect - Fendt\n\n7. Electric current flowing around the coil of the electric motor produces oppositely directed forces on each side of the coil. These forces cause the coil to rotate. Every half revolution the split ring commutator causes the current in the coil to reverse otherwise the coil would stop in the vertical position. The electric motor Electric motor - Fendt\n\n8. rotation axis N S + contact brush Brushes lose contact with the split ring commutator. Current no longer flows through the motor coil. The coil will continue to rotate clockwise due to its momentum. Brushes in contact with the split ring commutator. Current flows through the motor coil. Forces exert a clockwise turning effect on the coil Brushes regain contact with the split ring commutator. Current flows through the motor coil but in the opposite direction. Forces exert a clockwise turning effect on the coil. Brushes lose contact with the split ring commutator. Current no longer flows through the motor coil. The coil will continue to rotate clockwise due to its momentum. Brushes regain contact with the split ring commutator. Current flows through the motor coil but in the original direction. Forces exert a clockwise turning effect on the coil. split-ring commutator Electric motor - Fendt\n\n9. Model electric motor Electric motor - Fendt\n\n10. The sound signal consists of an alternating current supplied by the amplifier. This current flows through the coil of the loudspeaker. Due to the motor effect, the magnetic field around the coil causes the coil to vibrate in step with the alternating current. The coil causes the diaphragm (speaker cone) to vibrate in step with the original sound signal. The diaphragm causes air to vibrate and so produces a sound wave. The loudspeaker\n\n11. Question Choose appropriate words to fill in the gaps below: The motor effect occurs when a _______ carrying wire is placed inside a ________ field. The force exerted is __________ when the wire is at 90° to the magnetic field __________ but is zero if the wire is ________ to the field. The force increases with _________ or current strength, the force __________ in direction if either are reversed. Applications include the electric motor and ___________. current magnetic maximum direction parallel field reverses loudspeaker WORD SELECTION: parallel reverses loudspeaker direction field current magnetic maximum\n\n12. The motor effectNotes questions from pages 254 & 255 • What is the motor effect? • Copy out the bullet points at the bottom of page 254 listing the factors that affect the force on a current carrying wire inside a magnetic field. • Copy and answer question (a) on page 254. • Copy Figure 3 on page 255 and explain how a simple electric motor works. Your account should include the purpose of the split-ring commutator. • Copy and answer question (b) on page 255. • Copy Figure 4 on page 255 and explain how a moving coil loudspeaker works. • Copy and answer question (c) on page 255. • Copy the ‘Key points’ table on page 255. • Answer the summary questions on page 255. Motor effect - Fendt Electric motor - Fendt\n\n13. In text questions: No change, the actions cancel each other out. The material must conduct electricity. A direct current will not produce a changing magnetic field. Summary questions: (a) Current, coil, force, coil. (b) Current, force, coil. 2. (a) The direction of the current is reversed and so the force on the coil is in the opposite direction. (b) (i) Faster because the coil is lighter (ii) Faster because the field is much stronger due to the presence of iron. The motor effect ANSWERS\n\n14. The generator effect If an electrical conductor cuts through magnetic field lines, an electrical potential difference is induced across the ends of the conductor. If the wire is part of a complete circuit, a current is induced in the wire. This is also called electromagnetic induction. Generator - Fendt\n\n15. If a magnet is moved into a coil of wire, an electrical potential difference is induced across the ends of the coil. If the direction of motion, or the polarity of the magnet, is reversed, then the direction of the induced potential difference and the induced current are also reversed. The generator effect also occurs if the magnetic field is stationary and the coil is moved. Generator - Fendt\n\n16. The size of the induced potential difference increases when: – the speed of the movement increases – the strength of the magnetic field increases – the number of turns on the coil increases – the area of the coil is greater. Generator - Fendt\n\n17. Alternating Current Generators Most electricity is produced using the ‘generator effect’. The simplest generators and the types used in power stations produce alternating current (A.C.) Generator - Fendt\n\n18. Moving Coil A.C. Generator Generator - Fendt\n\n19. Generator - Fendt\n\n20. This like an electric motor in reverse. As the coil is rotated electromagnetic induction occurs. An alternating voltage is induced in the coil. An alternating current is drawn off through two slip rings. The faster the coil is rotated: - the greater is the amplitude of the voltage and current - the higher is the frequency of the a.c. Generator - Fendt\n\n21. When the wheel turns the magnet is made to rotate next to the fixed coil of wire. Electromagnetic induction occurs and a alternating potential difference is induced in the coil. This causes an alternating current to flow to the light bulb of the bicycle. Bicycle generator Generator - Fendt\n\n22. The graph opposite shows the potential difference of a generator varies in time. Using the same set of axes show how the potential difference would vary if the rotational speed of the generator was doubled. PD time Question 1 The new potential difference will have TWICE the amplitude AND frequency of the original.\n\n23. Question 2 Choose appropriate words to fill in the gaps below: The _________ effect occurs when a conductor is moved relative to a ____________ field. This is also known as electromagnetic ___________. The greater the relative __________ of the conductor and magnetic field the _______ is the potential difference ________. If the conductor is part of a ________ circuit an electric current will flow. ___________ current is produced if the direction of movement is continually _________. generator magnetic induction movement greater induced complete alternating reversed WORD SELECTION: alternating generator reversed magnetic complete movement induction induced greater\n\n24. Electromagnetic induction Notes questions from pages 256 & 257 • What is induced in a wire because of the dynamo effect? • Copy and answer question (a) on page 256. • Copy Figure 2 on page 256 and explain how a cycle dynamo works. • Copy and answer questions (b) and (c) on page 256. • Explain how the alternating current generator on page 257 works. Your explanation should include a copy of both parts of Figure 4. • Copy the ‘Key points’ table on page 257. • Answer the summary questions on page 257. Generator - Fendt\n\n25. In text questions: (a) (i) The current increases. (ii) The direction of the current reverses. (iii) No current is produced. The wires leading to the coil would get twisted up. No brushes are needed. (i) There is no current. (ii) A p.d. is produced in the opposite direction. Summary questions: 1. (a) The pointer would move to the right but not as far. (b) The pointer returns to zero. (c) The pointer would move rapidly to the left. 2. (a) Spin the coil faster, use more loops of coil, use stronger magnets. (b) The peak voltage would be lower and the period would be longer. Electromagnetic induction ANSWERS\n\n26. circuit symbol The transformer A transformer is a device that is used to change one alternating voltage level to another. Transformer - eChalk\n\n27. A transformer consists of at least two coils of wire wrapped around a laminated iron core. PRIMARY COIL of Np turns SECONDARY COIL of Ns turns PRIMARY VOLTAGE Vp SECONDARY VOLTAGE Vs laminated iron core Structure of a transformer Transformer - eChalk\n\n28. How a transformer works When an alternating voltage, Vp is applied to the primary coil of Np turns it causes an alternating current to flow in this coil. This current causes a changing magnetic field in the laminated iron core which cuts across the secondary coil of Ns turns. Electromagnetic induction occurs in this coil which produces an alternating voltage, Vs. Transformer - eChalk\n\n29. Why can a transformer not change the level of the voltage output of a battery? A battery produces a steady (DC) voltage. This voltage would cause a constant direct current in the primary coil of a transformer. This current would produce an unchanging magnetic field in the iron core. This unchanging magnetic field would NOT cause electromagnetic induction in the secondary coil. There would therefore be no secondary voltage. Question\n\n30. Transformers Notes questions from pages 258 & 259 • Copy Figure 1 on page 258 and (a) explain what a transformer is, (b) what a transformer does and (c) how a transformer works. • Copy and answer questions (a), (b) and (c) on page 258. • Copy the circuit symbol for a transformer on page 259 and explain why the electric current supplied to a transformer must be alternating in order for the transformer to function. • Copy and answer question (d) on page 259. • Copy the ‘Key points’ table on page 259. • Answer the summary questions on page 259. Transformer - eChalk\n\n31. In text questions: The magnetic field in the core would be much weaker because the core is not a magnetic material. The lamp would be brighter. The lamp would not light up with direct current in the primary coil. Iron is easier to magnetise and demagnetise as the alternating current increases and decreases each half cycle. Summary questions: 1. Current, primary, magnetic field, secondary, p.d., secondary. 2. (a) Direct current in the primary coil would not produce an alternating magnetic field, so no p.d. would be induced in the secondary coil. (b) The current would short-circuit across the wires instead of passing through them. This would cause the coil to overheat if it did not cause the fuse to blow. (c) Iron is a magnetic material, so it makes the magnetic field much stronger. It is easily magnetised and demagnetised when the current alternates. Transformers ANSWERS\n\n32. The transformer equation The voltages or potential differences across the primary and secondary coils of a transformer are related by the equation: primary voltage = primary turns secondary voltagesecondary turns Vp= Np Vs Ns Transformer - eChalk\n\n33. Step-up transformers In a step-up transformer the potential difference across the secondary coil is greater than the potential difference across the primary coil. The secondary turns must be greater than the primary turns. Use: To increase the voltage output from a power station from 25 kV (25 000 V) to up to 400 kV. Transformer - eChalk\n\n34. Step-down transformers In a step-down transformer the potential difference across the secondary coil is smaller than the potential difference across the primary coil. The secondary turns must be smaller than the primary turns. Use: To decrease the voltage output from the mains supply from 230V to 18V to power and recharge a lap-top computer. Transformer - eChalk\n\n35. Calculate the secondary voltage of a transformer that has a primary coil of 1200 turns and a secondary of 150 turns if the primary is supplied with 230V. primary voltage = primary turns secondary voltage secondary turns 230 / Vs = 1200 / 150 230 / Vs = 8 230 = 8 x Vs 230 / 8 = Vs Secondary voltage = 28.8 V Question 1 Transformer - eChalk\n\n36. Calculate the number of turns required for the primary coil of a transformer if secondary has 400 turns and the primary voltage is stepped up from 12V to a secondary voltage of 48V. primary voltage = primary turns secondary voltage secondary turns 12 / 48 = Np / 400 0.25 = Np / 400 0.25 x 400 = Np Primary has 100 turns Question 2 Transformer - eChalk\n\n37. Answers Complete: 50 46 V 200 9 V Transformer - eChalk\n\n38. Transformers and the National Grid The National Grid is the system of cables used to deliver electrical power from power stations to consumers. The higher the voltage used, the greater is the efficiency of energy transmission. Lower voltages result in higher electric currents and greater energy loss to heat due to the resistance of the cables.\n\n39. At power stations the output voltage of the generators is stepped up by transformers from 25kV to 132kV. The voltage may be further increased to up to 400 kV for transmission over long distance pylon lines.\n\n40. The voltage is reduced in stages by step-down transformers to different levels for different types of consumer. The lowest level is 230V for domestic use. The final step-down transformer will be at sub station within a few hundred metres of each group of houses.\n\n41. Why is electrical energy transmitted over the National Grid in the form of alternating current? To maximise efficiency high voltages must be used. Voltage therefore needs to be changed in level. Transformers are needed to change voltage levels. Transformers only work with alternating current. Question 1\n\n42. Question 2 Choose appropriate words to fill in the gaps below: Transformers are used to change one ___________ potential difference level to another. They do not work with ____________current. Step-up transformers _________ the voltage because their ___________ coil has more turns than the primary. Transformers are used in the __________ Grid. The _______ output of a power station is increased to up to _______. A high voltage reduces the ________ lost to heat due to the _________ of the power lines. alternating direct increase secondary 25 kV National 400 kV energy resistance WORD SELECTION: energy direct National secondary resistance alternating 25 kV increase 400 kV\n\n43. Transformers and the National GridNotes questions from pages 260 & 261 • (a) Why are transformers used in the National grid? (b) What is the advantage of using high voltages? • Copy the transformer equation on page 260. • Copy a version of the worked example on page 260 but in your version change the number of turns on the secondary coil from 60 to 30. • What is the purpose of (a) step-up and (b) step-down transformers? • Explain how the number of turns on the coils of a transformer determine whether a transformer is step-up or step-down. • State how the currents and voltages associated with the primary and secondary coils are related to each other with a 100% efficient transformer. • Copy and answer questions (a) and (b) on page 261. • Copy the ‘Key points’ table on page 261. • Answer the summary questions on page 261. Transformer - eChalk\n\n44. In text questions: 60 turns (i) 6A (ii) 0.26A Summary questions: 1. (a) (i) Secondary, primary. (b) Up, down. 2. (a) 2000 turns (b) (i) 3A (ii) 0.15A Transformers and the National GridANSWERS\n\n45. More power to youNotes questions from pages 262 & 263 • Answer questions 1 and 2 on page 263.\n\n46. Motor effect - Fendt Electric motor - Fendt Faraday Electromagnetic Lab – PhET Play with a bar magnet and coils to learn about Faraday's law. Move a bar magnet near one or two coils to make a light bulb glow. View the magnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field lines or use a meter to show the direction and magnitude of the current. You can also play with electromagnets, generators and transformers! Faraday's Law - PhET - Light a light bulb by waving a magnet. This demonstration of Faraday's Law shows you how to reduce your power bill at the expense of your grocery bill. Generator - Fendt Transformer - load can be changed but not turns ration - netfirms Transformer - eChalk Electromagnetism Simulations\n\n47. More power to youANSWERS • (a) They would not need heavy iron magnets. (b) There would be no power wasted in the wires, as the wires would have no resistance. 2. (a) Ionising radiation, carcinogenic (cancer-causing) substances. (b) People are at risk due to other causes. There is an extra risk to those exposed to these magnetic fields. (c) A hypothesis is put forward as an ‘unproven’ theory to be tested by scientific experiments. If lots of experiments are carried out and they all support the hypothesis, it gains scientific credibility and is accepted as a theory. But at any stage, it could be overthrown by any conflicting scientific evidence.\n\n48. The voltmeter was not sensitive enough. It would also not give a read-out of the voltage, so it would be impossible to get an accurate result even if it was sensitive enough. Height on the X-axis, voltage on the Y-axis. Axes fully labelled and plots correctly plotted. In part. The voltage increased as height increased, but it was not directly proportional. 0.01V Not at the greater heights. Improve the sensitivity of the oscilloscope. Repeat his results. By checking it against other data/other similar research/get someone else to repeat his work or calculate theoretical relationships. For example: Measuring the speed of an object through a tube. How Science WorksANSWERS" ]

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[ "# Get Algebras, Graphs and their Applications PDF", null, "By Ilwoo Cho\n\nISBN-10: 146659019X\n\nISBN-13: 9781466590199\n\nThis e-book introduces the learn of algebra triggered by way of combinatorial items known as directed graphs. those graphs are used as instruments within the research of graph-theoretic difficulties and within the characterization and answer of analytic difficulties. The publication provides contemporary examine in operator algebra thought hooked up with discrete and combinatorial mathematical items. It additionally covers instruments and strategies from various mathematical parts, together with algebra, operator conception, and combinatorics, and gives various functions of fractal conception, entropy thought, K-theory, and index theory.\n\nSimilar graph theory books\n\nThe Mathematical Coloring Book: Mathematics of Coloring and by Alexander Soifer PDF\n\nI have not encountered a ebook of this sort. the simplest description of it i will be able to supply is that it's a secret novel… i discovered it challenging to prevent studying earlier than i ended (in days) the full textual content. Soifer engages the reader's cognizance not just mathematically, yet emotionally and esthetically. may possibly you benefit from the ebook up to I did!\n\nThe software program package deal MuPAD is a working laptop or computer algebra process that enables to resolve computational difficulties in natural arithmetic in addition to in utilized parts resembling the normal sciences and engineering. This instructional explains the fundamental use of the approach and provides perception into its energy. the most beneficial properties and uncomplicated instruments are awarded in uncomplicated steps.\n\nTree Lattices by Hyman Bass PDF\n\nThis monograph extends this method of the extra basic research of X-lattices, and those \"tree lattices\" are the most item of research. The authors current a coherent survey of the consequences on uniform tree lattices, and a (previously unpublished) improvement of the speculation of non-uniform tree lattices, together with a few primary and lately proved lifestyles theorems.\n\nNew PDF release: Zero-symmetric Graphs: Trivalent Graphical Regular\n\nZero-Symmetric Graphs: Trivalent Graphical normal Representations of teams describes the zero-symmetric graphs with no more than a hundred and twenty vertices. The graphs thought of during this textual content are finite, hooked up, vertex-transitive and trivalent. This booklet is prepared into 3 components encompassing 25 chapters.\n\nAdditional info for Algebras, Graphs and their Applications\n\nExample text\n\nNow, let Gk be given as above, for k = 1, 2, and let G be the unioned graph G2 ∪ G1 . Then, by the very definition, the unioned graph G is graphisomorphic to the unioned graph G = G1 ∪ G2 . This guarantees the following proposition. 1 Let X = G1 + G2 be the sum of graph groupoids G1 and G2 . Then X is groupoid-isomorphic to G2 + G1 . Proof. Let X = G1 + G2 be the sum of the graph groupoids Gk of graphs Gk . We know that the groupoid X is groupoid-isomorphic to the graph groupoid G of the unioned graph G = G1 ∪ G2 .\n\nDefine a groupoid action α of G acting on M in B(K ⊗ HG ) by a nonunital partial representation satisfying αw (m)Lw L∗w = L∗w mLw = Lw−1 mLw , for all m ∈ M and w ∈ G. We call the above relation of α, the G-representation of G. Here, the operators Lw ’s are understood as 1K ⊗ Lw ’s in B(K ⊗ HG ). Remark that, in the G-representation, the operator Lw L∗w is Lww−1 , and hence it is a projection on K ⊗ HG . If θ : M → B(K) is a ∗-homomorphism (or a representation of M ), then αw (m) and m in the G-representation are understood as π (αw (m)) and π(m) in π(M ), respectively, for all m ∈ M.\n\nThis action L is called the canonical (left) groupoid action of G (acting on HG ). 3 Let G be a graph with its graph groupoid G, and let HG be the graph Hilbert space of G. Also, let L be the canonical groupoid action of G. Then the pair (HG , L) is said to be the canonical representation of G. We have seen that, if there is a countable directed graph G, then the graph groupoid G of G is embedded in an operator algebra B(HG ); moreover, the elements of G become partial isometries and their initial or final projections on HG , under the canonical representation (HG , L)." ]

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[ "# NAG Library Routine Document\n\n## 1Purpose\n\nf06fjf updates the Euclidean norm of real vector in scaled form.\n\n## 2Specification\n\nFortran Interface\n Subroutine f06fjf ( n, x, incx, scal,\n Integer, Intent (In) :: n, incx Real (Kind=nag_wp), Intent (In) :: x(*) Real (Kind=nag_wp), Intent (Inout) :: scal, sumsq\nC Header Interface\n#include <nagmk26.h>\n void f06fjf_ (const Integer *n, const double x[], const Integer *incx, double *scal, double *sumsq)\n\n## 3Description\n\nGiven an $n$-element real vector $x$, and real scalars $\\alpha$ and $\\xi$, f06fjf returns updated values $\\stackrel{~}{\\alpha }$ and $\\stackrel{~}{\\xi }$ such that\n $α~2ξ~=x12+x22+⋯+xn2+α2ξ.$\nf06fjf is designed for use in the safe computation of the Euclidean norm of a real vector, without unnecessary overflow or destructive underflow. An initial call to f06fjf (with $\\xi =1$ and $\\alpha =0$) may be followed by further calls to f06fjf and finally a call to f06bmf to complete the computation. Multiple calls of f06fjf may be needed if the elements of the vector cannot all be accessed in a single array x.\nNone.\n\n## 5Arguments\n\n1:     $\\mathbf{n}$ – IntegerInput\nOn entry: $n$, the number of elements in $x$.\n2:     $\\mathbf{x}\\left(*\\right)$ – Real (Kind=nag_wp) arrayInput\nNote: the dimension of the array x must be at least $\\mathrm{max}\\phantom{\\rule{0.125em}{0ex}}\\left(1,1+\\left({\\mathbf{n}}-1\\right)×{\\mathbf{incx}}\\right)$.\nOn entry: the $n$-element vector $x$. ${x}_{\\mathit{i}}$ must be stored in ${\\mathbf{x}}\\left(1+\\left(\\mathit{i}-1\\right)×{\\mathbf{incx}}\\right)$, for $\\mathit{i}=1,2,\\dots ,{\\mathbf{n}}$.\nIntermediate elements of x are not referenced.\n3:     $\\mathbf{incx}$ – IntegerInput\nOn entry: the increment in the subscripts of x between successive elements of $x$.\nConstraint: ${\\mathbf{incx}}>0$.\n4:     $\\mathbf{scal}$ – Real (Kind=nag_wp)Input/Output\nOn entry: the scaling factor $\\alpha$. On the first call to f06fjf ${\\mathbf{scal}}=0.0$.\nConstraint: ${\\mathbf{scal}}\\ge 0.0$.\nOn exit: the updated scaling factor $\\stackrel{~}{\\alpha }=\\underset{i}{\\mathrm{max}}\\phantom{\\rule{0.25em}{0ex}}\\left(\\alpha ,\\left|{x}_{i}\\right|\\right)$.\n5:     $\\mathbf{sumsq}$ – Real (Kind=nag_wp)Input/Output\nOn entry: the scaled sum of squares $\\xi$. On the first call to f06fjf ${\\mathbf{sumsq}}=1.0$.\nConstraint: ${\\mathbf{sumsq}}\\ge 1.0$.\nOn exit: the updated scaled sum of squares $\\stackrel{~}{\\xi }$, satisfying: $1\\le \\stackrel{~}{\\xi }\\le \\xi +n$.\n\nNone.\n\nNot applicable.\n\n## 8Parallelism and Performance\n\nf06fjf is not threaded in any implementation.\n\nNone.\n\nNone." ]

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[ "", null, "7.3* The Natural Exponential Function INVERSE FUNCTIONS In this section, we will learn about: The natural exponential function and its properties.\n\nPresentation on theme: \"7.3* The Natural Exponential Function INVERSE FUNCTIONS In this section, we will learn about: The natural exponential function and its properties.\"— Presentation transcript:\n\n7.3* The Natural Exponential Function INVERSE FUNCTIONS In this section, we will learn about: The natural exponential function and its properties.\n\nLAWS OF EXPONENTS If x and y are real numbers and r is rational, then 1. e x+y = e x e y 2. e x-y = e x /e y 3. (e x ) r = e rx Laws 7\n\nThe natural exponential function has the remarkable property that it is its own derivative. DIFFERENTIATION Formula 8\n\nThe function y = e x is differentiable because it is the inverse function of y = l n x.  We know this is differentiable with nonzero derivative. To find its derivative, we use the inverse function method. Formula 8—Proof DIFFERENTIATION\n\nLet y = e x. Then, l n y = x and, differentiating this latter equation implicitly with respect to x, we get: Formula 8—Proof DIFFERENTIATION\n\nDifferentiate the function y = e tan x.  To use the Chain Rule, we let u = tan x.  Then, we have y = e u.  Hence, DIFFERENTIATION Example 4\n\nIn general, if we combine Formula 8 with the Chain Rule, as in Example 4, we get: DIFFERENTIATION Formula 9\n\nFind y’ if y = e -4x sin 5x.  Using Formula 9 and the Product Rule, we have: DIFFERENTIATION Example 5\n\nDownload ppt \"7.3* The Natural Exponential Function INVERSE FUNCTIONS In this section, we will learn about: The natural exponential function and its properties.\"\n\nSimilar presentations" ]

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[ "Sunday, 05 December, 2004\n\nThe 20th of November issue of New Scientist ran a story about the search for negative mass. Having thought about negative mass it would have to have some strange properties to exist in the universe.\n\nFirstly, as mentioned in the article, it would be repulsed by gravity instead of being attracted. A blob of matter made out of negative mass dropped from your hand would rise like a helium balloon and eventually leave the atmosphere and drift out into space.\n\nI thought about this carefully and wondered where does the energy come from for this effect to happen? I quickly concluded that the energy must come from convertion of the negative mass to energy! How did I reach this conclusion?\n\nI borrowed Einstien's thought experiment with regard to photons climbing out of a gravitation field. Imagine a tower with a magical machine at both the bottom and the top. The machine at the bottom of the tower converts a given mass into one photon with exactly the energy contained in the mass and fires up towards the top of the tower. The machine at the top does the reverse, it turns the single photon back to a piece of matter equal in energy to the photon.\n\nFrom this simple set-up we can quickly determine that a photon must lose energy climbing out of a gravitational field. Why? Well, let's first assume the opposite; that a photon doesn't lose energy climbing out of a gravitational field. We start the process by dropping a weight over the side of the tower. It accelerates as it falls due to the influence of gravity and therefore gains kinetic energy. When it reaches the bottom of the tower the energy stored in the mass of the weight and the total kinetic energy are converted into a single photon. The photon climbs up the tower to the top where it is converted back into a mass. But this time the weight has more mass and therefore more energy than it did the first time it started the process. We've created energy out of nothing which is a clear violation of the first law of thermodynamics. The conclusion? The photon must lose energy as it climbs its way out of a gravitational field. If you don't like the idea of \"magic\" devices then consider the same experiment where the conversion process has an efficiently less than one hundred percent. I'll leave the analysis of that problem as an exercise to the reader.\n\nWhen light leaves the surface of the earth and goes off into space it loses exactly the amount of energy that a mass of equal energy would need to remove itself from earth and go off into space. This energy is not lost in the speed of the light since light always moves at the same speed, which is around three hundred million metres per second, but it is lost via a reduction in the light's frequency. This has a number of important consequences such as that objects with a large gravitational field appear colder and time runs more slowly the \"deeper\" into the field you go.\n\nI'll digress from the discussion of negative mass to justify these points because I feel they're quite cool. The reduction in temperature of an object in a large gravitational field is a fairly straightforward to prove. When the blackbody radiation of the object attempts to leave the gravitational field the light is all red shifted by an equal amount since the amount of energy lost is proportional to the amount of energy the photon had in the first place. This has the effect of shifting the blackbody-radiation profile downwards. Therefore, if we measure the tempreture of the object outside the gravitational field we'll conclude it's colder than it really is.\n\nThe time dilation effect is really suprising but is also pretty straightforward providing you know the background physics. Light is a physical clock of sorts. The frequency of light amounts to a certain number of wave crests per second so in principle we could build a clock to count these waves crests and therefore accurately measure time. Suppose we were viewing at a large distance a clock field that was placed on the surface of a Neutron star. When we looked at the clock we would count less wave crests per second than someone looking at the clock from the surface of the star.\n\nOkay, so that's one example of a clock running slow in a gravitational field. How do we know we get an identical effect for other types of clocks such as swinging pendulums or the decay of uranium atoms? The answer lies in the principle of relativity; that is, that there are no preferred frames of references in the universe, everything depends on your relative situation with regard to what your observing. If some processes slowed more than others then this would hint at a prefered and universal concept of time. The universe would have a master clock as such and our experience of time would be linked to that clock. Clearly this flies in the face of the principle of relativity so we must conclude that the world doesn't work this way. If one physical process slows down all of them do and they all slow by exactly the same amount.\n\nSo back to our discussion on negative mass. All we need to do to start exploring the physics of negative mass is to modify Einstien's thought experiment slightly. Replace a magical mass to light convertor with a magical mass to anti-mass convertor. If we drop our weight over the side it will fall down the tower towards the mass to anti-mass convertor. When the weight hits the bottom it is convertor to anti-mass and the mass starts to float back up the tower. At the top it bumps into the anti-mass to mass convertor and starts to fall down and we repeat the process ad infinitum.\n\nIf the anti-mass didn't lose energy when it climbed through the gravitational field we'd have a serious problem because the machine would be creating energy out of nothing. So it must lose energy from somewhere to be consistent with the first law of thermodynamics and the only store of energy the anti-mass has its mass. Therefore, when an object composed of anti-mass rises in a gravitational field it converts it's mass into kinetic energy.\n\nIt's interesting that anti-mass is permitted by the physics of today but it hasn't been observed. More over, there is no process known that can produce anti-mass. To me it implies theres a \"squaring\" going on somewhere that we don't know about. If I had to hedge my bet I'd say that theres some unknown quantity that's involved in the quantum description of gravity that you square to obtain the mass of a particle. That way a particle's mass would always have to be positive provided that it didn't make sense to represent the quantity as a complex number.\n\nSimon.\n\n14:21:51 GMT | #Randomness | Permalink" ]

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[ "# VMs: Proposed method\n\n```Firstly the universal percentages.\n\nTake a sample of input texts in various languages. The larger the number of\nsamples the better. Calculate the percentage occurance for each letter.\nUsing these percentages calculate the weighted mean for each percentage.\nThese will be the universal percentages for the respective languages.\n\nVMS universal percentages.\n\nAs with the above method calculate universal percentages by comparing\ndifferent sections of the text both overlapping and non overlapping.\n\nIndexed percentages.\n\nFor each EVA glyph assign a fixed number of substitutes that will be used in\na similar fashion as a vigenere key. These can be selected from both upper\nand lower case letters, digits or special characters. As long as no one\nglyph can be substituted with the same symbol as any other glyph. So if we\ntake ch as an example. Substitute the first occurance for a, the second for\nb, the third for c etc. Taking the universal percentages for these\nsubstitutions should indicate whether or not a shifting cipher could be the\nsolution by analysing the effects on the text.\n\nI know this sounds wrong but it would be interesting to try it. Let me know\nwhat you think. I'd hate to spend time on a dumb idea.\n\njeff\n\n______________________________________________________________________\nTo unsubscribe, send mail to majordomo@xxxxxxxxxxx with a body saying:\nunsubscribe vms-list\n\n```" ]

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[ "# Depth of a deep learning model\n\nWhat is the depth of a deep learning model", null, "Emroj Hossain\nTue Dec 24 2019\n\n## Deep learning model\n\nA deep learning model is a machine learning model in which the complex representations are represented in terms of relatively simple representation. For example, if a machine learning algorithm wants to detect a human face from an image, it can first detect edges in the first layer. In the next layer, the edges and other simple features can represent relatively more complex features such as the outline of eyes nose, etc. These complex features together represent higher-level features such as a human face.\n\n## Depth of a deep learning model\n\nThe depth of a deep learning model can be described in many ways-\n\nOne of the methods of determining the depth of a neural network or a machine learning model is based on the number of sequential instructions to be performed to evaluate the architecture. In this method, the depth of the model is determined by the longest chain in the calculation path for the model. For the example of the above deep learning model to detect the human face each layer will need several calculations. For example, there are three layers and each layer needs n calculation then. The depth of the model will be 3n. The number of calculation steps depends on the programming language used to code the model, And as a result this type of depth of model is dependent on the programming language used and hence, not preferred by many\n\nIn another method the depth of the deep learning model layers are assumed to be no of layer is used to built the model. For the above example three layers are used to determine the face. So, in this case, the depth of the model will be considered as three. I personally prefer this method to determine the depth of the graph.\n\nFor more complex problems, no of building blocks or no of separate blocks can be also considered as depth." ]

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[ "The VAR.P function is a statistical function that can calculate and return the variance of a population. This formula was essentially introduced as a replacement for the outdated VARP function. Variance provides a general idea of how much data points are spread in a data set, and can be used to calculate the standard deviation or determine the risk of an investment. In this guide, we’re going to show you how to use the VAR.P function and also go over some tips and error handling methods.\n\n## Supported versions\n\n• Excel 2010 and newer versions\n\n## VAR.P Function Syntax\n\nVAR.P(number1,[number2],...)\n\n## Arguments\n\n number1 The first number corresponding to the entire population. [number2] Optional. Any other number arguments. Up to 254 can be entered corresponding to a sample of a population.\n\n## VAR.P Function Example\n\nThe VAR.P function accepts numeric values as its arguments. Any type of text or logical values will be ignored. You can use range references or static values just like in any other formula.\n\nThe function calculates the variance using the following function:", null, "• x: population mean (average)\n• x ̅: element of the population\n• n: population size\n\nHere is the comparison of the two methods:\n\n=VAR.P(population)\n\n=SUMPRODUCT(POWER(AVERAGE(population)-population,2))/COUNT(population)\n\nThe example below uses a formula with the named range sample (B5:B9).", null, "## VAR.P Function Tips\n\n• The VAR.P function assumes that its arguments represent the entire population. If your data is a sample of a population, prefer using VAR.S instead.\n• The VAR.P function ignores text and logical values like TRUE and FALSE. If you want to evaluate text values and FALSE as 0 and TRUE as 1, use the VARPA function instead.\n• Empty cells are ignored.\n• In most cases, we recommend using the VAR.P instead of the outdated VARP function.\n\n## Issues\n\n• Any error in the arguments will cause the function to return an error." ]

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[ "📜  QA – 安置测验|工作和工资|问题 9\n\n📅  最后修改于: 2022-05-13 01:56:58.044000             🧑  作者: Mango\n\n# QA – 安置测验|工作和工资|问题 9\n\n(一) 4\n(乙) 6\n(三) 8\n(四) 10答案：（一）\n\n∑(M i E i ) D 1 H 1 / W 1 = ∑(M j E j ) D 2 H 2 / W 2 ，其中\n∑(M i E i ) = (3 xm) + (4 xw)\n∑(M j E j ) = (13 xm) + (24 xw)，其中“m”是每个男人的效率，“w”是每个女人的效率\nD 1 = 10 天\nD 2 = 2 天\nH 1 = 12 小时\nH 2 = 12 小时\nW 1 = W 2 = 要完成的工作所以，我们有\n(3m + 4w) x 10 x 12 = (13m + 24w) x 2 x 12\n=> 15m + 20w = 13m + 24w\n=> 2m = 4w\n=> 米 = 2w\n=> m: w = 2: 1\n\n∑(M i E i ) D 1 H 1 / W 1 = ∑(M j E j ) D 2 H 2 / W 2 ，其中\n∑(M i E i ) = (3 xm) + (4 xw)\n∑(M j E j ) = (12 xm) + (1 xw)\nD 1 = 10 天\nD 2 = 12 名男性和 1 名女性所需的天数\nH 1 = 12 小时\nH 2 = 12 小时\nW 1 = W 2 = 要完成的工作所以，我们有\n(3m + 4w) x 10 x 12 = (12m + w) x D 2 x 12\n=> 30m + 40w = (12m + w) x D 2\n=> 60k + 40k = (24k + k) x D 2\n=> 100k = 25k x D 2\n=> D 2 = 4" ]

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[ "# 关于windows下Tensorflow和pytorch安装教程\n\n1、Tensorflow介绍\n\n2、Tensorflow安装（cpu版本）\n\n```\n```\n1. pip install tensorflow==1.8.0\n\n3、新版本安装\n\n```\n```\n1. pip install tensorflow\n2. pip install -U --ignore-installed wrapt enum34 simplejson netaddr\n3. pip install tensorflow\n\n4、验证代码\n\n```\n```\n1. import tensorflow as tf\n2. hello = tf.constant('Hello, TensorFlow!')\n3. sess = tf.Session()\n4. print(sess.run(hello))\n\n5、警告Your CPU supports instructions that this TensorFlow binar......处理\n\n```\n```\n1. #忽略警告处理方法\n2. import os\n3. os.environ['TF_CPP_MIN_LOG_LEVEL'] = '2'\n\n6、GPU版本安装（win10+1050ti)1、下载安装CUDA（找到安装包一直下一步）", null, "2、安装CUDNN(需要登录才能下载)", null, "", null, "", null, "3、安装tensirflow-gup\n\n```\n```\n1. pip install tensorflow-gpu==1.8.0\n\npytorch官网：https://pytorch.org/\n\n1、pytorch介绍\n\npytorch是一个python优先的深度学习框架，是一个和tensorflow，Caffe，MXnet一样，非常底层的框架。\n\nTorch 自称为神经网络界的 Numpy, 因为他能将 torch 产生的 tensor 放在 GPU 中加速运算 (前提是你有合适的 GPU), 就像 Numpy 会把 array 放在 CPU 中加速运算. 所以神经网络的话, 当然是用 Torch 的 tensor 形式数据最好。 就像 Tensorflow 当中的 tensor 一样。pytorch是一个动态的建图的工具。不像Tensorflow那样，先建图，然后通过feed和run重复执行建好的图。相对来说，pytorch具有更好的灵活性。\n\n2、安装", null, "```\n```\n\n3、验证代码\n\n```\n```\n1. import torch\n2. print(torch.__version__)" ]

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[ "# How to determine the rate of input in a queue M/M/c?\n\nI know the exit rate ($$\\mu$$) and the average waiting time in the queue ($$W_q$$). I need solve to rate of input ($$\\lambda$$) in a queue.\n\n$$\\rho = \\frac{\\lambda}{c\\mu} < 1$$\n\n$$\\pi_0 = \\left[\\left(\\sum_{k=0}^{c-1}\\frac{(c\\rho)^k}{k!} \\right) + \\frac{(c\\rho)^c}{c!}\\frac{1}{1-\\rho}\\right]^{-1}$$\n\n$$W_q=\\frac{1}{(c-1)!}\\left(\\frac{\\lambda}{\\mu}\\right)^c\\pi_0\\frac{\\mu}{(c\\mu-\\lambda)^2}$$\n\nBut I do not know how to solve this sum. Is there a simple analytical solution?\n\n• Why not just add up the $c$ values $(c\\rho)^k/k!$? – jbowman Jan 29 at 17:47" ]

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[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A266195 Match-making permutation: start with a(1) = 1, then always choose for a(n) the least unused number such that multiplying a(n) by a(n-1) does not produce any carries when performed in base 2. 17\n 1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 17, 13, 32, 14, 18, 20, 19, 33, 15, 34, 22, 64, 21, 24, 36, 28, 65, 23, 66, 25, 40, 35, 72, 42, 48, 37, 68, 26, 128, 27, 129, 29, 130, 30, 132, 31, 256, 38, 80, 49, 73, 56, 136, 41, 96, 69, 144, 67, 84, 97, 137, 112, 145, 134, 160, 50, 133, 76, 161, 100, 257, 39, 258, 43, 260, 44 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,2 COMMENTS More formally: the lexicographically earliest injection of natural numbers such that for any n > 1, A061858(a(n), a(n-1)) = 0; a(1) = 1. By necessity also surjective on N (see below for why), thus a bijection. Less formally: In this context we say that two positive natural numbers x and y \"match\", when they will not produce any carries when multiplied in binary system (see the Examples). The purpose of this sequence is with a simple greedy algorithm to form pairs of natural numbers that \"match to each other\" according to that criterion. Note that each number after 1 will satisfy the matching condition both with its predecessor and its successor. For the sake of this discussion, we call a natural number n \"dense\" if the density of 1-bits in its binary representation (cf., e.g., A265917) is over a certain threshold, whose exact value we leave undefined, but can be subjectively gauged. In contrast, we call a number \"ethereal\" if its base-2 representation consists mostly of zeros. E.g., 258 = 100000010_2 is clearly one of the \"ethereals\", while 43 = 101011_2, is definitely on the denser side. When running the algorithm, we note that after a while, for long stretches of time, it mostly matches \"dense\" numbers with \"ethereal\" numbers, like 258 and 43, which occur next to each other in the sequence as a(76) and a(77), and also a(49)=31 and a(50)=256, which are the most dense and most ethereal members of their respective binary sizes (see the Example section). Also, it should be obvious that each number of the form 2^k (terms of A000079, the \"super-ethereals\") occur as the first representative of the numbers of the same binary length, and any number of the form (2^k)-1 (A000225, \"super-dense\") comes as the last of the numbers of binary length k. No matter how dense some number might look to us, there is always a sufficiently ethereal number with which it can be mated (that is, the algorithm is never stuck, because it can always try the next unused super-ethereal 2^k if everything else fails). Moreover, whenever that next 2^k has appeared, it also always immediately picks up from the backlog of (more or less dense) numbers the least unmatched number so far, which proves that no number is left out, and the sequence is indeed a permutation of the natural numbers. However, certain numbers intuitively feel to be much better matches to each other, like 10 and 12 (cf. Examples), because they are not so distant from each other. We define \"good matches\" to be such pairs that the binary length (A070939) of the numbers is equal. As 10 and 12 are both four bits long, they are one instance of such a good match. Note that 10 is also a good match with the immediately preceding number in the sequence, 9 = 1001_2. Sequence A266197 gives the positions of these good matches, and A265748 & A265749 give their first and second members respectively. It is an open question whether the algorithm generates an infinite number of good matches or not. LINKS Antti Karttunen, Table of n, a(n) for n = 1..2694 Eric Angelini, a(n)*a(n+1) shows at least twice the same digit, Posting on SeqFan-list Dec 21 2015. [Source of inspiration for this sequence.] Rémy Sigrist, Logarithmic scatterplot of the first 500000 terms EXAMPLE For n=11, we first note that a(10) = 10, and the least unused number after a(1) .. a(10) is 11. Trying to multiply 10 (= 1010_2) and 11 (= 1011_2), in the binary system results in      1011   *  1010   -------    c1011   1011   -------   1101110 = 110, and we see that there's a carry-bit (marked c) affecting the result, thus A048720(10,11) < 10*11 and A061858(10,10) > 0, thus we cannot select 11 for a(11). The next unused number is 12, and indeed, for numbers 10 and 12 (= 1100_2), the binary multiplication results in      1100   *  1010   -------     1100   1100   -------   1111000 = 120, which is a clean product without carries (i.e., A061858(10,12) = 0), thus 12 is selected to be a match for 10, and we set a(11) = 12. For a(49) = 31 (= 11111_2) and a(50) = 256 (= 100000000_2) the multiplication results in       100000000     *     11111   -------------       100000000      100000000     100000000    100000000   100000000   -------------   1111100000000 = 7936, and we see that the carryless product is this time obtained almost trivially, as the other number is so much larger and more spacious than the other that they can easily avoid any clashing bits that would produce carries. PROG (Scheme, with defineperm1-macro from Antti Karttunen's IntSeq-library) ;; Warning: this algorithm is quite \"dense\": (defineperm1 (A266195 n) (cond ((= 1 n) n) (else (let ((prev (A266195 (- n 1)))) (let loop ((k 1)) (cond ((and (not-lte? (A266196 k) (- n 1)) (zero? (A061858bi k prev))) k) (else (loop (+ 1 k))))))))) ;; In above code (A061858bi x y) is two-argument function returning the difference between x*y - A048720(x, y). See entries A048720 and A061858. ;; We consider a > b (i.e. not less than b) also in case a is #f. ;; (Because of the stateful caching system used by defineperm1-macro): (define (not-lte? a b) (cond ((not (number? a)) #t) (else (> a b)))) CROSSREFS Inverse permutation: A266196. Cf. A000079, A000225, A048720, A061858, A070939, A265917. Cf. A266194 (products of these pairs). Cf. A266197 (indices of good matches), Cf. A265748, A265749 (give the first and second members of good matches). Cf. A266186 (when 2^n appears), A266187 (when (2^n)-1 appears). Cf. A266191, A266351 (similar permutations). Cf. also A235034, A235035. Sequence in context: A334433 A334435 A334436 * A102530 A266196 A215501 Adjacent sequences:  A266192 A266193 A266194 * A266196 A266197 A266198 KEYWORD nonn,base AUTHOR Antti Karttunen, Dec 26 2015 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified August 9 20:09 EDT 2020. Contains 336326 sequences. (Running on oeis4.)" ]

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[ "Definition\n\nAdds two integers and replaces the first integer with the sum, as an atomic operation.\n\n Add(Int32, Int32) Adds two 32-bit integers and replaces the first integer with the sum, as an atomic operation. Add(Int64, Int64) Adds two 64-bit integers and replaces the first integer with the sum, as an atomic operation. Add(UInt32, UInt32) Adds two 32-bit unsigned integers and replaces the first integer with the sum, as an atomic operation. Add(UInt64, UInt64) Adds two 64-bit unsigned integers and replaces the first integer with the sum, as an atomic operation.\n\nAdds two 32-bit integers and replaces the first integer with the sum, as an atomic operation.\n\npublic:\nstatic int Add(int % location1, int value);\npublic static int Add (ref int location1, int value);\nstatic member Add : int * int -> int\nPublic Shared Function Add (ByRef location1 As Integer, value As Integer) As Integer\n\nParameters\n\nlocation1\nInt32\n\nA variable containing the first value to be added. The sum of the two values is stored in location1.\n\nvalue\nInt32\n\nThe value to be added to the integer at location1.\n\nReturns\n\nInt32\n\nThe new value that was stored at location1 by this operation.\n\nExceptions\n\nThe address of location1 is a null pointer.\n\nRemarks\n\nThis method handles an overflow condition by wrapping: if the value at location1 is Int32.MaxValue and value is 1, the result is Int32.MinValue; if value is 2, the result is (Int32.MinValue + 1); and so on. No exception is thrown.\n\nApplies to\n\nAdds two 64-bit integers and replaces the first integer with the sum, as an atomic operation.\n\npublic:\nstatic long Add(long % location1, long value);\npublic static long Add (ref long location1, long value);\nstatic member Add : int64 * int64 -> int64\nPublic Shared Function Add (ByRef location1 As Long, value As Long) As Long\n\nParameters\n\nlocation1\nInt64\n\nA variable containing the first value to be added. The sum of the two values is stored in location1.\n\nvalue\nInt64\n\nThe value to be added to the integer at location1.\n\nReturns\n\nInt64\n\nThe new value that was stored at location1 by this operation.\n\nExceptions\n\nThe address of location1 is a null pointer.\n\nRemarks\n\nThis method handles an overflow condition by wrapping: if the value at location1 is Int64.MaxValue and value is 1, the result is Int64.MinValue; if value is 2, the result is (Int64.MinValue + 1); and so on. No exception is thrown.\n\nApplies to\n\nImportant\n\nThis API is not CLS-compliant.\n\nAdds two 32-bit unsigned integers and replaces the first integer with the sum, as an atomic operation.\n\npublic:\nstatic System::UInt32 Add(System::UInt32 % location1, System::UInt32 value);\n[System.CLSCompliant(false)]\npublic static uint Add (ref uint location1, uint value);\n[<System.CLSCompliant(false)>]\nstatic member Add : uint32 * uint32 -> uint32\nPublic Shared Function Add (ByRef location1 As UInteger, value As UInteger) As UInteger\n\nParameters\n\nlocation1\nUInt32\n\nA variable containing the first value to be added. The sum of the two values is stored in location1.\n\nvalue\nUInt32\n\nThe value to be added to the integer at location1.\n\nReturns\n\nUInt32\n\nThe new value that was stored at location1 by this operation.\n\nAttributes\n\nExceptions\n\nThe address of location1 is a null pointer.\n\nApplies to\n\nImportant\n\nThis API is not CLS-compliant.\n\nAdds two 64-bit unsigned integers and replaces the first integer with the sum, as an atomic operation.\n\npublic:\nstatic System::UInt64 Add(System::UInt64 % location1, System::UInt64 value);\n[System.CLSCompliant(false)]\npublic static ulong Add (ref ulong location1, ulong value);\n[<System.CLSCompliant(false)>]\nstatic member Add : uint64 * uint64 -> uint64\nPublic Shared Function Add (ByRef location1 As ULong, value As ULong) As ULong\n\nParameters\n\nlocation1\nUInt64\n\nA variable containing the first value to be added. The sum of the two values is stored in location1.\n\nvalue\nUInt64\n\nThe value to be added to the integer at location1.\n\nReturns\n\nUInt64\n\nThe new value that was stored at location1 by this operation.\n\nAttributes\n\nExceptions\n\nThe address of location1 is a null pointer." ]

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[ "# Interface IndependentIncrements\n\nAll Known Subinterfaces:\nBrownianMotion\nAll Known Implementing Classes:\nBrownianBridge, BrownianMotionFromMersenneRandomNumbers, BrownianMotionFromRandomNumberGenerator, BrownianMotionLazyInit, BrownianMotionView, BrownianMotionWithControlVariate, CorrelatedBrownianMotion, GammaProcess, IndependentIncrementsFromICDF, JumpProcessIncrements, MertonJumpProcess, VarianceGammaProcess\n\npublic interface IndependentIncrements\nInterface description of a time-discrete n-dimensional stochastic process $$X = (X_{1},\\ldots,X_{n})$$ provided by independent increments $$\\Delta X(t_{i}) = X(t_{i+1})-X(t_{i})$$. Here the dimension n is called factors since this process is used to generate multi-dimensional multi-factor processes and there one might use a different number of factors to generate processes of different dimension.\nVersion:\n1.3\nAuthor:\nChristian Fries\n• ## Method Summary\n\nModifier and Type\nMethod\nDescription\nIndependentIncrements\ngetCloneWithModifiedSeed​(int seed)\nReturn a new object implementing BrownianMotion having the same specifications as this object but a different seed for the random number generator.\nIndependentIncrements\ngetCloneWithModifiedTimeDiscretization​(TimeDiscretization newTimeDiscretization)\nReturn a new object implementing BrownianMotion having the same specifications as this object but a different time discretization.\ndefault RandomVariable[]\ngetIncrement​(int timeIndex)\nReturn the increment for a given timeIndex.\nRandomVariable\ngetIncrement​(int timeIndex, int factor)\nReturn the increment for a given timeIndex and given factor.\nint\ngetNumberOfFactors()\nReturns the number of factors.\nint\ngetNumberOfPaths()\nReturns the number of paths.\nRandomVariable\ngetRandomVariableForConstant​(double value)\nReturns a random variable which is initialized to a constant, but has exactly the same number of paths or discretization points as the ones used by this BrownianMotion.\nTimeDiscretization\ngetTimeDiscretization()\nReturns the time discretization used for this set of time-discrete Brownian increments.\n• ## Method Details\n\n• ### getIncrement\n\ndefault  getIncrement(int timeIndex)\nReturn the increment for a given timeIndex. The method returns the random variable vector Δ X(ti) := X(ti+1)-X(ti) for the given time index i.\nParameters:\ntimeIndex - The time index (corresponding to the this class's time discretization)\nReturns:\nThe vector-valued increment (as a vector (array) of random variables).\n• ### getIncrement\n\nRandomVariable getIncrement(int timeIndex, int factor)\nReturn the increment for a given timeIndex and given factor. The method returns the random variable Δ Xj(ti) := Xj(ti+1)-X(ti) for the given time index i and a given factor (index) j\nParameters:\ntimeIndex - The time index (corresponding to the this class's time discretization)\nfactor - The index of the factor (independent scalar increment)\nReturns:\nThe factor (component) of the increments (a random variable)\n• ### getTimeDiscretization\n\nTimeDiscretization getTimeDiscretization()\nReturns the time discretization used for this set of time-discrete Brownian increments.\nReturns:\nThe time discretization used for this set of time-discrete Brownian increments.\n• ### getNumberOfFactors\n\nint getNumberOfFactors()\nReturns the number of factors.\nReturns:\nThe number of factors.\n• ### getNumberOfPaths\n\nint getNumberOfPaths()\nReturns the number of paths.\nReturns:\nThe number of paths.\n• ### getRandomVariableForConstant\n\nRandomVariable getRandomVariableForConstant(double value)\nReturns a random variable which is initialized to a constant, but has exactly the same number of paths or discretization points as the ones used by this BrownianMotion.\nParameters:\nvalue - The constant value to be used for initialized the random variable.\nReturns:\nA new random variable.\n• ### getCloneWithModifiedSeed\n\nIndependentIncrements getCloneWithModifiedSeed(int seed)\nReturn a new object implementing BrownianMotion having the same specifications as this object but a different seed for the random number generator. This method is useful if you like to make Monte-Carlo samplings by changing the seed.\nParameters:\nseed - New value for the seed.\nReturns:\nNew object implementing BrownianMotion.\n• ### getCloneWithModifiedTimeDiscretization\n\nIndependentIncrements getCloneWithModifiedTimeDiscretization(TimeDiscretization newTimeDiscretization)\nReturn a new object implementing BrownianMotion having the same specifications as this object but a different time discretization.\nParameters:\nnewTimeDiscretization - New time discretization\nReturns:\nNew object implementing BrownianMotion." ]

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[ "# k-fold crossvalidation: how does MSE go with k? [duplicate]\n\nThis question already has an answer here:", null, "I'm trying to get an intuition about choosing the right \"k\" for K-Fold validation. Is the following right?\n\nAverage of the OOS MSEs should generally decrease as k increases. Because, a bigger \"k\" means the training sets are larger, so we have more data to fit the model (assuming we are using the \"right\" model).\n\nVariance of the OOS MSEs should generally increase as k increases. A bigger \"k\" means having more validation sets. So we will have have more individual MSEs to average out. Since the MSEs of many small folds will be more sparse than MSEs of few large folds, variance will be higher.\n\nThank you! :)\n\n## marked as duplicate by Michael Chernick, kjetil b halvorsen, mkt, Peter Flom♦Sep 6 '18 at 11:39\n\nThis question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.\n\n## 1 Answer\n\nI would argue that your claims, and diagram, are incorrect generalizations and can be misleading.\n\n## Definitions and terminology\n\nBased on the terminology and definitions from Cross Validation and Confidence Interval of the True Error we can use an $L_2$ loss function and define the Cross Validation estimator $CV$ as follows:\n\nThe data set $D$ is chunked into $K$ disjoint subsets of the same size with $m = n / K$. Let us write $T_k$ for the $k$-th such block and $D_k$ for the training set obtained by removing the elements in $T_k$ from $D$.\n\nThe Cross Validation Estimator is the average of the errors on test block $T_k$ obtained when training the algorithm $A$ on $D_k$ $$CV(D) = \\frac{1}{K} \\sum_{k=1}^K \\frac{1}{m} \\sum_{z_i \\in T_k} L(A(D_k), z_i)$$\n\n## Bias of the CV estimator\n\nThe effect of $K$ on the bias of $CV$ depends on the shape of the learning curve:\n\n• If the learning curve has considerable slope at the given training set size, increasing $K$ tends to reduce the bias as the algorithm will train on a larger data set which will improve its bias.\n\n• If the learning curve is flat at the given training set size, then increasing $K$ tend to not impact the bias significantly\n\n## Variance of the CV estimator\n\nThe impact of $K$ on the variance of the CV estimator is even more subtle as several different, and opposing effects come in play.\n\n• If cross-validation were averaging independent estimates: then leave-one-out CV one should see relatively lower variance between models since we are only shifting one data point across folds and therefore the training sets between folds overlap substantially.\n• This is not true when training sets are highly correlated: Correlation may increase with K and this increase is responsible for the overall increase of variance in the second scenario.\n• In cases of instability of the algorithm leave-one-out CV may be blind to instabilities that exist, but may not be triggered by changing a single point in the training data, which makes it highly variable to the realization of the training set.\n\n### Example taken from the Yves Grandvalet and Yoshua Bengio (2004) paper", null, "On the left an experiment with no outliers, variance falls with $K$, on the right an experiment with outliers, variance increases with $K$" ]

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[ "Applying Lienard–Schipar’s method to solving of homogeneous fractional differential Euler-type equations on an interval\n\n• Zhukovskaya Natalia V., [email protected] Belarusian State University 4 Nezavisimost’ Avenue, Minsk 220030, Belarus\n• Sitnik Sergey M., [email protected] Belgorod State National Research University, 85 Pobeda st., Belgorod 308015, Russia\nKeywords: fractional differential Euler-type equation, Riemann–Liouville fractional integral, Riemann–Liouville fractional derivative, method of Hermitian forms, Hermite’s theorem, Lienard–Schipar’s method\n\nAbstract\n\nWe present the solution of the homogeneous fractional differential Euler-type equation on the half-axis in the class of functions representable by the fractional integral of order $\\alpha$ with the density of $L_1(0; 1)$. Using the method of Hermitian forms (Lienard–Schipar’s method), solvability conditions are obtained for the cases of two, three and a finite number of derivatives. It is shown that in the case when the characteristic equation has multiple roots original equation admits a solution with logarithmic singularities.\n\nReferences\n\n\nKilbas A. A., Srivastava H. M., and Trujillo J. J., Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam (2006). (Math. Stud.; V. 204).\n\n\nPodlubny I., Fractional Differential Equations, Acad. Press, San Diego (1999).\n\n\nSamko S. G., Kilbas A. A., and Marichev O. I., Fractional Integrals and Derivatives: Theory and Applications, Gordon and Breach Sci. Publ., Switzerland (1993).\n\n\nKrein M. G. and Naimark M. A., “The method of symmetric and Hermitian forms in the theory of separation of the roots of algebraic equations,” Linear Multilin. Algebra, 10, No. 4, 265–308 (1981). Published online: 02 Apr 2008 DOI: 10.1080/03081088108817420.\n\n\nMatveev N. M., New Methods of Integration of Ordinary Differential Equations [in Russian], Vysshaya Shkola, Moscow (1967).\n\n\nBateman H. and Erdelyi A., Higher Transcendental Functions, V. 1, McGraw–Hill (1953).\n\n\nGantmakher F. R., Matrix Theory [in Russian], Nauka, Moscow (1988).\n\n\nPostnikov M. M., Stable Polynomials [in Russian], Nauka, Moscow (1981).\n\n\nSitnik S. M., “Transmutations and applications: a survey,” arXiv: 1012.3741 [math.CA] (2010).\n\n\nKatrakhov V. V. and Sitnik S. M., “Transmutation method and new boundary-value problems for singular elliptic equations [in Russian],” Modern Math. Fund. Res., 64, No. 2, 211–426 (2018).\n\n\nSitnik S. M. and Shishkina E. L., Transmutation Method for Differential Equations with Bessel Operators [in Russian], Fizmatlit, Moscow (2018).\nHow to Cite\nZhukovskaya, N. and Sitnik, S. (&nbsp;) “Applying Lienard–Schipar’s method to solving of homogeneous fractional differential Euler-type equations on an interval”, Mathematical notes of NEFU, 25(3), pp. 33-42. doi: https://doi.org/10.25587/SVFU.2018.99.16949.\nIssue\nSection\nMathematics" ]

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[ "# compute $\\iint_De^{-x^2-4y^2} \\ dxdy$\n\nThe question is: $$\\iint_De^{-x^2-4y^2} \\ dxdy, \\quad D=\\{(x,y):0\\leq x\\leq2y\\}$$ This is how i've tried to attack this but i was getting nowhere with it : $$\\int_{0}^{\\infty}e^{-4y} \\ dy \\int_{0}^{2y}e^{-x^2}dx$$ Am i on the right path? how should i proceed\n\nAny suggestion?\n\nThe substitution $$x=r\\cos\\theta,\\,y=\\tfrac12r\\sin\\theta$$ of Jacobian $$r/2$$ writes the integral as$$\\int_{\\pi/4}^{\\pi/2}d\\theta\\int_0^\\infty\\tfrac12re^{-r^2}dr=\\tfrac{\\pi}{16}.$$\n• Can i ask you how did you find angles $\\pi/4$ and $\\pi/2$ ? Jan 10, 2021 at 15:43\n• @simon Note $\\tan\\theta=2y/x\\ge1$.\n• @simon as far as the limits, please note that your original integral is over the line $x = 2y$ and with the substitution it would mean $\\tan \\theta = 1$. Jan 10, 2021 at 16:11" ]

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[ "NumWords.com\n\n# How to write Two thousand seven hundred thirty-one in numbers in English?\n\nWe can write Two thousand seven hundred thirty-one equal to 2731 in numbers in English\n\n< Two thousand seven hundred thirty :||: Two thousand seven hundred thirty-two >\n\nFive thousand four hundred sixty-two = 5462 = 2731 × 2\nEight thousand one hundred ninety-three = 8193 = 2731 × 3\nTen thousand nine hundred twenty-four = 10924 = 2731 × 4\nThirteen thousand six hundred fifty-five = 13655 = 2731 × 5\nSixteen thousand three hundred eighty-six = 16386 = 2731 × 6\nNineteen thousand one hundred seventeen = 19117 = 2731 × 7\nTwenty-one thousand eight hundred forty-eight = 21848 = 2731 × 8\nTwenty-four thousand five hundred seventy-nine = 24579 = 2731 × 9\nTwenty-seven thousand three hundred ten = 27310 = 2731 × 10\nThirty thousand forty-one = 30041 = 2731 × 11\nThirty-two thousand seven hundred seventy-two = 32772 = 2731 × 12\nThirty-five thousand five hundred three = 35503 = 2731 × 13\nThirty-eight thousand two hundred thirty-four = 38234 = 2731 × 14\nForty thousand nine hundred sixty-five = 40965 = 2731 × 15\nForty-three thousand six hundred ninety-six = 43696 = 2731 × 16\nForty-six thousand four hundred twenty-seven = 46427 = 2731 × 17\nForty-nine thousand one hundred fifty-eight = 49158 = 2731 × 18\nFifty-one thousand eight hundred eighty-nine = 51889 = 2731 × 19\nFifty-four thousand six hundred twenty = 54620 = 2731 × 20\n\nSitemap" ]

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[ "# 1014 Moles to Kilomoles\n\nMole\n• Kilomole\n• Millimole\n• Mole\n• Pound-Mole\n=\nKilomole\n• Kilomole\n• Millimole\n• Mole\n• Pound-Mole\nFormula 1,014 mol = 1014 / 1000 kmol = 1.014 kmol\n1,014 mol = 1.014 kmol\n\nExplanation:\n• 1 mol is equal to 0.001 kmol, therefore 1014 mol is equivalent to 1.014 kmol.\n• 1 Mole = 1 / 1000 = 0.001 Kilomoles\n• 1,014 Moles = 1014 / 1000 = 1.014 Kilomoles\n\n## 1014 Moles to Kilomoles Conversion Table\n\nMole (mol) Kilomole (kmol)\n1,014.1 mol 1.0141 kmol\n1,014.2 mol 1.0142 kmol\n1,014.3 mol 1.0143 kmol\n1,014.4 mol 1.0144 kmol\n1,014.5 mol 1.0145 kmol\n1,014.6 mol 1.0146 kmol\n1,014.7 mol 1.0147 kmol\n1,014.8 mol 1.0148 kmol\n1,014.9 mol 1.0149 kmol\n\n## Convert 1014 mol to other units\n\nUnit Unit of Chemical Amount\nMillimole 1,014,000 mmol\nPound-Mole 2.2355 lb-mol\nKilomole 1.014 kmol" ]

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[ "ama-team/cookbook-linux-user-management\n\nShowing 9 of 26 total issues\n\nMethod eql? has a Cognitive Complexity of 8 (exceeds 5 allowed). Consider refactoring. Open\n\ndef eql?(other)\nreturn false unless other.is_a?(self.class) || is_a?(other.class)\nreturn false unless other.instance_variables == instance_variables\ninstance_variables.each do |variable|\nvalue = other.instance_variable_get(variable)\nFound in files/default/lib/mixin/entity.rb - About 45 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod process_public_keys has a Cognitive Complexity of 7 (exceeds 5 allowed). Consider refactoring. Open\n\ndef process_public_keys(current_state, desired_state)\nreturn [] if desired_state.nil? && !current_state.policy.remove?\naccount = desired_state || current_state\ncurrent_keys = current_state ? current_state.public_keys : {}\ndesired_keys = desired_state ? desired_state.public_keys : {}\nFound in files/default/lib/planner/account.rb - About 35 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod process has a Cognitive Complexity of 7 (exceeds 5 allowed). Consider refactoring. Open\n\ndef process(account, current_state, desired_state)\ncurrent_remotes = current_state ? current_state.remotes : {}\ndesired_remotes = desired_state ? desired_state.remotes : {}\nkey = desired_state || current_state\nactions = @remote.plan(\nFound in files/default/lib/planner/account/private_key.rb - About 35 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod process_privileges has a Cognitive Complexity of 7 (exceeds 5 allowed). Consider refactoring. Open\n\ndef process_privileges(current_state, desired_state)\nreturn [] if desired_state.nil? && !current_state.policy.remove?\naccount = desired_state || current_state\ncurrent_privileges = current_state ? current_state.privileges : {}\ndesired_privileges = desired_state ? desired_state.privileges : {}\nFound in files/default/lib/planner/account.rb - About 35 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod privilege_actions has a Cognitive Complexity of 7 (exceeds 5 allowed). Consider refactoring. Open\n\ndef privilege_actions(current_state, desired_state)\nreturn [] if desired_state.nil? && !current_state.policy.remove?\ngroup = desired_state || current_state\ncurrent = current_state.nil? ? {} : current_state.privileges\ndesired = desired_state.nil? ? {} : desired_state.privileges\nFound in files/default/lib/planner/group.rb - About 35 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod process_private_keys has a Cognitive Complexity of 7 (exceeds 5 allowed). Consider refactoring. Open\n\ndef process_private_keys(current_state, desired_state)\nreturn [] if desired_state.nil? && !current_state.policy.remove?\naccount = desired_state || current_state\ncurrent_keys = current_state ? current_state.private_keys : {}\ndesired_keys = desired_state ? desired_state.private_keys : {}\nFound in files/default/lib/planner/account.rb - About 35 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod contains has a Cognitive Complexity of 6 (exceeds 5 allowed). Consider refactoring. Open\n\ndef contains(path)\ncursor = @root\npath.each do |segment|\nreturn false unless cursor.respond_to?(:key?)\ncandidates = [segment.to_s, segment.to_sym]\nFound in files/default/lib/model/client/role_tree.rb - About 25 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod process_partition has a Cognitive Complexity of 6 (exceeds 5 allowed). Consider refactoring. Open\n\ndef process_partition(state, partition, clients)\nunless partition.policy.group == Model::Policy::NONE\nstate.groups[partition.id] = extract_group(partition)\nend\nclients.values.each do |client|\nFound in files/default/lib/state/builder.rb - About 25 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"\n\nMethod extract_account has a Cognitive Complexity of 6 (exceeds 5 allowed). Consider refactoring. Open\n\ndef extract_account(client, partition)\naccount = Model::Account.new(client.id)\naccount.policy = partition.policy.account\nunless client.public_keys.empty?\naccount.public_keys!(client.id).merge!(client.public_keys)\nFound in files/default/lib/state/builder.rb - About 25 mins to fix\n\nCognitive Complexity\n\nCognitive Complexity is a measure of how difficult a unit of code is to intuitively understand. Unlike Cyclomatic Complexity, which determines how difficult your code will be to test, Cognitive Complexity tells you how difficult your code will be to read and comprehend.\n\nA method's cognitive complexity is based on a few simple rules:\n\n• Code is not considered more complex when it uses shorthand that the language provides for collapsing multiple statements into one\n• Code is considered more complex for each \"break in the linear flow of the code\"\n• Code is considered more complex when \"flow breaking structures are nested\"" ]

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[ "Вы находитесь на странице: 1из 22\n\n# CHAPTER 1\n\nMathcad is a product of MathSoft inc. The Mathcad can help us to calculate, graph, and communicate technical ideas. It lets us work with mathematical expressions using standard math notation - but with the added ability to recalculate, view, present, and publish with ease, even to the Web. We shall explore the following functionalities provided by Mathcad , Numeric operators perform summations, products, derivatives, integrals and Boolean operations. Numeric functions apply trigonometric, exponential, hyperbolic and other functions and transforms. Symbolics simplify, dierentiate, integrate, and transform expressions algebraically. Vectors and Matrices manipulate arrays and perform various linear algebra operations, such as nding eigenvalues and eigenvectors, and looking up values in arrays. Dierential Equation Solvers support ordinary dierential equations, systems of dierential equations, and boundary value problems both at the command line and in solve blocks that use natural notation to specify the dierential equations and constraints. In this chapter, we give a brief introduction to its user interface and the basic usage of MathCad in dening, evaluating, and graphing of variables and functions.\n\n1. User Interface: Menus and Toolbars Since MathCad is a window program as shown in the following screen shot after run the Mathcad program.\n\nFigure 1. Mathcad Window The blank area is the work area that allow us typing and evaluating mathematical expression, graphing, symbolic computing, entering and running Mathcad scripting codes, and entering ordinary text. It has standard main menu bar at the top of the window as most window programs do that is shown in the following screen shot.\n\nFigure 2. Main Menu Bar But some menus deserve a little more attention. They are View, Insert, and Symbolic menus. 1.1. The View Menu. The View menu, among other features, allows a user to bring up popup menu bars for using with Mathcad .\n\n## 1. USER INTERFACE: MENUS AND TOOLBARS\n\nFigure 3. Menu Bar Mathcad has nine popup toolbars that for entering math formulas, mathematical symbol, graphing, programming, etc. We can choose to bring up each one at a time or bring up the top level popup menu toolbar by selecting Math as shown in the following screen shot,\n\nFigure 4. Top Level Popup Menu Toolbar By click each menu icon at this top level menu bar, we can bring each individual one as needed.\n\nPractice (1) Start the Mathcad program and explore it top menu system by click on each menu item. (2) Bring up the Math popup menu toolbar, move mouse over each icon to see the caption. (3) Click each icon of the Math popup toolbar to bring up other popup toolbars and explore them. 1.1.1. Graph Toolbar. The graphic toolbar contains toolbars for\n\nFigure 5. Graph Toolbar creating 2D-graph and 3D-graph that are most useful for our purpose. First click on any blank spot in the work area, which tells Mathcad where to place the graphic region; then click on the icon\n\nor type [Shift]( @ sign) Mathcad creates a 2D graphic region at the selected location that can be used to graph any equation y = f (x) or vector pair {Y, X }. We will give examples to illustrate how to graph functions and vector pair in later section. 1.1.2. Vector and Matrix Toolbar. The Vector and Matrix Toolbar contains toolbars for creating and manipulating matrices.\n\n## 1. USER INTERFACE: MENUS AND TOOLBARS\n\nThe most frequently use menu selection in the matrix popup toolbar is the create matrix icon,\n\nwhich upon click will bring up a dialog box that allow us to create a matrix with specied number of rows and columns. Another way to bring up the matrix creating dialog box is press [Ctrl][M] (hold [Ctrl] key and press [M] key)). matrix with one row or column is a vector. The matrix popup toolbar also contains menu selections that help us to nd eigenvalues and eigenvectors, to access each individual row or column, or nd inverse. However once we remember the hot-key or function-call for each operation, we dont really need these toolbars. Using hot-key or function-call is a much quick way to invoke the menu selection. For example, to nd inverse of an given matrix A, we only need to type A^{-1} (which will show as A1 in the workplace) and = (which tells Mathcad to execute the operation) to nd it. 1.1.3. Evaluation Toolbar. The Evaluation Toolbar contains tool-\n\nFigure 7. Evaluation Toolbar bar for assignment operator :=, global denition operator , and the evaluation command =. Notice in Mathcad , equal sign = is used as evaluation command that tells Mathcad to carry out the computation. There are two execution mode, one is automatic, in this mode, Mathcad carries out execution when we press [=]; another is manual, in this mode you will have rst press [=] followed by [F9] key to let Mathcad to carry out the execution. The manual mode is desirable when we want to delay the execution and dont want to be bothered by many error message Mathcad would produce in automatic mode. You turn o/on the automatic mode from the Math menu on the Main Menu bar.\n\nIn the automatic execution mode, after we type 3+2= we get 3 + 2 = 5. If we want to assign 2 to variable x we will have to type x:2 , which will be displayed as x:=2 in the work area. Therefore, a mathematical expression y =x+3 will be enter dierently according to the meaning of the expression, (1) Entering it as y:x+3 when x value has been dened before this expression and you want y to be the value of x plus 3, i.e. assign the value of x + 3 to y . (2) Entering it as y(x):x+3 when we want to set y as a function of x dened as x + 3. (3) Entering it as y[Ctrl]=x+3 when we like to compare y and x + 3, i.e. to logically determine if y has the same value as x + 3, here [Ctrl]= mean enter = while hold the [Ctrl] key. We can get the logic equal sign = from the Boolean Toolbar as shown below. Notice you also get , , =, and other logic operation from the Boolean Toolbar.\n\nFigure 8. Boolean Toolbar The global assignment operator (shortcut ) is to assign an value (expression) to a variable (a function) so it can be referred in any place of the work area. The following screen shot illustrates the dierence between =, :=, =, and ,\n\n## 1. USER INTERFACE: MENUS AND TOOLBARS\n\nThe reason behind these dierences and using of Mathcad is that scans your input from top to bottom and left to right as illustrated in the following diagram.\nScan Start Top-Left to\n\u000f Bottom\n\nto\n\n/ Right\n\nSo a variable or a function must be dened before it is used. The will give us a leverage to defy this scan order so we can dene and use a variable (function) without worrying about the proper order demanded by the Mathcad as illustrated in the following two screen shots. Notice red font indicate a error.\n\n1.1.4. Calculus Toolbar. The Calculus Toolbar as shown below allow us to perform computations found in Calculus. We will give detail description in later section about the symbolic computation using Mathcad\n\nFigure 9. Calculus Toolbar 1.2. The Insert Menu. The Insert menu,among other features, allows a user insert a text region in the Mathcad work place or math region inside text region,\n\nFigure 10. Insert Menu In side the text region, user can enter any text as she/he would do in a normal text editor with any format setting that can be specied through the Format->style menu. Another way to insert a text region is to type double quotation mark . Mathcad also allow a user to insert a math region inside a text region so users can insert a mathematical expression in the text region.\n\n## 1. USER INTERFACE: MENUS AND TOOLBARS\n\nA third usage we are interested in the Insert menu is to insert a graphic region, a matrix, and a special build in function. The following screen shot shows the case of inserting a matrix.\n\nFigure 11. Insert a Matrix 1.3. The Symbolic Menu. The Symbolic menu, allows us to\n\nFigure 12. Insert Menu carry out symbolic computation. We can factor polynomial, simplify an complex expression, nd antiderivative, Laplace transform and inverse Laplace transform, etc.\n\n10\n\nThe following screen shot shows the case of evaluating an expression symbolically.\n\nFigure 13. Evaluate Another screen shot to show nding Laplace transform symbolically.\n\nFigure 14. Laplace Transform Practice (1) Find dierence between =, := ([Shift][;], :), =([Ctrl][=]) and ([Shit][], sign) with mathematical express y = x + 3. (2) Dene some matrix in you working area and place with Matrix toolbar. (3) Place graph region in the work area, the typing something in the place hold .\n\n## 1. USER INTERFACE: MENUS AND TOOLBARS\n\n11\n\n1.4. Shortcut keystrokes. Besides using menus or popup menu toolbars to insert item into working place, we can (should) use shortcut keystrokes (or hot keys) to accomplish the same goals. One important concept in entering expression in Mathcad is the place holder. It is a little dark rectangle box . For example, when you type + (addition operator) in any blank area we get + . In the place holders we enter any valid mathematics expression. Mathcad uses arithmetic operators +, , etc in the same way as standard calculator. The following table list the keystrokes for arithmetic operators, root operators, and subscription (indexing). Keystroke + * / \\ [Ctrl]\\ [ Usage Type Display Remark add a+ a+ type an expression in subtract aa type an expression in times a* a type an expression in a divide a/ type an expression in power a a type an expression in square root \\ type an expression in nth root [Ctrl]\\ type index in the\n\nrst\n\nand\n\n## subscript a[ a type subscripts separated by , in Table 1. Arithmetic And Radical Operators\n\nSince in mathematics x = y has four dierent meanings, (1) denition: denes functions or notation, for example, in the expression f (x) = x2 + 1, one denes a function f by an expression x2 + 1. (2) assignment: assigns value of y to x; like in the sentence set x = y... (3) logical equal: x is equal to y logically; like in the sentence if x = y .... (4) result of computation: like 4 + 5 = 9 here 9 is result of adding 5 to 4. Mathcad uses = as execution command, that is to tell Mathcad to compute the value. It uses (:=) for assignment and denition, and (boldface =) for logic equal.\n\n12\n\nThe following table list keystrokes for assignment plus relational operators. Keystroke Usage Entering text : assignment ; range\nvariable\n\n## Display a := 3 a := 1 . . . a3 a=1 a< a> a a\n\nRemark\ntype any text in assign 3 to a if enter 4 in , a will have values 1, 2, 3, 4 assign 3 to a globally a is logically equal to 1 type an expression in type an expression in type an expression in type an expression in\n\n## [Shift][]() [Ctrl][=] < > [Ctrl] [Ctrl] [Ctrl]\n\nglobal assignment logic equal less than greater than less or equal greater equal or\n\nnot equal a [Ctrl] a= type an expression in Table 2. Assignment And Relational Operators\n\nThe following table lists keystrokes to bring up derivative operators, integration operators and matrix operators. Keystroke Usage [Shift][/](?) derivative [Shift][Ctrl][/] [Ctrl][I] [Shift](&) [Ctrl][M] [Ctrl]\nnth derivative indenite integral denite integral add matrix get a row of a matrix\n\n## Type [Shift][/] [Shift][Ctrl][/] [Ctrl][I] [Shift] [Ctrl][M] M[Ctrl]\n\nDisplay\nd d d d\n\nRemark\nvariable in at denominator and function in other index at two exponents able in d and function in last vari-\n\nat denominator after\n\nd d A dialog box M\n<>\n\nintegrant in rst and integral variable in second integrant in rst and integral variable in after d specify rows and columns in the dialog box specify row number in , which is between 0 and rows(M)-1 inclusive.\n\n## 2. BASIC USAGE OF Mathcad\n\n13\n\nPractice Start Mathcad and practice all shortcut keys (hot-key) listed in the four tables. 2. Basic Usage of Mathcad One can use Mathcad as ordinary graphic calculator in doing arithmetic computing like 3 + 5 = 8 , dening and graphing functions such as f (x) = 3x2 +3. When perform arithmetic computation, you just need to click any blank area and enter the expression, then press = or = and [F9] key. The result will be immediately display. For example the following key strokes 3+5= will produce 3 + 5 = 8 in the working area when aut-computation mode is on, or 3 + 5 = if the aut-computation mode is o. In later case, you will have to press [F9] key to generate the result. The dierence between = and [F9] is the following, = only evaluate expressions when the mode of execution is set to be automatic. If aut-computation mode is turn o, = will only create a . after the expression. F9 causes Mathcad to carry out all computation in a given workplace. You will have to use [F9] to tell Mathcad to do computation if you turn o the automatic compute mode, which is desirable if you dont want to be bothered by Mathcad warning message. You can turn on/o the aut-computation mode through Math menu. 2.1. Some General Editing Methods. As in most window based text editors, we can use mouse to choose input starting point by moving mouse cursor to a desired point and click the right button. Every Mathcad equation, text paragraph, and plot in a worksheet is a separate object called region. We can also use mouse to select several regions and drag them around. To select one region, just move mouse over the region and click. A light bounding box will appear, which indicates the region is selected, as in the following screen shot for a selected math region,\n\n14\n\n## and another screen shot for selected text region,\n\nFigure 15. Select Text Region To select one or more regions on the screen, move the mouse to the starting point and hold down right button and move the mouse. A selecting box appears, any item inside the selecting box is surrounding by a dashed box as shown in the following screen shot.\n\nFigure 16. Select Multiple Locations After releasing the right mouse button, all regions selected items are bounded by dash box and the mouse cursor changes to a small hand. Click and hold mouse on any of selected region, you can move the selected regions to any wanted location. To deselect, simply click on any blank region outside the selecting box. We can also select items inside each region. When we click a math region, we see a blue selecting line under the currently selected item. The blue selecting line has three dierent shape depends on where we click at in a math region. If we click at the beginning of the math region (or before a math operator like +, -, ) etc.), the blue selecting line is look like ; if we click at the middle of some text, the blue selecting line is look like ; if we click at the end of the region (or after a math operator like +, -,) etc.), the blue selecting line is look like . The following screen shot displays some of the cases, By push space bar, we can increase number of item selected, arrow keys are used to change selection and shape of the blue selection line. To select texts in a text region, we just click at any desired point inside the region, and either hold the right button down while dragging\n\n## 2. BASIC USAGE OF Mathcad\n\n15\n\nFigure 17. Click at Dierent Location or hold the [Shift] down and using the arrow keys to choose the selection as in most text editors. Furthermore, we can use [Ctrl]C (copy), [Ctrl]V (paste), and [Ctrl]X (cut) as in most text editors to copy, paste, and cut selected region(s). 2.2. Dene Variables and Functions. As in all computational program, when a computation involving a variable of functions, the variable or function must be dened before its use unless the function is predened in Mathcad . Mathcad predenes many functions, such as all trigonometric function, exponential function and logarithm function. The following table list most frequently used predened functions Function sin(x) cos(x) tan(x) exp(x) Type sin(x) cos(x) tan(x) exp(x) Display Remark sin(x) when type sin() = , you will\nget 0, since sin( ) = 0\n\n## cos(x) tan(x) exp(x)\n\nwhen type cos( ) = , you will get 1, since cos( ) = 1 when type tan( ) = , you will get 0, since tan( ) = 0 this is exponential function, you can also type e , you will get e in you can enter an valid expression\n\n(x)\n\nF[Ctrl]g(x)\n\n(x)\n\n## Heaviside function dened by 0 if x < 0 (x) = 1 if x 0\n\nln(x) ln(x) ln(x) natural logarithm function Table 4. Mathcad Some Predened Functions 2.2.1. Dene Variables. There are two type variables in Mathcad , a normal variable which holds one value, and a range variable which, like vectors in math, holds multiple values. The variable name is any sequence of letters and digits that begin with a letter such as a, a1, bx12, etc. To dene a normal variable you just click any clear region and type the name of the variable followed by :, which bring up\n\n16\n\nthe assignment operator :=, and followed by the numeral value. For example the keystrokes x:23 denes a variable named x whose value is 23, which is displayed as a := 23 in the worksheet. To dened a range variable, usually an index variable, we use ;, semicolon. The following table list several type of range variables, Variable a ax ab1 a1e ac Type a:1;10 ax:1,3;10 Display a := 1..10 ax := 1, 3..10 Remark\na takes values 1, 2, . . . , 10 ax takes values 1, 3, . . . , 10, incremented by 2. Notice the user of ,\n\n## ab1:0,.1;10 ab1 := 0, .1..10 a1e:10,9;1 a1e := 10, 9..1\n\nab1 takes values 0, 0.1,0.2, . . . , 10, incremented by 0.1. a1e takes values 10, 9, . . . , 1, decremented by 1. ac takes values 10, 9.8,9.6, . . . , 1, decremented by 0.2.\n\n## ac:10,9.8;1 ac := 10, 9.8..1\n\nTable 5. Mathcad Example of Range Variables 2.2.2. Dene and graph Functions. To dene a function in Mathcad is same as to write down a function in a piece of paper. You just type in the function name, followed by open parenthesis ( and the argument(s) that are separated by comma(,) and the close parenthesis ), then type the assignment operator colon(:) followed by typing in the expression(formula) for the given function. For example to dene a function f (x) = 4x + 3, we would type f(x):4x+3, notice you dont need to type the multiplication operator between 4 and x as Mathcad automatically insert it for you. However, if you want to dene a function such as f (t, y ) = 3t + ty , you would need to specically type in the multiplication operator *, as shown, f(t,y):3t^2+t*y , also notice that we should type no space between terms as Mathcad will insert it for us when we type in a arithmetic operator. When typing in an expression involving more than one terms, space character serves as grouping operator. Hitting space bar will cause Mathcad to group +3 , here term together. This is useful when entering expression like 3x x5 we will need to group x + 3 together by press space bar before we type the division operator(/). Another usage of space is to move out the superscript or subscript mode. The following table gives several example to illustrate the usage of space. Notice, in order to move out\n\n17\n\n## Display Remark x2+3x No space is typed x2 + 3 x\n\nafter ^ so\n\nMathcad continue entering the expression as exponents. A space is typed after ^ so Mathcad exists the power mode and enter 3x as another term.\n\nx[2+3x\n\nx2+3x\n\nNo space is typed after [, the subindex operator so Mathcad continue entering the expression as subindex.\n\nx[2 +3x\n\nx2 + 3 x\n\nA space is typed after [ so Mathcad exists the power mode and enter 3x as another term.\n\nx+\n\n3 4x+5\n\nx+3 4x+5\n\n## Table 6. Mathcad Eect of Space Character\n\nof the denominator, we need to press space too. For example, if we +3 want enter 4x + 7x, we would type in Mathcad as x+3 /4x+5 +7x x+5 that is entering one space after 3 and another space after 5 to move out the denominator. Play with the space character (space bar) to see if Mathcad can give us more surprises! To nd value of a function, such as f (3), or h(4, 2), you rst dene the function f and g and type f(3)= or h(4,2)=, Mathcad will happily nd the value for you. To graph a function of one variable after dening it, just click any blank space after the denition of the function and type @ or from Insert menu choose to insert xy-plot. A box will appear with several place holders. Enter the function name on the left center place holder and the variable on the bottom center place holder. If the variable name is not used before, the Mathcad will display a graph over default interval [-10, 10]. But if we has assign values for the variable, depending on how we assign value to variable, we might be surprised to nd no curve\n\n18\n\nis displayed. That might be due to our variable only has one value, so only a dot is plotted. To see if this is the case, double click the graph box, a popup window will show, by specifying the thickness of the trace we will see the dot if it is graphed. Another reason we dont see a curve is that the range for y is not properly specied. We can change the lower and upper limit for y variable by clicking the lower and upper left corners, place holders(or number) will appear, change it to the desired value, the graph will appear. So to graph a function, we do: (1) Dene the function. (2) Dene range value for the variable of your function. (3) Press @ and enter the function name and variable name in proper place holder. (4) Adjust the thickness of the trace. (5) Adjust the y ranges by changing the numbers on the far left (see the graph below). We have two screen shots here, the rst one shows a blanket graphing box with place holder and another graphing box display graph of f (x) = 1 x(3 x) over interval[-4, 4]. 3\n\nFigure 18. Graph of one function and blank graphing box The second one shows how to graph more than one functions in the same graph box; it also shows the conguration box and change of the weight of the curve. To graph two or more functions in one graphing box, you just type comma(,) after each function name as f(x),g(x), h(t). If two or more variable name (as in the case of screen cut) you need specify them in the place holder for variable comma separated as x,t.\n\n## 2. BASIC USAGE OF Mathcad\n\n19\n\nFigure 19. Graph of two function in one graphing box Notice, in this second screen shot t has only one value 2, so the graph displayed as one point(green). We have changed the weight of the trace to 5 so the point is big green dot. Mathcad is capable to display many type of graphs. In the later chapters we will learn how to plot 3-D graphs, such surface plot, contour plot, vector eld plot, etc. Practice Start Mathcad program, (1) Type the following expression in the work area (a) x + 3y (b) x2 + 2x 5 (c) e2x (d) x^2 *y^2 -x*y /2x*y-5, notice you need to hit space bar a few time to before you enter /, so you get expression (e) sin(3x)ex t3 e2t 3 4 (2) Dene matrices A = by entering A:[Ctrl][M] 1 18 and choose 2 for row and column numbers, entering the numbers in proper place holder. (a) Find inverse of A by type A^{-1} 2 (a) Find A by entering A*[Ctrl][M] and set row=2 4 and col=1 in the dialog box, entering the number then press =. 2 Dene b = , and nd Ab. 5 (3) Dene the following functions\nx2 y 2 xy 2xy 5\n\n20\n\n(a) f (t) = 2t et (4) (b) f (t, x) = t2 4t sin(x) 2xex (5) (c) f (t, x) = 2tx . cos(x) (6) Graph function f (t) = t2 2t 1 (a) without specifying value for t. (b) dene t = 0, 0.1 5 (c) Graph g (t) = 4 sin(t) in the same chart. (d) Graph a point (4, 5) in the same plot, now you need put t, t, 4 in the bottom place holder and f (t), g (t), 5 in the left middle place holder. You also need to double click the graph and change the weight so the point is shown as in the following screen shot.\n\nProject At beginning you should enter: Project title, your name, ss#, and due date in the following format Project One: Dene and Graph Functions John Doe SS# 000-00-0000 Due: Mon. Nov. 23rd, 2003\n\n## 2. BASIC USAGE OF Mathcad\n\n21\n\nYou should format the text region so that the color of text is dierent than math expression. You can choose color for text from Format >Style select normal and click modify, then change the settings for font. You can do this for headings etc. (1) Dene and Graph functions In the project you will do: Dene f (x) = (x 3)(x 1)(x + 2) Dene a range variable x = 5, 4 5 and nd function value at range variable. Find zero of its derivative by perform the following task, [-] dene df (x) with input df(x):[Shift][/] and enter f (x) and x in the place holder so you have df (x) := d f (x). dx [-] Type keyword Given , entering df(x)[Ctrl][=] in next line, and keyword Find following by (x), then press [Ctrl][.] [-] Find zeros of its second derivative. Determine intervals where the function is concave upward and concave downward. Type your explanation in text regions. Graph the function, its rst derivative, and second derivative in the same plot. Explain the relationship between the signs of its derivatives and concavities of its graph. (2) Matrix Calculation 3 2 4 Dene the matrix A = 1 2 1 5 8 22 Find determinant of A by type |A, notice when you type | you will get | | and in enter A. Find inverse of A. Find the solution of the following system of equation 3x1 2x2 + 4x3 = 1 x1 + 2x2 + x3 = 3 5x1 8x2 + 22x3 = 11 1 by computing A-1 b where b = 3 Solve the same 20 system of equations by [-] Enter keyword Given ,\n\n22" ]

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[ "# Using R To Estimate Spatial HAC Errors Per Conley\n\ntl;dr: Fast computation of standard errors that allows for serial and spatial auto-correlation.\n\nEconomists and political scientists often employ panel data that track units (e.g., firms or villages) over time. When estimating regression models using such data, we often need to be concerned about two forms of auto-correlation: serial (within units over time) and spatial (across nearby units). As Cameron and Miller (2013) note in their excellent guide to cluster-robust inference, failure to account for such dependence can lead to incorrect conclusions: “[f]ailure to control for within-cluster error correlation can lead to very misleadingly small standard errors…” (p. 4).\n\nConley (1999, 2008) develops one commonly employed solution. His approach allows for serial correlation over all (or a specified number of) time periods, as well as spatial correlation among units that fall within a certain distance of each other. For example, we can account for correlated disturbances within a particular village over time, as well as between that village and every other village within one hundred kilometers. As with serial correlation, spatial correlation can be positive or negative. It can be made visually obvious by plotting, for example, residuals after removing location fixed effects.", null, "Example Visualization of Spatial Correlation from Radil, S. Matthew, Spatializing Social Networks: Making Space for Theory In Spatial Analysis, 2011.\n\nWe provide a new function that allows `R` users to more easily estimate these corrected standard errors. (Solomon Hsiang (2010) provides code for STATA, which we used to test our estimates and benchmark speed.) Moreover using the excellent `lfe`, `Rcpp`, and `RcppArmadillo` packages (and Tony Fischetti’s Haversine distance function), our function is roughly 20 times faster than the STATA equivalent and can scale to handle panels with more units. (We have used it on panel data with over 100,000 units observed over 6 years.)\n\nThis demonstration employs data from Fetzer (2014), who uses a panel of U.S. counties from 1999-2012. The data and code can be downloaded here.\n\n#### STATA Code:\n\nWe first use Hsiang’s STATA code to compute the corrected standard errors (spatHAC in the output below). This routine takes just over 25 seconds.\n\n``````cd \"~/Dropbox/ConleySEs/Data\"\nclear; use \"new_testspatial.dta\"\n\ntab year, gen(yy_)\ntab FIPS, gen(FIPS_)\n\ntimer on 1\nols_spatial_HAC EmpClean00 HDD yy_*FIPS_2-FIPS_362, lat(lat ) lon(lon ) t(year) p(FIPS) dist(500) lag(5) bartlett disp\n\n# *-----------------------------------------------\n# * Variable | OLS spatial spatHAC\n# *-------------+---------------------------------\n# * HDD | -0.669 -0.669 -0.669\n# * | 0.608 0.786 0.838\n\ntimer off 1\ntimer list 1\n# 1: 24.8 / 3 = 8.2650``````\n\n#### R Code:\n\nUsing the same data and options as the STATA code, we then estimate the adjusted standard errors using our new R function. This requires us to first estimate our regression model using the `felm` function from the `lfe` package.\n\n``````# Loading sample data:\ndta_file <- \"~/Dropbox/ConleySEs/Data/new_testspatial.dta\"\nsetnames(DTA, c(\"latitude\", \"longitude\"), c(\"lat\", \"lon\"))\n\nsource(\"~/Dropbox/ConleySEs/ConleySEs_17June2015.R\")\n\nptm <-proc.time()\n\n# We use the felm() from the lfe package to estimate model with year and county fixed effects.\n# Two important points:\n# (1) We specify our latitude and longitude coordinates as the cluster variables, so that they are included in the output (m).\n# (2) We specify keepCx = TRUE, so that the centered data is included in the output (m).\n\nm <-felm(EmpClean00 ~HDD -1 |year +FIPS |0 |lat +lon,\ndata = DTA[!is.na(EmpClean00)], keepCX = TRUE)\n\ncoefficients(m) %>%round(3) # Same as the STATA result.``````\n`````` HDD\n-0.669 ``````\n\nWe then feed this model to our function, as well as the cross-sectional unit (county `FIPS` codes), time unit (`year`), geo-coordinates (`lat` and `lon`), the cutoff for serial correlation (5 years), the cutoff for spatial correlation (500 km), and the number of cores to use.\n\n``````SE <-ConleySEs(reg = m,\nunit = \"FIPS\",\ntime = \"year\",\nlat = \"lat\", lon = \"lon\",\ndist_fn = \"SH\", dist_cutoff = 500,\nlag_cutoff = 5,\ncores = 1,\nverbose = FALSE)\n\nsapply(SE, sqrt) %>%round(3) # Same as the STATA results.``````\n`````` OLS Spatial Spatial_HAC\n0.608 0.786 0.837 ``````\n``proc.time() -ptm``\n`````` user system elapsed\n1.619 0.055 1.844 ``````\n\nEstimating the model and computing the standard errors requires just over 1 second, making it over 20 times faster than the comparable STATA routine.\n\n#### R Using Multiple Cores:\n\nEven with a single core, we realize significant speed improvements. However, the gains are even more dramatic when we employ multiple cores. Using 4 cores, we can cut the estimation of the standard errors down to around 0.4 seconds. (These replications employ the Haversine distance formula, which is more time-consuming to compute.)\n\n``````pkgs <-c(\"rbenchmark\", \"lineprof\")\ninvisible(sapply(pkgs, require, character.only = TRUE))\n\nbmark <-benchmark(replications = 25,\ncolumns = c('replications','elapsed','relative'),\nConleySEs(reg = m,\nunit = \"FIPS\", time = \"year\", lat = \"lat\", lon = \"lon\",\ndist_fn = \"Haversine\", lag_cutoff = 5, cores = 1, verbose = FALSE),\nConleySEs(reg = m,\nunit = \"FIPS\", time = \"year\", lat = \"lat\", lon = \"lon\",\ndist_fn = \"Haversine\", lag_cutoff = 5, cores = 2, verbose = FALSE),\nConleySEs(reg = m,\nunit = \"FIPS\", time = \"year\", lat = \"lat\", lon = \"lon\",\ndist_fn = \"Haversine\", lag_cutoff = 5, cores = 4, verbose = FALSE))\nbmark %>%mutate(avg_eplased = elapsed /replications, cores = c(1, 2, 4))``````\n`````` replications elapsed relative avg_eplased cores\n1 25 23.48 2.095 0.9390 1\n2 25 15.62 1.394 0.6249 2\n3 25 11.21 1.000 0.4483 4``````\n\nGiven the prevalence of panel data that exhibits both serial and spatial dependence, we hope this function will be a useful tool for applied econometricians working in `R`.\n\n#### Feedback Appreciated: Memory vs. Speed Tradeoff\n\nThis was Darin’s first foray into C++, so we welcome feedback on how to improve the code. In particular, we would appreciate thoughts on how to overcome a memory vs. speed tradeoff we encountered. (You can email Darin at darinc[at]stanford.edu.)\n\nThe most computationally intensive chunk of our code computes the distance from each unit to every other unit. To cut down on the number of distance calculations, we can fill the upper triangle of the distance matrix and then copy it to the lower triangle. With [math]N[/math] units, this requires only  [math](N (N-1) /2)[/math] distance calculations.\n\nHowever, as the number of units grows, this distance matrix becomes too large to store in memory, especially when executing the code in parallel. (We tried to use a sparse matrix, but this was extremely slow to fill.) To overcome this memory issue, we can avoid constructing a distance matrix altogether. Instead, for each unit, we compute the vector of distances from that unit to every other unit. We then only need to store that vector in memory. While that cuts down on memory use, it requires us to make twice as many   [math](N (N-1))[/math]  distance calculations.\n\nAs the number of units grows, we are forced to perform more duplicate distance calculations to avoid memory constraints – an unfortunate tradeoff. (See the functions `XeeXhC` and `XeeXhC_Lg` in ConleySE.cpp.)\n\n``sessionInfo()``\n``````R version 3.2.2 (2015-08-14)\nPlatform: x86_64-apple-darwin13.4.0 (64-bit)\nRunning under: OS X 10.10.4 (Yosemite)\n\nlocale:\n en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8\n\nattached base packages:\n stats graphics grDevices utils datasets methods\n base\n\nother attached packages:\n geosphere_1.4-3 sp_1.1-1\n lfe_2.3-1709 Matrix_1.2-2\n ggplot2_1.0.1 foreign_0.8-65\n data.table_1.9.4 dplyr_0.4.2\n knitr_1.11\n\nloaded via a namespace (and not attached):\n Formula_1.2-1 magrittr_1.5 MASS_7.3-43\n munsell_0.4.2 xtable_1.7-4 lattice_0.20-33\n colorspace_1.2-6 R6_2.1.1 stringr_1.0.0\n plyr_1.8.3 tools_3.2.2 parallel_3.2.2\n grid_3.2.2 gtable_0.1.2 DBI_0.3.1\n htmltools_0.2.6 yaml_2.1.13 assertthat_0.1\n digest_0.6.8 reshape2_1.4.1 formatR_1.2\n evaluate_0.7.2 rmarkdown_0.8 stringi_0.5-5\n compiler_3.2.2 scales_0.2.5 chron_2.3-47\n proto_0.3-10 ``````" ]

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[ "# What is NPV in project management?\n\nContents\n\n*A project’s net present value (hereafter NPV) is defined as the sum of the discounted value of all receipts minus the sum of the discounted value of all expenditures. All discounting is to the beginning of the project. A rate frequently used for discounting is the firm’s cost of capital.\n\n## How do you calculate NPV in project management?\n\nIt is calculated by taking the difference between the present value of cash inflows and present value of cash outflows over a period of time. As the name suggests, net present value is nothing but net off of the present value of cash inflows and outflows by discounting the flows at a specified rate.\n\n## What is the NPV of the project?\n\nNet present value (NPV) is a method used to determine the current value of all future cash flows generated by a project, including the initial capital investment. It is widely used in capital budgeting to establish which projects are likely to turn the greatest profit.\n\n## What is NPV example?\n\nPut another way, it is the compound annual return an investor expects to earn (or actually earned) over the life of an investment. For example, if a security offers a series of cash flows with an NPV of \\$50,000 and an investor pays exactly \\$50,000 for it, then the investor’s NPV is \\$0.\n\n## Is a higher or lower NPV better?\n\nHigher discount rates, lower NPV. Net present value is the benchmark metric. It is our best capital budgeting tool. It incorporates the timing of the cash flows and it takes into account the opportunity cost, because the discount rate quantifies, in essence, what else could we do with the money.\n\n## What does NPV mean?\n\n“Net present value is the present value of the cash flows at the required rate of return of your project compared to your initial investment,” says Knight. In practical terms, it’s a method of calculating your return on investment, or ROI, for a project or expenditure.\n\n## How do you solve for NPV?\n\nNPV can be calculated with the formula NPV = ⨊(P/ (1+i)t ) – C, where P = Net Period Cash Flow, i = Discount Rate (or rate of return), t = Number of time periods, and C = Initial Investment.\n\n## How do we calculate cash flow?\n\nCash flow formula:\n\n1. Free Cash Flow = Net income + Depreciation/Amortization – Change in Working Capital – Capital Expenditure.\n2. Operating Cash Flow = Operating Income + Depreciation – Taxes + Change in Working Capital.\n3. Cash Flow Forecast = Beginning Cash + Projected Inflows – Projected Outflows = Ending Cash.\n\n3 апр. 2019 г.\n\n## How do you calculate the value of a project?\n\nIt is calculated by deducting the expected costs or investment of a project from its expected revenue and then dividing this (net profit) by the expected costs in order to get a return rate.\n\n## How do you calculate NPV scrap value?\n\nAnswer: The net present value (NPV) It is calculated by adding the present value of all cash inflows and subtracting the present value of all cash outflows. method of evaluating investments adds the present value of all cash inflows and subtracts the present value of all cash outflows.\n\nIT IS INTERESTING:  Frequent question: What is the purpose of SDLC?\n\n## What is a good NPV?\n\nNPV > 0: The PV of the inflows is greater than the PV of the outflows. The money earned on the investment is worth more today than the costs, therefore, it is a good investment. … NPV < 0: The PV of the inflows is less than the PV of the outflows.\n\n## What is NPV 10?\n\nPV10 is a calculation of the present value of estimated future oil and gas revenues, net of forecasted direct expenses, and discounted at an annual rate of 10%. The resulting figure is used in the energy industry to estimate the value of a corporation’s proven oil and gas reserves.\n\n## What is NPV and IRR methods?\n\nNPV and IRR are two discounted cash flow methods used for evaluating investments or capital projects. NPV is is the dollar amount difference between the present value of discounted cash inflows less outflows over a specific period of time.\n\n## What happens if NPV is positive?\n\nA positive net present value indicates that the projected earnings generated by a project or investment – in present dollars – exceeds the anticipated costs, also in present dollars. It is assumed that an investment with a positive NPV will be profitable, and an investment with a negative NPV will result in a net loss.\n\n## What does NPV 0 mean?\n\nIf a project’s NPV is positive (> 0), the company can expect a profit and should consider moving forward with the investment. If a project’s NPV is neutral (= 0), the project is not expected to result in any significant gain or loss for the company.\n\nIT IS INTERESTING:  You asked: Why is agility important in everyday life?\n\n## Does higher NPV mean higher IRR?\n\nWhenever an NPV and IRR conflict arises, always accept the project with higher NPV. It is because IRR inherently assumes that any cash flows can be reinvested at the internal rate of return. … The risk of receiving cash flows and not having good enough opportunities for reinvestment is called reinvestment risk.", null, "" ]

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[ "Quickcharging an EV, how much power do I need?\n\nThere are two points on which people criticise Electric Vehicles (EV):\n\n• Their range\n• The time it takes to charge them\n\nThe first can be solved by ‘simply’ adding a larger battery, this can be in physical size or having more Wh’s (What Hours) per Kilogram.\n\nFilling the tank of a car with a ICE (Internal Combustion Engine) takes about 3 minutes, it is something we are used to. But charging a EV can take up to several hours.\n\nA lot of people say that they will start driving an EV as soon as the range gets better or charging can be done fast, like they are used to right now.\n\nCharging a EV really quick has a few problems which can not be solved that easily:\n\n• The batteries can’t be charged that fast (Yet)\n• It takes a lot, really A LOT of energy to charge that fast\n\nTake a Tesla Roadster for example, this car has a 53kWh battery pack. 53kWh equals to 190800000 Joule (53 * 1000 * 3600). If we want to charge this battery in 5 minutes, we would need to put 636000 Joules per second into that battery. 636000 Joule equals to a current of 636kW (636000 / 1000).\n\nA simple micro-wave in your kitchen uses about 1kW of energy, charging a EV that fast would use the energy of 636 micro-waves! That would put a lot of stress in the grid, too much stress.\n\nIf we charge the EV in 10 minutes we would ‘only’ require 318kW of energy, 20 minutes 159kW and 30 minutes would take 106kW of energy. Those are still high numbers, but they come closer to what is possible.\n\nTake the Nissan Leaf for example, this car has a 24kWh battery which can be charged to 80% in 30 minutes, let’s calculate how much energy we would need.\n\n80% of 24kWh is 19.2kWh, that equals to 69120000 Joule (See my calculations above). 30 minutes equals to 1800 seconds, so charging in 30 minutes requires 38400 Joule per second, or 38.4kW of energy.\n\nCharging that quick will mostly be done at 480 Volt. 38400W / 480V = 80A, that is how much we need to charge a Leaf that fast.\n\n3-phase 480 Volt is not that hard to find / get here in Europe, so charging a Leaf that fast is feasible on a lot of locations.\n\nNot only will quickcharging put a lot of stress on the grid, it would also be unsafe for humans to connect such cables. If the current which flows through that cable would be exposed to a human, you would instantly be killed, no doubt.\n\nQuickcharging a EV has a few drawbacks, let’s sum them up:" ]

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[ "# std::comp_ellint_3,std::comp_ellint_3f,std::comp_ellint_3l (3) - Linux Man Pages\n\n## NAME\n\nstd::comp_ellint_3,std::comp_ellint_3f,std::comp_ellint_3l - std::comp_ellint_3,std::comp_ellint_3f,std::comp_ellint_3l\n\n## Synopsis\n\ndouble comp_ellint_3( double k, double ν );\nfloat comp_ellint_3f( float k, float ν ); (1) (since C++17)\nlong double comp_ellint_3l( long double k, long double ν );\nPromoted comp_ellint_3( Arithmetic k, Arithmetic ν ); (2) (since C++17)\n\n1) Computes the complete_elliptic_integral_of_the_third_kind of the arguments k and ν.\n2) A set of overloads or a function template for all combinations of arguments of arithmetic type not covered by (1). If any argument has integral_type, it is cast to double. If any argument is long double, then the return type Promoted is also long double, otherwise the return type is always double.\n\n## Parameters\n\nk - elliptic modulus or eccentricity (a value of a floating-point or integral type)\nν- elliptic characteristic (a value of floating-point or integral type)\n\n## Return value\n\nIf no errors occur, value of the complete elliptic integral of the third kind of k and ν, that is std::ellint_3(k,ν,π/2), is returned.\n\n## Error handling\n\nErrors may be reported as specified in math_errhandling\n\n* If the argument is NaN, NaN is returned and domain error is not reported\n* If |k|>1, a domain error may occur\n\n## Notes\n\nImplementations that do not support C++17, but support ISO_29124:2010, provide this function if __STDCPP_MATH_SPEC_FUNCS__ is defined by the implementation to a value at least 201003L and if the user defines __STDCPP_WANT_MATH_SPEC_FUNCS__ before including any standard library headers.\nImplementations that do not support ISO 29124:2010 but support TR 19768:2007 (TR1), provide this function in the header tr1/cmath and namespace std::tr1\nAn implementation of this function is also available_in_boost.math\n\n## Example\n\n// Run this code\n\n#include <cmath>\n#include <iostream>\nint main()\n{\ndouble hpi = std::acos(-1)/2;\nstd::cout << \"Π(0.5,0) = \" << std::comp_ellint_3(0.5, 0) << '\\n'\n<< \"K(0.5) = \" << std::comp_ellint_1(0.5) << '\\n'\n<< \"Π(0,0) = \" << std::comp_ellint_3(0, 0) << '\\n'\n<< \"π/2 = \" << hpi << '\\n'\n<< \"Π(0.5,1) = \" << std::comp_ellint_3(0.5, 1) << '\\n';\n}\n\n## Output:\n\nΠ(0.5,0) = 1.68575\nK(0.5) = 1.68575\nΠ(0,0) = 1.5708\nπ/2 = 1.5708\nΠ(0.5,1) = inf\n\nWeisstein,_Eric_W._\"Elliptic_Integral_of_the_Second_Kind.\" From MathWorld--A Wolfram Web Resource." ]

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[ "# Grassmannian.info\n\nA periodic table of (generalised) Grassmannians.\n\n## Orthogonal Grassmannian $\\OGr(6,15)$\n\nBasic information\ndimension\n33\nindex\n8\nEuler characteristic\n448\nBetti numbers\n$\\mathrm{b}_{ 0 } = 1$, $\\mathrm{b}_{ 2 } = 1$, $\\mathrm{b}_{ 4 } = 2$, $\\mathrm{b}_{ 6 } = 3$, $\\mathrm{b}_{ 8 } = 4$, $\\mathrm{b}_{ 10 } = 6$, $\\mathrm{b}_{ 12 } = 8$, $\\mathrm{b}_{ 14 } = 10$, $\\mathrm{b}_{ 16 } = 12$, $\\mathrm{b}_{ 18 } = 15$, $\\mathrm{b}_{ 20 } = 17$, $\\mathrm{b}_{ 22 } = 20$, $\\mathrm{b}_{ 24 } = 22$, $\\mathrm{b}_{ 26 } = 24$, $\\mathrm{b}_{ 28 } = 25$, $\\mathrm{b}_{ 30 } = 27$, $\\mathrm{b}_{ 32 } = 27$, $\\mathrm{b}_{ 34 } = 27$, $\\mathrm{b}_{ 36 } = 27$, $\\mathrm{b}_{ 38 } = 25$, $\\mathrm{b}_{ 40 } = 24$, $\\mathrm{b}_{ 42 } = 22$, $\\mathrm{b}_{ 44 } = 20$, $\\mathrm{b}_{ 46 } = 17$, $\\mathrm{b}_{ 48 } = 15$, $\\mathrm{b}_{ 50 } = 12$, $\\mathrm{b}_{ 52 } = 10$, $\\mathrm{b}_{ 54 } = 8$, $\\mathrm{b}_{ 56 } = 6$, $\\mathrm{b}_{ 58 } = 4$, $\\mathrm{b}_{ 60 } = 3$, $\\mathrm{b}_{ 62 } = 2$, $\\mathrm{b}_{ 64 } = 1$, $\\mathrm{b}_{ 66 } = 1$\n$\\mathrm{Aut}^0(\\OGr(6,15))$\n$\\mathrm{SO}_{ 15 }$\n$\\pi_0\\mathrm{Aut}(\\OGr(6,15))$\n$1$\n$\\dim\\mathrm{Aut}^0(\\OGr(6,15))$\n105\nProjective geometry\nminimal embedding\n\n$\\OGr(6,15)\\hookrightarrow\\mathbb{P}^{ 5004 }$\n\ndegree\n1922769522272501760\nHilbert series\n1, 5005, 3539536, 779502360, 77470542240, 4296933421704, 152258285110800, 3779087169010725, 70173170834641875, 1023649487261651250, 12176878653505238400, 121619460263939164800, 1043915339123131216896, 7847370495635040332160, 52471882793310820295040, 316149213060178411241874, 1735208951800106788650645, 8756440327813213380183839, 40951316491999034291601200, 178711268655478773276755000, ...\nExceptional collections\n\nNo full exceptional collection is known for $\\mathbf{D}^{\\mathrm{b}}(\\OGr(6,15))$. Will you be the first to construct one? Let us know if you do!\n\nKuznetsov–Polishchuk have constructed an exceptional collection of maximal length in MR3463417. Can you prove it's full?\n\nQuantum cohomology\n\nThe small quantum cohomology is generically semisimple.\n\nThe big quantum cohomology is generically semisimple.\n\nThe eigenvalues of quantum multiplication by $\\mathrm{c}_1(\\OGr(6,15))$ are given by:\n\nHomological projective duality" ]

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[ "", null, "# 1864\n\n## The book of science\n\nTom Sharp\n\n James Clerk Maxwell electromagnetism\n\n## Maxwell’s equations\n\n• Like little rafts, an electric current\n• runs along a river of force vectors\n• oriented perpendicular to a surface.\n• Bundles of magnetic lines of force\n• are like small tubes that carry a magnetic flux\n• in arcs above and below a charged surface.\n• The work needed to circulate a charge\n• around a closed curve in an electric field is related to\n• the work needed to circulate a magnetic monopole,\n• if one were to exist, in a magnetic field.\n• The imputed magnetic monopole\n• is a magnet with only one end,\n• like an electron, a point in time\n• whose negative charge is in our world\n• and whose positive charge is in a deformation\n• of the field in a minuscule invisible dimension.\n• Modern physics can explain everything\n• using the properties of fields. Fields, rafts, tubes,\n• magnetic monopoles, and numerous invisible dimensions\n• are, to those who understand them, like poetry.\n\n## Fields\n\n• The properties of a field can be described\n• by a set of numbers associated\n• with each point of the field.\n• First, mathematicians say a point is a location\n• with no extent.\n• Second, any extent, however small,\n• contains an infinite number of points.\n• Third, a field is an extent.\n• The relation between reality and a mathematical system\n• cannot be determined mathematically.\n• The relation rests on the faith\n• that each step from the known to the unknown\n• might help us recognize something new or useful.\n• In particular, it would be useful to explain\n• things that our senses\n• cannot determine—how gravity\n• acts at a distance, why light\n• travels at a certain speed, and whether the universe\n• will end in fire or in ice.\n\n## James Clerk Maxwell\n\n• James Clerk, named Maxwell after his mother died,\n• studied at Edinburgh, Cambridge, and Trinity,\n• earned honors, prizes, and a mathematics degree at 23;\n• taught at Marischal College and married the daughter of the principal,\n• planned the famous Cavendish laboratory\n• and became the first Cavendish professor at Cambridge,\n• was the first to use probability and statistics\n• to explain the behavior of gases,\n• studied color blindness and color vision,\n• resulting in the first color photograph,\n• extended Michael Faraday’s theories of electromagnetism\n• suggesting that “light consists in the transverse undulations\n• of the same medium which is the cause\n• of electric and magnetic phenomenon.”\n• Nine years after he died in 1879 at 48,\n• according to his theory, Heinrich Hertz\n• demonstrated that an electric disturbance\n• is transmitted through space in waves,\n• and, nearly one-hundred years after he died,\n• NASA’s Voyager space probe\n• confirmed Maxwell’s theory that the rings of Saturn\n• are composed of many small particles.\n\nJames Clerk Maxwell was the first to realize how electricity, magnetism, and light are manifestations of the same physical forces—electromagnetic radiation—described as photons or waves with different energies, wavelength, and frequencies. As Einstein wrote, Maxwell realized “that electromagnetic fields spread in the form of polarised waves, and at the speed of light!” Maxwell wrote that “light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.”\n\nElectric charges and currents create electric and magnetic fields (physical lines of force) that can be described mathematically. Maxwell’s equations describe and improve work by Carl Friedrich Gauss, Micael Faraday, and André-Marie Ampère, that is, Gauss’s law, Gauss’s law for magnetism, Faraday’s law of induction, and Ampère’s circuital law." ]

[ null, "https://sharpgiving.com/thebookofscience/iimgs/i1864.png", null ]

[ "# 10ticks Friday Puzzle", null, "Welcome to Ian’s Friday Puzzle! Dust off those Friday cobwebs with a little manipulation of the old grey matter. Perplexing puzzles, logical, illogical, and sometimes just plain stupid. Be prepared to be bewildered, befuddled and bedazzled!\n\nSee the latest Friday Puzzle below and look at all the past puzzles. Get the Friday Puzzle straight to your phone, follow us on Twitter. The first 5 correct answers join our Hall of Fame!\n\n#### This week's Friday Puzzle\n\nA circle has three equal smaller circles whose centres\n\nare placed on its diameter as shown.", null, "Which area is bigger, the three shaded circles or\n\none of the two equal unshaded areas?\n\n#### Friday Puzzle (26/11/2021)\n\nColin and his daughters Abigail and Brenda have the same birthday.\n\nToday, Colin is 28, Abigail is 6 and Zoey is 2.\n\nHow old will Colin be when his age is the sum of the ages of Abigail and Brenda?\n\n#### Friday Puzzle (19/11/2021)\n\nThe diagram shows an equilateral triangle inside a regular hexagon.", null, "The hexagon has sides of length 10 cm.\n\nThe vertices of the triangle are midpoints of sides of the hexagon.\n\nWhat is the length of the perimeter of the triangle?\n\n#### Friday Puzzle (12/11/2021)\n\nTwo circles of radius 2 cm fit exactly between two parallel lines, as shown in the diagram.", null, "The centres of the circles are 6 cm apart.\n\nWhat is the area that is shaded yellow?\n\n#### Friday Puzzle (05/11/2021)\n\nA positive integer has exactly eight factors.\n\nTwo of these factors are 14 and 22.\n\nWhat is the sum of all eight factors of the positive integer?\n\n#### Friday Puzzle (29/10/2021)\n\nThe England T20 Cricket team are messing about.\n\nJason thinks of a positive integer, which Jos then doubles.\n\nDawid then trebles Jos’ number.\n\nFinally Jonny multiplies Dawid’s number by six.\n\nEoin notices that the sum of the four numbers is a perfect square!\n\nWhat is the smallest number that Jason could have thought of?\n\n#### Friday Puzzle (22/10/2021)", null, "The diagram shows a regular hexagon with an area of 60 cm2.\n\nWhat is the area of the shaded triangle?\n\n#### Friday Puzzle (15/10/2021)\n\nThere are 243 players taking part in a knock-out tournament.\n\nEach match in the tournament involves 3 players and only the winner of the match remains in the tournament − the other two players are knocked out.\n\nHow many matches are required until there is an overall winner?\n\n#### Friday Puzzle (08/10/2021)\n\nMy biscuit barrel contains jammy dodgers and chocolate digestives.\nThe ratio of jammy dodgers to chocolate digestives in the barrel is 3 : 8.\nWhen I remove one jammy dodger the ratio changes to 1 : 3.\nHow many chocolate digestives are in the basket?\n\n#### Friday Puzzle (01/10/2021)\n\nSolve (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)2 − (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)2\n\nby starting with something smaller and looking for patterns.\n\n(1 + 2)2 − (1 )2\n(1 + 2 + 3)2 − (1 + 2)2\n(1 + 2 + 3 + 4)2 − (1 + 2 + 3)2" ]

[ null, "https://www.10ticks.co.uk/images/ian-about-us.jpg", null, "https://www.10ticks.co.uk/mlink/Circles3circles.png", null, "https://www.10ticks.co.uk/mlink/TriangleinHexagon.png", null, "https://www.10ticks.co.uk/mlink/CircleParallelLines.png", null, "https://www.10ticks.co.uk/mlink/shadedhexagon.png", null ]

[ "## § A slew of order theoretic and graph theoretic results\n\nI've been trying to abstract out the activity selection problem from the lens of order theory. For this, I plan on studying the following theorems/algebraic structures:\n• Intransitive indifference with unequal indifference intervals\n• Mirsky's theorem\n• Dilworth's theorem\n• Gallai–Hasse–Roy–Vitaver theorem\n• Dirac's theorem\n• Ore's theorem\nNaively, the solution goes as follows, which can be tested against CSES' movie festival question\n// https://cses.fi/problemset/task/1629\nint main() {\nint n;\ncin >> n;\nvector<pair<int, int>> ms(n);\nfor (int i = 0; i < n; ++i) {\ncin >> ms[i].first >> ms[i].second;\n}\n\nstd::sort(ms.begin(), ms.end(), [](pair<int, int> p1, pair<int, int> p2) {\nreturn (p1.second < p2.second) ||\n(p1.second == p2.second && p1.first < p2.first);\n});\n\nint njobs = 0;\nint cur_end = -1;\nfor (int i = 0; i < n; ++i) {\nif (cur_end <= ms[i].first) {\ncur_end = ms[i].second;\nnjobs++;\n}\n}\ncout << njobs << \"\\n\";\nreturn 0;\n}\n\n\n#### § Explanation 1: exchange argument\n\n• The idea is to pick jobs greedily , based on quickest finishing time .\n• The argument of optimality is strategy stealing. Think of the first job in our ordering O versus the optimal ordering O*.\n• If we both use the same job, ie, O = O*, recurse into the second job.\n• If we use different jobs then O != O*.\n• Since O ends quickest [acc to our algorithm ], we will have that end(O) < end(all other jobs), hence end(O) < end(O*).\n• Since O* is a correct job schedule, we have that end(O*) < start(O*).\n• Chaining inequalities, we get that end(O) < end(O*) < start(O*).\n• Thus, we can create O~ which has O~ = O and O~[rest] = O*[rest]. ( ~ for \"modified\").\n• Now recurse into O~ to continue aligning O* with O. We continue to have the same length between O~, O and O*." ]

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[ "# How do you find the slope that is perpendicular to the line -25 + 8x = 5y?\n\nMay 28, 2018\n\n$m ' = - \\frac{5}{8}$\n\n#### Explanation:\n\nYou need to do the algebra to get the equation in the slope intercept form:\n\n$y = m x + b$\n\nm is the slope.\n\n$- 25 + 8 x = 5 y$\n\n5y = 8x -25\n\n$y = \\frac{8}{5} y - 5$\n\n$m = \\frac{8}{5}$\n\nThe rule for perpendicular slope is:\n\n$m ' = - \\frac{1}{m}$\n\n$m ' = - \\frac{1}{\\frac{8}{5}}$\n\n$m ' = - \\frac{5}{8}$" ]

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[ "", null, "", null, "", null, "\u001b\\$B%[!<%`\u001b(B \u001b\\$B!d\u001b(B \u001b\\$B%a!<%k%^%,%8%s\u001b(B \u001b\\$B!d\u001b(B \u001b\\$B%a!<%k%^%,%8%s\\$N\\$40FFb!JEPO?!&2r=|!K\u001b(B", null, "###### \u001b\\$B%a!<%k%^%,%8%s\\$N\\$40FFb!JEPO?!&2r=|!K\u001b(B\n\u001b\\$B%P%C%/%J%s%P!<0lMw\u001b(B\n\u001b\\$B%5%s%W%k!J%F%-%9%H!K\u001b(B", null, "# \u001b\\$B%a!<%k%^%,%8%s\\$N\\$40FFb!JEPO?!&2r=|!K\u001b(B\n\n## \u001b\\$BEl5~ET?)Lt\u001b(Be\u001b\\$B%^%,%8%s\\$H\\$O\u001b(B\n\n\u001b\\$B!JNaOB\u001b(B2\u001b\\$BG/\u001b(B9\u001b\\$B7n\u001b(B8\u001b\\$BF|99?7!K\u001b(B\n\n\u001b\\$B!!El5~ET\\$N?)IJ1R@8\\$HLt;v1R@8\\$r=j4I\\$9\\$kIt=p\\$N%[!<%`%Z!<%8\\$KBP>]4|4VFb\\$K7G:\\\\$5\\$l\\$?pJs\\$r\\$*CN\\$i\\$;\\$9\\$k\\$H\\$H\\$b\\$K!\";~@a\\$NOCBj\\$dET\\$N;v6H\\$J\\$I\\$K\\$D\\$\\$\\$FEE;R%a!<%k\\$G\\$4>R2p\\$9\\$k%5!<%S%9!J%a!<%k%^%,%8%s!K\\$G\\$9!#\u001b(B\n\n\u001b\\$B!!\u001b(B \u001b\\$B\\$3\\$N%a!<%k%^%,%8%s\\$NG[?.\\$K\\$O!\"\u001b(B\u001b\\$B3t<02q\u001b\\$B\\$NDs6!\\$9\\$k%5!<%S%9\\$rMxMQ\\$7\\$F\\$\\$\\$^\\$9!#\\$=\\$N\\$?\\$a!\"0J2<\\$NE@\\$K\\$4Cm0U\\$/\\$@\\$5\\$\\$!#\\$J\\$*!\"J8Cf\\$N%j%s%/@h\\$O!\"3t<02q\n \u001b\\$B!!!!\u001b(B\u001b\\$B\"(\u001b(B \u001b\\$BFI\u001b\\$B3t<02q\u001b\\$B\\$K\\$h\\$j!\"!V\u001b(B\u001b\\$B%*%U%#%7%c%k%a%k%^%,\u001b(B\u001b\\$B!W\\$H\\$\\$\\$&\u001b(B\u001b\\$BL5NA\u001b(B\u001b\\$B%a!<%k%^%,%8%s\u001b(B(\u001b\\$B!V\\$^\\$0\\$^\\$0Am9gM<4)HG!WEy!\"\u001b(B\u001b\\$BJ#?t\u001b(B\u001b\\$B\\$N%a!<%k%^%,%8%s\\$G9=@.\\$5\\$l\\$F\\$\\$\\$^\\$9!#!K\\$b<+F0E*\\$KFIl9g\\$O!\"EPO?8e\\$KFO\\$/!V%a%k%^%,FI \u001b\\$B\"(\u001b(B \u001b\\$B!V%*%U%#%7%c%k%a%k%^%,!W\\$rEPO?2r=|\\$7\\$F\\$b!\"El5~ET?)Lt\u001b(Be\u001b\\$B%^%,%8%s\\$O0z\\$-B3\\$-G[?.\\$5\\$l\\$^\\$9!#\u001b(B \u001b\\$B!!!!\"(\u001b(B \u001b\\$B!V%*%U%#%7%c%k%a%k%^%,!W\\$NEPO?2r=|8e\\$b!\"!V%*%U%#%7%c%k%a%k%^%,!W\\$,FO\\$/>l9g\\$K\\$O!\"\\$^\\$0\\$^\\$0!*%X%k%W\u001b(B\u001b\\$B!V\\$3\\$NA0%*%U%#%7%c%k%a%k%^%,\\$r2r=|\\$7\\$?\\$N\\$K!\"\\$^\\$?FO\\$/\\$h\\$&\\$K\\$J\\$j\\$^\\$7\\$?!W\u001b(B\u001b\\$B\\$r\\$4Mw\\$/\\$@\\$5\\$\\$!#\u001b(B\n \u001b\\$B\"(\u001b(B \u001b\\$B%a!<%k%\"%I%l%9JQ99Ey!\"EPO?>pJs\\$NJQ99\\$O\u001b(B\u001b\\$B!V\\$^\\$0\\$^\\$0!*%X%k%W!W\u001b(B\u001b\\$B\\$+\\$i\\$*4j\\$\\$\\$7\\$^\\$9!#\u001b(B\n\n### \u001b\\$BH/9TF|\u001b(B\n\n\u001b\\$BKh7nBh#2!\"Bh#46bMKF|\u001b(B\n\n### \u001b\\$BFbMF\u001b(B\n\n• \u001b\\$B%[!<%`%Z!<%8\\$NpJs\u001b(B\n• \u001b\\$BJsF;H/I=\u001b(B\n• \u001b\\$B;v6H\\$N>R2p\u001b(B\n• \u001b\\$B%H%T%C%/%9!!\\$J\\$I\u001b(B\n\n## \u001b\\$B%a!<%k%^%,%8%sEPO?\\$O\\$3\\$A\\$i\\$+\\$i\u001b(B\n\n\u001b\\$BEl5~ET?)Lt\u001b(Be\u001b\\$B%^%,%8%s\\$NG[?.\\$r4uK>\\$5\\$l\\$kJ}\\$O!\"2<\\$N6uMs\\$KEE;R%a!<%k%\"%I%l%9!JH>3Q!K\\$rF~NO\\$7\\$FEPO?%\\%?%s\\$r%/%j%C%/\\$7\\$F\\$/\\$@\\$5\\$\\$!#\u001b(B\n\u001b\\$BCm0U!'<+F0E*\\$K3t<02q\n\n## \u001b\\$B%a!<%k%^%,%8%s2r=|\\$O\\$3\\$A\\$i\\$+\\$i\u001b(B\n\n\u001b\\$BEl5~ET?)Lt\u001b(Be\u001b\\$B%^%,%8%s\\$N2r=|\\$r4uK>\\$5\\$l\\$kJ}\\$O!\"2<\\$N6uMs\\$KEE;R%a!<%k%\"%I%l%9!JH>3Q!K\\$rF~NO\\$7\\$F2r=|%\\%?%s\\$r%/%j%C%/\\$7\\$F\\$/\\$@\\$5\\$\\$!#\u001b(B\n\n\u001b\\$BCm0U!'<+F0E*\\$K3t<02q\n\n\u001b\\$B\"%\\$3\\$N%Z!<%8\\$N%H%C%W\\$X\u001b(B\n\n##### \u001b\\$B\"'\u001b(B \u001b\\$B\\$*Ld\\$\\$9g\\$o\\$;@h\u001b(B\n\n\u001b\\$B\\$3\\$N%Z!<%8\\$O\u001b(B\u001b\\$BEl5~ETJ!;cJ]7r6I\u001b(B \u001b\\$B7r9/0BA48&5f%;%s%?!<\u001b(B \u001b\\$B4k2hD4@0It\u001b(B \u001b\\$B7r9/4m5!4IM}>pJs2]\u001b(B \u001b\\$B?)IJ0eLtIJ>pJsC4Ev\u001b(B\u001b\\$B\\$,4IM}\\$7\\$F\\$\\$\\$^\\$9!#\u001b(B\n\n\u001b\\$B\"%\\$3\\$N%Z!<%8\\$N%H%C%W\\$X\u001b(B", null, "" ]

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[ "Definition: finite type vs finitely generated\n\nThe mathematical term \"finite type\" appears more and more in the modern articles nowadays. But it is still hard to be found in the standard textbooks. I learned the definition of it from Stacks Project http://stacks.math.columbia.edu/tag/00F2 , it is defining on the ring maps. What is the right point of view when we say \"ring of finite type\", \"group of finite type\", \"module of finite type\"? Would the definition in each case be exactly equivalent to the definition of \"finitely generated\"?\n\n2. If $C$ is a variety in the sense of universal algebra, then an object $M \\in C$ is called finitely generated if there are elements $a_1,\\dotsc,a_n$ such that $M = \\langle a_1,\\dotsc,a_n \\rangle$, where the right hand side is the smallest subobject of $M$ containing the $a_1,\\dotsc,a_n$. This yields the usual notion when $C=\\mathsf{Set},\\, \\mathsf{Grp},\\,R\\mathsf{-Mod},\\, R\\mathsf{-CAlg}$ etc.\n3. Even more generally, an object $M$ of an arbitrary category $C$ is called finitely generated if for every directed diagram $\\{N_i\\}$ of objects whose transition maps are monomorphisms the canonical map $\\varinjlim_i \\hom(M,N_i) \\to \\hom(M,\\varinjlim_i N_i)$ is bijective. This coincides with the definition above if $C$ is a variety (easy exercise)." ]

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[ "", null, "FACTOID # 26: Delaware is the latchkey kid capital of America, with 71.8% of households having both parents in the labor force.\n\n Home Encyclopedia Statistics States A-Z Flags Maps FAQ About\n\n WHAT'S NEW\n\nSEARCH ALL\n\nSearch encyclopedia, statistics and forums:\n\n(* = Graphable)\n\nEncyclopedia > Riemann surface\n\nIn mathematics, particularly in complex analysis, a Riemann surface, named after Bernhard Riemann, is a one-dimensional complex manifold. Riemann surfaces can be thought of as \"deformed versions\" of the complex plane: locally near every point they look like patches of the complex plane, but the global topology can be quite different. For example, they can look like a sphere or a torus or a couple of sheets glued together. Image File history File links Riemann_sqrt. ... Image File history File links Riemann_sqrt. ... For other meanings of mathematics or math, see mathematics (disambiguation). ... Complex analysis is the branch of mathematics investigating functions of complex numbers, and is of enormous practical use in many branches of mathematics, including applied mathematics. ... Bernhard Riemann. ... In differential geometry, a complex manifold is a manifold such that every neighborhood looks like the complex n-space. ... In mathematics, the complex plane is a way of visualising the space of the complex numbers. ... A Möbius strip, a surface with only one side and one edge; such shapes are an object of study in topology. ... A sphere (< Greek σφαίρα) is a perfectly symmetrical geometrical object. ... A torus. ...\n\nThe main point of Riemann surfaces is that holomorphic functions may be defined between them. Riemann surfaces are nowadays considered the natural setting for studying the global behavior of these functions, especially multi-valued functions such as the square root or the logarithm. Holomorphic functions are the central object of study of complex analysis; they are functions defined on an open subset of the complex number plane C with values in C that are complex-differentiable at every point. ... This diagram does not represent a true function; because the element 3, in X, is associated with two elements b and c, in Y. In mathematics, a multivalued function is a total relation; i. ... In mathematics, a square root of a number x is a number whose square (the result of multiplying the number by itself) is x. ... The natural logarithm, formerly known as the hyperbolic logarithm, is the logarithm to the base e, where e is equal to 2. ...\n\nEvery Riemann surface is a two-dimensional real analytic manifold (i.e., a surface), but it contains more structure (specifically a complex structure) which is needed for the unambiguous definition of holomorphic functions. A two-dimensional real manifold can be turned into a Riemann surface (usually in several inequivalent ways) if and only if it is orientable. So the sphere and torus admit complex structures, but the Möbius strip, Klein bottle and projective plane do not. On a sphere, the sum of the angles of a triangle is not equal to 180°. A sphere is not a Euclidean space, but locally the laws of the Euclidean geometry are good approximations. ... An open surface with X-, Y-, and Z-contours shown. ... In differential geometry, a complex manifold is a manifold such that every neighborhood looks like the complex n-space. ... This article or section should be merged with Orientable manifold. ... A Möbius strip made with a piece of paper and tape. ... The Klein bottle immersed in three-dimensional space. ... Projective plane - Wikipedia, the free encyclopedia /**/ @import /skins-1. ...\n\nGeometrical facts about Riemann surfaces are as \"nice\" as possible, and they often provide the intuition and motivation for generalizations to other curves, manifolds or varieties. The Riemann-Roch theorem is a prime example of this influence. In mathematics, specifically in complex analysis and algebraic geometry, the Riemann-Roch theorem is an important tool in the computation of the dimension of the space of meromorphic functions with prescribed zeroes and allowed poles. ...\n\n## Contents\n\nLet X be a Hausdorff space. A homeomorphism from an open subset UX to a subset of C is called a chart. Two charts f and g whose domains intersect are said to be compatible if the maps f o g−1 and g o f −1 are holomorphic over their domains. If A is a collection of compatible charts and if any x in X is in the domain of some f in A, then we say that A is an atlas. When we endow X with an atlas A, we say that (X, A) is a Riemann surface. If the atlas is understood, we simply say that X is a Riemann surface. In topology and related branches of mathematics, a Hausdorff space is a topological space in which points can be separated by neighbourhoods. ... In the mathematical field of topology a homeomorphism or topological isomorphism (from the Greek words homeos = identical and morphe = shape) is a special isomorphism between topological spaces which respects topological properties. ... In topology, an atlas describes how a complicated space is glued together from simpler pieces. ... Holomorphic functions are the central object of study of complex analysis; they are functions defined on an open subset of the complex number plane C with values in C that are complex-differentiable at every point. ... In topology, an atlas describes how a complicated space is glued together from simpler pieces. ...\n\nDifferent atlases can give rise to essentially the same Riemann surface structure on X; to avoid this ambiguity, one sometimes demands that the given atlas on X be maximal, in the sense that it is not contained in any other atlas. Every atlas A is contained in a unique maximal one by Zorn's lemma. Zorns lemma, also known as the Kuratowski-Zorn lemma, is a proposition of set theory that states: Every non-empty partially ordered set in which every chain (i. ...\n\n## Examples\n\n• The complex plane C is perhaps the most trivial Riemann surface. The map f(z) = z (the identity map) defines a chart for C, and {f} is an atlas for C. The map g(z) = z* (the conjugate map) also defines a chart on C and {g} is an atlas for C. The charts f and g are not compatible, so this endows C with two distinct Riemann surface structures. In fact, given a Riemann surface X and its atlas A, the conjugate atlas B = {f* : f ∈ A} is never compatible with A, and endows X with a distinct, incompatible Riemann structure.\n• In an analogous fashion, every open subset of the complex plane can be viewed as a Riemann surface in a natural way. More generally, every open subset of a Riemann surface is a Riemann surface.\n• Let S = C ∪ {∞} and let f(z) = z where z is in S {∞} and g(z) = 1 / z where z is in S {0} and 1/∞ is defined to be 0. Then f and g are charts, they are compatible, and { f, g } is an atlas for S, making S into a Riemann surface. This particular surface is called the Riemann sphere because it can be interpreted as wrapping the complex plane around the sphere. Unlike the complex plane, it is compact.\n• The theory of compact Riemann surfaces can be shown to be equivalent to that of projective algebraic curves that are defined over the complex numbers and non-singular. Important examples of non-compact Riemann surfaces are provided by analytic continuation (see below.)\n\nIn mathematics, the complex plane is a way of visualising the space of the complex numbers. ... In mathematics, the complex conjugate of a complex number is given by changing the sign of the imaginary part. ... In topology and related fields of mathematics, a set U is called open if, intuitively speaking, you can wiggle or change any point x in U by a small amount in any direction and still be inside U. In other words, if x is surrounded only by elements of U... A rendering of the Riemann Sphere. ... In mathematics, a subset of Euclidean space Rn is called compact if it is closed and bounded. ... In algebraic geometry, an algebraic curve is an algebraic variety of dimension equal to 1. ...\n\n## Properties and further definitions\n\nA function f : MN between two Riemann surfaces M and N is called holomorphic if for every chart g in the atlas of M and every chart h in the atlas of N, the map h o f o g-1 is holomorphic (as a function from C to C) wherever it is defined. The composition of two holomorphic maps is holomorphic. The two Riemann surfaces M and N are called conformally equivalent if there exists a bijective holomorphic function from M to N whose inverse is also holomorphic (it turns out that the latter condition is automatic and can therefore be omitted). Two conformally equivalent Riemann surfaces are for all practical purposes identical. Partial plot of a function f. ... In topology, an atlas describes how a complicated space is glued together from simpler pieces. ... In mathematics, a bijection, bijective function, or one-to-one correspondence is a function that is both injective (one-to-one) and surjective (onto), and therefore bijections are also called one_to_one and onto. ...\n\nEvery simply connected Riemann surface is conformally equivalent to C or to the Riemann sphere C ∪ {∞} or to the open disk {zC : |z| < 1}. This is the main step in the uniformization theorem. A geometrical object is called simply connected if it consists of one piece and doesnt have any circle-shaped holes or handles. Higher-dimensional holes are allowed. ... In mathematics, the uniformization theorem for surfaces says that any surface admits a Riemannian metric of constant Gauss curvature. ...\n\nThe uniformization theorem states that every connected Riemann surface admits a unique complete 2-dimensional real Riemann metric with constant curvature -1, 0 or 1. The Riemann surfaces with curvature -1 are called hyperbolic; the open disk with the Poincar�-metric of constant curvature -1 is the local model. Examples include all surfaces with genus g>1. The Riemann surfaces with curvature 0 are called parabolic; C and the 2-torus are typical parabolic Riemann surfaces. Finally, the surfaces with curvature +1 are known as elliptic; the Riemann sphere C ∪ {∞} is the only example. In mathematical analysis, a metric space M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M. Intuitively, a space is complete if it doesnt have any holes, if there arent any points missing. For... In Riemannian geometry, a Riemannian manifold is a real differentiable manifold in which each tangent space is equipped with an inner product in a manner which varies smoothly from point to point. ... Curvature refers to a number of loosely related concepts in different areas of geometry. ... A rendering of the Riemann Sphere. ...\n\nFor every closed parabolic Riemann surface, the fundamental group is isomorphic to a rank 2 lattice, and thus the surface can be constructed as C/Γ, where C is the complex plane and Γ is the lattice. The set of representatives of the cosets are called fundamental domains. In mathematics, the fundamental group is one of the basic concepts of algebraic topology. ... In mathematics, especially in geometry and group theory, a lattice in Rn is a discrete subgroup of Rn which spans the real vector space Rn. ... In mathematics, given a lattice &#915; in a Lie group G, a fundamental domain is a set D of representatives for the cosets G/&#915;, that is also a well-behaved set topologically, in a sense that can be made precise in one of several ways. ...\n\nSimilarly, for every hyperbolic Riemann surface, the fundamental group is isomorphic to a Fuchsian group, and thus the surface can be modelled by a Fuchsian model H/Γ where H is the upper half-plane and Γ is the Fuchsian group. The set of representatives of the cosets of H/Γ are free regular sets and can be fashioned into metric fundamental polygons. In mathematics, a Fuchsian group is a particular type of group of isometries of the hyperbolic plane. ... In mathematics, a Fuchsian model is a representation of a hyperbolic Riemann surface R. By the uniformization theorem, every Riemann surface is either elliptic, parabolic or hyperbolic. ... In mathematics, the upper half plane H is the set of complex numbers x + iy such that y > 0. ... In mathematics, a free regular set is a subset of a topological space that is acted upon disjointly under a given group action. ... In mathematics, each closed surface in the sense of geometric topology can be constructed from an even-sided oriented polygon, called a fundamental polygon, by pairwise identification of its edges. ...\n\nWhen a hyperbolic surface is compact, then the total area of the surface is 4π(g − 1), where g is the genus of the surface; the area is obtained by applying the Gauss-Bonnet theorem to the area of the fundamental polygon. In mathematics, the genus has few different meanings Topology The genus of a connected, oriented surface is an integer representing the maximum number of cuttings along closed simple curves without rendering the resultant manifold disconnected. ... The Gauss-Bonnet theorem in differential geometry is an important statement about surfaces which connects their geometry (in the sense of curvature) to their topology (in the sense of the Euler characteristic). ...\n\nWe noted in the preamble that all Riemann surfaces, like all complex manifolds, are orientable as a real manifold. The reason is that for complex charts f and g with transition function h = f(g-1(z)) we can consider h as a map from an open set of R2 to R2 whose Jacobian in a point z is just the real linear map given by multiplication by the complex number h'(z). However, the real determinant of multiplication by a complex number α equals |α|^2, so the Jacobian of h has positive determinant. Consequently the complex atlas is an oriented atlas. This article or section should be merged with Orientable manifold. ... In vector calculus, the Jacobian is shorthand for either the Jacobian matrix or its determinant, the Jacobian determinant. ... In algebra, a determinant is a function depending on n that associates a scalar det(A) to every n×n square matrix A. The fundamental geometric meaning of a determinant is as the scale factor for volume when A is regarded as a linear transformation. ...\n\n## Other Examples\n\n• As noted above, the Riemann sphere is the only elliptic Riemann surface.\n• The only parabolic, simply connected Riemann surface is the complex plane. All parabolic surfaces can be obtained as a quotient of the plane. All parabolic surfaces are homeomorphic to either the plane, the annulus, or the torus. However it does not follow that all tori are biholomorphic to each other. This is the first appearance of the problem of moduli. The modulus of a toris can be captured by a single complex number with positive imaginary part. In fact, the marked moduli space (Teichmuller space) of the torus is biholomorphic to the open unit disk.\n• The only hyperbolic, simply connected Riemann surface is the open unit disk. The celebrated Riemann Mapping Theorem states that any simply connected strict subset of the complex plane is biholomorphic to the unit disk. All hyperbolic surfaces are quotients of the unit disk. Unlike elliptic and parabolic surfaces, no classification of the hyperbolic surfaces is possible. Any connected open strict subset of the plane gives a hyperbolic surface; consider the plane minus a Cantor set. A classification is possible for surfaces of finite type: those with finitely generated fundamental group. Any one of these has a finite number of moduli and so a finite dimensional Teichmuller space. The problem of moduli (solved by Ahlfors and extended by Bers) was to justify Riemann's claim that for a closed surface of genus g, 3g - 3 complex parameters suffice.\n\nIn mathematics, given a Riemann surface X, the Teichmüller space of X, notated TX or Teich(X), is a complex manifold whose points represent all complex structures of Riemann surfaces whose underlying topological structure is the same as that of X. It is named after the German mathematician Oswald... The Cantor set, introduced by German mathematician Georg Cantor, is a remarkable construction involving only the real numbers between zero and one. ...\n\n## Functions\n\nEvery non-compact Riemann surface admits non-constant holomorphic functions (with values in", null, "$mathbb{C}$). In fact, every non-compact Riemann surface is a Stein manifold. In mathematics, a Stein manifold in the theory of several complex variables and complex manifolds is a closed, complex submanifold of the vector space of n complex dimensions. ...\n\nIn contrast, on a compact Riemann surface every holomorphic function with value in", null, "$mathbb{C}$ is constant due to the maximum principle. However, there always exists non-constant meromorphic functions (=holomorphic functions with values in the Riemann sphere C ∪ {∞}). A meromorphic function is a function that is holomorphic on an open subset of the complex number plane C (or on some other connected Riemann surface) except at points in a set of isolated poles, which are certain well-behaved singularities. ... A rendering of the Riemann Sphere. ...\n\n## History\n\nRiemann surfaces were first studied by Bernhard Riemann and were named after him. Bernhard Riemann. ...\n\n## In art and literature\n\n• One of M.C. Escher's works, Print Gallery, is laid out on a cyclically growing grid that has been described as a Riemann surface.\n• In Aldous Huxley's novel Brave New World, \"Riemann Surface Tennis\" is a popular game.\n\nHand with Reflecting Sphere (Self-Portrait in Spherical Mirror), 1935. ... This article or section may contain original research or unverified claims. ... Brave New World is a dystopian novel by Aldous Huxley, first published in 1932. ...\n\nAlgebraic geometry is a branch of mathematics which, as the name suggests, combines abstract algebra, especially commutative algebra, with geometry. ... In mathematics, conformal geometry is the study of the set of angle-preserving (conformal) transformations on a Euclidean-like space with a point added at infinity, or a Minkowski-like space with a couple of points added at infinity. That is, the setting is a compactification of a familiar space... In mathematics, a Kähler manifold is a complex manifold which also carries a Riemannian metric and a symplectic form on the underlying real manifold in such a way that the three structures (complex, Riemannian, and symplectic) are all mutually compatible. ... In mathematics, given a Riemann surface X, the Teichmüller space of X, notated TX or Teich(X), is a complex manifold whose points represent all complex structures of Riemann surfaces whose underlying topological structure is the same as that of X. // Relation to moduli space The Teichmüller space... A rendering of the Riemann Sphere. ... In mathematics, the branching theorem is a theorem about Riemann surfaces. ... In mathematics, Hurwitzs automorphisms theorem bounds the group of automorphisms, via conformal mappings, of a compact Riemann surface of genus g > 1, telling us that the order of the group of such automorphisms is bounded by 84(g &#8722; 1). ... In mathematics, the identity theorem for Riemann surfaces is a theorem that states that a holomorphic function is completely determined by its values on any subset of its domain that has a limit point. ... In mathematics, specifically in complex analysis and algebraic geometry, the Riemann-Roch theorem is an important tool in the computation of the dimension of the space of meromorphic functions with prescribed zeroes and allowed poles. ... In mathematics, the Riemann-Hurwitz formula describes the relationship of the Euler characteristics of two surfaces when one is a ramified covering of the other. ... The Riemann mapping theorem in complex analysis states the following: if U is a simply connected open subset of the complex number plane C which is not all of C, then there exists a bijective holomorphic conformal map f : U -> D, where D = { z in C : |z| < 1 } denotes the... In mathematics, the uniformization theorem for surfaces says that any surface admits a Riemannian metric of constant Gauss curvature. ...", null, "Results from FactBites:\n\n PlanetMath: Riemann surface (1824 words)", null, "The simplest example of a Riemann surface which is not a subset of the complex plane is the Riemann sphere.", null, "A Riemann surface in the narrower sense is a branched covering of the complex plane.", null, "When one has constructed this surface and convinced oneself that it has the topology of a torus, one is well on one's way to developing an intuitive understanding of Riemann surfaces.\nMore results at FactBites »\n\nShare your thoughts, questions and commentary here\nPress Releases |", null, "Feeds | Contact" ]

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[ "# Questions tagged [lowpass-filter]\n\nA low-pass filter is an electronic filter that passes low-frequency signals but attenuates signals with frequencies higher than the cutoff frequency.\n\n497 questions\nFilter by\nSorted by\nTagged with\n118 views\n\nI have two cascaded one-pole sections with the same coefficient b for both sections. $$y_A[n]=x_A[n]+b\\cdot(y_A[n−1]−x_A[n])\\\\ y_B[n]=y_A[n]+b\\cdot(y_B[n−1]−y_A[n])$$ The final output of the system ...\n299 views\n\n### Cascaded one-pole sections and relationship to time constant\n\nI have implemented a typical one-pole lowpass filter to smooth out some control signals. The form of the filter is: $$y1 = x0 + b1\\centerdot(y1 - x0)$$ x0 is the current input sample. y1 is the past ...\n1k views\n\n### calculating the stop-band edge frequency of a low-pass FIR\n\nI have three information about a low-pass FIR filter. pass-band edge frequency pass-band ripple stop-band ripple I want to get the stop-band edge frequency I've been googling for the last hour ...\n2k views\n\n### Integration of square wave\n\ntrying to program an integrator. My input is a square wave and my expected output should be a triangle wave. However, whenever I pass it through my low pass filter algorithm (just a 2nd order ...\n75 views\n\n42 views\n\n### Denoising approach for a combination of several ADC voltage channles\n\nI have 2 ADC channels of constant voltage measurement with small amount of high frequency noise only and no low frequency oscillations. The final signal should be the simple sum of those and denoised ...\n290 views\n\n### Possible to implement simple IIR filter with greater than 6 db/octave slope?\n\nI am using the following IIR code as a low-pass filter in a software project, and it works great. However, I would like the slope of the filter to be much steeper as the one here seems to be very ...\n361 views\n\n### Implementation of a cascaded IIR filter\n\nI am attempting to implement a cascaded IIR filter in MATLAB. I used fdatool to design the filter and exported it as an object, Hd. In my script, I import a log file of data I captured and store my ...\n4k views\n\n### Why are ideal filters not particularly useful/ hard to do?\n\nWhy is a LPF ideal filter: a) Not useful? b) Hard to make? I was asked to make a Low pass filter (simulink block) that will take an input of a cutoff frequency, and for a signal with a frequency ...\n322 views\n\n1k views\n\n### Don't understand coefficients in MATLAB “butter” low pass result\n\nI don't unstand the nature of the output in Matlab's \"butter\" command. Calling [b,a] = butter(4, 0.5, 'low'); gives me two vectors, which the documentation says ...\n3k views\n\n### Is a high-passed signal the same as a signal minus a low-passed signal?\n\nMy question is, if I want to high-pass a signal, is it the same as low-passing a signal and subtracting it from the signal? Is it theoretically the same? Is it practically the same? I have searched (...\nI have implemented an IIR low pass filter in direct form 2, the input to the filter is a unit impulse $\\delta[n]$. I have calculated the output of the filter for the mentioned input using the ..." ]

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[ "## Saturday, April 23, 2016\n\n### Control of font sizes in plotting an object of class `\"cophylo\"` with phytools\n\nI received a user request recently about changing the font size in a plotted object of class `\"cophylo\"`. This is straightforward using the optional argument `fsize`, which is passed to the phytools to internal function `phylogram`. This is different from the more standard `cex` (character expansion factor); however it is kind of a legacy attribute of phytools' plotting functions, which are many.\n\nWhen I saw his plotted `\"cophylo\"` object, however, it was apparent to me that the problem was not merely that he didn't know how to control font size, but that the font sizes of the two plotted phylogenies needed to be able to be controlled independently. This is because it is allowed to plot trees in which the right tree & the left tree have different numbers of tips.\n\nThe solution to this is a little bit more complicated than it might seem because the optional arguments are passed internally in `plot.cophylo` using the three-dot `...` argument. This makes it a little bit harder than it otherwise might be because we first have to put the `...` arguments in a list, then we can check the length of the list vector `fsize` (if it exists), then we can supply the first or second of these numeric values to the first or second call of `phylogram`, respectively, by modifying our argument list values & then calling the function with a list of arguments using the R base function `do.call`.\n\nHere is a code-chunk to sort of illustrate what that looks like:\n\n``````obj<-list(...)\n## ...\nleftArgs<-rightArgs<-obj\nif(!is.null(obj\\$fsize)){\nif(length(obj\\$fsize)>1){\nleftArgs\\$fsize<-obj\\$fsize\nrightArgs\\$fsize<-obj\\$fsize\n## ...\n}\n}\nx1<-do.call(\"phylogram\",c(list(tree=x\\$trees[],part=0.4),leftArgs))\n## ...\nx2<-do.call(\"phylogram\",c(list(tree=x\\$trees[],part=0.4,\ndirection=\"leftwards\"),rightArgs))\n``````\n\nMore details of the updates can be seen here and here.\n\nThe following is a quick demo to illustrate its application using simulated data & trees:\n\n``````library(phytools)\nt1\n``````\n``````##\n## Phylogenetic tree with 26 tips and 25 internal nodes.\n##\n## Tip labels:\n## A, B, C, D, E, F, ...\n##\n## Rooted; includes branch lengths.\n``````\n``````t2\n``````\n``````##\n## Phylogenetic tree with 52 tips and 51 internal nodes.\n##\n## Tip labels:\n## t16, t17, t9, t8, t7, t45, ...\n##\n## Rooted; includes branch lengths.\n``````\n``````head(assoc) ## top of association matrix\n``````\n``````## [,1] [,2]\n## I1 \"I\" \"t16\"\n## I2 \"I\" \"t17\"\n## A \"A\" \"t9\"\n## B \"B\" \"t8\"\n## F \"F\" \"t7\"\n## G1 \"G\" \"t45\"\n``````\n``````obj<-cophylo(t1,t2,assoc)\n``````\n``````## Rotating nodes to optimize matching...\n## Done.\n``````\n``````obj\n``````\n``````## Object of class \"cophylo\" containing:\n##\n## (1) 2 (possibly rotated) phylogenetic trees in an object of class \"multiPhylo\".\n##\n## (2) A table of associations between the tips of both trees.\n``````\n\nNow if we plot using a single font size, then we either have text that is too small in one tree or too big in the other, for instance:\n\n``````plot(obj) ## fsize=1\n``````", null, "``````plot(obj,fsize=0.8)\n``````", null, "Let's try with two different font sizes:\n\n``````plot(obj,fsize=c(1.1,0.8))\n``````", null, "We can even add a scale bar with the same or different font size:\n\n``````h1<-max(nodeHeights(t1))\nh2<-max(nodeHeights(t2))\nplot(obj,fsize=c(1.2,0.7,1),scale.bar=round(0.5*c(h1,h2),2))\n``````", null, "Neat.\n\nThis version of phytools can be installed from GitHub as follows:\n\n``````library(devtools)\ninstall_github(\"liamrevell/phytools\")\n``````\n\nThe trees & associations were simulated as follows:\n\n``````library(phangorn)\nt1<-pbtree(n=26,tip.label=LETTERS)\nt2<-pbtree(n=52)\ntmp<-ltt(t2,plot=FALSE)\nh<-tmp\\$times[tmp\\$ltt==26]\nnn<-sapply(treeSlice(t2,h,trivial=TRUE),function(x) x\\$tip.label)\nN<-sapply(nn,length)\nmm<-mapply(function(x,n) rep(x,n),\n1.", null, "" ]

[ null, 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", null, 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", null, 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", null, 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", 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[ "Research Article | Open Access\n\nVolume 2019 |Article ID 8769428 | 11 pages | https://doi.org/10.1155/2019/8769428\n\n# On the Wiener Indices of Trees Ordering by Diameter-Growing Transformation Relative to the Pendent Edges\n\nAccepted22 Jan 2019\nPublished06 Feb 2019\n\n#### Abstract\n\nThe Wiener index of a graph is defined as the sum of distances between all unordered pairs of its vertices. We found that finite steps of diameter-growing transformation relative to vertices can not always enable the Wiener index of a tree to increase sharply. In this paper, we provide a graph transformation named diameter-growing transformation relative to pendent edges, which increases Wiener index of a tree sharply after finite steps. Then, twenty-two trees are ordered by their Wiener indices, and these trees are proved to be the first twenty-two trees with the first up to sixteenth smallest Wiener indices.\n\n#### 1. Introduction\n\nGraph theory has been applied in many areas of engineering extensively, such as mechanical design and manufacturing and chemical engineering. The link in the mechanism can be regarded as the vertex in the graph theory, kinematic pair can be regarded as an edge, and then the topological configuration of the mechanism is abstracted as the graph in graph theory. Therefore, relevant theories of graph theory can be used to analyse and discuss the nature and characteristics of the mechanism and especially be used to explore the movement patterns of the mechanism. Analogously in chemical engineering, a chemical molecule is usually regarded as a two-dimensional graph, and the graph’s vertices represent atoms and edges represent chemical bonds. Thus, the graph determines the topological properties of the given molecule. In fact, chemists are often interested in the Wiener indices of certain trees which represent certain molecular structures. In this paper, we study the trees with the first up to sixteenth smallest Wiener indices.\n\nAll graphs considered in this paper are simple and connected. Let be a graph that vertex set is and edge set is . For any , the distance between and in is the length of a shortest path. As the oldest graph-based structure descriptor (topological index), Wiener index was first provided by Wiener in and defined as Its mathematical properties and chemical applications have been extensively studied in many literatures. In , Wiener provided an alternative calculating form of Wiener index for each tree ; i.e., where (resp., ) is the number of vertices of the component of containing (resp., ). It should be mentioned that the above equation is valid only for trees.\n\nThere have been many related achievements on Wiener index and they can be roughly divided into the following categories.\n\n(i) Basic Theory and Calculation of Wiener Index (See ). Many known results about Wiener index of trees were introduced in , including the calculation of Wiener index, combinatorial expressions for Wiener index, and the relationship between Wiener index and some characteristic indicators of graph. Manuel et al. in presented a method for calculating Wiener index of certain chemical graphs without using distance matrix. Graovac et al. provided a modification of Wiener index which considers the symmetry of a graph in .\n\n(ii) The Wiener Index for Several Types of Trees (See ). Fuchs et al. in studied the Wiener index for several types of random digital trees. Wagner in demonstrated that every integer () is the Wiener index of a star-like tree. derived formulas for calculating the Wiener indices when a tree was modified.\n\n(iii) Other Topological Indices Associated with Wiener Index (See ). The Steiner Wiener index, its inverse problem and the Steiner hyper-Wiener index of a graph were studied in . References investigated the terminal Wiener index of graphs. For the results about the trees’ reverse Wiener indices and ordering problem by their reverse Wiener indices, one can refer to [15, 16].\n\nSince every atom has a certain valency, chemists especially focus on the trees with the maximum (resp., minimum) Wiener index. References characterized some -vertex trees which minimize or maximize the Wiener index over the set of trees with some degree and diameter restrictions.\n\nIt is natural to consider not only the maximum and minimum Wiener indices of given trees, but also the order of trees by Wiener indices. Ordering trees by their Wiener indices can help us to analyse and evaluate the relationship between the Wiener indices and the structures of chemical molecules represented by the certain trees. Actually, Dong et al. in ordered fifteen trees with in terms of the number of nonpendent edges. The trees with given matching number have been ordered by their Wiener indices in .\n\nMotivated by the results above, we extend the conclusion in to general case in this paper. In Section 2, we present an example which shows that the result of Lemma 5 in is not always true. As a by-product, we provide diameter-growing transformation relative to pendent edges; this transformation increases Wiener index sharply after finite steps. Then, we amend the proof of Theorem 2 in by applying this transformation. In Section 3, we order twenty-two trees by their Wiener indices and prove that these trees on are the first twenty-two trees with the first to sixteenth smallest Wiener indices.\n\nThroughout the paper, we denote the set of trees on vertices by , and the set of trees on vertices with diameter by .\n\n#### 2. The Trees’ Diameter-Growing Transformation Relative to Pendent Edges\n\nThe following example shows the conclusion of Lemma 5 in that is not always true. In fact, finite steps of diameter-growing transformation of relative to vertices can not always enable to increase sharply.\n\nExample 1. For each , the trees and are defined as in Figure 1. Then\n\nProof. From equality (2), it follows that and . So, it is obvious that holds for all .\n\nTo prove that Theorem 2 in is still true, we firstly show that the Wiener index increases sharply after finite steps of diameter-growing transformation of trees relative to pendent edges.\n\nLemma 2. Let with the longest path . Suppose . Then . In addition, the procedure from to shown in Figure 2 is called a diameter-growing transformation of relative to the pendent edge .\n\nProof. Let be the pendent edge of , be the nonpendent edge of , and and be the two components of containing and . Then and . Thus, it follows that\n\nIn fact, from the proof of Lemma 2, we can get the diameter-growing transformation relative to pendent edges in such general case as follows.\n\nRemark 3. For any tree defined as in Figure 3 and being a pendent edge of , it follows that .\n\nLet be the set of trees of order with exactly vertices of maximum degree and be the tree shown in Figure 4.\n\nTheorem 4 (Theorem 2 in ). Let , where . Then with equality if and only if .\n\nActually, the proof of Theorem 4 (Theorem 2 in ) is just needed to be modified by replacing Lemma 5 in with Lemma 2 in our paper.\n\n#### 3. The First Twenty-Two Trees in with the First up to Sixteenth Smallest Wiener Indices\n\nIn this section, firstly, we give the order of twenty-two trees on vertices according to Wiener indices by using the diameter-growing transformation relative to pendent edges.\n\nTheorem 5. For any , let , ) be some trees on vertices listed as in Figure 5, where is the diameter of each tree. Then we have\n\nProof. From equality (2), we can easily get All the above completes proof of the theorem.\n\nFor convenience, we denote\n\nNext, we are going to show that holds for any with . The proof process is organized according to the diameters of trees.\n\nTheorem 6. For each , let with . Then where tree is defined as in Figure 6.\n\nProof. On the one hand, from we get that for any . On the other hand, for any with and , is just shown as in Figure 7.\nIt is easy to see that holds for any with . Therefore, we get for any with and .\n\nNext, we are going to prove that is always true for each with and .\n\nLemma 7. For any , , and , let be the tree defined as in Figure 8. Then holds for any .\n\nProof. For each , we have It follows that Then the proof will be divided into three cases as follows.\nCase  1 (). For each , we have . Then, from for any , we get Meanwhile, we have , , and .\nCase  2 (). For each , we have . From , it follows that Moreover, .\nCase  3 (). For each , we have , which results in . Together with inequality (15), we get Since , then, for each and , from , it follows that All the above implies that holds for any .\n\nLemma 8. For each , , , and , let be the tree defined as in Figure 9. Then holds for any .\n\nProof. The proof will be divided into the following cases.\nCase  1 (). Then , and for each . Using Lemma 7, it is obvious that holds for any .\nCase  2 ()\nCase  2.1 (). Then . We get So, it follows that However, .\nFor any with , this case can be divided into two subcases as follows.\nCase  2.1.1 (). If , from and , it follows that , and hence Note that , so for each .\nIf , then . So . In fact, for each , from , we conclude that Thus, we have for any .\nCase  2.1.2 (). For each , from , we can get . Then it follows that .\nCase  2.2 ()\nCase  2.2.1 (). Since , for any , then for some . So, we have for .\nActually, from and , we get . Combining with , it follows that Case  2.2.2 (). Then we have . Indeed, from and , we get that , and hence From all of the above cases, it turns out that holds for any .\n\nLemma 9. Let be a tree on vertices and be a tree transformed from just as shown in Figure 10. Then, for each given , holds for any with . It is equivalent to say that holds for , where is the order of .\n\nProof. Actually, for any . In addition, from , it follows that , if and only if .\n\nActually, the following theorem is a result concluded from Lemmas 7, 8, and 9.\n\nTheorem 10. For each , and , let . Then\n\nProof. The proof will be divided into four cases as follows.\nFor each given , let be the set of all branching vertices of (where is the degree of vertex in ). From the assumption that and , it is obvious that .\nCase  1 (). Being from , together with Lemmas 8 and 9, we get .\nCase  2 (). The assumption that and implies that has at least two nonpendent vertices. If has exactly two nonpendent vertices, from Lemmas 7 and 9, together with , it follows that . Otherwise, from Lemmas 8 and 9, together with , it follows that .\nCase  3 (). The assumption that and implies that has at least two nonpendent vertices. If has exactly two nonpendent vertices, from Lemmas 7 and 9, together with , it follows that . Otherwise, from Lemmas 8 and 9, it follows that .\nCase  4 (). From Lemmas 8 and 9, we get that .\n\nTheorem 10 demonstrated the Wiener indices of the trees with diameter 4. Next, we will study the Wiener indices of the trees with diameters .\n\nIt is well-known that obtained the trees with minimum and second-minimum Wiener indices among all the trees with vertices and diameter . To be more precise, the authors in pointed out that the tree which attains the minimum Wiener index in is and that the tree which attains the second-minimum Wiener index in is . The trees and are shown as in Figure 11.\n\nLemma 11 (Theorem 3.6 in ). Let be the set of all trees on order with diameter .\n(i) For any with , the tree which attains the minimum Wiener index in the set is .\n(ii) For being odd and , the tree which attains the second-minimum Wiener index in the set is .\n(iii) For being even and , the tree which attains the second-minimum Wiener index in the set is or .\n\nLemma 12. For any , we have\n\nProof. In fact, holds if is even. Meanwhile, holds if is odd. From Remark 3, it follows that Thus, for any tree in Figure 12 and any in Figures 13 or 14, we get for each with whenever is even or odd.\n\nFrom Lemmas 11 and 12, it is easy to get the following conclusion.\n\nTheorem 13. Let with . Then .\n\nProof. Since we get for any . From Lemma 12, it follows that Moreover, from Lemma 11, we get for any tree on vertices with diameter . By using Lemma 11 again, we get for any tree on vertices with diameter , except for and . Thus, we conclude that holds for any tree on vertices with diameter .\n\nTheorem 14. For each , let . Then\n\nProof. By combining the previous arguments of Theorems 5, 6, 10, and 13, the proof is completed.\n\n#### 4. Conclusion\n\nIn this paper, we provide a transformation of trees named diameter-growing transformation relative to pendent edges. By using this transformation, we amend the proof of Theorem 2 in , then give the order of twenty-two trees by their Wiener indices, and prove that these trees are the first twenty-two trees with the first up to sixteenth smallest Wiener indices.\n\n#### Data Availability\n\nNo data were used to support this study.\n\n#### Conflicts of Interest\n\nThe authors declare that there are no conflicts of interest regarding the publication of this paper.\n\n#### Acknowledgments\n\nThis work is supported by the Natural Science Foundation of Shanxi Province (no. 201601D011003).\n\n1. H. Wiener, “Structural determination of paraffin boiling points,” Journal of the American Chemical Society, vol. 69, no. 1, pp. 17–20, 1947. 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Tratnik, “On the Steiner hyper-Wiener index of a graph,” Applied Mathematics and Computation, vol. 337, pp. 360–371, 2018. View at: Publisher Site | Google Scholar | MathSciNet\n12. I. Gutman, B. Furtula, and M. Petrović, “Terminal Wiener index,” Journal of Mathematical Chemistry, vol. 46, no. 2, pp. 522–531, 2009. View at: Publisher Site | Google Scholar | MathSciNet\n13. J. B. Babujee and J. Senbagamalar, “On Wiener and terminal Wiener index of graphs,” International Journal of Biomathematics, vol. 8, no. 5, Article ID 1550066, 2015. View at: Publisher Site | Google Scholar | MathSciNet\n14. Y.-H. Chen and X.-D. Zhang, “The terminal Wiener index of trees with diameter or maximum degree,” Ars Combinatoria, vol. 120, pp. 353–367, 2015. View at: Google Scholar | MathSciNet\n15. R. Xing and B. Zhou, “Ordering trees having small reverse Wiener indices,” Filomat, vol. 26, no. 4, pp. 637–648, 2012. View at: Publisher Site | Google Scholar | MathSciNet\n16. W. Luo and B. Zhou, “On ordinary and reverse Wiener indices of non-caterpillars,” Mathematical and Computer Modelling, vol. 50, no. 1-2, pp. 188–193, 2009. View at: Publisher Site | Google Scholar | MathSciNet\n17. R. H. Shi, “The average distance of trees,” Journal of Systems Science and Mathematical Sciences, vol. 6, no. 1, pp. 18–24, 1993. View at: Google Scholar | MathSciNet\n18. S. Mukwembi and T. Vetrík, “Wiener index of trees of given order and diameter at most 6,” Bulletin of the Australian Mathematical Society, vol. 89, no. 3, pp. 379–396, 2014. View at: Publisher Site | Google Scholar\n19. K. C. Das and M. J. Nadjafi-Arani, “On maximum Wiener index of trees and graphs with given radius,” Journal of Combinatorial Optimization, vol. 34, no. 2, pp. 574–587, 2017. View at: Publisher Site | Google Scholar | MathSciNet\n20. H. Lin, “A note on the maximal Wiener index of trees with given number of vertices of maximum degree,” MATCH - Communications in Mathematical and in Computer Chemistry, vol. 72, no. 3, pp. 783–790, 2014. View at: Google Scholar | MathSciNet\n21. M. Song and H. Lin, “Minimizing Wiener index of trees containing a prescribed subtree,” Ars Combinatoria, vol. 138, pp. 211–222, 2018. View at: Google Scholar | MathSciNet\n22. M. Goubko, “Maximizing Wiener index for trees with given vertex weight and degree sequences,” Applied Mathematics and Computation, vol. 316, pp. 102–114, 2018. View at: Publisher Site | Google Scholar | MathSciNet\n23. H. Dong and X. Guo, “Ordering trees by their Wiener indices,” MATCH - Communications in Mathematical and in Computer Chemistry, vol. 56, no. 3, pp. 527–540, 2006. View at: Google Scholar | MathSciNet\n24. S.-W. Tan, N.-N. Wei, Q.-L. Wang, and D.-F. Wang, “Ordering trees with given matching number by their Wiener indices,” Applied Mathematics and Computation, vol. 49, no. 1-2, pp. 309–327, 2015. View at: Publisher Site | Google Scholar | MathSciNet\n25. H. Liu and X.-F. Pan, “On the Wiener index of trees with fixed diameter,” MATCH - Communications in Mathematical and in Computer Chemistry, vol. 60, no. 1, pp. 85–94, 2008. View at: Google Scholar | MathSciNet\n\n#### More related articles\n\nWe are committed to sharing findings related to COVID-19 as quickly and safely as possible. Any author submitting a COVID-19 paper should notify us at [email protected] to ensure their research is fast-tracked and made available on a preprint server as soon as possible. We will be providing unlimited waivers of publication charges for accepted articles related to COVID-19. Sign up here as a reviewer to help fast-track new submissions." ]

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[ "", null, "Fifth grade geometry make sure your fifth grader is ready for middle school with these worksheets that guarantee they will have the skills needed to succeed in their high level math classes. With activities such as calculating the amount of flooring needed to remodel a room or the number of items that will fit into a moving box these fifth grade worksheets provide practice with measuring area and volume.", null, "Grade 5 Math Worksheets Add Subtract Multiply Fractions Decimals Measure Geometry And Word Problems Multiplying Decimals Math Worksheets Decimals\n\n### Below you will find a wide range of our printable worksheets in chapter planes and solids of section geometry these worksheets are appropriate for fifth grade math we have crafted many worksheets covering various aspects of this topic triangles and quadrilaterals drawing plane figures solid figures draw and construct solid figures and views and many more.", null, "Geometry fifth grade math worksheets grade 5. Free printable grade 5 math worksheets. These geometry worksheets give students practice in 2 d geometry such as classifying angles and triangles classifying quadrilaterals calculating perimeters and areas and working with circles. A brief description of the worksheets is on each of the worksheet widgets.\n\n3d geometry is introduced with rectangular prisms. This math pack includes word problems coordinate graphs calculation operation problems and answer keys that align to the 5th grade common core math standards. We also introduce variables and expressions into our word problem worksheets all worksheets are printable pdf documents.\n\n17 pages of printable problems 6 problems on every page with room to complete work for each problem. Worksheets math grade 5 geometry. Worksheets math grade 5.\n\nReading and plotting points on a coordinate grid is also covered. This page contains all our printable worksheets in section geometry of fifth grade math. 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[ "FAQ\n\n# what percent of 12 is 3\n\n## What Percent Of 12 Is 3?\n\nPercentage Calculator: 3 is what percent of 12? = 25.\n\n## How do you get 3 out of 12 as a percentage?\n\nNow we can see that our fraction is 25/100, which means that 3/12 as a percentage is 25%.\n\n## What is a percentage of 12?\n\nSolution for 12 is what percent of .12:\n\n12 is 100% since it is our output value.\n\n## What is 2 as a percentage of 12?\n\nPercentage Calculator: 2 is what percent of 12? = 16.67.\n\n## What percent is 4 out of 12?\n\nWhat is this? Now we can see that our fraction is 33.333333333333/100, which means that 4/12 as a percentage is 33.3333%.\n\n## What is 3/12 as a decimal?\n\n3/12 as a decimal is 0.25.\n\n## What percentage is a 8 out of 12?\n\nWhat is this? Now we can see that our fraction is 66.666666666667/100, which means that 8/12 as a percentage is 66.6667%.\n\n## What percent is 3 out of 11?\n\nPercentage Calculator: . 3 is what percent of 11? = 2.73.\n\n## How do you find a 12th mark percentage?\n\n12th mark percentage calculator\n1. Step 1: To calculate 12th marks percentage – first divide the scored mark divide by out of mark.\n2. Step 2: Then, Multiply the value by 100.\n3. Step 3: Normally 12th out of mark is 1200.\n4. Step 4: For Example, if person scored 910 out of 1200.\n5. Step 5: Hence, ((410/500)*100) = 75.83%\n\n## How do you find the percentage of an amount?\n\n1. How to calculate percentage of a number. Use the percentage formula: P% * X = Y\n1. Convert the problem to an equation using the percentage formula: P% * X = Y.\n2. P is 10%, X is 150, so the equation is 10% * 150 = Y.\n3. Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.\n\n## What is the percentage of 1 out of 12?\n\nWhat is this? Now we can see that our fraction is 8.3333333333333/100, which means that 1/12 as a percentage is 8.3333%.\n\n## How do I find 2/3 of a number?\n\nTo find 2/3 of a whole number we have to multiply the number by 2 and divide it by 3. To find two-thirds of 18, multiply 2/3 x 18/1 to get 36/3.\n\n## What is the percentage 9 out of 12?\n\nNow we can see that our fraction is 75/100, which means that 9/12 as a percentage is 75%.\n\n## What percent is 6 out of 12?\n\nNow we can see that our fraction is 50/100, which means that 6/12 as a percentage is 50%.\n\n## What is 7 out of 12 as a percentage?\n\n58.3%\nThe fraction 7/12 as a percent is 58.3%.\n\n## What percentage is 5 out of 13?\n\nPercentage Calculator: 5 is what percent of 13? = 38.46.\n\n## What is the simplest form 4 12?\n\nSteps to simplifying fractions\n\nTherefore, 4/12 simplified to lowest terms is 1/3.\n\n## What fraction is three quarters?\n\nThe fraction 3/4 or three quarters means 3 parts out of 4. The upper number, 3, is called the numerator and the lower number, 4, is the denominator.\n\n## What percent is a 10 out of 12?\n\nNow we can see that our fraction is 83.333333333333/100, which means that 10/12 as a percentage is 83.3333%.\n\n## What percentage is 4 out of 6?\n\n66.67%\nAnswer: 4 out of 6 is represented as 66.67% in the percentage form.\n\n67 – 69 D+\n63 – 66 D\n60 – 62 D-\n< 60 F\n\n## What percent is 1 out of 3?\n\nFraction to percent conversion table\nFraction Percent\n1/3 33.33%\n2/3 66.67%\n1/4 25%\n2/4 50%\n\n## What is 5/12 as a percentage?\n\nNow we can see that our fraction is 41.666666666667/100, which means that 5/12 as a percentage is 41.6667%.\n\n## What percent is a 8 out of 11?\n\nNow we can see that our fraction is 72.727272727273/100, which means that 8/11 as a percentage is 72.7273%.\n\n## How do you find the percentage of 3 subjects?\n\nA percentage is a number that is shown in terms of 100.To find the percentage of the marks obtained, one shall divide the total scores by marks obtained and then multiply the result with 100. Example: If 79 is the score obtained in the examination out of 100 marks, then divide 79 by 100, and then multiply it by 100.\n\n## How is 12th result calculated?\n\nThe CBSE included all the marks from the student’s historical performances to calculate the final result. What is 40:30:30 ratio? Through the 40:30:30 formula, 40 per cent weightage was given to the Class 12 category, 30 per cent to Class 11, and 30 per cent to Class 10, to calculate theory marks.\n\n## How do you calculate 2 percent?\n\nTo find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage. So, 0.27 multiplied by 100 equals 27 or 27%.\n\n## How much is percentage of a number?\n\nIf we have to calculate percent of a number, divide the number by whole and multiply by 100. Hence, the percentage means, a part per hundred. The word per cent means per 100. It represented by the symbol “%”.\n\n## What percentage of a number is another number?\n\nDivide the first number the second. For instance, if you want to find what percentage 43 is out of 57, divide 43 by 57 to get 0.754386. Multiply the result by 100 — 0.754386 x 100 = 75.4386. Round the result.\n\n## What percent is a 11 out of 12?\n\nWhat is this? Now we can see that our fraction is 91.666666666667/100, which means that 11/12 as a percentage is 91.6667%.\n\n## What percent is 2 out of 13?\n\nWhat is this? What is this? Now we can see that our fraction is 15.384615384615/100, which means that 2/13 as a percentage is 15.3846%.\n\n## What is 3/4 as a percentage?\n\n75%\nAnswer: 3/4 is expressed as 75% in terms of percentage.\n\n## What is a two third of 12?\n\n8\nAnswer: 2/3 of 12 is 8.\n\n## Fraction to Percent Conversion\n\nRelated Searches\n\nwhat is 3 out of 12 as a grade\n4 is what percent of 12\n2 is what percent of 12\npercentage calculator\n3 of 12\n1 is what percent of 12\n6 is what percent of 10\n5 out of 12 percentage\n\nSee more articles in category: FAQ\nCheck Also\nClose" ]

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[ "Pure and Applied Mathematics Journal\nVolume 4, Issue 5, October 2015, Pages: 219-224\n\nSolving Fourth Order Parabolic PDE with Variable Coefficients Using Aboodh Transform Homotopy Perturbation Method\n\nKhalid Suliman Aboodh\n\nDepartment of Mathematics, Faculty of Science & technology, Omdurman Islamic University, Khartoum, Sudan", null, "Khalid Suliman Aboodh. Solving Fourth Order Parabolic PDE with Variable Coefficients Using Aboodh Transform Homotopy Perturbation Method. Pure and Applied Mathematics Journal. Vol. 4, No. 5, 2015, pp. 219-224. doi: 10.11648/j.pamj.20150405.13\n\nAbstract: Here, a new method called Aboodh transform homotopy perturbation method (ATHPM) is used to solve one dimensional fourth order parabolic linear partial differential equations with variable coefficients. The proposed method is a combination of the new integral transform \"Aboodh transform\" and the homotopy perturbation method. Some cases of one dimensional fourth order parabolic linear partial differential equations are solved to illustrate ability and reliability of mixture of Aboodh transform and homotopy perturbation method. We have compared the obtained analytical solution with the available Aboodh decomposition solution and homotopy perturbation method solution which is found to be exactly same. The results obtained reveal that the combination of Aboodh transform and homotopy perturbation method is quite capable, practically well appropriate for use in such problems.\n\nKeywords: Aboodh Transform, Homotopy Perturbation Method, Linear Partial Differential Equation\n\nContents\n\n1. Introduction\n\nMany problems of physical interest are described by linear partial differential equations with initial and boundary conditions. One of them is fourth order parabolic partial differential equations with variable coefficients; these equations arise in the transverse vibration problems . In recent years, many research workers have paid attention to find the solution of these equations by using various methods. Among these are the variational iteration method [Biazar and Ghazvini (2007)], Adomian decomposition method [Wazwaz (2001) and Biazar et al (2007)], homotopy perturbation method [Mehdi Dehghan and Jalil Manafian (2008)], homotopy analysis method [Najeeb Alam Khan, Asmat Ara, Muhammad Afzal and Azam Khan (2010)] and Laplace decomposition algorithm [Majid Khan, Muhammad Asif Gondal and Yasir Khan (2011)]\n\nPrem Kiran G. Bhadane, and V. H. Pradhan (2013)]. In this paper we use coupling of new integral transform \"Aboodh transform\" and homotopy perturbation method. This method is a useful technique for solving linear and nonlinear differential equations. The main aim of this paper is to consider the effectiveness of the Aboodh transform homotopy perturbation method in solving higher order linear partial differential equations with variable coefficients. This method provides the solution in a rapid convergent series which leads the solution in a closed form.\n\n2. Aboodh Transform Homotopy Perturbation Method [1, 2, 3, 4]\n\nLet us consider the a one dimensional linear nonhomogeneous fourth order parabolic partial differential equation with variable coefficients as", null, "(1)\n\nwhere", null, "is a variable coefficient, with initial conditions", null, "and", null, "and boundary conditions as", null, ",", null, "(3)\n\nMaking use the Aboodh transform on both sides of\n\nEq. (1), we obtain", null, "(4)\n\nand, Aboodh transform of partial derivative", null, "With the aid of this property, Eq. (3) as redas", null, "(5)\n\nBy using of initial conditions in Eq. (5), we have", null, "or", null, "(6)\n\nWhere", null, "Applying Aboodh inverse on both sides of Eq. (6), admits to", null, "(7)\n\nWhere", null, ", represents the term\n\nArising from the source term and the prescribed initial conditions.\n\nBy apply the homotopy perturbation method.", null, "(8)\n\nBy substituting Eq. (8) into Eq. (7), we get", null, "(9)\n\nThis is the coupling of the Aboodh transform and the homotopy perturbation method. collectiy the coefficient of like powers of", null, ", we have", null, "And so on\n\nIn general recursive relation is given by", null, "Then the solution can be expressed as", null, "(10)\n\n3. Application\n\nIn this section, to demonstrate the effectiveness of the proposal method we consider homogeneous and non homogeneous one dimensional fourth order linear partial differential equations with initial and boundary conditions.\n\nExample 1.\n\nLet us first consider fourth order homogeneous partial differential equation as [1,5]", null, "(11)\n\nWith initial conditions", null, "and", null, "(12)\n\nand boundary conditions", null, "(13)\n\nApplying Aboodh transform to Eq. (11), admits to", null, "", null, "Using initial conditions from Eq. (12), we have", null, "By used Aboodh inverse on both sides of above Eq. (15),", null, "To solve Eq.(12) using  homotopy perturbation method,\n\nwe assume that", null, "inserty Eq ( 17) of u(x, t) into Eq. (16), we get", null, "Collecting the coefficient of like powers of", null, "in Eq. (18) we have", null, "", null, "and so on in the same manner the rest of the components of iteration formula can be obtained and thus solution can be written in closed form as", null, "which is an exact solution of Eq. (11) and can be verified through substitution\n\nExample 2.\n\nA second instructive model is the fourth order homogeneous partial differential equation as", null, "with the initial conditions", null, "and", null, "and boundary conditions", null, "Applying Aboodh transform to Eq. (20), we get", null, "", null, "Using initial conditions from Eq. (20), we get", null, "Now taking Aboodh inverse on both sides of above\n\nEq. (24), we have", null, "Now, we apply the homotopy perturbation method.", null, "Putting this value of u(x, t) into Eq. (25), we get", null, "collecty the coefficient of like powers of p , in Eq. (26) the following approximations are obtained", null, "", null, "and so on in the same manner the rest of the components of iteration formula can be obtained and thus solution can be written in closed form as", null, "Which is an exact solution of Eq. (20) and can be\n\nVerified  through substitution\n\nExample 3.\n\nin this case we consider the  fourth order nonhomogeneous partial differential equation as [1,5]", null, "with the following initial conditions", null, "and", null, "and boundary conditions", null, "Applying Aboodh transform to Eq. (28), we get", null, "", null, "Using initial conditions from Eq. (29), we get", null, "now taking Aboodh inverse on both sides of above Eq. (31),we have", null, "Now, we apply the homotopy perturbation method.", null, "Putting this value of u(x, t) into Eq. (32), we get", null, "Here, we assume that", null, "Can be divided into the sum of two parts namely", null, "and", null, "therefore we get", null, "Under this assumption, we propose a slight variation only in the components", null, ". The variation we propose is that only the part", null, "be assigned to the", null, "where as the remaining part", null, ", be combined with the other terms to define", null, ".", null, "In view of these, we formulate the modified recursive algorithm as follows", null, "And so on in the same manner the rest of the components of iteration formula can be obtained", null, "for", null, "Thus solution can be written in closed form as", null, "Which is an exact solution of Eq. (28) and can be verified through substitution\n\n4. Conclusion\n\nThe aim of this paper is to show the applicability of the mixture of new integral transform \"Aboodh transform\" with the homotopy perturbation method to solve one dimensional fourth order homogeneous and nonhomogeneous linear partial differential equations with variable coefficients. This combination of two methods successfully worked to give very reliable and exact solutions to the equation. This proposed method provides an analytical approximation in a rapidly convergent sequence with in exclusive manner computed terms. Its rapid convergence shows that the method is trustworthy and introduces a significant improvement in solving linear partial differential equations over existing methods.\n\nFinally it's worthwhile to mention that the proposed method can be applied to other linear and nonlinear partial differential equations arising in mathematical physics. This aim task in future.\n\nReferences\n\n1. Majid Khan, Muhammad Asif Gondal and Yasir Khan, An Efficient Laplace Decomposition algorithm for Fourth order Parabolic Partial Differential Equations with variable coefficients, World Applied Sciences Journal 13 (12), 2011, pp2463-2466.\n2. Tarig M. Elzaki and Eman M. A. Hilal, Homotopy Perturbation and ELzaki Transform for solving Nonlinear Partial Differential equations, Mathematical Theory and Modeling, 2(3),2012, pp33-42.\n3. Khalid Suliman Aboodh, Application of New Transform \"Aboodh Transform\" to Partial Differential Equations ''Global Journal of Pure and Applied Mathematics ISSN 0973-1768 Volume 10, Number 2 (2014), pp. 249-254.\n4. Hradyesh Kumar Mishra and Atulya K Nagar, He-Laplace Method for Linear and Nonlinear Partial Differential Equations, Journal of Applied Mathematics, Vol.2012,Article ID 180315, 16 pages.\n5. Mehdi Dehghan and Jalil Manafian, The Solution of the Variable Coefficients Fourth-Order Parabolic Partial Differential Equations by the Homotopy Perturbation Method,Verlag der Zeitschrift fur Naturforschung, Tu¨bingen, 64a, 2009, pp420-430.\n6. M. Hussain and Majid Khan, Modified Laplace Decomposition Method, Journal of Applied Mathematical Sciences, 4(36), 2010, pp 1769-1783.\n7. Najeeb Alam Khan, Asmat Ara, Muhammad Afzal and Azam Khan, Analytical Aspect of Fourth order Parabolic Partial differential Equations with variable coefficients, Mathematical and Computational Applications, 15(3), 2010, pp. 481-489.\n8. Khalid Suliman Aboodh, The New Integral Transform ''AboodhTransform''GlobalJournalofPure and Applied Mathematics ISSN0973-1768 Volume 9, Number 1 (2013), pp. 35-43.\n9. Tarig M. Elzaki and Salih M. Elzaki, Applications of New Transform \"ELzaki Transform\" to Partial Differential Equations,Global Journal of Pure and Applied Mathematics, (7)1,2011,pp65-70.\n10. He J. Homotopy-perturbation method for solving boundary value problem, Phys Lett A, 350 (2006), 87-88.\n\n Contents 1. 2. 3. 4.\nArticle Tools", null, "Abstract", null, "PDF(185K)", null, "" ]

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[ "## Course Outline\n\n• segmentGetting Started (Don't Skip This Part)\n• segmentStatistics and Data Science: A Modeling Approach\n• segmentPART I: EXPLORING VARIATION\n• segmentChapter 1 - Welcome to Statistics: A Modeling Approach\n• segmentChapter 2 - Understanding Data\n• segmentChapter 3 - Examining Distributions\n• segmentChapter 4 - Explaining Variation\n• segmentPART II: MODELING VARIATION\n• segmentChapter 5 - A Simple Model\n• segmentChapter 6 - Quantifying Error\n• segmentChapter 7 - Adding an Explanatory Variable to the Model\n• segmentChapter 8 - Models with a Quantitative Explanatory Variable\n• segmentFinishing Up (Don't Skip This Part!)\n• segmentResources\n\n### list High School / Statistics and Data Science I (AB)\n\nBook\n• High School / Statistics and Data Science I (AB)\n• College / Statistics and Data Science (ABC)\n• High School / Advanced Statistics and Data Science I (ABC)\n• College / Advanced Statistics and Data Science (ABCD)\n• High School / Statistics and Data Science II (XCD)\n\n## 2.6 Sampling From a Population\n\nTurning variation in the world into data that we can analyze requires us to engage in two fundamental processes: measurement and sampling. Now that we’ve talked a little bit about measurement, let’s shift our attention to sampling.\n\nThere are two major decisions that underlie a sampling scheme. First, we must decide what to include in our universe of objects to be measured. This seems straightforward. But, it is not. When we decide which objects to include, we are saying, in effect, that these objects are all examples of the same category of things. (See the wonderful book, Willful Ignorance, for an in-depth discussion of this point.) This category of things can be referred to as the population. The population is usually conceptualized as large, which is why we study only a sample.\n\nLet’s say we want to do a study of cocaine users to figure out the best ways to help them quit. What counts as a cocaine user? Someone who uses cocaine every day obviously would be included in the population. But what about someone who has used cocaine once? What about someone who used to frequently use cocaine but no longer does? You can see that our understanding of cocaine users will be impacted by how we define “cocaine user.”\n\nSecond, we must decide how to choose which objects from the defined population we actually want to select and measure. Most of the statistical techniques you will learn about in this course assume that your sample is independently and randomly chosen from the population you have defined.\n\n### Random Sampling\n\nFor a sample to be random, every object in the population must have an equal probability of being selected for the study. In practice, this is a hard standard to reach. But if we can’t have a truly random sample, we can at least try to feel confident that our sample is representative of the population we are trying to learn about.\n\nFor example, let’s say we want to measure how people from different countries rate their feelings of well-being. Which people should we ask? People from cities? Rural areas? Poor? Rich? What ages? Even though it might be difficult to select a truly random sample from the population, we might want to make sure that there are enough people in different age brackets, different socioeconomic status (SES), and different regions to be representative of that country.\n\n### Independent Sampling\n\nClosely related to randomness is the concept of independence. Independence in the context of sampling means that the selection of one object for a study has no effect on the selection of another object; if the two are selected, they are selected independently, just by chance.\n\nLike randomness, the assumption of independence is often violated in practice. For example, it might be convenient to sample two children from the same family for a study—think of all the time you’d save not having to go find yet another independently and randomly selected child!\n\nBut this presents a problem: if two children are from the same family, they also are likely to be similar to each other on whatever outcome variable we might be studying, or at least more similar than would a randomly selected pair. This lack of independence could lead us to draw erroneous conclusions from our data.\n\n### Sampling Variation\n\nEvery sample we take—especially if they are small samples—will vary. That is, samples will be different from one another. In addition, no sample will be perfectly representative of the population. This variation is referred to as sampling variation or sampling error. Similar to measurement error, sampling error can either be biased or unbiased.\n\nFor example, we’ve created a vector called fake_pop (for fake population) that has 100 0s, 100 1s, 100 2s, 100 3s, and so on for the numbers 0 to 9. Here is a graph (called a histogram) that helps you see that there are 100 of each digit in this fake population.", null, "Now let’s take a random sample of 10 numbers from fake_pop and save the result in a vector called random_sample. R provides a function for this, called sample().\n\nRun the code to take a random sample of 10 numbers from fake_pop and save them in random_sample. We’ve also included some code to create a histogram but don’t worry about that part. We’ll build histograms together in Chapter 3.\n\nrequire(coursekata) fake_pop <- rep(0:9, each = 100) # This takes a random sample of 10 numbers # from fake_pop and saves them in random_sample random_sample <- sample(fake_pop, 10) # This makes a histogram of your sample gf_histogram(~random_sample) %>% gf_refine(scale_x_continuous(breaks = 0:9)) random_sample <- sample(fake_pop, 10) # this will make a histogram of your sample gf_histogram(~random_sample) %>% gf_refine(scale_x_continuous(breaks = 0:9)) ex() %>% { check_object(., \"random_sample\") check_function(., \"gf_histogram\") }\nCK Code: ch2-17\n\nYou can run the R code above a few times to see a few different random samples. You’ll observe that these samples differ from each other and differ from the population they came from. Below we have depicted one example of a random sample of 10 numbers from fake_pop.", null, "Notice that even a random sample won’t “look” just like the population it came from. This fundamental idea of sampling variation is a tough problem. We’re going to meet up with sampling variation a lot in statistics.\n\nBias in the context of sampling means that some objects in the population are more likely than others to be included in the sample even though there aren’t actually more of those objects in the population. This violates the assumptions of an independent random sample. Because of sampling variation, no sample will perfectly represent the population. But if we select independent random samples, we can at least know that our sample is unbiased—no more likely to be off in one direction than in another. So even though our random sample did not look just like the population, it was unbiased in that it wasn’t more likely to be off in a particular way (e.g., only sampling even numbers or large numbers)." ]

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[ "# Arctan(x) Calculator\n\nAn online calculator to calculate inverse tangent function arctan(x) in radians and degrees.\n\n## How to use the arctan (x) calculator\n\n1 - Enter x as a real number and the number of decimal places desired then press \"enter\". Two answers; one in radians and the other in degrees are displayed.\n\n x = 1 Decimal Places = 3 arctan(x)= radians arctan(x) = degrees\n\nYou may use it to explore the domain and range of arctan(x).\n\nMore Math Calculators and Solvers." ]

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[ "Last Updated On: December 28, 2016\nEditor: Adeeb C\n\n# Decision Making Using If….Then Statements in VB.Net\n\nCategory: Programing,VB.NET\nTags\n\nVB.Net supports five types of decision making statements. Among them the simplest one is the If…Then statements. Below are the five types of decision making statements using in VB.Net.\n\nHere we are discussing the decision making in VB.Net using the If Then statements.\n\nThe simple if then statement checks whether the condition is true or false, and if the condition is true then the [statement x] will be executed.\n\nBasic Syntax\n\n```If condition Then\n[Statement x]\nEnd If```\n\nExample\n\n```If (a>0) Then\nconsole.WriteLine(\"Postive\")\nEnd If```\n\nIn the given example, if the value of a is greater than 0 statement x will be executed and will provide the output message “Positive”.\n\nThe below program will give you more explanation.\n\nExample Program\n\n```Module Module1\n\nSub Main()\nDim num As Integer\nConsole.WriteLine(\"Enter a positive number:\")\nnum = Console.ReadLine()\nIf num > 0 Then\nConsole.WriteLine(\"Number is valid\")\nEnd If\nEnd Sub\n\nEnd Module```" ]

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[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\nMain content\n\n# Geometric constructions: triangle-inscribing circle\n\nCCSS.Math:\n\n## Video transcript\n\nconstruct a circle inscribing the triangle so this would be a circle that's inside this triangle where each of the sides of the triangle are tangent to the circle and one way to or the one the probably the easiest way to think about it is the center of that circle is going to be at the in center of the triangle now what is the in center of the triangle the in center of the triangle is the intersection of the angle bisectors so if I were to make a line that perfectly splits an angle in - so I'm a balling it right over here this would be an angle bisector but to be a little bit more precise about angle bisectors I could actually use a compass so let me make this a little bit smaller and what I can do is I could put this the center of the circle on one of the sides of this angle right over here now let me get another circle and I want to I want to make it the same size so let me Center it there I want to make it the exact same size and now let me put it on the other one on the other side of this angle I'll put it right over here I want to put it so that the center of the circle is on the other side of the angle and the circle itself or the vertex sits on the circle itself and what this does is I can now look at the intersection of this point the vertex and this point and that's going to be an angle bisector or the angle bisector so let me go I'm going to go through there and I'm going to go through there now let me move these circles over to to here so I can take the angle bisector of this side as well so I can put this one over here and I could put this one see I want to be on the side of the angle and I want to go right through I want the circle to go right through the vertex and let me add another straight edge here so I want to go through this point and I want to bisect the angle so I want to bisect the angle go right through the other point of intersection of these two circles now let me get rid of one of these let me get rid of one of these two circles I don't need it that anymore and let me use this one to actually construct the circle inscribing the triangle so I'm going put it at the center right over there actually this one's already pretty close in terms of dimensions and with this tool you don't have to be 100% precise it has some margin for error it has some margin for error and so let's just go with this this actually should be touching but this has some margin for error let's see if this was good enough it was" ]

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[ "# What are all the factors of 955?\n\nHere you can find the answer to questions related to: What are all the factors of 955? or list the factors of 955. By using our online calculator to find the prime factors of any composite number and check if a number is prime or composite. This tool also draws the prime factor tree if the number is factorable and smaller than 16000.\n\n### Prime Factors Calculator\n\nEnter a natural number to get its prime factors:\n\nEx.: 4, 8, 9, 26, 128, etc.\nThis calculator accepts numbers up to 10,000,000,000 (10 billion)\nResults:\nThe number 955 is a composite number because 955 can be divided by one, by itself and at least by 5 and 191. A composite number is an integer that can be divided by at least another natural number, besides itself and 1, without leaving a remainder (divided exactly).\n\nThe factorization or decomposition of 955 = 5•191. Notice that here, it is written in exponential form.\n\nThe prime factors of 955 are 5 and 191. It is the list of the integer's prime factors.\n\nThe number of prime factors of 955 is 2.\n\n### Factor tree or prime decomposition for 955\n\nAs 955 is a composite number, we can draw its factor tree:\n\n## What is prime number? How to factorize a number?\n\nWatch this video below to learn more on prime numbers, prime factors and how you can draw a factor tree.\n\nYou can also find this video about factorization at mathantics.com\n\n## Other way people search this question\n\n• Is 955 a prime number?\n• Is 955 prime or composite?\n• Is 955 a composite number?\n• How to find the prime decomposition of 955?\n• how many factors does 955 have?" ]

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[ "# ANALOG-TO-DIGITAL CONVERSION\n\n## Review of the Nature of Digital Data\n\n• A digital bit has a concrete value that can be accurately stored reproduced.\n• Only 2 values means long binary numbers, but easier handling and less error-prone.  This is fine for computers, because they are stoopid, but fast.\n• Encoded sound is just one form of digital data, which overlaps with information technology.\n• Terms in digital audio also have similar meanings in digital images and video.\n\n## Analog-to-Digital Conversion\n\n• The process of taking a continuous analog signal and encoding it into discrete values.\n• Sometimes called ADC or A/D\n• Reverse process is digital-to-analog conversion (DAC or D/A)\n• Most common method uses PCM (pulse code modulation), but there are other ways to do it.\n• Foundation in Harry Nyquist's theorem (Bell Labs, 1928!)\n• Claude Shannon proved algebra could be done with electronics, making computing and information theory a reality (Bell Labs, 1949)\n\n# Digital Sound Quality\n\n## Meaningful Measurements?\n\n\"CD Quality\"\n\n\"Audiophile Quality\"\n\nSince we care about how it sounds, better go back to the Acoustic World...\n\nWhat are the most basic components of a sound, again?\n\n# Frequency Response\n\n## Sampling Rate = How Frequently to Measure?\n\n• Nyquist Theorem:  Sample twice as often as the highest frequency:\n• To capture 4 kHz, sample 8 kHz (8000 times per second).\n• 8 kHz = 16 kHz sample rate\n• 20 kHz = 40 kHz sample rate\n\n## Common Sampling Rates\n\n• 44.1 kHz = CD\n• 48 kHz = Video\n• 88.2, 96, 192, 384 kHZ!\n• Sample rate conversion is not a trivial process, and should be avoided unless necessary.\n\n# Amplitude\n\n## Bit Resolution = How Precisely to Measure?\n\n• Measuring amplitude means measuring dynamic range.\n• Bits combine to form 'words' (16-bit word, 24-bit word, etc.)\n• 1 Bit = 6 dB of dynamic range (meaning it lowers the noise floor by 6 dB)\n• More bits = more possible amplitude values = less quantization error\nWORD LENGTH DYNAMIC RANGE AMPLITUDE VALUES\n2 Bits 12 dB\n(6dB/bit x 2 bits)\n4\n8 Bits 48 dB 256\n16 Bits–CD 96 dB 65,536\n24 Bits 144 dB 16,777,216\n\n(\\( 2 ^2 \\))\n\n(\\( 2 ^8 \\))\n\n(\\( 2 ^{16} \\))\n\n(\\( 2 ^{24} \\))\n\n1 Bit (2 values)\n\n2 Bit (4 values)\n\n8 Bit (256 values)\n\nIt is difficult to appreciate the accuracy achieved by a 16-bit measurement. An analogy might help: If sheets of typing paper were stacked to a height of 22 feet, a single sheet of paper would represent one quantization level in a 16-bit system.\n\nLonger word lengths are even more impressive. In a 20-bit system, the stack would reach 352 feet. In a 24-bit system, the stack would tower 5632 feet in height-- Over a mile high. The quantizer could measure that mile to an accuracy equaling the thickness of one piece of paper. If a single page were removed, the least significant bit would change from 1 to 0.\n\nLooked at in another way, if the distance between New York and Los Angeles were measured with 24-bit accuracy, the measurement would be accurate to within 9 inches. A high-quality digital audio system thus requires components with similar tolerances-not a trivial feat.\n\n–from Fundamentals of Digital Audio by John Watkinson\n\nBy Brian\n\n• 69" ]

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[ "# Explanations of mathematical explanation\n\nI recently read Mathematical Explanation [gated], by Mark Steiner (1978). My summary follows, and my commentary follows that. I am aware that others have written things since 1978 on this topic, but I don’t have time to read them right now.\n\n***\n\nWe seem to think there is a distinction between explaining a mathematical fact and merely demonstrating it to be the case. We have proofs that do both things, and perhaps a sliding scale of explanatoriness between them. One big question then is what makes a proof actually explain the thing it proves? Or at least what makes it seem that way to us?\n\nOne suggestion has been the level of generality or abstractness. Perhaps if we show a particular fact follows from some much bigger theory, the fact feels more explained. But then consider this fact:\n\n1+2+3+…+n = n(n+1)/2\n\nThere is an inductive proof of this:\n\nS(1) = 1(1+1)/2 = 1\n\nS(n+1) = S(n) + (n + 1) = n(n+1)/2 + 2(n+1)/2 = (n + 1)(n+2)/2\n\nThis is not taken to be very explanatory. Whereas this is:\n\nO O O O O\nO O O O O\nO O O O O\nO O O O\n\n[the black circles make a triangle of 1+2+3+4. Any such triangle can be made into a rectangle of area n x (n+1) with another identical triangle. So the triangle is half of n(n+1).]\n\nIt seems the latter is if anything less general, yet it seems a much better explanation (I remember learning it this way as a preteen in book about fun math magic). There are other examples.\n\nThis case and others, suggest being able to visualize a proof is key to its seeming to be an explanation. Steiner discards this immediately as being too subjective, and claims there are also counterexamples.\n\nHe also quickly dismisses a third hypothesis that others have forwarded: that a proof is explanatory if it could have been used to discover the fact, rather than just to verify it. His counterexample is the Eulerian identity, which I shan’t go into here. I take it this hypothesis isn’t very plausible anyway, since often we discover a fact first then hope to explain it better.\n\nSteiner offers his own theory: that a proof is explanatory if it makes use of a ‘characterizing property’ of an entity that is mentioned in the theorem. ‘Characterizing properties’ characterize an entity relative to other entities in some similar family. For instance, 18 might be characterized as 2*3*3, since other numbers don’t have that property. 18 might also be characterized as being one more than 17, or in a huge number of other ways.\n\nIf I understand, the idea is that if we are clear on how a result depends on a particular characterizing property, we will feel that the result has been explained. If we don’t see how something unique about the entities in question ‘caused’ the outcome, the outcome seems arbitrary. He explains further that this means we can see that if we change the properties of the entity, perhaps swapping out 18 for 20, we would get a different result.\n\nSteiner explains how the many proofs he has presented that we have considered explanatory do in fact depend on characterizing properties, thus considers his theory to be quite supported.\n\nPerhaps I misunderstand this notion of ‘characterizing properties’. It seems to me that of course all proofs depend on properties specific to the entities they are about (relative to whatever entities the proof is not about). So to distinguish the explanatory proofs, Steiner needs a narrower notion of a characterizing property. For instance, a property that is particularly saliently related to the entity in question. Or he needs to claim that explanatoriness requires the observer to actually notice or understand the connection between the explanatory property and the outcome. In which case the explanatoriness of a proof would be a function of the observer’s psychology as well as the proof. Any proof would be perfectly explanatory if the reader followed it carefully enough.\n\nAt any rate, he doesn’t seem to be thinking of either of those things (though again I may be misunderstanding just what he is claiming at the end here). He rather claims that the various proofs he examines do in fact rely on properties that characterize the entities involved. The class seemed to agree with me here.\n\nMy tentative theory of when we feel something has been explained, which goes for scientific explanations as well as mathematical ones, is as follows. We feel like we understand a bunch of things that we are very familiar with: chunks of matter moving through space and knocking into each other, liquids, shapes, basic agenthood, that sort of thing.\n\nAnything that happens that only involves these things acting in their usual ways doesn’t feel like it needs any extra explanation. It is obvious. To ‘explain’ less familiar things, we can do one of two things. We can frame them in terms of something we already intuitively grasp in the above way. This is what is usually called an explanation. For instance we can think of electricity as being like water, or of the first n integers as being like bits of a triangle. Or of the mysterious murder being like a waitress putting poison in the soup. Alternatively we can just keep interacting with the entity in question until we become familiar with it’s properties, and then we think them obvious and not requiring explanation. For instance I no longer feel like I need an explanation for x^2 making a parabola shape, because I’m so familiar with it.\n\nThis arguably fits with many of the characteristics we have noted are associated with explanatoriness. Instances of generalizations that we understand feel explanatory. Pictures tend to be explanatory, especially diagrams with simple shapes. We feel like we could have discovered a thing ourselves if it follows from behavior of entities we can manipulate intuitively.\n\nWhile this seems to me a decent characterization of what feels explanatory, I can’t see that it is a particularly useful category outside of psychology, for instance for use in saying what it is that science is meant to be doing. Something like unification seems more apt there, but that’s a topic for another time.\n\n### 13 responses to “Explanations of mathematical explanation”\n\n1.", null, "r\n\nI think the explanations that feel the most explanatory are the ones we encounter closest in time to our coming into an understanding of that topic, or ones we come up with just after.\n\n2.", null, "Paul Gowder\n\nWait a minute, doesn’t the second step of the inductive proof just assume the theorem to be proven? Or is my rusty algebra missing something? How do we get S(n) + (n + 1) = n(n+1)/2 + 2(n+1)/2 without assuming the theorem?\n\n3.", null, "Jordan Rastrick\n\nPaul, inductive proofs always assume the proposition holds for the inductive step. Induction looks like:\n\n* Prove S(K) where K is some starting number (typically, 1)\n* Prove that S(N) implies S(N+1)\nThis is the step that assumes the proposition holds, but the reasoning is not circular: since its only trying to demonstrate that if it holds for N, then it holds for N+1. Assuming P is generally taken for granted when trying to show P implies Q.\n\nAfter this, then axiomatically the proposition holds for all integers greater than or equal to K.\n\n•", null, "Paul Gowder\n\nRight, but the step I’m looking at also looks to me like it assumes S(n+1)=(n+1)(n+2)/2. That is, it assumes the truth of the proposition for the general case, not just for the case of S(n). But I’m probably misreading it.\n\n•", null, "Doug S.\n\nS(n+1) = S(n) + (n+1) [definition of S(n)]\nS(n) = n(n+1)/2 [inductive assumption]\nS(n+1) = n(n+1)/2 + (n+1) [substitution]\nS(n+1) = (n(n+1)+ 2(n+1))/2 [common denominator]\nS(n+1) = (n+1)(n+2)/2 [factoring]\n\n4.", null, "tjic (@tjic)\n\nI agree with your theory; it matches something I’ve long thought: that (a) any mathematical proof is effectively a tautology [ x = something that is equal to x = something else that is equal to X = a third thing that is equal to x, therefore x = x ] (b) “explanation” is a concept that only makes sense given a model of human [ as opposed to a general thinking engine ].\n\nI first came across the model that the human brain is not a general purpose computer, but in fact has specialized “coprocessors” that are optimized for things like biology, social relations, counting the number of bears that enter a cave and then leave it, etc. via Steven Pinker almost 20 years ago (Language Instinct ch 13).\n\nThat insight (although obvious in retrospect it was new to me at the time – perhaps a statement about the prevailing orthodoxy re the plasticity of the human mind and human culture of the day) made me realize that the word “explanation” really means “a tautology that chains back to some concept that we’re hardwired to find “obvious”.\n\n5.", null, "Rich and Co.\n\nFew things:\n– Prediction (data) seems the main value and goal. The reset seem local word play.\n– There is no such thing as “science” or the scientific methodology independent of specific studies and sets of data. “Science” is a rhetorical straw man.\n– Animal studies should slove these matters. Human knowledge wouldn’t seem any different from other mammals. Language and consiousness appear trivial and of little information value.\n\n6.", null, "Henrico Otto\n\nProofs and Refutations by Imre Lakatos has a nice treatment of these issues.\n\n7.", null, "Greg Muller\n\nI have noticed a subjective difference between types of proofs, which may be interpreted as a measure of how well the result is `explained’. The central question is, after having read and understood the proof, do counterexamples seem *impossible* or merely *prohibited*?\n\nSome results have proofs which are so natural and inevitable that, after having seen the proof, it becomes impossible to imagine the existence of a counterexample. By this I mean, the brain regards the idea of a counterexample as an absurdity, and cannot entertain such a thing even in the abstract. These results are rarely cited when used; rather, they are incorporated into the mathematician’s understanding of the object at hand. Simple proofs and visual proofs often have this quality Examples (for me, individual results may vary):\n1) If xy=0, then either x=0 or y=0.\n2) A triangle is determined (up to congruence) by the lengths of its sides.\n3) Any finite group of prime order is commutative.\n4) The proof of the Pythagorean Theorem in which a square of size c is placed inside a square of size a+b in two different ways.\n\nOtherwise, a result seems to only prohibit the existence of a counterexample; that is, one could imagine them existing in the abstract, but the result assures us that this is not the case. Proofs by contradiction, classification, multiple cases or the axiom of choice often feel this way. Examples (again, for me):\n1) the infinitude of primes\n2) (Cantor’s diagonal theorem) the impossibility of a bijection between Z (the integers) and R (the reals)\n3) there are 2 groups of order 6 (up to isomorphism)\n4) Euclid’s original proof of the Pythagorean Theorem\n\n8.", null, "Stephen R. Diamond\n\nThis question seems to belong to psychology. For one thing, who’s to say people find the same things explanatory? Shouldn’t we try to learn to find explanatory what is explanatory, rather than try to produce explanations that feel right?\n\n9.", null, "Sigivald\n\nexplanatoriness\nExplanationality?\n\n10.", null, "silver price\n\nThe argument from introspection is a much more interesting argument, since it tries to appeal to the direct experience of everyman. But the argument is deeply suspect, in that it assumes that our faculty of inner observation or introspection reveals things as they really are in their innermost nature. This assumption is suspect because we already know that our other forms of observation — sight, hearing, touch, and so on — do no such thing. The red surface of an apple does not look like a matrix of molecules reflecting photons at certain critical wave-lengths, but that is what it is. The sound of a flute does not sound like a sinusoidal compression wave train in the atmosphere, but that is what it is. The warmth of the summer air does not feel like the mean kinetic energy of millions of tiny molecules, but that is what it is. If one’s pains and hopes and beliefs do not introspectively seem like electrochemical states in a neural network, that may be only because our faculty of introspection, like our other senses, is not sufficiently penetrating to reveal such hidden details. Which is just what one would expect anyway. The argument from introspection is therefore entirely without force, unless we can somehow argue that the faculty of introspection is quite different from all other forms of observation.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "Applied Mathematics and Mechanics\n\n, Volume 38, Issue 1, pp 39–56\n\nQuantum enigma hidden in continuum mechanics\n\nArticle\n\nAbstract\n\nIt is reported that there exist deformable media which display quantum effects just as quantum entities do. As such, each quantum entity usually treated as a point particle may be represented by a deformable medium, the dynamic behavior of which is prescribed by four dynamic state variables, including mass density, velocity, internal pressure, and intrinsic angular momentum. In conjunction with the finding of the characteristic equation characterizing the physical nature of such media, it is found that a complex field quantity may be introduced to uncover a perhaps unexpected correlation, i.e., the governing dynamic equations for such media may be exactly reduced to the Schrödinger equation, from which the closed-form solutions for all the four dynamic state variables can be obtained. It turns out that this complex field quantity is just the wavefunction in the Schrödinger equation. Moreover, the dynamic effects peculiar to spin are derivable as direct consequences. It appears that these results provide a missing link in quantum theory, in the sense of disclosing the physical origin and nature of both the wavefunction and the wave equation. Now, the inherent indeterminacy in quantum theory may be rendered irrelevant. The consequences are explained for certain long-standing fundamental issues.\n\nKeywords\n\nquantum entity deformable medium nonlinear dynamic equation new interpretation Schrödinger equation\n\nO33 O413\n\n2010 Mathematics Subject Classification\n\n74A99 81P05 81S99\n\nReferences\n\n1. \nMehra, J. and Rechnberg, H. The Historical Development of Quantum Theory I–VI, Springer, Berlin (2000)Google Scholar\n2. \nHealey, R. The Philosophy of Quantum Mechanics, Cambridge University Press, Cambridge (1989)\n3. \nOmnès, R. The Interpretation of Quantum Mechanics, Princeton University Press, Princeton (1994)\n4. \nWeinberg, S. Dreams of a Final Theory, Vaintage, London (1994)Google Scholar\n5. \nBohm, D. A suggested interpretation of the quantum theory in terms of hidden variables, I and II. Physical Review, 85, 166–193 (1952)\n6. \nBohm, D. and Hiley, B. J. The Undivided Universe: An Ontological Interpretation of Quantum Mechanics, Routledge & Kegan Paul, London (1993)Google Scholar\n7. \nValentini, A. Pilot-Wave Theory of Physics and Cosmology, Cambridge University Press, Cambridge (2007)Google Scholar\n8. \nEverett, H. III: relative state formulation of quantum mechanics. Reviews of Modern Physics, 29, 459–462 (1957)Google Scholar\n9. \nDeWitt, B. S. and Graham, R. N. The many-worlds interpretation of quantum mechanics. Prince-ton Series in Physics, Princeton University Press, Princeton (1973)Google Scholar\n10. \nDeutsch, D. The Fabric of Reality, Allen Lane, New York (1997)Google Scholar\n11. \nBallentine, L. B. Quantum Mechanics: A Modern Development, World Scientific, Singapore (1998)\n12. \nZeh, H. D, On the interpretation of measurement in quantum theory. On the interpretation of measurement in quantum theory 1, 69–76 (1970)Google Scholar\n13. \nZurek, W. H, Environment-induced superselection rules. Environment-induced superselection rules 26, 1862–1880 (1982)\n14. \nHartle, J. B, Spacetime coarse-grainings in non-reletivistic quantum mechanics. Spacetime coarse-grainings in non-reletivistic quantum mechanics 44, 3173–3196 (1991)Google Scholar\n15. \nGell-Mann, M. The Quarks and the Jaguar, Freeman, New York (1994)\n16. \nGell-Mann, M. and Hartle, J. B, Classical equations for quantum systems. Classical equations for quantum systems 47, 3345–3382 (1993)\n17. \nGriffiths, R. B. Consistent Quantum Theory, Cambridge University Press, Cambridge (2003)Google Scholar\n18. \nSchlosshauer, M, Decoherence, the measurement problem, and interpretations of quantum mechanics. Decoherence, the measurement problem, and interpretations of quantum mechanics 76, 1267–1305 (2005)Google Scholar\n19. \nGhirardi, G. C., Rimini, A., and Weber, T, Unified dynamics for subatomic and macroscopic systems. Unified dynamics for subatomic and macroscopic systems 34, 470–489 (1986)\n20. \nBassi, A. and Ghirardi, G. C, Dynamic reduction models. Dynamic reduction models 379, 257–426 (2003)\n21. \nCrane, L. Clock and catogary: is quantum gravity algebraic? Journal of Mathematical Physics, 36, 6180–6193 (1995)\n22. \nPenrose, R, Quantum computation, entanglement and state reduction. Quantum computation, entanglement and state reduction 356, 1927–1939 (1998)\n23. \nPenrose, R. The Emporor’s New Mind, Oxford University Press, Oxford (1999)Google Scholar\n24. \nT’Hooft, G. A mathematical theory for deterministic quantum mechanics. Journal of Physics: Conference Series 67, 012015 (2007)Google Scholar\n25. \nFeynman, R. P. The Character of Physical Law, MIT Press, Cambridge (1967)Google Scholar\n26. \nSmolin, L. The Trouble with Physics, Spin Networks, Ltd., New York (2006)\n27. \nDirac, P. A. M. The Principles of Quantum Mechanics, Oxford University Press, Oxford (1958)\n28. \nGriffiths, D. J. Introduction to Quantum Mechanics, Pearson Prentice Hall, New Jersey (2005)Google Scholar" ]

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[ "# PyAutoDiff: automatic differentiation for NumPy\n\nWe are excited to have a guest post discussing a new tool that is freely available for the Python community. Welcome, Jeremiah Lowin, the Chief Scientist of the Lowin Data Company, to the growing pool of Data Community DC bloggers. We are very excited to announce an early release of PyAutoDiff, a library that allows automatic differentiation in NumPy, among other useful features. A quickstart guide is available here.\n\nAutodiff can compute gradients (or derivatives) with a simple decorator:\n\nfrom autodiff import gradient\n\ndef f(x):\nreturn x ** 2\n\ndef g(x):\nreturn x ** 2\n\nprint f(10.0) # 100.0\nprint g(10.0) # 20.0\n\nMore broadly, autodiff leverages Theano's powerful symbolic engine to compile NumPy functions, allowing features like mathematical optimization, GPU acceleration, and of course automatic differentiation. Autodiff is compatible with any NumPy operation that has a Theano equivalent and fully supports multidimensional arrays. It also gracefully handles many Python constructs (though users should be very careful with control flow tools like if/else and loops!).\n\nIn addition to the  @gradient decorator, users can apply  @function to compile functions without altering their return values. Compiled functions can automatically take advantage of Theano's optimizations and available GPUs, though users should note that GPU computations are only supported for float32 dtypes. Other decorators, classes, and high-level functions are available; see the docs for more information.\n\nIt is also possible for autodiff to trace NumPy objects through multiple functions. It can then compile symbolic representations of all of the traced operations (or their gradients) -- even with respect to objects that were purely local to the function(s) scope.\n\nimport numpy as np\nfrom autodiff import Symbolic, tag\n\n# -- a vanilla function\ndef f1(x):\nreturn x + 2\n\n# -- a function referencing a global variable\ny = np.random.random(10)\ndef f2(x):\nreturn x * y\n\n# -- a function with a local variable\ndef f3(x):\nz = tag(np.ones(10), 'local_var')\nreturn (x + z) ** 2\n\n# -- create a general symbolic tracer\nx = np.random.random(10)\ntracer = Symbolic()\n\n# -- trace the three functions\nout1 = tracer.trace(f1, x)\nout2 = tracer.trace(f2, out1)\nout3 = tracer.trace(f3, out2)\n\n# -- compile a function representing f(x, y, z) = out3\nnew_fn = tracer.compile_function(inputs=[x, y, 'local_var'],\noutputs=out3)\n\nassert np.allclose(new_fn(x, y, np.ones(10)), f3(f2(f1(x))))\n\nOne of the original motivations for autodiff was working with SVMs that were defined purely in NumPy. The following example (also available at autodiff/examples/svm.py) fits an SVM to random data, using autodiff to compute parameter gradients for SciPy's L-BFGS-B solver:\n\nimport numpy as np\nfrom autodiff.optimize import fmin_l_bfgs_b\n\nrng = np.random.RandomState(1)\n\n# -- create some fake data\nx = rng.rand(10, 5)\ny = 2 * (rng.rand(10) > 0.5) - 1\nl2_regularization = 1e-4\n\n# -- define the loss function\ndef loss_fn(weights, bias):\nmargin = y * (np.dot(x, weights) + bias)\nloss = np.maximum(0, 1 - margin) ** 2\nl2_cost = 0.5 * l2_regularization * np.dot(weights, weights)\nloss = np.mean(loss) + l2_cost\nreturn loss\n\n# -- call optimizer\nw_0, b_0 = np.zeros(5), np.zeros(())\nw, b = fmin_l_bfgs_b(loss_fn, init_args=(w_0, b_0))\n\nfinal_loss = loss_fn(w, b)\n\nassert np.allclose(final_loss, 0.7229)\n\nSome members of the scientific community will recall that James Bergstra began the PyAutoDiff project a year ago in an attempt to unify NumPy's imperative style with Theano's functional syntax. James successfully demonstrated the project's utility, and this version builds out and on top of that foundation. Standing on the shoulders of giants, indeed!\n\nPlease note that autodiff remains under active development and features may change. The library has been performing well in internal testing, but we're sure that users will find new and interesting ways to break it. Please file any bugs you may find!" ]

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[ "KIITEE Syllabus 2021: Kalinga Institute of Industrial Technology (KIIT) release the syllabus of KIITEE every year. Hence, the candidates can access the syllabus @kiit.ac.in. KIITEE Information Brochure contains the detailed KIITEE Syllabus. Every candidate who is going to appear for the KIITEE entrance exam must be familiar with the syllabus. With the help of the KIITEE Exam Syllabus, candidates will be able to prepare well for the entrance examination. In addition to this, it will also help the candidates to develop a preparation strategy.\n\nSubscribed Successfully.\n\nAlong with the syllabus of KIITEE, candidates must be familiar with the KIITEE Exam Pattern too. It helps them to understand the question paper pattern in detail. For more details regarding KIITEE Entrance Exam Syllabus, keep on reading the article.\n\n## KIITEE Subject-Wise B.Tech Syllabus 2020\n\nBelow is the break-down of the subject-wise syllabus of KIITEE. Candidates can go through the topics and make a preparation plan.\n\n### KIITEE Mathematics syllabus\n\n Chapters Topics Sets, Relations and Functions Sets and their Representations, Union, intersection, and complements of sets, and their algebraic properties, Relations, equivalence relations, mappings, one-one, into and onto mappings, the composition of mappings Complex Numbers Complex numbers in the form a+ib and their representation in a plane. Argand diagram. Algebra of complex numbers, Modulus and Argument (or amplitude) of a complex number, square root of a complex number. Cube roots of unity, triangle inequality. Quadratic Equations Quadratic equations in real and complex number system and their solutions. Relation between roots and co-efficients, nature of roots, formation of quadratic equations with given roots; Symmetric functions of roots, equations reducible to quadratic equations-application to practical problems. Permutations and Combinations Fundamental principle of counting; Permutation as an arrangement and combination as selection, Meaning of P (n,r) and C (n,r). Simple applications. Binomial Theorem and Its Applications Binomial Theorem for a positive integral index; general term and middle term; Binomial Theorem for any index. Properties of Binomial Co-efficients. Simple applications for approximations. Sequences and Series Arithmetic, Geometric and Harmonic progressions. Insertion of Arithmetic Geometric and Harmonic means between two given numbers. Relation Between A.M., G.M. and H.M. Special series: Sn,Sn2 ,Sn3 . ArithmeticoGeometric Series, Exponential and Logarithmic series. Differential Calculus Polynomials, rational, trigonometric, logarithmic and exponential functions, Inverse functions. Graphs of simple functions. Limits, Continuity; differentiation of the sum, difference, product and quotient of two functions: differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions; derivatives of order upto two. Applications of derivatives: Rate of change of quantities, monotonic-increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals, Rolle’s and Lagrange’s Mean Value Theorems. Integral Calculus Integral as an anti-derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and partial fractions. Integration using trigonometric identities. Integral as limit of a sum. Properties of definite integrals. Evaluation of definite integrals; Determining areas of the regions bounded by simple curves. Differential Equations Ordinary differential equations, their order and degree. Formation of differential equations. Solution of differential equations by the method of separation of variables. Solution of homogeneous and linear differential equations, and those of the type d2 y/ dX2= f(x) Two Dimensional Geometry Cartesian system of rectangular coordinates in a planeThe straight line and pair of straight linesCircles and Family of CirclesConic Sections Three Dimensional Geometry Coordinates of a point in space Vector Algebra Vectors and Scalars, Application of vectors to plane geometry Measures of Central Tendency and Dispersion Calculation of Mean, median and mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data. Probability Probability of an event, addition and multiplication theorems of probability and their application; Conditional probability; Bayes’ Theorem, probability distribution of a random variate; Binomial and Poisson distributions and their properties. Trigonometry Trigonometrical identities and equations. Inverse trigonometric functions and their properties. Properties of triangles, including centroid, incentre, circum-centre and orthocenter, solution of triangles. Heights and Distances." ]

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[ "# Equations With Like Terms Worksheet\n\nEquations With Like Terms Worksheet – The aim of Expressions and Equations Worksheets is to help your child learn more effectively and efficiently. They include interactive activities as well as problems dependent on the order in which operations are performed. These worksheets make it easy for children to master complex concepts as well as simple concepts in a short time. These PDFs are free to download and can be used by your child in order to learn math concepts. These resources are useful for students who are in the 5th through 8th grades.", null, "These worksheets can be utilized by students from the 5th-8th grades. The two-step word problems were created with fractions or decimals. Each worksheet contains ten problems. You can access them through any website or print source. These worksheets can be used to exercise rearranging equations. In addition to practicing rearranging equations, they also aid your student in understanding the properties of equality and inverse operations.\n\nThese worksheets are targeted at fifth and eight grade students. They are ideal for students who have difficulty calculating percentages. It is possible to select three types of problems. You can select to solve single-step issues that involve decimal or whole numbers or use word-based methods for solving decimals or fractions. Each page will contain ten equations. These worksheets for Equations can be used by students from the 5th to 8th grades.", null, "These worksheets are a great way for practicing fraction calculations and other concepts in algebra. You can choose from many kinds of challenges with these worksheets. It is possible to select the one that is word-based, numerical or a mix of both. It is important to choose the problem type, because every problem is different. Each page has ten questions that make them a fantastic aid for students who are in 5th-8th grade.\n\nThese worksheets help students understand the relationship between numbers and variables. The worksheets give students the opportunity to practice solving polynomial problems as well as solving equations and understanding how to apply them in everyday situations. These worksheets are a great way to get to know more about equations and formulas. They will assist you in learning about different types of mathematical problems as well as the different types of symbols that are used to communicate them.", null, "These worksheets are great for students who are in the 1st grade. These worksheets teach students how to graph equations and solve them. These worksheets are excellent for practicing with polynomial variable. They also assist you to learn how to factor and simplify these variables. There are many worksheets to teach kids about equations. The best method to learn about equations is by doing the work yourself.\n\nThere are numerous worksheets that teach quadratic equations. Each level has its own worksheet. The worksheets are designed for you to help you solve problems in the fourth degree. After you’ve completed the required level, you can continue working on other kinds of equations. You can then work on the same level of problems. For example, you can identify a problem that has the same axis in the form of an elongated number." ]

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[ "# 6: How to scale up\n\nSo far we’ve figured out how to make a dilution, which we can then plate and count. But this means we are only counting a fraction of what was originally there. How do we account for that? By scaling up.\n\n“Scaling up” means starting from the number or concentration in a sample, and figuring out how many were in the original brew of coffee, or original sample of a chemical or a bacterial culture.\n\nFor example, we can start with a cup of weak coffee and figure out how much caffeine was in the original brew. All we need to know is what the overall dilution factor was. In the case of weak coffee, it was 1/50.\n\nSo, when we count the caffeine molecules in a cup of weak coffee, we know that we’ve got 1/50th of what was in a cup of the original brew — or in other words, there was 50 times as much in a cup of the original. So we multiply the number we counted or measured by the inverse of the dilution – in this case, 50. This inverse of the dilution is called the dilution factor.\n\n• If the dilution is in the form of a fraction,\nyou can “flip” the fraction\n(i.e., 1/50 becomes multiply by 50/1).\n• If the dilution is written in scientific notation (eg 10-5)\nyou can make the exponent positive\n(i.e., 10-5 becomes multiply by 105).\n\nFinally, notice that I’m telling you the total number of caffeine molecules in ONE CUP of the original brew. If I don’t know the actual amount of original brew, I won’t know the total amount of caffeine that there was  in the whole brew.\n\n##", null, "Back to the lab\n\nHere are some problems, ranging from easy to a bit hard…" ]

[ null, "http://mathbench.org.au/wp-content/uploads/2015/microbiology_viable-plate-count/lab.gif", null ]

[ "# To Calculate Compound Interest in Java\n\nThis program finds the compound interest for the given inputs. This source code is written in Java.\n\n## To Calculate Compound Interest\n\nThis source code takes the inputs of principle, rate of interest and time. By entering the basic formulas we get the requires amount and compound interest by calculation.\n\nImplemenation of the source code\n\n`import java.util.Scanner;public class Main{ public static void main(String args[]){double A=0,Pri,Rate,Time,t=1,CI;Scanner S=new Scanner (System.in);System.out.print(\"Enter Principal : \" );Pri=S.nextFloat();System.out.print(\"Enter Rate : \" );Rate=S.nextFloat();System.out.print(\"Enter Time : \" );Time=S.nextFloat();Rate(1 + Rate/100);for(int i=0;i<Time;i++)t*=Rate;A=Pri*t;System.out.print(\"Amount : \" +A);CI=A-Pri;System.out.print(\"\\nCompound Interest : \" + CI);`\n\nOUTPUT:", null, "Submitted by Kajjayam Priya (Priya1414)" ]

[ null, "https://coderspacket.com/uploads/user_files/2021-07/Compound_interest_output-1626417442-823.jpeg", null ]

[ "# 1999 AHSME Problems/Problem 13\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\n## Problem\n\nDefine a sequence of real numbers", null, "$a_1$,", null, "$a_2$,", null, "$a_3$,", null, "$\\dots$ by", null, "$a_1 = 1$ and", null, "$a_{n + 1}^3 = 99a_n^3$ for all", null, "$n \\geq 1$. Then", null, "$a_{100}$ equals", null, "$\\textbf{(A)}\\ 33^{33} \\qquad \\textbf{(B)}\\ 33^{99} \\qquad \\textbf{(C)}\\ 99^{33} \\qquad \\textbf{(D)}\\ 99^{99} \\qquad \\textbf{(E)}\\ \\text{none of these}$\n\n## Solution\n\nWe rearrange to get", null, "$\\dfrac{a_{n+1}}{a_n} = \\sqrt{99}$. Thus we get", null, "$\\dfrac{a_{n+1}}{a_n} = \\sqrt{99}$,", null, "$\\dfrac{a_{n}}{a_{n-1}} = \\sqrt{99}$, and so on. Multiplying them all gives", null, "$\\dfrac{a_{n+1}}{a_1} = (\\sqrt{99})^{n}$. Plugging in", null, "$n = 99$ and", null, "$a_1 = 1$,", null, "$a_{100} = (\\sqrt{99})^{99} = 99^{33}$, so the answer is", null, "$\\textbf{(C)}$.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]

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[ "# 6.5: s Orbitals are Spherically Symmetric\n\nThe hydrogen atom wavefunctions, $$\\psi (r, \\theta , \\varphi )$$, are called atomic orbitals. An atomic orbital is a function that describes one electron in an atom. The wavefunction with $$n = 1$$, $$l$$ = 0 is called the 1s orbital, and an electron that is described by this function is said to be “in” the ls orbital, i.e. have a 1s orbital state. The constraints on n, $$l$$, and $$m_l$$ that are imposed during the solution of the hydrogen atom Schrödinger equation explain why there is a single 1s orbital, why there are three 2p orbitals, five 3d orbitals, etc. We will see when we consider multi-electron atoms, these constraints explain the features of the Periodic Table. In other words, the Periodic Table is a manifestation of the Schrödinger model and the physical constraints imposed to obtain the solutions to the Schrödinger equation for the hydrogen atom.\n\nVisualizing the variation of an electronic wavefunction with r, $$\\theta$$, and $$\\varphi$$ is important because the absolute square of the wavefunction depicts the charge distribution (electron probability density) in an atom or molecule. The charge distribution is central to chemistry because it is related to chemical reactivity. For example, an electron deficient part of one molecule is attracted to an electron rich region of another molecule, and such interactions play a major role in chemical interactions ranging from substitution and addition reactions to protein folding and the interaction of substrates with enzymes.\n\nWe can obtain an energy and one or more wavefunctions for every value of $$n$$, the principal quantum number, by solving Schrödinger's equation for the hydrogen atom. A knowledge of the wavefunctions, or probability amplitudes $$\\psi_n$$, allows us to calculate the probability distributions for the electron in any given quantum level. When n = 1, the wavefunction and the derived probability function are independent of direction and depend only on the distance r between the electron and the nucleus. In Figure $$\\PageIndex{1}$$, we plot both $$\\psi_1$$ and $$P_1$$ versus $$r$$, showing the variation in these functions as the electron is moved further and further from the nucleus in any one direction. (These and all succeeding graphs are plotted in terms of the atomic unit of length, $$a_0 = 0.529 \\times 10^{-8}\\, cm$$.)", null, "Figure $$\\PageIndex{1}$$: The wavefunction and probability distribution as functions of $$r$$ for the $$n = 1$$ level of the H atom. The functions and the radius r are in atomic units in this and succeeding figures.\n\nTwo interpretations can again be given to the $$P_1$$ curve. An experiment designed to detect the position of the electron with an uncertainty much less than the diameter of the atom itself (using light of short wavelength) will, if repeated a large number of times, result in Figure $$\\PageIndex{1}$$ for $$P_1$$. That is, the electron will be detected close to the nucleus most frequently and the probability of observing it at some distance from the nucleus will decrease rapidly with increasing $$r$$. The atom will be ionized in making each of these observations because the energy of the photons with a wavelength much less than 10-8 cm will be greater than $$K$$, the amount of energy required to ionize the hydrogen atom. If light with a wavelength comparable to the diameter of the atom is employed in the experiment, then the electron will not be excited but our knowledge of its position will be correspondingly less precise. In these experiments, in which the electron's energy is not changed, the electron will appear to be \"smeared out\" and we may interpret $$P_1$$ as giving the fraction of the total electronic charge to be found in every small volume element of space. (Recall that the addition of the value of Pn for every small volume element over all space adds up to unity, i.e., one electron and one electronic charge.)\n\nVisualizing wavefunctions and charge distributions is challenging because it requires examining the behavior of a function of three variables in three-dimensional space. This visualization is made easier by considering the radial and angular parts separately, but plotting the radial and angular parts separately does not reveal the shape of an orbital very well. The shape can be revealed better in a probability density plot. To make such a three-dimensional plot, divide space up into small volume elements, calculate $$\\psi ^* \\psi$$ at the center of each volume element, and then shade, stipple or color that volume element in proportion to the magnitude of $$\\psi ^* \\psi$$.\n\nWe could also represent the distribution of negative charge in the hydrogen atom in the manner used previously for the electron confined to move on a plane (Figure $$\\PageIndex{2}$$), by displaying the charge density in a plane by means of a contour map. Imagine a plane through the atom including the nucleus. The density is calculated at every point in this plane. All points having the same value for the electron density in this plane are joined by a contour line (Figure $$\\PageIndex{2}$$). Since the electron density depends only on r, the distance from the nucleus, and not on the direction in space, the contours will be circular. A contour map is useful as it indicates the \"shape\" of the density distribution.", null, "Figure $$\\PageIndex{2}$$: (a) A contour map of the electron density distribution in a plane containing the nucleus for the $$n = 1$$ level of the H atom. The distance between adjacent contours is 1 au. The numbers on the left-hand side on each contour give the electron density in au. The numbers on the right-hand side give the fraction of the total electronic charge which lies within a sphere of that radius. Thus 99% of the single electronic charge of the H atom lies within a sphere of radius 4 au (or diameter = $$4.2 \\times 10^{-8}\\; cm$$). (b) This is a profile of the contour map along a line through the nucleus. It is, of course, the same as that given previously in Figure $$\\PageIndex{1}$$ for $$P_1$$, but now plotted from the nucleus in both directions.\n\nWhen the electron is in a definite energy level we shall refer to the $$P_n$$ distributions as electron density distributions, since they describe the manner in which the total electronic charge is distributed in space. The electron density is expressed in terms of the number of electronic charges per unit volume of space, e-/V. The volume V is usually expressed in atomic units of length cubed, and one atomic unit of electron density is then e-/a03. To give an idea of the order of magnitude of an atomic density unit, 1 au of charge density e-/a03 = 6.7 electronic charges per cubic Ångstrom. That is, a cube with a length of $$0.52917 \\times 10^{-8}\\; cm$$, if uniformly filled with an electronic charge density of 1 au, would contain 6.7 electronic charges.\n\nFor every value of the energy En, for the hydrogen atom, there is a degeneracy equal to $$n^2$$. Therefore, for n = 1, there is but one atomic orbital and one electron density distribution. However, for n = 2, there are four different atomic orbitals and four different electron density distributions, all of which possess the same value for the energy, E2. Thus for all values of the principal quantum number n there are n2 different ways in which the electronic charge may be distributed in three-dimensional space and still possess the same value for the energy. For every value of the principal quantum number, one of the possible atomic orbitals is independent of direction and gives a spherical electron density distribution which can be represented by circular contours as has been exemplified above for the case of n = 1. The other atomic orbitals for a given value of n exhibit a directional dependence and predict density distributions which are not spherical but are concentrated in planes or along certain axes. The angular dependence of the atomic orbitals for the hydrogen atom and the shapes of the contours of the corresponding electron density distributions are intimately connected with the angular momentum possessed by the electron.\n\nMethods for separately examining the radial portions of atomic orbitals provide useful information about the distribution of charge density within the orbitals. Graphs of the radial functions, $$R(r)$$, for the 1s and 2s orbitals plotted in Figure $$\\PageIndex{3}$$. The 1s function in Figure $$\\PageIndex{3; left}$$ starts with a high positive value at the nucleus and exponentially decays to essentially zero after 5 Bohr radii. The high value at the nucleus may be surprising, but as we shall see later, the probability of finding an electron at the nucleus is vanishingly small.", null, "Figure $$\\PageIndex{3}$$: Radial function, $$R(r)$$, for the 1s and 2s orbitals. For an interactive graph click here.\n\nNext notice how the radial function for the 2s orbital, Figure $$\\PageIndex{3; right}$$, goes to zero and becomes negative. This behavior reveals the presence of a radial node in the function. A radial node occurs when the radial function equals zero other than at $$r = 0$$ or $$r = ∞$$. Nodes and limiting behaviors of atomic orbital functions are both useful in identifying which orbital is being described by which wavefunction. For example, all of the s functions have non-zero wavefunction values at $$r = 0$$.\n\nExercise $$\\PageIndex{1}$$\n\nExamine the mathematical forms of the radial wavefunctions. What feature in the functions causes some of them to go to zero at the origin while the s functions do not go to zero at the origin?\n\nExercise $$\\PageIndex{2}$$\n\nWhat mathematical feature of each of the radial functions controls the number of radial nodes?\n\nExercise $$\\PageIndex{3}$$: Radial Nodes\n\nAt what value of $$r$$ does the 2s radial node occur?\n\nExercise $$\\PageIndex{4}$$\n\nMake a table that provides the energy, number of radial nodes, and the number of angular nodes and total number of nodes for each function with $$n = 1$$, $$n=2$$, and $$n=3$$. Identify the relationship between the energy and the number of nodes. Identify the relationship between the number of radial nodes and the number of angular nodes.\n\nRadial probability densities for the 1s and 2s atomic orbitals are plotted in Figure $$\\PageIndex{4}$$.", null, "Figure $$\\PageIndex{4}$$: Radial densities ($$R (r) ^* R(r)$$) for the 1s and 2s orbitals.\n\nRather than considering the amount of electronic charge in one particular small element of space, we may determine the total amount of charge lying within a thin spherical shell of space. Since the distribution is independent of direction, consider adding up all the charge density which lies within a volume of space bounded by an inner sphere of radius $$r$$ and an outer concentric sphere with a radius only infinitesimally greater, say $$r + \\Delta r$$. The area of the inner sphere is $$4\\pi r^2$$ and the thickness of the shell is $$\\Delta r$$. Thus the volume of the shell is $$4\\pi r^2 \\Delta r$$ and the product of this volume and the charge density P1(r), which is the charge or number of electrons per unit volume, is therefore the total amount of electronic charge lying between the spheres of radius $$r$$ and $$r + \\Delta r$$. The product $$4\\pi r^2P_n$$ is given a special name, the radial distribution function.\n\nVolume Element for a Shell in Spherical Coordinates\n\nThe reader may wonder why the volume of the shell is not taken as:\n\n$\\dfrac{4}{3} \\pi \\left[ (r + \\Delta r)^3 -r^3 \\right]$\n\nthe difference in volume between two concentric spheres. When this expression for the volume is expanded, we obtain\n\n$\\dfrac{4}{3} \\pi \\left(3r^2 \\Delta r + 3r \\Delta r^2 + \\Delta r^3\\right)$\n\nand for very small values of $$\\Delta r$$ the $$3r \\Delta r^2$$ and $$\\Delta r^3$$ terms are negligible in comparison with $$3r^2\\Delta r$$. Thus for small values of $$\\Delta r$$, the two expressions for the volume of the shell approach one another in value and when $$\\Delta r$$ represents an infinitesimal small increment in $$r$$ they are identical.", null, "The volume element of a box in spherical coordinates. (CC BY; OpenStax).\n\nThe radial distribution function is plotted in Figure $$\\PageIndex{5}$$ for the ground state of the hydrogen atom.", null, "Figure $$\\PageIndex{5}$$: The radial distribution function for an H atom. The value of this function at some value of r when multiplied by $$\\delta r$$ gives the number of electronic charges within the thin shell of space lying between spheres of radius $$r$$ and $$r + \\delta r$$.\n\nThe curve passes through zero at $$r = 0$$ since the surface area of a sphere of zero radius is zero. As the radius of the sphere is increased, the volume of space defined by $$4 \\pi r^2Dr$$ increases. However, as shown in Figure $$\\PageIndex{4}$$, the absolute value of the electron density at a given point decreases with $$r$$ and the resulting curve must pass through a maximum. This maximum occurs at $$r_{max} = a_0$$. Thus more of the electronic charge is present at a distance $$a_o$$, out from the nucleus than at any other value of $$r$$. Since the curve is unsymmetrical, the average value of $$r$$, denoted by $$\\bar{r}$$, is not equal to $$r_{max}$$. The average value of $$r$$ is indicated on the figure by a dashed line. A \"picture\" of the electron density distribution for the electron in the $$n = 1$$ level of the hydrogen atom would be a spherical ball of charge, dense around the nucleus and becoming increasingly diffuse as the value of $$r$$ is increased.\n\nThe radial distribution function gives the probability density for an electron to be found anywhere on the surface of a sphere located a distance $$r$$ from the proton. Since the area of a spherical surface is $$4 \\pi r^2$$, the radial distribution function is given by $$4 \\pi r^2 R(r) ^* R(r)$$.\n\nRadial distribution functions are shown in Figure $$\\PageIndex{6}$$. At small values of $$r$$, the radial distribution function is low because the small surface area for small radii modulates the high value of the radial probability density function near the nucleus. As we increase $$r$$, the surface area associated with a given value of $$r$$ increases, and the $$r^2$$ term causes the radial distribution function to increase even though the radial probability density is beginning to decrease. At large values of $$r$$, the exponential decay of the radial function outweighs the increase caused by the $$r^2$$ term and the radial distribution function decreases.", null, "Figure $$\\PageIndex{6}$$: The radial distribution function ($$4 \\pi r^2 R (r) ^* R(r)$$) for the 1s and 2s orbitals. Compare to the radial functions in Figure $$\\PageIndex{3}$$ or the radial densities in Figure $$\\PageIndex{4}$$. For an interactive graph click here.\n\nExample $$\\PageIndex{1}$$:\n\nCalculate the probability of finding a 1s hydrogen electron being found within distance $$2a_o$$ from the nucleus.\n\nSolution\n\nNote the wavefunction of hydrogen 1s orbital which is\n\n$ψ_{100}= \\dfrac{1}{\\sqrt{π}} \\left(\\dfrac{1}{a_0}\\right)^{3/2} e^{-\\rho} \\nonumber$\n\nwith $$\\rho=\\dfrac{r}{a_0}$$.\n\nThe probability of finding the electron within $$2a_0$$ distance from the nucleus will be:\n\n$prob= \\underbrace{\\int_{0}^{\\pi} \\sin \\theta \\, d\\theta}_{over\\, \\theta} \\, \\overbrace{ \\dfrac{1}{\\pi a_0^3} \\int_{0}^{2a_0} r^2 e^{-2r/a_0} dr}^{over\\, r} \\, \\underbrace{ \\int_{0}^{2\\pi} d\\phi }_{over\\, \\phi } \\nonumber$\n\nSince $$\\int_0^{\\pi} \\sin \\theta d\\theta=2$$ and $$\\int_0^{2\\pi} d\\phi=2\\pi$$, we have\n\n\\begin{align*} prob &=2 \\times 2\\pi \\times \\dfrac{1}{\\pi a_0^3} \\int_0^2a_0 (-a_0/2)r^2 d e^{-2r/a_0} \\\\[4pt]&=\\dfrac{4}{a_0^3}\\left(-\\dfrac{a_0}{2}\\right) (r^2 e^{-2r/a_0} |_0^{2a_0} - \\int_0^{2a_0} 2r e^{-2r/a_0} dr) \\\\[4pt]&= -\\dfrac{2}{a_0^2} [(2a_0)^2 e^{-4}-0-2\\int_0^{2a_0} r \\left(-\\dfrac{a_0}{2}\\right) d e^{-2r/a_0} ] \\\\[4pt]&=-\\dfrac{2}{a_0^2}4a_0^2 e^{-4} +\\dfrac{4}{a_0^2}(-\\dfrac{a_0}{2}) (r e^{-2r/a_0} |_0^{2a_0}-\\int_0^{2a_0} e^{-2r/a_0} dr ) \\\\[4pt]&=-8e^{-4}-\\dfrac{2}{a_0} \\left[2a_0e^{-4}-0-(-\\dfrac{a_0}{2})e^{-2r/a_0} |_0^{2a_0} \\right] \\\\[4pt]&=-8e^{-4}-4e^{-4}-e^{2r/a_0} |_0^{2a_0} \\\\[4pt]&=-12 e^{-4}-(e^{-4}-1)=1-13e^{-4}=0.762 \\end{align*}\n\nThere is a 76.2% probability that the electrons will be within $$2a_o$$ of the nucleus in the 1s eigenstate.\n\n## Summary\n\nThis completes the description of the most stable state of the hydrogen atom, the state for which $$n = 1$$. Before proceeding with a discussion of the excited states of the hydrogen atom we must introduce a new term. When the energy of the electron is increased to another of the allowed values, corresponding to a new value for $$n$$, $$y_n$$ and $$P_n$$ change as well. The wavefunctions $$y_n$$ for the hydrogen atom are given a special name, atomic orbitals, because they play such an important role in all of our future discussions of the electronic structure of atoms. In general the word orbital is the name given to a wavefunction which determines the motion of a single electron. If the one-electron wavefunction is for an atomic system, it is called an atomic orbital.\n\nDo not confuse the word orbital with the classical word and notion of an orbit. First, an orbit implies the knowledge of a definite trajectory or path for a particle through space which in itself is not possible for an electron. Secondly, an orbital, like the wavefunction, has no physical reality but is a mathematical function which when squared gives the physically measurable electron density distribution." ]

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[ "« Home » « Learn » « Download » « Github »", null, "## Cello High Level C\n\n### Examples\n\nUsage\n\n``````var prices = new(Table, String, Int);\nset(prices, \\$S(\"Apple\"), \\$I(12));\nset(prices, \\$S(\"Banana\"), \\$I( 6));\nset(prices, \\$S(\"Pear\"), \\$I(55));\n\nforeach (key in prices) {\nvar price = get(prices, key);\nprintln(\"Price of %\\$ is %\\$\", key, price);\n}\n``````\n\nManipulation\n\n``````var t = new(Table, String, Int);\nset(t, \\$S(\"Hello\"), \\$I(2));\nset(t, \\$S(\"There\"), \\$I(5));\n\nshow(\\$I(len(t))); /* 2 */\nshow(\\$I(mem(t, \\$S(\"Hello\")))); /* 1 */\n\nrem(t, \\$S(\"Hello\"));\n\nshow(\\$I(len(t))); /* 1 */\nshow(\\$I(mem(t, \\$S(\"Hello\")))); /* 0 */\nshow(\\$I(mem(t, \\$S(\"There\")))); /* 1 */\n\nresize(t, 0);\n\nshow(\\$I(len(t))); /* 0 */\nshow(\\$I(mem(t, \\$S(\"Hello\")))); /* 0 */\nshow(\\$I(mem(t, \\$S(\"There\")))); /* 0 */\n``````\n\n# Table\n\nHash table\n\nThe `Table` type is a hash table data structure that maps keys to values. It uses an open-addressing robin-hood hashing scheme which requires `Hash` and `Cmp` to be defined on the key type. Keys and values are copied into the collection using the `Assign` class and intially have zero'd memory.\n\nHash tables provide `O(1)` lookup, insertion and removal can but require long pauses when the table must be rehashed and all entries processed.\n\nThis is largely equivalent to the C++ construct std::unordered_map\n\n### Derives\n\n• Alloc`\\$` `alloc` `dealloc`\n• Cast`cast`\n• Copy`copy`\n• Size`size`\n• Swap`swap`\n\n### Implements\n\n• Assign`assign`\n• Cmp`cmp` `eq` `neq` `gt` `lt` `ge` `le`\n• Doc`name` `brief` `description` `definition`\n• Get`get` `set` `mem` `rem` `key_type` `val_type`\n• Hash`hash` `hash_data`\n• Iter`foreach` `iter_init` `iter_next` `iter_type`\n• Len`len`\n• Mark`mark`\n• New`new` `del` `construct` `destruct`\n• Resize`resize`\n• Show`show` `look` `print` `scan`" ]

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[ "# 2.1 Factorisation and Algebraic Fractions\n\n2.1 Factorisation and Algebraic Fractions\n\n2.1.1 Expansion\n1. The product of an algebraic term and an algebraic expression:\n• a(b + c) = ab + ac\n•  a(bc) = ab ac\n2. The product of an algebraic expression and another algebraic expression:\n• (a + b) (c + d)  = ac + ad + bc + bd\n• (a + b)2= a2 + 2ab + b2\n• (ab)2= a2 – 2ab + b2\n• (a + b) (ab) = a2b2\n\n2.1.2 Factorisation\n1. Factorize algebraic expressions:\n•  ab + ac = a(b + c)\n• a2b2 = (a + b) (ab)\n• a2+ 2ab + b2 = (a + b)2\n• ac + ad + bc + bd = (a + b) (c + d)\n2. Algebraic fractions are fractions where both the numerator and the denominator or either the numerator or the denominator are algebraic terms or algebraic expressions.\nExample:\n$\\frac{3}{b},\\frac{a}{7},\\frac{a+b}{a},\\frac{b}{a-b},\\frac{a-b}{c+d}$\n\n3(a) Simplification of algebraic fractions by using common factors:\n$\\begin{array}{l}•\\text{}\\frac{{}^{1}\\overline{)4}\\overline{)b}c}{{}^{{}_{{}^{3}}}\\overline{)12}\\overline{)b}d}=\\frac{c}{3d}\\\\ •\\text{}\\frac{bm+bn}{em+en}=\\frac{b\\overline{)\\left(m+n\\right)}}{e\\overline{)\\left(m+n\\right)}}\\\\ \\text{}=\\frac{b}{e}\\end{array}$\n\n3(b) Simplification of algebraic fractions by using difference of two squares:\n$\\begin{array}{l}\\frac{{a}^{2}-{b}^{2}}{an+bn}=\\frac{\\overline{)\\left(a+b\\right)}\\left(a-b\\right)}{n\\overline{)\\left(a+b\\right)}}\\\\ \\text{}=\\frac{a-b}{n}\\end{array}$\n\n2.1.3 Addition and Subtraction of Algebraic Fractions\n1. If they have a common denominator:\n$\\frac{a}{m}+\\frac{b}{m}=\\frac{a+b}{m}$\n\n2.\nIf they do not have a common denominator:\n$\\frac{a}{m}+\\frac{b}{n}=\\frac{an+bm}{nm}$\n\n2.1.4 Multiplication and Division of Algebraic Fractions\n1. Without simplification:\n$\\begin{array}{l}•\\text{}\\frac{a}{m}×\\frac{b}{n}=\\frac{ab}{mn}\\\\ •\\text{}\\frac{a}{m}÷\\frac{b}{n}=\\frac{a}{m}×\\frac{n}{b}\\\\ \\text{}=\\frac{an}{bm}\\end{array}$\n\n2.\nWith simplification:\n$\\begin{array}{l}•\\text{}\\frac{a}{c\\overline{)m}}×\\frac{b\\overline{)m}}{d}=\\frac{ab}{cd}\\\\ •\\text{}\\frac{a}{cm}÷\\frac{b}{dm}=\\frac{a}{c\\overline{)m}}×\\frac{d\\overline{)m}}{b}\\\\ \\text{}=\\frac{ad}{bc}\\end{array}$" ]

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[ "#### A code-free guide to Artificial Intelligence\n\nIf you didn’t click the link, do it now.\n\nDid you? Ok, good.\n\nNow here’s the thing — The article requires you to know a little bit of python, which you probably do.\n\nOn the off-chance that you’re interested in Neural Networks (if that phrase sounds utterly foreign to you, watch this YouTube playlist) and haven’t learned python yet, congratulations, you’re in the right spot.\n\nBut regardless of where you are in the vast landscape of deep learning, I think that once in a while, it’s nice to go back to the basics and revisit the fundamental mathematical ideas that brought us Siri, Alexa, and endless hours of Netflix binge-watching.\n\nSo without any further ado, I present to you the three equations that make up what I’ll call the “fundamental theorem of deep learning.”\n\n### 1. Linear Regression\n\nThe first equation is pretty basic. I guess the others are as well, but we’ll get to them in due time.\n\nFor now, all we’re doing is computing a vector z (from the equation above), where W is a matrix that is initially just filled with a bunch of random numbers, b is a vector that is initially just filled with a bunch of random numbers, and x vector that is not initially filled with a bunch of random numbers.\n\nx is a training example from our dataset. For example, if you’re training a neural net to predict someone’s age given their gender and height, you’d first need a few (or preferably a lot, the more, the merrier) examples of data in the form `[[height, gender], age]`. The vector `[height, gender]` is what we’re calling x.\n\n### 2. Activation Functions\n\nOn the left-hand side, we have our predicted values of y, which is the variable that I’m using to denote the labels on our data.\n\nThe hat on top means that this value of y is a predicted value, as opposed to the ground truth labels from our dataset.\n\nThe z in this equation is the same one that we computed above. The sigma represents the sigmoid activation function, which looks like this:\n\nSo in plain English, we’re taking z, a vector of real numbers that can be arbitrarily large or small, and squishing its components to be between 0 and 1.\n\nHaving a number between 0 and 1 is useful because if we’re trying to build a classifier, let’s say that predicts if an image is a cat or a dog, we can let 1 represent dogs, and we can let 0 be for cats. Or the other way around if you like cats more.\n\nBut suppose we’re not doing dogs and cats (yeah right, like there’s any other better use case for machine learning). Let’s go back to our age predictor. Over there, we can’t merely predict 1’s and 0’s.\n\nIn general, you could use whatever function you like, not necessarily just a sigmoid. But a bunch of smart people noticed that the sigmoid worked pretty well. So we’re stuck with it.\n\nHowever, it’s a different story when we’re dealing with labels that are actual numbers, and not classes. For our age predictor, we need to use a different activation function.\n\nEnter ReLU.\n\nLet me say upfront that I think that this is the most boring part of deep learning. I mean, seriously, just a boring ol’ straightforward-looking function? Where’s the fun in that?\n\nLooks can be deceiving though. While it’s pretty dull — ReLU(x) is just `max(0,x)` — the ReLU function works really well in practice. So hey, live with it.\n\n### 3. Back-propagation And Gradient Descent\n\nOk, you got me. I cheated. It’s technically four lines of math. But hey, you could condense steps 1 and 2 into a single step, so I guess I come out victorious.\n\nNow to digest all of that (literal) Greek stuff.\n\nIn the first equation, we’re doing that fancy stuff to y and y-hat to compute a single number called the loss, denoted by L.\n\nAs can be inferred by the name, the loss measures how badly we’ve lost in our vicious battle to conquer the machine learning grimoire.\n\nIn particular, our L here is measuring something called the binary cross entropy loss, which is a shortcut to sounding like you have a math Ph.D. when you’re actually just measuring how far y is from y-hat. Nevertheless, there’s a lot more under the surface of the equation, so check out Daniel Godoy’s article on the topic.\n\nAll you need to know to get the intuition behind this stuff is that L gets big if our predicted values are far away from the ground truth values, and L gets tiny when our predictions and reality match.\n\nThe sum is there so that we can add up all the messed-up-ness for each of the training examples, so that our neural net understands how messed-up it is overall.\n\nNow, the actual “learning” part of deep learning begins.\n\nThe final step in our stack is to update the matrix W and the vector b so that our loss goes down. By doing this, we are effectively minimizing how far are predictions are from the ground truth values, and thus, our model is getting more accurate.\n\nHere’s the equation again:\n\nW’ is the matrix with updated numbers that gets us closer to the ground truth. Alpha is a constant that we get to choose. That last term you’re looking at is the gradient of the loss with respect to a parameter. Put simply, it’s a measure of much our loss changes for a small tweak in the numbers in the W matrix.\n\nAgain, I’m not going to go too in-depth into gradient descent (the process of updating our numbers in the matrix) since there are already a lot of great resources on the topic. I’d highly recommend this article by Sebastian Ruder.\n\nBy the way, we can do the same thing for the initially random values in the b vector. Just tweak them by the right amount in the right direction, and BOOM! We just got closer to an all time low loss.\n\n### Conclusion\n\nAnd there you have it. The three great equations that make up the foundations of the neural networks that we use today.\n\nPause and ponder for a second. What you just saw is a compilation of humanity’s understanding of the intricacies of intelligence.\n\nSure, this is a pretty basic vanilla neural net that we just looked at, and there have been countless improvements in learning algorithms over the years that have resulted in significant breakthroughs. When coupled with the unprecedented explosion of data and computing power over the years, it seems, to a degree, almost inevitable that well-thought out mathematics is able to grasp the subtle art of distinguishing cats and dogs.\n\nBut still. This is where it all began.\n\nIn a way, the heart and soul of this decade’s (arguably) most significant technological advancement lie right before your eyes. So take a second. Pause and ponder." ]

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[ "# Gauss's Law\n\nGauss's law is able to give the relationship between the electric field at every point on the surface and charge enclosed by that surface. Gauss's law can be used to find the electric field from a given charge distribution (total net charge) and total net charge from a given field if the electric field is uniform on a highly symmetric surface so that the integral $\\int {E\\cos \\theta {\\kern 1pt} dA}$ can be evaluated easily. According to Gauss's law the total electric flux $\\Phi$ through a closed surface is equal to the total charge (net charge) $q$ enclosed by that surface divided by $\\epsilon_0$. Here $q$ represents the magnitude of electric charge and inward and outward flux is determined by the sign of $q$.\n\n${\\rm{ }}\\Phi {\\rm{ }} = \\int {E\\cos \\theta {\\mkern 1mu} dA} = \\frac{q}{{{\\epsilon _0}}} \\tag{3} \\label{3}$\n\nWe know that the total electric flux produced by a net charge $q$ through a closed surface is given by the integration $\\int {E\\cos \\phi {\\mkern 1mu} dA}$. Here ${E\\cos \\theta {\\mkern 1mu} }$ is the perpendicular component of electric field through the plane of area $dA$. The integration $\\int {E\\cos \\phi {\\mkern 1mu} dA}$ can also be defined as $\\int {dA\\cos \\theta {\\mkern 1mu} E}$ where ${dA\\cos \\theta {\\mkern 1mu} }$ is the projection of $dA$ on a plane perpendicular to the direction of electric field.", null, "Figure 3 The electric flux is independent of the size of the enclosing surface.\n\nAs an example we apply Eq. \\eqref{3} on a sphere of radius $r$ which encloses a net charge $q$ shown in Figure 3. The sphere is symmetric and therefore the electric field is uniform throughout the sphere. Hence, the electric field lines passing through the surface of the sphere are perpendicular to the corresponding area elements $dA$. So, $\\cos \\theta = \\cos 0 = 1$ and,\n\n\\begin{align*} \\Phi {\\rm{ }} &= \\int {EdA} {\\rm{ }} = E\\int {dA} = EA\\\\ \\quad &= E(4\\pi {r^2}) = \\left( {\\frac{q}{{4\\pi \\epsilon_0{r^2}}}} \\right)(4\\pi {r^2}) = \\frac{q}{\\epsilon_0} \\end{align*}\n\nThis is the expression obtained for a symmetric surface where the electric field is uniform on the surface. You'll see later that Gauss's law is valid for any kind of closed surface where the electric field may not be the same at every point on the surface. Consider two concentric spheres enclosing the same charge $q$ and we now calculate the electric flux through both of the spheres produced by the same charge. Now we determine the electric flux through the sphere of radius $r_1$ which can be obtained as,\n\n\\begin{align*} {\\rm{ }}\\Phi {\\rm{ }} &= E\\int {dA} = EA = E(4\\pi {r_1}^2)\\\\ {\\rm{or,}}\\quad \\Phi &= \\left( {\\frac{q}{{4\\pi {\\epsilon_0}{r_1}^2}}} \\right)(4\\pi {r_1}^2) = \\frac{q}{{{\\epsilon_0}}} \\end{align*}\n\nAnd the electric flux through the concentric sphere of radius ${{r}_{2}}$ is,\n\n\\begin{align*} {\\rm{ }}\\Phi {\\rm{ }} &= E\\int {dA} = EA = E(4\\pi {r_2}^2)\\\\ {\\rm{or,}}\\quad \\Phi &= \\left( {\\frac{q}{{4\\pi {\\epsilon_0}{r_2}^2}}} \\right)(4\\pi {r_2}^2) = \\frac{q}{{{\\epsilon_0}}} \\end{align*}\n\nSo the electric flux is independent of the size of the surface enclosing the charge but only depends on the magnitude of charge enclosed by the surface. Suppose that $r_2=2r_1$. Now the electric field at the surface of the sphere of radius $r_2$ decreases by a factor of $\\frac{1}{4}$ which is $k\\frac{q}{{{r_2}^2}} = k\\frac{q}{{{{(2{r_1})}^2}}} = k\\frac{q}{{4{r_1}^2}}$. On the other hand the surface area of the sphere increases by a factor of 4 that is $4\\pi {r_2}^2 = 4\\pi {(2{r_1})^2} = 4(4\\pi {r_1}^2)$. So the total electric flux through both of the surfaces remain the same.\n\nThe Gauss's Law is still true even if the enclosing surface is not symmetric; it means the electric field may not be the same at every point on the surface. It is because we obtain Gauss Law from the integration in Eq. \\eqref{3} in which we determine the electric flux. As already noted the value ${dA\\cos \\theta {\\mkern 1mu} }$ is the projection of $dA$ which is always perpendicular to the electric field and for any kind of closed surface the projection is alwyas the spherical surface. So, Gauss's Law is still valid for irregular surface. So a general form of Gauss's Law can be given as,\n\n$\\Phi {\\rm{ }} = \\oint {E\\cos \\theta dA} = \\frac{q}{{{\\epsilon_0}}} \\tag{4} \\label{4}$\n\nThe sign of electric flux is determined by the sign of $q$. A circle in the integral sign in \\eqref{4} is a reminder that the integration is always taken over a closed surface. Gauss's Law requires a net charge should be enclosed by a surface and if there are multiple charges enclosed by the surface we determine the net charge and use the Gauss' law. We can determine the electric field from a given charge distribution and charge distribution from the electric field but the integral in \\eqref{3} or \\eqref{4} is difficult to evaluate for irregular surfaces. So for the highly symmetric closed surface the integral is much easier to evaluate and the result can be obtained easily.\n\nWe apply Gauss's law to find the electric field due to a given charge distribution and we can also find the charge distribution from a given electric field if the enclosing surface is symmetric so that the integral $\\Phi {\\rm{ }} = \\oint {E\\cos \\theta dA} = \\frac{q}{{{\\epsilon_0}}}$ can be evaluated easily. In this article we find the electric field due to various charge distributions using Gauss's law.\n\n## Charges on Conductors\n\nThe charges always lie on the outer surface of a conductor. When excess charge is added to a conductor the charge always lies at rest on the outer surface of the conductor. This can be explained in terms of the electrostatic situation of the charge. In electrostatic situation the charge remains at rest not in motion.\n\nSince the excess charge added to the conductor is at rest, there shouldn't be any electric field inside the conductor, and if there is any electric field inside the conductor the charge will move which disturbs the electrostatic situation. So, there should not be any electric field inside the conductor, and no electric field means no charge. Which concludes that if there is no charge inside the conductor, there is no electric field which disturbs the electrostatic situation and this is valid only if the charge lies on the outer surface of the conductor.", null, "Figure 1 The charge lies on the outer surface of a conductor.\n\nLet's check this with Gauss's Law. We make a Gaussian surface exactly like the one shown in Figure 1. For the electrostatic situation there shouldn't be any electric field inside the conductor, so there shouldn't be any electric field inside the Gaussian surface which means there does not exist any charge inside the Gaussian surface. Therefore, there is no charge inside the Gaussian surface and it means the charge should lie on the conductor's outer surface.\n\n## Gauss's Law: Electric Field of a Uniformly Charged Conducting Sphere\n\nConsider a uniformly charged conducting sphere (Figure 2) of radius $r$ and charge $q$. We calculate the electric field produced by the charge on the sphere at various points inside or outside the sphere. First we calculate the electric field due to the charge distribution on the sphere outside the sphere and therefore enclose the sphere of radius $r$ by a concentric Gaussian sphere of radius $R$.", null, "Figure 2 The Gaussian sphere inside the sphere does not enclose any charge and therefore the electric field inside the sphere is zero.\n\nIn each case we create a Gaussian surface, the point where we are calculating the electric field always lies on the Gaussian surface. Note that the sphere is symmetric and the electric field is radially outward at each point of the sphere. And we know from Gauss's law (applying Gauss's law),\n\n${\\rm{ }}\\Phi = \\oint {E\\cos \\theta dA} {\\rm{ }} = \\frac{q}{{{\\epsilon_0}}}{\\rm{ }}$\n\nSince the sphere is symmetric and the electric field is radially outward the electric field at every point on the Gaussian surface is uniform and perpendicular to the area element $dA$. And $\\cos \\theta =\\cos 0=1$. So,\n\n\\begin{align*} \\oint {EdA} {\\rm{ }} &= \\frac{q}{{{\\epsilon_0}}}\\\\ {\\rm{or, }}\\quad E\\oint {dA} &= \\frac{q}{{{\\epsilon_0}}}\\\\ {\\rm{or,}}\\quad {\\rm{ }}EA &= \\frac{q}{{{\\epsilon_0}}}\\\\ {\\rm{or,}}\\quad E(4\\pi {R^2}) &= \\frac{q}{{{\\epsilon_0}}}\\\\ \\therefore E &= \\frac{q}{{4\\pi {\\epsilon_0}{R^2}}} = k\\frac{q}{{{R^2}}} \\tag{5} \\label{5} \\end{align*}\n\nIn this way we can determine the electric field at any distance $R$ form the centre of the charged sphere. Our result shows that the electric field outside the sphere is the same as if all the charge were concentrated at the centre of the sphere, that is the electric field is the same as that of a point charge at the centre of the sphere.\n\nNow we determine the electric field inside the charged sphere and in this case we make a Gaussian sphere of radius $R'$ inside the sphere. In electrostatic situation i.e. charges not in motion the electric field inside the charged sphere is zero. The Gaussian surface which is inside the conductor ($R' < r$) encloses no charge and electric field inside the conductor is zero.\n\nAt the surface of the charged sphere, the Gaussian sphere has radius $r$, the same as the radius of the charged sphere. Therefore the electric field is:\n\n$E = \\frac{q}{{4\\pi {\\epsilon_0}{r^2}}}{\\rm{ = }}k\\frac{q}{{{r^2}}} \\tag{6} \\label{6}$\n\nThe electric field at the surface of the charged sphere is also the same as though all the charge were concentrated at the centre.\n\nWe recently determined the electric field of an uniformly charged conducting sphere. Now we determine the same thing for a uniformly charged insulating sphere where the charge is distributed uniformly throughout its volume.\n\n## Gauss's Law: Electric Field of a Uniformly Charged Insulating Sphere\n\nConsider an insulating sphere of radius $r$ with net charge $q$ distributed uniformly throughout its volume in Fig:11.18. Now we determine the electric field due to that charge distribution at various points inside, on the surface and outside the sphere.", null, "Figure 3 In case of an insulating uniformly charged sphere the Gaussian surface inside the sphere encloses some charge and therefore has electric field.\n\nFirst we attempt to find the electric field inside the sphere and therefore make a concentric Gaussian sphere of radius $R'$. Here $\\rho$ is the volume charge density which is the total charge divided by the volume of the sphere. So $\\rho$ is,\n\n$\\rho =\\frac{q}{\\tfrac{4}{3}\\pi {{r}^{3}}}=\\frac{3q}{4\\pi {{r}^{3}}} \\nonumber$\n\nNow the charge inside the Gaussian surface inside the sphere($r > R'$) is the volume of the Gaussian surface ${\\textstyle{4 \\over 3}}\\pi {{R'}^3}$ multiplied by the volume charge density. So the new charge enclosed by the Gaussian surface q' is,\n\n$q' = \\left( {\\frac{{3q}}{{4\\pi {r^3}}}} \\right)\\left( {\\frac{4}{3}\\pi {{R'}^3}} \\right) = q\\frac{{{{R'}^3}}}{{{r^3}}}{\\rm{ }}$\n\nAnd the electric flux through the Gaussian surface inside the sphere is:\n\n$\\Phi = \\oint {E\\cos \\theta dA} = \\frac{{q'}}{{{\\epsilon_0}}}$\n\nThe sphere is symmetric and charge is distributed uniformly throughout its volume so the electric field is radially outward and also uniform at every point on the Gaussian surface. And you know $\\cos \\theta =\\cos 0=1$,\n\n\\begin{align*} \\oint {EdA} {\\rm{ }} &= \\frac{{q'}}{{{\\epsilon_0}}}\\\\ {\\rm{or, }}\\quad E\\oint {dA} &= \\frac{{q'}}{{{\\epsilon_0}}}\\\\ {\\rm{or,}}\\quad {\\rm{ }}EA &= \\frac{{q'}}{{{\\epsilon_0}}}\\\\ {\\rm{or,}}\\quad E(4\\pi R{'^2}) &= \\frac{{q\\frac{{R{'^3}}}{{{r^3}}}}}{{{\\epsilon_0}}}\\\\ \\therefore E &= \\frac{{qR{'^3}}}{{4\\pi {\\epsilon_0}R{'^2}{r^3}}} = k\\frac{{qR'}}{{{r^3}}} \\tag{7} \\end{align*}\n\nThis is the expression for the electric field at every point on the Gaussian surface inside the sphere at a distance $R'$ form the centre. At surface of the sphere $R' = r$ and the electric field is,\n\n$E = k\\frac{q}{{{r^2}}} \\tag{8}$\n\nThis is the same expression as that of conducting sphere in the previous application of Gauss's law. And the electric field is the same as if all the charge were concentrated at the centre of the sphere i.e. the charge on the sphere behaves like a point charge at the centre of the sphere.\n\nTo determine the electric field outside the sphere we again make a Gaussian sphere of radius $R$($R>r$) outside the sphere. Note that the Gaussian sphere is concentric with the original sphere. In this case the charge enclosed by the Gaussian surface is $q$ and we can use Gauss's law to calculate the electric field at any point on the Gaussian surface outside the sphere. So, Gauss's law gives\n\n\\begin{align*} E(4\\pi {R^2}) &= \\frac{q}{{{\\epsilon_0}}}\\\\ \\therefore E &= \\frac{q}{{4\\pi {\\epsilon_0}{R^2}}} = k\\frac{q}{{{R^2}}} \\tag{9} \\end{align*}\n\nIf you noticed the electric fields inside, on the surface and outside the sphere, you'll find that the electric field increases as $R'$ increases from $R'$ to $r$ inside the sphere and decreases as the distance increases outside the sphere ($R > r$).\n\n## Gauss's Law: Electric Field of a Line of Charge\n\nIn this case we consider that the charge $q$ is distributed uniformly in a line and forms a line of charge. Here we find the electric field at a perpendicular distance from the line of charge. We apply symmetry considerations and use Gauss's law to find the electric field. When you make a Gaussian surface to solve problems using Gauss's law you can make any kind of Gaussian surface either regular or irregular but a trick is that we make the Gaussian surface symmetrical with the charge distribution so that we can easily evaluate the Gauss's law equation (see Figure 4). Here the line of charge has cylindrical symmetry, so we apply the same cylindrical symmetry in our Gaussian surface.", null, "Figure 4 The line of charge has cylindrical symmetry, so we use a cylinder as the Gaussian surface. The electric field is perpendicularly outward from the line of charge.\n\nWe first determine the electric flux through each ends of the cylinder and then through the curved surface. Finally we get the total electric flux by adding them together. The electric field of the line of charge is radially outward. It is because there is no such thing as the component of electric field parallel to the line of charge or tangent to the curved Gaussian surface or any other component and can not be concluded that the electric field is not radially outward. The charge is distributed uniformly throughout the line or wire, the electric field is uniform.\n\nSince the electric field is radially outward, it is parallel to the ends of the Gaussian cylinder and hence the electric flux through the ends of the cylinder is zero. The electric field due to the line of charge is perpendicular to the curved surface. The total charge enclosed by the Gaussian surface is the liner charge density (charge per unit length) $\\lambda$ multiplied by the length of the Gaussian cylinder $l$. If the Gaussian cylinder has radius $r$, the area of the curved surface of the Gaussian surface is $2\\pi rl$. Now the electric field can be determined by using Gauss's law,\n\n\\begin{align*} E(2\\pi rl) &= \\frac{{\\lambda l}}{{{\\epsilon_0}}}\\\\ {\\rm{or, }}\\quad E &= \\frac{\\lambda }{{2\\pi {\\epsilon_0}r}} = k\\frac{{2\\lambda }}{r} \\tag{10} \\end{align*}\n\nThis expression is the same as the expression of the electric field of an infinite length line of charge we obtained in the electric field calculation without using Gauss's law. This is because we have considered a small portion of the line of charge in our Gaussian cylinder where all the electric field lines are radially outward which satisfies the condition of being the perpendicular distance $r$ is small enough in comparison to the length of the line of charge (also satisfies the point where we are calculating the Electric field (Gaussian surface) is close enough to the line of charge).\n\n## Gauss's Law: Electric Field of an Infinite Plane Sheet of Charge\n\nConsider an infinite plane sheet of charge. The infinite plane sheet of charge is an idealization which works only if the point where the electric field to be calculated is close enough to the sheet compared to the sheet's dimensions and not too near the edges. The charge is distributed uniformly throughout the sheet and produces the electric field radially outward on either side of the sheet. If there is any component of the electric field parallel to the sheet, then we need to explain why the electric field has parallel component. So, the electric field is perpendicularly outward from the sheet. Now we make a Gaussian surface, a cylinder with its ends each having area $A$. The cylinder's ends are parallel to the sheet and the sheet is at the middle point of the axis of the cylinder. See Figure 5.", null, "Figure 5 The sheet produces electric field radially outward perpendicular to the ends of the Gaussian cylinder but parallel to the curved surface.\n\nLet the sheet has total charge $q$ and the surface charge density i.e. charge per unit area be $\\sigma$. Now we divide the Gaussian surface into different parts and find the electric flux through each of those parts and obtain the total flux by adding them. But the electric field is parallel to the curved surface and perpendicular to the ends of the cylinder. Therefore the electric flux through the curved Gaussian surface is zero. Let the cylinder has radius $r$ and the surface charge density of the sheet is $\\sigma$. The charge enclosed by the Gaussian surface is $\\sigma A$ (surface charge density multiplied by the sheet area enclosed by the Gaussian surface). Now the electric flux through the Gaussian surface is\n\n\\begin{align*} EA + EA + 0 &= \\frac{{\\sigma A}}{{{\\epsilon_0}}}\\\\ \\therefore E &= \\frac{\\sigma }{{2{\\epsilon_0}}} \\tag{11} \\end{align*}\n\nHere 0 the electric flux is the electric flux through the curved Gaussian surface and $EA$ through each ends of the Gaussian cylinder. The electric field is independent of the distance from the sheet so the electric field is uniform and perpendicular to the sheet.\n\n## Gauss's Law: Electric Field between Two Charged Parallel Plates\n\nConsider two oppositely charged conducting plates parallel to each other and we are going to find the electric field between those plates as shown in Figure 6. Note that both plates have the same surface charge density $\\sigma$, that is charge per unit area. There is electrostatic attraction between the charge on opposite faces of the plates and the electric field has a direction from positive face towards negative face, so the electric field is zero on the left side of the negatively charged plate and on the right side of the positively charged plate.\n\nNote that the charges are accumulated on the opposite faces of the plates , that is the charges are accumulated at one face of each plate. Now take a look at the Gaussian surface $G_1$ which is a cylinder where the electric field is parallel to the curved surface and perpendicular to its left end. There is no electric field to the right end of $G_1$ because the electric field due to positively charged plate $P_2$ is equal and opposite to the electric field due to negatively charged plate $P_1$. Note that for an infinite plane sheet of charge the electric field is independent of the distance from the sheet as we obtained in the previous application of Gauss's law. Therefore, the electric flux through the right end and the curved surface of $G_1$ is zero. If $A$ is the area of one of the ends, the total electric flux through the Gaussian surface $G_1$ which encloses the charge $\\sigma A$ is\n\\begin{align*} 0 + 0 + EA = \\frac{{\\sigma A}}{{{\\epsilon_0}}}\\\\ \\therefore E = \\frac{\\sigma }{{{\\epsilon_0}}} \\tag{12} \\end{align*}\nYou can also obtain the same result by making a Gaussian surface ${{G}_{2}}$ on the negatively charged plate. You can also make a Gaussian surface $G_3$ as shown in Figure 6 to make sure that the electric field is zero on the right side of the positively charged plate. The Gaussian surface $G_3$ does not enclose any charge because the charges are accumulated on the opposite faces of the plates due to electrostatic interaction, so the electric field on the right side of the positively charged plate is zero. The electric field is also zero on the left side of the negatively charged plate due to the same reason as that with the positively charged plate." ]

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[ "# Applying Newtons Second Law to Vertical Motion\n\n18. An elevator that weighs 3.5 x 103 N is accelerated upward at 1.0 m/s2. What force does the cable exert to give it this acceleration?\nA. 357 N\nB. 0 N\nC. 3500 N\nD. 3857 N\n(For this quesion I was really confused and I guessed that the force of tension in the cable was what the cable weighed )\nIs that answer correct if not how do I find the correct answer?\n\ncepheid\nStaff Emeritus\nGold Member\n(For this quesion I was really confused and I guessed that the force of tension in the cable was what the cable weighed )\n\nThis would only be true if the forces on the elevator car were balanced. However, since the elevator car has a non-zero acceleration, Newton's Second Law tells you that the forces on the elevator car are NOT balanced i.e. there is a net force acting. Hint: you can use Newton's Second Law to calculate the net force from the mass and acceleration, and then use the result to figure out how the tension compares to the weight.\n\n\"What force does the cable exert to give it this acceleration?\" is this asking for net force if it is could I use the equation Fnet=Ft+Fg and If I used this Would it look like this?:\nFnet=Ft+Fg Fnet=(3500N/9.81m/s2)(1.00m/s2)+3500N so would the answer be D. 3857 N\n\ncepheid\nStaff Emeritus" ]

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[ "# CMX Lunch Seminar\n\nWednesday January 29, 2020 12:00 PM\n\nStatistical Guarantees for MAP Estimators in PDE-Constrained Regression Problems\n\nSpeaker: Sven Wang, Department of Pure Mathematics and Mathematical Statistics, University of Cambridge\nLocation: Annenberg 213\n\nThe main topic of the talk are convergence rates for penalised least squares (PLS) estimators in non-linear statistical inverse problems, which can also be interpreted as Maximum a Posteriori (MAP) estimators for certain Gaussian Priors. Under general conditions on the forward map, we prove convergence rates for PLS estimators.\n\nIn our main example, the parameter f is an unknown heat conductivity function in a steady state heat equation [a second order elliptic PDE]. The observations consist of a noisy version of the solution u[f] to the boundary value corresponding to f. The PDE-constrained regression problem is shown to be solved a minimax-optimal way.\n\nThis is joint work with S. van de Geer and R. Nickl. If time permits, we will mention some related work on the non-parametric Bayesian approach, as well as computational questions for the Bayesian posterior.\n\nSeries CMX Lunch Series\n\nContact: Jolene Brink at 6263952813 [email protected]" ]

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[ "# Yahoo Web Search\n\nrelated to ACP\n\n1. Sort by\n\n1. ### Trigonometry. Is this an ambiguous triangle?\n\n.... AP = 9*sin(40°) = 5.79 cm Now apply Pythagoras to ACP. 6^2 = 5.79^2 + CP^2 CP = 1.59 cm (this would have to be 4 to make...\n\n1 Answers · Science & Mathematics · 16/09/2017\n\n2. ### Height and area of circular segment?\n\nThe formula to find the area of the segment is given by: ½ R² (π/180 * θ − sinθ) Where: θ = the central angle in degrees R = the radius of the circle General formula for circle center to chord midpoint distance d: d = √[(R²...\n\n1 Answers · Science & Mathematics · 05/07/2017\n\n3. ### Trngle ABC is an isosceles with AB=AC such that P is a point inside it and angle BCP=30,APB=150,CAP=39 (degrees), find angle BAP in degrees.?\n\nLet ∠ACP=θ then by simple geometry ∠APC=141−θ , ∠ABP=2θ−51 and ∠...\n\n2 Answers · Science & Mathematics · 28/11/2016\n\n4. ### Math - homework help?\n\n... x, that is CP = 10/3 feet vi) Again the triangle ACP is a right triangle with same <C = 90 deg as before. ...\n\n2 Answers · Science & Mathematics · 10/01/2016\n\n5. ### Can someone explain this question to me?\n\n... look at the sector whose central angle is < ACP= 180-(30+60) = 120* for this sector area Sector Area A= (θ/360...\n\n2 Answers · Science & Mathematics · 20/04/2014\n\n6. ### How to solve Systems of Linear Equation with variables?\n\nIt's not really six variables. You have three variables x, y, z and three unknown constants a, b, c. Note: I assume the last equation should be 2ax - by + cz = 14. The main thing we have to be careful about is not dividing by 0. So...\n\n1 Answers · Science & Mathematics · 10/04/2014\n\n7. ### Proving geometric theorems?\n\n...common tangent at P. As before by alternate segment theorem, <ACP = <APX; as well <BNP = <...\n\n1 Answers · Science & Mathematics · 25/02/2014\n\n8. ### Triangle inequality problem?\n\n...; angle ABC ...................(1) angle PCB = angle ACB -- angle ACP whence angle PCB < angle ACB ...................(2) from (1...\n\n2 Answers · Science & Mathematics · 04/09/2013\n\n9. ### how to find the angle of elevation?\n\n... AC = 12.81cm Now, try to imagine the right triangle ACP 12.81^2 + 6^2 = (Distance PC)^2 200 = (Distance PC...\n\n5 Answers · Science & Mathematics · 01/08/2013\n\n10. ### Find the area of an equilateral triangle which has an internal pt P such PA=p,PB=q,PC=r?\n\n... + q² - r²)/(2pq) The same law, applied to ∆ACP' yields |AC|² = p² + q² - 2pq cos(θ + 60...\n\n1 Answers · Science & Mathematics · 12/07/2013" ]

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[ "# Decares to square inches (daa to sq in)\n\n## Convert decares to square inches\n\nDecares to square inches converter above calculates how many square inches are in 'X' decares (where 'X' is the number of decares to convert to square inches). In order to convert a value from decares to square inches (from daa to sq in) simply type the number of daa to be converted to sq in and then click on the 'convert' button.\n\n## Decares to square inches conversion factor\n\n1 decare is equal to 1550003.1000062 square inches\n\n## Decares to square inches conversion formula\n\nArea(sq in) = Area (daa) × 1550003.1000062\n\nExample: Pressume there is a value of area equal to 170 decares. How to convert them in square inches?\n\nArea(sq in) = 170 ( daa ) × 1550003.1000062 ( sq in / daa )\n\nArea(sq in) = 263500527.00105 sq in or\n\n170 daa = 263500527.00105 sq in\n\n170 decares equals 263500527.00105 square inches\n\n## Decares to square inches conversion table\n\ndecares (daa)square inches (sq in)\n710850021.700043\n913950027.900056\n1117050034.100068\n1320150040.300081\n1523250046.500093\n1726350052.700105\n1929450058.900118\n2132550065.10013\ndecares (daa)square inches (sq in)\n150232500465.00093\n200310000620.00124\n250387500775.00155\n300465000930.00186\n350542501085.00217\n400620001240.00248\n450697501395.00279\n500775001550.0031\n\nVersions of the decares to square inches conversion table. To create a decares to square inches conversion table for different values, click on the \"Create a customized area conversion table\" button.\n\n## Related area conversions\n\nBack to decares to square inches conversion\n\nTableFormulaFactorConverterTop" ]

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[ "", null, "Download", null, "Download Presentation", null, "Chapter 6. Mechanical Behavior\n\n# Chapter 6. Mechanical Behavior\n\nDownload Presentation", null, "## Chapter 6. Mechanical Behavior\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Chapter 6. Mechanical Behavior • Stress versus Strain • Elastic Deformation • Plastic Deformation • Hardness • Creep and Stress Relaxation • Viscoelastic Deformation\n\n2. Stress versus Strain • Mechanical Properties • Deal directly with behavior of materials under applied forces. • Properties are described by applied stress and resulting strain, or applied strain and resulting stress. • Example: 100 lb force applies to end of a rod results in a stress applied to the end of the rod causing it to stretch or elongate, which is measured as strain. • Strength: ability of material to resist application of load without rupture. • Ultimate strength- maximum force per cross section area. • Yield strength- force at yield point per cross section area. • Other strengths include rupture strength, proportional strength, etc. • Stiffness: resistance of material to deform under load while in elastic state. • Stiffness is usually measured by the Modulus of Elasticity (Stress/strain) • Steel is stiff (tough to bend). Some beds are stiff, some are soft (compliant)\n\n3. Testing Procedures • Mechanical Testing • Properties that deal with elastic or inelastic behavior of a material under load • Primary measurements involved are load applied and effects of load application • Two classification of tests; method of loading and the condition of the specimen during the test • Primary types of tests • Tensile • Compression • Shear • Torsion • Flexure\n\n4. Mechanical Test Considerations shear • Principle factors are in three main areas • manner in which the load is applied • condition of material specimen at time of test • surrounding conditions (environment) during testing • Tests classification- load application • kind of stress induced. Single load or Multiple loads • rate at which stress is developed: static versus dynamic • number of cycles of load application: single versus fatigue • Primary types of loading compression tension torsion flexure\n\n5. Standardized Testing Conditions • Moisture • 100F, 100% R.H. • 1 Day, 7 Days, 14 Days • Temperature • Room Temperature: Most common • Elevated Temperature: Rocket engines • Low Temperature: Automotive impact • Salt spray for corrosion • Rocker Arms on cars subject to immersion in NaCl solution for 1 Day and 7 Days at Room Temperature and 140 F. • Acid or Caustic environments • Tensile tests on samples after immersion in acid/alkaline baths.\n\n6. Stress • Stress: Intensity of the internally distributed forces or component of forces that resist a change in the form of a body. • Tension, Compression, Shear, Torsion, Flexure • Stress calculated by force per unit area. Applied force divided by the cross sectional area of the specimen. • Stress units • Pascals = Pa = Newtons/m2 • Pounds per square inch = Psi Note: 1MPa = 1 x106 Pa = 145 psi • Example • Wire 12 in long is tied vertically. The wire has a diameter of 0.100 in and supports 100 lbs. What is the stress that is developed? • Stress = F/A = F/r2 = 100/(3.1415927 * 0.052 )= 12,739 psi = 87.86 MPa\n\n7. Stress 0.1 in 1 in 10in • Example • Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the stress that is developed? What is the Load? • Stress = F/A = F/(width*thickness)= 100lbs/(1in*.1in )= 1,000 psi = 1000 psi/145psi = 6.897 Mpa • Load = 100 lbs • Block is 10 cm x 1 cm x 5 cm is mounted on its side in a test machine. The block is pulled with 100 N on both sides. What is the stress that is developed? What is the Load? • Stress = F/A = F/(width*thickness)= 100N/(.01m * .10m )= 100,000 N/m2 = 100,000 Pa = 0.1 MPa= 0.1 MPa *145psi/MPa = 14.5 psi • Load = 100 N 100 lbs 1 cm 5cm 10cm\n\n8. Strain lF l0 • Strain: Physical change in the dimensions of a specimen that results from applying a load to the test specimen. • Strain calculated by the ratio of the change in length and the original length. (Deformation) • Strain units (Dimensionless) • When units are given they usually are in/in or mm/mm. (Change in dimension divided by original length) • % Elongation = strain x 100%\n\n9. Strain 0.1 in 1 in 10in • Example • Tensile Bar is 10in x 1in x 0.1in is mounted vertically in test machine. The bar supports 100 lbs. What is the strain that is developed if the bar grows to 10.2in? What is % Elongation? • Strain = (lf - l0)/l0 = (10.2 -10)/(10)= 0.02 in/in • Percent Elongation = 0.02 * 100 = 2% • Block is 10 cm x 1 cm x 5 cm is mounted on its side in a test machine. The block is pulled with 1000 kN on bone side. If the material elongation at yield is 1.5%, how far will it grow at yield? • Strain = Percent Elongation /100 = 1.5%/100 = 0.015 cm /cm • Strain = (lf - l0)/l0 = (lf -5)/(5)= 0.015 cm/cm • Growth = 5 * 0.015 = 0.075 cm • Final Length = 5.075 cm 100 lbs 1 cm 5cm 10cm\n\n10. Strain • Permanent set is a change in form of a specimen once the stress ends. • Axial strain is the strain that occurs in the same direction as the applied stress. • Lateral strain is the strain that occurs perpendicular to the direction of the applied stress. • Poisson’s ratio is ratio of lateral strain to axial strain. Poisson’s ratio = lateral strain axial strain • Example • Calculate the Poisson’s ratio of a material with lateral strain of 0.002 and an axial strain of 0.006 • Poisson’s ratio = 0.002/0.006 = 0.333 Lateral Strain Axial Strain • Note: For most materials, Poisson’s ratio is between 0.25 and 0.5 • Metals: 0.29 (304 SS) to 0.3 (1040 steel) to 0.35 (Mg) • Ceramics and Glasses: 0.19 (TiC) to 0.26 (BeO) to 0.31 (Cordierite) • Plastics: 0.35 (Acetals) to 0.41 (Nylons)\n\n11. Stress-Strain Diagrams • Equipment • Strainometers: measures dimensional changes that occur during testing • extensometers, deflectometers, and compressometers measure changes in linear dimensions. • load cells measure load • data is recorded at several readings and the results averaged, e.g., 10 samples per second during the test.\n\n12. Stress-Strain Diagrams Linear (Hookean) Stress Non-Linear (non-Hookean) Strain • Stress-strain diagrams is a plot of stress with the corresponding strain produced. • Stress is the y-axis • Strain is the x-axis\n\n13. Stiffness • Stiffness is a measure of the materials ability to resist deformation under load as measured in stress. • Stiffness is measures as the slope of the stress-strain curve • Hookean solid: (like a spring) linear slope • steel • aluminum • iron • copper • All solids (Hookean and viscoelastic) • metals • plastics • composites • ceramics\n\n14. Modulus • Modulus of Elasticity (E) or Young’s Modulus is the ratio of stress to corresponding strain (within specified limits). • A measure of stiffness • Stainless Steel E= 28.5 million psi (196.5 GPa) • Aluminum E= 10 million psi • Brass E= 16 million psi • Copper E= 16 million psi • Molybdenum E= 50 million psi • Nickel E= 30 million psi • Titanium E= 15.5 million psi • Tungsten E= 59 million psi • Carbon fiber E= 40 million psi • Glass E= 10.4 million psi • Composites E= 1 to 3 million psi • Plastics E= 0.2 to 0.7 million psi\n\n15. Modulus Types Stress Strain • Modulus: Slope of the stress-strain curve • Initial Modulus: slope of the curve drawn at the origin. • Tangent Modulus: slope of the curve drawn at the tangent of the curve at some point. • Secant Modulus: Ratio of stress to strain at any point on curve in a stress-strain diagram. It is the slope of a line from the origin to any point on a stress-strain curve. Initial Modulus Tangent Modulus Secant Modulus\n\n16. Compression Testing l0 lF • Principles • Compression results from forces that push toward each other. • Specimens are short and large diameter. • Circular cross section is recommended. • Length to diameter ratio is important consideration • Universal test machine (UTM) • Size and load of compression machine are specially built. • Load and compression amount are measured. • Stress • Force per unit area. Applied force divided by the cross sectional area of the specimen. • Strain calculated by the ratio of the change in length and the original length. (Deformation)\n\n17. Expected Results Stress Strain • Similar Stress-strain curve as tensile testing\n\n18. Shear Testing • Principles • Direct shear occurs when parallel forces are applied in the opposite direction. • Single shear occurs on a single plane. • Double shear occurs on two planes simultaneously.\n\n19. Shear Testing • Principles • Torsional shearing forces occur when the forces applied lie in parallel but opposite directions. Twisting motion. • Torsional forces developed in a material are the result of an applied torque. • Torque is Forces x distance.. • Universal test machine (UTM) • Special fixtures are needed to hold the specimen. • One end of the specimen is placed in a fixture that applies torsional load and the other end is connected to a tropometer, which measures the detrusion (load and deflection or twist)\n\n20. Expected Results Stress Strain • Similar Stress-strain curve as tensile testing\n\n21. Bend of Flexure Testing • Principles • Bending forces occur when load is applied to a beam or rod that involves compression forces on one side of a beam and tensile forces on the other side. • Deflection of a beam is the displacement of a point on a neutral surface of a beam from its original position under action of applied loads. • Flexure is the bending of a material specimen under load. • Strength that material exhibits is a function of the flexural modulus of the material and the cross-sectional geometry. • Example, rectangular beam of 1” x 4” (W) will exhibit higher flexural strength than a 2” by 2” square beam of the same material modulus. • Properties are the same as in tensile testing. • Strength, deflection, modulus, ultimate strength, etc. • Specimen is loaded in a 3-point bending test • bottom goes in tension and the top goes in compression. • Failure analysis can provide information as the type of failure, • either tension or compression failure, • buckle prior to failure, • condition of fracture, e.e., rough, jagged, or smooth.\n\n22. Equipment • Universal test machine (UTM) • Special fixtures are needed to hold the specimen. • Precautions • Specimen length should be 6 to 12 times the width to avoid shear failure or buckling. • Areas of contact with the material under test should be such that unduly high stress concentrations are avoided. • Longitudinal adjustments are necessary for the supports. • Lateral rotational adjustments should be provided to prevent torsional stresses. • The parts should be arranged to be stable under load.\n\n23. Expected Results Stress Strain • Similar Stress-strain curve as tensile testing\n\n24. Impact Testing • Principles • Materials exhibit different properties depending on the rate at which a load is applied and the resulting strain that occurs. • If a load is applied over a long period of time (static test)the material can withstand greater loads than if the test is applied rapidly (dynamic). • Properties of materials are stain dependent. • Standardized tests are used to determine the amount of energy required to break a material in impact tests. • Outcome of impact tests is to determine the amount of energy needed to break a sample.\n\n25. Impact Testing • Principles • Energy absorbed in several ways • Elastic deformation of the members or parts of a system. • Plastic deformation. • Hysteresis effects. • Frictional action • effects of inertia on moving parts. • Energy is defined as the ability to do work. E =W = F*D • Work is Force times distance moved. • Energy of a dropped object hitting a specimen is • E = w*h Energy is weight times height dropped. • E = m*g*h (metric) Energy is mass times gravity acceleration times height.\n\n26. Equipment • Impact Testing Equipment • Izod and Charpy are the most common tests. • Both employ a swinging pendulum and conducted on small notched specimens. The notch concentrated the load at a point causing failure. Other wise without the notch the specimen will plastically deform throughout. • They are different in the design of the test specimen and the velocity at which the pendulum strikes the specimen. • Charpy: the specimen is supported as a single beam and held horizontally. Impacted at the back face of the specimen. • Izod: the specimen is supported as a cantilever and help vertically. Impacted at front face of the specimen. • Figure 19-1\n\n27. Impact Test • In standard testing, such as tensile and flexural testing, the material absorbs energy slowly. • In real life, materials often absorb applied forces very quickly: falling objects, blows, collisions, drops, etc. • A product is more likely to fail when it is subjected to an impact blow, in comparison to the same force being applied more slowly. • The purpose of impact testing is to simulate these conditions.\n\n28. Impact Test • Impact testing is testing an object's ability to resist high-rate loading. • An impact test is a test for determining the energy absorbed in fracturing a test piece at high velocity. • Most of us think of it as one object striking another object at a relatively high speed. • Impact resistance is one of the most important properties for a part designer to consider, and without question the most difficult to quantify. • The impact resistance of a part is, in many applications, a critical measure of service life. More importantly these days, it involves the perplexing problem of product safety and liability. • One must determine: • 1.the impact energies the part can be expected to see in its lifetime, 2.the type of impact that will deliver that energy, and then 3.select a material that will resist such assaults over the projected life span. • Molded-in stresses, polymer orientation, weak spots (e.g. weld lines or gate areas), and part geometry will affect impact performance. • Impact properties also change when additives, e.g. coloring agents, are added to plastics.\n\n29. Impact Test • Most real world impacts are biaxial rather than unidirectional. • Plastics, being anisotropic, cooperate by divulging the easiest route to failure. • Complicated choice of failure modes: Ductile or brittle. • Brittle materials take little energy to start a crack, little more to propagate it to a shattering climax. • Highly ductile materials fail by puncture in drop weight testing and require a high energy load to initiate and propagate the crack. • Many materials are capable of either ductile or brittle failure, depending upon the type of test and rate and temperature conditions. • They possess a ductile/brittle transition that actually shifts according to these variables. • For example, some plastic food containers are fine when dropped onto the floor at room temperature but a frozen one can crack when dropped.\n\n30. Expected Results • Charpy Test • Capacity of 220 ft-lb for metals and 4 ft-lbs for plastics • Pendulum swings at 17.5 ft/sec. • Specimen dimensions are 10 x 10 x 55 mm, notched on one side. • Procedure • Pendulum is set to angle, , and swings through specimen and reaches the final angel, . If no energy given then  = . • Energy is\n\n31. Expected Results • Izod Test • Capacity of 120 ft-lb for metals and 4 ft-lbs for plastics • Impacted at the front face of the specimen. • Specimen dimensions are 10 x 10 x 75 mm, notched on one side. • Procedure • Pendulum is set to angle, , and swings through specimen and reaches the final angel, . If no energy given then  = . • Energy is\n\n32. Fundamentals of Hardness • Hardness is thought of as the resistance to penetration by an object or the solidity or firmness of an object • Resistance to permanent indentation under static or dynamic loads • Energy absorption under impact loads (rebound hardness) • Resistance toe scratching (scratch hardness) • Resistance to abrasion (abrasion hardness) • Resistance to cutting or drilling (machinability) • Principles of hardness (resistance to indentation) • indenter: ball or plain or truncated cone or pyramid made of hard steel or diamond • Load measured that yields a given depth • Indentation measured that comes from a specified load • Rebound height measured in rebound test after a dynamic load is dropped onto a surface\n\n33. Hardness Mechanical Tests • Brinell Test Method • One of the oldest tests • Static test that involves pressing a hardened steel ball (10mm) into a test specimen while under a load of • 3000 kg load for hard metals, • 1500 kg load for intermediate hardness metals • 500 kg load for soft materials • Various types of Brinell • Method of load application:oil pressure, gear-driven screw, or weights with a lever • Method of operation: hand or electric power • Method of measuring load: piston with weights, bourdon gage, dynamoeter, or weights with a lever • Size of machine: stationary (large) or portable (hand-held)\n\n34. Brinell Test Conditions • Brinell Test Method (continued) • Method • Specimen is placed on the anvil and raised to contact the ball • Load is applied by forcing the main piston down and presses the ball into the specimen • A Bourbon gage is used to indicate the applied load • When the desired load is applied, the balance weight on top of the machine is lifted to prevent an overload on the ball • The diameter of the ball indentation is measured with a micrometer microscope, which has a transparent engraved scale in the field of view\n\n35. Brinell Test Example • Brinell Test Method (continued) • Units: pressure per unit area • Brinell Hardness Number (BHN) = applied load divided by area of the surface indenter Where: BHN = Brinell Hardness Number L = applied load (kg) D = diameter of the ball (10 mm) d = diameter of indentation (in mm) • Example: What is the Brinell hardness for a specimen with an indentation of 5 mm is produced with a 3000 kg applied load. • Ans:\n\n36. Brinell Test Method (continued) • Range of Brinell Numbers • 90 to 360 values with higher number indicating higher hardness • The deeper the penetration the higher the number • Brinell numbers greater than 650 should not be trusted because the diameter of the indentation is too small to be measured accurately and the ball penetrator may flatten out. • Rules of thumb • 3000 kg load should be used for a BHN of 150 and above • 1500 kg load should be used for a BHN between 75 and 300 • 500 kg load should be used for a BHN less than 100 • The material’s thickness should not be less than 10 times the depth of the indentation\n\n37. Advantages & Disadvantages of the Brinell Hardness Test • Advantages • Well known throughout industry with well accepted results • Tests are run quickly (within 2 minutes) • Test inexpensive to run once the machine is purchased • Insensitive to imperfections (hard spot or crater) in the material • Limitations • Not well adapted for very hard materials, wherein the ball deforms excessively • Not well adapted for thin pieces • Not well adapted for case-hardened materials • Heavy and more expensive than other tests (\\$5,000)\n\n38. Rockwell Test • Hardness is a function of the degree of indentation of the test piece by action of an indenter under a given static load (similar to the Brinell test) • Rockwell test has a choice of 3 different loads and three different indenters • The loads are smaller and the indentation is shallower than the Brinell test • Rockwell test is applicable to testing materials beyond the scope of the Brinell test • Rockwell test is faster because it gives readings that do not require calculations and whose values can be compared to tables of results (ASTM E 18)\n\n39. Rockwell Test Description • Specially designed machine that applies load through a system of weights and levers • Indenter can be 1/16 in hardened steel ball, 1/8 in steel ball, or 120° diamond cone with a somewhat rounded point (brale) • Hardness number is an arbitrary value that is inversely related to the depth of indentation • Scale used is a function of load applied and the indenter • Rockwell B- 1/16in ball with a 100 kg load • Rockwell C- Brale is used with the 150 kg load • Operation • Minor load is applied (10 kg) to set the indenter in material • Dial is set and the major load applied (60 to 100 kg) • Hardness reading is measured • Rockwell hardness includes the value and the scale letter\n\n40. Rockwell Values • B Scale: Materials of medium hardness (0 to 100HRB) Most Common • C Scale: Materials of harder materials (> 100HRB) Most Common • Rockwell scales divided into 100 divisions with each division (point of hardness) equal to 0.002mm in indentation. Thus difference between a HRB51 and HRB54 is 3 x 0.002 mm - 0.006 mm indentation • The higher the number the harder the number\n\n41. Rockwell and Brinell Conversion • For a Rockwell C values between -20 and 40, the Brinell hardness is calculated by • For HRC values greater than 40, use • For HRB values between 35 and 100 use\n\n42. Rockwell and Brinell Conversion • For a Rockwell C values, HRC, values greater than 40, • Example, • Convert the Rockwell hardness number HRc 60 to BHN\n\n43. Form of Polymers Melt Temp Tm Rubbery Increasing Temp Tg Glassy Polymer Form • Thermoplastic Material: A material that is solid, that possesses significant elasticity at room temperature and turns into a viscous liquid-like material at some higher temperature. The process is reversible • Polymer Form as a function of temperature • Glassy: Solid-like form, rigid, and hard • Rubbery: Soft solid form, flexible, and elastic • Melt: Liquid-like form, fluid, elastic\n\n44. Glass Transition Temperature, Tg • Glass Transition Temperature, Tg: The temperature by which: • Below the temperature the material is in animmobile(rigid) configuration • Above the temperature the material is in a mobile (flexible) configuration • Transition is called “Glass Transition” because the properties below it are similar to ordinary glass. • Transition range is not one temperature but a range over a relatively narrow range (10 degrees). Tg is not precisely measured, but is a very important characteristic. • Tg applies to all polymers (amorphous, crystalline, rubbers, thermosets, fibers, etc.)\n\n45. Glass Transition Temperature, Tg Amorphous Modulus (Pa) or (psi) Crystalline Tg Tg Tg -50C 50C 100C 150C 200C 250C -50C 50C 100C 150C 200C 250C Temperature Temperature • Glass Transition Temperature, Tg: Defined as • the temperature wherein a significant the loss of modulus (or stiffness) occurs • the temperature at which significant loss of volume occurs Vol.\n\n46. Crystalline Polymers: Tm • Tm: Melting Temperature • T > Tm,The order of the molecules is random (amorphous) • Tm >T>Tg,Crystallization begins at various nuclei and the order of the molecules is a mixture of crystals and random polymers (amorphous). Crystallization continues as T drops until maximum crystallinity is achieved. The amorphous regions are rubbery and don’t contribute to the stiffness. The crystalline regions are unaffected by temperature and are glassy and rigid. • T < Tg,The amorphous regions gain stiffness and become glassy Melt Tm Temp Rubbery Decreasing Temp Tg Glassy Polymer Form\n\n47. Crystalline Polymers Tg -50C 50C 100C 150C 200C 250C Temperature • Tg: Affected by Crystallinity level • High Crystallinity Level = high Tg • Low Crystallinity Level = low Tg Modulus (Pa) or (psi) High Crystallinity Medium Crystallinity Low Crystallinity Tg\n\n48. Temperature Effects on Specific Volume Amorphous Specific Volume Crystalline Tg Tg -50C 50C 100C 150C 200C 250C • T > Tm, The amorphous polymer’s volume decreases linearly with T. • Tm > T>Tg, As crystals form the volume drops since the crystals are significantly denser than the amorphous material. • T < Tg, the amorphous regions contracts linearly and causes a change in slope Temperature\n\n49. Elastomers • Elastomers are rubber like polymers that are thermoset or thermoplastic • butyl rubber: natural rubber • thermoset: polyurethane, silicone • thermoplastic: thermoplastic urethanes (TPU), thermoplastic elastomers (TPE), thermoplastic olefins (TPO), thermoplastic rubbers (TPR) • Elastomers exhibit more elastic properties versus plastics which plastically deform and have a lower elastic limit. • Rubbers have the distinction of being stretched 200% and returned to original shape. Elastic limit is 200%\n\n50. Rubbers • Rubbers have the distinction of being stretched 200% and returned to original shape. Elastic limit is 200% • Natural rubber (isoprene) is produced from gum resin of certain trees and plants that grow in southeast Asia, Ceylon, Liberia, and the Congo. • The sap is an emulsion containing 40% water & 60% rubber particles • Vulcanization occurs with the addition of sulfur (4%). • Sulfur produces cross-links to make the rubber stiffer and harder. • The cross-linkages reduce the slippage between chains and results in higher elasticity. • Some of the double covalent bonds between molecules are broken, allowing the sulfur atoms to form cross-links. • Soft rubber has 4% sulfur and is 10% cross-linked. • Hard rubber (ebonite) has 45% sulfur and is highly cross-linked." ]

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[ "# Question b1da4\n\nMay 7, 2017\n\nElement $\\text{X}$ is chlorine.\n\n#### Explanation:\n\nWe can use the Ideal Gas Law to solve this problem:.\n\n$\\textcolor{b l u e}{\\overline{\\underline{| \\textcolor{w h i t e}{\\frac{a}{a}} p V = n R T \\textcolor{w h i t e}{\\frac{a}{a}} |}}} \\text{ }$\n\nSince $n = \\frac{m}{M}$, we can rearrange this equation to get\n\n$p V = \\left(\\frac{m}{M}\\right) R T$\n\nAnd we can solve this equation to get\n\n$M = \\frac{m R T}{p V}$\n\n$m = \\text{283.3 g}$\n$R = \\text{0.082 06 L·atm·K\"^\"-1\"\"mol\"^\"-1}$\n$T = \\text{(27 + 273.15) K\" = \"300.15 K}$\n$p = \\text{3.2 atm}$\n$V = \\text{30 L}$\n\nM = (\"283.3 g\" × \"0.082 06\" color(red)(cancel(color(black)(\"L·atm·K\"^\"-1\")))\"mol\"^\"-1\" × 300.15 color(red)(cancel(color(black)(\"K\"))))/(3.2 color(red)(cancel(color(black)(\"atm\"))) ×30 color(red)(cancel(color(black)(\"L\")))) = \"72.7 g/mol\"#\n\nThe molecular mass of ${\\text{X}}_{2}$ is 72.7u.\n\nThe atomic mass of $\\text{X}$ is 36.3 u.\n\nThe atomic mass of $\\text{Cl}$ is 35.45 u, and the atomic mass of $\\text{Ar}$ is 39.95 u.\n\nThe atomic mass of $\\text{X}$ is closer to that of $\\text{Cl}$, and $\\text{Cl}$ also forms diatomic ${\\text{Cl}}_{2}$ molecules.\n\nElement $\\text{X}$ is chlorine." ]

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[ "# Pi", null, "", null, "Pi is a mathematical constant whose value is the ratio of any circle's circumference to its diameter in Euclidean space; this is the same value as the ratio of a circle's area to the square of its radius. It is approximately equal to 3.14159265 in the usual decimal notation. Many formulae from mathematics, science, and engineering involve ą, which makes it one of the most important mathematical constants.\n\nPi is an irrational number, which means that its value cannot be expressed exactly as a fraction m/n, where m and n are integers. Consequently, its decimal representation never ends or repeats. It is also a transcendental number, which implies, among other things, that no finite sequence of algebraic operations on integers (powers, roots, sums, etc.) can be equal to its value; proving this was a late achievement in mathematical history and a significant result of 19th century German mathematics. Throughout the history of mathematics, there has been much effort to determine ą more accurately and to understand its nature; fascination with the number has even carried over into non-mathematical culture.\n\nProbably because of the simplicity of its definition, the concept of ą has become entrenched in popular culture to a degree far greater than almost any other mathematical construct. It is, perhaps, the most common ground between mathematicians and non-mathematicians.Reports on the latest, most-precise calculation of Pi (and related stunts) are common news items. The current record for the decimal expansion of ą, if verified, stands at 5 trillion digits.\n\nThe Greek letter ą, often spelled out pi in text, was first adopted for the number as an abbreviation of the Greek word for perimeter (as an abbreviation for \"perimeter/diameter\") by William Jones in 1706. The constant is also known as Archimedes' Constant, after Archimedes of Syracuse, although this name is uncommon in modern English-speaking contexts.\n\nPi   Wikipedia\n\nMATHEMATICS INDEX" ]

[ null, "https://www.crystalinks.com/pi.jpg", null, "https://www.crystalinks.com/pi3.14.jpg", null ]

[ "# How do you evaluate \\cot (45^\\@ )-\\cos (30^\\@ )?\n\nMay 22, 2018\n\n#### Answer:\n\n$\\cot 45 - \\cos 30 = 0.134$\n\n#### Explanation:\n\nshow that\n\n$\\cot x = \\cos \\frac{x}{\\sin} x$\n\n$\\cos 45 = \\frac{1}{\\sqrt{2}}$\n\n$\\sin 45 = \\frac{1}{\\sqrt{2}}$\n\n$\\cos 30 = \\frac{\\sqrt{3}}{2}$\n\n$\\cot 45 - \\cos 30 = \\cos \\frac{45}{\\sin} 45 - \\cos 30$\n\n$= \\frac{\\frac{1}{\\sqrt{2}}}{\\frac{1}{\\sqrt{2}}} - \\frac{\\sqrt{3}}{2} = 1 - \\frac{\\sqrt{3}}{2} = 1 - 0.866 = 0.134$" ]

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[ "# Polygon Analysis: Overview and Explanation", null, "## Polygon Analysis\n\nOur aim is to build a module which is enough to analyze a polygon and return some fruitful information like an angle, slope, sides, corner coordinates etc. So we build py2pyAnalysis. Two line of code is sufficient to tear down a polygon of upto 15 sides.\n\n### Installation\n\nUse the package manager pip to install py2pyAnalysis.\n\n```pip install py2pyAnalysis\n```\n\n### Samples\n\n• Pentagon\n\n• Nonagon\n\n• Irregular Decagon\n\n• Regular Decagon\n\n• Parallelogram\n\n### Documents\n\nLet’s take a look at it,\n\n```def polygon_analysis(file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN)\n\n```\n• file_name : as clear from the name, it takes a file name, if the file is in the same directory then you can simply type name, else type the full path to the file.\n• show_and _save_contour : it allows you to see the various shapes that program detects in the image, it will also save the image file in a folder to Data/file_name_analysis/contour image. Everything that saves during the program related to this file will save in this particular folder only.\n• show_and_save_analysis : it allows you to see the numerical data on the image itself. the data includes Sides, Angles, and Slope. also, the name of an image will also be written there, The image will also save in the same directory as above.\n• Show_angles : This will show the angles on the image, changing it to NO will result to not showing angles on the image.\n• Show_slope: This will show the Slope on the image, changing it to NO will result to not showing Slope on the image.\n• font: you can change the Font if you want, Here is the list of some fonts\n```cv2.FONT_HERSHEY_SIMPLEX\ncv2.FONT_HERSHEY_PLAIN\ncv2.FONT_HERSHEY_DUPLEX\ncv2.FONT_HERSHEY_COMPLEX\ncv2.FONT_HERSHEY_TRIPLEX\ncv2.FONT_HERSHEY_COMPLEX_SMALL\ncv2.FONT_HERSHEY_SCRIPT_SIMPLEX\ncv2.FONT_HERSHEY_SCRIPT_COMPLEX\n```\n\n#### This program returns\n\n` return (len(sides),sides,distance,slope,angles,Name) `\n• len(sides): Number of sides in a detected Polygon.\n• Sides: Tit returns the coordinates of the shape. Side is the first [x,y] set. Similarly side & side will give you first x and y point respectivelly.\n• Distances: It returns the distance between two points in pixels.\n• Slope: It returns back the slope values corresponds to each side.\n• Angle: It returns back an array of angles.\n• Name: It returns back the name of the shape.\n\nIf anytime you forgot the which come after which then just type\n\n```import py2pyAnalysis as py\npy.help()\n```\n\nand it will bring you the code below\n\n``` print(\"\"\"#https://pypi.org/project/py2pyAnalysis/\n#https://github.com/Pushkar-Singh-14/Polygon-Analysis\n#http://py2py.com/polygon-analysis-overview-and-explaination/\n\nNumber_of_sides,Coordinates,Distance_in_pixels,Slopes,Angles,Names= py.polygon_analysis ( file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN\n) \"\"\")\n```\n\n# Explanation\n\nLets understand the in depth working of the module.\n\nThis module detects the contour in the image, a contour is like the closed shape path, as soon as program find the contour it saves its coordinates in the variable (sides) and proceeds further if sides pass through the checkpoints which we will discuss later,\n\nthen it analyzes the polygon and shows the needful data, the whole program is about how to effectively use these coordinates. because there can be many contours in the image, like if something is written on it then there will be some small contours in the alphabets. so it should be cared of. Let’s tear down the program and understand how it works.\n\n`I am adding the full code here, if you want to understand the specific function or specific line then just navigate to the particular line in the explaination`\n```import cv2\nimport csv\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport image\nfrom PIL import Image\nimport os\nimport math\nfrom decimal import Decimal\nfrom scipy import ndimage\n\ndef polygon_analysis(file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN):\n\ncwd = os.getcwd()\nname_file=os.path.splitext(file_name)\ncounter3=0\nlimit=3 #detection_limit #3\n\nif ((show_and_save_analysis=='yes') or (show_and_save_contour=='yes')or (save_data_to_csv=='yes')):\n\npath_save_temp=os.path.join(cwd,'Data')\npath_save=os.path.join(path_save_temp,f'{name_file}_analysis')\n\nif not os.path.exists(path_save):\nos.makedirs(path_save)\n\nimage = Image.open(file_name, 'r')\nimage_size = image.size\nwidth_old = image_size\nheight_old = image_size\n\nbigside=int(max(width_old,height_old)*1.5)\nbackground = Image.new('RGBA', (bigside, bigside), (255, 255, 255, 255))\noffset = (0,0)\nbackground.paste(image, offset)\nfile_name2=f'{width_old*2}X{height_old*2}_{name_file}.png'\nsave_image=os.path.join(cwd,file_name2)\nsave_image_in_data=os.path.join(path_save,file_name2)\n\nif ((show_and_save_analysis=='yes') or (show_and_save_contour=='yes') or (save_data_to_csv=='yes')):\nbackground.save(save_image_in_data)\nimage = Image.open(save_image_in_data)\nwidth, height = image.size\nblur = cv2.GaussianBlur(img,(5,5),0)\nplt.imshow(img)\n\nelse:\nbackground.save(save_image)\nimage = Image.open(save_image)\nwidth, height = image.size\nblur = cv2.GaussianBlur(img,(5,5),0)\nplt.imshow(img)\n\nfont_of_name=cv2.FONT_HERSHEY_TRIPLEX\nfont_size_name=max(height,width)*0.002\nfont=cv2.FONT_HERSHEY_TRIPLEX\nfont_size=font_size_name/1.5\n\ncolors = 10*['r', 'b', 'y','g','k','c', 'm', 'seagreen','navy','gold','coral', 'violet', 'crimson','skyblue','hotpink','slateblue', 'b', 'y','g','k','r', 'b', 'y','g','k']\nmarkers = 10*['*', '+', 'o', 'P', 'x','s', 'p', 'h', 'H', '&amp;lt;','&amp;gt;', 'd', 'D', '^', '1']\n\nabc=[]\nsides=[]\ndistance=[]\nm=[]\nangles=[]\nslope=[]\nName=[]\n\ndef error_detection(abc):\nerror = []\nfor i in range(len(abc)):\nif (i== len(abc)-1):\nerror.append(abs((abc[i]-abc)/abc))\nelse:\nerror.append(abs((abc[i]-abc[i+1])/abc[i+1]))\nreturn (abs(np.mean(error)*100))\n\ndef error_detection_alternate(abc):\nerror = []\nfor i in range(int(len(abc)/2)):\nalt_error= (abs((abc[i]-abc[i+2])/abc[i+2]))\nerror.append(alt_error)\nreturn (abs(np.mean(error)*100))\n\ndef sides_length_and_slope(sides):\nsides= np.reshape(sides,(len(sides),2))\nx=[]\ny=[]\nm=[]\ndeg_tan=[]\nside_len=[]\n\nfor a,b in sides:\nx.append(a)\ny.append(b)\n\nfor i in range(len(sides)):\nif (i == (len(sides)-1)):\nside_len.append(round((math.sqrt(((x[i]-x)**2)+((y[i]-y)**2))),2))\nif ((x-x[i])==0):\nm.append(round(((y-y[i])/1),2))\nelse:\nm.append(round(((y-y[i])/(x-x[i])),2))\n\nelse:\nside_len.append(round((math.sqrt(((x[i]-x[i+1])**2)+((y[i]-y[i+1])**2))),2))\nif ((x[i+1]-x[i])==0):\nm.append(round(((y[i+1]-y[i])/1),2))\nelse:\nm.append(round((((y[i+1]-y[i])/(x[i+1]-x[i]))),2))\n#print(side_len)\nreturn side_len,m,x,y\n\ndef allow(sides=sides,width=width,height=height):\nside,_,x,y = sides_length_and_slope(sides)\n\nfor i in range(len(sides)):\nif (side[i]&amp;lt;(width_old*0.05)) or (x[i]&amp;lt;(max(width_old,height_old)*0.010))or (y[i]&amp;lt;(max(width_old,height_old)*0.010))or (x[i]&amp;gt;(max(width_old,height_old)*0.98))or(y[i]&amp;gt;(max(width_old,height_old)*0.98)) :\n\n#height-height*0.02\n## if(x[i]==0)or(y[i]==0)or(x[i]&amp;gt;(height-5))or(y[i]&amp;gt;(width-5))or(side[i]&amp;lt;(width/20)):\nflag=0\nbreak\nelse:\nflag=1\nif(flag==1):\n\nreturn (np.reshape(sides,(len(sides),2)))\n\ndef angle(sides,m):\n\nfor i in range(len(sides)):\nif (i == (len(sides)-1)):\nif math.degrees(math.atan(m)-math.atan(m[i]))&amp;lt; 0:\nangles.append(round(math.degrees(math.atan(m)-math.atan(m[i]))+180,2))\n\nelse:\nangles.append(round(math.degrees(math.atan(m)-math.atan(m[i])),2))\n\nelse:\nif math.degrees(math.atan(m[i+1])-math.atan(m[i]))&amp;lt; 0:\nangles.append(round(math.degrees(math.atan(m[i+1])-math.atan(m[i]))+180,2))\n\nelse:\nangles.append(round(math.degrees(math.atan(m[i+1])-math.atan(m[i])),2))\n\n## print(angles)\nreturn angles\n\ndef Fiveto15shape(sides):\n\nfor i in range(11):\nif len(sides) == i+5:\n\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\nif (error_detection(angles)&amp;lt;limit):\n\nprint (f'Regular {shapes[i]}')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name(f'Regular {shapes[i]}')\nsave_to_csv(sides,side,angles,name=f&amp;quot;Regular {shapes[i]}&amp;quot;, m=m)\n\nelse:\n\nprint (f'{shapes[i]}')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name(f'{shapes[i]}')\nsave_to_csv(sides,side,angles,name=f'{shapes[i]}',m=m)\n\ndef show_and_save_fig_data(sides,counter3):\n\nfor i in range(len(sides)):\ncounter2=0\nplt.scatter(np.reshape(sides,(len(sides),2))[i][counter2],np.reshape(sides,(len(sides),2))[i][counter2+1],marker= markers[counter3], c=colors[counter3])\n\ndef write_angle_slope_and_sides(sides,side,angles,m,show_angles=show_angles,show_sides=show_sides):\nmiddle_point_X=[]\nmiddle_point_Y=[]\nfor j in range(len(sides)):\nd=0\nif (j == (len(sides))-1):\nmiddle_point_X.append(int((((sides[j][d]+sides[d])/2))))\nmiddle_point_Y.append(int(((sides[j][d+1]+sides[d+1])/2)))\nelse:\nmiddle_point_X.append(int((((sides[j][d]+sides[j+1][d])/2))))\nmiddle_point_Y.append(int(((sides[j][d+1]+sides[j+1][d+1])/2)))\n## print(middle_point_X)\n## print(middle_point_Y)\n## print(sides)\n\nif (show_angles=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f&amp;quot;{angles[j]}&amp;quot;, (sides[j], sides[j]), font, font_size, ((183,9,93)))\nif(show_sides=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f&amp;quot;{side[j]}&amp;quot;, (middle_point_X[j], middle_point_Y[j]), font, font_size, ((0,0,255))) #blue green red\nif(show_slope=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f&amp;quot;{(m[j])}&amp;quot;, (middle_point_X[j], int(middle_point_Y[j]+(max(height,width)*0.05))), font, font_size, ((0,255,0))) #blue green red\n\ndef save_to_csv(sides,side,angles,name,m):\nslope.append(m)\ndistance.append(side)\nName.append(name[:])\n\nif save_data_to_csv=='yes':\nx= 'csv_data_'+file_name[:(len(file_name)-4)]+'.csv'\n\nsave_csv=os.path.join(path_save,f'csv_data_{name_file}.csv')\nwith open(save_csv, mode='w') as data_file:\ndata_writer = csv.writer(data_file, delimiter=';')\nfieldname=[['x_coordinate','y_coordinate','distance_in_pixels', 'angles', 'name', 'slope']]\ndata_writer.writerows(fieldname)\nfor i in range(len(side)):\nc=0\ndata_writer.writerow([sides[i],sides[i],side[i], angles[i], name, m[i]])\n\ndef write_name(name):\nif(show_name=='yes'):\ncv2.putText(img1, name, (int(max(height,width)*0.20), int(max(height,width)*0.80)), font_of_name, font_size_name, ((255,0,0))) #blue green red\nif(show_angles=='yes'):\ncv2.putText(img1, '# - Angles', (int(max(height,width)*0.70), int(max(height,width)*0.75)), font_of_name, font_size_name*0.40, ((183,9,93)))\nif(show_sides=='yes'):\ncv2.putText(img1, '# - Distance(in px)', (int(max(height,width)*0.70), int(max(height,width)*0.80)), font_of_name, font_size_name*0.40, ((0,0,255)))\nif(show_slope=='yes'):\ncv2.putText(img1, '# - Slope', (int(max(height,width)*0.70), int(max(height,width)*0.85)), font_of_name, font_size_name*0.40, ((0,255,0)))\n\n_, threshold = cv2.threshold(blur, 240, 255, cv2.THRESH_BINARY)\n_, contours, _ = cv2.findContours(threshold, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)\n\nfor cnt in contours:\nsides = cv2.approxPolyDP(cnt, 0.01*cv2.arcLength(cnt, True), True)\ncv2.drawContours(img, [sides], 0, (0), 5)\nx = sides.ravel()\ny = sides.ravel()\n\nif (show_and_save_contour=='yes'):\ncounter3+=1\nshow_and_save_fig_data(sides,counter3)\n\n## print(sides)\n\nsides=allow(sides)\n## print(len(sides))\nif (sides is not None):\n## print(sides)\n\nif len(sides) == 3:\n\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\nif (error_detection(angles)&amp;lt;limit):\nif (error_detection(side)&amp;lt;limit):\nprint ('Eq Tri')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name('Eq Tri')\nsave_to_csv(sides,side,angles,name='Eq Tri', m=m)\nbreak\n\nelse:\nprint ('Triangle')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name('Triangle')\nsave_to_csv(sides,side,angles,name='Triangle',m=m)\nbreak\n\nif len(sides) == 4:\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\n\nif (error_detection(angles)&amp;lt;limit):\nif (error_detection(side)&amp;lt;limit):\nprint('square')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('square')\nsave_to_csv(sides,side,angles,name='square',m=m)\ndistance.append(side)\nbreak\nelif (error_detection_alternate(side)&amp;lt;limit):\nprint('rectangle')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('rectangle')\nsave_to_csv(sides,side,angles,name='rectangle',m=m)\nbreak\n\nelif (error_detection_alternate(angles)&amp;lt;limit):\n\nif (error_detection(side)&amp;lt;limit):\nprint('Rhombus')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('Rhombus')\nsave_to_csv(sides,side,angles,name='Rhombus',m=m)\nbreak\n\nelif (error_detection_alternate(side)&amp;lt;limit):\nprint('Parallelogram')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('Parallelogram')\nsave_to_csv(sides,side,angles,name='Parallelogram',m=m)\nbreak\n\nelse:\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\n\nbreak\n\nif(len(sides)&amp;gt;4):\n\nFiveto15shape(sides)\n\nbreak\nelse:\npass\n\nif (show_and_save_contour=='yes'):\nsave_conotur=os.path.join(path_save,f&amp;quot;contour_{file_name}&amp;quot;)\nplt.savefig(save_conotur)\nim= Image.open(save_conotur)\nim.show()\nplt.show()\n\nif (show_and_save_analysis=='yes'):\nsave_analysis=os.path.join(path_save,f&amp;quot;analysis_{file_name}&amp;quot;)\ncv2.imwrite(save_analysis,img1)\nim= Image.open(save_analysis)\nim.show()\n\nreturn len(sides),sides,distance,slope,angles,Name\n\ndef help():\nprint(&amp;quot;&amp;quot;&amp;quot;#https://pypi.org/project/py2pyAnalysis/\n#https://github.com/Pushkar-Singh-14/Polygon-Analysis\n#https://py2py.com/polygon-analysis-overview-and-explanation/\n\nNumber_of_sides,Coordinates,Distance_in_pixels,Slopes,Angles,Names= py.polygon_analysis ( file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN\n) &amp;quot;&amp;quot;&amp;quot;)\n\n```\n\nFirst of all we are importing the necessary packages\n\n```import cv2\nimport csv\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport image\nfrom PIL import Image\nimport os\nimport math\n```\n\nand here is our function\n\n```def polygon_analysis(file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN):\n```\n\nHere we are doing some initial work which we may require in the program later here we are saving the folder path in a variable in which the program is running which helps the program to save the files if allowed.\n\nIn the next step we are extracting the file name without extention, For example we have file named square.jpeg then it extracts “square” and then initializing some variables, and limit,which we will discuss\nlater\n\n``` cwd = os.getcwd()\nname_file=os.path.splitext(file_name)\ncounter3=0\nlimit=3 #detection_limit #3\n```\n\nOur plan is if we want to save any of these data or all of these, then a new separate folder should be created which has named data and within folder and new folder should also be created with the name of ({name of file}_Analysis) in which all the files are stored,\n\nThus if the program run again to detect another polygon then it will create folder next to the previously created (name of file_analysis) folder, it ensures the accessibility of data and cleanliness at the same time,\n\n``` if ((show_and_save_analysis=='yes') or (show_and_save_contour=='yes')or (save_data_to_csv=='yes')):\n\npath_save_temp=os.path.join(cwd,'Data')\npath_save=os.path.join(path_save_temp,f'{name_file}_analysis')\n\nif not os.path.exists(path_save):\nos.makedirs(path_save)\n```\n\nHere we are opening the image with an intention to read. We are extracting the width and height of the file. We saved the data in as width_old and height_old.\n\nIn the later step we increase the background by pasting the current image on a white square canvas whose side is determined by the max side between width_old and height_old. in the last step a path is saved as a new file with the name of (twice the width x twice the height)_name of file.\n\n``` image = Image.open(file_name, 'r')\nimage_size = image.size\nwidth_old = image_size\nheight_old = image_size\n\nbigside=int(max(width_old,height_old)*1.5)\nbackground = Image.new('RGBA', (bigside, bigside), (255, 255, 255, 255))\noffset = (0,0)\nbackground.paste(image, offset)\nfile_name2=f'{width_old*2}X{height_old*2}_{name_file}.png'\nsave_image=os.path.join(cwd,file_name2)\n```\n\nHere if any of these conditions satisfied then we will save that new image in the data folder else we will save it somewhere outside and access from there, then we will read the image in grayscale because color doesnt matter here and monotone makes it fast to process the data.\n\nHere notice that we are opening the image again in another variable, this is because we show have to the contour data on separate file, i.e. one fo the name, sides, angle and slopes, and one for the contour.\n\nWidth and height are saved using ‘Image save function’, also Gaussian blur is used, this is because we want to smooth our image a little but just to blend the corner pixels in one another so that some unnecessary contours can be ignored.\n\n``` if ((show_and_save_analysis=='yes') or (show_and_save_contour=='yes') or (save_data_to_csv=='yes')):\nbackground.save(save_image_in_data)\nimage = Image.open(save_image_in_data)\nwidth, height = image.size\nblur = cv2.GaussianBlur(img,(5,5),0)\nplt.imshow(img)\n\nelse:\nbackground.save(save_image)\nimage = Image.open(save_image)\nwidth, height = image.size\nblur = cv2.GaussianBlur(img,(5,5),0)\nplt.imshow(img)\n```\n\nIn this step we are specifying the fonts, font_of_name is the font we used to show the name of the polygon, so, it should be bigger then the normal font, but if we hard-coded our code here then we end up bigger font even on the small resolution image or smaller font on higher resolution image.\n\nSo, to prevent this we define the font size as the 0.2% of max between height and width similarly for the normal font we described as the 3/4th of font_size_name. in order to maintain the image size to font size ratio.\n\n``` font_of_name=cv2.FONT_HERSHEY_TRIPLEX\nfont_size_name=max(height,width)*0.002\nfont=cv2.FONT_HERSHEY_TRIPLEX\nfont_size=font_size_name/1.5\n```\n\nHere we are initializing the colors, marker, and shapes which we will use later. along with this we also initializing some lists.\n\n```colors = 10*['r', 'b', 'y','g','k','c', 'm', 'seagreen','navy','gold','coral', 'violet', 'crimson','skyblue','hotpink','slateblue', 'b', 'y','g','k','r', 'b', 'y','g','k']\nmarkers = 10*['*', '+', 'o', 'P', 'x','s', 'p', 'h', 'H', '<','>', 'd', 'D', '^', '1']\n\nabc=[]\nsides=[]\ndistance=[]\nm=[]\nangles=[]\nslope=[]\nName=[]\n```\n\nThis error_detection() function is used to measure the mean error in a list. like if we take an example of a square then all sides and angles should be equal but this program measures the distance in pixels so there is a possibility that angles will not equal it maybe like 89.5 or 91.0 and same with the sides.\n\nSo, to tackle this kind of problem we will use this function. it only returns the mean error value.\n\n```def error_detection(abc):\nerror = []\nfor i in range(len(abc)):\nif (i== len(abc)-1):\nerror.append(abs((abc[i]-abc)/abc))\nelse:\nerror.append(abs((abc[i]-abc[i+1])/abc[i+1]))\nreturn (abs(np.mean(error)*100))\n```\n\nThis error_detection_alternate() is the same function as above but it’s for alternate sides like in rectangle. As we know in the parallelogram or rhombus the opposite interior angles are equal. so to check those we will this function.\n\n``` def error_detection_alternate(abc):\nerror = []\nfor i in range(int(len(abc)/2)):\nalt_error= (abs((abc[i]-abc[i+2])/abc[i+2]))\nerror.append(alt_error)\nreturn (abs(np.mean(error)*100))\n```\n\nHere this function sides_length_and_slope() takes the contour coordinates of the polygon as we learn earlier, here length of sides denote the number of sides, like 4 for Rectangle, 5 for Pentagon and so on\n\nSo first we reshaped the array in (whatever no of sides X 2), in this method we tried to figure out the distance between two points and the slope of that line.\nNote: here side_len is the length of individual sides and m is the slope.\n\nFormula for distance between two points is sqrt((x2-x2)^2 + (y2-y1)^2)\n\nFormula for slope is ((y2-y1)/(x2-x1))\n\nRound function is used to round off the decimal digit up to 2, like 89.7367264774 will become 89.73 and then we store the values in their respective variables.\n\nNote: Since the array is unidirectional but a polygon is cyclic so we have to define that if the value of ‘i’ is at the last point then it must be paired with the first coordinate to make a closed loop\n\n``` def sides_length_and_slope(sides):\nsides= np.reshape(sides,(len(sides),2))\nx=[]\ny=[]\nm=[]\ndeg_tan=[]\nside_len=[]\n\nfor a,b in sides:\nx.append(a)\ny.append(b)\n\nfor i in range(len(sides)):\nif (i == (len(sides)-1)):\nside_len.append(round((math.sqrt(((x[i]-x)**2)+((y[i]-y)**2))),2))\nif ((x-x[i])==0):\nm.append(round(((y-y[i])/1),2))\nelse:\nm.append(round(((y-y[i])/(x-x[i])),2))\n\nelse:\nside_len.append(round((math.sqrt(((x[i]-x[i+1])**2)+((y[i]-y[i+1])**2))),2))\nif ((x[i+1]-x[i])==0):\nm.append(round(((y[i+1]-y[i])/1),2))\nelse:\nm.append(round((((y[i+1]-y[i])/(x[i+1]-x[i]))),2))\n# print(side_len)\nreturn side_len,m,x,y\n```\n\nNext is allow() function, If the image is not clean or if anything is written on it then the program will also treat it as a contour and take it in consideration, also the 4 corners of an image will also contribute in the shape detection so we always get a square in an image, Thus allow function tackle all these problems.\n\nHere we specify few conditions and if the array of contour/corner points passes these condition then only it will allow for the further analysis.\n\nThe first condition is to check for any contour due to any alphabets or any such short sides which is not actually the desired polygon, here the sides or the length of the detected closed curve should be greater than the 5% of the width of an image,\n\nSecond and third condition checks that square detection problem which I mentioned above, here the x,y should greater than the 1 percent of whatever max between width or height, this eliminate the x,y coordinate which have zero in it.\n\n``` def allow(sides=sides,width=width,height=height):\nside,_,x,y = sides_length_and_slope(sides)\n\nfor i in range(len(sides)):\nif (side[i]<(width_old*0.05)) or (x[i]<(max(width_old,height_old)*0.010))or (y[i]<(max(width_old,height_old)*0.010))or (x[i]>(max(width_old,height_old)*0.98))or(y[i]>(max(width_old,height_old)*0.98)) :\n\n#height-height*0.02\n## if(x[i]==0)or(y[i]==0)or(x[i]>(height-5))or(y[i]>(width-5))or(side[i]<(width/20)):\nflag=0\nbreak\nelse:\nflag=1\nif(flag==1):\n\nreturn (np.reshape(sides,(len(sides),2)))\n```\n\nHere is the angle() function which we use to detect and angle.\n\nThe formula used here is atan(slope2)-atan(slope1) this gives us a radian of the angle, to convert into degree we use math.degrees() function.\n\nIf this gives us a negative value then 180 should be added there in order to make it positive, then all the angles will store in the list named angles.\n\n``` def angle(sides,m):\n\nfor i in range(len(sides)):\nif (i == (len(sides)-1)):\nif math.degrees(math.atan(m)-math.atan(m[i]))< 0:\nangles.append(round(math.degrees(math.atan(m)-math.atan(m[i]))+180,2))\n\nelse:\nangles.append(round(math.degrees(math.atan(m)-math.atan(m[i])),2))\n\nelse:\nif math.degrees(math.atan(m[i+1])-math.atan(m[i]))< 0:\nangles.append(round(math.degrees(math.atan(m[i+1])-math.atan(m[i]))+180,2))\n\nelse:\nangles.append(round(math.degrees(math.atan(m[i+1])-math.atan(m[i])),2))\n\n## print(angles)\nreturn angles\n```\n\nThe polygon with the length 4 has many possibilities, like a square, rectangle, parallelogram, and rhombus. We can say every polygon with 4 sides is a quadrilateral but if we need to specify between these 4 shapes then we have to classify them individually.\n\nThus after 4th sided polygon, every polygon can be classified by simple conditions, like if error detection ( which we learn above ) for an angle is in the limit (we can specify it, but I choose it as 3 percent) then the shape is said to be a Regular polygon, thus we use this Fiveto15shape() function\n\nHere we are not checking for the sides because if all angle are equal then side will also be equal, it only increases the weight on the code.\n\nNote: Here some functions like write_angle_slope_and_sides, write_name, or save_to_csv are also there, we will learn about them as we proceed further,\n\n``` def Fiveto15shape(sides):\n\nfor i in range(11):\nif len(sides) == i+5:\n\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\nif (error_detection(angles)<limit):\n\nprint (f'Regular {shapes[i]}')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name(f'Regular {shapes[i]}')\nsave_to_csv(sides,side,angles,name=f\"Regular {shapes[i]}\", m=m)\n\nelse:\n\nprint (f'{shapes[i]}')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name(f'{shapes[i]}')\nsave_to_csv(sides,side,angles,name=f'{shapes[i]}',m=m)\n```\n\nHere show_and_save_fig_data() shows the contour data on the figure, if it selected yes during function calling then whenever a set of a contour is found, it plots all corner points with same marker and color so we can examine visually that where is the possible contour in the image.\n\n```def show_and_save_fig_data(sides,counter3):\n\nfor i in range(len(sides)):\ncounter2=0\nplt.scatter(np.reshape(sides,(len(sides),2))[i][counter2],np.reshape(sides,(len(sides),2))[i][counter2+1],marker= markers[counter3], c=colors[counter3])\n```\n\nHere write_angle_slope_and_sides() writes the angles, sides, and slope on the image if it is selected yes individually.\n\nInitially, we detected the middle of sides so that we can show sides and slope there, we will do some adjustment with the slope position just to ensure that side and slope data will not overlap.\n\nWhen show angles, show side and show slope is set to yes respectively at the time of function call, this function writes the required data on the image using the put_text function.\n\n``` def write_angle_slope_and_sides(sides,side,angles,m,show_angles=show_angles,show_sides=show_sides):\nmiddle_point_X=[]\nmiddle_point_Y=[]\nfor j in range(len(sides)):\nd=0\nif (j == (len(sides))-1):\nmiddle_point_X.append(int((((sides[j][d]+sides[d])/2))))\nmiddle_point_Y.append(int(((sides[j][d+1]+sides[d+1])/2)))\nelse:\nmiddle_point_X.append(int((((sides[j][d]+sides[j+1][d])/2))))\nmiddle_point_Y.append(int(((sides[j][d+1]+sides[j+1][d+1])/2)))\n## print(middle_point_X)\n## print(middle_point_Y)\n## print(sides)\n\nif (show_angles=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f\"{angles[j]}\", (sides[j], sides[j]), font, font_size, ((183,9,93)))\nif(show_sides=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f\"{side[j]}\", (middle_point_X[j], middle_point_Y[j]), font, font_size, ((0,0,255))) #blue green red\nif(show_slope=='yes'):\nfor j in range(len(sides)):\nc=0\ncv2.putText(img1, f\"{(m[j])}\", (middle_point_X[j], int(middle_point_Y[j]+(max(height,width)*0.05))), font, font_size, ((0,255,0))) #blue green red\n```\n\nwhen set to yes then save_to_csv() will save all the data to the CSV file which will save in the data>filename folder with other images,\n\nIt stores the x,y coordinate of the corner of the polygon, the distance between two points (in pixels), angles, name of the shape detected and the slope.\n\n```def save_to_csv(sides,side,angles,name,m):\nslope.append(m)\ndistance.append(side)\nName.append(name[:])\n\nif save_data_to_csv=='yes':\nx= 'csv_data_'+file_name[:(len(file_name)-4)]+'.csv'\n\nsave_csv=os.path.join(path_save,f'csv_data_{name_file}.csv')\nwith open(save_csv, mode='w') as data_file:\ndata_writer = csv.writer(data_file, delimiter=';')\nfieldname=[['x_coordinate','y_coordinate','distance_in_pixels', 'angles', 'name', 'slope']]\ndata_writer.writerows(fieldname)\nfor i in range(len(side)):\nc=0\ndata_writer.writerow([sides[i],sides[i],side[i], angles[i], name, m[i]])\n```\n\nWe use this write_name() function to write the name of the polygon on the image, here we determine the position, in which x coordinate is 20% of max among height and width, and y coordinate is the 80 percent of max among height and width.\n\nNote: Height and Width should not be confused with height_old and width_old as previous, and also\n\nNote that the font size of name is 40% for sides marker and slope marker. also the position is toggled too. just to maintian the space.\n\n```def write_name(name):\nif(show_name=='yes'):\ncv2.putText(img1, name, (int(max(height,width)*0.20), int(max(height,width)*0.80)), font_of_name, font_size_name, ((255,0,0))) #blue green red\nif(show_angles=='yes'):\ncv2.putText(img1, '# - Angles', (int(max(height,width)*0.70), int(max(height,width)*0.75)), font_of_name, font_size_name*0.40, ((183,9,93)))\nif(show_sides=='yes'):\ncv2.putText(img1, '# - Distance(in px)', (int(max(height,width)*0.70), int(max(height,width)*0.80)), font_of_name, font_size_name*0.40, ((0,0,255)))\nif(show_slope=='yes'):\ncv2.putText(img1, '# - Slope', (int(max(height,width)*0.70), int(max(height,width)*0.85)), font_of_name, font_size_name*0.40, ((0,255,0)))\n```\n\nNow from here our main program starts, here we take the file in which we used the Gaussian blur function to smooth the edges and we saved that file in blur variable.\n\nIn the next step are using a threshold function on that image, threshold function works here like threshold(img,A,B,operation) if the pixel intensity is greater then A then it is converted to B else zero, here 255 means white and zero means black, also the operation is important, here we use cv2.THRESH_BINARY, if we used THRESH_BINARY_INV then the mechanism become opposite, or inverted.\n\nNext we use findContour function to find obviously the contours, we pass the threshold image and the RETR_TREE is used for hierarcical differentiation of contour. next is cv2.CHAIN_APPROX_SIMPLE, since the contour is the full closed path, but we are only interested in the corners, so this CHAIN_APPROX_SIMPLE method only shows the approx coordinates, i.e extreme ones or corner ones,\n\n```_, threshold = cv2.threshold(blur, 240, 255, cv2.THRESH_BINARY)\n_, contours, _ = cv2.findContours(threshold, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)\n```\n\nThen we extract each contour one by one and change the sides using cv2.approxPolyDP. here note that one sides variable is the list of all the corner points which found in the closed shape,\n\nHere we comment out the function which we can use to draw the contours in the form of sides. We only show the contours in term of the corner point, then we saved the coordinates in x,y coordinate for further use.\n\n```for cnt in contours:\nsides = cv2.approxPolyDP\n##cv2.drawContours(img, [sides], 0, (0), 5)\nx = sides.ravel()\ny = sides.ravel()\n```\n\nHere if show_and_save_contour is set to yes then we pass the sides and a counter which is always initialized as zero.\n\n``` if (show_and_save_contour=='yes'):\ncounter3+=1\nshow_and_save_fig_data(sides,counter3)\n```\n\nthe allow() function is acting as a barrier to pass only legitimate contour when there is no contour left in the contours list then allow() function returns None, which then throw an error so to tackle this we use the if statement.\n\n``` sides=allow(sides)\n## print(len(sides))\nif (sides is not None):\n## print(sides)\n```\n\nNow here len(sides) means the number of sides, if it is 3, then obviously its a triangle, but if the error detection is within the limit i.e 3% then its called equilateral triangle else just a triangle then we pass the sides angle and slope to the write angle_slope_and_sides() function. and name to the write_name() funtion and same data to the csv file.\n\n``` if len(sides) == 3:\n\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\nif (error_detection(angles)<limit):\nif (error_detection(side)<limit):\nprint ('Eq Tri')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name('Eq Tri')\nsave_to_csv(sides,side,angles,name='Eq Tri', m=m)\nbreak\n\nelse:\nprint ('Triangle')\nwrite_angle_slope_and_sides(sides,side,angles,m)\nwrite_name('Triangle')\nsave_to_csv(sides,side,angles,name='Triangle',m=m)\nbreak\n```\n\nas the code is simple now if the error in angles is within the limit then it can be a square or rectangle because, in both cases, all angles are 90 deg, so we check if error_detection() function for sides is less then limit, if so then we conclude its a square and break the loop if not then we again check if error_detection_alternate() for sides is within the limit. if so then we conclude it a rectangle else we conclude its a quadrilateral.\n\nElse if the error_detection_alternate() for sides is less than a limit then we have two options because, in both rhombus and parallelogram, alternate angles are equal. so we check if error_detection for sides is less then limit then its rhombus. else if error_detection_alternate() for sides is less then limit then its parallelogram else it’s a quadrilateral\n\n``` if len(sides) == 4:\nside,m,_,_= sides_length_and_slope(sides)\nangles =angle(sides,m)\n\nif (error_detection(angles)<limit):\nif (error_detection(side)<limit):\nprint('square')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('square')\nsave_to_csv(sides,side,angles,name='square',m=m)\ndistance.append(side)\nbreak\nelif (error_detection_alternate(side)<limit):\nprint('rectangle')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('rectangle')\nsave_to_csv(sides,side,angles,name='rectangle',m=m)\nbreak\n\nelif (error_detection_alternate(angles)<limit):\n\nif (error_detection(side)<limit):\nprint('Rhombus')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('Rhombus')\nsave_to_csv(sides,side,angles,name='Rhombus',m=m)\nbreak\n\nelif (error_detection_alternate(side)<limit):\nprint('Parallelogram')\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\nwrite_name('Parallelogram')\nsave_to_csv(sides,side,angles,name='Parallelogram',m=m)\nbreak\n\nelse:\nwrite_angle_slope_and_sides(sides,side,angles,m=m)\n\nbreak\n```\n\nfor the shape having sides more then 4, we pass the sides to Fiveto15shape()\n\n``` if(len(sides)>4):\n\nFiveto15shape(sides)\n\nbreak\nelse:\npass\n```\n\nIf show_and_save_contour is set to yes then we save the file at the [name_file]_analysis folder within Data folder and then open that image from there and show it, using Image.\n\nWe did exactly the same with show_and_save_analysis. in the end, we return some variable like which might be useful in some other program\n\n```if (show_and_save_contour=='yes'):\nsave_conotur=os.path.join(path_save,f\"contour_{file_name}\")\nplt.savefig(save_conotur)\nim= Image.open(save_conotur)\nim.show()\n## plt.show()\n\nif (show_and_save_analysis=='yes'):\nsave_analysis=os.path.join(path_save,f\"analysis_{file_name}\")\ncv2.imwrite(save_analysis,img1)\nim= Image.open(save_analysis)\nim.show()\n\nreturn len(sides),sides,distance,slope,angles,Name\n```\n\nHere is the help() function in which we print all the function attributes so that if someone forgot which come after which then they can just type py2pyAnalysis.help() and it will display the following bunch of lines.\n\n```def help():\nprint(\"\"\"#https://pypi.org/project/py2pyAnalysis/\n#https://github.com/Pushkar-Singh-14/Polygon-Analysis\n#http://py2py.com/polygon-analysis-overview-and-explaination/\n\nNumber_of_sides,Coordinates,Distance_in_pixels,Slopes,Angles,Names= py.polygon_analysis ( file_name,\nshow_and_save_contour='yes',\nshow_and_save_analysis='yes',\nshow_sides='yes',\nshow_angles='yes',\nshow_slope='yes',\nshow_name='yes',\nsave_data_to_csv='yes',\nfont=cv2.FONT_HERSHEY_PLAIN\n) \"\"\")\n```\n\nThats all from my side, please feel free to share your feedback. If you have any doubt then please comment below." ]

[ null, "https://i0.wp.com/py2py.com/wp-content/uploads/2019/01/Screenshot-2019-01-25-at-4.51.37-PM.png", null ]

[ "# What is a possible set of four quantum numbers (n,l,ml,ms) in order, for the highest energy electron in gallium?\n\nJun 18, 2016\n\nGallium ($\\text{Ga}$) is atomic number $31$ and it is on column 13, row 4.", null, "So its electron configuration involves the $1 s$, $2 s$, $2 p$, $3 s$, $3 p$, $4 s$, $3 d$, and $4 p$ orbitals.\n\n$\\implies 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{1}$\n\nor\n\n$\\implies \\textcolor{b l u e}{\\left[A r\\right] 3 {d}^{10} 4 {s}^{2} 4 {p}^{1}}$\n\nThe highest energy electron in $\\text{Ga}$ is the single $4 p$ electron, which can either be in the $4 {p}_{x}$, $4 {p}_{y}$, or $4 {p}_{z}$ orbital, and it can either be spin up or spin down. So, there are $2 \\times 3 = 6$ possible sets of quantum numbers.\n\nFor the $4 p$ orbital:\n\n• $n = 1 , 2 , . . . , N \\implies \\textcolor{b l u e}{4}$ for the principal quantum number.\n• $l = 0 , 1 , 2 , . . . , n - 1 \\implies \\textcolor{b l u e}{1}$ for the angular momentum quantum number.\n• ${m}_{l} = \\left\\{0 , 1 , . . . , \\pm l\\right\\} = \\left\\{0 , \\pm 1\\right\\}$ for the magnetic quantum number, so there exist $2 l + 1 = 2 \\left(1\\right) + 1 = 3$ total $4 p$ orbitals.\n\nFor a $4 p$ electron:\n\n• The quantum numbers $n$ and $l$ are fixed.\n• ${m}_{l}$ will vary as $\\textcolor{b l u e}{- 1}$, $\\textcolor{b l u e}{0}$, or $\\textcolor{b l u e}{+ 1}$ as mentioned above, and tells you that there are three $4 p$ orbitals.\n• ${m}_{s}$, the spin quantum number, can be $\\textcolor{b l u e}{\\pm \\frac{1}{2}}$.\n\nThus, the 6 possible sets of quantum numbers are:\n\n1. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,-1,+1/2\")}}$\n2. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,0,+1/2\")}}$\n3. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,+1,+1/2\")}}$\n4. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,-1,-1/2\")}}$\n5. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,0,-1/2\")}}$\n6. $\\left(n , l , {m}_{l} , {m}_{s}\\right) = \\textcolor{b l u e}{\\text{(\"4,1,+1,-1/2\")}}$\n\nAnother way to represent this each one of these, respectively, is:\n\n1. color(white)([(\" \",color(black)(uarr),color(black)(ul(\" \")),color(black)(ul(\" \"))), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])\n\n2. color(white)([(\" \",color(black)(ul(\" \")),color(black)(uarr),color(black)(ul(\" \"))), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])\n\n3. color(white)([(\" \",color(black)(ul(\" \")),color(black)(ul(\" \")),color(black)(uarr)), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])\n\n4. color(white)([(\" \",color(black)(darr),color(black)(ul(\" \")),color(black)(ul(\" \"))), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])\n\n5. color(white)([(\" \",color(black)(ul(\" \")),color(black)(darr),color(black)(ul(\" \"))), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])\n\n6. color(white)([(\" \",color(black)(ul(\" \")),color(black)(ul(\" \")),color(black)(darr)), (color(black)(\"orbital\": ), color(black)(4p_x),color(black)(4p_y),color(black)(4p_z)), (color(black)(m_l: ),color(black)(-1),color(black)(0),color(black)(+1))])" ]

[ null, "https://d2jmvrsizmvf4x.cloudfront.net/wXk8GXGSQmQREAyzkBC9_PERIODIC_-_Gallium.png", null ]

[ "Search Results\n\nYou are looking at 1 - 10 of 19 items for :\n\n• \"combustion energy\"\n• Refine by Access: All Content\nClear All\n\nConstant-volume combustion energy of the lead salts of 2HDNPPb and 4HDNPPb and their effect on combustion characteristics of RDX–CMDB propellant\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: Z. Fengqi, G. Hongxu, L. Yang, H. Rongzu, C. Pei, G. Sheng-li, Y. Xu-wu, and S. Qizhen\n\nAbstract\n\nThe constant-volume combustion energies of the lead salts of 2-hydroxy-3,5-dinitropyridine (2HDNPPb) and 4-hydroxy-3,5-dinitropyridine (4HDNPPb), ΔU c (2HDNPPb(s) and 4HDNPP(s)), were determined as –4441.922.43 and –4515.741.92 kJ mol–1 , respectively, at 298.15 K. Their standard enthalpies of combustion, Δc m H θ(2HDNPPb(s) and 4HDNPPb(s), 298.15 K), and standard enthalpies of formation, Δr m H θ(2HDNPPb(s) and 4HDNPPb(s), 298.15 K) were as –4425.812.43, –4499.631.92 kJ mol–1 and –870.432.76, –796.652.32 kJ mol–1 , respectively. As two combustion catalysts, 2HDNPPb and 4HDNPPb can enhance the burning rate and reduce the pressure exponent of RDX–CMDB propellant.\n\nRestricted access\n\nThermochemical studies on the thioproline\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: L. Qiang-Guo, L. Xu, D. Bin, and Y. Xu-Wu\n\nAbstract\n\nThe combustion energy of thioproline was determined by the precision rotating-bomb calorimeter at 298.15 K to be Δc U= –2469.301.44 kJ mol–1. From the results and other auxiliary quantities, the standard molar enthalpy of combustion and the standard molar enthalpy of formation of thioproline were calculated to be Δc H m θC4H7NO2S, (s), 298.15 K= –2469.921.44 kJ mol–1 and Δf H m θC4H7NO2S, (s), 298.15K= –401.331.54 kJ mol–1.\n\nRestricted access\n\nThermochemistry of copper complex of 6-benzylaminopurine\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: Y. Xu-Wu, Z. Hang-Guo, S. Wu-Juan, W. Xiao-Yan, and G. Sheng-Li\n\nAbstract\n\nThe copper(II) complex of 6-benzylaminopurine (6-BAP) has been prepared with dihydrated cupric chloride and 6-benzylaminopurine. Infrared spectrum and thermal stabilities of the solid complex have been discussed. The constant-volume combustion energy, Δc U, has been determined as −12566.92±6.44 kJ mol−1 by a precise rotating-bomb calorimeter at 298.15 K. From the results and other auxiliary quantities, the standard molar enthalpy of combustion, Δc H m θ, and the standard molar of formation of the complex, Δf H m θ, were calculated as −12558.24±6.44 and −842.50±6.47 kJ mol−1, respectively.\n\nRestricted access\n\nThermochemistry of the complexes of some microelements and histidine\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: S. Chen, Sh Gao, X. Yang, R. Hu, and Q. Shi\n\nAbstract\n\nSolid complexes of M(His)2Cl2 nH2O (M=Mn, Co, Ni, Cu) of MnCl26H2O, CoCl26H2O, NiCl26H2O, CuCl22H2O and L-α-histidine (His) have been prepared in 95% ethanol solution and characterized by elemental analyses, chemical analyses, IR and TG-DTG. The constant-volume combustion energies of the complexes have been determined by a rotating-bomb calorimeter. And the standard enthalpies of formation of the complexes have been calculated as well.\n\nRestricted access\n\nMicro-combustion Calorimetry Aiming At 1 mg Samples\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: Yatsuhisa Nagano and Takehiko Sugimoto\n\nAbstract\n\nA micro-combustion calorimeter was developed. The small energy equivalent (ca. 68 JK–1) of this calorimeter makes it possible to measure combustion energies of very small samples. The energy equivalent was determined by burning 2 mg of benzoic acid. The standard deviation of the mean energy equivalent was reduced to 0.014% in 5 experiments. The standard massic energy of combustion of salicylic acid and the standard deviation of the mean were determined to be –21871±5 J g–1, which agrees well with the literature values. The standard molar enthalpy of formation of salicylic acid was derived as –591.2±1.7 kJ mol–1.\n\nRestricted access\n\nThermochemical study of the solid complex Ho(PDC)3 (o-phen)\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: G. Xie, S. Chen, S. Gao, X. Meng, and Q. Shi\n\nAbstract\n\nA novel solid complex, formulated as Ho(PDC)3 (o-phen), has been obtained from the reaction of hydrate holmium chloride, ammonium pyrrolidinedithiocarbamate (APDC) and 1,10-phenanthroline (o-phenH2O) in absolute ethanol, which was characterized by elemental analysis, TG-DTG and IR spectrum. The enthalpy change of the reaction of complex formation from a solution of the reagents, Δr H m θ (sol), and the molar heat capacity of the complex, c m, were determined as being –19.1610.051 kJ mol–1 and 79.2641.218 J mol–1 K–1 at 298.15 K by using an RD-496 III heat conduction microcalorimeter. The enthalpy change of complex formation from the reaction of the reagents in the solid phase, Δr H m θ(s), was calculated as being (23.9810.339) kJ mol–1 on the basis of an appropriate thermochemical cycle and other auxiliary thermodynamic data. The thermodynamics of reaction of formation of the complex was investigated by the reaction in solution at the temperature range of 292.15–301.15 K. The constant-volume combustion energy of the complex, Δc U, was determined as being –16788.467.74 kJ mol–1 by an RBC-II type rotating-bomb calorimeter at 298.15 K. Its standard enthalpy of combustion, Δc H m θ, and standard enthalpy of formation, Δf H m θ, were calculated to be –16803.957.74 and –1115.428.94 kJ mol–1, respectively.\n\nRestricted access\n\nLow-temperature heat capacities and standard molar enthalpy of formation of aspirin\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: F. Xu, L.-X. Sun, Z.-C. Tan, J.-G. Liang, Y.-Y. Di, Q.-F. Tian, and T. Zhang\n\nAbstract\n\nMolar heat capacities (C p,m) of aspirin were precisely measured with a small sample precision automated adiabatic calorimeter over the temperature range from 78 to 383 K. No phase transition was observed in this temperature region. The polynomial function of C p,m vs. T was established in the light of the low-temperature heat capacity measurements and least square fitting method. The corresponding function is as follows: for 78 K≤T≤383 K, C p,m/J mol-1 K-1=19.086X 4+15.951X 3-5.2548X 2+90.192X+176.65, [X=(T-230.50/152.5)]. The thermodynamic functions on the base of the reference temperature of 298.15 K, {ΔH TH 298.15} and {S T-S 298.15}, were derived. Combustion energy of aspirin (Δc U m) was determined by static bomb combustion calorimeter. Enthalpy of combustion (Δc H o m) and enthalpy of formation (Δf H o m) were derived through Δc U m as - (3945.262.63) kJ mol-1 and - (736.411.30) kJ mol-1, respectively.\n\nRestricted access\n\nThermochemistry of the solid complex Gd(Et2dtc)3(phen)\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: S. Gao, S. Chen, G. Xie, G. Fan, and Q. Shi\n\nSummary A ternary solid complex Gd(Et2dtc)3(phen) has been obtained from reactions of sodium diethyldithiocarbamate (NaEt2dtc), 1,10-phenanthroline (phen) and hydrated gadolinium chloride in absolute ethanol. The title complex was described by chemical and elemental analyses, TG-DTG and IR spectrum. The enthalpy change of liquid-phase reaction of formation of the complex, Δr H Θ m(l), was determined as (-11.628±0.0204) kJ mol-1 at 298.15 K by a RD-496 III heat conduction microcalorimeter. The enthalpy change of the solid-phase reaction of formation of the complex, Δr H Θ m(s), was calculated as (145.306±0.519) kJ mol-1 on the basis of a designed thermochemical cycle. The thermodynamics of reaction of formation of the complex was investigated by changing the temperature of liquid-phase reaction. Fundamental parameters, the apparent reaction rate constant (k), the apparent activation energy (E), the pre-exponential constant (A), the reaction order (n), the activation enthalpy (Δr H Θ ), the activation entropy (Δr S Θ ), the activation free energy (Δr G Θ ) and the enthalpy (Δr H Θ ), were obtained by combination of the thermodynamic and kinetic equations for the reaction with the data of thermokinetic experiments. The constant-volume combustion energy of the complex, Δc U, was determined as (-18673.71±8.15) kJ mol-1 by a RBC-II rotating-bomb calorimeter at 298.15 K. Its standard enthalpy of combustion, Δc H Θ m, and standard enthalpy of formation, Δf H Θ m, were calculated to be (-18692.92±8.15) kJ mol-1 and (-51.28±9.17) kJ mol-1, respectively.\n\nRestricted access\n\nThermochemistry of europium and dithiocarbamate complex Eu(C5H8NS2)3(C12H8N2)\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: S. Chen, X. Meng, Q. Shuai, B. Jiao, S. Gao, and Q. Shi\n\nAbstract\n\nA solid complex Eu(C5H8NS2)3(C12H8N2) has been obtained from reaction of hydrous europium chloride with ammonium pyrrolidinedithiocarbamate (APDC) and 1,10-phenanthroline (o-phen⋅H2O) in absolute ethanol. IR spectrum of the complex indicated that Eu3+ in the complex coordinated with sulfur atoms from the APDC and nitrogen atoms from the o-phen. TG-DTG investigation provided the evidence that the title complex was decomposed into EuS. The enthalpy change of the reaction of formation of the complex in ethanol, Δr H m θ(l), as –22.2140.081 kJ mol–1, and the molar heat capacity of the complex, c m, as 61.6760.651 J mol–1 K–1, at 298.15 K were determined by an RD-496 III type microcalorimeter. The enthalpy change of the reaction of formation of the complex in solid, Δr H m θ(s), was calculated as 54.5270.314 kJ mol–1 through a thermochemistry cycle. Based on the thermodynamics and kinetics on the reaction of formation of the complex in ethanol at different temperatures, fundamental parameters, including the activation enthalpy (ΔH θ), the activation entropy (ΔS θ), the activation free energy (ΔG θ), the apparent reaction rate constant (k), the apparent activation energy (E), the pre-exponential constant (A) and the reaction order (n), were obtained. The constant-volume combustion energy of the complex, Δc U, was determined as –16937.889.79 kJ mol–1 by an RBC-II type rotating-bomb calorimeter at 298.15 K. Its standard enthalpy of combustion, Δc H m θ, and standard enthalpy of formation, Δf H m θ, were calculated to be –16953.379.79 and –1708.2310.69 kJ mol–1, respectively.\n\nRestricted access\n\nThermodynamics of sodium 5-methylisophthalic acid monohydrate and sodium isophthalic acid hemihydrate\n\nJournal of Thermal Analysis and Calorimetry\nAuthors: Xin Jin, Zhen Wang, San-Ping Chen, Zhu-Jun Wang, and Sheng-Li Gao\n\nThe constant-volume combustion energy of the compound was determined by an RBC-type II type precise rotating-bomb combustion calorimeter. The main experimental procedures were described previously [ 16 , 17 ]. The correct value of the heat exchange\n\nRestricted access" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Solve stuck at 243\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg124220] Re: Solve stuck at 243\n• From: DrMajorBob <btreat1 at austin.rr.com>\n• Date: Sat, 14 Jan 2012 02:54:30 -0500 (EST)\n• Delivered-to: [email protected]\n• References: <[email protected]>\n\n```I believe there are two such numbers less than 500,000:\n\nlimit = 500000;\nTiming[can =\nUnion@First@\nLast@Reap@\nDo[Sow[Prime@m + 2 k^2], {m, 2, PrimePi@limit}, {k, 1, Sqrt[(\nlimit - Prime@m)/2]}];\ncannot =\nComplement[Range[3, limit, 2], Prime@Range@PrimePi@limit, can]]\n\n{44.8573, {5777, 5993}}\n\nThat makes 5777 the smallest.\n\nThis also works, but it may be slower:\n\nlimit = 500000;\nTiming[can =\nUnion@Flatten@\nTable[Prime@m + 2 k^2, {m, 2, PrimePi@limit}, {k, 1, Sqrt[(\nlimit - Prime@m)/2]}];\ncannot =\nComplement[Range[3, limit, 2], Prime@Range@PrimePi@limit, can]]\n\n{48.9388, {5777, 5993}}\n\nBest of all is knowing how far to look:\n\nlimit = 6000;\nTiming[can =\nUnion@First@\nLast@Reap@\nDo[Sow[Prime@m + 2 k^2], {m, 2, PrimePi@limit}, {k, 1, Sqrt[(\nlimit - Prime@m)/2]}];\ncannot =\nComplement[Range[3, limit, 2], Prime@Range@PrimePi@limit, can]]\n\n{0.187722, {5777, 5993}}\n\nBobby\n\nOn Fri, 13 Jan 2012 03:53:15 -0600, Ralph Dratman\n<ralph.dratman at gmail.com> wrote:\n\n> Project Euclid asks, \"What is the smallest odd composite\n> that cannot be written as the sum of a prime and twice a\n> square?\"\n>\n> I tried the following equation, not really expecting it to\n> work:\n>\n> oddComposite == Prime[m] + 2 k^2\n>\n> Surprisingly, the above actually does work for all the odd\n> composite numbers through 237.\n>\n> solveInstance[oddComposite_] := Solve[{oddComposite ==\n> Prime[m] + 2*k^2, k > 0, m > 0}, {k, m}, Integers];\n> For[i = 9, i < 300, i = i + 2,\n> If[Not[PrimeQ[i]], Print[i,\": \", solveInstance[i]]]]\n>\n> 9: {{k->1,m->4}}\n> 15: {{k->1,m->6},{k->2,m->4}}\n> 21: {{k->1,m->8},{k->2,m->6},{k->3,m->2}}\n> 25: {{k->1,m->9},{k->2,m->7},{k->3,m->4}}\n> 27: {{k->2,m->8}}\n> 33: {{k->1,m->11}}\n> 35: {{k->3,m->7},{k->4,m->2}}\n> 39: {{k->1,m->12},{k->2,m->11},{k->4,m->4}}\n> 45: {{k->1,m->14},{k->2,m->12},{k->4,m->6}}\n> 49: {{k->1,m->15},{k->2,m->13},{k->3,m->11},{k->4,m->7}}\n> 51: {{k->2,m->14},{k->4,m->8}}\n>\n> - - - - - - snip - - - - - -\n>\n> 217: {{k->3,m->46},{k->5,m->39},{k->8,m->24},{k->10,m->7}}\n> 219: {{k->2,m->47},{k->10,m->8}}\n> 221: {{k->6,m->35},{k->9,m->17}}\n> 225: {{k->1,m->48},{k->4,m->44},{k->7,m->31},{k->8,m->25}}\n> 231:\n> {{k->1,m->50},{k->2,m->48},{k->4,m->46},{k->5,m->42},{k->8,m\n> ->27},{k->10,m->11}}\n> 235:\n> {{k->1,m->51},{k->2,m->49},{k->6,m->38},{k->7,m->33},{k->8,m\n> ->28},{k->9,m->21}}\n> 237: {{k->2,m->50},{k->7,m->34},{k->8,m->29},{k->10,m->12}}\n>\n> - - - - - - but then, at 243, something changes - - - - -\n>\n> 243: {{k->1,m->53},{k->4,m->47},{k->5,m->44},{k->10,m->14}}\n> Solve::nsmet: This system cannot be solved with the methods\n> available to Solve. >>\n>\n> 245: Solve[{245==2 k^2+Prime[m],k>0,m>0},{k,m},Integers]\n> Solve::nsmet: This system cannot be solved with the methods\n> available to Solve. >>\n>\n> 247: Solve[{247==2 k^2+Prime[m],k>0,m>0},{k,m},Integers]\n> Solve::nsmet: This system cannot be solved with the methods\n> available to Solve. >>\n>\n> ... and so on. Strange.\n>\n> Does anyone know why such a threshold might appear?\n>\n> Thank you.\n>\n> Ralph Dratman\n>\n\n--\nDrMajorBob at yahoo.com\n\n```\n\n• Prev by Date: Re: Solve stuck at 243\n• Next by Date: Re: Solve stuck at 243\n• Previous by thread: Re: Solve stuck at 243\n• Next by thread: Re: Solve stuck at 243" ]

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[ "## Non-leaf Polymorphic Table Functions (PTF)\n\n### Table Semantics\n\nApplies to: Oracle 18c, Tested against: Oracle 18.1\n\nSo far in this tutorial we have only explored \"Row Semantics\" polymorphic table functions, i.e. those defined with a PIPELINED ROW clause. In this module we explore \"Table Semantics\" polymorphic table functions, i.e. those defined with a PIPELINED TABLE clause. Table Semantics PTFs allow us to specify PARTITION BY and ORDER BY clauses (like those available with Oracle analytic functions) when invoking a PTF.\n\nLet's create both types of functions and take them for a spin.\n\n```create or replace package ptf_demo as\n\nfunction describe( p_input_table_metadata in out dbms_tf.table_t )\nreturn dbms_tf.describe_t ;\n\n-- this is the Table Semantics PTF\n\nfunction table_semantics( p_table table )\nreturn table PIPELINED TABLE polymorphic using ptf_demo ;\n\n-- we'll also create a row semantics version for comparison\n\nfunction row_semantics( p_table table )\nreturn table PIPELINED ROW polymorphic using ptf_demo ;\n\nend ptf_demo ;\n/\n\ncreate or replace package body ptf_demo as\n\nfunction describe( p_input_table_metadata in out dbms_tf.table_t )\nreturn dbms_tf.describe_t as\nbegin\nreturn null;\nend describe ;\n\nend ptf_demo;\n/\n```\n\nNow we run both PTFs to see what they return.\n\n```select deptno, ename\nfrom ptf_demo.row_semantics( scott.emp );```\n``` DEPTNO ENAME\n---------- ----------\n10 KING\n10 MILLER\n10 CLARK\n20 SMITH\n20 JONES\n20 FORD\n20 SCOTT\n30 ALLEN\n30 WARD\n30 MARTIN\n30 BLAKE\n30 TURNER\n30 JAMES```\n```select deptno, ename\nfrom ptf_demo.table_semantics( scott.emp );```\n``` DEPTNO ENAME\n---------- ----------\n10 KING\n10 MILLER\n10 CLARK\n20 SMITH\n20 JONES\n20 FORD\n20 SCOTT\n30 ALLEN\n30 WARD\n30 MARTIN\n30 BLAKE\n30 TURNER\n30 JAMES```\n\nNote how the output of both SELECT commands is the same.\n\nLet's try using PARTITION BY and ORDER BY clauses with ROW_SEMANTICS().\n\n```select deptno, ename\nfrom ptf_demo.row_semantics( scott.emp PARTITION BY DEPTNO ORDER BY ENAME )\norder by deptno, ename asc ;```\n```Error starting at line : 1 in command -\nselect deptno, ename\nfrom ptf_demo.row_semantics( scott.emp PARTITION BY DEPTNO ORDER BY ENAME )\norder by deptno, ename asc\nError at Command Line : 2 Column : 55\nError report -\nSQL Error: ORA-62568: invalid use of PARTITION BY or ORDER BY clause```\n\nIt does not work with ROW_SEMANTICS(). We will have better luck with TABLE_SEMANTICS().\n\n```select deptno, ename\nfrom ptf_demo.table_semantics( scott.emp PARTITION BY DEPTNO ORDER BY ENAME DESC )\norder by deptno, ename asc ;```\n``` DEPTNO ENAME\n---------- ----------\n10 CLARK\n10 KING\n10 MILLER\n20 FORD\n20 JONES\n20 SCOTT\n20 SMITH\n30 ALLEN\n30 BLAKE\n30 JAMES\n30 MARTIN\n30 TURNER\n30 WARD```\n\nIn the absence of any useful FETCH_ROWS() logic the output is unremarkable however. Let's continue exploring.\n\n#### PARTITION BY\n\nTo better understand how the PARTITION BY clause affects execution we take a look at what happens internally.\n\n```create or replace package ptf_demo as\n\nfunction describe( p_input_table_metadata in out dbms_tf.table_t )\nreturn dbms_tf.describe_t ;\n\nprocedure fetch_rows ;\n\nfunction table_semantics( p_table table )\nreturn table PIPELINED TABLE polymorphic using ptf_demo ;\n\nfunction row_semantics( p_table table )\nreturn table PIPELINED ROW polymorphic using ptf_demo ;\n\nend ptf_demo ;\n/\n\ncreate or replace package body ptf_demo as\n\nfunction describe(\nreturn\ndbms_tf.describe_t\nas\n\nv_new_column_desc dbms_tf.describe_t ;\n\nbegin\n\n--\n-- make ENAME available to get_row_set()\n--\n\n--\n-- create a new output column\n--\nv_new_column_desc.new_columns(1).name := 'INTROSPECTION' ;\nv_new_column_desc.new_columns(1).type := dbms_tf.type_varchar2 ;\n\nreturn v_new_column_desc ;\n\nend describe ;\n\nprocedure fetch_rows as\n\nv_input_column dbms_tf.tab_varchar2_t ;\nv_output_column dbms_tf.tab_varchar2_t ;\n\ni pls_integer ;\n\nbegin\n\ndbms_tf.get_col( 1, v_input_column );\n\nfor i in 1 .. dbms_tf.get_env().row_count loop\n\nv_output_column(i) :=\n'Row Count = ' || dbms_tf.get_env().row_count || ', ' ||\n'First ENAME in the row set = ' || v_input_column(1) ;\n\nend loop ;\n\ndbms_tf.put_col( 1, v_output_column );\n\nend fetch_rows ;\n\nend ptf_demo ;\n/```\n```select deptno, ename, introspection\nfrom ptf_demo.row_semantics( scott.emp ) ;```\n``` DEPTNO ENAME INTROSPECTION\n---------- ---------- ------------------------------------------------------------\n10 KING Row Count = 14, First ENAME in the row set = KING\n10 MILLER Row Count = 14, First ENAME in the row set = KING\n10 CLARK Row Count = 14, First ENAME in the row set = KING\n20 SMITH Row Count = 14, First ENAME in the row set = KING\n20 JONES Row Count = 14, First ENAME in the row set = KING\n20 ADAMS Row Count = 14, First ENAME in the row set = KING\n20 FORD Row Count = 14, First ENAME in the row set = KING\n20 SCOTT Row Count = 14, First ENAME in the row set = KING\n30 ALLEN Row Count = 14, First ENAME in the row set = KING\n30 WARD Row Count = 14, First ENAME in the row set = KING\n30 MARTIN Row Count = 14, First ENAME in the row set = KING\n30 BLAKE Row Count = 14, First ENAME in the row set = KING\n30 TURNER Row Count = 14, First ENAME in the row set = KING\n30 JAMES Row Count = 14, First ENAME in the row set = KING\n```\n\nThe output above tells us that FETCH_ROWS() is only called once when executing ROW_SEMANTICS().\n\n```select deptno, ename, introspection\nfrom ptf_demo.table_semantics( scott.emp ) ;```\n``` DEPTNO ENAME INTROSPECTION\n---------- ---------- ------------------------------------------------------------\n10 KING Row Count = 14, First ENAME in the row set = KING\n10 MILLER Row Count = 14, First ENAME in the row set = KING\n10 CLARK Row Count = 14, First ENAME in the row set = KING\n20 SMITH Row Count = 14, First ENAME in the row set = KING\n20 JONES Row Count = 14, First ENAME in the row set = KING\n20 ADAMS Row Count = 14, First ENAME in the row set = KING\n20 FORD Row Count = 14, First ENAME in the row set = KING\n20 SCOTT Row Count = 14, First ENAME in the row set = KING\n30 ALLEN Row Count = 14, First ENAME in the row set = KING\n30 WARD Row Count = 14, First ENAME in the row set = KING\n30 MARTIN Row Count = 14, First ENAME in the row set = KING\n30 BLAKE Row Count = 14, First ENAME in the row set = KING\n30 TURNER Row Count = 14, First ENAME in the row set = KING\n30 JAMES Row Count = 14, First ENAME in the row set = KING\n```\n\nThe query above tells us FETCH_ROWS() is also only called once when executing TABLE_SEMANTICS() with no PARTITION BY clause.\n\n```select deptno, ename, introspection\nfrom ptf_demo.table_semantics( scott.emp PARTITION BY DEPTNO )\norder by deptno, ename ;```\n``` DEPTNO ENAME INTROSPECTION\n---------- ---------- ------------------------------------------------------------\n10 CLARK Row Count = 3, First ENAME in the row set = CLARK\n10 KING Row Count = 3, First ENAME in the row set = CLARK\n10 MILLER Row Count = 3, First ENAME in the row set = CLARK\n20 ADAMS Row Count = 5, First ENAME in the row set = ADAMS\n20 FORD Row Count = 5, First ENAME in the row set = ADAMS\n20 JONES Row Count = 5, First ENAME in the row set = ADAMS\n20 SCOTT Row Count = 5, First ENAME in the row set = ADAMS\n20 SMITH Row Count = 5, First ENAME in the row set = ADAMS\n30 ALLEN Row Count = 6, First ENAME in the row set = ALLEN\n30 BLAKE Row Count = 6, First ENAME in the row set = ALLEN\n30 JAMES Row Count = 6, First ENAME in the row set = ALLEN\n30 MARTIN Row Count = 6, First ENAME in the row set = ALLEN\n30 TURNER Row Count = 6, First ENAME in the row set = ALLEN\n30 WARD Row Count = 6, First ENAME in the row set = ALLEN```\n\nThe third query reveals that adding a PARTITION BY clause causes FETCH_ROWS() to be called once per distinct partition key value.\n\n#### ORDER BY\n\nAs with analytic functions the ORDER BY clause in the PTF table argument controls the input row set order. The SELECT command's ORDER BY clause controls the output row set order. The following example demonstrates both clauses.\n\n```select deptno, ename, introspection\nfrom ptf_demo.table_semantics( scott.emp partition by deptno ORDER BY ENAME DESC )\nORDER BY DEPTNO, ENAME ASC ;\n```\n``` DEPTNO ENAME INTROSPECTION\n---------- ---------- ------------------------------------------------------------\n10 CLARK Row Count = 3, First ENAME in the row set = MILLER\n10 KING Row Count = 3, First ENAME in the row set = MILLER\n10 MILLER Row Count = 3, First ENAME in the row set = MILLER\n20 ADAMS Row Count = 5, First ENAME in the row set = SMITH\n20 FORD Row Count = 5, First ENAME in the row set = SMITH\n20 JONES Row Count = 5, First ENAME in the row set = SMITH\n20 SCOTT Row Count = 5, First ENAME in the row set = SMITH\n20 SMITH Row Count = 5, First ENAME in the row set = SMITH\n30 ALLEN Row Count = 6, First ENAME in the row set = WARD\n30 BLAKE Row Count = 6, First ENAME in the row set = WARD\n30 JAMES Row Count = 6, First ENAME in the row set = WARD\n30 MARTIN Row Count = 6, First ENAME in the row set = WARD\n30 TURNER Row Count = 6, First ENAME in the row set = WARD\n30 WARD Row Count = 6, First ENAME in the row set = WARD```\n\n#### Replication\n\nThe fact that FETCH_ROWS() executes once per distinct partition key value means we can set the replication factor to a different value for each partition, as demonstrated next.\n\n```create or replace package body ptf_demo as\n\nfunction describe(\nreturn\ndbms_tf.describe_t\nas\n\nv_new_column_desc dbms_tf.describe_t ;\n\nbegin\n\n--\n-- make DEPTNO available to get_row_set()\n--\n\n--\n-- create a new output column\n--\nv_new_column_desc.new_columns(1).name := 'INTROSPECTION' ;\nv_new_column_desc.new_columns(1).type := dbms_tf.type_varchar2 ;\n\nv_new_column_desc.row_replication := true ;\n\nreturn v_new_column_desc ;\n\nend describe ;\n\nprocedure fetch_rows as\n\nv_deptno dbms_tf.tab_number_t ;\nv_introspection dbms_tf.tab_varchar2_t ;\n\nv_input_row_number pls_integer ;\nv_output_row_number pls_integer := 1 ;\n\nv_replication_factor naturaln := 1 ;\n\nbegin\n\ndbms_tf.get_col( 1, v_deptno );\n\n--\n-- let's replicate once for each DEPTNO 10 row,\n-- twice for each DEPTNO 20 row, and three times for\n-- each DEPTNO 30 row\n--\n\nv_replication_factor := v_deptno(1) / 10 ;\n\n--\n-- assemble the output rows\n--\nv_input_row_number := v_deptno.first ;\n\nwhile v_input_row_number is not null loop\n\nfor v_copy_number in 1 .. v_replication_factor loop\n\nv_introspection(v_output_row_number) :=\n'Replication factor = ' || v_replication_factor ;\n\nv_output_row_number := v_output_row_number + 1 ;\n\nend loop ;\n\nv_input_row_number := v_deptno.next(v_input_row_number) ;\n\nend loop ;\n\ndbms_tf.row_replication( v_replication_factor ) ;\n\ndbms_tf.put_col( 1, v_introspection );\n\nend fetch_rows ;\n\nend ptf_demo ;\n/\n\nselect deptno, ename, introspection\nfrom ptf_demo.table_semantics( scott.emp partition by deptno order by ename )\norder by deptno, ename ;\n```\n``` DEPTNO ENAME INTROSPECTION\n---------- ---------- ------------------------------------------------------------\n10 CLARK Replication factor = 1\n10 KING Replication factor = 1\n10 MILLER Replication factor = 1\n20 ADAMS Replication factor = 2\n20 ADAMS Replication factor = 2\n20 FORD Replication factor = 2\n20 FORD Replication factor = 2\n20 JONES Replication factor = 2\n20 JONES Replication factor = 2\n20 SCOTT Replication factor = 2\n20 SCOTT Replication factor = 2\n20 SMITH Replication factor = 2\n20 SMITH Replication factor = 2\n30 ALLEN Replication factor = 3\n30 ALLEN Replication factor = 3\n30 ALLEN Replication factor = 3\n30 BLAKE Replication factor = 3\n30 BLAKE Replication factor = 3\n30 BLAKE Replication factor = 3\n30 JAMES Replication factor = 3\n30 JAMES Replication factor = 3\n30 JAMES Replication factor = 3\n30 MARTIN Replication factor = 3\n30 MARTIN Replication factor = 3\n30 MARTIN Replication factor = 3\n30 TURNER Replication factor = 3\n30 TURNER Replication factor = 3\n30 TURNER Replication factor = 3\n30 WARD Replication factor = 3\n30 WARD Replication factor = 3\n30 WARD Replication factor = 3```\n\nBeware of bug 28165108. Code that encounters a replication factor of 0 may produce incorrect output until this bug is fixed.\n\n### Revision Notes\n\nDate Category Note" ]

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[ "Courses\nCourses for Kids\nFree study material\nOffline Centres\nMore\n\n# Multiplication and Division", null, "Last updated date: 28th Nov 2023\nTotal views: 331.2k\nViews today: 4.31k", null, "", null, "", null, "", null, "", null, "## Multiplying and Dividing Decimals\n\nThe multiplication division process is a simple step-by-step format used daily in all fields of Mathematics. The multiplication and division of decimals represent the fraction of a number and taking ten as the base of the decimal system of numbers. Point differentiates or decimal notations is an integer part extracted from the part of a fraction. Decimals can undergo addition, subtraction, multiplication, and division. Multiplication and division processes are easier to use with decimals as compared to addition and subtraction. These functions are normally used in computational platforms and tasks.\n\n### Multiplication of Decimals\n\nIn normal multiplication, the numbers indicate also the addition of the number 20 twice. Similarly, this can be applied to decimals as well. Taking the example of 0.22 x 2 = 0.22 + 0.22, the multiplication of whole numbers and the multiplication of decimal numbers go hand-in-hand.  Some simple steps used in the multiplication of decimals are provided below with an example.\n\n## Example of Multiplying Decimals\n\nLet’s take the following example to understand how multiplication of decimals is done.\n\nLet us consider the multiplication of two numbers, 4.42 and 2.\n\n### Step 1:\n\nCount the total number of digits to the right of the decimal point in both numbers provided. We notice that in 4.42, to the right of the decimal point there are two digits and 4 is a whole number. Therefore, the total number of digits to the right of the decimal is 2.\n\n### Step 2:\n\nWithout considering the decimal point,  multiply the numbers found.\n\n442 x 2 = 884\n\n### Step 3:\n\nAfter multiplication, add the decimal point two places to the right of the answer calculated.\n\n## Division of Decimals\n\nThe division of decimals is almost similar to the multiplication of decimals. When dividing a decimal directly, it can be confusing where to place the decimal point. But, with a little practice, one can use the same trick of multiplying decimals with ease. When dividing the decimal as a numerator with a whole number as a denominator of two given numbers, it is easier to obtain a result.\n\n## Examples of Dividing Decimals By Decimals\n\nThere are two methods used to divide a given set of decimal numbers. Let us consider the numbers given below, to understand the mechanism of dividing decimals.\n\nDivide the following decimals 0.398 by 0.20.\n\n### Method 1:\n\nConversion of the decimals to whole numbers by multiplying each with a common factor to make the denominator one. It is important to note that the denominator has to always be a whole number for the division to take place.\n\n0.398 ÷ 0.20 = $\\frac{0.398}{0.20}$ = $\\frac{0.398 \\times 5}{0.20 \\times 5}$ = $\\frac{1.99}{1}$ = 1.99\n\n(Multiply both the numerator and denominator by 5)\n\n### Method 2:\n\nAnother method to use is to convert the decimal numbers into whole numbers. This can be done by multiplying with numbers having powers of 10 (10, 100, 1000, etc.).\n\nLet us take the above example and count the number of digits right to the decimal point in the denominator.\n\nThe denominator is 0.20 and the number of digits right to the decimal point is 2.\n\n0.398 ÷ 0.20 = $\\frac{0.398}{0.20}$\n\n• The power of 10 is taken common depending on the number of digits present to the right of the decimal point.( 102 = 100)\n\n• Multiply both numerator and denominator by 100.\n\n0.398 ÷ 0.20 = $\\frac{0.398 \\times 100}{0.20 \\times 100}$ = $\\frac{39.8}{20}$ = 1.99\n\n### Multiplying and Dividing Decimals Using 10, 100 and 1000\n\nWe have read that decimals are a form of expression fractions having their base as 10. Let us consider a couple of examples with fractions of decimals having the base of 100 and 1000.  Some rules applied for multiplication and division of decimal numbers by 10, 100, and 1000 are as follows:\n\n Arithmetic Operation Rule Example Multiply by 10(101) The numerical moves one place to the left 8.37 x 10 = 83.7 Multiply by 100  (102) The numerical moves two places to the left 8.37 x 100 = 837 Multiply by 1000 (103) The numerical moves three places to the left 8.37 x 1000 = 8370 Divide by 10(101) The numerical moves one place to the right 8.37 ÷ 10 = 0. 837 Divide by 100(102) The numerical moves two places to the right 8.37 ÷  100 = 0.0837 Divide by 1000(103) The numerical moves three places to the right 8.37 ÷  1000 = 0.00837\n\n## FAQs on Multiplication and Division\n\n1. What are Some of the Applications of Multiplying and Dividing Decimals?\n\nAns: The basic and most simplistic methods of solving numerical problems involve addition, subtraction, multiplication, and division. Both multiplication and division are used in our daily lives in a variety of scenarios. They also help us solve complex numerical problems involving algebra and calculus. More than addition and subtraction, multiplication and division are easier to work with and obtain an accurate result.\n\nThe concepts can be taught in a variety of ways using illustrations, drawings, simulations of certain scenarios, and the usual pen to paper method. One uses a variety of numbers like whole numbers, integers, fractions, and so on to represent various events and results. Therefore, multiplication and division aid to assemble and breakdown complex results in simplistic ones.\n\n2. What is the Difference Between Long Division and Short Division in Terms of Using Decimals?\n\nAns: Division by whole numbers or decimals can be done with ease provided one has practised tremendously various problems regarding the two types of numerical as well as the methods that can be used. In the cases of using long or short division, both methods yield the same result. The question arises only in terms of speed. It is known that short division would yield faster results than long division but the long division is known for its accuracy.\n\nThe choice of using any of these two division methods lies with the questions or data provided. In terms of dividing by decimals, the short division method is faster and accurate. The placement of the decimal point is the key factor to take note of." ]

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[ "Before understanding the wave motion, it is important to understand what is a wave. Wave is defined as the disturbance which transfers energy through matter or space. It consists of oscillations or vibrations. Know Characteristics Of Wave Motion in detail here.\n\n## What are the types of waves?\n\nThere are two fundamental types of waves:\n\n• Longitudinal waves: These waves consist of compressions and rarefactions and the particles are displaced parallel to the direction in which the wave travels. The waves observed in a slinky and sound wave in the air are examples of longitudinal waves.\n• Transverse waves: These waves particles are displaced perpendicular to the direction in which the wave travels. A wave on a string and a ripple on a pond are examples of transverse waves.\n\n## What is wave motion?\n\nWave motion is defined as the propagation of disturbances or vibrations from the equilibrium with a constant velocity. The understanding of wave motion began with the study of sound and vacuums.\n\n## What are the factors of wave motion?\n\nIrrespective of the types of waves, the factors of transverse and longitudinal waves are the same. Following are the list of factors of wave motions:\n\n• Frequency: Frequency(f) of a wave is defined as the number of complete cycles taken by the waves in a specific period of time. As the frequency increases, the number of compressions and expansions in the waves increases. It is measured in Hertz and mathematically it is given as f=1/T\n• Time period: The time period (T) of a wave is the time required to produce one complete cycle.\n• Amplitude: The amplitude (A) of a wave is the maximum displacement of the vibrating particles. It is measured in meter (m).\n• Wavelength: Wavelength (λ) of a wave is the distance between two points in a wave that is close to each other and in the same phase. It is denoted by λ.\n\n## What are the characteristics of wave motion?\n\nFollowing are the list of characteristics of wave motion;\n\n• There is no transfer of matter through the medium, only the energy is transferred.\n• The velocity at which the wave propagates is the same throughout the medium and is dependent on the nature of the medium.\n• The medium must possess elasticity and inertia for the propagation of the waves.\n• The vibration of the particles takes place at their equilibrium position.\n• The particles possess kinetic energy as well as potential energy at their mean position and at extreme positions respectively.\n\nInterested to learn more about what causes wave motion and other Physics related concepts like Simple Harmonic Motion, unit of magnetic flux, types of optical fibresetc, visit BYJU’S." ]

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