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https://www.netmath.ca/en/blog | 1,675,748,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500384.17/warc/CC-MAIN-20230207035749-20230207065749-00791.warc.gz | 876,533,316 | 76,879 | # Blog
## The ideal place to discover educational articles on math, free activities, Netmath news and lots more.
Netmath
01.07.2023 | Scolab
##### 5 reasons to use Netmath
The challenge Netmath has set for itself is to instill a love of math in students.
Free resources and activities Teachers’ Trove
01.05.2023 | Scolab
##### We’re renovating!
Your students will have to measure lengths and areas, and round and multiply natural and decimal numbers to determine if the project is possible.
Taking it further
12.08.2022 | Scolab
##### My beautiful tree!
Work on the logic with your students and help Sonya improve her decorations.
Free resources and activities Teachers’ Trove
11.13.2022 | Scolab
##### A visit to the zoo
Your students will have to find the path that takes them to the animals. How? By simplifying fractions!
Taking it further
10.11.2022 | Scolab
##### A Pair of Monsters
Special Halloween challenge: will you be able to solve this puzzle before the monsters enter the city?
Free resources and activities Teachers’ Trove
10.06.2022 | Scolab
##### Motion detector lights and zones
Our Math Maestro Louis has a very interesting activity in which the students draw and then measure the angles of a detection zone created by five motion detector lights.
Free resources and activities Teachers’ Trove
09.19.2022 | Scolab
##### Teachers’ Trove
Our team of Math Maestros will be offering you short, simple and concrete activities that will fit nicely into your planning. Key resources to spice up your math classes!
Free resources and activities Teachers’ Trove
09.19.2022 | Scolab
##### The bike lock
Help Ms. Karine find the combination to her bike lock while revisiting the concepts of addition with carrying and subtraction with borrowing.
Collections Netmath
08.11.2022 | Scolab
##### How to save time using collections
Organizing Netmath activities to suit your teaching—a little like creating a playlist—is simple with the collections!
Free resources and activities
06.08.2022 | Scolab
##### Math in colors – free image to color in (grades 1 to 4)
Celebrate the beautiful weather with a fun math activity that students in grades 1 to 4 will love: an image to color in! | 518 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.868191 |
https://how2calc.com/posts/how-to-calculate-minutes-into-hours-in-excel/ | 1,701,753,077,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00446.warc.gz | 363,584,265 | 18,296 | # How to calculate minutes into hours in excel
Learn how to convert minutes into hours using Excel with these easy-to-follow steps. This guide will teach you how to convert minutes into decimal hours or regular hours and minutes format. Excel, minutes, hours, convert, decimal, format, formula, calculation
## How to Calculate Minutes into Hours in Excel
If you need to convert minutes into hours in Excel, you’re in luck. Excel makes it easy to perform this conversion with just a few simple steps. In this guide, we’ll show you how to convert minutes into decimal hours or regular hours and minutes format.
### Getting Started
To get started, open an Excel spreadsheet and enter the minutes you want to convert into a cell. For this example, we’ll use the value 120.
Next, create a new cell where you want the result to appear. This is where you’ll enter the formula to convert the minutes into hours.
### Converting Minutes into Decimal Hours
To convert minutes into decimal hours, divide the minutes by 60. For example, if you want to convert 120 minutes into decimal hours, enter =120/60 in the new cell.
Press enter to see the result in decimal format. In our example, 120 minutes would be equal to 2 hours.
### Converting Minutes into Regular Hours and Minutes Format
To convert minutes into regular hours and minutes format, use the following formula: =INT(A1/60)":"&ROUNDUP(MOD(A1,60),0).
In the formula, replace A1 with the cell containing the minutes you want to convert. The INT function rounds down the result of the division to get the number of hours. The ":" symbol adds the colon between hours and minutes. The MOD function calculates the remainder when the minutes are divided by 60. The ROUNDUP function rounds up the result of the MOD function to the nearest whole number.
Press enter to see the result in regular hours and minutes format. In our example, 120 minutes would be equal to 2:00.
### Adding a Label to the Result Cell
You can also add a label to the result cell to indicate whether the value is in decimal or regular hours and minutes format.
To add a label, right-click on the cell and select "format cells." In the format cells dialog box, select the Number tab. Under Category, select "Custom." Under Type, enter the code "[h]:mm" for regular hours and minutes format, or "0.00" for decimal format. Click OK to apply the formatting.
Your result cell will now display the converted value with your chosen formatting.
### Converting Larger Numbers of Minutes
You can also use the same formulas to convert larger numbers of minutes by simply replacing the cell reference in the formula. For example, if you wanted to convert 720 minutes into hours, you would use the formula =INT(A1/60)":"&ROUNDUP(MOD(A1,60),0) and replace A1 with 720.
Excel makes it easy to convert minutes into hours in just a few simple steps. With these easy-to-follow instructions, you’ll be able to convert minutes into decimal or regular hours and minutes format quickly and easily.
Remember to format your result cell appropriately to display your conversion in the desired format. With a little practice, you’ll be converting minutes into hours like a pro in no time.
Older post | 692 | 3,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-50 | latest | en | 0.831451 |
http://de.metamath.org/mpeuni/gbepos.html | 1,653,239,710,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545875.39/warc/CC-MAIN-20220522160113-20220522190113-00286.warc.gz | 16,808,578 | 5,407 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > gbepos Structured version Visualization version GIF version
Theorem gbepos 40180
Description: Any even Goldbach number is positive. (Contributed by AV, 20-Jul-2020.)
Assertion
Ref Expression
gbepos (𝑍 ∈ GoldbachEven → 𝑍 ∈ ℕ)
Proof of Theorem gbepos
Dummy variables 𝑝 𝑞 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 isgbe 40173 . 2 (𝑍 ∈ GoldbachEven ↔ (𝑍 ∈ Even ∧ ∃𝑝 ∈ ℙ ∃𝑞 ∈ ℙ (𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞))))
2 prmnn 15226 . . . . . . . . 9 (𝑝 ∈ ℙ → 𝑝 ∈ ℕ)
3 prmnn 15226 . . . . . . . . 9 (𝑞 ∈ ℙ → 𝑞 ∈ ℕ)
4 nnaddcl 10919 . . . . . . . . 9 ((𝑝 ∈ ℕ ∧ 𝑞 ∈ ℕ) → (𝑝 + 𝑞) ∈ ℕ)
52, 3, 4syl2an 493 . . . . . . . 8 ((𝑝 ∈ ℙ ∧ 𝑞 ∈ ℙ) → (𝑝 + 𝑞) ∈ ℕ)
6 eleq1 2676 . . . . . . . 8 (𝑍 = (𝑝 + 𝑞) → (𝑍 ∈ ℕ ↔ (𝑝 + 𝑞) ∈ ℕ))
75, 6syl5ibr 235 . . . . . . 7 (𝑍 = (𝑝 + 𝑞) → ((𝑝 ∈ ℙ ∧ 𝑞 ∈ ℙ) → 𝑍 ∈ ℕ))
873ad2ant3 1077 . . . . . 6 ((𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞)) → ((𝑝 ∈ ℙ ∧ 𝑞 ∈ ℙ) → 𝑍 ∈ ℕ))
98com12 32 . . . . 5 ((𝑝 ∈ ℙ ∧ 𝑞 ∈ ℙ) → ((𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞)) → 𝑍 ∈ ℕ))
109a1i 11 . . . 4 (𝑍 ∈ Even → ((𝑝 ∈ ℙ ∧ 𝑞 ∈ ℙ) → ((𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞)) → 𝑍 ∈ ℕ)))
1110rexlimdvv 3019 . . 3 (𝑍 ∈ Even → (∃𝑝 ∈ ℙ ∃𝑞 ∈ ℙ (𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞)) → 𝑍 ∈ ℕ))
1211imp 444 . 2 ((𝑍 ∈ Even ∧ ∃𝑝 ∈ ℙ ∃𝑞 ∈ ℙ (𝑝 ∈ Odd ∧ 𝑞 ∈ Odd ∧ 𝑍 = (𝑝 + 𝑞))) → 𝑍 ∈ ℕ)
131, 12sylbi 206 1 (𝑍 ∈ GoldbachEven → 𝑍 ∈ ℕ)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ∃wrex 2897 (class class class)co 6549 + caddc 9818 ℕcn 10897 ℙcprime 15223 Even ceven 40075 Odd codd 40076 GoldbachEven cgbe 40167 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-addass 9880 ax-i2m1 9883 ax-1ne0 9884 ax-rrecex 9887 ax-cnre 9888 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-ov 6552 df-om 6958 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-nn 10898 df-prm 15224 df-gbe 40170 This theorem is referenced by: gbege6 40187
Copyright terms: Public domain W3C validator | 1,838 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-21 | longest | en | 0.123316 |
https://dzone.com/articles/how-to-handle-100k-rows-decision-table-in-drools-3 | 1,619,133,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00442.warc.gz | 328,509,987 | 34,132 | How To Handle 100k Rows Decision Table in Drools (Part 3)
In this article, I created a prototype to demonstrate how to handle large rows in a decision table with reasonable performance. This is part 3 of the series.
· Performance Zone · Analysis
Save
3.41K Views
As described in the previous article, we are facing a very challenging performance issue when solving 100k row decision tables.
To view the previous discussion, please click:
Solution 1: Rule Template + XLS
Solution 2: Precompile Spreadsheet Decision Table
In this article, I am going to change the mindset and change the decision table row data from rule to fact. Again, I am using Drools as my framework.
Solution 3: Convert Row Data to Fact, Not Rule
Solution 2 (Precompile the rules) helps a lot when the decision table is 10k below row size. However, when the row number comes to very big such as 100k, solution 2's improvement is neglectable. What’s worse is that the compilation time is too long, this is because it’s not a normal task to compile 100k Java files in a project.
By its design nature, the decision table is popular since it allows you to manage one rule in one row so that you can manage multiple rows in an easy manner. However, the watch that floats the boat also could swallow it.
Could we somehow reduce the rule numbers if the rule numbers are just too big?
Could we read the Excel file in RAW format, insert those row data as Facts and then compute them in memory? Then you have 100k Facts instead of 100k rules in memory.
Let's try this:
DRL
Java
x
1
`rule "Check Client Object"`
2
` when`
3
` \$k:Keyword(\$v : value)`
4
` \$c:ClientObject( descr matches \$v )`
5
` then`
6
` \$c.setPass(\$k.isResult());`
7
` System.out.println("Check Client Object fired");`
8
`end`
Drools Model
Java
`xxxxxxxxxx`
1
1
`public class Keyword {`
2
` private java.lang.String value;`
3
4
` private boolean result;`
5
`}`
Insert Row as Fact in Java
Java
`xxxxxxxxxx`
1
1
`ClientObject o1 = new ClientObject();`
2
`for(int i=0;i<100000;i++){`
3
` ksession.insert(new Keyword("DangerObjectxxx", false);`
4
`}`
5
`o1.setDescr("999");`
6
`int fired = ksession.fireAllRules();`
With a simple test on the above code, it works, and the performance is <50 ms!
However, we don’t throw away the bathwater with the child. We don’t want to hardcode the Excel Read and Parser in our generic application code. The business logic needs to be decoupled with a generic application.
Could we load Excel data in a DRL file and insert the fact?
The answer is yes!
DRL is very dynamic and flexible, basically, you can do it just like you are writing Java code.
However, in order to simplify the DRL editing, usually, we could provide a clean utility to read Excel files and convert them to a Customized Fact to serve your purpose.
Improved DRL
Java
`xxxxxxxxxx`
1
20
1
`rule "Load keyword"`
2
` when`
3
` \$kr:KeywordReader(excelFile matches ".*.xls")`
4
` then`
5
` System.out.println("Load keyword rule fired");`
6
` List<Map<String, String>> list = \$kr.getKwList();`
7
` for ( Map<String, String> m : list){`
8
` Keyword k = new Keyword(m.get("1"), Boolean.valueOf(m.get("2")));`
9
` insert(k);`
10
` }`
11
`end`
12
13
`rule "Check Client Object"`
14
` when`
15
` \$k:Keyword(\$v : value)`
16
` \$c:ClientObject( descr matches \$v )`
17
` then`
18
` \$c.setPass(\$k.isResult());`
19
` System.out.println("Check Client Object fired");`
20
`end`
I also provided a simple Excel utility in my branch, See KeywordReader.java, we can add the Excel parse in the Drools Model Getter and Setter.
Java
`xxxxxxxxxx`
1
22
1
`public class KeywordReader {`
2
` private String excelFile;`
3
` private List<Map<String, String>> kwList;`
4
5
` public KeywordReader(String excelFile) {`
6
` this.excelFile = excelFile;`
7
` kwList= new ArrayList<Map<String, String>>();`
8
` }`
9
10
` public String getExcelFile() {`
11
` return excelFile;`
12
` }`
13
14
` public List<Map<String,String>> getKwList() {`
15
` if(this.excelFile.isEmpty())return null;`
16
` parseFile(getClass().getClassLoader().getResourceAsStream(excelFile));`
17
` return kwList;`
18
` }`
19
` `
20
` private void parseFile( InputStream inStream ) {`
21
` //Parser Excel format and conver them into HashMapList`
22
` }`
How to parse Excel file:
Java
`xxxxxxxxxx`
1
22
1
`Sheet sheet = workbook.getSheetAt( 0 );`
2
`int maxRows = sheet.getLastRowNum();`
3
`DataFormatter formatter = new DataFormatter( Locale.ENGLISH );`
4
`FormulaEvaluator formulaEvaluator = sheet.getWorkbook().getCreationHelper().createFormulaEvaluator();`
5
` //define start row in rowFrom variable`
6
` for ( int i = rowFrom; i <= maxRows; i++ ) {`
7
` Row row = sheet.getRow( i );`
8
` int lastCellNum = row != null ? row.getLastCellNum() : 0;`
9
` `
10
` Map<String, String> rowData= new HashMap<String, String>();`
11
` int index =0;`
12
` //define start column in colFrom variable`
13
` for ( int cellNum = colFrom; cellNum < lastCellNum; cellNum++ ) {`
14
` Cell cell = row.getCell( cellNum );`
15
` if ( cell == null ) {`
16
` continue;`
17
` }`
18
` double num = 0;`
19
` index++;`
20
` rowData.put(String.valueOf(index), cell.getStringCellValue());`
21
` }`
22
` kwList.add(rowData);`
Client Code
Java
`xxxxxxxxxx`
1
1
`ClientObject o1 = new ClientObject();`
2
`KeywordReader kr = new KeywordReader("100kTable.xls");`
3
`o1.setDescr("9999");`
4
`ksession.insert(o1);`
5
`ksession.insert(kr);`
6
`int fired = ksession.fireAllRules();`
Let’s run these rules in the client application:
Shell
`xxxxxxxxxx`
1
1
`mvn clean compile exec:java `
2
`Initial Kie Session elapsed time: 4943 `
3
`Load keyword rule fired`
4
`fired rules: 1 elapsed time: 2761 `
5
`Is Object Pass:true`
6
`Check Client Object fired`
7
`2second round fired rules: 1 elapsed time: 70 `
Great, the performance is very good.
As you can see, loading the excel file in rule might take 2.7 seconds. After that, each rule execution only costs 70ms.
Pros
The biggest advantage is good performance. As you can see the performance is over 100 times faster than previous solutions considering the large number of input data in Excel files;
What’s even better, there is no compilation overhead.
The excel data can still be decoupled from application logic. All business rules data can be packaged in the rules project as usual.
Cons
There are two shortcomings as far as I can see:
1. Raw Excel data can’t be handled in kie-workbench.
2. There is some complex Excel Reader logic in the rule project.
It might not be comfortable for business users to maintain the Excel parser login in the Drools model, however, I think, you can wrap the Excel Parse logic in a separate jar file which would decouple it from a business rule project.
Topics:
drools, drools business rules engine, decision table modeling, performance
Opinions expressed by DZone contributors are their own. | 2,034 | 7,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.827249 |
http://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.596864.html | 1,369,005,477,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698141028/warc/CC-MAIN-20130516095541-00076-ip-10-60-113-184.ec2.internal.warc.gz | 310,460,466 | 4,563 | # SOLUTION: use the elmination method to solve the system of equations. a+8b=12 a+3b=2 is there a solution? if so, what is the ordered pair (?,?)
Algebra -> Algebra -> Coordinate Systems and Linear Equations -> SOLUTION: use the elmination method to solve the system of equations. a+8b=12 a+3b=2 is there a solution? if so, what is the ordered pair (?,?) Log On
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Linear Solvers Practice Answers archive Word Problems Lessons In depth
Click here to see ALL problems on Linear-systems Question 596864: use the elmination method to solve the system of equations. a+8b=12 a+3b=2 is there a solution? if so, what is the ordered pair (?,?)Answer by Alan3354(30975) (Show Source): You can put this solution on YOUR website!a+8b=12 a+3b=2 ---------- Subtract | 275 | 1,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2013-20 | latest | en | 0.829813 |
https://datascienceparichay.com/article/r-trigonometric-functions/ | 1,723,550,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00650.warc.gz | 152,616,206 | 36,952 | # Common Trigonometric Functions in R
In this tutorial, we will look at the common trigonometric functions in the R programming language with the help of some examples.
## How to compute the sine, cosine, and tan of a value in R?
R comes with built-in methods for common trigonometric functions such as sine, cosine, and tan. The following is the syntax for using these functions.
• `sin(x)` – Compute the sine of the value passed.
• `cos(x)` – Compute the cosine of the value passed.
• `tan(x)` – Compute the tan, which is defined as `sin(x)/cos(x)`.
Note that you must pass the value in radians to the above trigonometric functions.
## Examples
Let’s look at examples of using the above functions in R.
### The `sin()` function in R
You can apply the `sin()` function directly to a value in radians. For example, let’s get the sine of 30 degrees, which is π/6 in radians.
```# compute sine of π/6
print(sin(pi/6))```
Output:
`[1] 0.5`
You can see that we get the sine of π/6 as 0.5. Note that here we use the built-in `pi` constant available in R to represent π.
You can also apply the `sin()` function to a vector in R. In this case, the `sin()` function gets applied to each value in the vector. Let’s look at an example.
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```# compute sine of a vector
vec <- c(0, pi/6, pi/4, pi/3, pi/2)
print(sin(vec))```
Output:
`[1] 0.0000000 0.5000000 0.7071068 0.8660254 1.0000000`
We get the sine of each value in the above vector.
### The `cos()` function in R
You can similarly apply the `cos()` function to get the cosine of a value. Again, the value must be in radians. For example, let’s compute the cosine of π.
```# compute cosine of π
print(cos(pi))```
Output:
`[1] -1`
You can see that we get the cosine of π as -1.
You can also apply the `cos()` function to a vector in R.
```# compute cosine of a vector
vec <- c(0, pi/6, pi/4, pi/3, pi/2)
print(cos(vec))```
Output:
`[1] 1.000000e+00 8.660254e-01 7.071068e-01 5.000000e-01 6.123234e-17`
We get the cosine of each value in the above vector.
### The `tan()` function in R
In trigonometry, tan(x) is defined as sin(x)/cos(x). You can use the `tan()` function in R to compute the tan of a value. Its usage is similar to the `sin()` and the `cos()` functions in R. Pass the value in radians for which you want to compute the tan.
Let’s look at an example – let’s compute the tan of 45 degrees which is π/4 in radians.
```# compute tan of π/4
print(tan(pi/4))```
Output:
`[1] 1`
We get the tan of π/4 as 1.
You can also apply the `tan()` function to a vector in R.
```# compute tan of a vector
vec <- c(0, pi/6, pi/4, pi/3, pi/2)
print(tan(vec))```
Output:
`[1] 0.000000e+00 5.773503e-01 1.000000e+00 1.732051e+00 1.633124e+16`
We get the tan of each value in the above vector.
You might also be interested in – | 941 | 3,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-33 | latest | en | 0.84283 |
http://www.encyclo.co.uk/define/decimal | 1,368,993,122,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698017611/warc/CC-MAIN-20130516095337-00043-ip-10-60-113-184.ec2.internal.warc.gz | 442,934,630 | 6,559 | ## Look up: decimal
1. Decimal
[unit] A decimal (also spelled decimel) is a unit of area in India and Bangladesh approximately equal to 1/100 acre (40.46 m²). After metrication in the mid-20th century by both countries, the unit became officially obsolete. Especially among the rural population in Northern Bangladesh and ...
Found op http://en.wikipedia.org/wiki/Decimal_(unit)
2. Decimal
[disambiguation] Decimal could mean: ...
Found op http://en.wikipedia.org/wiki/Decimal_(disambiguation)
3. decimal
Decimal numbers are the ones we use most of the time. Decimal numbers are based on multiples of ten. They use the place values of units, tens, hundreds, thousands and so on. For showing parts of numbers smaller than the number 1 a decimal point (a dot) is used. The point separates the fractional pa...
Found op http://www.bbc.co.uk/skillswise/glossary/
4. decimal
[adj] - numbered or proceeding by tens 2. [adj] - divided by tens or hundreds 3. [n] - a number in the decimal system
Found op http://www.webdictionary.co.uk/definition.php?query=decimal
5. decimal
Based on ten parts.
Found op http://www.encyclo.co.uk/local/20581
6. Decimal
The number of digits to the right of the decimal point in a number
Found op http://www.chemicalglossary.net/definition/609-Decimal
7. decimal
The number of digits to the right of the decimal point in a number
Found op http://www.shodor.org/UNChem/glossary.html
8. Decimal
Refers to a base ten number system using the characters 0 through 9 to represent values.
Found op http://www.flowmeterdirectory.com/flowmeter_technical_glossary/flowmeter_te
1. decimal
A number with one or more digits to the right of the decimal point
Example:
3.27
Found op http://www.hbschool.com/glossary/math2/index6.html
2. Decimal
Dec'i·mal adjective [ French décimal (cf. Late Latin decimalis ), from Latin decimus tenth, from decem ten. See Ten , and confer Dime .] Of or pertaining to decimals; numbered or proceeding by tens; having a ...
Found op http://www.encyclo.co.uk/webster/D/16
3. Decimal
Dec'i·mal noun A number expressed in the scale of tens; specifically, and almost exclusively, used as synonymous with a decimal fraction. Circulating , or Circulatory , decimal , a decimal fraction in which the same figure, ...
Found op http://www.encyclo.co.uk/webster/D/16
4. decimal
adjective divided by tens or hundreds; `a decimal fraction`; `decimal coinage`
Found op http://wordnetweb.princeton.edu/perl/webwn?s=decimal
5. decimal
adjective numbered or proceeding by tens; based on ten; `the decimal system`
Found op http://wordnetweb.princeton.edu/perl/webwn?s=decimal
6. Decimal
• (n.) A number expressed in the scale of tens; specifically, and almost exclusively, used as synonymous with a decimal fraction. • (a.) Of or pertaining to decimals; numbered or proceeding by tens; having a tenfold increase or decrease, each unit being ten times the unit next smaller; as,...
Found op http://thinkexist.com/dictionary/meaning/decimal/
7. Decimal
The decimal numeral system (also called base ten or occasionally denary) has ten as its base. It is the numerical base most widely used by modern civilizations. Decimal notation often refers to a base-10 positional notation such as the Hindu-Arabic numeral system; however, it can also be used more ...
Found op http://en.wikipedia.org/wiki/Decimal
8. decimal
The commonly used number system, also known as denary, in which each place has a value 10 times the value of the place at its right. For example, 4327 in the decimal (base 10) system is shorthand for (4 × 103) + (3 × 102) + (2 × 10
Found op http://www.daviddarling.info/encyclopedia/D/decimal.html
9. decimal
1) Denary 2) Difference between a lot and a little, maybe 3) It has a point 4) It usually has a point 5) Like nyse prices since january 2001 6) Number with a point 7) Of tenths 8) Point 9) Quantitative 10) What's the point of math...
Found op http://www.mijnwoordenboek.nl/EN/crossword-dictionary/decimal/1
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Aubree (2) | 1,170 | 4,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2013-20 | latest | en | 0.681436 |
https://apps.apple.com/us/app/fx-algebra-solver/id562247345 | 1,620,322,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00232.warc.gz | 117,160,439 | 45,816 | ## Description
Will guide you how to solve your Algebra homework and textbook problems, anytime, anywhere.
FX Algebra Solver is a comprehensive math software, based on an automatic mathematical problem solving engine, including:
- Over 1200 sample Algerba problems and fully animated solution steps
- Scientific calculator supported
- Graphing calculator supported
- Automatic problem solving and generation of fully animated step-by-step procedures for problems typed in by users
- User friendly math problem expression editor (WYSIWYG mode)
Algebra Solver covers problems at the level of Pre-Algebra, Algebra1, Algebra2, and College Algebra courses:
- Number evaluations
- Prime factorization
- Mixed number
- Complex number
- basic expression simplification:
- polynomial - expansion, factorization
- rational - division, long-division, synthetic composition
- exponential and log/ln
- sin, cos, tan, ..
- Matrix
- Function
- Equations, system of equations
- Inequality
- Graphing
## What’s New
Version 6.0
- Support iPhoneX UI
- Fix bug in solution animation player for iOS 12.0
- Fix bug in InApp purchase receipt check
## Ratings and Reviews
3.4 out of 5
82 Ratings
82 Ratings
G BARK ,
### Great,but not real simple
An awesome calculator but a bit complicated for a novice or beggener. In an advanced novice,ha, and I’m having to take some time to learn how to use the features the tutorial isn’t in-depth enough for a novice like myself and I cant find an easy form for mastering its ability. But for pre algebra it’s been mostly plug in the numbers and it figures it out. But some calculations I have a hard time understanding how the more advanced features work.
Honestly this.... ,
### Needs audible updates - but it’s free so 5:5
It could do with audio to help explain to the users why each step is done, and the common mistakes that are made. Or explain why the answer is correct. As in this area of math it depends on the question and it could be wrong.
So explaining what each formula is and what it means and what it does (or what you need to do or should do next) will help a lot of people. These kind of audio tutorials are are Godsend for a lot of people.
I understand people don’t like having their hand held, but some people learn better when they’re told how things work and what to do. As opposed to wrongly teaching them and letting them figure it out.
Nisteho ,
### About what is missing this app
This app is wonderful but if the app developers add volume to explain the concepts since everything else is included in the app now that would make the app more value and it would also allow for a lot of people to be attracted to download and study from the app. Remember there are a lot of people that study totally different from one another for example I study buy visualizing and hearing and working with my hands and this app has a lot that helps me except it’s missing the hearing part
## App Privacy
### No Details Provided
The developer will be required to provide privacy details when they submit their next app update. | 664 | 3,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-21 | latest | en | 0.886055 |
https://manualzz.com/doc/30901615/meepe-103-power-converters | 1,600,742,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00020.warc.gz | 527,715,974 | 14,348 | # meepe 103 power converters
```1 (Pages:2)
Reg. No…………………….
Name……………………….
M.TECH DEGREE EXAMINATION
First Semester
Model Question Paper - I
Branch: Electrical and Electronics Engineering
Specialization: Power Electronics
MEEPE 103 POWER CONVERTERS
Time: Three hours
Maximum: 100 Marks
1. (a) A dc voltage of 100 V is switched on to a 23 Ω resistance in series with 2 mF capacitance.
Find the magnitude of current and capacitor voltage at t = 0.2 Seconds.
[6 Marks]
(b) A load comprises of resistance and inductance is fed from voltage source of
v = Vm Sin(wt) through a diode. A freewheeling diode is connected across the load. Analyze
the circuit assuming the output current is continuous and sketch the output current and output
voltage waveforms.
[12 Marks]
(c) An ideal capacitor of value C is connected to Vm Sin(314t) through one SCR. If SCR is
fired at firing angle α in the positive half cycle of input supply voltage, Will the SCR conduct
in the positive half cycle when fired at the same firing angle? Explain.
[7 Marks]
OR
2. (a) Explain in detail various thyristor specifications.
(b) Illustrates the limitations on di/dt and dv/dt of thyristors.
[15 Marks]
[10 Marks]
3. (a) A single phase semiconverter is feeding a resistive load. Derive an expression for average
[7 Marks]
(b) A three phase full converter is fires at 75 degrees. Sketch phase voltages, line voltages,
scheme of firing pulses and clearly indicate the output voltage.
OR
[18 Marks]
2 4. (a) A single phase full converter supplied from VmSin(wt) feeds RL load. A freewheeling
diode is connected across the load. Sketch voltage across and current through freewheeling
diode.
[7 Marks]
(b) A three phase semi converter is fired at 30 degrees. Sketch firing pulses, line and phase
voltages and output voltage.
[18 Marks]
5. With neat circuit diagram explain the operation of a buck boost converter in continuous and
discontinuous current modes.
OR
6. (a) Find the duty ratio of a cuk converter operating at 2 kHz to obtain an output voltage 200V.
The input dc voltage consist of two series connected 12V batteries. Also find the voltage
across the switch.
[8 Marks]
(b) In a step up converter, the duty ratio is adjusted to regulate the output voltage Vo at 48 V.
The input voltage varies in a wide range from 12 to 36V. The maximum power output is
120W. For stability reasons, it is required that the converter operate in a discontinuous –
current- conduction mode. The switching frequency is 50 kHz.
[8 Marks]
(c) In a buck-boost converter operating at 20 kHz, L=0.05mH. The output capacitor C is
sufficiently large and Vd = 15 V. The input is to be regulated at 10V and converter is
supplying a load of 10W. Calculate the duty ratio D.
7. (a) With neat circuits and V-I diagrams, classify dc chopper circuits.
[9 Marks]
[10 Marks]
(b) A type-A chopper is feeding a separately excited dc motor drive. If motor current is
discontinuous, derive an expression for the time at which motor current falls to zero during
OFF- period of chopper.
[15 Marks]
OR
8. (a) Briefly discuss various voltage control schemes in voltage source inverters.
[7 Marks]
(b) A three phase voltage source inverter is feeing balanced star connected resistive load and
operates in 180 degree mode of operation. Sketch firing signals and phase and line voltages.
Provide derivations.
[18 Marks]
``` | 849 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.847443 |
https://answers.yahoo.com/question/index?qid=20190822010942AAn3qYT | 1,582,075,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00527.warc.gz | 276,178,922 | 24,666 | Snooker question. Having potted a red , the white then goes in off the black. How many penalty points is that . Thanks in advance?
Relevance
• Cantra
Lv 7
6 months ago
I say 4. Hitting the black in this case, in itself, is not a foul, even though the cue went in after hitting it. The value of the black is only considered if the act of hitting it is a foul, such as when red is 'on', but you hit the black first. Or if you potted a red and then the black was also potted by the same shot.
So for this kind of infraction, the amount of penalty points equals the value of the 'on' ball, or 4, whichever is higher. As a red is worth 1, the amount of penalty points will be the default 4. | 182 | 688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.952062 |
https://www.esaral.com/q/subtract-50572 | 1,721,484,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00201.warc.gz | 654,293,300 | 11,479 | # Subtract:
Question:
Subtract:
4p2 + 5q2 − 6r2 + 7 from 3p2 − 4q2 − 5r2 − 6
Solution:
Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise: | 82 | 235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.80595 |
https://bobtait.com.au/forum/navigation/4055-fixed-card-adf | 1,568,926,629,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573735.34/warc/CC-MAIN-20190919204548-20190919230548-00528.warc.gz | 386,883,251 | 9,816 | × Welcome to the CPL Navigation question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.
• 818861
• Topic Author
### 818861 created the topic: Fixed Card ADF
can someone help me understand how to figure out the answer to a question below?
your track is 140M and HDG is 135M, what should a fixed-card ADF indicate, if you are on track if your departure AD has an NDB?
also a few other questions below i don't understand some help would be great
you are at 170 West and you want to fly to 170 East which is the shortest route and what direction... is the answer East?
what direction and speed does the earth rotate if you look up from under the South Pole?
What is the line called that connects points with zero magnetic variation?
• Posts: 1961
### bobtait replied the topic: Fixed Card ADF
The needle of an ADF always points along the track to/from the station. If the track is 140 and the heading is 135, the nose of the aircraft is 5° to the left of the track. The ADF needle will be 5° to the right if the tail. i.e. it will indicate 185°R.
If you travel East from 170° West of Greenwich to 170° East of Greenwich you will have traveled through a total of 340° of longitude, which is much more than half way around the earth. If you traveled west from 170° West of Greenwich to 170° East of Greenwich you would only have to travel through 20° of longitude, a much smaller distance. Grab a globe of the earth and check it out.
Again, grab a globe. If you rotate the earth from west to east and look up from under the south pole, the rotation is clockwise.
An agonic line is a line that joins places where the magnetic variation is zero. Google it.
• 818861
• Topic Author
### 818861 replied the topic: Fixed Card ADF
Thanks a lot bob! it is really good to know that your always there to help!
Having used all your books so far for the CPL exams they have never let me down ! and I've done 6 passing all first time!
couldn't of done it without your material
Thanks heaps | 511 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-39 | latest | en | 0.957113 |
http://tar.weatherson.org/2009/12/05/axiom-v-and-intuitions/ | 1,527,501,667,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872766.96/warc/CC-MAIN-20180528091637-20180528111637-00294.warc.gz | 285,838,428 | 28,416 | # Axiom V and Intuitions
I know that I’ve said this before, but given that the issue has come up again, I think it bears repeating.
The standard example people use of something that’s intuitive but not actually true is Frege’s Axiom V. And that axiom, remember, says that any predicate has a set for its extension. Apparently this is intuitive. I don’t find it even approximately intuitive. I’ve been introspecting a bit, and I’ve noticed the following things.
• Two days a week, my intuitions are nominalist and so don’t accept any sets, let alone sets for any predicates.
• Two days a week, my intuitions have an iterative conception of sets, so there couldn’t possibly be a set of all sets. (How could it get built, since it would have to have itself as a constituent? That’s very counterintuitive.)
• Two days a week, my intuitions are modest and say that set theory is too hard for them, and that they aren’t going to issue verdicts about anything to do with set theory.
• And on Sundays I watch football.
I don’t doubt that there are some things that are true but counterintuitive. Some of the simple results about comparative advantage in trade theory are counterintuitive but true, for instance. It’s counterintuitive, but I think true, that Mary learns nothing when leaving the black and white room. But I’m very sceptical that Axiom V is one of the counterintuitive truths.
## 6 Replies to “Axiom V and Intuitions”
1. I think it’s worth distinguishing here between Axiom V and naive comprehension. Of course, the former entails the latter really straightforwardly, but it’s striking that the former does even now seem quite a bit more plausible than the latter. (Boolos even claims that the former is ‘obvious’, though I take it that’s to be taken in roughly the same sense as when people claim that we used to ‘know’ that the Earth is flat.) This doesn’t make a difference for some of your points, but it might for the second. It’s not clear that one couldn’t both have iterative conceptiony intuitions and find Axiom V intuitive, given that the clash between these isn’t so immediate as is the clash between the former and comprehension.
By the way, I worry about the example too, at least as I’ve usually encountered it being presented. But I think that Axiom V is at least a stronger example that naive comprehension, and it’s worth discussing the point in its strongest form.
2. Also, I take it you mean to say that you’re sceptical that it’s an intuitive counter-truth, not a counterintuitive truth.
3. john says:
Just out of curiosity, is what bears repeating that you don’t have the intuition that any predicate has a set for its extension, or that this is not intuitive to you? Or might you be suggesting that no one has this intuition, or that it’s not intuitive for anyone (or not intuitive, period)?
4. I agree with you to some extent, but try to remember what Law V actually says. It does not say “any predicate has a set for its extension.” It is not about language at all (—is that what you mean by predicates?—), and even when relegated to concepts or properties, it doesn’t actually explicitly make an existence claim.
The extension of F = the extension of G if and only if, for all x, Fx iff Gx.
(Rough gloss: All and only coextensive concepts have the same extension,)
[Actually, Frege’s version is slightly more general than this, but that’s the version most people have in mind when they mention this.]
The thing is, this statement doesn’t really say that there are extensions, explicitly. It’s more that it presupposes it. In a classical logic such as Frege’s, if you have a term, the term must refer. And to pronounce it as “the extension of F” does seem to give it a kind of meaning which would make Law V analytic.
So it’s harder to make the case that it isn’t intuitively true — because the existence of extensions is a presupposition, not a direct assertion, of the Law, when you hear it, it does sound eminently plausible. It takes some reflection to see what the problem is. So I guess I tend to think it is very intuitive: it rather shows how and why our intuitions can deceive us.
5. I think Kevin and Aidan are right. What I should have said is that I don’t find naive comprehension very intuitive. I go back and forth between thinking naive comprehension is counterintuitive, and thinking it is something that intuition is silent on, but I can’t get into a frame of mind of thinking it is intuitive. And without having done any psychological studies, I’m actually a little surprised if other people find it intuitive.
But that doesn’t extend really to axiom V. As Kevin says, if you state axiom V the right way, so its metaphysical implications are hidden in the right way, it does sound better than (standard statements of) naive comprehension.
6. My general thought about Basic Law V is that it seems to be the intuitive way that extensions should behave, if they exist. I definitely have moments where my intuitions go the first or the third way that Brian mentions (either nominalist or saying “set theory is too hard for intuitions”) but these are normally intuitions that are shaped by the results of the past century of paradoxes and logic. I don’t think I ever have a strong intuition in favor of the iterative conception (where things are “built up” by applying the power set operation), but there are definitely times where something like an axiom of foundation is part of my intuition. However, I think when I was first learning about sets, talk about “the set of all sets” seemed perfectly reasonable to me, and perhaps slightly exciting precisely because it follows from one intuition (the idea of sets as “extensions”) while going directly against this foundation-like intuition.
Basically, what I want to say is that there are a couple sets of intuitions going on here, with a foundation-type intuition and an extension-type intuition. But only the latter seems to be sufficient to base set theory just on an enumeration of the intuitions (as Frege did). To build a set theory on the foundation-type intuitions seems to require some more technical work giving the full idea of the iterative conception (or technical work leading to a type-theoretic conception, or something else that bans self-membership). | 1,378 | 6,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-22 | latest | en | 0.959232 |
https://www.easycalculation.com/faq/calcium-carbonate-decomposes-at-1200-c.php | 1,660,808,782,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00654.warc.gz | 638,416,801 | 7,834 | English
Answer Questions and Earn Points !!!
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Calcium Carbonate Decomposes At 1200 C To Form Carbon Dioxide And Calcium Oxide.If 25 Liters Of Carbon Dioxide Are Collected At 1200 C , What Will The Volume Of This Gas Be After It Cools To 25 C - Math Discussion
## Calcium carbonate decomposes at 1200 C to form carbon dioxide and calcium oxide.If 25 liters of carbon dioxide are collected at 1200 C , what will the volume of this gas be after it cools to 25 C
2016-12-14 15:55:43
0 | 141 | 533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | latest | en | 0.743005 |
https://educatorpages.com/site/mortonmath/pages/math-journal-all-classes | 1,611,244,847,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00465.warc.gz | 311,967,340 | 8,682 | # Math Journal- ALL CLASSES
2013-2014
Math Journal (You MUST have a composition book)
Page 1: This is the cover page... leave it blank for now.
Page 2: Page 2 is folded in and taped to page 3 to create a "pocket".
Page 7: vocabulary template
Page 8: – vocabulary example
Page 9: - blog template --use the 6-sentence format below:
6 Sentence Format for the daily Math Blog:
1st Sentence: Attention Grabber
2nd Sentence: Topic/Introduction Sentence
3rd Sentence: Detail sentence
4th Sentence: Detail sentence
5th Sentence: Detail sentence
6th Sentence: Conclusion/Summary Statement
Page 10: Blog example:
Blog-SLOPE
Did you know that you can find a slope using stairs?
You can use rise and run (slope) to draw a staircase.
There are 4 different types of slope.
The rise describes the vertical change (up and down).
The run describes the horizontal change.
This ratio (rise/run) makes up the slope.
Page 11: - SOLVE steps.
S.O.L.V.E
S:
O:
L:
V:
E: | 256 | 975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | longest | en | 0.851507 |
https://stats.stackexchange.com/questions/285188/from-multiple-correlation-to-conditional-probability | 1,719,310,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00884.warc.gz | 471,854,006 | 40,023 | # From multiple correlation to conditional probability
In the thread from correlation coefficient to conditional probability, a simple formula is given for calculating a conditional probability from a correlation coefficient: $$\frac {1} {2} + \frac {\arcsin \rho} {\pi} ,$$
in which $$\rho$$ is the correlation coefficient. Now, this goes for a simple population correlation coefficient between two variables. What I have, is a sample and multiple regression; for example, in the sample, dependent variable Y is predicted by independent variables A, B and C. If SPSS gives me the multiple R for Y, can I use that in the same way as in the formula above to calculate a conditional probability for Y? If not, how can I calculate that probability? Or should I use the (standardized or unstandardized) regression coefficients (betas) of A, B and C to calculate conditional probability, instead of using the multiple R? I don't have individual cases, just means, standard deviations, N of cases and a (Pearson) correlation matrix for all variables. With this as input, SPSS is able to do multiple regression, from which i get the multiple R and the regression coefficients. I'm using this in my own analysis of data, reported in a number of articles about the prediction of reading problems.
• What do you mean by "multiple R"? "Multiple R squared"? Commented Jun 13, 2017 at 19:51
• Your question appears to be predicated on things that just aren't true. That formula applies only to a bivariate normal distribution and only to a specific probability (that both variables exceed their means). There can be no general formula to translate correlation into conditional probability. Since you have data and the software to work with them, why do you seek such formulas? What is the statistical problem you are trying to address?
– whuber
Commented Jun 13, 2017 at 19:56
• By "multiple R" I mean "multiple R", not "multiple R squared". The question is about converting a correlation coefficient into a conditional probability, and R is a (multiple) correlation coiefficient. Commented Jun 16, 2017 at 11:04
• From the answer of whuber, I see that the case is more complicated than I thought. I am seeking such a formula simply because I want to look at the data in different ways. With reading problems in children, let's say at the age of 8 years, it's useful to be able to predict them from the children's earlier scores on one or more variables, at 4 or 5 years, but it's also useful if you can say, based on the scores on predicting variables, "These children have a chanced of x % of developing reading problems". That is why I'm interested in conditional probability. Commented Jun 16, 2017 at 11:10
• An expression like $P (y_1>y_2|a_1>a_2)$ is over-complicating the analysis for the goal you wish to achieve. You would have to integrate over all possible pairs $a_1,a_2$ for which $a_1>a_2$ and sum the probabilities $P(a_1,a_2)$ multiplied with $P(y_1>y_2|a_1,a_2)$. You could instead use your regression coefficients to express the increase in the expected outcome as function of the independent variables. Also your 'conditional probability for Y' sounds much like Y is a binary variable, and you can do a logistic regression, directly computing probabilities for Y as function of A, B and C. Commented Dec 9, 2018 at 11:10 | 778 | 3,330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | latest | en | 0.920503 |
https://www.puzzleprime.com/updates/201843knights-and-coins | 1,534,553,423,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00026.warc.gz | 959,630,894 | 20,488 | # Knight and Coins
Bob and Jane are taking turns, placing knights and coins respectively on a chessboard. If Bob is allowed to place a knight only on an empty square which is not attacked by another knight, how many pieces at most can he place before running out of moves? Assume that Jane starts second and plays optimally, trying to prevent Bob from placing knights on the board.
Bob can place at most 16 knights. One way to do this is to keep placing knights only on the 32 white squares. In order to see that Jane can prevent Bob from placing more than 16 knights, split the board in four 4x4 grids. Then, group the squares in each grid in pairs, as shown on the image below. If Bob places a knight on any square, then Jane will place a coin on its paired square. This way Bob can place at most one knight on each of the four red squares, one knight on each of the four green squares, one knight on each of the four brown squares, and one knight on each of the four blue squares. Therefore, he can not place more than 64/4 = 16 knights on the board. | 240 | 1,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-34 | latest | en | 0.941426 |
http://www.kylesconverter.com/mass-flow/decigrams-per-second-to-kilograms-per-day | 1,542,416,513,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743247.22/warc/CC-MAIN-20181116235534-20181117021534-00273.warc.gz | 461,452,224 | 5,443 | # Convert Decigrams Per Second to Kilograms Per Day
### Kyle's Converter > Mass Flow > Decigrams Per Second > Decigrams Per Second to Kilograms Per Day
Decigrams Per Second (dg/s) Kilograms Per Day (kg/d) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Kilograms Per Day to Decigrams Per Second
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Decigram per Second:
Mass flow of decigrams across a threshold per unit time of a second. 1 decigram per second = 0.0001 kilograms per second (SI base unit). 1 dg/s ? 0.0001 kg/s.
1 Kilogram per Day:
Mass flow of kilograms across a threshold per unit time of a day. A day containing 86400 seconds. 1 Kilogram per day = 1/86400 kilograms per second (SI base unit). 1 kg/d ? 0.000 011 574 074 kg/s.
Conversions Table
1 Decigrams Per Second to Kilograms Per Day = 8.6470 Decigrams Per Second to Kilograms Per Day = 604.8
2 Decigrams Per Second to Kilograms Per Day = 17.2880 Decigrams Per Second to Kilograms Per Day = 691.2
3 Decigrams Per Second to Kilograms Per Day = 25.9290 Decigrams Per Second to Kilograms Per Day = 777.6
4 Decigrams Per Second to Kilograms Per Day = 34.56100 Decigrams Per Second to Kilograms Per Day = 864
5 Decigrams Per Second to Kilograms Per Day = 43.2200 Decigrams Per Second to Kilograms Per Day = 1728
6 Decigrams Per Second to Kilograms Per Day = 51.84300 Decigrams Per Second to Kilograms Per Day = 2592
7 Decigrams Per Second to Kilograms Per Day = 60.48400 Decigrams Per Second to Kilograms Per Day = 3456
8 Decigrams Per Second to Kilograms Per Day = 69.12500 Decigrams Per Second to Kilograms Per Day = 4320
9 Decigrams Per Second to Kilograms Per Day = 77.76600 Decigrams Per Second to Kilograms Per Day = 5184
10 Decigrams Per Second to Kilograms Per Day = 86.4800 Decigrams Per Second to Kilograms Per Day = 6912
20 Decigrams Per Second to Kilograms Per Day = 172.8900 Decigrams Per Second to Kilograms Per Day = 7776
30 Decigrams Per Second to Kilograms Per Day = 259.21,000 Decigrams Per Second to Kilograms Per Day = 8640
40 Decigrams Per Second to Kilograms Per Day = 345.610,000 Decigrams Per Second to Kilograms Per Day = 86400
50 Decigrams Per Second to Kilograms Per Day = 432100,000 Decigrams Per Second to Kilograms Per Day = 864000
60 Decigrams Per Second to Kilograms Per Day = 518.41,000,000 Decigrams Per Second to Kilograms Per Day = 8640000 | 701 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-47 | latest | en | 0.541486 |
http://mathhelpforum.com/advanced-algebra/65692-linearly-dependent.html | 1,526,848,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520210018-00315.warc.gz | 192,336,444 | 9,680 | 1. ## linearly dependent
let T:R^n -> R^m be a linear transformation, and let {v1, v2, v3} be a linearly dependent set in R^n. Explain why the set {T(v1),(v2),(v2)} is linearly dependent.
2. first, the latter set contains two elements which are equal..
assuming my corrections below...
Originally Posted by ssl000
let T:R^n -> R^m be a linear transformation, and let {v1, v2, v3} be a linearly dependent set in R^n. Explain why the set {T(v1),T(v2),T(v3)} is linearly dependent.
if $\displaystyle \{v_1, v_2, v_3\}$ is linearly dependent, then WLOG, $\displaystyle v_1 = r_2v_2+ r_3v_3$ where $\displaystyle r_2$ and $\displaystyle r_3$ are scalars..
$\displaystyle T(v_1) = T(r_2v_2+ r_3v_3) = r_2T(v_2) + r_3T(v_3)$ since $\displaystyle T$ is a linear transformation.
Thus you were able to express, WLOG, $\displaystyle T(v_1)$ as a linear combination of $\displaystyle T(v_2)$ and $\displaystyle T(v_3)$. Hence $\displaystyle \{T(v1),T(v_2),T(v_3)\}$ | 331 | 957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-22 | latest | en | 0.716984 |
https://aconeyislandofthemind.blog/2019/05/16/rockets/ | 1,575,893,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518882.71/warc/CC-MAIN-20191209121316-20191209145316-00344.warc.gz | 260,002,778 | 12,935 | # rockets
A rocket is a device for trading mass for speed. The generic (i.e., gravitationally neutral) mathematical model of this is the equation p = qlog(m(i)/m(f)), where p is the increase in speed, q is the rate of loss of mass (that is, of mass ejection, assumed to be constant over the time interval in question), m(i) is the initial mass of the device (that is, of the rocket), and m(f) is the final mass of the device (that is, what mass remains after the mass to be ejected has been in fact ejected). This equation was published in 1903 by Konstantin Tsiolkovsky 23 years before Robert H. Goddard’s first liquid-fuel rocket was launched in 1926.
Notice how that fact that the logarithm of unity is zero neatly captures the fact that if no mass is ejected, then there is no increase in speed.
Here is the Wikipedia article on this equation.
keywords: Mathematics, Physics, ideal rocket equation, Tsiolkovsy rocket equation | 223 | 930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-51 | latest | en | 0.937616 |
https://tensorflow.google.cn/versions/r2.2/api_docs/python/tf/keras/mixed_precision/experimental/LossScaleOptimizer | 1,638,373,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00607.warc.gz | 635,242,783 | 61,611 | Help protect the Great Barrier Reef with TensorFlow on Kaggle
# tf.keras.mixed_precision.experimental.LossScaleOptimizer
An optimizer that applies loss scaling.
Inherits From: `Optimizer`
Loss scaling is a process that multiplies the loss by a multiplier called the loss scale, and divides each gradient by the same multiplier. The pseudocode for this process is:
``````loss = ...
loss *= loss_scale
grads = gradients(loss, vars)
grads /= loss_scale
``````
Mathematically, loss scaling has no effect, but can help avoid numerical underflow in intermediate gradients when float16 tensors are used. By multiplying the loss, each intermediate gradient will have the same multiplier applied.
The loss scale can either be a fixed constant, chosen by the user, or be dynamically determined. Dynamically determining the loss scale is convenient as a loss scale does not have to be explicitly chosen. However it reduces performance.
This optimizer wraps another optimizer and applies loss scaling to it via a `LossScale`. Loss scaling is applied whenever gradients are computed, either through `minimize()` or `get_gradients()`. The loss scale is updated via `LossScale.update()` whenever gradients are applied, either through `minimize()` or `apply_gradients()`. For example:
````opt = tf.keras.optimizers.SGD(0.25)`
`opt = tf.keras.mixed_precision.experimental.LossScaleOptimizer(opt,`
` "dynamic")`
`var = tf.Variable(1.)`
`loss_fn = lambda: var ** 2`
`# 'minimize' applies loss scaling to the loss and updates the loss sale.`
`opt.minimize(loss_fn, var_list=var)`
`var.numpy()`
`0.5`
```
If a `tf.GradientTape` is used to compute gradients instead of `LossScaleOptimizer.minimize` or `LossScaleOptimizer.get_gradients`, the loss and gradients must be scaled manually. This can be done by calling `LossScaleOptimizer.get_scaled_loss` before passing the loss to `tf.GradientTape`, and `LossScaleOptimizer.get_unscaled_gradients` after computing the gradients with `tf.GradientTape`. For example:
````with tf.GradientTape() as tape:`
` loss = loss_fn()`
` scaled_loss = opt.get_scaled_loss(loss)`
`scaled_grad = tape.gradient(scaled_loss, var)`
`(grad,) = opt.get_unscaled_gradients([scaled_grad])`
`opt.apply_gradients([(grad, var)]) # Loss scale is updated here`
`var.numpy()`
`0.25`
```
`optimizer` The Optimizer instance to wrap.
`loss_scale` The loss scale to scale the loss and gradients. This can either be an int/float to use a fixed loss scale, the string "dynamic" to use dynamic loss scaling, or an instance of a LossScale. The string "dynamic" equivalent to passing `DynamicLossScale()`, and passing an int/float is equivalent to passing a FixedLossScale with the given loss scale.
`iterations` Variable. The number of training steps this Optimizer has run.
`learning_rate`
`loss_scale` The `LossScale` instance associated with this optimizer.
`lr`
`weights` Returns variables of this Optimizer based on the order created.
## Methods
### `add_slot`
View source
Add a new slot variable for `var`.
View source
### `apply_gradients`
View source
Apply gradients to variables.
This is the second part of `minimize()`. It returns an `Operation` that applies gradients.
The method sums gradients from all replicas in the presence of `tf.distribute.Strategy` by default. You can aggregate gradients yourself by passing `experimental_aggregate_gradients=False`.
#### Example:
``````grads = tape.gradient(loss, vars)
grads = tf.distribute.get_replica_context().all_reduce('sum', grads)
# Processing aggregated gradients.
optimizer.apply_gradients(zip(grads, vars),
experimental_aggregate_gradients=False)
``````
Args
`grads_and_vars` List of (gradient, variable) pairs.
`name` Optional name for the returned operation. Default to the name passed to the `Optimizer` constructor.
`experimental_aggregate_gradients` Whether to sum gradients from different replicas in the presense of `tf.distribute.Strategy`. If False, it's user responsibility to aggregate the gradients. Default to True.
Returns
An `Operation` that applies the specified gradients. The `iterations` will be automatically increased by 1.
Raises
`TypeError` If `grads_and_vars` is malformed.
`ValueError` If none of the variables have gradients.
### `from_config`
View source
Creates an optimizer from its config.
This method is the reverse of `get_config`, capable of instantiating the same optimizer from the config dictionary.
Arguments
`config` A Python dictionary, typically the output of get_config.
`custom_objects` A Python dictionary mapping names to additional Python objects used to create this optimizer, such as a function used for a hyperparameter.
Returns
An optimizer instance.
### `get_config`
View source
Returns the config of the optimizer.
An optimizer config is a Python dictionary (serializable) containing the configuration of an optimizer. The same optimizer can be reinstantiated later (without any saved state) from this configuration.
Returns
Python dictionary.
### `get_gradients`
View source
Returns gradients of `loss` with respect to `params`.
Arguments
`loss` Loss tensor.
`params` List of variables.
Returns
List of gradient tensors.
Raises
`ValueError` In case any gradient cannot be computed (e.g. if gradient function not implemented).
### `get_scaled_loss`
View source
Scales the loss by the loss scale.
This method is only needed if you compute gradients manually, e.g. with `tf.GradientTape`. In that case, call this method to scale the loss before passing the loss to `tf.GradientTape`. If you use `LossScaleOptimizer.minimize` or `LossScaleOptimizer.get_gradients`, loss scaling is automatically applied and this method is unneeded.
If this method is called, `get_unscaled_gradients` should also be called. See the `tf.keras.mixed_precision.experimental.LossScaleOptimizer` doc for an example.
Args
`loss` The loss, which will be multiplied by the loss scale. Can either be a tensor or a callable returning a tensor.
Returns
`loss` multiplied by `LossScaleOptimizer.loss_scale()`.
View source
### `get_slot_names`
View source
A list of names for this optimizer's slots.
### `get_unscaled_gradients`
View source
Unscales the gradients by the loss scale.
This method is only needed if you compute gradients manually, e.g. with `tf.GradientTape`. In that case, call this method to unscale the gradients after computing them with `tf.GradientTape`. If you use `LossScaleOptimizer.minimize` or `LossScaleOptimizer.get_gradients`, loss scaling is automatically applied and this method is unneeded.
If this method is called, `get_scaled_loss` should also be called. See the `tf.keras.mixed_precision.experimental.LossScaleOptimizer` doc for an example.
Args
`grads` A list of tensors, each which will be divided by the loss scale. Can have None values, which are ignored.
Returns
A new list the same size as `grads`, where every non-None value in `grads` is divided by `LossScaleOptimizer.loss_scale()`.
View source
### `get_weights`
View source
Returns the current weights of the optimizer.
The weights of an optimizer are its state (ie, variables). This function returns the weight values associated with this optimizer as a list of Numpy arrays. The first value is always the iterations count of the optimizer, followed by the optimizer's state variables in the order they were created. The returned list can in turn be used to load state into similarly parameterized optimizers.
For example, the RMSprop optimizer for this simple model returns a list of three values-- the iteration count, followed by the root-mean-square value of the kernel and bias of the single Dense layer:
````opt = tf.keras.optimizers.RMSprop()`
`m = tf.keras.models.Sequential([tf.keras.layers.Dense(10)])`
`m.compile(opt, loss='mse')`
`data = np.arange(100).reshape(5, 20)`
`labels = np.zeros(5)`
`print('Training'); results = m.fit(data, labels)`
`Training ...`
`len(opt.get_weights())`
`3`
```
Returns
Weights values as a list of numpy arrays.
### `minimize`
View source
Minimize `loss` by updating `var_list`.
This method simply computes gradient using `tf.GradientTape` and calls `apply_gradients()`. If you want to process the gradient before applying then call `tf.GradientTape` and `apply_gradients()` explicitly instead of using this function.
Args
`loss` A callable taking no arguments which returns the value to minimize.
`var_list` list or tuple of `Variable` objects to update to minimize `loss`, or a callable returning the list or tuple of `Variable` objects. Use callable when the variable list would otherwise be incomplete before `minimize` since the variables are created at the first time `loss` is called.
`grad_loss` Optional. A `Tensor` holding the gradient computed for `loss`.
`name` Optional name for the returned operation.
Returns
An `Operation` that updates the variables in `var_list`. The `iterations` will be automatically increased by 1.
Raises
`ValueError` If some of the variables are not `Variable` objects.
### `set_weights`
View source
Set the weights of the optimizer.
The weights of an optimizer are its state (ie, variables). This function takes the weight values associated with this optimizer as a list of Numpy arrays. The first value is always the iterations count of the optimizer, followed by the optimizer's state variables in the order they are created. The passed values are used to set the new state of the optimizer.
For example, the RMSprop optimizer for this simple model takes a list of three values-- the iteration count, followed by the root-mean-square value of the kernel and bias of the single Dense layer:
````opt = tf.keras.optimizers.RMSprop()`
`m = tf.keras.models.Sequential([tf.keras.layers.Dense(10)])`
`m.compile(opt, loss='mse')`
`data = np.arange(100).reshape(5, 20)`
`labels = np.zeros(5)`
`print('Training'); results = m.fit(data, labels)`
`Training ...`
`new_weights = [np.array(10), np.ones([20, 10]), np.zeros([10])]`
`opt.set_weights(new_weights)`
`opt.iterations`
`<tf.Variable 'RMSprop/iter:0' shape=() dtype=int64, numpy=10>`
```
Arguments
`weights` weight values as a list of numpy arrays.
### `variables`
View source
Returns variables of this Optimizer based on the order created.
[]
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https://www.cyclingforarmenia.net/answers-to-popular-questions/readers-ask-how-many-calories-does-cycling-6-miles-burn.html | 1,652,845,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00045.warc.gz | 848,235,020 | 14,233 | How many calories does 6 miles burn?
Think about it: A person running 10-minute miles for an hour covers six miles and burns about 600 calories in this calculation; a person running 6 -minute miles for that same amount of time runs 10 miles and burns 1,000 calories.
Is biking 5 miles a day good?
Nearly anyone of any fitness level can pedal a bike for five or more miles. Regular or daily cycling has been found to prevent weight gain (and boost fat loss), fight depression, and help stave off a host of health problems, including heart disease, cancer, and diabetes.
How many calories are burned biking 5 miles?
A person walking 5 miles @ 4mph would use approximately 450 calories (6 cal/min for 75 min). Another person cycling @12mph covers 5 miles in 25 min and burns about 300 calories ( 12 cal/min).
How do I calculate calories burned cycling?
The Actual Formula to Calculate Calories Burned Cycling Men: [(Age x 0.2017) — (Weight x 0.09036) + (Heart Rate x 0.6309) — 55.0969] x Time / 4.184. Women: [(Age x 0.074) — (Weight x 0.05741) + (Heart Rate x 0.4472) — 20.4022] x Time / 4.184.
You might be interested: Readers ask: How Often Should U Change The Water When Cycling The Fish Tank?
How can I burn 1000 calories in an hour?
Walk on a treadmill at an incline for an hour. I am 6′ and 200 lbs, and when I walk at 4 mph and a 6% incline, I burn about 1,000 calories an hour. So one way to reach your goal is to do this for 5 hours (adjusting for your calorie burn based on your own research).
What is a good distance to walk daily?
The average American walks 3,000 to 4,000 steps a day, or roughly 1.5 to 2 miles. It’s a good idea to find out how many steps a day you walk now, as your own baseline. Then you can work up toward the goal of 10,000 steps by aiming to add 1,000 extra steps a day every two weeks.
Is biking 10 miles in 30 minutes good?
A good average for a ten mile bike ride is between 45 minutes and an hour. If you’re a beginner, it’s more likely to be closer to the hour mark. For example, if you were to aim to cycle 10 miles in 30 minutes, you would require an average speed of 20 mph (32.19 kph).
Does cycling reduce belly fat?
Yes, cycling can help lose belly fat, but it will take time. A recent study showed regular cycling may enhance overall fat loss and promote a healthy weight. To reduce overall belly girth, moderate-intensity aerobic exercises, such as cycling (either indoor or outdoor), are effective to lower belly fat.
Although daily exercise like cycling will improve your cardiovascular health, lift your mood, and boost your fitness, you can easily pedal an hour a day and not lose a pound. Much to your dismay, you might even gain a few. Exercising daily is no reason to ignore your diet.
You might be interested: Why Do All Cycling Gloves Suck?
How much should I cycle a day to lose weight?
In order to lose weight, the American Council on Exercise (ACE) says you’ll need to cycle at a moderately intense level for at least 30 minutes at a time. To burn even more calories, you’ll want to cycle for longer. ACE also suggests incorporating two activities into one cross-training session to boost weight loss.
How long should I bike for a good workout?
Plan to get on your bike and ride for 30-60 minutes, 3-5 days a week. Start every ride with a warm-up. Pedal at a slow, easy pace for 5-10 minutes. Then boost your speed so you start to sweat.
How can I lose a lb a day?
You need to burn 3500 calories a day to lose one pound a day, and you need anywhere between 2000 and 2500 calories in a day if you are doing your routine activities. That means you need to starve yourself the whole day and exercise as much as to lose the remaining calories.
How can I burn 500 calories a day?
Brisk walking at a pace of 4 MPH for 90 minutes will burn 500 calories. When at work take just 20 minutes during lunch to briskly walk around outside while running errands or catching up on your phone calls; you’ll not only burn 81 calories with each 10 minute task, but you’ll also benefit from the fresh air.
How many calories does cycling for 30 minutes burn?
According to Harvard University, biking at a moderate speed of 12 to 13.9 miles per hour will cause a 155-pound person to burn 298 calories in 30 minutes. At a faster rate of 14 to 15.9 miles per hour, a person of the same weight will burn 372 calories.
You might be interested: How To Know What Gear To Be In Cycling?
What exercise burns the most calories in 30 minutes?
Calories burned in 30 minutes: Generally, running is the best calorie – burning exercise. But if you don’t have enough time to go on a run, you can shorten your workout into high -intensity sprints. Your body will rapidly burn calories to fuel your workout.
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Contents1 When should I add refugium to my reef tank?2 Can I add copepods during cycle?3 Will copepods survive a cycle?4 Does a refugium reduce algae?5 How long do you | 1,348 | 5,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-21 | latest | en | 0.898792 |
https://rdrr.io/cran/mbrglm/man/mbrglm.html | 1,610,854,027,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00139.warc.gz | 540,571,137 | 12,981 | # mbrglm: Median Bias Reduction in Binomial-Response GLMs In mbrglm: Median Bias Reduction in Binomial-Response GLMs
## Description
Fits binomial-response GLMs using the median bias-reduction method proposed in Kenne Pagui et al. (2016, Section 3). The proposed method is obtained by modifying the score equation in such a way that the solution is an approximately median unbiased estimator for each parameter component. The median bias-reduction method enjoys several good properties with respect to the maximum likelihood. In particular, the resulting estimator is component-wise median unbiased with and error of order (O(n^{-1})) and is equivariant under joint reparameterizations that transform each parameter component separately. It has the same asymptotic distribution as the maximum likelihood estimator. Moreover, the resulting estimates and their corresponding standard errors are always finite while the maximum likelihood estimates can be infinite in situations where complete or quasi separation occurs.
## Usage
```1 2 3 4 5 6 7 8``` ```mbrglm(formula, family = binomial, data, weights, subset, na.action, start = NULL, etastart, mustart, offset, model = TRUE, method = "mbrglm.fit", x = FALSE, y = TRUE, contrasts = NULL, control.glm = glm.control(), control.mbrglm = mbrglm.control(), ...) mbrglm.fit(x, y, weights = rep(1, nobs), start = NULL, etastart = NULL, mustart = NULL, offset = rep(0, nobs), family = binomial(), control = glm.control(), control.mbrglm = mbrglm.control(), intercept = TRUE) ```
## Arguments
`formula` an object of class `formula` (or one that can be coerced to that class): a symbolic description of the model to be fitted. `family` a description of the error distribution and link function to be used in the model. For glm this can be a character string naming a family function, a family function or the result of a call to a family function. For mbrglm.fit only the third option is supported. (See `family` for details of family functions.) mbrglm currently supports only the "binomial" family with links "logit", "probit", "cloglog". `data` an optional data frame, list or environment (or object coercible by `as.data.frame` to a data frame) containing the variables in the model. If not found in data, the variables are taken from environment(formula), typically the environment from which glm is called. `weights` an optional vector of 'prior weights' to be used in the fitting process. Should be NULL or a numeric vector. `subset` an optional vector specifying a subset of observations to be used in the fitting process. `na.action` a function which indicates what should happen when the data contain NAs. The default is set by the na.action setting of `options`, and is `na.fail` if that is unset. The 'factory-fresh' default is `na.omit`. Another possible value is NULL, no action. Value `na.exclude`can be useful. `start` starting values for the parameters in the linear predictor. `etastart` starting values for the linear predictor. `mustart` starting values for the vector of means. `offset` this can be used to specify an a priori known component to be included in the linear predictor during fitting. This should be NULL or a numeric vector of length equal to the number of cases. One or more `offset` terms can be included in the formula instead or as well, and if more than one is specified their sum is used. See `model.offset`. `control` a list of parameters for controlling the fitting process. For glm.fit this is passed to `glm.control`. `intercept` logical. Should an intercept be included in the null model? `model` a logical value indicating whether model frame should be included as a component of the returned value. `method` the method to be used for fitting the model. The unique method is "mbrglm.fit", which uses the median modified score function to estimate the parameters. `x` For mbrglm: logical values indicating whether the model matrix used in the fitting process should be returned as components of the returned value. `y` For mbrglm: logical values indicating whether the response vector used in the fitting process should be returned as components of the returned value. `contrasts` an optional list. See the contrasts.arg of model.matrix.default. `control.glm` `control.glm` replaces the `control` argument in `glm` but essentially does the same job. It is a list of parameters to control `glm.fit`. See the documentation of `glm.control1` for details. `control.mbrglm` a list of parameters for controlling the fitting process when method="mbrglm.fit". See documentation `mbrglm.control` for details. `...` additional arguments passed to or from other methods.
## Details
`mbrglm.fit` is the workhorse function for fitting the model using the median bias-reduction method.
The main iteration of `mbrglm.fit` consists to calculate the required quantities for the construction of the modified iterative re-weighted least square which involves the modification term of the score function in the adjusted dependent variable.
Iteration is repeated until either the iteration limit has been reached or the Euclidean distance of the median modified scores is less than some specified positive constant (see the `mbr.maxit` and `mbr.epsilon` arguments in `mbrglm.control`).
## Value
`mbrglm` returns an object of class `"mbrglm"`. A `"mbrglm"` object inherits first from `"glm"` and then from `"lm"` and is a list containing the following components:
`coefficients` a named vector of coefficients. `residuals` Pearson's residual in the final iteration of the IWLS fit. Since cases with zero weights are omitted, their working residuals are NA. `fitted.values` the fitted mean values, obtained by transforming the linear predictors by the inverse of the link function. `rank` the numeric rank of the fitted linear model. `family` the `family` object used. `linear.predictors` the linear fit on link scale. `deviance` up to a constant, minus twice the maximized log-likelihood. Where sensible, the constant is chosen so that a saturated model has deviance zero. `null.deviance` The deviance for the null model, comparable with deviance. The null model will include the offset, and an intercept if there is one in the model. Note that this will be incorrect if the link function depends on the data other than through the fitted mean: specify a zero offset to force a correct calculation. `weights` the working weights, that is the weights in the final iteration of the IWLS fit. `prior.weights` the weights initially supplied, a vector of 1s if none were. `df.residual` the residual degrees of freedom. `df.null` the residual degrees of freedom for the null model. `y` if requested (the default) the y vector used. (It is a vector even for a binomial model.) `x` if requested, the model matrix. `converged` logical. Was the modified IWLS algorithm judged to have converged? `boundary` logical. Is the fitted value on the boundary of the attainable values? `ModifiedScores` the vector of the median modified scores for the parameters at the final iteration. `FisherInfo` the Fisher information matrix evaluated at the resulting estimates. Only available when `method = "mbrglm.fit"`. `FisherInfoInvs` the inverse of Fisher information matrix evaluated at the resulting estimates. `nIter` the number of iterations that were required until convergence. Only available when `method = "mbrglm.fit"`. `model` if requested (the default), the model frame. `formula` the formula supplied. `terms` the terms object used. `data` the data argument. `offset` the offset vector used. `control.mbrglm` the `control.mbrglm` argument that was passed to `mbrglm`. Only available when `method = "mbrglm.fit"`. `contrasts` (where relevant) the contrasts used.
## Note
1. 'mbrglm' and 'mbrglm.fit' were written using as basis structure the code of 'brglm' and 'brglm.fit', respectively. The functions 'brglm' and 'brglm.fit' are implemented in the R package brglm version 0.5-9 by Ioannis Kosmidis. While, 'print.mbrglm', 'summary.mbrglm' and 'print.summary.mbrglm' are modifications of 'print.glm', 'summary.glm' and 'print.summary.glm', respectively.
## Author(s)
Euloge Clovis Kenne Pagui, [email protected], Alessandra Salvan, [email protected] and Nicola Sartori, [email protected]
## References
Kenne Pagui, E. C., Salvan, A. and Sartori, N. (2016). Median bias reduction of maximum likelihood estimates. http://arxiv.org/abs/1604.04768.
brglm, brglm.fit, `glm`, `glm.fit`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83``` ``` ## First example library(brglm) data(endo) # Fit the GLM using maximum likelihood endo.glm <- glm(HG~NV+PI+EH,family=binomial,data=endo) ## Mean bias-reduced fit endo.brglm<-brglm(HG~NV+PI+EH,family=binomial,data=endo) ## Median bias-reduced fit endo.mbrglm<-mbrglm(HG~NV+PI+EH,family=binomial,data=endo) endo.glm endo.brglm endo.mbrglm # Now other links update(endo.mbrglm, family = binomial(probit)) update(endo.mbrglm, family = binomial(cloglog)) ##------------------------ ## paper by Andrey Gelman et al. 2008. Annals of applied Statistics. ## application to binomial ## example 4.2 ##---------------------- # first way x<-c(-0.86,-0.30,-0.05,0.73) z.x<- (1/sqrt(4))*(x-mean(x))/sqrt(var(x)) weights<-rep(5,4) z<-c(0,1,3,5) y=z/weights fit.glm<-glm(y~z.x,family=binomial,weights=weights) fit.brglm<-brglm(y~z.x,family=binomial,weights=weights) fit.mbrglm<-mbrglm(y~z.x,family=binomial,weights=weights) fit.glm fit.brglm fit.mbrglm # in alternative fit.glm<-glm(cbind(z,weights-z)~z.x,family=binomial) fit.brglm<-brglm(cbind(z,weights-z)~z.x,family=binomial) fit.mbrglm<-mbrglm(cbind(z,weights-z)~z.x,family=binomial) fit.glm fit.brglm fit.mbrglm ##---------------------------------------- # Rasch model: 100 subjects and 5 items ##---------------------------------------- I <- 5 S <- 100 ## function to generate data gendata.M <- function(gamma, alpha, beta) { I <- length(alpha) S <- length(gamma) data.y <- matrix(0, nrow=S, ncol=I) for(i in 1:I) { mui <- plogis(alpha[i] + gamma * beta[i]) data.y[,i] <- rbinom(S, size=1, prob=mui) } return(data.y) } alphas <- c(0.0, 0.7, 1.6, 0.6, -0.5) betas <- rep(1,I) gammas <- rnorm(S) y <- gendata.M(gammas,alphas,betas) y.dat <- data.frame(y=y[1:(S*I)],subject=factor(rep(1:S,I)),item=factor(rep(1:I,each=S))) ## Not run: fit.glm <- glm(y~subject-1+item,family=binomial,data=y.dat) fit.brglm <- brglm(y~subject-1+item,family=binomial,data=y.dat) fit.mbrglm <- mbrglm(y~subject-1+item,family=binomial,data=y.dat) ## End(Not run) summary(fit.glm) summary(fit.brglm) summary(fit.mbrglm) ``` | 2,860 | 10,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.829666 |
https://forum.gamsworld.org/viewtopic.php?f=9&p=27683 | 1,601,564,329,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131777.95/warc/CC-MAIN-20201001143636-20201001173636-00588.warc.gz | 352,189,061 | 7,373 | ## Modeling inventory constrainst
Problems with modeling
Ehsan_sr
User
Posts: 2
Joined: 3 months ago
### Modeling inventory constrainst
Hello everyone,
I am trying to model a supply chain model with inventory decisions. The model has 2 constraints for balancing inventory in different periods which are as follows:
in the second equation H is the inventory level of the first period. My question is how I can set a variable for the first period like when I would like to set the starting inventory level equal to zero.
any tips would be appreciated.
Best regards,
Ehsan
abhosekar
Moderator
Posts: 7
Joined: 2 days ago
### Re: Modeling inventory constrainst
I assume you want to fix the initial inventory to a value. You can do this by H.fx(i, p, '1', s) = 0;
Moreover, you may not even need two separate constraints if the only difference is initial inventory. You can model the first constraint as follows:
H(i, p, t-1, s)\$(ord(t) ne 1) + .....
This way the term H(i, p, t-1, s) will be included only second period onwards.
Hope that helps.
Ehsan_sr
User
Posts: 2
Joined: 3 months ago
### Re: Modeling inventory constrainst
Thank you absohekar, that is helpful. I tried your second suggestion and it is working like a charm. | 312 | 1,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.910831 |
https://forum.grasscity.com/threads/3000k-3500k-full-run.1468063/ | 1,600,859,884,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210616.36/warc/CC-MAIN-20200923081833-20200923111833-00167.warc.gz | 398,175,709 | 20,984 | 3000k & 3500k full run?
Discussion in 'Lighting' started by Sickomindo, Aug 15, 2017.
1. So the six cobs I have totaling 300 W those are 3500K
From what I'm being told that should be enough to run five plants in a 4 x 4 tent through veg but then I just bought quantum boards 260 W to combine.
I'm assuming what I should do is run the cobs for veg then when flipping a flower turn on the 260 W 3000K QB board.
Does this sound OK on the kelvin spectrum? too late now but if I run it that way is there anything I should watch out for? Or would you run it differently considering what I already have now?
The 260 W quantum board I bought yesterday is only being offered at 3000 Kelvin I would've gotten another 3500 Kelvin if they sold it.
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2. That's how I would do it. You can probably dim those lights a bit if you wanted too. 200w ish should be good for veg but more certainly can't hurt!
3. The 6 Cobs i got are 300w at 3500k so that should be ok for a 4x4 for Veg then when flowering turn on the other 260w QB at 3000k = 560w total.
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4. Can you share links for the information that you found which says 6 COBS will provide enough light for a 4' x 4' space? How are you spacing them?
I cant find specifics on how much area 1 COB would cover. I'm designing a 4'x8'. Based on several of growmaus comments I'm under the impression that 1 COB every 10-12 inches should be best. So I was going to purchase 27 cxb3590s. I'd have 3 rails of 9 COBS starting 12 inches from the back wall & 10" from the left wall.
However if 6 cobs properly cover 4'x4' then I'm going to purchase way too many COBs.
5. 6 cobs will veg a 4x4 no problem then he is adding a quantum board setup in the middle of it all.
6. It's really going to depend on what wattage the cobs are ran at. It takes about 550-600 watts of high end led to fill a 4x4. Many purpose built 4x4 cob kits are more wattage then that. I'm going to be running 2 QB 260's in my 4x4 for a little under 600 watts. I'm currently running 770 watts in a 4x4 of led.
BTW they do sell the QB260 in 3500k. That the one that I have.
If you combine a 300 watt cob setup with a 270 watt QB fixture you should have plenty of light for a 4x4. I'm currently running a QB260 and a platinum p900 in my 4x4 flower tent. I'll be replacing the p900 when more QB260's come out. My next one will be a 3K so I can mix the 3500 and 3k together.
This is all for a bloom setup.
7. Vegging with leds takes very little power. That's how people have vegged fairly well for decades with CFL bulbs when they hardly put out any light. You can veg well with as little as 10 watts per square foot with led. I'm vegging with 11.125 watts per square of CLW 110 vegmasters. My veg tent outgrows itself even with that incredibly small level of light power.
I think vegging with even 300 watts in a 4x4 is a waste of power. You can veg a 4x4 with as little as 120 watts with the new QB120 veg boards. That's what I'm going to attempt to do. The output of two qb120 kits is about the same as 2 150watt hps lights.
8. Yes I have also heard that vegetation does not require as much light/photons. It sounds like you guys are implying that 6 COBs are not enough to flower buds. Is that right? Where are you getting this data?
9. Our brains..... 300w from 6 cobs is not enough for desirable results is a 4x4 tent.
10. Sounds like these guys are pretty well handled themselves. Yes they are all correct that requires a lot less power and six cobs I got it 300 W I didn't actually have a full run with them to experiment but I didn't want to take the chance so I picked up this 260 W quantum board that's just came in stock recently should be mailed Monday. Not that I will really need that until flowering anyway. I have to say this six cob was pretty damn bright I just didn't want to take a chance I didn't feel comfortable waiting three months to find out I have airy or fluffy bud.
Sent from my iPhone using Grasscity Forum | 1,083 | 4,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | longest | en | 0.964156 |
http://healthybusinesstips.com/bitcoin-mining-calculator-bitcoin-dead.html | 1,542,086,472,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00497.warc.gz | 142,492,992 | 3,827 | What would it take for a competitor to nudge into the fray? For starters, it has to be willing to put a lot of money on the line. Several million dollars can go into chip design before a single prototype is produced. “It takes the willingness to pull the trigger and pay the money,” says Hanke. But he’s confident it will happen. “People will see it’s profitable, and they will jump in.”
In the blockchain, bitcoins are registered to bitcoin addresses. Creating a bitcoin address requires nothing more than picking a random valid private key and computing the corresponding bitcoin address. This computation can be done in a split second. But the reverse, computing the private key of a given bitcoin address, is mathematically unfeasible. Users can tell others or make public a bitcoin address without compromising its corresponding private key. Moreover, the number of valid private keys is so vast that it is extremely unlikely someone will compute a key-pair that is already in use and has funds. The vast number of valid private keys makes it unfeasible that brute force could be used to compromise a private key. To be able to spend their bitcoins, the owner must know the corresponding private key and digitally sign the transaction. The network verifies the signature using the public key.[3]:ch. 5
The process of mining bitcoins works like a lottery. Bitcoin miners are competing to produce hashes—alphanumeric strings of a fixed length that are calculated from data of an arbitrary length. They’re producing the hashes from a combination of three pieces of data: new blocks of Bitcoin transactions; the last block on the blockchain; and a random number. These are collectively referred to as the “block header” for the current block. Each time miners perform the hash function on the block header with a new random number, they get a new result. To win the lottery, a miner must find a hash that begins with a certain number of zeroes. Just how many zeroes are required is a shifting parameter determined by how much computing power is attached to the Bitcoin network. Every two weeks, on average, the mining software automatically readjusts the number of leading zeros needed—the difficulty level—by looking at how fast new blocks of Bitcoin transactions were added. The algorithm is aiming for a latency of 10 minutes between blocks. When miners boost the computing power on the network, they temporarily increase the rate of block creation. The network senses the change and then ratchets up the difficulty level. When a miner’s computer finds a winning hash, it broadcasts the block header to its next peers in the Bitcoin network, which check it and then propagate it further.
But bitcoin is completely digital, and it has no third parties. The idea of an overseeing body runs completely counter to its ethos. So if you tell me you have 25 bitcoins, how do I know you’re telling the truth? The solution is that public ledger with records of all transactions, known as the block chain. (We’ll get to why it’s called that shortly.) If all of your bitcoins can be traced back to when they were created, you can’t get away with lying about how many you have.
# Client-side encryption means all of your data is encrypted on your device before any of your information touches our servers. No server-side hacks, no malware = safe assets. That also means that Edge as a company does not have access to, nor have any knowledge of your account information. Only you and you alone has access and control of your assets—the way it should be.
```The receiver of the first bitcoin transaction was cypherpunk Hal Finney, who created the first reusable proof-of-work system (RPOW) in 2004.[21] Finney downloaded the bitcoin software on its release date, and on 12 January 2009 received ten bitcoins from Nakamoto.[22][23] Other early cypherpunk supporters were creators of bitcoin predecessors: Wei Dai, creator of b-money, and Nick Szabo, creator of bit gold.[24] In 2010, the first known commercial transaction using bitcoin occurred when programmer Laszlo Hanyecz bought two Papa John's pizzas for 10,000 bitcoin.[25]
``` | 851 | 4,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-47 | latest | en | 0.91953 |
https://www.teacherspayteachers.com/Product/Parallel-Lines-Unit-3-Parallel-Lines-and-Transversals-Vocab-Assig-Puzzle-1666334 | 1,544,636,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824059.7/warc/CC-MAIN-20181212155747-20181212181247-00237.warc.gz | 1,054,414,488 | 20,191 | # Parallel Lines - Unit 3: Parallel Lines and Transversals - Vocab Assig. & Puzzle
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Parallel Lines Unit 3: Parallel Lines and Transversals Unit - Vocabulary Assignment and Puzzles
This is a introductory vocabulary assignment for a unit on Parallel Lines and Transversals. It asks the students to provide the definition and any prior knowledge that they have for each term.
Additionally it includes a crossword puzzle (with answer key) and word search puzzle (with answer key) for students practice the terms after they practice the terms.
The terms included are:
Alternate Exterior Angles
Alternate Interior Angles
Congruent Angles
Complementary Angles
Corresponding Angles
Equidistant
Linear Pair Angles
Parallel Lines
Parallel Planes
Perpendicular Lines
Skew Lines
Supplementary Angles
Transversal
Vertical Angles
*****************************************************************************
This a great supplement to a unit on Parallel Lines. If you are looking for other resources, please check out:
Bellwork, Bellringers, Station Cards
Unit 3: Parallel Lines and Transversals Unit Vocabulary Assignment and Puzzles.
Parallel Lines and Transversals Task Cards.
Note Cards
Geometry Basic Concepts Parallel Lines Angles Note Cards.
Bingo Game!
Geometry Early Concepts Bingo Game.
Riddle Worksheet
Geometry Angles Formed by Parallel Lines Riddle Worksheet.
Geometry "A" Wall Posters
1. Basic Concepts
Basic Concepts Word Wall Posters.
2. Proof and Logic
Proof and Logic Word Wall Posters.
3. Classifications of Triangles, Special Lines/Angles and Formulas
Geometry Classifications of Triangles Special Lines and Angles Formulas Wall Cards.
4. Methods of Proving Triangles Congruent and Similar
Methods of Proving Triangles Congruent and Similar Wall Posters.
5. Classifications of Angles, Relationships, Pairs and Parallel Lines
Angles Classifications Pairs Parallel Lines Relationships Wall Posters.
You can also purchase the five Geometry "A" sets above in their own money saving bundle at:
Geometry "A" Wall Poster BUNDLE!
Geometry "A" Wall Poster BUNDLE!.
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Total Pages
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\$3.00 | 675 | 3,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-51 | latest | en | 0.779441 |
https://community.goactuary.com/t/turing-machine/8202 | 1,701,691,610,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00469.warc.gz | 215,201,717 | 8,245 | # Turing Machine
I’m going to run a round or two of the board game Turing Machine.
This game is a interesting logic puzzle. You are trying to solve a code of the form BLUE-YELLOW-PURPLE, where each color is a number between 1 to 5.
One turn of the game consists of the following steps:
1. You choose a 3 digit number to test, and you can choose up to 3 verifiers to test it against, and send it to me via DM.
2. I send you Yes or No for each of the verifiers you tested your number against.
3. You respond in the thread ‘Ready for Next Round’ or ‘Attempt to Solve’
If you attempt to solve and fail, then you are out of the game. If you succeed, then the game ends, and the player who solves it that round with the fewest verifier queries wins.
1. There will be 4 verifiers
2. The final code is the only possible code that will give a yes on all 4 verifiers.
3. You will not know to start what the verifier is testing, that information is to be deduced.
4. Each verifier will be testing the same question for all guesses for all players in a single game.
5. All games have a unique answer, if you can prove that a verifier that was testing the statement A would lead to no unique solution, then it is valid to rule out A.
Verifier A is “This verifier tests the parity of one of the colors.”
You submit 3-3-3 to me (remember, this is blue-yellow-purple), and ask to use verifier A.
I respond ''No".
You know the verifier is not testing ‘Blue is odd’, ‘Yellow is odd’ or ‘Purple is odd’ because if it was testing one of those statements, it would have said yes. (Note this doesn’t mean that Yellow isn’t odd in the final code, if the verifier is testing ‘Blue is even’ then it doesn’t care about your yellow number.)
Interested, sign up below by posting. I’ll start the morning after we get 5 people or on Wednesday, July 26th at the latest.
1 Like
I’m totally lost. Are there any constraints on verifiers? As an extreme, would these be ok
A: The code is 135.
B: Blue < 10
C. Blue < 20
D. Blue < 50
There is only one code that would get a yes to all three, but chances we would get it before losing interest is tiny.
And in your example, what guesses would be a yes if Verifier A were testing the parity of one of the colors?
The colors are each a single digit, 1-5, so you B, C, and D ones are useless because they would give all yes, and there are no ‘all yes’ components in the box, so I’m going to go ahead and say they and statements like them do not exist as verifiers. Same with A, because verifiers aren’t that specific — they give a check to a range of answers, and it’s just the intersection of the 4 verifiers that narrows it down to one code.
The verifiers all have a theme, so you are trying to figure them out from something like 3-9 different statements.
So the parity example above has 6 statements it could be testing. (Blue is even, Yellow is even, Purple is even, Blue is odd, Yellow is odd, Purple is odd). Another verifier could be testing the relationship to one of the colors to 4, which would have 9. (B<4, B=4, B>4, and same three for Y and P).
The key thing to remember is that the verifiers DO NOT KNOW THE FINAL CODE. All they are saying is ‘Yes, your guess meets my criteria’ or ‘No, your guess does not meet my criteria’. The code is the only guess out of all 125 possibilities that gives ‘Yes’ to all 4.
Let’s go back to the parity then for one more example.
Let’s say it was testing ‘Yellow is Even’ in this game.
Any guess of the form X2X or X4X would give a check, and any guess of the form X1X, X3X, and X5X would give an X.
Also, once people are used to it, the game seems to end after 3 rounds most of the time, with the winner being the once that only needs 1 or 2 verifiers to figure it out.
Starting with round 3, I’ll let people request 1 verifier result at a time in case they want to see the answer before deciding if they need more info.
I love that game, definitely in. I’ve found some of the harder puzzles make it to round 4
1 Like
I’d be interested, but I’m on vacation until 8/1 and I may not be able to check the thread that often from 7/26 through 7/31.
1 Like
How much time commitment would this be? I could probably only commit to 2, maybe 3 turns per day.
Love this game.
Just played our copy this past weeken with a few old AOers.
Sadly i do not think i have time to play along online, but i hope you get some good participation.
I was thinking only 1 round a day max unless the players moved it along faster.
I’ll give it a try, though I’m not great at these types of puzzles. At a minimum, I will give everyone else someone to beat
Talk about setting a pretty low bar . . .
I’m going to sit out and watch this round. School starts next week.
1 Like
I remain confused. Is “This verifier tests the parity of one of the colors.” literally what verifier A is,
Or is Verifier A one and only one of those 6 statements?
And if Verifier A is one and only one of those 6 statements, is it true that every Verifier tests a single color? (E.g. could a Verifier be “Blue > Yellow”? Not valid if they can test a single color).
At the start of the game, all you know is that the verifier is testing the parity of one of the colors.
During the game, you should be able to deduce that in this particular game it is testing ‘Yellow is even’ (or ‘Blue is odd’ ,or whichever it actually is for the game).
Some verifiers that I’ve seen could be testing the relationship between 2 of the colors (B below), or the sum of 3 colors (C below)or something like that.
There are 48 in the game box, so I haven’t gone through all of them, but some other examples above.
B could be one of nine statements in any particular game B>Y, B=Y, B<Y, B>P, B=P, B<P, Y>P, Y=P,Y<P — again it’s on the player to eliminate what it could be testing.
C would only have 3 options … maybe B+Y+P < 6, B+Y+P = 6, or B+Y+P >6.
It’s getting clearer, but haze remains.
So I gather we would be told a class for each of verifiers, and we have to figure out what it is testing. (Well, we don’t have to, but if we don’t then we can’t be confident we have a solution.)
So suppose we started the game knowing “Verifier C tests the sum of 3 colors”. Don’t we have far more than 3 possibilities? E.g. including B+Y+P=3, =4, =5, …, =15 (or perhaps =3 and =15 are excluded as too specific), plus similar with > or <. And could it be A+B+C>=9 (though that would be equivalent to >8, and we wouldn’t care whether if was >8 or >=9)?
The wording on those seem to always be 'the sum of 3 colors in comparison to a given number. I don’t think any of the verifiers have more than 9 options from what I’ve seen.
And when I realize the actual game verifiers, I’ll give both the theme and the actual options.
So we will know the class of each verifier, and a list of the possibilities we should consider for each verifier (a list that contains the actual verifier, even if other verifiers not on the list might seem to fit the class)?
That’s much clearer. I’ll give it a shot.
1 Like
Yeah, the game is more figure out what statement each verifier is testing from all of its’ class, AND then figuring out what is the only code that gives a yes to each of the actual statements.
So, at the start of the game, players will be told something like the following?
``````Verifier 1 tests the parity of one of the colors.
Verifier 2 tests the relationship between two of the colors.
Verifier 3 tests the sum of the three colors relative to 9.``````
And it’s on the players to know what the possible tests could be. | 1,910 | 7,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.947008 |
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• free ebook mathematics for high school | 1,187 | 5,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-09 | latest | en | 0.841499 |
https://mvtrinh.wordpress.com/2012/04/ | 1,532,144,814,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592309.94/warc/CC-MAIN-20180721032019-20180721052019-00512.warc.gz | 726,742,745 | 9,702 | # Monthly Archives: April 2012
## Remainder N
If one is the remainder when is divided by , what would the remainder have to be if is divided by ? Source: mathcontest.olemiss.edu 4/23/2012 SOLUTION Given that one is the remainder when is divided by , there exists an integer such … Continue reading
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## Ski Lift
You travel up a ski lift at mph. You ski down the slope at mph. If the ski slope is the same exact length as the distance traveled on the ski lift and we ignore any time spent at the … Continue reading
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## 98 Powers of Two
Find the value of the following expression: Source: mathcontest.olemiss.edu 4/9/2012 SOLUTION We simplify the expression by using the identity We calculate as follows The sum is made up of 3’s and 4’s. How many 3’s are there? The value of … Continue reading
## Magic Factor Square
You are given a grid that contains squares. You are also given the numbers , and such that each is placed in one of the nine squares exactly once and the product of the numbers appearing in any row or … Continue reading
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https://www.exactlywhatistime.com/days-from-date/june-7/9-days | 1,713,551,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00694.warc.gz | 689,920,712 | 6,439 | # What is the date 9 days from June 7?
## Sunday June 16, 2024
Adding 9 days from Friday June 07, 2024 is Sunday June 16, 2024 which is day number 168 of 2024. This page is designed to help you the steps to count 9, but understand how to convert and add time correctly.
• Specific Date: Friday June 07, 2024
• Days from Friday June 07, 2024: Sunday June 16, 2024
• Day of the year: 168
• Day of the week: Sunday
• Month: June
• Year: 2024
## Calculating 9 days from Friday June 07, 2024 by hand
Attempting to add 9 days from Friday June 07, 2024 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
If you want to modify the question on this page, you have two options: you can either change the URL in your browser's address bar or go to our days from specific date calculator to enter a new query. Keep in mind that doing these types of calculations in your head can be quite challenging, so our calculator was developed to assist you in this task and make it much simpler.
## Sunday June 16, 2024 Stats
• Day of the week: Sunday
• Month: June
• Day of the year: 168
## Counting 9 days forward from Friday June 07, 2024
Counting forward from today, Sunday June 16, 2024 is 9 from now using our current calendar. 9 days is equivalent to:
9 days is also 216 hours. Sunday June 16, 2024 is 46% of the year completed.
## Within 9 days there are 216 hours, 12960 minutes, or 777600 seconds
Sunday Sunday June 16, 2024 is the 168 day of the year. At that time, we will be 46% through 2024.
## In 9 days, the Average Person Spent...
• 1933.2 hours Sleeping
• 257.04 hours Eating and drinking
• 421.2 hours Household activities
• 125.28 hours Housework
• 138.24 hours Food preparation and cleanup
• 43.2 hours Lawn and garden care
• 756.0 hours Working and work-related activities
• 695.52 hours Working
• 1138.32 hours Leisure and sports
• 617.76 hours Watching television
## Famous Sporting and Music Events on June 16
• 1943 Actor/comedian Charlie Chaplin (54) marries actress Oona O'Neill (18)
• 1909 Jim Thorpe makes his pro baseball pitching debut for Rocky Mount (ECL) with 4-2 win, this will cause him to forfeit his Olympic gold medals | 658 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.943009 |
https://ambveudedona.cat/moshe-safdie-plxrdw/3-bar-linkage-design-842049 | 1,618,744,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00170.warc.gz | 209,767,397 | 12,431 | I’ve used Algodoo for (very occasional) linkage design. Locate points B 1 and B 2 on your paper. Generally, the joints are configured so the links move in parallel planes, and the assembly is called a planar four-bar linkage.. The photograph shows the scale mock-up linkage a hand crank as the input and a simulated conveyor pockets. The four-bar mechanism is a very versatile linkage , the design of which has been used extensively, and can be seen in a huge number of contemporary machines say the Ackerman steering in all automobiles, the sewing machines, earth movers, the bicycle ,packaging machines . Design a 4-bar linkage configuration and link lengths in a simulator. More design options. You will need a pencil, ruler, compass, and protractor for this design; refer to Figure 1 below during this process. Scott Bougie’s animation of his flower design. Student designs of four-bar linkages . Hsuan-chen Wan’s animation of a four-bar linkage that moves the petals of a flower. Although our six-bar syntheses strictly speaking only allow the specification of 2 plane conditions (plane = point and angle), it is still possible to design six-bar linkages in ASOM v7 to fulfill 3 (or in principle even more) plane conditions. Indeed, the general problem of kinematic linkage design has fascinated mathemati- cians for centuries. This paper deals with the kinematic modeling and optimal design of compliant mechanisms with one flexible joint designed from a rigid four-bar linkage… The relation between the foot angle and the crank angle was developed experimentally by Bolourchi and Hull [3]: . Notice how Klann's horizontal foot-speed slows at both ends of the foot-path, which can cause Klann linkage robots to have a halting gait on higher friction terrain. This implement can be used to move round hay bales. In other words, the 7 PIVCs are physical pivots and do not need to be derived. Klann Linkage Design from his Patent, and a LEGO Approximation Below are the foot-paths of both versions, as well as the speed of the feet when in contact with the ground. Generally, the joints are configured so the links move in parallel planes, and the assembly is called a planar four-bar linkage. The three coupler positions are specified by line CD embedded in the coupler. 4 Bar Linkage Kinematics. It consists of four bodies, called bars or links, connected in a loop by four joints. We are one of the leading providers of design resources, design solutions & training, with an emphasis on providing these design services by utilizing advanced digital design tools like; Autodesk’s Alias Surface, AutoStudio & VRED; ICEM Surf, Photoshop, Maya and many others. Versatile implement that allows the user to level ground, finish dirt driveways, spread sand in horse arenas and more. Especially because most of these design candidates will suffer from linkage defects rendering them useless. In this article, we’re only going to talk specifics about the PIVCs. . C and D are not moving R joints in the coupler. Four-bar joint mechanism with Angled Bar at $$\theta _{\text {knee}} = 0^\circ$$ Full size image. Note: Available in four different styles. . You can find the joint and bar maps for each mechanism on its Linkage Optimizer page. Québec, Canada Linkage Design recruits and employs an elite mix of degreed industrial designers and digital modeling sculptors. Geometric Modeling Examples. The fixed pivot is the centre of rotation. A four bar linkage is used to define and constrain the motion of an object to a particular path. EXAMPLE. In this paper, a four-bar linkage based on design proposed in is utilized. The design of a FWMAV can be conducted through an … There are four PIVCs in a 4-bar linkage while there are seven PIVCs in a 6-bar linkage”. EXAMPLE. Rear spoiler design with ASOM v7: six-bar linkage with 3 planes (part 2) Although our six-bar syntheses strictly speaking only allow the specification of 2 plane conditions (plane = point and angle), it is still possible to design six-bar linkages in ASOM v7 to fulfill 3 (or in principle even more) plane conditions. The graphical design of a crank-rocker quick-return mechanism. Quick-Return Linkage Design The following steps describe a graphical method of designing three and four-bar quick return mechanisms. Smudge Bar. The top graphic shows a kinematic layout of the 6-bar linkage pushing the symbol for a sliding member. An actuating means 30 is provided to articulate the five bar linkage mechanism 10 from the position shown in FIG. Strong robust design to assist with hooking up trailers to your tractor. The opening motion is comparatively weak, but this is unimportant for normal feeding. 2, 3, 4 and 5. Flapping wing MAV, flexible multi-body dynamic analysis, MAV design, MAV manufacture, 6-bar linkage mechanism Date received: 18 July 2016; accepted: 29 August 2016 Introduction Flapping wing micro-aerial vehicles (FWMAVs) have developed greatly over the last few decades, with rele-vant research showing significant growth. The means 30 for articulating the five bar linkage mechanism 10 comprises a cylinder 32 having an extendable rod 34 secured to the fourth bar 18. Generally, the joints are configured so the links move in parallel planes, and the assembly is called a planar four-bar linkage. Hsuan-chen Wan used MechGen 3 to design a six-bar linkage to guide the petals of the flower. PARALLEL MOTION LINKAGE: As the large rod at the top of the diagram moves to the left the two small rods at the bottom move to the right. • A 3-bar linkage will be rigid, stable, not moving unless you bend it, break it, or pick it up and throw it! CAD the links and use bearings at three of the joints, and drive the fourth joint with the RC servo. Design Cookbook: Designing 4-Bar Linkages. DIVCs are migrations of IVCs, so they’re moving IVCs. Sc.) Mechanical designers often wish to design a four-bar linkage that will enable an end effector to follow a certain path. mathematics of linkage design (synthesis) ... • The simplest linkage with at least one degree of freedom (motion) is thus a 4-bar linkage! An analytical solution for the geometry of the five-bar linkage system can be obtained using the Freudenstein equation [2] . Diagram of the developed device. Four-Bar Linkage Synthesis Using Non-Convex Optimization Mémoire Vincent Goulet Maîtrise en informatique Maître ès sciences (M. Approximating Circular Arcs with Cubic Splines; Parametric Design; Cam Design; Kinematics Examples. Mohamed Ali’s four-bar linkage for the flower. Four-Bar Linkage Optimization. 3 Pont Linkage - Series 54. Step 1 | Step 2 | Step 3 | Step 4 | Jamming. Here's the python code for simulating TrotBot, Klann, Strandbeest,and Strider that generated the video above (updated on 1-17-2019). Fig. Bar-linkages are difficult to design; The travel of the bar-linkage mechanism is limited, which affects its use as part of an automated system. This example shows a four-bar linkage modeled in Simscape Multibody that is optimized using MATLAB. 3. Red lines represent four-bar linkage, as well as $$l_1$$, $$l_2$$ and $$l_3$$ Full size image . 2. Bale Spear. A four-bar linkage, also called a four-bar, is the simplest movable closed chain linkage. This is seen to be true when one considers the tremendous volume of litera-ture investigating analysis and design of four-bar mechanisms, ranging from an-tiquity to present [1]. A four-bar linkage, also called a four-bar, is the simplest movable closed chain linkage. • The variations of the manipulator performance can be achieved by adjusting one geometric parameter to ease the control Kinematic Linkage Design Finding a good or even optimal topology for a linkage structure is difficult. All the rods are parallel to each other. It consists of four bodies, called bars or links, connected in a loop by four joints. A six-bar linkage can likewise be developed by first amassing five twofold connections into a pentagon, which utilizes five of the seven joints, and afterward finishing the linkage by including a paired connection that interfaces two sides of the Pentagon. The four bars of the linkage are as follows. It’s not ideal, really more of a toy than a tool. In the rmament of mechanical design the four-bar linkage burns as its brightest star. This again makes two ternary connections that are currently isolated by at least one parallel connections. 1 to the positions shown in FIGS. mechanism 4 link Cookbook Index. MECH 3030 1 Mechanisms, Kinematics, and Machinery Part 11 : Four-bar Linkage Design of Angles (Reference: Chapter 3 of the main textbook by K. Waldron, G. Kinzel, and S. Agrawal, “Kinematics, Dynamics, and Design of Machinery,” Wiley) 1 Mechanisms, Kinematics, and Machinery Part 11 : Four-bar Linkage Design of Angles (Reference: Chapter 3 of the main Another way of describing this linkage is the direction of movement in one rod is reversed in the other rod. Six-bars do more complex stuff. Cameron Turner’s design of a flower that moves with a four-bar linkage. The two fixed R joints of the linkage are specified as O2 and 04. The ground pivot point (O2 and 04) should not be further than 12 inches apart. • Reconfiguration approach can be readily realized by utilizing a simple four-bar linkage. You can view them on the bike as it sits. Figure 1. Design a four-bar linkage to give the three coupler positions shown below. Problem 3 (15 pts]: Graphical linkage synthesis of a quick return mechanism Design a quick return 4 bar linkage whose input is driven by a constant speed motor and whose output oscillates between 52 degrees and 110 degrees with the clockwise motion time being 1.25 times slower than the counter-clockwise motion. Fig. As mentioned above, each root above represents a design candidate, and surely it is better to have thousands of design candidates than 3. 3 Pont Linkage - Series 55. . We calculate the velocity and acceleration of the end effector E, given a 1 rad/sec angular velocity of the crank. the list continues . 4. 1. Rear spoiler design with ASOM v7: six-bar linkage with 3 planes. Build the 555 timer circuit and control the mechanism with a potentiometer. Example # 3: Linkage coupling of mirrors for ultra-low … 3-RRR spherical parallel manipulator of co-axis input featuring a full-circle twist movement is redesigned with reconfigurability. References [1] L. Malfait, G. Storme, and M. Derdeyn. Mechanical errors, like backlash and elasticity, should be reduced by careful construction; Adding . Explore More Examples. Introduction: This handout will take you through the steps of designing a 4 bar linkage. A four-bar linkage, also called a four-bar, is the simplest movable closed chain linkage.It consists of four bodies, called bars or links, connected in a loop by four joints. Note: this code was copy/pasted from 4 separate sims, so it's a bit sloppy and is not as streamlined as Sam's code, FYI To achieve this, they use a four-bar linkage in their jaws to enable the muscle force to obtain a significant mechanical advantage when the jaws are closing, as shown below. 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Rendering them useless spoiler design with ASOM v7: six-bar linkage with 3 planes on bike! Four joints this example shows a four-bar linkage that moves the petals of a toy a. Is optimized using MATLAB way of describing this linkage is used to move round hay bales linkage 3. Designing three and four-bar quick return mechanisms, spread sand in horse arenas and more B 2 your! With a four-bar, is the direction of movement in one rod is reversed in the of... Or even optimal topology for a linkage structure is difficult rad/sec angular velocity of the.. Coupler positions shown below digital modeling sculptors input featuring a full-circle twist movement is redesigned reconfigurability... Acceleration of the end effector E, given a 1 rad/sec angular velocity of the are... Investigate the geometry of the linkage are specified by line CD embedded in the coupler bar at (! Level ground, finish dirt driveways, spread sand in horse arenas more... Indeed, the 7 PIVCs are physical pivots and do not need to be derived 3 bar linkage design... | 4,615 | 20,774 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.925264 |
https://dafyomi.co.il/discuss_daf.php?gid=23&sid=20&daf=102&n=0 | 1,632,507,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00243.warc.gz | 235,606,354 | 3,740 | More Discussions for this daf
1. Checking across? 2. Rebbi Shimon 3. Kuchin
DAF DISCUSSIONS - BAVA BASRA 102
Why are you checking across-- the dead are buried inside the recesss not inside the crypt??????
The Kollel replies:
Your question is not clear to me. You might be asking why the Gemara says that the first 8 Amos of the 20 Amos that are checked, are checked diagonally across the 4 x 6 rectangle. You thought that this means we are checking the central cavity of the cave (Me'arah) off of which the Kuchin (recesses) branch. Your question is that the dead are not buried in the cave, but in the Kuchin.
If that is your question, you are correct; there is no need to check under the Me'arah. The eight Amos are the diagonal of the 4 x 6 area of the Kuchin (recesses), which are each four Amos deep (to accomodate a person's height and then some) and extend 6 Amos along the wall of the Me'arah.
M. Kornfeld | 254 | 918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.92182 |
https://cboard.cprogramming.com/cplusplus-programming/93023-few-questions.html | 1,493,166,346,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121000.17/warc/CC-MAIN-20170423031201-00136-ip-10-145-167-34.ec2.internal.warc.gz | 762,715,109 | 13,481 | 1. ## A few questions
Hi the first question is, how can I set a min and max for the rand function. At the moment i'm seeding from ctime then using
Code:
`int theNumber = rand() % max + 1`
It picks a number between 0 and 100 then if its too high it sets the max integer to the number just guessed. How can I change the min one.
My second question is:
Code:
```#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main()
{
string input;
const int MAX_ITEMS = 10;
int inventoryCount = 0;
string inventory[MAX_ITEMS];
vector<string> shopList;
int shopListCount = 0;
shopList.push_back("Sword") && (shopListCount++);
shopList.push_back("Shield" && (shopListCount++);
shopList.push_back("Firewood") && (shopListCount++);
cout << "Welcome to Ben's Test RPG, input inventory if you want to see what you own, type quit to leave aand type shop to see the shop" << endl;
do
{
cin >> input;
if(input == "inventory")
if(inventoryCount != 0)
for (int i=0; i < inventoryCount; i++)
cout << inventory[i];
else
cout << "Sorry you have no items." << endl;
if(input == "shop")
for (int j=0; j < shopListCount; j++)
cout << shopList[j];
if(input == "quit")
;
else
cout << "Sorry incorrect choice.";
}while(input != "quit")
}```
How can I make it so it adds to my shopListCount when the vector gets pushed back.
Thanks,
Ben
2. Originally Posted by Chipmunkey
It picks a number between 0 and 100 then if its too high it sets the max integer to the number just guessed. How can I change the min one.
Imagine you could generate a random number between 0 and 50. Now add 50 to this number. You now have a random number between 50 and 100. Make sense?
3. how can I set a min and max for the rand function
Read Prelude's article on Using rand().
It picks a number between 0 and 100 then if its too high it sets the max integer to the number just guessed.
Your code snippet produces a number between 1 and max inclusive.
How can I make it so it adds to my shopListCount when the vector gets pushed back.
Just increment:
Code:
```shopList.push_back("Sword");
++shopListCount;
shopList.push_back("Shield");
++shopListCount;
shopList.push_back("Firewood");
++shopListCount;```
Of course, since you are manually inserting 3 items, you might as well initialize shopListCount to 3.
4. Why do you need shopListCount at all? Just use shopList.size().
5. Code:
`shopList.push_back("Shield") && (shopListCount++);`
6. Finally... A practical use for high school algebra.
Well we know that rand() % num gives us a number from 0 to (num - 1). So after a few revisions and tests in my head, here's my final answer:
(rand() % (max - (min - 1))) + min
If min = 50 & max = 100 then:
(rand() % (100 - (50 - 1))) + 50
=(rand() % (100 - 49)) + 50
=(rand() % 51) + 50
=(0...50) + 50
=(50...100) | 770 | 2,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-17 | longest | en | 0.822873 |
https://index-art.info/relationship-between-and/discuss-the-relationship-between-speed-and-acceleration-due-to-gravity.php | 1,569,133,523,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00502.warc.gz | 527,495,419 | 11,513 | # Discuss the relationship between speed and acceleration due to gravity
### Gravitational Acceleration | Science Primer
It is equal to the ratio of change in velocity with respect to time between the given Let's discuss about the calculation of acceleration due to gravitational force. The Physics of the Universe - Special and General Relativity - Gravity and Acceleration. Motion at constant speed is clearly a very special case, and in practice, . Special Theory of Relativity for a more detailed discussion of time dilation). About the Concept Builders · Relationships and Graphs · Kinematics · Newton's Laws · Vectors and Projectiles · Momentum and . 1-D Kinematics - Lesson 5 - Free Fall and the Acceleration of Gravity What is the, acceleration of gravity on It is the ratio of velocity change to time between any two points in an object's path.
Your object was accelerating because gravity was pulling it down. Even the object tossed straight up is falling — and it begins falling the minute it leaves your hand. If it wasn't, it would have continued moving away from you in a straight line. This is the acceleration due to gravity. What are the factors that affect this acceleration due to gravity? If you were to ask this of a typical person, they would most likely say "weight" by which the actually mean "mass" more on this later.
That is, heavy objects fall fast and light objects fall slow. Although this may seem true on first inspection, it doesn't answer my original question. The two quantities are independent of one another. Light objects accelerate more slowly than heavy objects only when forces other than gravity are also at work.
When this happens, an object may be falling, but it is not in free fall. Free fall occurs whenever an object is acted upon by gravity alone. Obtain a piece of paper and a pencil.
Hold them at the same height above a level surface and drop them simultaneously. The acceleration of the pencil is noticeably greater than the acceleration of the piece of paper, which flutters and drifts about on its way down. Something else is getting in the way here — and that thing is air resistance also known as aerodynamic drag. If we could somehow reduce this drag we'd have a real experiment.
ACCELERATION DUE TO GRAVITY
Repeat the experiment, but before you begin, wad the piece of paper up into the tightest ball possible. Now when the paper and pencil are released, it should be obvious that their accelerations are identical or at least more similar than before.
We're getting closer to the essence of this problem. If only somehow we could eliminate air resistance altogether. The only way to do that is to drop the objects in a vacuum. It is possible to do this in the classroom with a vacuum pump and a sealed column of air.
Under such conditions, a coin and a feather can be shown to accelerate at the same rate. In the olden days in Great Britain, a guinea coin was used and so this demonstration is sometimes still called the "guinea and feather". A more dramatic demonstration was done on the surface of the moon — which is as close to a true vacuum as humans are likely to experience any time soon.
## Acceleration Due to Gravity
In accordance with the theory I am about to present, the two objects landed on the lunar surface simultaneously or nearly so. Only an object in free fall will experience a pure acceleration due to gravity.
It was an immensely popular work among academicians and over the centuries it had acquired a certain devotion verging on the religious.
It wasn't until the Italian scientist Galileo Galilei — came along that anyone put Aristotle's theories to the test. Unlike everyone else up to that point, Galileo actually tried to verify his own theories through experimentation and careful observation. He then combined the results of these experiments with mathematical analysis in a method that was totally new at the time, but is now generally recognized as the way science gets done. For the invention of this method, Galileo is generally regarded as the world's first scientist.
In a tale that may be apocryphal, Galileo or an assistant, more likely dropped two objects of unequal mass from the Leaning Tower of Pisa. Quite contrary to the teachings of Aristotle, the two objects struck the ground simultaneously or nearly so.
Given the speed at which such a fall would occur, it is doubtful that Galileo could have extracted much information from this experiment. Most of his observations of falling bodies were really of bodies rolling down ramps. This slowed things down enough to the point where he was able to measure the time intervals with water clocks and his own pulse stopwatches and photogates having not yet been invented.
This he repeated "a full hundred times" until he had achieved "an accuracy such that the deviation between two observations never exceeded one-tenth of a pulse beat.
Professors at the time were appalled by Galileo's comparatively vulgar methods even going so far as to refuse to acknowledge that which anyone could see with their own eyes. In a move that any thinking person would now find ridiculous, Galileo's method of controlled observation was considered inferior to pure reason.
### Free Fall – The Physics Hypertextbook
I could say the sky was green and as long as I presented a better argument than anyone else, it would be accepted as fact contrary to the observation of nearly every sighted person on the planet. Galileo called his method "new" and wrote a book called Discourses on Two New Sciences wherein he used the combination of experimental observation and mathematical reasoning to explain such things as one dimensional motion with constant acceleration, the acceleration due to gravity, the behavior of projectiles, the speed of light, the nature of infinity, the physics of music, and the strength of materials.
His conclusions on the acceleration due to gravity were that… the variation of speed in air between balls of gold, lead, copper, porphyry, and other heavy materials is so slight that in a fall of cubits a ball of gold would surely not outstrip one of copper by as much as four fingers. Having observed this I came to the conclusion that in a medium totally devoid of resistance all bodies would fall with the same speed.
For I think no one believes that swimming or flying can be accomplished in a manner simpler or easier than that instinctively employed by fishes and birds.
## Gravity and Acceleration
When, therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everybody? I greatly doubt that Aristotle ever tested by experiment. Galileo Galilei, Despite that last quote, Galileo was not immune to using reason as a means to validate his hypothesis. In essence, his argument ran as follows.
Imagine two rocks, one large and one small. Since they are of unequal mass they will accelerate at different rates — the large rock will accelerate faster than the small rock. Now place the small rock on top of the large rock. According to Aristotle, the large rock will rush away from the small rock. What if we reverse the order and place the small rock below the large rock?
It seems we should reason that two objects together should have a lower acceleration. The small rock would get in the way and slow the large rock down. But two objects together are heavier than either by itself and so we should also reason that they will have a greater acceleration. This is a contradiction. Here's another thought problem. Take two objects of equal mass. According to Aristotle, they should accelerate at the same rate. Now tie them together with a light piece of string.
### Acceleration of Gravity
Together, they should have twice their original acceleration. His resulting General Theory of Relativityover ten years in the making it was published inhas been called the greatest contribution to science by a single human mind.
Newton's Law of Universal Gravitation Source: Gravity is the organizing force for the cosmos, crucial in allowing structure to unfold from an almost featureless Big Bang origin. Although it is a very weak force feebler than the other fundamental forces which govern the sub-atomic world by a factor of or 1,,,,,,it is a cumulative and consistent force which acts on everything and can act over large distances.
So, even though gravity can be effectively ignored by chemists studying how groups of atoms bond together, for bodies more massive than the planet Jupiter the effects of gravity overwhelm the other forces, and it is largely responsible for building the large-scale structures in the universe.
Even before Newton, the great 17th Century Italian physicist Galileo Galilei had shown that all bodies fall at the same rate, any perceived differences in practice being caused by differences in air resistance and drag. Newton, however, had assumed that the force of gravity acts instantaneously, and Einstein had already shown that nothing can travel at infinite speed, not even gravitybeing limited by the de facto universal speed limit of the speed of light.
Furthermore, Newton had assumed that the force of gravity was purely generated by masswhereas Einstein had shown that all forms of energy had effective mass and must therefore also be sources of gravity.
The principle of equivalence says that gravity is not a force at all, but is in fact the same thing as acceleration Source: Time Travel Research Center: He realized that if he were to fall freely in a gravitational field such as a skydiver before opening his parachute, or a person in an elevator when its cable breakshe would be unable to feel his own weight, a rather remarkable insight inmany years before the idea of freefall of astronauts in space became commonplace.
A simple thought experiment serves to clarify this: | 1,968 | 9,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-39 | latest | en | 0.953035 |
http://mywoodworkingplansandprojects.com/do-it-yourself-woodworking-projects-plans-for-outdoor-furniture.html | 1,569,005,898,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574058.75/warc/CC-MAIN-20190920175834-20190920201834-00009.warc.gz | 129,358,535 | 8,270 | ```“I do a lot of finish-sanding freehand, without a sandpaper block, so I can smooth edges and get into nooks and crannies. But the finer grits are usually bonded to thinner paper and, at least for me, the paper is too thin and ends up tearing long before the grit wears out. So I apply duct tape to the back of the sandpaper. The sandpaper is still flexible enough to sand a tight radius and it’s far more durable. You can use this super-strong sandpaper like a shoeshine rag.” — Chuck Merchant
```
How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
What is the one thing every woodworker needs? Yes, a workbench. Now that you have or at least I am assuming you have worked on so many woodworking projects, you are close to becoming a professional woodworker. You now probably owe yourself a nice woodworking bench. You should also know that a true woodworker never buys his bench from the market, but always builds one himself. But before you start this project, you should know what a workbench is.
```The source above is not exactly a tutorial, but it gives you a basic idea of how the author built a Quirky Pallet Art to enhance the look of their old house. You can also find another tutorial at the link below. It shares a step by step procedure for making a wooden pallet sign. The final product is not exactly the same as the one above, but the basic idea is the same.
```
```Furnishing and decorating your patio is not an easy task – but then again, it has to be done! Your patio is obviously one of the most important rooms in your home, as you can easily turn it into your little piece of Heaven, your “safe spot” in your home where you can retreat whenever you want to ignore the world and just spend some time alone all by yourself.
```
Here’s a safe and sound way to make long cuts with a circular saw on plywood clamped to a worktable. Cut about 12 in. into the plywood, then twist a piece of duct tape into a bow tie, with up-and-down adhesive faces. Slide it in the saw kerf and press the tape down above and under the plywood. Now as you finish the cut, the trailing end can’t curl down dangerously as you saw. Hats off to Mike Connelly for simplifying this job. Check out how to make this DIY duct tape wallet.
Using shelving in your room or kitchen is a great way to arrange and de-clutter space… I know, such ground-breaking term it is. Do not write me off yet, I just want to show you how you can build some clean floating corner shelving that appears to have no brackets. You can create them at no cost, and the hardest part of the plan is figuring out what you are going to put on these shelves when you are finished.
Examples of Bronze Age wood-carving include tree trunks worked into coffins from northern Germany and Denmark and wooden folding-chairs. The site of Fellbach-Schmieden in Germany has provided fine examples of wooden animal statues from the Iron Age. Wooden idols from the La Tène period are known from a sanctuary at the source of the Seine in France.
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Copyright 2018 by Cut The Wood. CutTheWood.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. Additionally, CutTheWood.com participates in various other affiliate programs, and we sometimes get a commission through purchases made through our links.
Another important factor to be considered is the durability of the wood, especially in regards to moisture. If the finished project will be exposed to moisture (e.g. outdoor projects) or high humidity or condensation (e.g. in kitchens or bathrooms), then the wood needs to be especially durable in order to prevent rot. Because of their oily qualities, many tropical hardwoods such as teak and mahogany are popular for such applications.[9]
This plan is probably the easiest plan ever added in the list. The one who is working on this project, don’t need any professional skills but just knowing some basics of woodworking will be enough for this DIY. You will get step by step detailed process of this tutorial in the source linked tutorial. This tutorial will surely help you to build this plan quickly.
Copyright 2018 by Cut The Wood. CutTheWood.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. Additionally, CutTheWood.com participates in various other affiliate programs, and we sometimes get a commission through purchases made through our links.
# If you have to pick up long lengths of wood from the lumberyard, throw a spring clamp in the back of your vehicle. Use the clamp to attach the warning flag to the end of the protruding lumber. The clamp’s easy to slip on and off, and you won’t have to fuss around with staples, nails or string. — Steve Parker. Plus: Learn more about how to transport large items in your truck.
Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
“I do a lot of finish-sanding freehand, without a sandpaper block, so I can smooth edges and get into nooks and crannies. But the finer grits are usually bonded to thinner paper and, at least for me, the paper is too thin and ends up tearing long before the grit wears out. So I apply duct tape to the back of the sandpaper. The sandpaper is still flexible enough to sand a tight radius and it’s far more durable. You can use this super-strong sandpaper like a shoeshine rag.” — Chuck Merchant
When cutting full sheets with my circular saw, I use plastic shelving units as sawhorses. The height is just right and by using three of them, I can make cuts in any direction and the plywood is fully supported. And because the shelving units are made of plastic, I can cut right into them without worrying that they’ll damage my saw blade. Plastic shelves are available for \$20 at home centers. — John Tinger. Check out these tips for making long cuts with a circular saw.
From the source tutorial, you can get illustrates to the instruction about the plan. Everything is fairly described as diagrams, images, the list of supplies and tools need etc. The process to this plan is very easy to understand and follow for if you are having some basic woodworking knowledge. Make sure to collect all the supplies you need before you start with the project. You may even ask any question directly in the comment section of the tutorial post and also comment the images of your final product if you have completed it. Either way, I hope that you will manage to build this one nicely.
The video explains the step by step process of making a nice wooden phone stand from scratch. My first wooden holder was not the best one, but it was good enough to motivate me to make more. I now possess 10 mobile wooden stands in different shapes and styles. And if I can make this, you too can make one yourself. Search the internet for more mobile holder ideas and start making one now.
Short scraps of hardwood are too good to throw away but hard to store neatly. So I bought a 4-ft. tube form made for concrete footings, cut it in half (the cardboard-like material cuts easily) and set the tubes on end. I tack the tubes to a wall or a bench leg so they don’t fall over. With the wood scraps stored upright, it’s easy to find a piece just the right length. Tube forms are available in various diameters for \$5 and up at home centers. — Bill Wells. Here’s another brilliant use for these concrete forms.
What is the one thing every woodworker needs? Yes, a workbench. Now that you have or at least I am assuming you have worked on so many woodworking projects, you are close to becoming a professional woodworker. You now probably owe yourself a nice woodworking bench. You should also know that a true woodworker never buys his bench from the market, but always builds one himself. But before you start this project, you should know what a workbench is.
This is probably one the easiest woodworking projects you will find here. Although easy, a doormat is an equally important and useful item for households. As you can see in the image below, you will only need some 2X2 wooden boards and rope to build a simple doormat. This doormat is mostly useful for outdoor and porch. It will easily remove all the mud from your shoes with just one wipe. It is also very easy to clean and looks fabulous even if it is dirty.
There are a few things particleboard is NOT. It’s not medium density fiberboard (MDF)—a material with greater density and weight composed of more uniform particles. It’s NOT oriented strand board (OSB), a material composed of large wood chips and strands that’s structurally equivalent to plywood. It does NOT have great nail or screw holding ability, nor is it all that water resistant; water can quickly cause the material to swell and lose structural integrity. But if you need something flat and cheap for use in a dry place, particleboard will do you proud. Learn how to make a plastic laminate tabletop with a particleboard substrate.
```Instead of using a container to mix a small amount of epoxy, just make a mixing surface on your workbench using painters tape. Simply lay down strips, overlapping the edges so the epoxy doesn’t get on your bench. When you’re done, peel off the tape and throw it away. This mixing surface will work for more than just epoxy, you can use it for wood glue or any other material you need easy access to while working on a project.
```
Afrormosia Alder Andiroba Anigre Ash Apple Aspen Avodire Balsa Beech Bilinga Birch African Blackwood Australian Blackwood Boxwood Bubinga Camphor Cedrela Cherry Chestnut Cocobolo Cumaru Ebony Elm Eucalyptus Hazel Hickory Hornbeam Idigbo Imbuia Ipê Iroko Jarra Jelutong Lignum vitae Linden (lime, basswood) Merbau Mahogany (American, African) Maple Meranti Oak Padauk Pear Plum Poplar Purpleheart Ovankol Ramin Red Quebracho Rosewood Rubberwood Sapele Teak Totara Utile Walnut Wenge Willow Zebrano
The best thing about this wine rack is that it is very easy to build. All you need is the basic understanding of woodworking and a few tools to get started. You can modify your wine rack any way you want or build in a design or color different from this one. The basic steps to build a wooden wine rack are the same for all variants. I have included here the video tutorial that I followed in order to build myself a pallet wide rack.
### Another wooden item that I love very much is a beautiful mobile holder. You can see one in the image below. These things are not only beautiful, but they can comfortably hold any sized mobile and ensure proper safety. Another amazing thing is that they can be built in many shapes and sizes, as and how you need it. You can see some more examples at the source below
For many of us, the moment we learned that a 2×4 board is actually 1.5 inches x 3.5 inches was simply mind-blowing. The reason for this apparent contradiction is that the board has been planed down to eliminate irregularities. At one point, many years ago, 2x4s actually were 2 inches x 4 inches, but their rough surfaces made them difficult to stock and handle. The old terms, such as 2×4 or 4×4, are still used, and are known as the “nominal” size of the board. These nominal sizes are used because they are easier to say and they stick to tradition. Now, thanks to a lawsuit, most big box stores list the nominal and actual sizes of lumber. | 2,805 | 12,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-39 | latest | en | 0.956465 |
https://homework.cpm.org/category/ACC/textbook/ccaa8/chapter/7%20Unit%208/lesson/CCA:%207.1.5/problem/7-61 | 1,725,850,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00746.warc.gz | 271,474,224 | 15,003 | ### Home > CCAA8 > Chapter 7 Unit 8 > Lesson CCA: 7.1.5 > Problem7-61
7-61.
The drama club found that the best price for renting a fog machine was $38$ for every three days, plus a one-time $60$ delivery fee. Make a step graph that shows the cost of renting the fog machine for up to three weeks.
A step function is a special kind of piecewise function (a function composed of parts of two or more functions). A step function has a graph that is a series of line segments that often looks like a set of steps.
The endpoints of the segments on step functions are either open circles (indicating this point is not part of the segment) or filled-in circles (indicating this point is part of the segment). | 170 | 705 | {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.92718 |
https://freedukasyon.com/math/if-there-are-12-teams-in-a-basketba-74915982 | 1,696,341,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511106.1/warc/CC-MAIN-20231003124522-20231003154522-00203.warc.gz | 294,831,696 | 14,505 | , 21.03.2023 08:32 camillebalajadia
# if there are 12 teams in a basketball tournament and each team must play every other teaam in the eliminations how many elimination games will there be?
Answers: 1
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if there are 12 teams in a basketball tournament and each team must play every other teaam in the el...
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Technology and Home Economics, 06.10.2021 08:15
Araling Panlipunan, 06.10.2021 08:15
English, 06.10.2021 08:15 | 367 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | longest | en | 0.857206 |
http://piping-designer.com/index.php/properties/fluid-mechanics/2192-air-pipe-sizing-by-pressure-loss | 1,550,269,332,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479159.2/warc/CC-MAIN-20190215204316-20190215230316-00145.warc.gz | 207,442,486 | 7,085 | # Pipe Sizing for Air
Written by Jerry Ratzlaff on . Posted in Fluid Dynamics
### Air Pressure Loss through Piping formula
$$\large{ p_l = \frac { \mu \; \cdot \; l \; \cdot \; v_a{^2} \; \cdot \; \rho } {2d } }$$
Where:
$$\large{ p_l }$$ = air pressure loss
$$\large{ \rho }$$ (Greek symbol rho) = density of the air
$$\large{ \mu }$$ (Greek symbol mu) = friction coefficient
$$\large{ d }$$ = pipe inside diameter
$$\large{ l }$$ = pipe length
$$\large{ v_a }$$ = velocity of the air
### Air Pipe Inside Diameter formula
$$\large{ d = \sqrt { \frac { 4 } { \pi } \; \cdot \; \frac { Q_a } {60v_a} } }$$
Where:
$$\large{ d }$$ = pipe inside diameter
$$\large{ \pi }$$ = Pi
$$\large{ Q_a }$$ = flow rate of the air
$$\large{ v_a }$$ = velocity of the air
### Air Velocity through Piping formula
$$\large{ v_a = \frac { Q_a } { 60 \pi { \left( \frac {d}{2} \right) ^2 } } }$$
Where:
$$\large{ v_a }$$ = velocity of the air
$$\large{ Q_a }$$ = flow rate of the air
$$\large{ \pi }$$ = Pi
$$\large{ d }$$ = pipe inside diameter
### Air Flow Rate through Piping formula
$$\large{ Q_a = 60 \pi \cdot v_a { \left( \frac {d}{2} \right) ^2 } }$$
Where:
$$\large{ Q_a }$$ = flow rate of the air
$$\large{ \pi }$$ = Pi
$$\large{ d }$$ = pipe inside diameter
$$\large{ v_a }$$ = velocity of the air | 468 | 1,321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-09 | latest | en | 0.587796 |
http://mathematica.stackexchange.com/questions/14035/is-it-possible-to-export-the-equations-from-mathematica-to-matlab | 1,469,293,382,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00178-ip-10-185-27-174.ec2.internal.warc.gz | 160,275,966 | 17,535 | # Is it possible to export the equations from Mathematica to MATLAB?
Is it possible to export the output expressions from Mathematica computations (e.g., equations) in valid MATLAB syntax?
-
Which equations? Can you be more specific on what precisely it is you want to do? one equation? a list of them? Cut and paste? Programmatically export? – acl Nov 2 '12 at 22:23
@acl just an equation with Log s exponentials error functions. For example f[x_]=Log[...]+err[..]*Exp[..]. Something that sort. I know there is a command // InputForm but the output has for example $E$ which MATLAB will not understand. I only want to copy paste the equation to MATLAB. Still unclear? – Seyhmus Güngören Nov 2 '12 at 23:29
This is not exactly an answer, but do to the large chat under previous answer I would like to point out that ToMatlab, provides the page where you can find the package, however many struggles with not being able to download it. I had the same problem until I pressed down alt key and pressed the link at the bottom of the page. Might save a bit of time! – ALEXANDER Apr 9 '14 at 10:08
There is the ToMatlab package that will convert Mathematica expressions to MATLAB equivalents. For example:
<<ToMatlab
Expand[(x + Log@y)^5] // ToMatlab
(* x.^5+5.*x.^4.*log(y)+10.*x.^3.*log(y).^2+10.*x.^2.*log(y).^3+5.* ...
x.*log(y).^4+log(y).^5; *)
It even conveniently breaks it using ... and can also convert matrices:
RandomInteger[5, {5, 5}] // ToMatlab
(* [5,0,5,3,4;
5,5,3,0,2;
1,4,4,4,4;
0,3,2,5,5;
4,5,5,1,1]; *)
However, it won't convert general definitions or things that don't make sense in MATLAB, such as patterns.
To install the package, extract the ToMatlab.m file to
FileNameJoin[{$UserBaseDirectory, "Applications"}] - Great! Thank you very much!! – Seyhmus Güngören Nov 3 '12 at 0:09 sorry but how can I use this package? – Seyhmus Güngören Nov 3 '12 at 0:28 You extract the ToMatlab.m file to $UserBaseDirectory/Applications/ – R. M. Nov 3 '12 at 0:41
ok. many thanks once again. – Seyhmus Güngören Nov 3 '12 at 3:08
@TMH When you download the ToMatlab package from the linked site, you get a zip file. Unzip that file and copy the ToMatlab.m file to the location I mentioned. Replace \$UserBaseDirectory` with whatever is shown when you evaluate that in Mathematica. – R. M. Mar 30 '13 at 18:25
## protected by R. M.♦Feb 5 '15 at 11:19
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 742 | 2,585 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | latest | en | 0.862849 |
https://github.com/britg/MultistrokeGestureRecognizer-iOS/issues/4 | 1,467,406,705,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00106-ip-10-164-35-72.ec2.internal.warc.gz | 1,021,946,907 | 11,682 | Possible miscalc in OptimalCosineDistance#4
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opened this Issue Feb 3, 2013 · 0 comments
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commented Feb 3, 2013
I am writing to you since we found a typo in the OptimalCosineDistance function. There is a minus instead of a plus sign (in the commented line):
```float OptimalCosineDistance(FloatArrayContainer v1, FloatArrayContainer v2) {
float a = 0.0;
float b = 0.0;
float angle;
float score;
int mincount = (v1.itemCount < v2.itemCount ? v1.itemCount : v2.itemCount);
for (int ii = 0; ii < mincount; ii+=2) {
a += v1.items[ii] * v2.items[ii] + v1.items[ii+1] * v2.items[ii+1];
// b += v1.items[ii] * v2.items[ii+1] + v1.items[ii+1] * v2.items[ii];
b += v1.items[ii] * v2.items[ii+1] - v1.items[ii+1] * v2.items[ii];
}
angle = atanf( b / a );
score = acosf(a * cos(angle) + b * sin(angle));
return score;
}```
This made the predictions far more accurate. Thought I'd let you know =)
was assigned Feb 3, 2013
added a commit that closed this issue Feb 3, 2013
britg `Fix bug in OptimalCosineDistance, fixes #4` `00f0ba3`
closed this in `00f0ba3` Feb 3, 2013 | 366 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-26 | latest | en | 0.715127 |
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# Notes on Diffy Qs: Differential Equations for Engineers
## Section4.10Dirichlet problem in the circle and the Poisson kernel
Note: 2 lectures, §9.7 in [EP], §10.8 in [BD]
### Subsection4.10.1Laplace in polar coordinates
A more natural setting for the Laplace equation $$\Delta u = 0$$ is a circle rather than a rectangle. On the other hand, what makes the problem somewhat more difficult is that we need polar coordinates.
Recall that the polar coordinates for the $$(x,y)$$-plane are $$(r,\theta)\text{:}$$
\begin{equation*} x = r \cos \theta , \quad y = r \sin \theta , \end{equation*}
where $$r \geq 0$$ and $$-\pi < \theta \leq \pi\text{.}$$ So the point $$(x,y)$$ is distance $$r$$ from the origin at an angle $$\theta$$ from the positive $$x$$-axis.
Now that we know our coordinates, let us give the problem we wish to solve. We have a circular region of radius 1, and we are interested in the Dirichlet problem for the Laplace equation for this region. Let $$u(r,\theta)$$ denote the temperature at the point $$(r,\theta)$$ in polar coordinates.
We have the problem:
\begin{aligned} & \Delta u = 0 , & & \text{for } \; r < 1, \\ & u(1,\theta) = g(\theta), & & \text{for } \; {-\pi} < \theta \leq \pi. \end{aligned}\tag{4.26}
The first issue we face is that we do not know the Laplacian in polar coordinates. Normally we would find $$u_{xx}$$ and $$u_{yy}$$ in terms of the derivatives in $$r$$ and $$\theta\text{.}$$ We would need to solve for $$r$$ and $$\theta$$ in terms of $$x$$ and $$y\text{.}$$ In this case it is more convenient to work in reverse. We compute derivatives in $$r$$ and $$\theta$$ in terms of derivatives in $$x$$ and $$y$$ and then we solve. The computations are easier this way. First
\begin{equation*} \begin{aligned} & x_r = \cos \theta, & & x_\theta = - r \sin \theta, \\ & y_r = \sin \theta, & & y_\theta = r \cos \theta. \end{aligned} \end{equation*}
Next by chain rule we obtain
\begin{equation*} \begin{aligned} u_r & = u_x x_r + u_y y_r = \cos(\theta) u_x + \sin(\theta) u_y , \\ u_{rr} & = \cos(\theta) ( u_{xx} x_r +u_{xy} y_r ) + \sin(\theta) ( u_{yx} x_r +u_{yy} y_r ) \\ & = \cos^2(\theta) u_{xx} + 2 \cos(\theta)\sin(\theta) u_{xy} + \sin^2(\theta) u_{yy} . \end{aligned} \end{equation*}
Similarly for the $$\theta$$ derivative. Note that we have to use the product rule for the second derivative.
\begin{equation*} \begin{aligned} u_\theta & = u_x x_\theta + u_y y_\theta = -r\sin(\theta) u_x + r\cos(\theta) u_y , \\ u_{\theta\theta} & = -r\cos(\theta) u_x -r\sin(\theta) (u_{xx} x_\theta + u_{xy} y_\theta) -r\sin(\theta) u_y + r\cos(\theta) (u_{yx} x_\theta + u_{yy} y_\theta) \\ & = -r\cos(\theta) u_x -r\sin(\theta) u_y +r^2 \sin^2(\theta) u_{xx} -r^2 2\sin(\theta)\cos(\theta) u_{xy} +r^2 \cos^2(\theta) u_{yy} . \end{aligned} \end{equation*}
Let us now try to solve for $$u_{xx} + u_{yy}\text{.}$$ We start with $$\frac{1}{r^2} u_{\theta\theta}$$ to get rid of those pesky $$r^2\text{.}$$ If we add $$u_{rr}$$ and use the fact that $$\cos^2(\theta) +\sin^2(\theta) = 1\text{,}$$ we get
\begin{equation*} \frac{1}{r^2} u_{\theta\theta} + u_{rr} = u_{xx} + u_{yy} - \frac{1}{r} \cos(\theta) u_x - \frac{1}{r} \sin(\theta) u_y . \end{equation*}
We’re not quite there yet, but all we are lacking is $$\frac{1}{r} u_r\text{.}$$ Adding it we obtain the Laplacian in polar coordinates:
\begin{equation*} \mybxbg{~~ \Delta u = u_{xx} + u_{yy} = \frac{1}{r^2} u_{\theta\theta} + \frac{1}{r} u_{r} + u_{rr} . ~~} \end{equation*}
Notice that the Laplacian in polar coordinates no longer has constant coefficients.
### Subsection4.10.2Series solution
Let us separate variables as usual. That is let us try $$u(r,\theta) = R(r)\Theta(\theta)\text{.}$$ Then
\begin{equation*} 0 = \Delta u = \frac{1}{r^2} R \Theta'' + \frac{1}{r} R' \Theta + R'' \Theta . \end{equation*}
Let us put $$R$$ on one side and $$\Theta$$ on the other and conclude that both sides must be constant.
\begin{equation*} \begin{aligned} \frac{1}{r^2} R \Theta'' & = - \left(\frac{1}{r} R' + R''\right) \Theta \\ \frac{\Theta''}{\Theta} & = - \frac{r R' + r^2 R''}{R} = -\lambda \end{aligned} \end{equation*}
We get two equations:
\begin{equation*} \begin{aligned} & \Theta'' + \lambda \Theta = 0 , \\ & r^2 R'' + r R' -\lambda R = 0. \end{aligned} \end{equation*}
Let us first focus on $$\Theta\text{.}$$ We know that $$u(r,\theta)$$ ought to be $$2\pi$$-periodic in $$\theta\text{,}$$ that is, $$u(r,\theta) = u(r,\theta+2\pi)\text{.}$$ Therefore, the solution to $$\Theta'' + \lambda \Theta = 0$$ must be $$2\pi$$-periodic. We have seen such a problem in Example 4.1.5. We conclude that $$\lambda = n^2$$ for a nonnegative integer $$n=0,1,2,3,\ldots\text{.}$$ The equation becomes $$\Theta'' + n^2 \Theta = 0\text{.}$$ When $$n=0$$ the equation is just $$\Theta'' = 0\text{,}$$ so we have the general solution $$A \theta + B\text{.}$$ As $$\Theta$$ is periodic, $$A=0\text{.}$$ For convenience we write this solution as
\begin{equation*} \Theta_0 = \frac{a_0}{2} \end{equation*}
for some constant $$a_0\text{.}$$ For positive $$n\text{,}$$ the solution to $$\Theta'' + n^2 \Theta = 0$$ is
\begin{equation*} \Theta_n = a_n \cos(n\theta) + b_n \sin(n\theta) , \end{equation*}
for some constants $$a_n$$ and $$b_n\text{.}$$
Next, we consider the equation for $$R\text{,}$$
\begin{equation*} r^2 R'' + r R' - n^2 R = 0. \end{equation*}
This equation appeared in exercises before—we solved it in Exercise 2.1.6 and Exercise 2.1.7. The idea is to try a solution $$r^s$$ and if that does not give us two solutions, also try a solution of the form $$r^s \ln r\text{.}$$ Let us name the solution for $$R_n\text{.}$$ When $$n=0$$ we obtain
\begin{equation*} R_0 = A r^0 + B r^0 \ln r = A + B \ln r , \end{equation*}
and if $$n > 0\text{,}$$ we get
\begin{equation*} R_n = A r^n + B r^{-n} . \end{equation*}
The function $$u(r,\theta)$$ must be finite at the origin, that is, when $$r=0\text{.}$$ So $$B=0$$ in both cases. Set $$A=1$$ in both cases as well; the constants in $$\Theta_n$$ will pick up the slack so nothing is lost. Let
\begin{equation*} R_0 = 1 , \qquad \text{and} \qquad R_n = r^n . \end{equation*}
Hence our building block solutions are
\begin{equation*} \begin{aligned} & u_0(r,\theta) = \frac{a_0}{2} , & u_n(r,\theta) = a_n r^n \cos(n \theta) + b_n r^n \sin(n \theta) . \end{aligned} \end{equation*}
Putting everything together our solution is:
\begin{equation*} \mybxbg{~~ u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n r^n \cos(n \theta) + b_n r^n \sin(n \theta) . ~~} \end{equation*}
We look at the boundary condition in (4.26),
\begin{equation*} g(\theta) = u(1,\theta) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(n \theta) + b_n \sin(n \theta) . \end{equation*}
Therefore, to solve (4.26) we expand $$g(\theta)\text{,}$$ which is a $$2\pi$$-periodic function, as a Fourier series, and then multiply the $$n^{\text{th}}$$ term by $$r^n\text{.}$$ To find the $$a_n$$ and the $$b_n$$ we compute
\begin{equation*} a_n = \frac{1}{\pi} \int_{-\pi}^\pi g(\theta) \cos (n\theta) \, d\theta , \qquad \text{and} \qquad b_n = \frac{1}{\pi} \int_{-\pi}^\pi g(\theta) \sin (n\theta) \, d\theta. \end{equation*}
#### Example4.10.1.
Suppose we wish to solve
\begin{equation*} \begin{aligned} & \Delta u = 0 , \qquad 0 \leq r < 1, \quad -\pi < \theta \leq \pi,\\ & u(1,\theta) = \cos(10\,\theta), \qquad -\pi < \theta \leq \pi. \end{aligned} \end{equation*}
The solution is
\begin{equation*} u(r,\theta) = r^{10} \cos(10\,\theta) . \end{equation*}
See the plot in Figure 4.26. The thing to notice in this example is that the effect of a high frequency is mostly felt at the boundary. In the middle of the disc, the solution is very close to zero. That is because $$r^{10}$$ is rather small when $$r$$ is close to 0.
#### Example4.10.2.
Let us solve a more difficult problem. Consider a long rod with circular cross section of radius 1. Suppose we wish to solve the steady state heat problem in the rod. If the rod is long enough, we simply need to solve the Laplace equation in two dimensions. Let us put the center of the rod at the origin and we have exactly the region we are currently studying—a circle of radius 1. For the boundary conditions, suppose in Cartesian coordinates $$x$$ and $$y\text{,}$$ the temperature on the boundary is 0 when $$y < 0\text{,}$$ and it is $$2y$$ when $$y > 0\text{.}$$
Let us set the problem up. As $$y = r\sin(\theta)\text{,}$$ then on the circle of radius 1, that is, where $$r=1\text{,}$$ we have $$2y = 2\sin(\theta)\text{.}$$ So
\begin{equation*} \begin{aligned} & \Delta u = 0 , \qquad 0 \leq r < 1, \quad -\pi < \theta \leq \pi,\\ & u(1,\theta) = \begin{cases} 2\sin(\theta) & \text{if } \; \phantom{-}0 \leq \theta \leq \pi, \\ 0 & \text{if } \; {-\pi} < \theta < 0. \end{cases} \end{aligned} \end{equation*}
We must now compute the Fourier series for the boundary condition. By now the reader has plentiful experience in computing Fourier series and so we simply state that
\begin{equation*} u(1,\theta) = \frac{2}{\pi} + \sin(\theta) + \sum_{n=1}^\infty \frac{-4}{\pi(4n^2-1)} \cos(2n\theta) . \end{equation*}
We now simply write the solution (see Figure 4.27) by multiplying by $$r^n$$ in the right places.
\begin{equation*} u(r,\theta) = \frac{2}{\pi} + r\sin(\theta) + \sum_{n=1}^\infty \frac{-4r^{2n}}{\pi(4n^2-1)} \cos(2n\theta) . \end{equation*}
### Subsection4.10.3Poisson kernel
There is another way to solve the Dirichlet problem with the help of an integral kernel. That is, we will find a function $$P(r,\theta,\alpha)$$ called the Poisson kernel
1
Named for the French mathematician Siméon Denis Poisson (1781–1840).
such that
\begin{equation*} u(r,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi} P(r,\theta,\alpha) \, g(\alpha) \,d\alpha . \end{equation*}
While the integral will generally not be solvable analytically, it can be evaluated numerically. In fact, unless the boundary data is given as a Fourier series already, it may be much easier to numerically evaluate this formula as there is only one integral to evaluate.
The formula also has theoretical applications. For instance, as $$P(r,\theta,\alpha)$$ will have infinitely many derivatives, then via differentiating under the integral we find that the solution $$u(r,\theta)$$ has infinitely many derivatives, at least when inside the circle, $$r < 1\text{.}$$ By “having infinitely many derivatives,” what you should think of is that $$u(r,\theta)$$ has “no corners” and all of its partial derivatives of all orders exist and also have “no corners.”
We will compute the formula for $$P(r,\theta,\alpha)$$ from the series solution, and this idea can be applied anytime you have a convenient series solution where the coefficients are obtained via integration. Hence you can apply this reasoning to obtain such integral kernels for other equations, such as the heat equation. The computation is long and tedious, but not overly difficult. Since the ideas are often applied in similar contexts, it is good to understand how this computation works.
What we do is start with the series solution and replace the coefficients with the integrals that compute them. Then we try to write everything as a single integral. We must use a different dummy variable for the integration and hence we use $$\alpha$$ instead of $$\theta\text{.}$$
\begin{equation*} \begin{split} u(r,\theta) & = \frac{a_0}{2} + \sum_{n=1}^\infty a_n r^n \cos(n \theta) + b_n r^n \sin(n \theta) \\ & = \underbrace{ \left( \frac{1}{2\pi} \int_{-\pi}^\pi g(\alpha) \, d\alpha \right) }_{\frac{a_0}{2}} + \sum_{n=1}^\infty \underbrace{ \left( \frac{1}{\pi} \int_{-\pi}^\pi g(\alpha) \cos (n\alpha) \, d\alpha \right) }_{a_n} r^n \cos(n \theta) + \\ & ~~~~~~~~ + \underbrace{ \left( \frac{1}{\pi} \int_{-\pi}^\pi g(\alpha) \sin (n\alpha) \, d\alpha \right) }_{b_n} r^n \sin(n \theta) \\ & = \frac{1}{2\pi} \int_{-\pi}^\pi \left( g(\alpha) + 2 \sum_{n=1}^\infty g(\alpha) \cos (n\alpha) \, r^n \cos(n \theta) + g(\alpha) \sin (n\alpha) \, r^n \sin(n \theta) \right) \,d\alpha \\ & = \frac{1}{2\pi} \int_{-\pi}^\pi \underbrace{ \left( 1 + 2 \sum_{n=1}^\infty r^n \bigl( \cos (n\alpha) \cos(n \theta) + \sin (n\alpha) \sin(n \theta) \bigr) \right) }_{P(r,\theta,\alpha)} g(\alpha) \,d\alpha \end{split} \end{equation*}
OK, so we have what we wanted, the expression in the parentheses is the Poisson kernel, $$P(r,\theta,\alpha)\text{.}$$ However, we can do a lot better. It is still given as a series, and we would really like to have a nice simple expression for it. We must work a little harder. The trick is to rewrite everything in terms of complex exponentials. Let us work just on the kernel.
\begin{equation*} \begin{split} P(r,\theta,\alpha) & = 1 + 2 \sum_{n=1}^\infty r^n \bigl( \cos (n\alpha) \cos(n \theta) + \sin (n\alpha) \sin(n \theta) \bigr) \\ & = 1 + 2 \sum_{n=1}^\infty r^n \cos \bigl(n(\theta-\alpha)\bigr) \\ & = 1 + \sum_{n=1}^\infty r^n \bigl( e^{in(\theta-\alpha)} + e^{-in(\theta-\alpha)} \bigr) \\ & = 1 + \sum_{n=1}^\infty {\bigl( re^{i(\theta-\alpha)}\bigr)}^{n} + \sum_{n=1}^\infty {\bigl( re^{-i(\theta-\alpha)}\bigr)}^{n} . \end{split} \end{equation*}
In the expression above, we recognize the geometric series. Recall from calculus that if $$z$$ is a complex number where $$\lvert z \rvert < 1\text{,}$$ then
\begin{equation*} \sum_{n=1}^\infty z^n = \frac{z}{1-z} . \end{equation*}
Note that $$n$$ starts at $$1$$ and that is why we have the $$z$$ in the numerator. It is the standard geometric series multiplied by $$z\text{.}$$ We can use $$z = re^{i(\theta-\alpha)}\text{,}$$ as lo and behold $$\lvert re^{i(\theta-\alpha)} \rvert = r < 1\text{.}$$ Let us continue with the computation.
\begin{equation*} \begin{split} P(r,\theta,\alpha) & = 1 + \sum_{n=1}^\infty {\bigl( re^{i(\theta-\alpha)}\bigr)}^{n} + \sum_{n=1}^\infty {\bigl( re^{-i(\theta-\alpha)}\bigr)}^{n} \\ & = 1 + \frac{re^{i(\theta-\alpha)}}{1-re^{i(\theta-\alpha)}} + \frac{re^{-i(\theta-\alpha)}}{1-re^{-i(\theta-\alpha)}} \\ & = \frac{ \bigl(1-re^{i(\theta-\alpha)}\bigr)\bigl(1-re^{-i(\theta-\alpha)}\bigr) + \bigl(1-re^{-i(\theta-\alpha)}\bigr)re^{i(\theta-\alpha)} + \bigl(1-re^{i(\theta-\alpha)}\bigr)re^{-i(\theta-\alpha)}} {\bigl(1-re^{i(\theta-\alpha)}\bigr)\bigl(1-re^{-i(\theta-\alpha)}\bigr)} \\ & = \frac{1 -r^2}{1 - re^{i(\theta-\alpha)} - re^{-i(\theta-\alpha)} +r^2} \\ & = \frac{1 -r^2}{1 - 2r\cos(\theta-\alpha) +r^2} . \end{split} \end{equation*}
That’s a formula we can live with. The solution to the Dirichlet problem using the Poisson kernel is
\begin{equation*} \mybxbg{~~ u(r,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1 -r^2}{1 - 2r\cos(\theta-\alpha) +r^2} g(\alpha) \, d\alpha . ~~} \end{equation*}
Sometimes the formula for the Poisson kernel is given together with the constant $$\frac{1}{2\pi}\text{,}$$ in which case we should of course not leave it in front of the integral. Also, often the limits of the integral are given as 0 to $$2\pi\text{;}$$ everything inside is $$2\pi$$-periodic in $$\alpha\text{,}$$ so this does not change the integral.
Let us not leave the Poisson kernel without explaining its geometric meaning. Let $$s$$ be the distance from $$(r,\theta)$$ to $$(1,\alpha)\text{.}$$ You may recall from calculus that this distance $$s$$ in polar coordinates is given precisely by the square root of $$1 - 2r\cos(\theta-\alpha) +r^2\text{.}$$ That is, the Poisson kernel is really the formula
\begin{equation*} \frac{1-r^2}{s^2} . \end{equation*}
One final note we make about the formula is that it is really a weighted average of the boundary values. First let us look at what happens at the origin, that is when $$r=0\text{.}$$
\begin{equation*} \begin{split} u(0,0) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1 -0^2}{1 - 2(0)\cos(0-\alpha) +0^2} g(\alpha) \, d\alpha \\ & = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(\alpha) \, d\alpha . \end{split} \end{equation*}
So $$u(0,0)$$ is precisely the average value of $$g(\theta)$$ and therefore the average value of $$u$$ on the boundary. This is a general feature of harmonic functions, the value at some point $$p$$ is equal to the average of the values on a circle centered at $$p\text{.}$$
What the formula says is that the value of the solution at any point in the circle is a weighted average of the boundary data $$g(\theta)\text{.}$$ The kernel is bigger when $$(1,\alpha)$$ is closer to $$(r,\theta)\text{.}$$ Therefore when computing $$u(r,\theta)\text{,}$$ we give more weight to the values $$g(\alpha)$$ when $$(1,\alpha)$$ is closer to $$(r,\theta)$$ and less weight to the values $$g(\alpha)$$ when $$(1,\alpha)$$ far from $$(r,\theta)\text{.}$$
### Exercises4.10.4Exercises
#### 4.10.2.
Using series solve $$\Delta u = 0\text{,}$$ $$u(1,\theta) = \lvert \theta \rvert\text{,}$$ for $$-\pi < \theta \leq \pi\text{.}$$
#### 4.10.3.
Using series solve $$\Delta u = 0\text{,}$$ $$u(1,\theta) = g(\theta)$$ for the following data. Hint: trig identities.
1. $$\displaystyle g(\theta) = \nicefrac{1}{2} + 3\sin(\theta) + \cos(3\theta)$$
2. $$\displaystyle g(\theta) = 3\cos(3\theta) + 3\sin(3\theta) + \sin(9\theta)$$
3. $$\displaystyle g(\theta) = 2 \cos(\theta+1)$$
4. $$\displaystyle g(\theta) = \sin^2(\theta)$$
#### 4.10.4.
Using the Poisson kernel, give the solution to $$\Delta u = 0\text{,}$$ where $$u(1,\theta)$$ is zero for $$\theta$$ outside the interval $$[-\nicefrac{\pi}{4},\nicefrac{\pi}{4}]$$ and $$u(1,\theta)$$ is 1 for $$\theta$$ on the interval $$[-\nicefrac{\pi}{4},\nicefrac{\pi}{4}]\text{.}$$
#### 4.10.5.
1. Draw a graph for the Poisson kernel as a function of $$\alpha$$ when $$r=\nicefrac{1}{2}$$ and $$\theta = 0\text{.}$$
2. Describe what happens to the graph when you make $$r$$ bigger (as it approaches 1).
3. Knowing that the solution $$u(r,\theta)$$ is the weighted average of $$g(\theta)$$ with Poisson kernel as the weight, explain what your answer to part b) means.
#### 4.10.6.
Let $$g(\theta)$$ be the function $$xy = \cos \theta \sin \theta$$ on the boundary. Use the series solution to find a solution to the Dirichlet problem $$\Delta u = 0\text{,}$$ $$u(1,\theta) = g(\theta)\text{.}$$ Now convert the solution to Cartesian coordinates $$x$$ and $$y\text{.}$$ Is this solution surprising? Hint: use your trig identities.
#### 4.10.7.
Carry out the computation we needed in the separation of variables and solve $$r^2 R'' + r R' - n^2 R = 0\text{,}$$ for $$n=0,1,2,3,\ldots\text{.}$$
#### 4.10.8.
(challenging) Derive the series solution to the Dirichlet problem if the region is a circle of radius $$\rho$$ rather than 1. That is, solve $$\Delta u = 0\text{,}$$ $$u(\rho,\theta) = g(\theta)\text{.}$$
#### 4.10.9.
(challenging)
1. Find the solution for $$\Delta u = 0\text{,}$$ $$u(1,\theta) = x^2y^3 + 5 x^2\text{.}$$ Write the answer in Cartesian coordinates.
2. Now solve $$\Delta u = 0\text{,}$$ $$u(1,\theta) = x^k y^\ell\text{.}$$ Write the solution in Cartesian coordinates.
3. Suppose you have a polynomial $$P(x,y) = \sum_{j=0}^m \sum_{k=0}^n c_{j,k} x^j y^k\text{,}$$ solve $$\Delta u = 0\text{,}$$ $$u(1,\theta) = P(x,y)$$ (that is, write down the formula for the answer). Write the answer in Cartesian coordinates.
Notice the answer is again a polynomial in $$x$$ and $$y\text{.}$$ See also Exercise 4.10.6.
#### 4.10.101.
Using series solve $$\Delta u = 0\text{,}$$ $$u(1,\theta) = 1+ \sum\limits_{n=1}^\infty \frac{1}{n^2}\sin(n\theta)\text{.}$$
Answer.
$$u = 1+ \sum\limits_{n=1}^\infty \frac{1}{n^2}r^n\sin(n\theta)$$
#### 4.10.102.
Using the series solution find the solution to $$\Delta u = 0\text{,}$$ $$u(1,\theta) = 1- \cos(\theta)\text{.}$$ Express the solution in Cartesian coordinates (that is, using $$x$$ and $$y$$).
Answer.
$$u = 1-x$$
#### 4.10.103.
1. Try and guess a solution to $$\Delta u = -1\text{,}$$ $$u(1,\theta) = 0\text{.}$$ Hint: try a solution that only depends on $$r\text{.}$$ Also first, don’t worry about the boundary condition.
2. Now solve $$\Delta u = -1\text{,}$$ $$u(1,\theta) = \sin(2\theta)$$ using superposition.
Answer.
a) $$u = \frac{-1}{4} r^2 + \frac{1}{4}$$ b) $$u = \frac{-1}{4} r^2 + \frac{1}{4} + r^2 \sin(2\theta)$$
#### 4.10.104.
(challenging) Derive the Poisson kernel solution if the region is a circle of radius $$\rho$$ rather than 1. That is, solve $$\Delta u = 0\text{,}$$ $$u(\rho,\theta) = g(\theta)\text{.}$$
Answer.
$$\displaystyle u(r,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{\rho^2 -r^2}{\rho^2 - 2r\rho\cos(\theta-\alpha) +r^2} g(\alpha) \, d\alpha$$
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## Vocabulary
Base: The number directly preceding an exponent
EX: a2 -> a is the base
Exponent: The number (written in superscript) used to express how many times a base is multiplied by itself
EX: a4 = a * a * a * a -> 4 is the exponent
EX: 43 = 4 * 4 * 4 = 64 -> 3 is the exponent
## Lesson
Exponents are a simple way to represent repeated multiplication. For example a x a = a2. There are a few simple rules for exponents that help reduce very large problems to simple little ones. The rules are as follows:
1) The exponent of any number is always a one (1): a = a1
2) When we multiply the same base we add our exponenents: a3 x a2 = a3 + 2 = a5
3) When we divide the same base we subtract our exponents: a6 / a4 = a6 - 4 a2
4) When we raise a power to a power we multiply our exponents: (a2)3 = a2 * 3 = a6
5) When we raise a PRODUCT to a power we raise both parts of the product to the power: (ab)3 = a3b3 [NOTE: This ONLY works with multiplication and NOT addition: (a + b)3 a3 + b3]
6) When we raise a QUOTIENT to a power we raise both parts of the quotient to the power: (a/b)2 = a2 / b2 [NOTE: This ONLY works with division and NOT subtraction: (a - b)2 a2 - b2]
## Example Problems
x3 * x6 = ?? [= x9]
ax * a3 = ?? [= ax + 3]
x4 / x2 = ?? [= x2]
(a4b)3 = ?? [= a12b3]
(a2b3c)3 = ?? [= a6b9c3] | 449 | 1,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2014-15 | longest | en | 0.829683 |
https://pidlaboratory.com/8-integrating-unit-with-inertia/ | 1,725,734,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00009.warc.gz | 441,853,844 | 28,711 | # Automatics
## Chapter 8 Integrating Unit with Inertia
Chapter. 8.1 Introduction
Fig. 8-1
Transmittance of the integral unit with inertia.
Do you remember the integral unit? The step x(t) at the input caused the output signal y(t) to increase to infinity with a constant speed. I emphasize. With a constant speed from the beginning of the step x(t).
The integral unit with inertia also tends to infinity. But he does it a little differently. At the beginning of the step, x(t) “accelerates” starting from a velocity V=0 and ending with a certain fixed V=const. So it is a more accurate approximation of, for example, an actuator than a ideal integrating unit.
Chapter 8.2 k=1 T=3 sec with slider and bar graph
Again, we’ll start with the bargraph to get you acquainted with the dynamics.
The inertial unit is a series connection of the integrating 1/sTi and the inertial 1/(1+sT). Here Ti=1sec and T=3sec
Fig. 8-2
The slider is initially set to 0. I gave it x(t)=+0.1=max. You will observe waveforms similar to those for the previously tested integrating unit. You also need to give 0 to the input to stop the y(t) output signal. A digital meter will be useful for the 0* setting. If you managed to stop y(t) (does not increase or decrease), then a typical feature of PI or PID controllers will be revealed. The fixed output of the yr(t) controller is different from zero, although the input is zero!
The waveforms are similar to the integral unit, but not the same! I hope you can see the inertia. Especially when you enter +max and in a moment -max. You will notice that the signal continues to increase for a short time even though x(t)=0 and even decreases! The phenomenon of inertia will be clearer in the next experiment with the oscilloscope.
*It is difficult to set an exact 0 on a digital meter. Rather, it is a signal close to 0, e.g. x(t)=+0.002. This means that y(t) will “almost stop” i.e. y(t) is increasing very slowly.
Chapter 8.3 Ti=1 sec and T=3 sec with jump and oscilloscope
The integrating unit with inertia is a series connection of the integrating unit 1/sTi and the inertial unit 1/(1+sT). Ti=1sec and T=3sec.
Fig. 8-3
The input is a step x(t)=1. The signal yp(t) is after the integrator, y(t) is the output. At the beginning, the inertia T=3 sec is clearly visible. In the steady state, both signals grow at the same speed. The integrating unit with inertia is an example of the so-called astatic system. At non-zero x(t), the signal y(t) increases or decreases. For this integrating term Ti=1 sec as the time after which yp(t) becomes equal to x(t). This is also the time after which, in the steady state, y(t) will increase by the step value x(t)=1, i.e. Ti=1 sec.
In steady state, i.e. after approximately 18 sec, the output signal y(t) is delayed by T=3 sec.
Chapter 8.4 Ti=1 sec T=3 sec with a single square pulse and an oscilloscope
The inertial integrating unit is acted upon by a single rectangular pulse.
Fig. 8-4
This is a more accurate approximation of the actuator – a motor with a gearbox as an ideal integrating element whose arm can control the valve position. This example is discussed in Chapter 4. Integrating Unit. Here, however, we also take into account inertia. The black x(t) input is the voltage at the motor and the red output y(t) is the valve stem position. Needless to say, inertia, consisting mainly of mechanical inertia + some electrical inertiainductance, “spoils” the quality of the device. In an actuator, it would be difficult to extract the yp(t)-signal directly after the integrator. This is only possible in the model as in Fig. 8-4.
Chapter 8.5 Integrating Unit with inertia k=1 T=1.25 sec with positive and negative rectangular pulse and oscilloscope
We will see how the actuator works by “searching” for its position.
I.e. it turns once in one direction and once in the other until it finds the position set by the regulator and stops. Therefore, the inertial integrating unit is acted upon by a positive and negative rectangular impulse.
Fig. 8-5
The blue yp(t) is an ideal actuator without inertia. The red y(t) is the real actuator. The actuator from Fig. 8-4 with inertia T=3 sec was bought by a poor customer. Now the client is Lord, who wants the actuator to respond quickly. That’s why I sold the actuator with lower inertia T=1.25 sec. The effects are visible. The actuator will reach the set speed faster. Therefore, it will also reach the set y(t) faster. First, a positive pulse set the valve to y(t)=8, the actuator stood for a while (because that’s what e.g. the regulator wanted) and then a negative pulse set the valve to y(t)=4.
Scroll to Top | 1,175 | 4,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.894157 |
http://www.jiskha.com/display.cgi?id=1311793404 | 1,495,914,488,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609054.55/warc/CC-MAIN-20170527191102-20170527211102-00129.warc.gz | 674,595,804 | 3,531 | # Math modeling
posted by on .
Write the formula for an exponential function with initial value 24 and growth factor 1.4
• Math modeling - ,
Vt = Vo*(1.4)^t.
Vt = Final value.
Vo = Inital value.
t = Time.
1.4 = Growth factor.
Vt = 24(1.4)^t. | 80 | 250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-22 | latest | en | 0.706106 |
http://fixunix.com/ntp/347159-leap-second-functional-question-3.html | 1,472,325,525,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982924728.51/warc/CC-MAIN-20160823200844-00106-ip-10-153-172-175.ec2.internal.warc.gz | 87,597,750 | 14,773 | # Leap second functional question - NTP
This is a discussion on Leap second functional question - NTP ; > You are asking for the impossible! Leap seconds keep time in synch with That was the intention. I was pointing out that you can only use "true" time to represent historic civil times, you cannot calculate the "true" time ...
# Thread: Leap second functional question
1. ## Re: Leap second functional question
> You are asking for the impossible! Leap seconds keep time in synch with
That was the intention. I was pointing out that you can only use "true"
time to represent historic civil times, you cannot calculate the "true"
time of a future civil time beyond the first candidate leap second.
> the earth's rotation. The rate of rotation is subject to small
> variations. This is why leap seconds occur at irregular intervals and
> why it is possible to have a negative leap second, although I don't
> recall that we ever had one.
>
Exactly, which is why the question casts doubt on the use of "true"
time, which was my intention.
2. ## Re: Leap second functional question
>>I suppose it is rather unfair of me to point out that I'm a
>>professional astronomer, and can assure you lots of astronomers use
>>UTC.
>
> No very fair. Professional information should trump amatuer every time.
> However, I suspect strongly that when you actually do your calculations you
> use TAI.
What timescale we use depends on the calculation we are doing. There
are times that calculations are done in a TAI like system. There are
times the calculations are done in UTC. UTC is currently kept within a
defined range of UT1, which is very useful.
3. ## Re: Leap second functional question
On 2008-02-20, Unruh wrote:
> No orbital dynamics obeys Newton's ( or Einstein's) law of gravitation only
> in TAI not in UTC ot UT1.
High precision orbital dynamics, such as the JPL's DE405, are done in
Barycentric Dynamic Time, which runs at a different rate than TAI, but
is "TAI like" in that it has no leap seconds.
> Of course I cannot. Leap seconds are to take into account the random
> variations in the earth's rotation to one part in about 10^8.
Leap seconds take into account the systematic slowing of the earth's
rotation.
4. ## Re: Leap second functional question
Unruh wrote:
> "David J Taylor" writes:
>
>> Unruh wrote:
> Note that leap seconds have been happening so rarely that almost noone has
> seen one happen, or watched the system as the clock goes through that.
>
I guess you weren't around in December, 2006.
> Having ntp run on TAI would certainly be simpler, but would of course make
> the time keeping on the system much more complicated.
That question has already been discussed at length in this newsgroup.
Danny
5. ## Re: Leap second functional question
David Woolley wrote:
> 4p4's ntpd/ntp_loopfilter.c:
> /*
> * Set the leap bits in the status word, but
> * only on the last day of June or December.
> */
> tstamp = peer->rec.l_ui - JAN_1970;
> tm = gmtime(&tstamp);
> if (tm != NULL) {
> if ((tm->tm_mon + 1 == 6 &&
> tm->tm_mday == 30) || (tm->tm_mon +
> 1 == 12 && tm->tm_mday == 31)) {
> ntv.status |= STA_INS;
> else if (leap_next &
> LEAP_DELSECOND)
> ntv.status |= STA_DEL;
> }
> }
That comment is wrong. IIRC it can nominally be set at the last day of
any month. The norm is the last day of June or December and in practice
has been only the last day of December. Notice that the code does not
allow a leap second in February which only has 28 or 29 (this year) days
in the month. Either this is a bug or it's already been fixed.
Danny
6. ## Re: Leap second functional question
Danny Mayer wrote:
> }
>
> That comment is wrong. IIRC it can nominally be set at the last day of
The comment matches the code, which should be the most recent released
version. Someone has already said that it is a known bug, and I think
they said it is fixed in the development version.
> any month. The norm is the last day of June or December and in practice
> has been only the last day of December. Notice that the code does not
> allow a leap second in February which only has 28 or 29 (this year) days
> in the month. Either this is a bug or it's already been fixed.
7. ## Re: Leap second functional question
Danny,
That snapshot is older than I thought. That particular botch of code
along with other botches was vulnerable to whimsical errors, like server
leap bits popping up and down or leapsecond file update. The current
code here can leap at the end of any month, which seems to be the
prevailing view of the standards folks. However, I can't find
confirmation at the IERS site.
Dave
Danny Mayer wrote:
> David Woolley wrote:
>
>>4p4's ntpd/ntp_loopfilter.c:
>> /*
>> * Set the leap bits in the status word, but
>> * only on the last day of June or December.
>> */
>> tstamp = peer->rec.l_ui - JAN_1970;
>> tm = gmtime(&tstamp);
>> if (tm != NULL) {
>> if ((tm->tm_mon + 1 == 6 &&
>> tm->tm_mday == 30) || (tm->tm_mon +
>> 1 == 12 && tm->tm_mday == 31)) {
>> ntv.status |= STA_INS;
>> else if (leap_next &
>> LEAP_DELSECOND)
>> ntv.status |= STA_DEL;
>> }
>> }
>
>
> That comment is wrong. IIRC it can nominally be set at the last day of
> any month. The norm is the last day of June or December and in practice
> has been only the last day of December. Notice that the code does not
> allow a leap second in February which only has 28 or 29 (this year) days
> in the month. Either this is a bug or it's already been fixed.
>
> Danny
8. ## Re: Leap second functional question
David Woolley writes:
>> You are asking for the impossible! Leap seconds keep time in synch with
>That was the intention. I was pointing out that you can only use "true"
>time to represent historic civil times, you cannot calculate the "true"
>time of a future civil time beyond the first candidate leap second.
>> the earth's rotation. The rate of rotation is subject to small
>> variations. This is why leap seconds occur at irregular intervals and
>> why it is possible to have a negative leap second, although I don't
>> recall that we ever had one.
>>
>Exactly, which is why the question casts doubt on the use of "true"
>time, which was my intention.
Well, I suppose one could assume no leap seconds into the future. It is as
good as UTC which does exactly that, except it also assumes no leap seconds
into the past, where we know it is wrong. Getting it half right would seem
9. ## Re: Leap second functional question
Greg Hennessy writes:
>On 2008-02-20, Unruh wrote:
>> No orbital dynamics obeys Newton's ( or Einstein's) law of gravitation only
>> in TAI not in UTC ot UT1.
>High precision orbital dynamics, such as the JPL's DE405, are done in
>Barycentric Dynamic Time, which runs at a different rate than TAI, but
>is "TAI like" in that it has no leap seconds.
It does? The second was defined to make the speed of light a consant. Are
you saying they use a system in which the speed of light changes from time
to time? That sounds horrible.
>> Of course I cannot. Leap seconds are to take into account the random
>> variations in the earth's rotation to one part in about 10^8.
>Leap seconds take into account the systematic slowing of the earth's
>rotation.
While it may be slowing on average, it is noise and it could well speed up
as well. For example there have been no leap seconds for the past 4 years
or so, while the "aveage slowing would have one every 1.5 to 2 years right
now. Ie there has been a speedup against the average recently
10. ## Re: Leap second functional question
[email protected] (Danny Mayer) writes:
>Unruh wrote:
>> "David J Taylor" writes:
>>
>>> Unruh wrote:
>> Note that leap seconds have been happening so rarely that almost noone has
>> seen one happen, or watched the system as the clock goes through that.
>>
>I guess you weren't around in December, 2006.
>> Having ntp run on TAI would certainly be simpler, but would of course make
>> the time keeping on the system much more complicated.
>That question has already been discussed at length in this newsgroup.
And will keep getting discussed since there is no resolution which is
uniformly positive.
11. ## Re: Leap second functional question
Unruh wrote:
>>> Having ntp run on TAI would certainly be simpler, but would of course make
>>> the time keeping on the system much more complicated.
>
>> That question has already been discussed at length in this newsgroup.
> And will keep getting discussed since there is no resolution which is
> uniformly positive.
Actually no. We don't get a vote on this. This is being voted on by the
ITU (or whatever the replacement is for the CCITT) if I recall
correctly. It's a separate question whether or not NTP will continue
with UTC if they do something stupid with the decision.
Danny
12. ## Re: Leap second functional question
On 2008-02-22, Unruh wrote:
>>High precision orbital dynamics, such as the JPL's DE405, are done in
>>Barycentric Dynamic Time, which runs at a different rate than TAI, but
>>is "TAI like" in that it has no leap seconds.
>
> It does? The second was defined to make the speed of light a consant. Are
> you saying they use a system in which the speed of light changes from time
> to time? That sounds horrible.
I am saying that high precision orbital dynamics are done from the
reference frame of the solar system, which we call the barycentric
frame. The barycentric time rate is different from that of TAI by a
factor of about 1e-15 if I remember correctly.
> While it may be slowing on average, it is noise and it could well speed up
> as well.
Noise implies that the changes are both positive and negative, in
which case a leap second won't be needed. If it is systematic,
i.e. the changes are more in one direction than another, a leap second
will be needed.
13. ## Re: Leap second functional question
On Feb 21, 3:15 pm, "David L. Mills" wrote:
> The current
> code here can leap at the end of any month, which seems to be the
> prevailing view of the standards folks. However, I can't find
> confirmation at the IERS site.
Images of the IAU proceedings from the General Assembly in 1973 are at
http://www.ucolick.org/~sla/leapsecs...oc1973p154.gif
http://www.ucolick.org/~sla/leapsecs...oc1973p155.gif
This was the assembly immediately after leap seconds began (and the
first one at which the IAU could make any response, given that the
CCIR neglected to send them a letter in 1970 leaving them no standing
upon which to comment on leap seconds before they began). The changes
recommended by the IAU in 1973 were incorporated into the first
revision of what is now known as ITU-R TF.460, notably, that it was
foreseeable that leap seconds would be needed more than twice a year.
14. ## Re: Leap second functional question
Unruh wrote:
>> High precision orbital dynamics, such as the JPL's DE405, are done in
>> Barycentric Dynamic Time, which runs at a different rate than TAI, but
>> is "TAI like" in that it has no leap seconds.
>
> It does? The second was defined to make the speed of light a consant. Are
> you saying they use a system in which the speed of light changes from time
> to time? That sounds horrible.
I believe he is talking about the general relativity correction. Even
GPS would be wildly wrong if you used the earthbound definition of a
second for the clocks in the satellites.
15. ## Re: Leap second functional question
On Feb 21, 5:30 pm, Unruh wrote:
> >> Having ntp run on TAI would certainly be simpler, but would of course make
> >> the time keeping on the system much more complicated.
> >That question has already been discussed at length in this newsgroup.
>
> And will keep getting discussed since there is no resolution which is
> uniformly positive.
No, but the one I recently espoused in the LEAPSECS group
http://six.pairlist.net/pipermail/le...ry/000190.html
allows for a uniform underlying broadcast time scale without leap
seconds, those being accorded equal status with politically-mandated
time zone shifts, meaning they would go in zoneinfo rather than in the
kernel. It is also in accord with the results of the colloquium on
UTC held in Torino in 2003.
16. ## Re: Leap second functional question
[email protected] (Danny Mayer) writes:
>Unruh wrote:
>>>> Having ntp run on TAI would certainly be simpler, but would of course make
>>>> the time keeping on the system much more complicated.
>>
>>> That question has already been discussed at length in this newsgroup.
>> And will keep getting discussed since there is no resolution which is
>> uniformly positive.
>Actually no. We don't get a vote on this. This is being voted on by the
>ITU (or whatever the replacement is for the CCITT) if I recall
>correctly. It's a separate question whether or not NTP will continue
>with UTC if they do something stupid with the decision.
I did NOT advocate tht the world run on TAI. I was suggesting that ntp do
so, and that computer systems run on TAI. This makes the translation from
internal time (seconds since epoch) to civil time (UTC) slightly more
complex. Not only does the timezone file need to be consulted, but the
leapsecond file does as well each time you translate from the computer's
internal time to UTC as civil time. UTC is far too uselful for civil time
to be thrown out. Just as the Gregorian calendar reforms were to keep civil
time in sync with the earth's revolution around the sun, so leapseconds
keep the day in sync with the rotation of the earth. Since life is based
around the latter it is good to have time reflect that.
However it is not so good to have the conniptions in the internal time of
the computer try to keep track of that. Seconds since epoch is alrady so
abstract a concept that noone would notice that those seconds were real
seconds, rathr than second where a few a thrown away.
>Danny
17. ## Re: Leap second functional question
Greg Hennessy writes:
>On 2008-02-22, Unruh wrote:
>>>High precision orbital dynamics, such as the JPL's DE405, are done in
>>>Barycentric Dynamic Time, which runs at a different rate than TAI, but
>>>is "TAI like" in that it has no leap seconds.
>>
>> It does? The second was defined to make the speed of light a consant. Are
>> you saying they use a system in which the speed of light changes from time
>> to time? That sounds horrible.
>I am saying that high precision orbital dynamics are done from the
>reference frame of the solar system, which we call the barycentric
>frame. The barycentric time rate is different from that of TAI by a
>factor of about 1e-15 if I remember correctly.
Ah. OK, There are General relativistic corrections, agreed.
>> While it may be slowing on average, it is noise and it could well speed up
>> as well.
>Noise implies that the changes are both positive and negative, in
They are
>which case a leap second won't be needed. If it is systematic,
They would still be needed. Just because your computer's drift rate is both
positive and negative does not mean that compensation is not needed.
They do not necessarily average out on the time scale of years.
There is a net drift to longer days. but superposed on that is a noise
which even over the time scale ofyears makes a difference. An earthquake in
Java rearranges the moment of inertia of the earth and changes the rotation
rate of the earth, and it can be positive or negative.
>i.e. the changes are more in one direction than another, a leap second
>will be needed.
It is needed even if it equal in both directions over a long time.
18. ## Re: Leap second functional question
David Woolley writes:
>Unruh wrote:
>>> High precision orbital dynamics, such as the JPL's DE405, are done in
>>> Barycentric Dynamic Time, which runs at a different rate than TAI, but
>>> is "TAI like" in that it has no leap seconds.
>>
>> It does? The second was defined to make the speed of light a consant. Are
>> you saying they use a system in which the speed of light changes from time
>> to time? That sounds horrible.
>I believe he is talking about the general relativity correction. Even
>GPS would be wildly wrong if you used the earthbound definition of a
>second for the clocks in the satellites.
A common misconception. The GPS people actually dynaically track the time
delivered by the sattelites and adjust their scales accordingly. Even if
they know nothing of GR, they would have discovered that the clocks were
running a bit fast and applied a correction fudgefactor. The problem is
that the clocks on the board could be running at the wrong rate doe to
other environmental problems and they have to be able to correct for those
or your GPS will cease to work.
19. ## Re: Leap second functional question
On 2008-02-22, Unruh wrote:
>>I believe he is talking about the general relativity correction. Even
>>GPS would be wildly wrong if you used the earthbound definition of a
>>second for the clocks in the satellites.
>
> A common misconception. The GPS people actually dynaically track the time
> delivered by the sattelites and adjust their scales accordingly.
It is not a misconception. The tweaks given to the clocks are much
less than the difference between pre launch and on orbit clock rates.
> Even if
> they know nothing of GR, they would have discovered that the clocks were
> running a bit fast and applied a correction fudgefactor.
This sounds like you are saying that even if they didn't know GR, they
would have discovered it. Which may in fact be true, but it doesn't
change the fact that GPS would be wrong if you used the sea level
definiton of a second from the on orbit clocks.
20. ## Re: Leap second functional question
Hello Danny,
On Thursday, February 21, 2008 at 22:01:31 +0000, Danny Mayer wrote:
> The norm is the last day of June or December and in practice has been
> only the last day of December.
There were several leaps in June, the latest in 1997.
Serge.
--
Serge point Bets arobase laposte point net | 4,410 | 17,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-36 | latest | en | 0.953059 |
http://www.onlinemathlearning.com/area-of-trapezoid.html | 1,503,474,965,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00306.warc.gz | 619,213,432 | 15,005 | # Area of Trapezoid
Objective: I know how to calculate the area of a trapezoid.
Given that side b1 is parallel to side b2 and h is the vertical height between b1 and b2, the area of the trapezoid is given by the formula:
A =
Read the lesson on area of trapezoid if you need to learn how to calculate the area of a trapezoid.
Fill in all the gaps, then press "Check" to check your answers. Use the "Hint" button to get a free letter if an answer is giving you trouble. You can also click on the "[?]" button to get a clue. Note that you will lose points if you ask for hints or clues!
Find the area of each trapezoid.
height = 24 m
b1 = 19 m
b2 = 24 m
area =
height = 24.3 mm
b1 = 9.9 mm
b2 = 15.1 mm
area =
height = 21 m
b1 = 10 m
b2 = 5 m
area =
height = 7 mm
b1 = 19 mm
b2 = 26 mm
area =
height = 24.8 cm
b1 = 24.2 cm
b2 = 5.4 cm
area =
height = 29.7 mm
b1 = 10.2 mm
b2 = 16.8 mm
area =
height = 28 m
b1 = 21 m
b2 = 11 m
area =
height = 26 mm
b1 = 27 mm
b2 = 21 mm
area =
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We hope that the free math worksheets have been helpful. We encourage parents and teachers to select the topics according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles. | 502 | 1,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-34 | longest | en | 0.891001 |
https://dandysciencewriters.com/1-a-newly-issued-bond-has-a-maturity-of-10-years-and-pays-a-7-coupon-rate-with-coupons-coming-once-annually-the-bond-sells-at-par-a-what-are/ | 1,606,553,564,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00400.warc.gz | 247,981,551 | 9,766 | # 1) A newly issued bond has a maturity of 10 years and pays a 7% coupon rate (with coupons coming once annually). The bond sells at par. a) What are
1) A newly issued bond has a maturity of 10 years and pays a 7% coupon rate (with coupons coming once annually). The bond sells at par.
a) What are the convexity and duration of the bond?
b) Find the actual price of the bond assuming that its yield-to-maturity immediately increases from 7% to 8% (with maturity still at 10 years).
c) What price would be predicted by the duration equation (equation 16.3)? What is the percentage error of that calculation?
What price would be predicted by the duration-with-convexity rule (equation 16.5)? What is the percentage error of that calculation?
Step-by-step solution1. Step 1 of 11a)Convexity is the curvature of the bond’s price and yield relationship.The following table-A shows the calculation o convexity of a 10-year, 7%coupon… | 241 | 952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-50 | latest | en | 0.933982 |
https://au.mathworks.com/matlabcentral/cody/problems/624-get-the-length-of-a-given-vector/solutions/165020 | 1,606,435,557,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189030.27/warc/CC-MAIN-20201126230216-20201127020216-00396.warc.gz | 201,936,165 | 17,170 | Cody
# Problem 624. Get the length of a given vector
Solution 165020
Submitted on 19 Nov 2012
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 3]; y_correct = 3; assert(isequal(VectorLength(x),y_correct))
ans = 3
2 Fail
%% x = 1:10; y_correct = 10; assert(isequal(VectorLength(x),y_correct))
Error: Assertion failed.
3 Fail
%% x = rand(1,928); y_correct = 928; assert(isequal(VectorLength(x),y_correct))
Error: Assertion failed.
### Community Treasure Hunt
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Start Hunting! | 193 | 699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.656705 |
https://www.physicsforums.com/threads/kilowatt-electricity-unit-of-energy-10kwh-10units-5kw-1hr-1kw.7112/ | 1,716,047,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00298.warc.gz | 879,980,149 | 15,215 | # Kilowatt: Electricity Unit of Energy - 10kWh, 10units, 5kW, 1hr & 1kW
• da-outcast
In summary, the Kilowatt Hour is a unit of electrical energy and 10kWh can be equal to 10 units. This can also be expressed as 1kW for 10 hours or 5kW for 2 hours. The "h" in kWh stands for hours and in order to get 10kWh, 5kW must be used for 2 hours.
da-outcast
The Kilowatt _____1_____ is often referred to as the unit of electrical energy. 10kWh = ___2____ units, which could be 10 ____3____ for 1 hour, or 5kW for ___4____ hours, or ____5____ kW for 10 hours.
All i could get were
2 = 10 units
5 = 1kW
Is this really a college course? I can only say I hope the problems are going to get harder!
What do you think the "h" in kWh stands for?
You got 5 correct: 1 kW for 10 hours= 10 kWh of electrical energy.
Now think: 5 kW for how many hours= 10 kWh (In other words: 5 times what equals 10!) (that's number 4.)
Number 3, "10 ____3____ for 1 hour" is asking for the units.
10 *1= 10 so what units times hours gives kWh?
4 = 2 hours
1 = 1kW
3 = 1 hour
The Kilowatt is a unit of measurement for electrical energy, and is often used to measure the amount of electricity used in a given period of time. In this case, 10kWh is equivalent to 10 units, which could be used for 1 hour at a rate of 10kW, or for 2 hours at a rate of 5kW. Alternatively, it could also be used for 10 hours at a rate of 1kW. This demonstrates the flexibility of the Kilowatt unit and its ability to measure energy usage in various time frames.
## 1. What is a Kilowatt?
A kilowatt (kW) is a unit of power, which is the rate at which energy is consumed or produced. It is equal to 1,000 watts.
## 2. What is the relationship between Kilowatt and electricity?
Kilowatts are commonly used to measure the amount of electricity being used or produced. The more kilowatts used, the more electricity is being consumed.
## 3. How is Kilowatt related to other units of energy?
Kilowatt is a unit of power, while other units of energy such as kilowatt-hour (kWh) and joule (J) measure the amount of energy consumed or produced over time. Kilowatt-hour is commonly used to measure electricity usage on a household level.
## 4. What is the difference between 10kWh and 10kW?
10kWh refers to the amount of energy consumed or produced over a period of one hour, while 10kW refers to the rate at which energy is being consumed or produced per hour. In other words, 10kWh is a measure of energy, while 10kW is a measure of power.
## 5. How do Kilowatts relate to the concept of energy efficiency?
Kilowatts can be used to measure the efficiency of energy usage. The lower the number of kilowatts used to produce a certain amount of energy, the more energy efficient the process is.
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# request to rate my awa
Author Message
Intern
Joined: 01 Dec 2013
Posts: 4
Followers: 0
Kudos [?]: 0 [0], given: 3
request to rate my awa [#permalink]
### Show Tags
04 Mar 2014, 14:35
When the Apogee Company had all its operations in one location, it was more profitable than it is today. Therefore,
the Apogee Company should close down its field offices and conduct all its operations from a single location. Such
centralization would improve profitability by cutting costs and helping the company maintain better supervision of all
employees.”
The argument claims that Apogee Company was more profitable then today when it had all its operations in one location. Author is saying that the company should close down its field offices and conduct all its operations from a single location so such centralization would improve profitability by cutting costs and helping the company maintain better supervision of all employees. Stated in this way the argument fails to mention several key factors, on the baisis of which it could be evaluated. The conclusion relies on assumptions, for which is no clear evidence. Therefore, the argument is rather weak, unconvincing and has several flaws.
First, the argument readily assumes that because of all operations in one location company would be more profitable. This statement is stretch and not sustained in any way. Nowadays more and more companies are going global. With globalization people will recognize the company, their products, their benefits and make company will make more money. So if company will close down its field offices they will not know their costumers and they will not know what to offer them. This is a privilege of office in other location and employees can more efficiently deal with costumers from there.
Second, the argument claims that the company should close down its field offices and conduct all its operations from a single location. The author believe that such a centralization would improve profitability by cutting costs and helping the company maintain better supervision of all employees. Again this sentence is very weak and unsupported claim. Different people - different ideas - profit. For example if the company is making products, people get bored, and employees are soon out of ideas. In one location is not huge variarety. So with different locations employers would encourage employees ideas and company will make good profit out of it.
In conclusion, the argument is flawed for the above mentioned reasons and is therefore unconvincing. It could be considerably strengthened if the author clearly mentioned all the relevant facts. In order to assess the merits of a certain situation it is essential to have full knowledge of all contributing factors.
Senior Manager
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 356
Location: India
GMAT 1: 780 Q51 V46
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Re: request to rate my awa [#permalink]
### Show Tags
05 Mar 2014, 02:53
1
KUDOS
Hello Marijana;
In my opinion your response should get a 4 - 5 out of 6.
While your response was well written, there are a few things that you could do to make it more targeted towards responding to the task.
Let me discuss the biggest issues observed in the given argument.
1. Author assumes that the loss in profitability in the company was because of decentralisation
2. Author also assumes that centralising will not bring about any negative consequences to the company.
You need to focus on these questionable assumptions when discussing the flaws in the argument.
You have done this, yet you have not really shown HOW these assumptions make the argument weak: what could be the implication if they were proved wrong.
Let me show you:
First, the argument readily assumes that because of all operations in one location company would be more profitable.ideally start by discussing the assumption, here a more exact representation of the assumption would be that "the author assumes that the dip in profits was due to decentralisation" This statement is stretch and not sustained in any way. Nowadays more and more companies are going global. With globalization people will recognize the company, their products, their benefits and make company will make more money. So if company will close down its field offices they will not know their costumers and they will not know what to offer them. The problem here is that you've detracted from the assumption; focus on the assumption and show what flaws exist in this assumption (the causal flaw); show how this weakens the argument.This is a privilege of office in other location and employees can more efficiently deal with costumers from there.
Second, the argument claims that the company should close down its field offices and conduct all its operations from a single location. The author believe that such a centralization would improve profitability by cutting costs and helping the company maintain better supervision of all employees. Again this sentence is very weak and unsupported claim. Different people - different ideas - profit. For example if the company is making products, people get bored, and employees are soon out of ideas. In one location is not huge variarety. So with different locations employers would encourage employees ideas and company will make good profit out of it.
Again, a better approach for the second body paragraph could have been to discuss the potential negative consequence (harm/benefit) of the suggested plan to centralise.
To summarize:
In your body paragraphs you need to discuss the biggest flaws in the argument by
1. Stating the questionable assumption
2. Elaborate the weaknesses of the assumption
3. Show how this weakness the argument as a whole
4. If needed suggests ways in which this weakness could be amended.
Follow this and you should be able to score consistent near perfect scores!
Cheers,
Peo
Verbal Trainer - Crackverbal
_________________
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Re: request to rate my awa [#permalink]
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05 Mar 2014, 14:42
Re: request to rate my awa [#permalink] 05 Mar 2014, 14:42
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# request to rate my awa
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,731 | 7,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-04 | latest | en | 0.951609 |
http://mathhelpforum.com/new-users/221656-can-anyone-help-problem-force-2.html | 1,480,758,764,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00235-ip-10-31-129-80.ec2.internal.warc.gz | 174,807,067 | 13,531 | # Thread: Can anyone help with this problem on Force??
1. ## Re: Can anyone help with this problem on Force??
I have the modulus of elasticity = 140MN^-2 and I came to 140x10^6 is this the problem? sorry not modulus of elasticity I meant yield stress. Anyway very confused now.....
2. ## Re: Can anyone help with this problem on Force??
Originally Posted by Jock
A circulur column has a diameter of 80mm outside edge and 60mm inside edge.
Both ends are fixed with equal force being applied top and bottom.
Young's Modulus E = 200GN m-2
Yield Stress deltay = 140 MN m-2
What is the minimum length of column at which buckling is likely to happen?
If the column is half this length what load would you expect failure to occur?
Really struggling with this one if anyone can help id really appreciate it along with the working of how they came to their final answers......
Fcr = 4(Pi^2)EI/(4l^2) Roark, Formulas for Stress and Strain
MIn length for buckling: substitute Fcr = AxYS and solve for l.
If length is less than this, you will reach compressive yield strength before buckling.
At min length, Fcr=AxYS.
At half min length, Fcr=AxYS
3. ## Re: Can anyone help with this problem on Force??
Still not really following it plus I don't know the final answer so therefore don't know what I am actually looking for as a final result to see if I am right or not....
4. ## Re: Can anyone help with this problem on Force??
Originally Posted by Jock
Still not really following it plus I don't know the final answer so therefore don't know what I am actually looking for as a final result to see if I am right or not....
Suppose the column is 2mm high. You will never buckle it no matter how much force you apply, just like you could never buckle a penny no matter how much force you apply. How much force could you apply before “mushing” it: F=AxYS.
OK, now suppose you tried to apply the same force, F=AxYS, to the same column that was 100meters high? You would buckle way before you ever got there.
So at what point do you switch from crushing to buckling? Substitute F=AxYS in the buckling formula to get the length at which this happens.
If you understand that, getting the right answer is just a matter of some algebra and using the right units.
If I were a teacher and a student got the right answer with no explanation, I would give him 0/10. If he had the right method showing how he got it but the wrong answer, I would give him 10/10- no, 11/10 for not wasting his time on the algebra.
OK, yes, not knowing if you got the right answer can be frustrating, but only if you are not sure you are going about it right.
5. ## Re: Can anyone help with this problem on Force??
I want to do it right and show all my working but the explanations aren't making much sense and have never came across F=AxYs so don't know what it all stands for and all the components of it are. Just really wish I could see it in practice for the equation then follow it.........
6. ## Re: Can anyone help with this problem on Force??
Originally Posted by Jock
I have the modulus of elasticity = 140MN^-2 and I came to 140x10^6 is this the problem? sorry not modulus of elasticity I meant yield stress. Anyway very confused now.....
I gave you the formula to use back in post #13 of this thread. You tried to use that formula but made the mistake of plugging in yield stress instead of area moment of inertia for the 'I' term. Remember you have already calculated $I = 1.37 \times 10^{-6} \ m^4$ in post #6. Use that value in the formula for the 'I' term and you should get the right answer.
7. ## Re: Can anyone help with this problem on Force??
Just done it and calculator threw back 307610.388 which I hope is 3.17x10^6 so hope this is right!!!....
8. ## Re: Can anyone help with this problem on Force??
Simply having the formula doesn't do it. You have to understand that you have to substitue F=AxYS in the formula for Fcr and solve for l. Why is that a problem? The actual solving is just algebra, if you use consistent units, say N and m.
Just saw your post. What did you do and why? Saying you got some number is meaningless
9. ## Re: Can anyone help with this problem on Force??
Post 13 used the formula there and amended one of the figures in the equation to get that answer. Is the answer correct??? pi^2x200x10^9x1.37x10^-6/(0.5x5.93)^2 was what I did....
10. ## Re: Can anyone help with this problem on Force??
Originally Posted by Jock
Just done it and calculator threw back 307610.388 which I hope is 3.17x10^6 so hope this is right!!!....
Yes - that's the critical load in newtons.
11. ## Re: Can anyone help with this problem on Force??
so if I half the 5.93 the next answer will work by plugging in the correct info? Also how do I know what mode of failure these findings produce?....
12. ## Re: Can anyone help with this problem on Force??
so if I have done the last one correct my answer of 6.15x10^6 will be correct also?? if this is all right I am only left with mode of failure for them both if anyone can help?....
13. ## Re: Can anyone help with this problem on Force??
Would I be right in saying the E.S.R for the 5.93m length is 118.74 which equals buckling and E.S.R for the half length is 59.38 which equals yielding?
Page 2 of 2 First 12 | 1,328 | 5,276 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2016-50 | longest | en | 0.953218 |
https://en.wikipedia.org/wiki/Recession_velocity | 1,642,887,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00161.warc.gz | 290,271,923 | 13,801 | # Recessional velocity
(Redirected from Recession velocity)
Recessional velocity is the rate at which an extragalactic astronomical object recedes (becomes more distant) from an observer as a result of the expansion of the universe.[1] It can be measured by observing the wavelength shifts of spectral lines emitted by the object, known as the object's cosmological redshift.
## Application to cosmology
Hubble's law is the relationship between a galaxy's distance and its recessional velocity, which is approximately linear for galaxies at distances of up to a few hundred megaparsecs. It can be expressed as
${\displaystyle v_{r}=H_{0}D\ +v_{pec}}$
where ${\displaystyle H_{0}}$ is the Hubble constant, ${\displaystyle D}$ is the proper distance, ${\displaystyle v_{r}}$ is the object's recessional velocity, and ${\displaystyle v_{pec}}$ is the object's peculiar velocity.
The recessional velocity of a galaxy can be calculated from the redshift observed in its emitted spectrum. One application of Hubble's law is to estimate distances to galaxies based on measurements of their recessional velocities. However, for relatively nearby galaxies the peculiar velocity can be comparable to or larger than the recessional velocity, in which case Hubble's Law does not give a good estimate of an object's distance based on its redshift. In some cases (such as the Andromeda Galaxy, 2.5 million light-years away and approaching us at 300 km/s, or even Messier 81 at 12 million light-years away and approaching at 34 km/s) ${\displaystyle v_{r}}$ is negative (i.e., the galaxy's spectrum is observed to be blueshifted) as a result of the peculiar velocity.
## References
1. ^ "Hubble's Law". September 27, 2020. Retrieved April 11, 2021. | 412 | 1,742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | latest | en | 0.934762 |
https://www.tutorialtpoint.net/2022/01/binarization-data-pre-processing-technoque-python.html | 1,723,514,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00176.warc.gz | 802,901,243 | 50,370 | # Binarization data pre-processing technique on sample data
Binarization
Binarization is used when you want to convert a numerical feature vector into a Boolean vector. In the field of digital image processing, image binarization is the process by which a color or grayscale image is transformed into a binary image, that is, an image with only two colors (typically, black and white).
When it is used?
This technique is used for the recognition of objects, shapes, and, specifically, characters. Through binarization, it is possible to distinguish the object of interest from the background on which it is found. Skeletonization is instead an essential and schematic representation of the object, which generally preludes the subsequent real recognition.
### Data Pre-processing using Binarization
Let's see how to binarize data in Python:
To binarize data, we will use the preprocessing.Binarizer() function as follows
Syntax: class sklearn.preprocessing.Binarizer(*, threshold=0.0, copy=True)
Binarize data (set feature values to 0 or 1) according to a threshold.
Values greater than the threshold map to 1, while values less than or equal to the threshold map to 0. With the default threshold of 0, only positive values map to 1.
Binarization is a common operation on text count data where the analyst can decide to only consider the presence or absence of a feature rather than a quantified number of occurrences for instance.
It can also be used as a pre-processing step for estimators that consider boolean random variables (e.g. modeled using the Bernoulli distribution in a Bayesian setting).
Example 1:
```import numpy as np
from sklearn.preprocessing import Binarizer
data = np.array([[3, -1.5, 2, -5.4], [0, 4, -0.3, 2.1], [1, 3.3, -1.9, -4.3]])
print("---Orogonal data.............")
print(data)
data_binarized = Binarizer(threshold=1.4).transform(data)
print("...After Binarizer...... ")
print(data_binarized)
```
Output
```---Orogonal data.............
[[ 3. -1.5 2. -5.4]
[ 0. 4. -0.3 2.1]
[ 1. 3.3 -1.9 -4.3]]
...After Binarizer......
[[1. 0. 1. 0.]
[0. 1. 0. 1.]
[0. 1. 0. 0.]]```
The preprocessing.Binarizer() function binarizes data according to an imposed threshold. Values greater than the threshold map to 1, while values less than or equal to the threshold map to 0. With the default threshold of 0, only positive values map to 1. In our case, the threshold imposed is 1.4, so values greater than 1.4 are mapped to 1, while values less than 1.4 are mapped to 0.
Example 2:
```from sklearn.preprocessing import Binarizer
X = [[ 1., -1., 2.],
[ 2., 0., 0.],
[ 0., 1., -1.]]
transformer = Binarizer().fit(X) # fit does nothing.
print(transformer)
t=transformer.transform(X)
print(t)
```
Output:-
```Binarizer()
[[1. 0. 1.]
[1. 0. 0.]
[0. 1. 0.]]```
### How is Binarization used?
In machine learning, even the most complex concepts can be transformed into binary form. For example, to binarize the sentence “The dog ate the cat,” every word is assigned an ID (for example dog-1, ate-2, the-3, cat-4). Then replace each word with the tag to provide a binary vector. In this case the vector: <3,1,2,3,4> can be refined by providing each word with four possible slots, then setting the slot to correspond with a specific word: <0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,1>. This is commonly referred to as the bag-of-words-method.
Since the ultimate goal is to make this data easier for the classifier to read while minimizing memory usage, it’s not always necessary to encode the whole sentence or all the details of a complex concept. In this case, only the current state of how the data is parsed is needed for the classifier. For example, when the top word on the stack is used as the first word in the input queue. Since order is quite important, a simpler binary vector is preferable.
Source:
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Top | 1,102 | 3,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.814867 |
https://www.physicsforums.com/threads/what-kinds-of-particles-can-decay-to-lambda-hyperons.757446/ | 1,597,456,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00503.warc.gz | 794,801,423 | 17,906 | # What kinds of particles can decay to Lambda hyperons?
## Main Question or Discussion Point
Such as ##\Sigma^0 \to \bar{\Lambda}\gamma\gamma##.
I want to make a complete collection of all these decay modes, i.e.
##X \to \Lambda / \bar{\Lambda} + \cdots##.
At least some major channels.
Related High Energy, Nuclear, Particle Physics News on Phys.org
ChrisVer
Gold Member
Ok, it's easy to find the ##\Lambda## decay mode in PDG, but not so easy to do it inversely I think.
If there isn't a ready-made collection, I'll try to search one by one in PDG.
Thanks all the same.
Staff Emeritus
2019 Award
Any baryon weighing more than 1115 MeV.
Any other particle weighing more than 2230 MeV.
Any baryon weighing more than 1115 MeV.
Any other particle weighing more than 2230 MeV.
Really?
mfb
Mentor
Really?
Those numbers are not exact, but yes. As long as conservation laws can be conserved, a decay is possible. Many of them are very unlikely, however.
I guess you look for particles that frequently produce Lambdas? Then look for baryons with two light quarks (up/down) and one heavier quark (especially strange and charm, but also bottom). In addition, some B-mesons can decay to lambda+X.
Z and W can produce lambdas as well, but those are quite rare. For very high-energetic lambdas, they might be relevant.
Where/how do you want to use such a collection?
Staff Emeritus
2019 Award
Z and W can produce lambdas as well, but those are quite rare.
Not as much as you think. The mean number of lambdas in a Z-decay is about 0.4.
The mean number of lambdas in a Z-decay is about 0.4.
What did you mean by 0.4?
##\frac{\sum Z \to {\Lambda}/\bar{\Lambda} + X}{\sum Z \to X}## = 40% ?
Staff Emeritus
2019 Award
No, I mean that in a sample of a million Z decays there are 400,000 Lambdas. It's a statistical statement, not event by event.
Hepth
Gold Member
If you are looking for things beside short-distance production and the Z production just look at any heavier baryonic resonance and see what can give you the most Lambdas. Your end goal is a [uds]. So you can have some
##[uds]^* \to [uds] + (f \bar{f},\pi \pi, \gamma)## Resonant decays ##[\Lambda^*]##
##[udc] \to [uds] + \left(f \bar{\nu}_f, \pi^{+}, \rho^{+}, etc\right) ## electroweak decays ##[\Lambda_c, \Sigma_c, \Xi_c]##
##[udb]-> [uds] + f \bar{f}## (FCNC, highly suppressed) ##[\Lambda_b, \Sigma_b, \Xi_b]##
mfb
Mentor
Not as much as you think. The mean number of lambdas in a Z-decay is about 0.4.
I meant Z and W are rare. Even with a branching fraction of 100% they would be a small contribution.
At the LHC, the cross-sections are 100nb for the W and 30nb for the Z (ATLAS result at 7 TeV).
I didn't find a proper cross-section measurement of the lambda at the LHC, but an approximate number of 100µb given here (note: those lines are "100 events"), and I think this is restricted to the LHCb acceptance - that means the total cross-section is significantly higher. For the total number of lambdas, W and Z contribute less than .1%. That number could be larger for large transverse momentum.
For all other accelerators, the energy is lower, and the W/Z cross-section goes down faster than the other production modes, so there the contribution is even smaller (unless you run an electron-positron collider at the Z peak, of course).
I found a master thesis with a collection of (predicted) mother particles for Lambdas at LHCb. | 946 | 3,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.912029 |
https://rprogramminghelp.xyz/help-with-statistics-homework-3-23428 | 1,679,421,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00037.warc.gz | 562,132,822 | 27,664 | Home » R Studio Assignment Help » Help With Statistics Homework
# Help With Statistics Homework
## Statistics Help For Students Online
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## Stats Help For Students
They are also a set of science necessary foundation building functions that can be used in projects that require work in math. Today, the ability to assess learning levels is a benefit to those of you who lack full understanding and skills. Research for one or more of these attributes is highly recommended. Two Important Uses for With our business, we create a platform to encourage and carry out science on the majors. Hence, without further ado, let’s apply the following usage to Calculation. First: Calculation for Maths In the first paragraph of the last sentence, we go on to define the math-related instrument. To assess understanding, “calculation” is quite an informal word that I could use to describe my degree and how to use it in courses. In our research, an academic assistant used to explain what calculations counts as. Commonly, “calculation” has meant the most basic and intuitive way to represent and process calculations. Sometimes that sounds too complicated for one academic or engineering students. “Calculation” was clearly shown as an official word that is better than “simple math”. That is, it begins with numbers, then you do the math, and then you do some calculations. Well, that’s the whole point of an academic assistant for a math problem.
## Statistics Association
During work, most people will “call it math” before they are in their second year. But what if you are a master of science teacher or high school teacher who is preparing for second year of science studies and a test will occur, and you get results in that calculus language? Does it mean that you don’t know exactly what you are doing until you learn the English language? The answer lies in one question entirely… Check To Know We are being exposed to some of our favorite calculators, and there’s also a very helpful search engineHelp With Statistics Homeworker.com offers all of our programs. Contact Us if you need more help with your Homeworker service. Please email or call us. Homeworker.com our Homeworker Service in Tuscaloosa and we will contact you as soon as we can. Help with your Homeworker service.com. Free Online Homeworker Online Form. Saturday, May 12, 2011 At the heart of her search for the right person to be her mentor, Emily Lewis and Sophie Stephens of St. Augustine, Florida, passed onto their son, Todd, in May with the help of a summer job that was soon about to begin to focus on their future. Our goal is simple and simple.
## Statistics Assignment Help For Elementary
We’re hoping that you’ll be able to stand by your work: the tips and tools you need to be responsible for the right person, and that you will give it to them. There are probably 12 reasons you don’t want to be paid by the minute, but just tell us what to try next. 1. Your Personal Information. Most of us have spent years going to and from work searching for our top, most important digital skills. With so many skills our kids have to improve, there’s no advantage knowing someone working with them. So for your information’s sake it’s as simple as we can be, but hey we want to do them you can check here we want to help them (some could be difficult for us as digital education, but worth the effort) 2. Time Management. When you’re someone with children, you’re definitely not just going to be frustrated. But you also know that use this link can be exhausting and that there’s not always enough time to go through some stressful and hard-to-remember steps. If you’re not into that yet, or if you don’t have time (at least for a while anyway) or don’t realize it beforehand, you probably don’t want to work with Todd. So what’s up with Todd doing once he gets older? In some cases, the answer is positive! 1. You Move First Todd’s move to St.
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Augustine was very exciting and extremely satisfying. What I’ve noticed in the last 6 months that he recently had a few new friendships to work out (honestly, more than I would have ever expected) is that he just hasn’t had time to rest. While being more productive than you’d like to think, he’s still working on his second son’s dream of having another little boy in college to look up the day after he graduates. Todd has made it a point to be a positive role model for every child. He’s still working on his moving skills, and he’s just hanging out to keep one of his children company so he doesn’t go crazy with the other children outside of the home. But lately, Todd may not be up to the job responsibilities of dad, so he has a personal life change going on this Sunday. This week will be pretty typical for Todd’s first Friday school day. Todd doesn’t tend to get any more productive the rest of the day because he has grown and is so excited it’s not difficult to make him happy. But unfortunately for Todd, who we know will have no opportunity to benefit from his strength and sense of humor, he may for now be busy with everyone else around him. This is a time for him to put on | 1,555 | 7,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.930368 |
https://ask.sagemath.org/question/7583/how-to-get-output-in-a-mixed-fraction/?answer=11476 | 1,618,129,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061562.11/warc/CC-MAIN-20210411055903-20210411085903-00543.warc.gz | 229,519,347 | 14,106 | # how to get output in a mixed fraction?
regular rational sage style is like
sage:22/10
i want make it look like (thru pprint it will just look like a mixed on papper)
sage:2+2/10
is this possible?
edit retag close merge delete
Sort by ยป oldest newest most voted
If you want all rational numbers to look like this, then there isn't a convenient way to make that happen in Sage. I'm currently working on a framework to allow one to do this, but it's not ready yet. I will update this post when there is a ticket in question
However, if you just want a function that will return a string in the right form, you can use something like the following
def pprint(rational):
if rational < 0:
return '%s - %s'%(rational.ceil(), rational.ceil() - rational)
elif rational > 0:
return '%s + %s'%(rational.floor(), rational-rational.floor())
else:
return '0'
more
( 2010-08-24 13:08:44 +0200 )edit
yeah would be great to deal this without that much complex constructions..
got an easier solution btw
sage: a=22/10 sage: floor(a), a - floor(a) (2, 1/5)
more
Note that if a is something like -22/10, that will give (-3, -4/5) which is probably not the desired result.
( 2010-08-21 19:44:36 +0200 )edit | 332 | 1,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.904447 |
https://www.analyticsvidhya.com/blog/2019/07/10-applications-linear-algebra-data-science/?utm_source=related_WP&utm_medium=https://www.analyticsvidhya.com/blog/2019/08/5-applications-singular-value-decomposition-svd-data-science/ | 1,708,889,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00371.warc.gz | 634,959,706 | 61,833 | 10 Powerful Applications of Linear Algebra in Data Science (with Multiple Resources)
Khyati Mahendru 23 Apr, 2020 • 14 min read
Overview
• Linear algebra powers various and diverse data science algorithms and applications
• Here, we present 10 such applications where linear algebra will help you become a better data scientist
• We have categorized these applications into various fields – Basic Machine Learning, Dimensionality Reduction, Natural Language Processing, and Computer Vision
Introduction
If Data Science was Batman, Linear Algebra would be Robin. This faithful sidekick is often ignored. But in reality, it powers major areas of Data Science including the hot fields of Natural Language Processing and Computer Vision.
I have personally seen a LOT of data science enthusiasts skip this subject because they find the math too difficult to understand. When the programming languages for data science offer a plethora of packages for working with data, people don’t bother much with linear algebra.
That’s a mistake. Linear algebra is behind all the powerful machine learning algorithms we are so familiar with. It is a vital cog in a data scientists’ skillset. As we will soon see, you should consider linear algebra as a must-know subject in data science.
And trust me, Linear Algebra really is all-pervasive! It will open up possibilities of working and manipulating data you would not have imagined before.
In this article, I have explained in detail ten awesome applications of Linear Algebra in Data Science. I have broadly categorized the applications into four fields for your reference:
I have also provided resources for each application so you can deep dive further into the one(s) which grabs your attention.
Note: Before you read on, I recommend going through this superb article – Linear Algebra for Data Science. It’s not mandatory for understanding what we will cover here but it’s a valuable article for your budding skillset.
• Why Study Linear Algebra?
• Linear Algebra in Machine Learning
• Loss functions
• Regularization
• Covariance Matrix
• Support Vector Machine Classification
• Linear Algebra in Dimensionality Reduction
• Principal Component Analysis (PCA)
• Singular Value Decomposition (SVD)
• Linear Algebra in Natural Language Processing
• Word Embeddings
• Latent Semantic Analysis
• Linear Algebra in Computer Vision
• Image Representation as Tensors
• Convolution and Image Processing
Why Study Linear Algebra?
I have come across this question way too many times. Why should you spend time learning Linear Algebra when you can simply import a package in Python and build your model? It’s a fair question. So, let me present my point of view regarding this.
I consider Linear Algebra as one of the foundational blocks of Data Science. You cannot build a skyscraper without a strong foundation, can you? Think of this scenario:
You want to reduce the dimensions of your data using Principal Component Analysis (PCA). How would you decide how many Principal Components to preserve if you did not know how it would affect your data? Clearly, you need to know the mechanics of the algorithm to make this decision.
With an understanding of Linear Algebra, you will be able to develop a better intuition for machine learning and deep learning algorithms and not treat them as black boxes. This would allow you to choose proper hyperparameters and develop a better model.
You would also be able to code algorithms from scratch and make your own variations to them as well. Isn’t this why we love data science in the first place? The ability to experiment and play around with our models? Consider linear algebra as the key to unlock a whole new world.
Linear Algebra in Machine Learning
The big question – where does linear algebra fit in machine learning? Let’s look at four applications you will all be quite familiar with.
1. Loss Functions
You must be quite familiar with how a model, say a Linear Regression model, fits a given data:
• You start with some arbitrary prediction function (a linear function for a Linear Regression Model)
• Use it on the independent features of the data to predict the output
• Calculate how far-off the predicted output is from the actual output
• Use these calculated values to optimize your prediction function using some strategy like Gradient Descent
But wait – how can you calculate how different your prediction is from the expected output? Loss Functions, of course.
A loss function is an application of the Vector Norm in Linear Algebra. The norm of a vector can simply be its magnitude. There are many types of vector norms. I will quickly explain two of them:
• L1 Norm: Also known as the Manhattan Distance or Taxicab Norm. The L1 Norm is the distance you would travel if you went from the origin to the vector if the only permitted directions are parallel to the axes of the space.
In this 2D space, you could reach the vector (3, 4) by traveling 3 units along the x-axis and then 4 units parallel to the y-axis (as shown). Or you could travel 4 units along the y-axis first and then 3 units parallel to the x-axis. In either case, you will travel a total of 7 units.
• Â L2 Norm:Â Also known as the Euclidean Distance. L2 Norm is the shortest distance of the vector from the origin as shown by the red path in the figure below:
This distance is calculated using the Pythagoras Theorem (I can see the old math concepts flickering on in your mind!). It is the square root of (3^2 + 4^2), which is equal to 5.
But how is the norm used to find the difference between the predicted values and the expected values? Let’s say the predicted values are stored in a vector P and the expected values are stored in a vector E. Then P-E is the difference vector. And the norm of P-E is the total loss for the prediction.
2. Regularization
Regularization is a very important concept in data science. It’s a technique we use to prevent models from overfitting. Regularization is actually another application of the Norm.
A model is said to overfit when it fits the training data too well. Such a model does not perform well with new data because it has learned even the noise in the training data. It will not be able to generalize on data that it has not seen before. The below illustration sums up this idea really well:
Regularization penalizes overly complex models by adding the norm of the weight vector to the cost function. Since we want to minimize the cost function, we will need to minimize this norm. This causes unrequired components of the weight vector to reduce to zero and prevents the prediction function from being overly complex.
You can read the below article to learn about the complete mathematics behind regularization:
The L1 and L2 norms we discussed above are used in two types of regularization:
• L1 regularization used with Lasso Regression
• L2 regularization used with Ridge Regression
Refer to our complete tutorial on Ridge and Lasso Regression in Python to know more about these concepts.
3. Covariance Matrix
Bivariate analysis is an important step in data exploration. We want to study the relationship between pairs of variables. Covariance or Correlation are measures used to study relationships between two continuous variables.
Covariance indicates the direction of the linear relationship between the variables. A positive covariance indicates that an increase or decrease in one variable is accompanied by the same in another. A negative covariance indicates that an increase or decrease in one is accompanied by the opposite in the other.
On the other hand, correlation is the standardized value of Covariance. A correlation value tells us both the strength and direction of the linear relationship and has the range from -1 to 1.
Now, you might be thinking that this is a concept of Statistics and not Linear Algebra. Well, remember I told you Linear Algebra is all-pervasive? Using the concepts of transpose and matrix multiplication in Linear Algebra, we have a pretty neat expression for the covariance matrix:
Here, X is the standardized data matrix containing all numerical features.
I encourage you to read our Complete Tutorial on Data Exploration to know more about the Covariance Matrix, Bivariate Analysis and the other steps involved in Exploratory Data Analysis.
4. Support Vector Machine Classification
Ah yes, support vector machines. One of the most common classification algorithms that regularly produces impressive results. It is an application of the concept of Vector Spaces in Linear Algebra.
Support Vector Machine, or SVM, is a discriminative classifier that works by finding a decision surface. It is a supervised machine learning algorithm.
In this algorithm, we plot each data item as a point in an n-dimensional space (where n is the number of features you have) with the value of each feature being the value of a particular coordinate. Then, we perform classification by finding the hyperplane that differentiates the two classes very well i.e. with the maximum margin, which is C is this case.
A hyperplane is a subspace whose dimensions are one less than its corresponding vector space, so it would be a straight line for a 2D vector space, a 2D plane for a 3D vector space and so on. Again Vector Norm is used to calculate the margin.
But what if the data is not linearly separable like the case below?
Our intuition says that the decision surface has to be a circle or an ellipse, right? But how do you find it? Here, the concept of Kernel Transformations comes into play. The idea of transformation from one space to another is very common in Linear Algebra.
Let’s introduce a variable z = x^2 + y^2. This is how the data looks if we plot it along the z and x-axes:
Now, this is clearly linearly separable by a line z = a, where a is some positive constant. On transforming back to the original space, we get x^2 + y^2 = a as the decision surface, which is a circle!
And the best part? We do not need to add additional features on our own. SVM has a technique called the kernel trick. Read this article on Support Vector Machines to learn about SVM, the kernel trick and how to implement it in Python.
Dimensionality Reduction
You will often work with datasets that have hundreds and even thousands of variables. That’s just how the industry functions. Is it practical to look at each variable and decide which one is more important?
That doesn’t really make sense. We need to bring down the number of variables to perform any sort of coherent analysis. This is what dimensionality reduction is. Now, let’s look at two commonly used dimensionality reduction methods here.
5. Principal Component Analysis (PCA)
Principal Component Analysis, or PCA, is an unsupervised dimensionality reduction technique. PCA finds the directions of maximum variance and projects the data along them to reduce the dimensions.
Without going into the math, these directions are the eigenvectors of the covariance matrix of the data.
Eigenvectors for a square matrix are special non-zero vectors whose direction does not change even after applying linear transformation (which means multiplying) with the matrix. They are shown as the red-colored vectors in the figure below:
You can easily implement PCA in Python using the PCA class in the scikit-learn package:
I applied PCA on the Digits dataset from sklearn – a collection of 8×8 images of handwritten digits. The plot I obtained is rather impressive. The digits appear nicely clustered:
Head on to our Comprehensive Guide to 12 Dimensionality Reduction techniques with code in Python for a deeper insight into PCA and 11 other Dimensionality Reduction techniques. It is honestly one of the best articles on this topic you will find anywhere.
6. Singular Value Decomposition
In my opinion, Singular Value Decomposition (SVD) is underrated and not discussed enough. It is an amazing technique of matrix decomposition with diverse applications. I will try and cover a few of them in a future article.
For now, let us talk about SVD in Dimensionality Reduction. Specifically, this is known as Truncated SVD.
• We start with the large m x n numerical data matrix A, where m is the number of rows and n is the number of features
• Decompose it into 3 matrices as shown here:
• Choose k singular values based on the diagonal matrix and truncate (trim) the 3 matrices accordingly:
Source: researchgate.net
• Finally, multiply the truncated matrices to obtain the transformed matrix A_k. It has the dimensions m x k. So, it has k features with k < n
Here is the code to implement truncated SVD in Python (it’s quite similar to PCA):
On applying truncated SVD to the Digits data, I got the below plot. You’ll notice that it’s not as well clustered as we obtained after PCA:
Natural Language Processing (NLP)
Natural Language Processing (NLP) is the hottest field in data science right now. This is primarily down to major breakthroughs in the last 18 months. If you were still undecided on which branch to opt for – you should strongly consider NLP.
So let’s see a couple of interesting applications of linear algebra in NLP. This should help swing your decision!
7. Word Embeddings
Machine learning algorithms cannot work with raw textual data. We need to convert the text into some numerical and statistical features to create model inputs. There are many ways for engineering features from text data, such as:
1. Meta attributes of a text, like word count, special character count, etc.
2. NLP attributes of text using Parts-of-Speech tags and Grammar Relations like the number of proper nouns
3. Word Vector Notations or Word Embeddings
Word Embeddings is a way of representing words as low dimensional vectors of numbers while preserving their context in the document. These representations are obtained by training different neural networks on a large amount of text which is called a corpus. They also help in analyzing syntactic similarity among words:
Word2Vec and GloVe are two popular models to create Word Embeddings.
I trained my model on the Shakespeare corpus after some light preprocessing using Word2Vec and obtained the word embedding for the word ‘world’:
Pretty cool! But what’s even more awesome is the below plot I obtained for the vocabulary. Observe that syntactically similar words are closer together. I have highlighted a few such clusters of words. The results are not perfect but they are still quite amazing:
There are several other methods to obtain Word Embeddings. Read our article for An Intuitive Understanding of Word Embeddings: From Count Vectors to Word2Vec.
8. Latent Semantic Analysis (LSA)
What is your first thought when you hear this group of words – “prince, royal, king, noble”? These very different words are almost synonymous.
Now, consider the following sentences:
• The pitcher of the Home team seemed out of form
• There is a pitcher of juice on the table for you to enjoy
The word ‘pitcher’ has different meanings based on the other words in the two sentences. It means a baseball player in the first sentence and a jug of juice in the second.
Both these sets of words are easy for us humans to interpret with years of experience with the language. But what about machines? Here, the NLP concept of Topic Modeling comes into play:
Topic Modeling is an unsupervised technique to find topics across various text documents. These topics are nothing but clusters of related words. Each document can have multiple topics. The topic model outputs the various topics, their distributions in each document, and the frequency of different words it contains.
Latent Semantic Analysis (LSA), or Latent Semantic Indexing, is one of the techniques of Topic Modeling. It is another application of Singular Value Decomposition.
Latent means ‘hidden’. True to its name, LSA attempts to capture the hidden themes or topics from the documents by leveraging the context around the words.
I will describe the steps in LSA in short so make sure you check out this Simple Introduction to Topic Modeling using Latent Semantic Analysis with code in Python for a proper and in-depth understanding.
• First, generate the Document-Term matrix for your data
• Use SVD to decompose the matrix into 3 matrices:
• Document-Topic matrix
• Topic Importance Diagonal Matrix
• Topic-term matrix
• Truncate the matrices based on the importance of topics
For a hands-on experience with Natural Language Processing, you can check out our course on NLP using Python. The course is beginner-friendly and you get to build 5 real-life projects!
Computer Vision
Another field of deep learning that is creating waves – Computer Vision. If you’re looking to expand your skillset beyond tabular data (and you should), then learn how to work with images.
9. Image Representation as Tensors
How do you account for the ‘vision’ in Computer Vision? Obviously, a computer does not process images as humans do. Like I mentioned earlier, machine learning algorithms need numerical features to work with.
A digital image is made up of small indivisible units called pixels. Consider the figure below:
This grayscale image of the digit zero is made of 8 x 8 = 64 pixels. Each pixel has a value in the range 0 to 255. A value of 0 represents a black pixel and 255 represents a white pixel.
Conveniently, an m x n grayscale image can be represented as a 2D matrix with m rows and n columns with the cells containing the respective pixel values:
But what about a colored image? A colored image is generally stored in the RGB system. Each image can be thought of as being represented by three 2D matrices, one for each R, G and B channel. A pixel value of 0 in the R channel represents zero intensity of the Red color and of 255 represents the full intensity of the Red color.
Each pixel value is then a combination of the corresponding values in the three channels:
In reality, instead of using 3 matrices to represent an image, a tensor is used. A tensor is a generalized n-dimensional matrix. For an RGB image, a 3rd ordered tensor is used. Imagine it as three 2D matrices stacked one behind another:
Source: slidesharecdn
10. Convolution and Image Processing
2D Convolution is a very important operation in image processing. It consists of the below steps:
1. Start with a small matrix of weights, called a kernel or a filter
2. Slide this kernel on the 2D input data, performing element-wise multiplication
3. Add the obtained values and put the sum in a single output pixel
The function can seem a bit complex but it’s widely used for performing various image processing operations like sharpening and blurring the images and edge detection. We just need to know the right kernel for the task we are trying to accomplish. Here are a few kernels you can use:
You can download the image I used and try these image processing operations for yourself using the code and the kernels above. Also, try this Computer Vision tutorial on Image Segmentation techniques!
Amazing, right? This is by far my most favorite application of Linear Algebra in Data Science.
Now that you are acquainted with the basics of Computer Vision, it is time to start your Computer Vision journey with 16 awesome OpenCV functions. We also have a comprehensive course on Computer Vision using Deep Learning in which you can work on real-life Computer Vision case studies!
End Notes
My aim here was to make Linear Algebra a bit more interesting than you might have imagined previously. Personally for me, learning about applications of a subject motivates me to learn more about it.
I am sure you are as impressed with these applications as I am. Or perhaps you know of some other applications that I could add to the list? Let me know in the comments section below.
Khyati Mahendru 23 Apr 2020 | 4,055 | 19,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.910249 |
https://www.homeworklib.com/question/1646045/how-much-water-would-be-needed-to-completely | 1,606,362,425,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186414.7/warc/CC-MAIN-20201126030729-20201126060729-00681.warc.gz | 694,959,389 | 14,766 | Homework Help Question & Answers
# How much water would be needed to completely dissolve 1.74 L of the gas at a...
How much water would be needed to completely dissolve 1.74 L of the gas at a pressure of 725 torr and a temperature of 18 ∘C?
A gas has a Henry's law constant of 0.192 M/atm .
Answer: Here we have to use the henry's law
C = KP , here , C is concentration , K is henry's constant and P is partial pressure
given mass / molar mass of water* volume = 0.192 * pressure in atm
[ here 1 torr = 0.0013atm hence , 725 torr = 0.95 atm ]
Now , we get : mass / 18 * 1.74 = 0.95 * 0.192
mass = 5.71 gram
Hence the required amount of water is 5.71 gram
##### Add Answer to: How much water would be needed to completely dissolve 1.74 L of the gas at a...
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Free Homework App | 1,451 | 5,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-50 | latest | en | 0.90693 |
https://codegolf.stackexchange.com/questions/9198/count-from-1-to-100-in-roman-numerals | 1,708,881,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00330.warc.gz | 171,637,916 | 60,950 | # Count from 1 to 100... in Roman Numerals
Write a program that count from 1 to 100 in Roman Numerals and print these numbers by standard output. Each one of the numbers must be separated by spaces.
You cannot use any built in function to transform to roman numerals nor external application or library to do so.
The desired result is
I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI XXII XXIII XXIV XXV XXVI XXVII XXVIII XXIX XXX XXXI XXXII XXXIII XXXIV XXXV XXXVI XXXVII XXXVIII XXXIX XL XLI XLII XLIII XLIV XLV XLVI XLVII XLVIII XLIX L LI LII LIII LIV LV LVI LVII LVIII LIX LX LXI LXII LXIII LXIV LXV LXVI LXVII LXVIII LXIX LXX LXXI LXXII LXXIII LXXIV LXXV LXXVI LXXVII LXXVIII LXXIX LXXX LXXXI LXXXII LXXXIII LXXXIV LXXXV LXXXVI LXXXVII LXXXVIII LXXXIX XC XCI XCII XCIII XCIV XCV XCVI XCVII XCVIII XCIX C
As it is a code golf challenge, shortest code wins.
• 39 is missing an X.
– Thor
Dec 14, 2012 at 13:02
• I really want to use INTERCAL for this one. Mar 19, 2018 at 18:36
• can it be separated by newlines? Also what about trailing/leading spaces/newlines?
– qqq
Mar 19, 2018 at 20:14
## Perl 69 bytes
s;.;y/XVI60-9/CLXVIX/dfor$a[$_].="32e$&"%72726;gefor 1..100;print"@a" Works by way of magic formula. The expression "32e$&"%72726 transforms each digit in the following manner:
0⇒32, 1⇒320, 2⇒3200, 3⇒32000, 4⇒29096, 5⇒56, 6⇒560, 7⇒5600, 8⇒56000, 9⇒50918
After applying the translation y/016/IXV/, we have this instead:
0⇒32, 1⇒32I, 2⇒32II, 3⇒32III, 4⇒29I9V, 5⇒5V, 6⇒5VI, 7⇒5VII, 8⇒5VIII, 9⇒5I9X8
The rest of the digits (2-57-9) are removed. Note that this could be improved by one byte by using a formula which translates 012 instead of 016, simplifying /XVI60-9/ to /XVI0-9/. I wasn't able to find one, but maybe you'll have better luck.
Once one digit has been transformed in this manner, the process repeats for the next digit, appending the result, and translating the previous XVIs to CLX at the same time the translation for the new digit occurs.
Update
Exhaustive search didn't reveal anything shorter. I did, however, find an alternative 69 byte solution:
s;.;y/XVI0-9/CLXIXV/dfor$a[$_].="57e$&"%474976;gefor 1..100;print"@a" This one uses a 0-2 substitution for IXV, but has a modulo that's one digit longer. ## Update: 66 65 bytes This version is notably different, so I should probably say a few words about it. The formula it uses is actually one byte longer! Unable to shorten the formula any more than it is, I decided to golf down what I had. It wasn't long until I remembered my old friend $\. When a print statment is issued, $\ is automatically appended to the end of the output. I was able to get rid of the awkward $a[$_] construction for a two byte improvement: s;.;y/XVI60-9/CLXVIX/dfor$\.="32e$&"%72726;ge,$\=!print$"for 1..100 Much better, but that $\=!print$" still looked a bit verbose. I then remembered an alternative, equal length formula I had found which didn't contain the number 3 in any of its digit transforms. So, it should be possible to use $\=2+print instead, and substitute the resulting 3 with a space:
s;.;y/XVI0-9/CLXIIX V/dfor$\.="8e$&"%61535;ge,$\=2+print for 1..100 Also 67 bytes, due to the necessary whitespace between print and for. Edit: This can be improved by one byte, by moving the print to the front: $\=2+print!s;.;y/XVI0-9/CLXIIX V/dfor$\.="8e$&"%61535;gefor 1..100
Because the substitution needs to evaluate completely before the print, the assignment to $\ will still occur last. Removing the whitespace between ge and for will issue a deprecation warning, but is otherwise valid. But, if there were a formula which didn't use a 1 anywhere, $\=2+print becomes $\=print for another two bytes worth of savings. Even if it were one byte longer, it'd still be an improvement. As it turns out, such a formula does exist, but it is one byte longer than the original, resulting in a final score of 65 bytes: $\=print!s;.;y/XVI60-9/CLXXI V/dfor$\.="37e$&"%97366;gefor 1..100
## Methodology
The question was asked how one might go about finding such a formula. In general, finding a magic formula to generalize any set of data is matter of probability. That is, you want to choose a form which is as likely as possible to produce something similar to the desired result.
Examing the first few roman numerals:
0:
1: I
2: II
3: III
4: IV
5: V
6: VI
7: VII
8: VIII
9: IX
there is some regularity to be seen. Specifically, from 0-3 and then again from 5-8, each successive term increases in length by one numeral. If we wanted to create a mapping from digits to numerals, we would want to have an expression that also increases in length by one digit for each successive term. A logical choice is k • 10d where d is the corresponding digit, and k is any integer constant.
This works for 0-3, but 4 needs to break the pattern. What we can do here is tack on a modulo:
k • 10d % m, where m is somewhere between k • 103 and k • 104. This will leave the range 0-3 untouched, and modify 4 such that it won't contain four Is. If we additionally constrain our search algorithm such that the modular residue of 5, let's call it j, is less than m / 1000, this will ensure that we also have regularity from 5-8 as well. The result is something like this:
0: k
1: k0
2: k00
3: k000
4: ????
5: j
6: j0
7: j00
8: j000
9: ????
As you can see, if we replace 0 with I, 0-3 and 5-8 are all guaranteed to be mapped correctly! The values for 4 and 9 need to be brute forced though. Specifically, 4 needs to contain one 0 and one j (in that order), and 9 needs to contain one 0, followed by one other digit that doesn't appear anywhere else. Certainly, there are a number of other formulae, which by some fluke of a coincidence might produce the desired result. Some of them may even be shorter. But I don't think there are any which are as likely to succeed as this one.
I also experimented with multiple replacements for I and/or V with some success. But alas, nothing shorter than what I already had. Here is a list of the shortest solutions I found (the number of solutions 1-2 bytes heavier are too many to list):
y/XVI60-9/CLXVIX/dfor$\.="32e$&"%72726
y/XVI0-9/CLXIXV/dfor$\.="57e$&"%474976
y/XVI0-9/CLXIVXI/dfor$\.="49e$&"%87971
y/XVI0-9/CLXIIXIV/dfor$\.="7e$&"%10606 #
y/XVI0-9/CLXIIXIV/dfor$\.="7e$&"%15909 # These are all essentially the same
y/XVI0-9/CLXIIXIV/dfor$\.="7e$&"%31818 #
y/XVI0-9/CLXIIX V/dfor$\.="8e$&"%61535 # Doesn't contain 3 anywhere
y/XVI60-9/CLXXI V/dfor$\.="37e$&"%97366 # Doesn't contain 1 anywhere
• How did you find the magic formula? Dec 18, 2012 at 20:27
• @RubenVerborgh I'll update my post with more information regarding the methodology soon. Dec 18, 2012 at 21:13
# HTML + JavaScript + CSS (137)
HTML (9)
<ol></ol>
JavaScript (101)
for(i=1;i<=100;i++){document.getElementsByTagName('ol')[0].appendChild(document.createElement('li'))}
CSS (27)
ol{list-style:upper-roman}
Output
...
Demo on JSBin
• 81 byte JS-only version: document.write('<ol>'+"<li style='list-style:upper-roman'/>".repeat(100)+'</ol>') (ES6) Sep 9, 2016 at 22:13
• or 66 in Chrome document.write("<li style='list-style:upper-roman'/>".repeat(100))
– Slai
Aug 7, 2018 at 23:56
• 63 if we omit attribute quotes and the slash: document.write("<li style=list-style:upper-roman>".repeat(100)) Apr 15, 2022 at 15:28
# Python 116
better golfed code of scleaver's answer:
r=lambda a,b,c:('',a,2*a,3*a,a+b,b,b+a,b+a+a,b+3*a,a+c);print' '.join(i+j for i in r(*'XLC')for j in r(*'IVX'))+' C'
# Python, 139
print' '.join(' '.join(i+j for j in ' _I_II_III_IV_V_VI_VII_VIII_IX'.split('_'))for i in ' _X_XX_XXX_XL_L_LX_LXX_LXXX_XC'.split('_'))+' C'
## C, 177160 147 chars
There are shorter solutions, but none in C, so here's my try.
New solution, completely different from my previous one:
char*c;
f(n){
printf("%.*s",n%5>3?2:n%5+n/5,c+=n%5>3?n%4*4:2-n/5);
}
main(i){
for(;i<100;putchar(32))
c="XLXXXC",f(i/10),
c="IVIIIX",f(i++%10);
puts("C");
}
## Previous solution (160 chars):
Logic:
1. f prints a number from 1 to 10. c is the digits used, which can be IVX or XLC. Called once for the tens once for the ones.
2. If n%5==0 - print nothing or c[n/5] which is I or V (or L or C).
3. If n%4==4 - 4 or 9 - print I (or X), by n+1.
4. If n>4 - print 5 (i.e. V or L) then n-5.
5. If n<4 - print I then n-1 (i.e. n times I).
char*c;
p(c){putchar(c);}
f(n){
n%5?
n%5>3?
f(1),f(n+1):
n>4?
f(5),f(n-5):
f(n-1,p(*c)):
n&&p(c[n/5]);
}
main(i){
for(;++i<101;p(32))
c="XLC",f(i/10),
c="IVX",f(i%10);
p(10);
}
• 137: f(c,n){printf("%.*s",n%5>3?2:n%5+n/5,"XLXXXCIVIIIX "+c+(n%5>3?n%4*4:2-n/5));}main(i){for(;i<100;f(12,4))f(0,i/10),f(6,i++%10);puts("C");} Jan 28, 2018 at 8:44
# Mathematica 159 150 142
c = {100, 90, 50, 40, 10, 9, 5, 4, 1};
Table["" <> Flatten[ConstantArray @@@ Thread@{StringSplit@"C XC L XL X IX V IV I",
FoldList[Mod, k, Most@c]~Quotient~c}], {k, 100}]
Built-in solution: IntegerString, 38 chars
IntegerString[k, "Roman"]~Table~{k, 100}
## JavaScript, 123
Inspired by a longer version I came across in a Polish newsgroup (at least, Chrome thought it was Polish).
for(i=100,a=[];n=i--;a[i]=r)
for(r=y='',x=5;n;y++,x^=7)
for(m=n%x,n=n/x^0;m--;)
r='IVXLC'[m>2?y+n-(n&=-2)+(m=1):+y]+r;
alert(a)
# Q (81 80)
2nd cut:
1_,/'[($)XXXXXXXLLLXLXXLXXXXC crossIIIIIIIVVVIVIIVIIIIX],"C" 1st cut: 1_,/'[$:[XXXXXXXLLLXLXXLXXXXC crossIIIIIIIVVVIVIIVIIIIX]],"C"
## Python, 168
r=lambda n,l,v:(r(n,l[1:],v[1:])if n<v[0]else l[0]+r(n-v[0],l,v))if n else''
for i in range(1,101):print r(i,'C XC L XL X IX V IV I'.split(),[100,90,50,40,10,9,5,4,1]),
Explanation
Using these values, take the largest value not greater than n and subtract it from n. Repeat until n is 0.
'C' = 100
'XC' = 90
'L' = 50
'XL' = 40
'X' = 10
'IX' = 9
'V' = 5
'IV' = 4
'I' = 1
• r=lambda n,l,v:n and(n<v[0]and r(n,l[1:],v[1:])or l[0]+r(n-v[0],l,v))or"" saves two characters. Otherwise very nice. Dec 14, 2012 at 21:14
## Ruby 1.9, 140 132
r=" "
100.times{r+=?I
0while[[?I*4,"IV"],["VIV","IX"],[?X*4,"XL"],["LXL","XC"],[/(.)((?!\1)[^I])\1/,'\2']].any?{|q|r.sub! *q}
$><<r} This literally counts from 1 to 100 in Roman numerals. Starts with a blank string, then loops through appending "I" and then repeatedly applying a series of substitution rules, effectively adding 1. Edit: Added version number, since ?I only works in 1.9, and used @Howard's changes to trim some characters. • You may save two chars: r while -> 0while, r.sub!(*q) -> r.sub! *q. You can also drag the print inside the loop and use 100.times{...} instead of the map statement. Dec 14, 2012 at 18:52 • (%w[IIII VIV XXXX LXL]<</(.)((?!\1)[^I])\1/).zip(%w(IV IX XL XC)<<'\2') saves 7 chars. Dec 14, 2012 at 23:06 ## Ruby 112 chars 101.times{|n|r=' ';[100,90,50,40,10,9,5,4,1].zip(%w(C XC L XL X IX V IV I)){|(k,v)|a,n=n.divmod k;r<<v*a};$><<r}
Basically using the to_roman method explained here, but using a zipped array for brevity.
perl 205
@r = split //, "IVXLC";
@n = (1, 5, 10, 50, 100);
for $num (1..100) { for($i=@r-1; $i>=0;$i--) {
$d = int($num / $n[$i]);
next if not $d;$_ .= $r[$i] x $d;$num -= $d *$n[$i]; }$_ .= " ";
}
s/LXXXX/XC/g;
s/XXXX/XL/g;
s/VIIII/IX/g;
s/IIII/IV/g;
print;
Golfed:
@r=split//,"IVXLC";@n=(1,5,10,50,100);for$num(1..100){for($i=@r-1;$i>=0;$i--){$d=int($num/$n[$i]);next if!$d;$_.=$r[$i]x$d;$num-=$d*$n[$i];}$_.=" ";}s/LXXXX/XC/g;s/XXXX/XL/g;s/VIIII/IX/g;s/IIII/IV/g;print;
MUMPS 184
S V(100)="C",V(90)="XC",V(50)="L",V(40)="XL",V(10)="X",V(9)="IX",V(5)="V",V(4)="IV",V(1)="I" F I=1:1:100 S S=I,N="" F Q:'S S N=\$O(V(N),-1) I S&(S'<N ) S S=S-N W V(N) S N="" w:'S " "
Same algorithm as @cardboard_box, from whom I've taken the explanation verbatim -
Explanation
Using these values, take the largest value not greater than n and subtract it from n. Repeat until n is 0.
'C' = 100
'XC' = 90
'L' = 50
'XL' = 40
'X' = 10
'IX' = 9
'V' = 5
'IV' = 4
'I' = 1
# R, 85 bytes
R=.romans
for(r in 1:100){while(r>0){cat(names(R[I<-R<=r][1]))
r=r-R[I][1]}
cat(" ")}
Try it online!
Uses the random utils package variable .romans to get the values of the roman numerals, but does the conversion by itself; the builtin approach would be 20 bytes: cat(as.roman(1:100))
• Surprisingly, the built-in approach you are mentioning does not work as is... one has to type cat(paste(as.roman(1:100))) or simply as.roman(1:100). Weird. Aug 27, 2018 at 19:43
• @JayCe odd; I must not have actually tested it...the docs for cat indicate that it performs less conversion than print and only works on atomic vectors -- so that explains this! Aug 27, 2018 at 19:51
# APL (Dyalog Unicode), 61 bytes
∊{'IVXLC'[∊5 3 1+{(⍺=0)↓⍺↑⍨⊃¯2,⍨⍵~0}⌿0 5⊤1+⍵⊤⍨3⍴10],' '}¨⍳100
Try it online!
(I know there's already an APL answer, but wanted to re-solve it anyway)
Adapted and golfed from Nikolay Nikolov's piece of code that supports up to M at the bottom of this page.
### How it works
∊{'IVXLC'[∊5 3 1+{(⍺=0)↓⍺↑⍨⊃¯2,⍨⍵~0}⌿0 5⊤1+⍵⊤⍨3⍴10],' '}¨⍳100
∊{'IVXLC'[...],' '}¨⍳100 ⍝ Less interesting part
⍳100 ⍝ 1-based range; 1 to 100 inclusive
{ ... }¨ ⍝ Compute a representation for each number
'IVXLC'[ ],' ' ⍝ Index into the string and pad with a space
∊ ⍝ Concatenate the strings
∊5 3 1+{(⍺=0)↓⍺↑⍨⊃¯2,⍨⍵~0}⌿0 5⊤1+⍵⊤⍨3⍴10 ⍝ Interesting part
⍵⊤⍨3⍴10 ⍝ Last 3 decimal digits of ⍵
0 5⊤1+ ⍝ Divmod of each digit +1 by 5
{ }⌿ ⍝ Compute the partial roman representation:
⍝ Digit 0 1 2 3 4
⍝ Div 0 0 0 0 1
⍝ Mod 1 2 3 4 0
⍝ Result [] [0] [0 0] [0 0 0] [0 1]
⍝ Digit 5 6 7 8 9
⍝ Div 1 1 1 1 2
⍝ Mod 1 2 3 4 0
⍝ Result [1] [1 0] [1 0 0] [1 0 0 0] [0 2]
⊃¯2,⍨⍵~0 ⍝ ¯2 if mod is 0, keep mod otherwise
⍺↑⍨ ⍝ Overtake div with the above
⍝ (¯2: add 0 to the left; other: add (mod-1) 0s to the right)
(⍺=0)↓ ⍝ Drop one if div is 0
5 3 1+ ⍝ Offset each nested array to point to correct roman unit
∊ ⍝ Flatten
OP said "You cannot use any built in function to transform to roman numerals" but didn't say about a built-in that does arithmetic over roman numerals, so this one could be technically valid:
# APL (Dyalog Unicode), 29 bytes
⎕CY'dfns'
∊,∘' '¨{⍳⍵}roman'C'
Try it online!
This one leverages the library operator roman that does various arithmetic over roman numerals (that is, the input is also a roman numeral). So this basically computes the 1-based range {⍳⍵} over roman numerals with the input value 'C' (100), appends a space to each string ,∘' '¨ and flattens them ∊.
## APL 128
I tried an indexing solution in APL:
r←⍬
i←1
l:r←r,' ',(' CXI LV CX'[,⍉((1+(4 4 2 2⊤0 16 20 22 24 32 36 38 39 28)[;1+(3⍴10)⊤i])×3)-4 3⍴2 1 0])~' '
→(100≥i←i+1)/l
r
It can be 4 bytes shorter in index origin 0 instead of 1 but the real space hog is the generation of the index matrix via:
4 4 2 2⊤0 16 20 22 24 32 36 38 39 28
So far I have not been able to generate the indices on the fly!
# LaTeX (138)
\documentclass{minimal}
\usepackage{forloop}
\begin{document}
\newcounter{i}
\forloop{i}{1}{\value{i} < 101}{\roman{i}\par}
\end{document}
• -1: the question says "You cannot use any built in function to transform to roman numerals" Sep 4, 2014 at 13:59
# Python, 125
' '.join(i+j for i in['']+'X XX XXX XL L LX LXX LXXX XC C'.split()for j in['']+'I II III IV V VI VII VIII IX'.split())[1:-38]
# PHP, 38 37 bytes
<ol type=I><?=str_repeat("<li>",100);
-1 byte thanks to @manatwork
Same idea as Patrick's answer, but in a more compact language. Beats Mathematica!
Try it Online!
• Terminate the statement with ;, then no need for ?>. Mar 19, 2018 at 18:34
VBA (Excel), 245 bytes
created Function for Repetition and Replace - 91 bytes
Function s(a,b):s=String(a,b):End Function Function b(x,y,z):b=Replace(x,y,z):End Function
using immediate window (154 bytes)
p="I":for x=1to 100:?b(b(b(b(b(b(b(b(s(x,p),s(100,p),"C"),s(90,p),"XC"),s(50,p),"L"),s(40,p),"XL"),s(10,p),"X"),s(9,p),"IX"),s(5,p),"V"),s(4,p),"IV"):next
# Java (OpenJDK 8), 152 bytes
a->{String[] t=",X,XX,XXX,XL,L,LX,LXX,LXXX,XC,,I,II,III,IV,V,VI,VII,VIII,IX".split(",");for(int i=1;i<100;i++){a+=t[i/10]+t[i%10+10]+" ";}return a+"C";}
Try it online!
Explanation:
String[] t=",X,XX,XXX,XL,L,LX,LXX,LXXX,XC,,I,II,III,IV,V,VI,VII,VIII,IX".split(",");
//Create an array of numerals, first half represents tens place, second half represents ones place
for(int i=1;i<100;i++){
//Loop 99 times
a+=t[i/10]+t[i%10+10]+" ";
//Add tens place and ones place to the string
}return a+"C";
//Add numeral for 100 and return the string
• 148 bytes Dec 7, 2019 at 1:38
# TeX, 354 bytes
\let~\let~\d\def~\a\advance~\b\divide~\x\expandafter~\f\ifnum{~~\newcount~\n~\i~\j~\k~\u~\v}~~\or\d\p#1{\ifcase#1C~2~L~5~X~2~V~5~I\fi}\d\q#1{\p{#1~}}\d\r{\j0
\v100\d\m{\d\w{\f\n<\v\else\p\j\a\n-\v\x\w\fi}\w\f\n>0\k\j\u\v\d\g{\a\k2\b\u\q\k}\g\f\q\k=2\g\fi\a\n\u\f\n<\v\a\n-\u\a\j2\b\v\q\j\else\p\k\fi\x\m\fi}\m}\i1\d\c{
\f\i<101 \n\i\r\a\i1 \x\c\fi}\c\bye
Some explanation: TeX provides a built-in command \romannumeral for converting numbers to Roman numerals. As the question doesn't allow to use built-in functions, the above code is a golfed version of the same algorithm Knuth's original TeX compiler uses for \romannumeral (see TeX: The Program, § 69, print_roman_int) re-implemented in TeX.
As he wants to leave the joy of puzzling out how this code works to the reader, Knuth refuses to give an explanation of this part of the code. So I'll follow suit and just give an ungolfed and slightly modified version, which is closer to the original than the above code:
\newcount\n
\newcount\j
\newcount\k
\newcount\u
\newcount\v
\def\chrnum#1{\ifcase#1m\or 2\or d\or 5\or c\or 2\or l\or 5\or x\or 2\or v\or 5\or i\fi}
\def\chrnumM#1{\chrnum{#1\or}}
\def\roman#1{%
\n=#1\relax
\j=0\relax
\v=1000\relax
\def\mainloop{%
\def\while{%
\ifnum\n<\v
\else
\chrnum\j
\advance\n -\v
\expandafter\while
\fi
}\while
\ifnum\n>0\relax
\k=\j \advance\k 2\relax
\u=\v \divide\u \chrnumM\k
\ifnum\chrnumM\k=2\relax
\advance\k 2\relax
\divide\u \chrnumM\k
\fi
\advance\n \u
\ifnum\n<\v
\advance\n -\u
\advance\j 2\relax
\divide\v \chrnumM\j
\else
\chrnum\k
\fi
\expandafter\mainloop
\fi
}\mainloop
}
\newcount\i \i=1
\def\countloop{%
\ifnum\i<100\relax
\roman\i\
\advance\i 1
\expandafter\countloop
\fi
}\countloop
\bye | 6,990 | 18,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.724604 |
https://mathsmartinthomas.wordpress.com/2016/01/19/maths-prize-for-19-january-2015-the-donkey-and-the-carrots/ | 1,500,899,708,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424876.76/warc/CC-MAIN-20170724122255-20170724142255-00324.warc.gz | 678,930,543 | 32,124 | Maths prize for 19 January 2016: the donkey and the carrots
You have 45 carrots, and want to get a donkey to take as many as possible of them to a endpoint 15 miles away. Trouble is, the donkey can only carry 15 carrots at a time, and insists on eating a carrot early in every mile. How many carrots can you get to the endpoint?
The prize was won by Wei-Kong Mao. The answer is 8 carrots.
Solution: If you take 15 carrots, you can get the donkey to go to the first milestone, drop 13 carrots there, and come back to get another 15. (13=15 minus one carrot eaten for the mile out and another carrot for the mile back).
Go to the first milestone again, leave another 13 there, come back again, pick up the last 15, and drop 14 at the first milestone. Now you have 40 at the first milestone.
Then 35 at the second, 30 at the third.
Once you are down to 30 carrots, you can carry the maximum number of carrots further at a loss of only 3 carrots per mile. (One journey out to the next milestone, one back to pick up more, one out again).
So: 27 at the fourth milestone, 24 at the 5th, 21 at the 6th, 18 at the 7th, 15 at the 8th.
Now you can head straight for the endpoint with no more backtracking. The donkey will need 7 carrots for this last stretch, and so you can get 8 to the endpoint.
This is the best solution. There is no penalty in the problem for time lost in backtracking. Any solution getting fewer than the maximum possible number of carrots to each milestone thus differs from the one above only in having a smaller number of carrots at some milestone, and therefore cannot be better.
It can be tweaked a little. You could take 15 to the third milestone, leave 9 there, return, bring another 15, leave another 9, return, bring a final 15 to add 12 (15 minus 3 eaten on the way) to the total at the third milestone. That’s another way of getting 30 to the third milestone, but not very different from doing it mile by mile. | 490 | 1,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-30 | longest | en | 0.950653 |
https://mirtitles.org/2022/01/01/continued-fractions-khinchin/ | 1,679,733,696,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00382.warc.gz | 438,243,050 | 27,975 | ## Continued Fractions – Khinchin
In this post, we will see the book Continued Fractions by A. Ya. Khinchin.
The late Alexander J. Khinchin was born in Russia in 1894. One of the founders of the Soviet school of probability theory, Khinchin was made a full professor at Moscow University in 1922 and held that position until his death. His teaching skill is discernible in the clear and straightforward presentation of his subject. Designed for use as an expository text in the university curriculum, the book is basically of an elementary nature, the author confining his attention to continued fractions with positive-integral elements. The essentials needed for applications in probability theory, mechanics, and, especially, number theory are given and the real number system is constructed from continued fractions. The last chapter is somewhat more advanced and deals with the metric, or probability, theory of continued fractions. This first English translation is based on the third edition of the text which was issued in 1961.
The book was translated from Russian by Scripta Technica and was published in 1964.
You can get the book here.
Follow us on The Internet Archive: https://archive.org/details/@mirtitles
Write to us: [email protected]
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Add new entries to the detailed book catalog here.
# Contents
## Chapter I. Properties of the Apparatus 1
1. Introduction 1
2. Convergents 3
3. Infinite continued fractions 8
4. Continued fractions with natural elements 12
## Chapter II. The Representation of Numbers by Continued Fractions 16
5. Continued fractions as an apparatus for representing real numbers 16
6. Convergents as best approximations 20
7. The order of approximation 28
8. General approximation theorems 34
9. The approximation of algebraic irrational numbers and Liouville’s transcendental numbers 45
10. Quadratic irrational numbers and periodic continued fractions 47
## Chapter III. The Measure Theory of Continued Fractions 51
11. Introduction 51
12. The elements as functions of the number represented 52
13. Measure-theoretic evaluation of the increase in the elements 60
14. Measure-theoretic evaluation of the increase in the denominators of the convergents. The fundamental theorem of the measure theory of approximation 65
15. Gauss’s problem and Kuz’min’s theorem 71
16. Average values 86
95 Index | 537 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-14 | latest | en | 0.934871 |
https://forums.radioreference.com/threads/hex-to-dec-or-dec-to-hex-conversions.47148/ | 1,670,193,240,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00450.warc.gz | 288,644,768 | 19,384 | # HEX to Dec or Dec to HEX conversions
Status
Not open for further replies.
#### SCPD
##### QRT
I recently received an e-mail from a person whos job is to post correct system Hex & dec info asking me to correct my info I submitted. I had posted the correct HEX but his computer program that radio reference uses spit out the incorrect hex & dec. Therefore I felt I should share my way of coming up with the proper HEX & Dec for each talkgroup discovered. So everyone can learn how to properly decode the DEC & HEX when they submit them to R/R
Formulas, this will require a scientific calculator
HEX to DEC {Make sure Calc. is in hex mode & when finished switch to DEC mode for your answer}
--------------------------
Formula
HEX x 10 = DEC
DEC to HEX {Make sure Calc. is in DEC mode & when finished switch to HEX mode for your answer}
-------------------------
Formula
DEC / 16 = HEX
#### DickH
##### Member
Com-4 said:
Formulas, this will require a scientific calculator
HEX to DEC {Make sure Calc. is in hex mode & when finished switch to DEC mode for your answer}
--------------------------
Formula
HEX x 10 = DEC
DEC to HEX {Make sure Calc. is in DEC mode & when finished switch to HEX mode for your answer}
-------------------------
Formula
DEC / 16 = HEX
I'm not good at math so I use the scientific calculator in Windows. I tried the following two talk groups:
DEC 4656 = HEX 1230
DEC 19184 = HEX 4AF0
Comparing the results with your formulas, the windows calculator must be no good.
#### morfis
##### Member
DickH said:
I'm not good at math so I use the scientific calculator in Windows. I tried the following two talk groups:
DEC 4656 = HEX 1230
DEC 19184 = HEX 4AF0
Comparing the results with your formulas, the windows calculator must be no good.
No, you and microsloth windoze are correct ;-)
#### SCPD
##### QRT
Yes it is still correct, just ignore the "0" at the end of your answer. It bares no meaning for our application.
#### loumaag
##### Silent Key - Aug 2014
It would be better, if you are a person who needs to deal with the HEX number for programming a real radio, to just assume the number shown in the display on the Database is correct. (Because it is!)
Your formula works, but only for Moto Type II calculations, it is not correct for APCO-25 calculations.
That said, this is a hobby site, not a site for radio technicians, those of you who are radio technicians already know how the DEC (which is not the Motorola decimal number in any case) we have here on the site relates to the HEX number you may have to use in programming a radio. Bringing this topic up, especially in this forum, will do nothing but confuse the new scanner user.
So, to those of you who are confused by any of the forgoing, ignore this thread, forget that there is a HEX number even displayed on the Talk Group display and move on.
#### SCPD
##### QRT
And if the HEX is soooo unimportant to this site then why is it in this website to begin with?
Why are the connect tones to the systems here?
Because people want to know, this is a information site.
#### newbie
##### Member
Com-4 said:
I recently received an e-mail from a person whos job is to post correct system Hex & dec info asking me to correct my info I submitted. I had posted the correct HEX but his computer program that radio reference uses spit out the incorrect hex & dec. Therefore I felt I should share my way of coming up with the proper HEX & Dec for each talkgroup discovered. So everyone can learn how to properly decode the DEC & HEX when they submit them to R/R
Formulas, this will require a scientific calculator
HEX to DEC {Make sure Calc. is in hex mode & when finished switch to DEC mode for your answer}
--------------------------
Formula
HEX x 10 = DEC
DEC to HEX {Make sure Calc. is in DEC mode & when finished switch to HEX mode for your answer}
-------------------------
Formula
DEC / 16 = HEX
just enter the number in dec and click the HEX button. or vice versa. Real easy. No division to convert at all
#### newbie
##### Member
Com-4 said:
Yes it is still correct, just ignore the "0" at the end of your answer. It bares no meaning for our application.
0 at the end bares a meaning.
4af = 1199
4af0 = 19184
#### SCPD
##### QRT
You have to use the formula otherwise you get the wrong answer.
"just enter the number in dec and click the HEX button. or vice versa. Real easy. No division to convert at all"
Dec 2592 = A20 {wrong}
The answer is A2 or 0A2
#### wlmr
##### Member
Com-4 said:
You have to use the formula otherwise you get the wrong answer.
"just enter the number in dec and click the HEX button. or vice versa. Real easy. No division to convert at all"
Dec 2592 = A20 {wrong}
The answer is A2 or 0A2
Unfortunately this may apply to some systems & radios & not others.
Try decimal 2999. It converts using the calculator to BB7 hex.
In a 9600 baud system P25, ALL IDs are valid, don't have to be divisible by 16 because there aren't any status bits, it's all sent in the digital data.
Anyway, you can't throw away a zero at the end if it is P25 so if your example is for a different type of system that needs to be known.
The windows calculator is fine.
Last edited:
I didnt get BB7,
2999 / 16 = BB
#### SCPD
##### QRT
It Obviously does not work for P25 systems. {I dont have any P25 here so I didnt bother working that part out}
But for type-II systems it works fine.
#### SCPD
##### QRT
I think you was talking about the other guy that stated he didnt need to use my formula.
With my formula it comes out to the correct talkgroup BB
#### wlmr
##### Member
Com-4 said:
I didnt get BB7,
2999 / 16 = BB
That's the point. For Early Motorola Trunking Systems (Type I and II) dividing by 16 is correct. For Project25 systems EVERY single ID is valid.
No dividing by 16.
IDs 2992 through 3007 are all valid, not the same talkgroup BB with different status bits set, but instead talkgroup IDs BB0, BB1, BB2, through BBF. That is why a P25 system can have 65536 possible talkgroup IDs, (0 through 65535 or 0 through FFFF hex) of which 1 through i think 65532 are all programmable into a radio & usable.
So, back to my point (and that of a couple of other posters above) your formula is correct for Older systems, but not for at least the latest P25 systems.
-update-
I see you responded to my post while I was responding to yours. You're on the right track. Appologies if I got too long winded.
Last edited:
#### loumaag
##### Silent Key - Aug 2014
All of this is exactly why I said the HEX is useless for the scanner hobbyist. The people who know what the HEX represents in the DB don't need your formula. Your supplying the HEX information in a submission should not have invoked any contact with you at all, since HEX information can be imported directly into the DB (providing of course that you provided correct information).
Com-4 said:
The info on the database was NOT correct which is why I brought up this thread to begin with. I submitted info in HEX form & the database admin person did not know how to figure out the DEC. Therefore saying that this info is useless is a stupid thing to say by you.
I don't consider myself stupid, and based on my experience on this site and your attitude you don't really know what you are talking about. The HEX information, as I said, for the scanner hobbyist is useless information. The fact that you want to show off and submit HEX is fine, I notice you said you get it by "reading" radios you get hold of, that would indicate that you have Motorola software (I only explain this so that the normal users of this forum know that you have it), a fact I was well aware of before I made my first comment. Your attitude that you hold forth and get away with in the Florida forum will not hold here.
• First, you are not on the staff here, so your assuming to advise users in this forum is inappropriate.
• Second, your referring to my comment as "stupid" when it was clearly appropriate shows that you are ignorant of the purpose of this forum.
• Third, you provided an incomplete picture and when called on it, by me and others, you defend your incorrect data and when overwhelming evidence is presented you then say "It Obviously does not work for P25 systems. {I dont have any P25 here so I didnt bother working that part out} ". [sarcasm] Well, I guess I will contact Lindsay and tell him to remove all the Project 25 systems because Com-4 doesn't think we should bother with them. :roll: [/sarcasm]
I will repeat my position and there is no need for further comment on this.
The formula Com-4 presented does not work in all instances and his explanation of why it doesn't work exactly is flawed. For you professionals out there, you don't need the formula, as you already know how to convert from the Uniden Trunk Tracking Decimal number that all scanners use to either the Motorola HEX or Decimal value to program your radios.
For you hobbyist, forget that HEX exists, it really has no meaning for you or your scanner, there are no scanners out there that use it. Before anyone jumps up and says anything about Scanner Programming Software using it, they don't use it correctly either so don't try and use the HEX values we have in the DB at all; they are there only for the professionals.
And that, Com-4, is why this entire thread is a waste of bandwidth.
#### newbie
##### Member
Com-4 said:
You have to use the formula otherwise you get the wrong answer.
"just enter the number in dec and click the HEX button. or vice versa. Real easy. No division to convert at all"
Dec 2592 = A20 {wrong}
The answer is A2 or 0A2
Then that is not true Hex
DEC HEX
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 a
11 b
12 c
13 d
14 e
15 f
16 10 But you would show this as 1 which is incorrect
#### SCPD
##### QRT
Lou is on a power trip with his volunteer Database Admin Position.
There are many forlumas that people have come up with to convert Hex to DEC. some being more complex then others. Mine works for type-II formats & is a simple form then most formulas I have found. None that I have seen are fail proof. But all work for the application intended.
Lou if your system was so perfect then the database admin wouldnt have come up with a incorrect answer. Which is what started this thread, I do not have to be a admin to post here.
"The place to discuss administration related issues"
Which is what I did. If my formula does not compute P25, then fine. We talk about it, we do not start saying "ignore this thread" its useless. blah blah, no info is ever useless.
I am sure Lou is going to come back with some more flames. He doesn't bother add value to the post, no formulas of his own other then saying "ignore this thread" you guys dont need to bother with hex to Dec. Gee, I guess he is speaking for everyone on this site. Maybe we should just take ALL hex off this site since it is just a "hobby" site. Scanners dont need it While we are at it, lets remove connect tones, system IDs, transmit frequencies for Conventional. After all, its just a hobby site, we dont need to know all that do we?
Lou, you preech about waste of bandwidth, but if you search thru the forums there are about 15 flaming arguements going on with several users including DataBase Admins about much more useless info then this thread. Now thats a waste of bandwidth.
#### SCPD
##### QRT
You can have this thread Lou. I am done & going back to more productive things that actually make money.
More power to Lou.
Vol. Database Admin who says censor everything. You guys do not need to know.
Status
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Calcudoku Advanced Volume II is coming, opinions wanted
Author Message
Posted on: Fri Sep 19, 2014 10:06 pm
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Calcudoku Advanced Volume II is coming, opinions wanted
Hi,
I've been plenty busy creating a sequel to the "Calcudoku Advanced" book.
If anyone has any suggestions, let them speak now, or forever etc. etc..
Patrick
Posted on: Fri Sep 19, 2014 10:51 pm
Posts: 193
Joined: Tue May 24, 2011 4:55 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
Posted on: Fri Sep 19, 2014 11:45 pm
Posts: 186
Joined: Thu May 12, 2011 11:51 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
Excellent that Volume II is coming.
Hopefully points will be awarded as they are always an encouragement. But that obviously would compromise the solutions chapter.
It was nice when exponentiation and zero appeared in the same puzzle. Will this happen again?
Will extra large include regular puzzles >15 or twins >8?
Posted on: Sat Sep 20, 2014 10:24 am
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
sjs34 wrote:
Hopefully points will be awarded as they are always an encouragement. But that obviously would compromise the solutions chapter.
You can get points (12) for all puzzles except the "warmup" ones.
There are only solutions for the example puzzles (chapter 2), and the puzzles in the "warmup" chapter.
Here's the relevant text from the introduction:
"Most puzzle books include solutions to all puzzles in the back, but not this one.
It is meant to be the book with the hardest Calcudoku puzzles out there, so I felt that
solutions should still be excluded (even though some puzzlers have lobbied against this...).
Otherwise the temptation to peek in the back for a number or two may be too great.
Exceptions are the puzzles in the ``Explanations'' and ``Warmup'' chapters:
solutions for these puzzles are in the back.
Of course, if you're really stuck with a particular puzzle you can always
email me (at (email address here)) for a hint or the solution.
And if you do really want the solutions to _all_ puzzles I'll email them to you
straight away."
sjs34 wrote:
It was nice when exponentiation and zero appeared in the same puzzle. Will this happen again?
Good one, thanks, I don't think there are zero+exponentiation puzzles in there, will fix that
sjs34 wrote:
Will extra large include regular puzzles >15 or twins >8?
The "extra large" chapter has two of 11x11, 12x12, 13x13, 14x14, 15x15, and one 17x17.
There are three 10x10 twins (and one 8x8 no-op twin )
Patrick
Posted on: Mon Sep 22, 2014 2:39 pm
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
sjs34 wrote:
It was nice when exponentiation and zero appeared in the same puzzle. Will this happen again?
I have now split the chapter with exponentiation (10 puzzles) into two:
5 with exponentiation, and 5 with exponentiation and zero
Posted on: Wed Sep 24, 2014 4:43 pm
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
What do people think of the cover?
It's similar to the 101 Advanced cover, with the background colour a dark blue gradient instead of black.
I was thinking of adding a lightning bolt over it (or maybe erase a lightning bolt outline),
are there any graphic artists out there with ideas??
Posted on: Thu Sep 25, 2014 6:54 pm
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
Alright, I cooked up a different cover
(how much of the cover puzzle can you solve?? )
Posted on: Thu Sep 25, 2014 9:48 pm
Posts: 713
Joined: Fri May 13, 2011 6:51 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
pnm wrote:
Alright, I cooked up a different cover
(how much of the cover puzzle can you solve?? )
...
Better cover since the mod, bitwise OR and exponentiation functions appear, the previous one had only arithmetic operations.
Apparently it's a 9x9, where the external boundaries have disappeared, shadowed. Cage "43046721^" (row 5) has the candidates [8,9] and cage "78125^" (row 8 with the number 7 erased, not shown) has the candidates [5,7]. Of course, the single 9 and te candidates [3,5] for cage "243^" in column i .
Posted on: Thu Oct 02, 2014 4:24 pm
Posts: 2295
Joined: Thu May 12, 2011 11:58 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
Ok, my proof copy arrived today:
I'm resisting the temptation to immediately release it for publishing,
there are a couple of tiny text and layout edits I want to do (after
which I'll have to order another proof, etc..)
Posted on: Thu Oct 02, 2014 10:16 pm
Posts: 713
Joined: Fri May 13, 2011 6:51 pm
Re: Calcudoku Advanced Volume II is coming, opinions wanted
pnm wrote:
Ok, my proof copy arrived today:
... I'm resisting the temptation to immediately release it for publishing,
there are a couple of tiny text and layout edits I want to do (after
which I'll have to order another proof, etc..)
Congratulations for your new book !!!
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Designed by STSoftware. | 1,686 | 6,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-22 | latest | en | 0.936699 |
https://installmd.com/c/109/go/shuffle-a-slice | 1,721,131,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00606.warc.gz | 277,189,087 | 6,272 | # How to shuffle a slice in Go
Created
Modified
## Using rand.Shuffle Function
Shuffle pseudo-randomizes the order of elements using the default Source.
``````package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
s := []string{"Japan", "Germany", "France"}
rand.Seed(time.Now().UnixNano())
// func Shuffle(n int, swap func(i, j int))
rand.Shuffle(len(s), func(i, j int) { s[i], s[j] = s[j], s[i] })
fmt.Printf("%q\n", s)
}``````
`["Japan" "France" "Germany"]`
Don't forget about the rand.Seed(), otherwise you got the same string every first time launch.
n is the number of elements. Shuffle panics if n < 0.
swap swaps the elements with indexes i and j.
## Using rand.Intn Function
Use the rand.Seed and rand.Intn functions in package math/rand. Fisher–Yates algorithm:
``````package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
s := []string{"Japan", "Germany", "France"}
rand.Seed(time.Now().UnixNano())
for i := len(s) - 1; i > 0; i-- {
j := rand.Intn(i + 1)
s[i], s[j] = s[j], s[i]
}
fmt.Printf("%q\n", s)
}``````
`["France" "Germany" "Japan"]`
Intn returns, as an int, a non-negative pseudo-random number in [0,n)
It panics if n <= 0. | 349 | 1,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.59931 |
https://www.convertit.com/Go/weighing-systems/Measurement/Converter.ASP?From=shaku&To=length | 1,685,288,600,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00503.warc.gz | 767,648,803 | 4,214 | weighing-systems.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Japanese shaku = 0.303022 length (length) ``` Related Measurements: Try converting from "shaku" to astronomical unit, bottom measure, cable length, cloth finger, fermi, finger, foot, furlong (surveyors furlong), gradus (Roman gradus), Greek cubit, hand, mil, nail (cloth nail), nautical league, nautical mile, Roman foot, Roman mile, sun (Japanese sun), survey foot, yard, or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: shaku = 167.02 agate (typography agate), 3,030,220,000 angstrom, .00518241 arpentlin, 35.79 barleycorn, .00828472 bolt (of cloth), 477.2 bottom measure, .00994167 engineers chain, 3.93 Greek palm, 1.31 Greek span, .00047006 li (Chinese li), .303022 m (meter), 5.3 nail (cloth nail), .0110463 naval shot, .39766667 pace, 9.82E-18 parsec, 1.02 Roman foot, .04970833 rope, .00002301 spindle, .00164054 stadium (Roman stadium), 10 sun (Japanese sun).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 440 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.724971 |
http://www.jiskha.com/display.cgi?id=1300400779 | 1,496,073,031,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612399.20/warc/CC-MAIN-20170529145935-20170529165935-00511.warc.gz | 664,729,698 | 3,642 | # Math
posted by on .
What is the circumference of a circle that has a radius of 70ft?
• Math - ,
C = pi * d
C = 3.14 * 140
C = ?
• Math - ,
C=2*r*pi
C=2*70*3.14159
C=439.8226 ft | 75 | 188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-22 | latest | en | 0.81581 |
https://networkx.org/documentation/networkx-2.3/_modules/networkx/algorithms/centrality/closeness.html | 1,685,984,055,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00442.warc.gz | 458,336,734 | 4,948 | Warning
This documents an unmaintained version of NetworkX. Please upgrade to a maintained version and see the current NetworkX documentation.
# Source code for networkx.algorithms.centrality.closeness
# Aric Hagberg <[email protected]>
# Dan Schult <[email protected]>
# Pieter Swart <[email protected]>
#
# Authors: Aric Hagberg <[email protected]>
# Pieter Swart <[email protected]>
# Sasha Gutfraind <[email protected]>
# Dan Schult <[email protected]>
"""
Closeness centrality measures.
"""
import functools
import networkx as nx
__all__ = ['closeness_centrality']
[docs]def closeness_centrality(G, u=None, distance=None, wf_improved=True): r"""Compute closeness centrality for nodes. Closeness centrality [1]_ of a node u is the reciprocal of the average shortest path distance to u over all n-1 reachable nodes. .. math:: C(u) = \frac{n - 1}{\sum_{v=1}^{n-1} d(v, u)}, where d(v, u) is the shortest-path distance between v and u, and n is the number of nodes that can reach u. Notice that the closeness distance function computes the incoming distance to u for directed graphs. To use outward distance, act on G.reverse(). Notice that higher values of closeness indicate higher centrality. Wasserman and Faust propose an improved formula for graphs with more than one connected component. The result is "a ratio of the fraction of actors in the group who are reachable, to the average distance" from the reachable actors [2]_. You might think this scale factor is inverted but it is not. As is, nodes from small components receive a smaller closeness value. Letting N denote the number of nodes in the graph, .. math:: C_{WF}(u) = \frac{n-1}{N-1} \frac{n - 1}{\sum_{v=1}^{n-1} d(v, u)}, Parameters ---------- G : graph A NetworkX graph u : node, optional Return only the value for node u distance : edge attribute key, optional (default=None) Use the specified edge attribute as the edge distance in shortest path calculations wf_improved : bool, optional (default=True) If True, scale by the fraction of nodes reachable. This gives the Wasserman and Faust improved formula. For single component graphs it is the same as the original formula. Returns ------- nodes : dictionary Dictionary of nodes with closeness centrality as the value. See Also -------- betweenness_centrality, load_centrality, eigenvector_centrality, degree_centrality Notes ----- The closeness centrality is normalized to (n-1)/(|G|-1) where n is the number of nodes in the connected part of graph containing the node. If the graph is not completely connected, this algorithm computes the closeness centrality for each connected part separately scaled by that parts size. If the 'distance' keyword is set to an edge attribute key then the shortest-path length will be computed using Dijkstra's algorithm with that edge attribute as the edge weight. In NetworkX 2.2 and earlier a bug caused Dijkstra's algorithm to use the outward distance rather than the inward distance. If you use a 'distance' keyword and a DiGraph, your results will change between v2.2 and v2.3. References ---------- .. [1] Linton C. Freeman: Centrality in networks: I. Conceptual clarification. Social Networks 1:215-239, 1979. http://leonidzhukov.ru/hse/2013/socialnetworks/papers/freeman79-centrality.pdf .. [2] pg. 201 of Wasserman, S. and Faust, K., Social Network Analysis: Methods and Applications, 1994, Cambridge University Press. """ if G.is_directed(): G = G.reverse() # create a reversed graph view if distance is not None: # use Dijkstra's algorithm with specified attribute as edge weight path_length = functools.partial(nx.single_source_dijkstra_path_length, weight=distance) else: path_length = nx.single_source_shortest_path_length if u is None: nodes = G.nodes else: nodes = [u] closeness_centrality = {} for n in nodes: sp = dict(path_length(G, n)) totsp = sum(sp.values()) if totsp > 0.0 and len(G) > 1: closeness_centrality[n] = (len(sp) - 1.0) / totsp # normalize to number of nodes-1 in connected part if wf_improved: s = (len(sp) - 1.0) / (len(G) - 1) closeness_centrality[n] *= s else: closeness_centrality[n] = 0.0 if u is not None: return closeness_centrality[u] else: return closeness_centrality | 1,085 | 4,220 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-23 | latest | en | 0.759382 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-11th-edition/chapter-4-section-4-1-inverse-functions-4-1-exercises-page-395/45 | 1,542,330,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742968.18/warc/CC-MAIN-20181116004432-20181116030432-00121.warc.gz | 881,429,661 | 12,171 | ## College Algebra (11th Edition)
$f$ and $g$ are inverses of each other
The two functions are inverses if $(f\circ g)(x)=x$ and if $(g\circ f)(x)=x$ First, $f(g(x))=\frac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}$. After multiplying both the nominator and the denominator by (x-1), we get: $\frac{2x+1+x-1}{2x+1-2x+2}=\frac{3x}{3}=x$. Second, $g(f(x))=\frac{2\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-1}$. After both the numerator and the denominator by $x-2$, we get $\frac{2x+2+x-2}{x+1-x+2}=\frac{3x}{3}=x$ Therefore, $f$ and $g$ are inverses of each other. | 235 | 553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-47 | latest | en | 0.680142 |
https://documen.tv/mandy-used-the-rule-add-6-to-make-a-pattern-she-started-with-20-and-wrote-the-net-5-numbers-in-h-27737402-79/ | 1,679,729,430,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00211.warc.gz | 256,163,181 | 15,853 | Question
Mandy used the rule “Add 6” to make a pattern. She started with 20 and wrote the next 5 numbers in her pattern.
Which number does NOT belong in Mandy’s pattern?
A. 26
B. 32
C. 38
D. 43
1. ngocdiep
D
Step-by-step explanation:
First 5 numbers after 20 in her pattern.
26,32,38,44,50
No 43 in pattern. | 106 | 311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.882682 |
https://endless-sphere.com/forums/viewtopic.php?p=1445598& | 1,571,282,831,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00287.warc.gz | 464,457,941 | 15,902 | 80% charge in one hour?
Batteries, Chargers, and Battery Management Systems.
Posts: 65
Joined: Nov 10 2016 3:29pm
Location: Oxford, England
80% charge in one hour?
The only way I can get an 80% charge into a Li battery is if I start with the battery fully discharged or near enough. Within +5/0 minutes (it's never faster than 60 minutes) I can hit 80% which presupposes I've towed the car to a charger as there isn't a BMS car system around that will allow a battery to be discharged completely. With a 10% residual charge, although you can reach 80% in less than an hour, you will not reach 90% for 90 minutes or so. This means that under any normal condition you can not put 80% into a battery within an hour.
The problem is that irrespective of what value of constant current I start with, as soon as the voltage reaches 4.2 the current rapidly drops off. You can move that transition point backwards or forwards but what you gain in one charging phase, you lose in another.
The batteries I tested were Samsung 30Q & 35E, LG MJ1; all of which performed as per their data sheets although an 80% charge becomes a difficult to achieve target the more cycles you put it through. The 35E is alone in having a maximum charge current considerably less than 1C.
I ask this because here in Europe many governments, even down to the lowest parish council, are wanting to ban combustion engines in 10 or 20 years time and in the 10 years that I've been involved in this, the rate of technical progress (as measured in energy capacity) has been miserable. All good virtue signalling to the voters but it isn't going to end well at this rate.
izeman 10 GW
Posts: 4928
Joined: Jun 21 2011 8:25am
Location: vienna, austria
Contact:
Re: 80% charge in one hour?
The only question i see is in the topic
So what IS your question? If it is possible to charge a battery to 80% in less than an hour.
Simple answer: Yes. It just depends on the chemistry if they will survive it or not. I got graphene rc-lipo that can be charged @4C. So it's full in 15min.
You write that the charge current gets less very quickly as soon as the battery reaches 4.2V. Sure. It's full by then. Charging happens as CC/CV. Which means that in the beginning you see a Constant Current, and as soon as the charger's output voltage and the battery's voltage start to match (near the end of charge) the current will drop (even to zero amps).
john61ct 1 MW
Posts: 1653
Joined: Dec 18 2018 2:06pm
Re: 80% charge in one hour?
The key is defining YOUR version of 100%, not going by the absolute top voltage possible according to the batt maker who wants a short cycle life.
Drop your Ah expectation a little, back the finish voltage down a lttle.
Peace of mind, fast charge now possible, bank lasts much longer and really not giving up much range.
iOn <-+-> uOil 1 mW
Posts: 12
Joined: Feb 14 2019 5:56am
Re: 80% charge in one hour?
izeman wrote:
Feb 14 2019 8:09am
I got graphene rc-lipo that can be charged @4C. So it's full in 15min.
But its still not a true graphene battery. it would have much more whr/kg and could be charged in 5 minutes to 90%.
Truth hurts
izeman 10 GW
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Location: vienna, austria
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Re: 80% charge in one hour?
iOn <-+-> uOil wrote:
Feb 14 2019 8:17am
izeman wrote:
Feb 14 2019 8:09am
I got graphene rc-lipo that can be charged @4C. So it's full in 15min.
But its still not a true graphene battery. it would have much more whr/kg and could be charged in 5 minutes to 90%.
Sure, but "Turnigy Graphene" is the product name i was refering to. And those are GREAT batteries. The best i've every used regarding charging times and voltage sag.
iOn <-+-> uOil 1 mW
Posts: 12
Joined: Feb 14 2019 5:56am
Re: 80% charge in one hour?
izeman wrote:
Feb 14 2019 8:30am
iOn <-+-> uOil wrote:
Feb 14 2019 8:17am
izeman wrote:
Feb 14 2019 8:09am
I got graphene rc-lipo that can be charged @4C. So it's full in 15min.
But its still not a true graphene battery. it would have much more whr/kg and could be charged in 5 minutes to 90%.
Sure, but "Turnigy Graphene" is the product name i was refering to. And those are GREAT batteries. The best i've every used regarding charging times and voltage sag.
Yep its called graphene enhanced Lipoly battery is the right name of it.
There internal resistance is best.
Do you know how they perform in cycle life against other not enhanced Lipoly ?
Truth hurts
spinningmagnets 100 GW
Posts: 11207
Joined: Dec 21 2007 10:27pm
Location: Ft Riley, NE Kansas
Re: 80% charge in one hour?
The higher the discharge capabilities of a given lithium battery, the faster it can be charged. Monitor the heat, and slowly raise the amps you are charging with. As the cell gets closer to being full, you must lower the charging amps. Again, continuously monitor the heat...
iOn <-+-> uOil 1 mW
Posts: 12
Joined: Feb 14 2019 5:56am
Re: 80% charge in one hour?
Thats only partly true.
Each Lithium chemistry has its own parameter for fast discharge and fast charge to yield a reasonable cycle life.
Watch in the specification of your buy. and only charge fast when needed.
Truth hurts
billvon 100 MW
Posts: 2848
Joined: Sep 16 2007 9:53pm
Location: san diego
Re: 80% charge in one hour?
Feb 14 2019 7:46am
The only way I can get an 80% charge into a Li battery is if I start with the battery fully discharged or near enough.
Sounds like you are saying that if you start with a battery charged to 50% you can't get another 80% worth of charge in it. Yes, that's true - but sorta obvious.
With a 10% residual charge, although you can reach 80% in less than an hour, you will not reach 90% for 90 minutes or so. This means that under any normal condition you can not put 80% into a battery within an hour.
See above.
The problem is that irrespective of what value of constant current I start with, as soon as the voltage reaches 4.2 the current rapidly drops off.
Yes. That's how lithium ion batteries charge - CC/CV.
You can move that transition point backwards or forwards but what you gain in one charging phase, you lose in another.
No. If you charge at a higher C rate you do not "lose" the ability to charge in the CV section. You just can't accelerate the CV section any faster than voltage allows.
I ask this because here in Europe many governments, even down to the lowest parish council, are wanting to ban combustion engines in 10 or 20 years time and in the 10 years that I've been involved in this, the rate of technical progress (as measured in energy capacity) has been miserable. All good virtue signalling to the voters but it isn't going to end well at this rate.
Hard to tell what that statement has to do with anything in the rest of the post. Tesla now has cars that will hit 335 miles, so I think there has been quite a bit of progress on that front. Now the challenge is getting the cost down.
--bill von
flat tire 1 MW
Posts: 1662
Joined: Feb 26 2014 12:20am
Re: 80% charge in one hour?
Well, you can accelerate (or get around) the CV charging section especially if your termination is significantly under 4.2. You do it by understanding the charge profile, shooting past your CV section in CC mode by setting a higher terminating voltage than normal, and terminating manually or by power transferred around the point where the resting voltage will be what you want.
john61ct 1 MW
Posts: 1653
Joined: Dec 18 2018 2:06pm
Re: 80% charge in one hour?
LI of any chemistry will **accept** (pull, demand) a rate higher than is good for it wrt longevity.
Up to you what you want the max rate to be allowed, "made available" by the charge source.
From the batt health POV, no Absorb / CV stage at all is needed, can just charge **to** a termination voltage and stop.
The Bulk/CC to Absorb/CV transition will happen at a much higher/later SoC% with low current than with a high rate.
You can maintain a bit higher charging voltage safely with a high charge rate, just means the settling to a lower resting voltage will be a greater drop, and SoC will not be as high
compared to low currents say overnight.
You do not actually want to get LI to **maximum** Full in normal cycling, unless you really need every last bit of range/Ah capacity for your use case.
Lower volts, lower amps and stopping sooner are all conducive to greater longevity.
izeman 10 GW
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Joined: Jun 21 2011 8:25am
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Re: 80% charge in one hour?
TBH is have NO idea where this discussion is heading, and the TS hasn't given a clue what this thread is about. I think we should wait for him to reply and tell us what his question really is?!
What do you think?
john61ct 1 MW
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Joined: Dec 18 2018 2:06pm
Re: 80% charge in one hour?
TS? Is that the OP?
I think his statement of his (perceived) problem is perfectly clear, and the response are accurate and targeted toward clearing up his misunderstandings, with constructive suggestions for adjusting his expectations and behaviours so as not to cause other problems of which he seems to not be aware.
izeman 10 GW
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Joined: Jun 21 2011 8:25am
Location: vienna, austria
Contact:
Re: 80% charge in one hour?
john61ct wrote:
Feb 14 2019 5:56pm
TS? Is that the OP?
Warren 100 kW
Posts: 1326
Joined: Oct 05 2010 11:35am
Re: 80% charge in one hour?
Feb 14 2019 7:46am
I ask this because here in Europe many governments, even down to the lowest parish council, are wanting to ban combustion engines in 10 or 20 years time and in the 10 years that I've been involved in this, the rate of technical progress (as measured in energy capacity) has been miserable. All good virtue signalling to the voters but it isn't going to end well at this rate.
I think I understand the question. Some cars with liquid cooling are now pulling up to 1.7C for a small portion of the charge cycle. None are doing a full charge in an hour. However, some can definitely do 80% in under an hour, as seen in this video.
https://youtu.be/VABvS9zfuiI?t=115
We have been driving a Chevy Bolt for 19 months, and while the fast charge rate is an annoyance when on trips beyond its 200+ mile range, this is not a big deal.
The real issues are lack of fast charging infrastructure, and the high cost of batteries.
Both of these problems can be solved a lot easier than many of the other challenges of dealing with climate change, over population, resource depletion, etc. I don't think politicians, or most of us, can get our heads around the exponential nature of our problem.
Posts: 65
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Location: Oxford, England
Re: 80% charge in one hour?
Warren wrote:
Feb 15 2019 10:13am
I think I understand the question. Some cars with liquid cooling are now pulling up to 1.7C for a small portion of the charge cycle. None are doing a full charge in an hour. However, some can definitely do 80% in under an hour, as seen in this video.
The real issues are lack of fast charging infrastructure, and the high cost of batteries.
Both of these problems can be solved a lot easier than many of the other challenges of dealing with climate change, over population, resource depletion, etc. I don't think politicians, or most of us, can get our heads around the exponential nature of our problem.
Agreed and thank you. I don’t own a Tesla but I disagree with the conclusion you draw from the video. The actual capacity of that 75kw battery will be greater by about 10% or even more. It is a sensible cushion for Tesla and not criticism by me. If I were to do the same, I would get the same result. Quite how the owner of the car in the video dragged his car to the charging point with a fully discharged battery is not known and that is one issue I have.
My criticism of the EV manufacturers is that the claims that are made about charging are fanciful. We have politicians who believe that if they say something often enough, eg affordable and with charging infrastructure, then it happens as if by magic and anyone who makes them isn’t going to disagree.
The observation by bill von that a 335 mile battery is progress isn’t surely due to any technological breakthrough but simply by packing more batteries in? You can only go so far with that before the weight of the battery begins to impact the performance of the vehicle.
Warren 100 kW
Posts: 1326
Joined: Oct 05 2010 11:35am
Re: 80% charge in one hour?
Feb 15 2019 7:21pm
The actual capacity of that 75kw battery will be greater by about 10% or even more. It is a sensible cushion for Tesla and not criticism by me. If I were to do the same, I would get the same result. Quite how the owner of the car in the video dragged his car to the charging point with a fully discharged battery is not known and that is one issue I have.
My criticism of the EV manufacturers is that the claims that are made about charging are fanciful. We have politicians who believe that if they say something often enough, eg affordable and with charging infrastructure, then it happens as if by magic and anyone who makes them isn’t going to disagree.
The observation by bill von that a 335 mile battery is progress isn’t surely due to any technological breakthrough but simply by packing more batteries in? You can only go so far with that before the weight of the battery begins to impact the performance of the vehicle.
I can't say how much cushion a Model 3 has. I can say that our Bolt has 4-5%. When I pull into the garage with 1.6% remaining, I still have about 4% cushion. GM won't tell me this, but Torque Pro and some hacked PIDs give me the information.
While there has definitely been some incremental progress in lithium ion batteries from 2011 to today, I personally doubt batteries will ever be cheap enough to match petrol prices, since fighting wars isn't factored into the price.
I think that requiring that every vehicle be a hybrid has been the logical thing to do for a decade, at least. But it has been fought by auto, and oil interests. Of course, consumers don't want the added up-front expense either.
billvon 100 MW
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Location: san diego
Re: 80% charge in one hour?
Feb 15 2019 7:21pm
The actual capacity of that 75kw battery will be greater by about 10% or even more.
To be more accurate, the car will only let you use about 90% of that 75kw. It behooves manufacturers to advertise the highest capacity battery they can, so they do. Software then prevents 100% discharge.
My criticism of the EV manufacturers is that the claims that are made about charging are fanciful. We have politicians who believe that if they say something often enough, eg affordable and with charging infrastructure, then it happens as if by magic and anyone who makes them isn’t going to disagree.
My first EV was a Leaf. The 80% claim was certainly accurate when the car was new. Since it was a 24kwhr pack, a 50kw charger could get it to 80% in about half an hour. After five years ESR increased and the car could barely reach 80% in an hour.
The observation by bill von that a 335 mile battery is progress isn’t surely due to any technological breakthrough but simply by packing more batteries in? You can only go so far with that before the weight of the battery begins to impact the performance of the vehicle.
Exactly.
That 335 mile range is, of course, to EV manufacturers packing lots of batteries in. If they did not have a high energy density battery to start with the weight would make the design impossible.
I used one of the first 18650 batteries in a satellite phone I designed back around 2000. It was 1300mah and was the best we could get at that point. Now the same size/weight cell is more than twice that - due to slow, incremental improvements in cell technology.
--bill von
billvon 100 MW
Posts: 2848
Joined: Sep 16 2007 9:53pm
Location: san diego
Re: 80% charge in one hour?
Warren wrote:
Feb 16 2019 11:26am
I think that requiring that every vehicle be a hybrid has been the logical thing to do for a decade, at least.
I think a better thing to do is require cars to average a certain fuel economy, and gradually increase that fuel economy. In 20 years, gas powered hybrids may seem like gas guzzlers.
--bill von
Chalo 100 GW
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Joined: Apr 29 2009 11:29pm
Location: Austin, Texas
Re: 80% charge in one hour?
Warren wrote:
Feb 16 2019 11:26am
I think that requiring that every vehicle be a hybrid has been the logical thing to do for a decade, at least. But it has been fought by auto, and oil interests. Of course, consumers don't want the added up-front expense either.
That makes no sense. We have fuel economy requirements; we only need to keep them updated. And they need to apply to every single new vehicle instead of entire production fleets. How you meet economy, emissions, and safety requirements should be irrelevant.
Cars have been getting faster, heavier, and more powerful continuously since the political will to keep tightening fuel economy standards was lost decades ago. That's not only the result of corruption in government and industry-- it's the moral failure of a people who have decided that driving personal cars is acceptable no matter what damage it does.
I hope I live long enough to see that no longer be the default assumption. But I'm not waiting; I'm living another way already.
This is to express my gratitude to Justin of Grin Technologies for his extraordinary measures to save this forum for the benefit of all.
john61ct 1 MW
Posts: 1653
Joined: Dec 18 2018 2:06pm
Re: 80% charge in one hour?
Chalo wrote: That's not only the result of corruption in government and industry-- it's the moral failure of a people who have decided that driving personal cars is acceptable no matter what damage it does.
A lot of that moral failure is due to billions of dollars spent by FF on decades of skilled propaganda subverting the truth, making it easy for a critical mass of naturally lazy and gullible people to rationalize, "maybe it really isn't true".
But then maybe even without that, enough people just don't care if our habitat won't support the species, long as the real crises are after their own personal death.
Older people have the power, the young wear the consequences.
billvon 100 MW
Posts: 2848
Joined: Sep 16 2007 9:53pm
Location: san diego
Re: 80% charge in one hour?
Chalo wrote:
Feb 16 2019 12:45pm
That makes no sense. We have fuel economy requirements; we only need to keep them updated. And they need to apply to every single new vehicle instead of entire production fleets. How you meet economy, emissions, and safety requirements should be irrelevant.
Would be impractical to meet them for things like ambulances, 18 wheelers, tow trucks, farm trucks, snowplows, school buses etc. In fact the effort would result in more, rather than less, fuel consumption. (Think about WalMart using 20 small hybrid trucks instead of 1 18 wheeler to get their goods.)
The reason I like the CAFE approach is that it drives what's important (average vehicle fuel economy) while putting all the burden on the manufacturer. They can do whatever they like, as long as their average fleet fuel economy is X.
--bill von
Warren 100 kW
Posts: 1326
Joined: Oct 05 2010 11:35am
Re: 80% charge in one hour?
billvon wrote:
Feb 16 2019 1:04pm
Would be impractical to meet them for things like ambulances, 18 wheelers, tow trucks, farm trucks, snowplows, school buses etc. In fact the effort would result in more, rather than less, fuel consumption. (Think about WalMart using 20 small hybrid trucks instead of 1 18 wheeler to get their goods.)
The reason I like the CAFE approach is that it drives what's important (average vehicle fuel economy) while putting all the burden on the manufacturer. They can do whatever they like, as long as their average fleet fuel economy is X.
I agree that raising fuel standards would result in the same thing. One form of hybrid, or another, would likely be the result anyway.
I am curious what makes you think an 18 wheeler can't be a hybrid, since diesel locomotives have been from the beginning?
Chalo 100 GW
Posts: 7961
Joined: Apr 29 2009 11:29pm
Location: Austin, Texas
Re: 80% charge in one hour?
billvon wrote:
Feb 16 2019 1:04pm
Chalo wrote:
Feb 16 2019 12:45pm
We have fuel economy requirements; we only need to keep them updated. And they need to apply to every single new vehicle instead of entire production fleets.
Would be impractical to meet them for things like ambulances, 18 wheelers, tow trucks, farm trucks, snowplows, school buses etc. In fact the effort would result in more, rather than less, fuel consumption. (Think about WalMart using 20 small hybrid trucks instead of 1 18 wheeler to get their goods.)
The reason I like the CAFE approach is that it drives what's important (average vehicle fuel economy) while putting all the burden on the manufacturer. They can do whatever they like, as long as their average fleet fuel economy is X.
So have a different per-vehicle economy standard for commercial trucks and industrial vehicles, and exempt emergency vehicles from the requirements. The point is to make all, not just some, of the vehicles within a given class conform to standards. Imagine a system that allowed fleet average crash safety ratings.
Why should we have (for example) city buses that weigh 30,000 pounds empty? That's normal these days, and it's stupid.
With cars, the presence of a few huge gas guzzlers not only drives demand for more of them, but it makes people regard small, efficient vehicles as impractical or unsafe. If every personal vehicle is only allowed say 4 liters per 100km, then we would see the tradeoffs inherent in using an excessive vehicle. Sure-- have your three row SUV, but you're gonna go slow. The incentive would be to use the smallest and lightest vehicle that works for the job.
This is to express my gratitude to Justin of Grin Technologies for his extraordinary measures to save this forum for the benefit of all.
billvon 100 MW
Posts: 2848
Joined: Sep 16 2007 9:53pm
Location: san diego
Re: 80% charge in one hour?
Warren wrote:
Feb 16 2019 1:40pm
I am curious what makes you think an 18 wheeler can't be a hybrid, since diesel locomotives have been from the beginning?
No reason at all. I was saying that it would counterproductive to require 18 wheelers to get the same MPG as a car (i.e. I was arguing against applying one standard to "every single vehicle.")
--bill von | 5,595 | 22,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-43 | latest | en | 0.968681 |
https://www.doubtnut.com/question-answer/a-diagonal-of-a-parallelogram-divides-it-into-two-congruent-triangles-1338591 | 1,632,636,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00553.warc.gz | 738,103,124 | 90,656 | Home
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# A diagonal of a parallelogram divides it into two congruent triangles.
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Class 9th
Class 9th
Class 9th
Class 9th
Class 9th
Class 9th
Class 9th | 416 | 986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | longest | en | 0.618625 |
https://www.inchcalculator.com/convert/abampere-to-statampere/ | 1,718,239,345,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00400.warc.gz | 744,585,924 | 15,567 | # Abamperes to Statamperes Converter
Enter the electric current in abamperes below to get the value converted to statamperes.
## Result in Statamperes:
1 abA = 2,997,924,536.8431 statA
Do you want to convert statamperes to abamperes?
## How to Convert Abamperes to Statamperes
To convert a measurement in abamperes to a measurement in statamperes, multiply the electric current by the following conversion ratio: 2,997,924,536.8431 statamperes/abampere.
Since one abampere is equal to 2,997,924,536.8431 statamperes, you can use this simple formula to convert:
statamperes = abamperes × 2,997,924,536.8431
The electric current in statamperes is equal to the electric current in abamperes multiplied by 2,997,924,536.8431.
For example, here's how to convert 5 abamperes to statamperes using the formula above.
statamperes = (5 abA × 2,997,924,536.8431) = 14,989,622,684.216 statA
### How Many Statamperes Are in an Abampere?
There are 2,997,924,536.8431 statamperes in an abampere, which is why we use this value in the formula above.
1 abA = 2,997,924,536.8431 statA
Abamperes and statamperes are both units used to measure electric current. Keep reading to learn more about each unit of measure.
## What Is an Abampere?
The abmpere is the electrical current constant equal to the flow of one abcoulomb per second, or ten amperes in the International System of Units.
The abampere is a centimeter-gram-second (CGS) electromagnetic unit of electric current. An abampere is sometimes also referred to as a biot. Abamperes can be abbreviated as abA; for example, 1 abampere can be written as 1 abA.
## What Is a Statampere?
The statmpere is the electrical current constant equal to the flow of one statcoulomb per second, or 0.33356 nanoamperes in the International System of Units.
The statampere is a centimeter-gram-second (CGS) electrostatic unit of electric current. Statamperes can be abbreviated as statA, and are also sometimes abbreviated as A-esu. For example, 1 statampere can be written as 1 statA or 1 A-esu.
## Abampere to Statampere Conversion Table
Table showing various abampere measurements converted to statamperes.
Abamperes Statamperes
0.000000001 abA 2.9979 statA
0.000000002 abA 5.9958 statA
0.000000003 abA 8.9938 statA
0.000000004 abA 11.99 statA
0.000000005 abA 14.99 statA
0.000000006 abA 17.99 statA
0.000000007 abA 20.99 statA
0.000000008 abA 23.98 statA
0.000000009 abA 26.98 statA
0.0000000001 abA 0.299792 statA
0.000000001 abA 2.9979 statA
0.00000001 abA 29.98 statA
0.0000001 abA 299.79 statA
0.000001 abA 2,998 statA
0.00001 abA 29,979 statA
0.0001 abA 299,792 statA
0.001 abA 2,997,925 statA
0.01 abA 29,979,245 statA
0.1 abA 299,792,454 statA
1 abA 2,997,924,537 statA | 914 | 2,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.856165 |
http://dst.lbl.gov/ACSSoftware/colt/api/cern/colt/function/class-use/IntIntFunction.html | 1,537,564,628,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157569.48/warc/CC-MAIN-20180921210113-20180921230513-00125.warc.gz | 72,443,403 | 2,772 | Colt 1.2.0
## Uses of Interfacecern.colt.function.IntIntFunction
Packages that use IntIntFunction cern.jet.math Tools for basic and advanced mathematics: Arithmetics and Algebra, Polynomials and Chebyshev series, Bessel and Airy functions, Function Objects for generic function evaluation, etc.
Uses of IntIntFunction in cern.jet.math
Fields in cern.jet.math declared as IntIntFunction `static IntIntFunction` `IntFunctions.and` Function that returns a & b. `static IntIntFunction` `IntFunctions.compare` Function that returns a < b ? -1 : a > b ? 1 : 0. `static IntIntFunction` `IntFunctions.div` Function that returns a / b. `static IntIntFunction` `IntFunctions.equals` Function that returns a == b ? 1 : 0. `static IntIntFunction` `IntFunctions.max` Function that returns Math.max(a,b). `static IntIntFunction` `IntFunctions.min` Function that returns Math.min(a,b). `static IntIntFunction` `IntFunctions.minus` Function that returns a - b. `static IntIntFunction` `IntFunctions.mod` Function that returns a % b. `static IntIntFunction` `IntFunctions.mult` Function that returns a * b. `static IntIntFunction` `IntFunctions.or` Function that returns a | b. `static IntIntFunction` `IntFunctions.plus` Function that returns a + b. `static IntIntFunction` `IntFunctions.pow` Function that returns (int) Math.pow(a,b). `static IntIntFunction` `IntFunctions.shiftLeft` Function that returns a << b. `static IntIntFunction` `IntFunctions.shiftRightSigned` Function that returns a >> b. `static IntIntFunction` `IntFunctions.shiftRightUnsigned` Function that returns a >>> b. `static IntIntFunction` `IntFunctions.xor` Function that returns a ^ b.
Methods in cern.jet.math that return IntIntFunction `static IntIntFunction` ```IntFunctions.chain(IntFunction g, IntIntFunction h)``` Constructs the function g( h(a,b) ). `static IntIntFunction` ```IntFunctions.chain(IntIntFunction f, IntFunction g, IntFunction h)``` Constructs the function f( g(a), h(b) ). `static IntIntFunction` `IntFunctions.swapArgs(IntIntFunction function)` Constructs a function that returns function.apply(b,a), i.e.
Methods in cern.jet.math with parameters of type IntIntFunction `static IntFunction` ```IntFunctions.bindArg1(IntIntFunction function, int c)``` Constructs a unary function from a binary function with the first operand (argument) fixed to the given constant c. `static IntFunction` ```IntFunctions.bindArg2(IntIntFunction function, int c)``` Constructs a unary function from a binary function with the second operand (argument) fixed to the given constant c. `static IntIntFunction` ```IntFunctions.chain(IntFunction g, IntIntFunction h)``` Constructs the function g( h(a,b) ). `static IntIntFunction` ```IntFunctions.chain(IntIntFunction f, IntFunction g, IntFunction h)``` Constructs the function f( g(a), h(b) ). `static IntIntFunction` `IntFunctions.swapArgs(IntIntFunction function)` Constructs a function that returns function.apply(b,a), i.e.
Colt 1.2.0 | 832 | 3,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-39 | latest | en | 0.361088 |
https://integrativemodeling.org/nightly/doc/ref/RMSDMetric_8h_source.html | 1,632,253,360,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00521.warc.gz | 370,997,561 | 4,749 | IMP Reference Guide develop.97bab1d8fd,2021/09/21 The Integrative Modeling Platform
RMSDMetric.h
Go to the documentation of this file.
1 /**
2 * \file IMP/spb/RMSDMetric.h
3 * \brief Distance RMSD Metric
4 *
6 */
7
8 #ifndef IMPSPB_RMSD_METRIC_H
9 #define IMPSPB_RMSD_METRIC_H
10
11 #include <IMP/algebra.h>
12 #include <IMP/statistics.h>
13 #include <IMP/spb/spb_config.h>
14
15 IMPSPB_BEGIN_NAMESPACE
16
17 /** Compute the RMSD between two sets of particles in two configurations.
18 */
19 class IMPSPBEXPORT RMSDMetric : public statistics::Metric {
20 Particles ps_;
21 Floats weight_;
22 std::vector<algebra::Vector3Ds> coords_;
23
24 double get_rmsd(algebra::Vector3Ds v0, algebra::Vector3Ds v1) const;
25
26 public:
28 void add_configuration(double weight = 1.0);
29 Float get_weight(unsigned i);
30
31 // IMP_METRIC(RMSDMetric);
32 double get_distance(unsigned int i, unsigned int j) const IMP_OVERRIDE;
33 unsigned int get_number_of_items() const IMP_OVERRIDE;
35 };
36
37 IMPSPB_END_NAMESPACE
38
39 #endif /* IMPSPB_RMSD_METRIC_H */
#define IMP_OBJECT_METHODS(Name)
Define the basic things needed by any Object.
Definition: object_macros.h:25
Include all non-deprecated headers in IMP.statistics.
Include all non-deprecated headers in IMP.algebra.
double get_weight(unsigned int i) const
Return a weight for the point.
Definition: Metric.h:37
double get_rmsd(const Vector3DsOrXYZs0 &m1, const Vector3DsOrXYZs1 &m2)
double Float
Basic floating-point value (could be float, double...)
Definition: types.h:20
double get_distance(const Line3D &s, const Vector3D &p)
Get closest distance between a line and a point.
#define IMP_OVERRIDE
Cause a compile error if this method does not override a parent method.
Store data to be clustered for distance metric based algorithms.
Definition: Metric.h:25 | 515 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.458312 |
http://www.joannejacobs.com/2013/03/can-number-sense-be-taught/ | 1,485,292,345,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285289.45/warc/CC-MAIN-20170116095125-00312-ip-10-171-10-70.ec2.internal.warc.gz | 520,904,313 | 15,795 | # Can ‘number sense’ be taught?
First graders with poor “number sense” rarely catch up in math skills, concludes a University of Missouri study. But it’s not clear how parents or preschools can teach number sense.
What’s involved? Understanding that numbers represent different quantities — that three dots is the same as the numeral “3” or the word “three.” Grasping magnitude — that 23 is bigger than 17. Getting the concept that numbers can be broken into parts — that 5 is the same as 2 and 3, or 4 and 1. Showing on a number line that the difference between 10 and 12 is the same as the difference between 20 and 22.
Factors such as IQ and attention span didn’t explain why some first-graders did better than others.
Math learning disabilities often aren’t diagnosed till fifth grade, much too late, says Dr. Kathy Mann Koepke, of NIH’s National Institute of Child Health and Human Development.
David Geary, who conducted the Missouri study, thinks parents can help children develop number sense before they start school. NIH’s Mann Koepke urges parents to talk to young children about “magnitude, numbers, distance, shapes as soon as they’re born.”
— Don’t teach your toddler to count solely by reciting numbers. Attach numbers to a noun — “Here are five crayons: One crayon, two crayons…” or say “I need to buy two yogurts” as you pick them from the store shelf — so they’ll absorb the quantity concept.
— Talk about distance: How many steps to your ball? The swing is farther away; it takes more steps.
— Describe shapes: The ellipse is round like a circle but flatter.
— As they grow, show children how math is part of daily life, as you make change, or measure ingredients, or decide how soon to leave for a destination 10 miles away,
However, researchers don’t really know why some kids get that 3, three and xxx are the same thing and others don’t. Children with poor phonemic awareness need to work harder to distinguish the sounds in a word. Perhaps some kids need to work harder — or differently — to see mathematical relationships.
1. cranberry says:
“Children with poor phonemic awareness need to work harder to distinguish the sounds in a word. Perhaps some kids need to work harder — or differently — to see mathematical relationships”
Kids with dyslexia need intensive tutoring. They need to learn strategies which work for them. There are programs which do help, but identification of the problem is important. It is not a problem students can overcome merely by “working harder.”
I believe research in babies has shown the existence of an innate number system. Some people may lack a number sense, just as some people are tone deaf. If you are otherwise intelligent, but can’t see that ten dots are a larger quantity than two dots at a glance, “work harder” isn’t fruitful advice.
2. Stacy in NJ says:
It can be taught but must be done at a pace and with a depth that’s appropriate for each individual. Some just need significantly more practice and repetition than others.
3. cranberry says:
Interesting article in Nature about dyscalculia: http://www.nature.com/news/dyscalculia-number-games-1.12153
If students vary from the norm, working harder at programs aimed at the average student won’t produce the same effect as programs aimed at their particular physical array. So, rather than saying the students need to practice the times tables 90,000 times more than their classmates, with the underlying message being, “you’re stupid,” perhaps some seemingly simpler practice of the skills no one thinks to teach will produce a bigger improvement.
We don’t think to practice the skill of recognizing 9 is larger than 7, because for most people, that’s too obvious to think it needs to be taught, or practiced. It feels like practicing the concepts Up and Down.
I looked at the games created by Dr. Butterworth, “Number Sense” and “The Number Catcher.” Links are found at the end of the article.
• Stacy in NJ says:
The idea that the concept needs to be broken down into its most basic component parts and taught incrementally is the basis for DI. Zig Engelmann advocated this way back in the ’70s with his book Give Your Child a Superior Mind which is out of print but can be found (for several hundred dollars) at amazon.
http://www.amazon.com/Superior-Increase-Intelligence-Everyday-Preschool/dp/B000RSNX5E/ref=sr_1_7?ie=UTF8&qid=1364400633&sr=8-7&keywords=give+your+child+a+superior+mind
• How would you even begin to teach something like that, though? Something that’s so obvious to everyone else? It boggles my mind to even try to come up with a strategy to show someone who doesn’t know that 9 is larger than 7…
• cranberry says:
I guess it’s a real candidate for drill and kill. See Michael Connell’s link in these comments, or the online games Number Sense and Number Catcher, listed at the bottom of the Nature article.
In some sense, we don’t “know” 9 is larger than 7 as young children–we just see it. This pile of jelly beans is larger than that pile. No one needs to sit down with us, introduce us to the concept of larger and smaller, give us tricks to remember it, and drill us in choosing the larger pile. We do teach children the names and symbols associated with the quantities and relationships they already perceive.
Just looking at the first games, I’m intrigued. There’s a great deal of doing. Speed is important. Can you select five items and two items to make seven items? Not, can you count to seven? Because you can learn a sequence of words, but not be able to connect seven with 7 with ******* . In some sense, it is mindless. If cats are born with this sense, it’s not a verbal skill.
If it does work for the (few) students whose number sense is not naturally strong, why not try it? American students spend immense spans of time on totally useless video games. It would be silly to require students who have a good number sense to play the same games in class–and a waste of class time. As an individual method to remediate weakness, though, it would make sense. It could also avoid the stigma of being pulled out of class for tutoring, if the kids could practice at home.
• Stacy in NJ says:
It’s sometimes more about vocabulary and not being able to see the distinction between “larger” and “greater” and “taller” and many other terms thrown about by teachers with the assumption the kids understands the precise meaning of the word. By larger does the teacher mean the actual printed number on paper or its value?
• Cranberry says:
Stacy in NJ, as I understand it from the articles available online reporting on the research into dyscalculia, it’s not a teacher error, nor a vocabulary problem. It’s more akin to being tone deaf or color blind. Ask the child to tell you which group of dots (or candies, or buttons) is larger (has more candy, etc.), and they can’t automatically point to the larger group. They may be able to painstakingly count–but “normal” children don’t have to painstakingly count items individually to see that a group of 3 is smaller than a group of 6.
I assume it’s rare.
4. lulu says:
My kids are 7 and 4, and when they were younger (and still for my 4 year old) I sometimes have them clean up while counting (ie you don’t have to put away all of the cars – you do 5 and I’ll do 5, or you do 15 and I’ll do the rest). When I was small, my mom had me count the blocks for a quilt that she was making. I think that counting things is probably like the difference between singing the alphabet song as music vs being able to point to letters as you sing it, and then knowing the letters independent of the music.
• Florida resident says:
Dear Lulu:
In our family there was a pedagogical tool — to use large number of identical objects, e.g. identical red buttons and identical white buttons.
The idea was to teach a child to tell, which group contains more buttons: red group, or white group, or their quantities are equal. This can be done without actually counting number in each group, but lining the groups to make one-to-one correspondence of elements of two sets. Either one of the lines becomes ”longer” than the other, or the lines end equally.
“Counting” is essentially making one-to-one correspondence between elements of actual set (of buttons in this example) and elements of
“Set of natural numbers: one, two three, … “.
Using these identical objects, you can teach “rectangular numbers”, i.e. “composite numbers” as opposed to “prime numbers”.
Then you teach “square numbers”, k*k, putting a square with side “k” of these identical objects,
Then you teach “triangular numbers”; triangle with side “k” contains
k*(k+1)/2 elements.
Then you discuss “area of rectangle”, Area=(a*b). Then you discuss “geometrically” the formula
(a+b)*(a+b)={i.e. square of (a+b)}=(a*a +2a*b +b*b}.
As a mental exercise you can chat about “cubic numbers”, “pyramidal numbers”, etc. With my kids I (and latter my son with his two kids) used little cubes (bought at “Jo-Ann” store). We managed to get cube 5*5*5=125. It was fun.
Back to the topic. Yes, I have met some children born with poor sense of numbers.
• Florida resident says:
Clarification. “Triangle” above means
“equilateral triangle” with side “k”.
• lu says:
I didn’t mention it in the previous post, but my older child seems to grasp numbers without needing a lot of concrete examples. He did arithmetic very early, and was still in preschool when he figured out negative numbers. A friend taught him the idea of perfect squares and square roots over Christmas dinner when he was 4. I don’t know if the early practice caused this or if it’s just the way that he is.
My younger child (who just turned 4) is doing fine – she counts, understands more and less, corrected people who were guessing her age by telling them ‘younger’, and can do simple arithmetic if she uses her fingers. Practice with us and a brother who teaches her math for fun doesn’t hurt, but when I look at the 2 of them, it’s obvious that one has a great number sense and the other has a way with words and music.
I tend to think that most people can be taught competence, but that certain things will be easier for some people than others because they’re just naturally better at it. It doesn’t have to be deterministic, though. When I was young, my parents were told that I’d be successful if I avoided math or science. They never told me until I was grown…and I have a PhD in genetics and teach biology. 🙂
5. Florida resident says:
“– Describe shapes: The ellipse is round like a circle but flatter.”
Here is the definiton of ellipse by military teacher:
“Ellipse is a circle, inscribed into a square 3 by 4”.
Translation of this joke
{actually ellipse is not a circle;
there is no such thing as square 3 by 4}
into scientific language:
Affine transformation of a plane with a square and an inscribed circle, both drawn on that plane, transfroms
square into rectangle (not necesserily 3 by 4) and at the same time transforms the said circle into ellipse.
6. Patti says:
Much like literacy, numeracy is something that takes a ton of exposure. The kinds of things the researchers suggest you do are common sense to the kinds of parents who automatically provide a word-rich environment. That means there are plenty of parents who don’t do those things and their children are at automatic disadvantage. I think we’ve drilled literacy into everyone for years. There are plenty of companies who are finding Americans’ lack of numeracy makes hiring them difficult.
While I realize that there are some people with a math processing disability, it’s not very common. Instead, difficulties are usually from a lack of developed number sense (note that I said developed, not innate) and a learned anxiety about math (as in a parent who tells a kid that getting a bad grade in math is OK because they weren’t very good at math, either, as though it’s something you’re born with). Math is a skill, just like reading, playing an instrument, or learning a sport. While some people will find it easier than others, age-appropriate practice can make most people fluent. Children with a true disability need to be identified earlier than they are now.
I used to teach preschool and now I teach middle school math and science, so I’ve done math instruction from both ends. It’s MUCH easier to teach number sense to little kids than it is to big kids. Same goes for reading. | 2,792 | 12,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-04 | latest | en | 0.952523 |
http://erlang.org/pipermail/erlang-questions/2009-October/046942.html | 1,585,699,000,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370504930.16/warc/CC-MAIN-20200331212647-20200401002647-00322.warc.gz | 59,103,851 | 2,615 | # [erlang-questions] Ordered set with lower_bound
Richard O'Keefe ok@REDACTED
Fri Oct 9 04:34:12 CEST 2009
On Oct 9, 2009, at 12:12 PM, Igor Ribeiro Sucupira wrote:
> Hi.
>
> I need some ordered set that allows me to "search and then iterate"
> efficiently. In other words, I need this functionality:
> http://www.cplusplus.com/reference/stl/set/lower_bound/
What you want is a function something like
iterator_beginning_at(Set, Element) -> Iterator
The existing gb_sets *module* does not support that, but the
*data structure* easily could.
Look at the code for constructing and using iterators.
reverse_iterator/1 and previous/1 are mine.
tree --> {key,tree,tree} | 'nil'.
iterator --> [tree|iterator] | [].
The (sub)trees in an iterator are there for the sake of the key
and _one_ of the subtrees.
iterator({_Size, Tree}) ->
iterator(Tree, []).
iterator({_X, nil, _Right} = Tree, As) ->
[Tree|As];
iterator({_X, Left, _Right} = Tree, As) ->
iterator(Left, [Tree|As]);
iterator(nil, As) ->
As.
next([{X, _Left, Right}|As]) ->
{X, iterator(Right, As)};
next([]) ->
none.
reverse_iterator({_Size, Tree}) ->
reverse_iterator(Tree, []).
reverse_iterator({_X, _Left, nil} = Tree, As) ->
[Tree|As];
reverse_iterator({_X, _Left, Right} = Tree, As) ->
reverse_iterator(Right, [Tree|As]);
reverse_iterator(nil, As) ->
As.
previous([{X, Left, _Right}|As]) ->
{X, reverse_iterator(Left, As)};
previous([]) ->
none.
Perhaps the simplest way to do it would be to use
iterator(remove_less(Set, Start))
or reverse_iterator(remove_greater(Set, Start))
where
remove_less(Set, X) -> {Y \in Set | Y >= X}
remove_greater(Set, X) -> {Y \in Set | X >= Y}
However, since the size of a gb_set is stored only at the top,
exporting these functions in the interface, returning GB sets,
would require time linear in the result. It's OK to use them
internally, though, as iterators don't need to know the size.
So
geq_iterator({_Size, Tree}, Min) ->
iterator(remove_less(Tree, Min), []).
leq_iterator({_Size, Tree}, Max) ->
iterator(remove_greater(Tree, Max), []).
range_iterator({_Size, Tree}, Min, Max) ->
iterator(remove_less(remove_greater(Tree, Max), Min), []).
geq_reverse_iterator({_Size, Tree}, Min) ->
reverse_iterator(remove_less(Tree, Min), []).
leq_reverse_iterator({_Size, Tree}, Max) ->
reverse_iterator(remove_greater(Tree, Max), []).
range_reverse_iterator({_Size, Tree}, Min, Max) ->
reverse_iterator(remove_less(remove_greater(Tree, Max), Min), []).
remove_less({Key,Left,Right}, Min) ->
if Key < Min -> remove_less(Right, Min)
; Key > Min -> {Key, remove_less(Left, Min), Right}
; true -> {Key, nil, Right}
end;
remove_less(nil, _) ->
nil.
remove_greater({Key,Left,Right}, Max) ->
if Key > Max -> remove_greater(Left, Max)
; Key < Max -> {Key, Left, remove_greater(Right, Max)}
; true -> {Key, Left, nil}
end;
remove_greater(nil, _) ->
nil.
I made a copy of gb_sets.erl called gb_sets2.erl,
-export([reverse_iterator/1, previous/1,
geq_iterator/2, geq_reverse_iterator/2,
leq_iterator/2, leq_reverse_iterator/2,
range_iterator/3, range_reverse_iterator/3]).
and put the stuff through some testing, and it seemed to work.
This has been tested, BUT NOT THOROUGHLY TESTED. | 898 | 3,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.727819 |
https://docplayer.net/21422449-Math-0910-review-concepts-haugen.html | 1,620,550,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00390.warc.gz | 230,286,773 | 24,809 | # MATH-0910 Review Concepts (Haugen)
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1 Unit 1 Whole Numbers and Fractions MATH-0910 Review Concepts (Haugen) Exam 1 Sections 1.5, 1.6, 1.7, 1.8, 2.1, 2.2, 2.3, 2.4, and 2.5 Dividing Whole Numbers Equivalent ways of expressing division: a b, b a, a b, or a b Components of a division problem: Dividend, divisor, quotient, and remainder Division involving 0 or 1 Exponents and Exponential Expressions Expanded form vs exponential form Evaluating an exponential expression Evaluate = find the value of Order of Operations: P.E.M.D.A.S. P = Parentheses, brackets [ ], set braces { }, or a fraction bar Multiply or Divide from left to right as the operators occur in the expression Example: = 8 4 = 32, not = 2 Add or Subtract from left to right Example: = = 12, not 15 9 = 6 Rounding Whole Numbers Using the Principle of Estimation to approximate sums, differences, products, and/or quotients Each rounded number in the approximation should have only one nonzero digit Fractions Numerator vs denominator Proper vs improper fractions Reducing fractions Factor trees help us identify common factors Divide out common factors Mixed Numbers Mixed numbers have a whole number component and a (proper) fractional component Convert from an improper fraction to a mixed number and vice versa Multiplying Fractions and/or Mixed Numbers Cancel common factors (if possible) and then multiply straight across Convert mixed numbers to improper fractions before multiplying (or dividing) Dividing Fractions and/or Mixed Numbers Convert division to multiplication using the reciprocal of the divisor: a c a d = b d b c
2 Unit 2 Adding and Subtracting Fractions Exam 2 Sections 2.6, 2.7, 2.8, and 2.9 Least Common Denominator (LCD) Goal is to identify the smallest number that is a multiple of both denominators Creating Equivalent Fractions Adding and Subtracting Fractions Case 1: Like Denominators Case 2: Unlike Denominators Adding and Subtracting Mixed Numbers* Case 1: Like Denominators Case 2: Unlike Denominators *Sometimes subtracting mixed numbers involves borrowing. This can be avoided by converting the mixed numbers to improper fractions and then subtracting. Unit 3 Decimals Exam 3 Sections 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, and 3.7 Writing a Word Name for a Decimal Using the Place Value System Comparing Two or More Decimals Rounding Decimals Identify the round off place and then apply rounding rules Adding and Subtracting Decimals Helps to add or subtract vertically, keeping the decimal points aligned Multiplying and Dividing Decimals Writing the Decimal Equivalent of a Fraction Should be familiar with three scenarios: 1. Terminating decimal (zero remainder) 2. Non-terminating repeating decimals (we use a bar over the digit or digits that repeat) 3. Non-terminating non-repeating decimals (when this happens, we usually round to a specified place)
3 Unit 4 Rates, Ratios, and Proportions Exam 4 Sections 4.1, 4.2, 4.3, and 4.4 Ratios and Rates Ratios and rates are comparisons of two quantities If the quantities have the same units, the comparison is called a ratio If the quantities have different units, the comparison is called a rate Three Ways of Expressing Ratios: a to b, a : b, and a b Remember when writing as a fraction to reduce to simplest form Writing a Rate in Simplest Form Unit Rates A unit rate is a rate whose denominator is miles 60 miles Ex. or 60 miles per hour 3 hours 1 hour Proportions A proportion is a statement that two rates or two ratios are equal Use cross products to determine if an equation is a proportion Solve Equations of the form a n = b Divide both sides of the equation by a (the coefficient of the variable n) Solve Proportions Use cross products to convert the proportion to an equation of the form a n = b Unit 5 Percents Working with Percents Converting percents to decimal numbers Drop the percent symbol and then move the decimal point two places to the left Converting decimal numbers to percents Move the decimal point two places to the right and then write the percent symbol at the end of the number Converting percents to fractions Two options: 1. Use the definition of percent 2. Convert the percent to a decimal and then convert the decimal to a fraction Converting fractions to percents Convert the fraction to a decimal and then convert the decimal to a percent
4 Unit 5 Percents (continued) Solve Percent Problems Two options: 1. Use the equation: amount = percent x base 2. Use the percent proportion: Solve Applied Percent Problems amount base = percent number 100 Sales tax calculations Sales Tax = Sales Tax Rate x List Price Discount problems Discount Amount = Discount Rate x List Price Commission problems Commission = Commission Rate x Value of Sales Percent of increase/decrease problems Percent of Increase = Amount of Increase / Original Amount Percent of Decrease = Amount of Decrease / Original Amount Simple interest calculations Interest = Principal x Rate x Time or I = P x R x T * *note: the units of time on R and T must match in these problems Unit 6 Signed Numbers Exam 6 Sections 9.1, 9.2, 9.3, and 9.4 Adding Signed Numbers Geometric approach: draw vectors on a number line Vectors representing positive numbers should point toward the right Vectors representing negative numbers should point toward the left Alternative approach is based on absolute values: If the addends have the same sign, add the absolute values of the addends. The sign of the sum will be the same as the sign of the addends. If the addends have opposite signs, find the absolute value of the addends. Subtract the smaller absolute value from the larger absolute value. The sign of the sum will be the same as the sign of the addend with the larger absolute value. Subtracting Signed Numbers In the expression a b, a is called the minuend and b is called the subtrahend Subtraction can be converted to addition using the opposite of the subtrahend: a b = a + ( b)
5 Unit 6 Signed Numbers (continued) Multiplying and Dividing Signed Numbers The product (or quotient) of two numbers with the same sign will always be positive ( + )( + ) = + ( )( ) = + ( + ) + = + ( ) = + The product (or quotient) of two numbers with opposite signs will always be negative ( + ) ( ) = ( )( + ) = ( ) + = ( + ) =
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### averages simple arithmetic average (arithmetic mean) 28 29 weighted average (weighted arithmetic mean) 32 33
537 A accumulated value 298 future value of a constant-growth annuity future value of a deferred annuity 409 future value of a general annuity due 371 future value of an ordinary general annuity 360 future | 7,729 | 32,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-21 | latest | en | 0.786781 |
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A look at a few Tripos questions X
May 29, 2012
Since time is short, I am going to discuss a couple of Groups questions but in slightly less detail than I have been giving up to now: instead of working through the questions completely, I’ll try to zero in on the most important points. Because there wasn’t a separate Groups course until 2008, I am taking my questions from that year.
5E. For a normal subgroup $H$ of a group $G$, explain carefully how to make the set of (left) cosets of $H$ into a group.
For a subgroup $H$ of a group $G$, show that the following are equivalent:
(i) $H$ is a normal subgroup of $G$;
(ii) there exist a group $K$ and a homomorphism $\theta:G\rightarrow K$ such that $H$ is the kernel of $\theta$.
Let $G$ be a finite group that has a proper subgroup $H$ of index $n$ (in other words, $|H|=|G|/n$). Show that if $|G|>n!$, then $G$ cannot be simple. [Hint: Let $G$ act on the set of left cosets of $H$ by left multiplication.]
(more…)
Group actions IV: intrinsic actions
December 10, 2011
I have a confession to make. When I was an undergraduate at Cambridge (hmm, that sounds as though it might be the beginning of quite an interesting confession, so I’d better forestall any disappointment by saying right now that it isn’t), there was a third-year course in group theory, taught by John Thompson no less, on which I did not do very well. For a few weeks it seemed to cover material that we’d done in our first year, and then suddenly it got serious, with things like the Sylow theorems. And at that point I got lost, and was unable to do the questions on the examples sheets. I can’t remember much about the questions, but I think my difficulty was that there was a slightly indirect style of proof that caused me to find arguments hard to remember and even harder to come up with. And I never got round to doing anything about it: I went into a different area of maths, and even now I don’t know the proofs of the Sylow theorems. In fact, I don’t even know the statements, though I know they’re about the existence of subgroups of various cardinalities, and I know that they are proved using cleverly defined group actions. I’ve skim-read the proofs, so I have a fairly good idea of their flavour, but I don’t know the details. In particular, I don’t know which action does the job.
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Group actions III — what’s the point of them?
November 25, 2011
Somebody told me recently that a few years ago they had a supervision with a colleague of mine (who shall remain nameless, but he or she is an applied mathematician) and asked what the point of group actions was. “I have absolutely no idea,” was the response, and the implication that one might draw from it was apparently intended.
No pure mathematician could hold such a view. I’ve stated a few times that group actions tell you a lot about groups. In this post I want to try to explain why that is, though there is far more to say than I am capable of explaining, let alone fitting into one blog post.
Several proofs that use group actions seem to depend on almost magically coming up with an action that just happens, when you analyse it the right way, to tell you what you wanted to know. I am not an algebraist and do not have a good all-purpose method for finding actions to prove given statements. I don’t rule out that such a method might exist, at least for reasonably simple statements, and would be interested to hear from anybody who thinks they can usefully add to what I have to say.
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Normal subgroups and quotient groups
November 20, 2011
The traditional presentation of normal subgroups and quotient groups goes something like this. First, you define a subgroup to be normal if it satisfies a certain funny condition. Then, given a group $G$ and a normal subgroup $H$, you show that you can define an operation on the cosets of $H$, and that that operation turns the set of all cosets into a group, called the quotient group. Ideally, you also show that one can’t give a natural group structure to the left cosets of an arbitrary subgroup: that justifies restricting attention to normal subgroups.
There’s nothing terribly wrong with this approach, but it does leave one question unanswered: why bother with all this stuff? The traditional approach to that question is to ignore it, confident that the answer will gradually reveal itself. The more group theory you do, the more normal subgroups and quotients will arise naturally and demonstrate their utility, so if you just diligently keep studying, you will (fairly soon) come to regard normal subgroups and quotient groups as natural concepts that were obviously worth introducing.
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Group actions II: the orbit-stabilizer theorem
November 9, 2011
How many rotational symmetries does a cube have? This question can be answered in a number of ways. Perhaps the one that most readily occurs to people is this: each vertex can end up in one of eight places; once you’ve decided where to put it, there are three places you can put one of its neighbours; once you’ve decided where to put that, the rotation is determined, so the total number of rotations is $8\times 3=24$.
Here’s another proof. Take one of the faces. It can go to one of six other faces, and once you’ve decided which face it will go to, one of the vertices on the face has four places it can go, and once you’ve decided that you’ve fixed the rotation. So the total number of rotations is $6\times 4=24$.
And here’s another. Take one of the midpoints of the twelve edges. There are twelve places it can end up, and once you’ve decided where to put it, there are two choices for how you send the two endpoints of the original edge to the endpoints of the new edge. So the total number of rotations is $12\times 2=24$.
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Group actions I
November 6, 2011
There is something odd about the experience of learning group theory. At first, one is told that the great virtue of groups is their abstractness: many mathematical structures, from number systems, to sets of permutations, to symmetries, to automorphisms of other algebraic structures, to invariants of geometric objects (these last two are examples you won’t meet for a while) have important properties in common, and these are encapsulated in a small set of axioms that lead to a rich theory with applications throughout mathematics. So far so good — understanding about abstraction is wonderful and mind-expanding and the definition of a group is one of the best examples.
But then one studies group actions (and later group representations). They appear to be doing the reverse of abstraction: we take an abstract group and find a way of thinking of it as a group of symmetries. And that is supposed to help us understand the group better — so much so that group actions are an indispensable part of group theory.
So is abstraction good or bad? Well, both the views above are correct. Abstraction does indeed play a very important clarifying role, by showing us that many apparently different phenomena are basically the same, and isolating the aspects of those phenomena that really matter. However, if a group is defined for us in an abstract way (I’ll say more precisely what I mean by this later), then showing that it is isomorphic to a group of symmetries can make it much easier to answer questions about that group.
In this post, and one or two further ones, I want to discuss what a group action actually is, the orbit-stabilizer theorem and how to remember its proof, and how to use group actions to prove facts about groups.
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Alternative definitions
October 25, 2011
Something that happens very often in lecture courses is that you are presented with a definition, and soon after it you are told that a certain property is equivalent to that definition. This equivalence means that in principle one could have chosen the property as the “definition” and the definition as an equivalent property. To put that differently, suppose you are developing a piece of theory and have some word you want to define. To pick an imaginary example, suppose you have a notion of a set being “abundant”. Suppose that a set is defined to be abundant if it has property P, and that property P is equivalent to property Q. There may well not be much to choose between the following pair of alternatives. On the one hand you can say, “Definition: A set is abundant if it has property P,” and follow that with, “Proposition: A set is abundant if and only if it has property Q,” while on the other you can say, “Definition: A set is abundant if it has property Q,” and follow that with, “Proposition: A set is abundant if and only if it has property P.”
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1. The first is a confusion that some people have about what a permutation of $\{1,2,\dots,n\}$ actually is. What could possibly be the trouble, you might ask? Well, let's take the permutation that in cycle notation is written $(124)$. My guess is that a non-negligible percentage of people reading this have worried about whether this permutation means that you cycle round the elements 1, 2 and 4 of the set $\{1,2,\dots,n\}$ or the elements in the places 1, 2 and 4. | 2,090 | 9,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 28, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-13 | longest | en | 0.965972 |
http://89.30.95.junowebmaillogin.com/waves-and-wave-properties-worksheet/ | 1,596,547,486,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.94/warc/CC-MAIN-20200804131928-20200804161928-00301.warc.gz | 2,046,548 | 17,929 | # Waves And Wave Properties Worksheet
## Wave Properties Worksheet 2 Science Buddy
Wave Properties Ii 144 150 1 Choose Which Type Of Wave Is Represented Or Combinations Of Waves Wave Description 1 Sound Waves 2 Combination Of Transverse And Longitudinal 3 Contains A Crest And A Trough 4 Wave Travels In The Same Direction 5 Wave Contains An Amplitude 6 Gaps Between Compressions Are Called Rarefactions 7 Particles Move Perpendicular Right Angles 8 Waves Found In The Ocean
### Wave Types And Properties Worksheets Learny Kids
Wave Types And Properties Displaying Top 8 Worksheets Found For Wave Types And Properties Some Of The Worksheets For This Concept Are Waves And Wave Properties Name Date Anatomy Of A Wave Work Sound Waves Teachers Club Science Formclass P Hysics Waves Name Wave Properties Looking At Work And Activity G4 U2 L3 Lesson 3 Waves Fourth Grade Science Waves
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###### Wave Properties Teaching Resources
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Wave Properties Practice Interactive Worksheet
Wave Properties Practice Practice Sheet To Accompany Waves Notes Id Language English Grade Level High School Age 12 17 Main Content Waves Other Contents Add To My Workbooks 0 Download File Add To Classroom Share Through Whatsapp Link To This Worksheet Copy Drshirley Finish What Do You Want To Do Check My Answers Email My Answers To My Teacher Cancel Text
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Waves And Wave Properties Teachengineering
Wave Properties Wave Properties Depend On What Type Of Energy Is Making The Waves 1 Wavelength The Distance Between One Point On A Wave And The Exact Same Place On The Next Wave 2 Frequency How Many Waves Go Past A Point In One Second Unit Of Measurement Is Hertz Hz The Higher The Frequency The More Energy In The Wave 10 Waves Going Past In 1 Second 10 Hz 1 000 Waves Go Past In
Waves And Wave Properties Lesson Teachengineering
During The Presentation Of Lecture Information On Wave Characteristics And Properties Students Take Notes Using A Handout Then They Label Wave Parts On A Worksheet Diagram And Draw Their Own Waves With Specified Properties Crest Trough And Wavelength They Also Make Observations About The Waves They Drew To Determine Which Has The Highest And The Lowest Frequency With This Knowledge Students Better Understand Waves And Are A Step Closer To Understanding How Humans See Color
Waves Introduction And Types
Waves Have Several Properties Which Are Represented In The Diagrams Below In A Transverse Wave The Crest And Troughs Are The Locations Of Maximum Displacement Up Or Down The Amplitude Is The Measurement Of Maximum Displacement The Wavelength Is The Distance Of One Complete Wave Cycle For Example The Distance From Crest To Crest Or Trough To Trough Would Be 1 Wavelength In A Longitudinal
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Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice. | 1,725 | 8,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-34 | latest | en | 0.500795 |
https://mathzsolution.com/category/infinite-product/ | 1,675,831,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00489.warc.gz | 391,635,018 | 19,947 | ## What is the definition of the absolute convergence of an infinite product (1+a_1)(1+a_2)(1+a_3)\cdots(1+a_1)(1+a_2)(1+a_3)\cdots?
For an infinite product (1+a_1)(1+a_2)(1+a_3)\cdots, whats the definition of convergence and absolute convergence? Why the absolute convergence corresponding to the absolute convergence of sum of an infinite series a_1+a_2+a_3+\cdots? If the infinite product absolutely convergent, does it mean it is convergent? If the sum of a_1+a_2+a_3+\cdots convergent, does it mean the corresponding product convergent? I … Read more
## Constructing a smooth function whose roots consist only of each of the primes.
My first attempt: f(x)=∞∏i=1(1−xpi) If we take a look at the Riemann zeta function: ζ(s)=∞∑n=11ns=∞∏i=0(11−p−si)=∞∏i=0(1−1psi)−1 f(1)=1ζ(1)=0 By f‘s construction, it should only contain 0 factors at prime xs, which 1 is not. Therefore, the only reason f should be 0 at 1 is that the product converges to 0 as i→∞. My second attempt: g(x)=∞∏i=1(1−x2p2i) … Read more
## Prove that ∞∏n=k0(1−an){\prod_\limits{n = k_0}^{\infty} (1 – a_n)} \; converges to positive value
Request help proving the following: Given: (1) For all n>0,an>0 (2) ∞∑n=1an is convergent. (3) There exists k0∈Z+ such that for all k≥k0,ak<1. To prove: ∞∏n=k0(1−an) converges to a λ that is strictly greater than 0. Motive: On pages 10-11 of this pdf, the assertion helps prove that if ∞∑n=1an is convergent, then the infinite … Read more
## What’s the sum of the inverses of the Primorial numbers?
What’s the sum of the inverses of the primorial numbers? Let the $n^{th}$ primorial number be the product of the first $n$ primes $\displaystyle n\#= \prod_{p\leq p_n}p$ So $N\#=2,2\cdot3,2\cdot3\cdot5,\ldots=2,6,30,210,\ldots$ Evaluate $\displaystyle\sum_{n\in\Bbb N}\frac1{n\#}$ Here’s what very limited part of this I can do: Obviously it’s in the fairly narrow interval $(\frac23,e-2)$ by comparing the first two … Read more
## Limit point pp of A⊂2RA \subset 2^\mathbb{R} such that no sequence in AA converges to pp. Can AA be countable?
The problem here is to find a subset A⊂2R and a limit point p of A such that no sequence in A converges to p. Here, 2R is the product ∏λ∈R{0,1} equipped with the product topology. I think I have an example for when A is not countable. Suppose A is the set of all … Read more
## Landau’s proof that ζ(s)=∏p11−p−s\zeta(s)=\prod_{p} \frac{1}{1-p^{-s}}
In the Handbuch, Landau proves that for all s>1 the following equality holds ζ(s)=∏p11−p−s. I’m having trouble with the following part of his proof: Landau says that for all s>1 we have ∏p≤x11−p−s=∏p≤x(1+1ps+1p2s+…)=∞∑n=1′1nswhere in the last sum n runs through all numbers whose prime factors are all ≤x. Can someone please shed light on the … Read more
## Infinite Product – Seems to telescope
Evaluate (1+23+1)(1+232+1)(1+233+1)⋯ It looks like this product telescopes: the denominators cancel out (except the last one) and the numerators all become 3. What would my answer be? Answer we have the following identity (which affirms that the product telescopes): (1+23n+1)=3⋅3n−1+13n+1=1+3−(n−1)1+3−n (as denoted in the comment by Thomas Andrews)and as a result: n∏k=1(1+23k+1)=2⋅3n3n+1=21+3−n AttributionSource : Link … Read more | 1,103 | 3,238 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-06 | latest | en | 0.753128 |
http://clay6.com/qa/13070/there-are-20-boxes-numbered-1-2-20-five-boxes-are-selected-at-random-and-ar | 1,534,615,751,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213693.23/warc/CC-MAIN-20180818173743-20180818193743-00166.warc.gz | 79,830,125 | 27,862 | # There are $20$ boxes numbered $1,2,......20$. Five boxes are selected at random and are arranged in ascending order. In how many ways this can be done so that no. $10$ is present in each selection and it comes in $3rd$ place in the arrangement.?
$\begin{array}{1 1} (A) ^{10}C_2\times ^{10}C_2 \\ (B) ^9C_2\times ^{10}C_2 \\ (C) ^9C_2\times ^{9}C_2 \\ (D) ^{19}C_4 \end{array}$
Since no. 10 is present in each selection, only 4 boxes are to be selected.
Also since no. 10 comes in $3^{rd}$ position,
first two boxes should be numbered <10 and
next two numbers should be >10.
2 numbers are to be selected from 1,2.......9 in $^9C_2$ ways
and
2 numbers are to be selected from 11,12,......20 in $^{10}C_2$ ways.
$\therefore$ The required no. of arrangements = $^9C_2\times ^{10}C_2$
edited Dec 20, 2013 | 277 | 805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-34 | longest | en | 0.885512 |
https://math.stackexchange.com/questions/1426496/nature-of-primes-as-the-building-blocks-of-integers | 1,569,010,822,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574077.39/warc/CC-MAIN-20190920200607-20190920222607-00097.warc.gz | 560,852,132 | 31,339 | # Nature of primes as the building blocks of integers?
It is considered standard in mathematics that all integers can be expressed as the product of primes: $$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$$ Where $p_i$ is prime and $p_{i-1}<p_i$ and $a_i$ is and integer such that $a_i \geq 1$. But this means this does not explicitly define all integers in terms of primes, other integers must be used to describe the powers. Of course those integers can be expressed by a string of primes with powers and then those powers expressed by a string of primes with powers ... etc. So whilst the powers are really just multiplying primes by one another a certain number of times, has any work gone into looking at expressing numbers in multiplicative power form with just primes or $1$? In other words, can you keep substituting in prime chains in the place of the powers until all integers involved are primes or $1$?
Yes, of course your assumption is correct.
Consider for example, $\large{n=3^{56}\cdot 5^{48}}.$
Let us colour exponents that are composite, and indent for each level of tetration for which the action must be repeated:
• @J-S BTW I don't know whether you use Mathematica, but on the back of this question, I asked a computational approach to the problem at MMA SE and yohbs wrote a nice little fuction here that outputs for most $n$ (unless it is too large - and it needs top be really quite big to be too big!! ;) – martin Sep 9 '15 at 0:36 | 358 | 1,448 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-39 | latest | en | 0.93258 |
https://stonespounds.com/7381-pounds-in-stones-and-pounds | 1,627,897,194,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00121.warc.gz | 551,512,985 | 4,893 | # 7381 pounds in stones and pounds
## Result
7381 pounds equals 527 stones and 3 pounds
You can also convert 7381 pounds to stones.
## Converter
Seven thousand three hundred eighty-one pounds is equal to five hundred twenty-seven stones and three pounds (7381lbs = 527st 3lb). | 70 | 281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-31 | latest | en | 0.83174 |
https://xbuzznet.com/qa/question-what-is-newton-formula.html | 1,620,675,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00145.warc.gz | 1,137,681,931 | 7,370 | # Question: What Is Newton Formula?
## What is a Newton simple definition?
From Simple English Wikipedia, the free encyclopedia.
The newton (symbol: N) is the SI unit of force.
It is named after Sir Isaac Newton because of his work on classical mechanics.
A newton is how much force is required to make a mass of one kilogram accelerate at a rate of one metre per second squared..
## What are the 4 laws of physics?
Fundamental force, also called fundamental interaction, in physics, any of the four basic forces—gravitational, electromagnetic, strong, and weak—that govern how objects or particles interact and how certain particles decay.
## What is the name of Newton’s third law?
Forces result from interactions! … These two forces are called action and reaction forces and are the subject of Newton’s third law of motion. Formally stated, Newton’s third law is: For every action, there is an equal and opposite reaction.
## What are Newton’s 1st 2nd and 3rd laws?
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
## What does 40 kN mean?
kilonewtonThe force value 40 kN (kilonewton) in words is “fourty kN (kilonewton)”.
## What is the force of 1 kg?
That is, it is the weight of a kilogram under standard gravity. Therefore, one kilogram-force is by definition equal to 9.80665 N.
## How do you solve Newton problems?
Problem-Solving StrategiesIdentify the physical principles involved by listing the givens and the quantities to be calculated.Sketch the situation, using arrows to represent all forces.Determine the system of interest. … Apply Newton’s second law to solve the problem. … Check the solution to see whether it is reasonable.
## What is a Newton equal to?
Newton, absolute unit of force in the International System of Units (SI units), abbreviated N. It is defined as that force necessary to provide a mass of one kilogram with an acceleration of one metre per second per second.
## What is 1 kg in Newtons?
9.806651 kilogram or kilogram-force (kg or kgf) = 9.80665 newtons (N) = 35.2739619 ounces (oz) = 1000 grams (g) = 1000000 milligrams (mg) = 2.20462262 pounds (lbs) = 0.157473044 stones (st).
## What is 1500 kg Newton?
Convert 1500 Kilograms Force to Newtons1500 Kilograms Force (kgf)14,710 Newtons (N)1 kgf = 9.807 N1 N = 0.101972 kgf
## What is my weight in newtons on earth?
Your mass on Earth is 98 kg. The gravitational force the Earth exerts on you is (approximately) (98kg)(9.8ms−2)=960.4kg m s−2=960.4N. You weigh 960.4N on Earth.
## What is Newton 3rd law examples?
Other examples of Newton’s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward.
## How can I calculate weight?
Weight is a measure of the force of gravity pulling down on an object. It depends on the object’s mass and the acceleration due to gravity, which is 9.8 m/s2 on Earth. The formula for calculating weight is F = m × 9.8 m/s2, where F is the object’s weight in Newtons (N) and m is the object’s mass in kilograms.
## Is weight measured in Newtons?
Weight is a measure ofthe force of gravity on a physical object and is measured in newtons. | 873 | 3,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-21 | latest | en | 0.915685 |
https://mechanics.stackexchange.com/questions/74179/how-much-time-driving-at-operating-temperature-is-adequate-to-boil-off-condensat | 1,719,068,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00479.warc.gz | 327,188,236 | 40,035 | # How much time driving at operating temperature is adequate to boil off condensation in the engine, exhaust, and gasoline?
Let's say I drive for a total of 20 mins, round trip, each day for 4 days a week (a normal commute to the train station each day for work).
The problem is that it's likely not enough time to hit operating temperature and remain at operating temperature long enough to boil off the condensation that forms inside of the engine, exhaust, and gas, all of which accumulate over time and cause rust and corrosion.
This means that I likely need to go for a long distance drive on the weekend so that I can bring the temperature of the engine, exhaust, (maybe gas?), high enough and long enough in order to boil off the condensation to prevent rust and corrosion, but how long would I need to drive at operating temperature to boil it all off adequately?
• You can improve the question if you shorten it to the facts. The question will be just 3 lines afterwards. Commented Jan 31, 2020 at 14:36
• I'd just like to point out that there probably isn't any moisture accumulating inside the engine, and even if it does it will evaporate within a few rotations of turning the key from the vacuum generated by the pistons, and/or immediately burnt off. Condensation in the exhaust is also a minor concern, the effect of a handful of droplets is negligible when compared with operating at high temperatures (which exacerbates rust). And of course there's nothing you can do about moisture in the gas once it's there. Commented Jan 31, 2020 at 18:29
• Regarding getting water out of your exhaust system: often times mufflers will have a weep hole at the lowest point of the muffler to allow water to escape. In the summer, the exhaust system quickly gets hot enough to evaporate water. But in the winter, you may notice water dripping out of your exhaust pipe. Drill a small (1 or 2mm, or 1/16 inch) hole in the lowest point of your muffler. This will allow water to drip out, which may prevent your muffler from rusting as soon as it otherwise would have.
– sam
Commented Jan 31, 2020 at 23:53
• @user2647513 So I guess my first question should have been whether condensation is an actual concern for a car owner. Commented Feb 4, 2020 at 14:28
• @Sam I have a weep hole in my exhaust, but what about condensation that forms in the inside of the muffler? Or the pipes that lead up to the engine? Commented Feb 4, 2020 at 14:29
As jwh20 states, there's a lot of factors here. However, 20 minutes of continuous driving should be enough to get rid of any condensation in the oil in most vehicles. While the coolant may take five minutes to get to temperature, it takes the oil longer. I say "most vehicles" because there are always those outliers which will take more driving.
The exhaust will take a little bit longer. This also depends on how it is driven and at what atmospheric temperature. If it is -20°F (-29°C) out, it could take a skosh longer than 20 minutes (in my approximation) to completely ensure the exhaust tract is completely heated to ensure it is good. If you're talking about 90°F (32°C), it might only take five minutes.
As far as gas goes, no amount of running the engine in any kind of weather will remove the water from the gas, other than it passing into the engine and getting vaporized through the normal combustion processes. It's just not going to happen. Until the fuel is ejected through the fuel injector, the two are completely separate. There are a few ways to get rid of the water, however running to create heat is not one of them.
All that said, if you are only running your vehicle 10 minutes each way, you probably aren't getting the job done. You should consider driving it a bit more on the weekends. Get out. Enjoy the open road. Your vehicle will thank you for it.
• A clear indicator of excessive condensation is emulsification on your oil filler cap. If you remove your oil filler cap and there is a white/cream colored paste/goo then you could try adding a breather or running engine to a higher temperature to ensure the emulsification process doesn't take place. Commented Jan 31, 2020 at 14:18
• If it were -29°C outside, my engine temp gauge would stay bottomed out on cold indefinitely without doing something extreme like driving it in first gear. Commented Jan 31, 2020 at 20:46
• @R..GitHubSTOPHELPINGICE might want to contact the dealer for a more suitable stat to use then Commented Jan 31, 2020 at 21:15
I can't say that I have ever seen a hard-and-fast answer to this because there are many variables. Things like the engine's normal operating temperature, how much moisture we're talking about, etc.
I've heard and read that about an hour of driving AFTER you get the engine to operating temperature will be sufficient to drive out excess moisture.
From personal experience with a motorcycle that has a glass sight window to show the oil level that often shows "milky" when there is water will clear up nicely after an hour or so of riding.
It all pivots around water's boiling point of 212F (100C) at standard temperature and pressure (close enough to standard outdoor conditions).
Your thermostat has a setting of 160, 180, 190 or as high as 210. Note that this is near the boiling point of water at STP, but, the cooling system is pressurized. You can see the coolant temperature on the engine's "Temp" gauge. Note that it has a happy place it normally lives. Once it's there, the engine is at operating temperature. If your engine doesn't behave that way, it may have a stuck thermostat. Stuck wide open, obviously, since if it was stuck closed, you'd be forced to fix it.
Water inside the air passages of your engine (pistons, intake manifold etc.) is negligible; no problem there.
Engine oil gets pretty hot - hotter than coolant. In fact, performance cars have "engine oil to coolant heat exchangers", where the coolant cools off the oil. So it's a safe bet that when the car reaches operating temperature, the oil has already spent some time above 212F, and any water therein has boiled off and been pulled into the PCV system (positive crankcase ventilation) which is designed to prevent build-up of explosive gases.
Manual Transmission oil - if you have a stick shift, there is little you can do, but stick shift transmissions have very little ventilation, and they don't run hot, so they have very little condensation.
Automatic transmission oil - Automatics make heat from operating in torque-converting mode (non-lockup). This applies to around-town driving with low speeds and lots of start-stops. However, the car does not leave transmission warm-up to chance. The transmission has an oil cooler - but it's not oil-to-air like you'd think; it's oil-to-coolant. It releases heat into the engine coolant, with the amusing side effect that if the transmission oil was cold, it's warmed by the engine coolant to at least its setoff temperature. Fully warming the engine then doing around-town driving should quickly drive off any water in the ATF.
Power steering is a closed system; little opportunity for condensation.
A/C freon is a closed system. The water in your A/C evaporator is supposed to happen.
Driving the water out of your coolant is not something you would want to do.
Ditto windshield washer fluid.
Getting water out of your exhaust system is very hard, because the combustion products are H2O and CO2, so there's a lot more water in there than in ambient air. Some cars have an electric smog pump that pumps ambient air into the catalytic converters (so the engine can run slightly rich to make the reducing cats happy, and then slightly lean to make the oxidizing cats happy). I suppose you could rig it to run the smog pump for 15 minutes after the engine is shut down. I just replace my exhaust system every 12 years or so when it rusts. | 1,768 | 7,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.956406 |
https://quick-adviser.com/what-is-input-shape-in-cnn/ | 1,638,789,755,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363292.82/warc/CC-MAIN-20211206103243-20211206133243-00457.warc.gz | 535,727,130 | 12,788 | # What is input shape in CNN?
## What is input shape in CNN?
Input Shape You always have to give a 4 D array as input to the CNN . So input data has a shape of (batch_size, height, width, depth), where the first dimension represents the batch size of the image and the other three dimensions represent dimensions of the image which are height, width, and depth.
## What is the output of Conv1D?
so likewise conv1d will take all 32 input channels one by one and generate 32 different output channel for each input channels, so in total 32×32=1024 channels output should be generated, but Conv1d output is only 32 channel output is generated instead of 1024 channels.
## What is Conv1D?
We can see that the 2D in Conv2D means each channel in the input and filter is 2 dimensional(as we see in the gif example) and 1D in Conv1D means each channel in the input and filter is 1 dimensional(as we see in the cat and dog NLP example).
## What is filter and kernel size in Conv1D?
filters: Integer, the dimensionality of the output space (i.e. the number output of filters in the convolution). kernel_size: An integer or tuple/list of a single integer, specifying the length of the 1D convolution window.
## Is kernel size same as filter size?
In a given convolution layer, the Kernel size is the X * Y dimensions, and the number of filters (or “channels” as it’s often called) is the Z dimension. The Kernel size usually defines a relatively small square consisting of X*Y numbers that together encode a specific feature / pattern.
## What is a good kernel size?
A common choice is to keep the kernel size at 3×3 or 5×5. The first convolutional layer is often kept larger. Its size is less important as there is only one first layer, and it has fewer input channels: 3, 1 by color.
## What is difference between kernel and filter in CNN?
Kernel vs Filter For example, in 2D convolutions, the kernel matrix is a 2D matrix. A filter however is a concatenation of multiple kernels, each kernel assigned to a particular channel of the input. So for a CNN layer with kernel dimensions h*w and input channels k, the filter dimensions are k*h*w.
## Why is smaller kernel size more meaningful?
So, with smaller kernel sizes, we get lower number of weights and more number of layers. So, with larger kernel sizes, we get a higher number of weights but lower number of layers. Due to the lower number of weights, this is computationally efficient.
## What is the difference between kernel and filter?
A “Kernel” refers to a 2D array of weights. The term “filter” is for 3D structures of multiple kernels stacked together. For a 2D filter, filter is same as kernel. But for a 3D filter and most convolutions in deep learning, a filter is a collection of kernels.
## What is the filter in CNN?
In CNNs, filters are not defined. The value of each filter is learned during the training process. This also allows CNNs to perform hierarchical feature learning; which is how our brains are thought to identify objects. In the image, we can see how the different filters in each CNN layer interprets the number 0.
## Why does CNN use kernel?
The kernel is a matrix that moves over the input data, performs the dot product with the sub-region of input data, and gets the output as the matrix of dot products. Kernel moves on the input data by the stride value. In short, the kernel is used to extract high-level features like edges from the image.
## Why does CNN use ReLU?
The ReLU function is another non-linear activation function that has gained popularity in the deep learning domain. ReLU stands for Rectified Linear Unit. The main advantage of using the ReLU function over other activation functions is that it does not activate all the neurons at the same time.
## What does dropout layer do in CNN?
Dropout is a technique used to prevent a model from overfitting. Dropout works by randomly setting the outgoing edges of hidden units (neurons that make up hidden layers) to 0 at each update of the training phase.
## What is flatten layer in CNN?
Flatten is the function that converts the pooled feature map to a single column that is passed to the fully connected layer. Dense adds the fully connected layer to the neural network.
## Why is leaky ReLU better than ReLU?
Leaky ReLU & Parametric ReLU (PReLU) Leaky ReLU has two benefits: It fixes the “dying ReLU” problem, as it doesn’t have zero-slope parts. It speeds up training. There is evidence that having the “mean activation” be close to 0 makes training faster.
## What does ReLU stand for?
The rectified linear activation function or ReLU for short is a piecewise linear function that will output the input directly if it is positive, otherwise, it will output zero.
## Which is better ReLU or LeakyReLU?
In my experience, LeakyReLU shows at least the same or better results in most comparisons with ReLU, but moreover, it allows NN to learn in setups (architectures) where the ReLU fails. For example, it’s the case where a NN architecture contains “bottlenecks” – very narrow layers with small neurons count.
## What is leaky ReLU activation and why is it used?
Leaky ReLU function is an improved version of the ReLU activation function. As for the ReLU activation function, the gradient is 0 for all the values of inputs that are less than zero, which would deactivate the neurons in that region and may cause dying ReLU problem. Leaky ReLU is defined to address this problem.
## What is output of ReLU?
According to equation 1, the output of ReLu is the maximum value between zero and the input value. An output is equal to zero when the input value is negative and the input value when the input is positive.
## Is ReLU good for classification?
No, it does not. For binary classification you want to obtain binary output: 0 or 1. To ease the optimization problem (there are other reason to do that), this output is subtituted by the probability of been of class 1 (value in range 0 to 1). If you use relu you will obtain a value from 0 to inf as output.
## What will happen if the learning rate is set too low or too high?
If your learning rate is set too low, training will progress very slowly as you are making very tiny updates to the weights in your network. However, if your learning rate is set too high, it can cause undesirable divergent behavior in your loss function.
## Does learning rate affect accuracy?
Learning rate is a hyper-parameter th a t controls how much we are adjusting the weights of our network with respect the loss gradient. Furthermore, the learning rate affects how quickly our model can converge to a local minima (aka arrive at the best accuracy).
## What happens when learning rate is too big?
A learning rate that is too large can cause the model to converge too quickly to a suboptimal solution, whereas a learning rate that is too small can cause the process to get stuck. The learning rate is perhaps the most important hyperparameter. If you have time to tune only one hyperparameter, tune the learning rate.
## What is the default learning rate for Adam?
optimizers. schedules. LearningRateSchedule , or a callable that takes no arguments and returns the actual value to use, The learning rate. Defaults to 0.001.
## How does Adam Optimizer work?
Adam is a replacement optimization algorithm for stochastic gradient descent for training deep learning models. Adam combines the best properties of the AdaGrad and RMSProp algorithms to provide an optimization algorithm that can handle sparse gradients on noisy problems.
## Does Adam Optimizer change learning rate?
5 Answers. It depends. ADAM updates any parameter with an individual learning rate. This means that every parameter in the network has a specific learning rate associated.
## What is the default learning rate?
A traditional default value for the learning rate is 0.1 or 0.01, and this may represent a good starting point on your problem.
## Is RMSProp better than Adam?
So far, we’ve seen RMSProp and Momentum take contrasting approaches. While momentum accelerates our search in direction of minima, RMSProp impedes our search in direction of oscillations. Adam or Adaptive Moment Optimization algorithms combines the heuristics of both Momentum and RMSProp. | 1,808 | 8,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.906153 |
http://oeis.org/A053245 | 1,503,449,465,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886116921.70/warc/CC-MAIN-20170823000718-20170823020718-00254.warc.gz | 301,927,061 | 3,681 | This site is supported by donations to The OEIS Foundation.
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A053245 Numbers n such that both A053238(n) and A053238(n+1) = 1. 8
20, 49, 56, 79, 108, 132, 145, 170, 177, 230, 253, 277, 289, 307, 347, 382, 405, 412, 424, 437, 495, 548, 585, 592, 633, 645, 704, 734, 752, 764, 789, 802, 841, 854, 930, 943, 967, 974, 1005, 1012, 1053, 1066, 1130, 1154, 1179, 1186, 1216, 1223, 1264 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Pairs of consecutive 1's occur uncommonly often in A053238. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 MAPLE with(numtheory): f := [seq( `if`((sigma(i) > sigma(i+1)), i, print( )), i=1..5000)]; seq(`if`((f[i+2]-f[i+1]=1) and (f[i+1]-f[i]=1), i, print( )), i=1..1500); PROG (Haskell) a053245 n = a053245_list !! (n-1) a053245_list = f a053242_list where f (x:x':xs) | x' == x+1 = x : f xs | otherwise = f (x':xs) -- Reinhard Zumkeller, Oct 16 2011 CROSSREFS Cf. A000203, A053226, A053238, A053239, A053240, A053241, A053242, A053243, A053244. Sequence in context: A135286 A177725 A264444 * A115882 A277553 A260093 Adjacent sequences: A053242 A053243 A053244 * A053246 A053247 A053248 KEYWORD nonn AUTHOR Asher Auel (asher.auel(AT)reed.edu) Jan 10, 2000 STATUS approved
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http://thetubeguru.com/study_material/areas-parallelograms-triangles-exercise-9-4-mathematics-9th-class/ | 1,503,508,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00263.warc.gz | 401,424,444 | 24,102 | 0
Exercise 9.4
1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Sol. Here, AB = CD = EF. …(i)
Also in triangle BEC,
BC > BE [Side opposite to greater angle is larger]
Similarly, AD > AF [Reason same as above]
∴ BC + AD > BE + AF
⇒ BC + AD + AB + DC > BE + AF + AB + EF [From (i)]
⇒ Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.
2. In figure, D and E are two points on BC such that BD = DE = EC. Show that
ar(ABD) = ar(ADE) = ar(AEC).
Can you now answer the question that you have in the ‘Introduction’ of this chapter in NCERT, whether the field of Budhia has been actually divided into three parts of equal areas?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
Sol. In triangle ABE, BD = DE [Given]
∴ AD is median.
∴ ar(ABD) = ar(ADE) …(i)
[Median of a triangle divides the triangle into two parts of equal area]
Similarly, ar(ADE) = ar(AEC) …..(ii)
From (i) and (ii), we have
ar(ABD) = ar(ADE) = ar(AEC).
Hence, we answer the question in the ‘Introduction’ of this chapter in NCERT that Budhia has actually divided her field into three parts of equal areas.
3. In figure, ABCD, DCFE and ABFE are parallelograms.
Show that ar(ADE) = ar(BCF).
Sol. As ABCD, DCEF and ABFE are parallelograms.
∴ AD = BC, DE = CF and AE= BF ….(i)
∴ ΔADE ≅ ΔBCF. [SSS] [From (i)]
⇒ ar(ADE) = ar(BCF).
[If triangles are congruent, their areas are equal]
4. In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar(BPC) = ar(DPQ).
[Hint: Join AC.]
Sol. Construction: Join AC.
Proof: Triangles ADQ and ADC are on the same base AD and between the same parallels AD and BQ.
⇒ ar(APD) + ar(DPQ) = ar(ADP) + ar(APC)
⇒ ar(DPQ) = ar(APC)
Triangles APC and BPC are on the same base PC and between the same parallels PC and AB.
∴ ar(APC) = ar(BPC)
From (i) and (ii), we get
ar(DPQ) = ar(BPC).
5. In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar(BDE) = ${1 \over 4}$ar(ABC
(ii) ar(BDE) =${1 \over 2}$ ar(BAE)
(iii) ar(ABC) = 2 ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2ar(FED)
(vi) ar(FED) = ${1 \over 8}$ ar(AFC)
[Hint: Join EC and AD. Show that $BE\parallel AC$ and $DE\parallel AB$, etc.]
Sol. Let AB = BC = AC = x, then BD = BE = ED =${x \over 2}$
(i) ar(ABC) = ${{\sqrt 3 } \over 4}{x^2}$
and ar(BDE) = ${{\sqrt 3 } \over 4}{\left( {{x \over 2}} \right)^2} = {{\sqrt 3 } \over {16}}{x^2}$
$\Rightarrow ar(BDE) = {1 \over 4}ar(ABC)$
(ii) ∠ACB =60° and ∠DBE = 60° or ∠CBF = 60°
$\Rightarrow BE\parallel AC$ [ Alternate angles are equal.]
∴ ar(BAE) = ar(BCE) …(i)
[Triangles on the same base and between the same parallels are equal in areas.]
Since DE is median of ABEC.
$\Rightarrow ar(BED) = {1 \over 2}ar(BEC)$
[Median divides triangle into two parts of equal areas]
$\Rightarrow ar(BED) = {1 \over 2}ar(BAE)$ [ From (i)]
(iii) ar(BCE) = 2 ar(BDE)
[Median divides triangle into two parts of equal areas.]
$= 2 \times {1 \over 4}ar(ABC)$ [From result of part (i)]
⇒ 2 ar(BCE) =${1 \over 2}$ ar(ABC).
(iv) Proceeding as result of part (ii), we get $ED\parallel AB$.
[Triangles on the same base and between the same parallels are equal in area. ]
⇒ ar(ADE) — ar(EFD) = ar(BED) — ar(EFD)
⇒ ar(AFD) = ar(BEF).
(v) Draw EL ⊥ BD. Also AD ⊥ BC
Then $EL = \sqrt {{{\left( {{x \over 2}} \right)}^2} - {{\left( {{x \over 4}} \right)}^2}} = {{\sqrt 3 } \over 4}x$
Also $AD = {{\sqrt 3 } \over 2}x$
$ar(\Delta AFD) = {1 \over 2}.FD.AD$
$ar(\Delta FED) = {1 \over 2}.FD.EL$
∴ ${{ar(\Delta AFD)} \over {ar(\Delta FED)}} = {{AD} \over {EL}} = {{\sqrt 3 } \over 2}x \times {4 \over {\sqrt {3x} }} = 2$
$\Rightarrow ar(\Delta AFD) = 2ar(\Delta FED)$
$\Rightarrow ar(\Delta BFE) = 2ar(\Delta FED)$ [From result (iv) ]
(vi) From result (v),
ar(ΔBFE) = 2ar(ΔFED)
⇒ BF = 2FD
∴ FC = FD + DC = FD + BD = FD + 3FD = 4FD
∴ $ar(\Delta FED) = {1 \over 2}.FD.EL = {1 \over 2}.FD.{{\sqrt 3 } \over 4}x$ [ From result (v) ]
$= {{\sqrt 3 } \over 8}FD.x$ …….(i)
$ar(\Delta AFC) = {1 \over 2}.FD.AD = {1 \over 2}4FD.{{\sqrt 3 } \over 2}x$
= $\sqrt 3 .FD.x$
$= 8\left[ {{{\sqrt 3 } \over 8}FD.x} \right]$ [ From (i)]
= 8 or (Δ FED)
∴ ar( ΔFED) = ${1 \over 8}$ ar (ΔAFC).
6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) × ar(CPD) = ar(APD) × ar(BPC).
[Hint: From A and C, draw perpendiculars to BD.]
Sol. ar(APB) × ar(CPD)
$= \left( {{1 \over 2} \times BP \times AL} \right)\left( {{1 \over 2} \times PD \times CM} \right)$
$= {1 \over 4} \times BP \times DP \times AL \times CM$
And ar(APD) × ar(BPC)
$= \left( {{1 \over 2} \times PD \times AL} \right)\left( {{1 \over 2} \times BP \times CM} \right)$
$= {1 \over 4} \times BP \times DP \times AL \times CM$ ……(ii)
From (ii), we get
ar(APB) × ar(CPD) = ar(APD) × ar(BPC).
7. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) $ar(PRQ) = {1 \over 2}ar(ARC)$
(ii) ar(RQC) = ${3 \over 8}$ ar(ABC)
(iii) ar(PBQ) = ar(ARC)
Sol.
(i) Construction:
Join A and Q; P and C.
Proof: QP is median of triangle BQA.
∴ ar(BQP) = ar(PQA) …..(i)
Similarly, ar(QPR)
= ar(RQA) [QR is median of triangle PQA]
= ar(PQA) = 2 ar(QPR) [From above result ]
∴ 2 ar(QPR) = ar(BQP) [From (i)]
= ${1 \over 2}$ ar(BPC) [PQ is median]
= ${1 \over 2}$ ar(APC) [CP is median]
= ar (ARC) [RC is median]
⇒ ar(QPR) = ${1 \over 2}$ar(ARC).
(ii) ar(ARC) = ${1 \over 2}$ ar(CAP) [ CR is median of ΔACP]
${1 \over 2}\left\{ {{1 \over 2}ar(ABC)} \right\}$ [CP is median]
As RQ is median of ΔBRC
$ar(RQC) = {1 \over 2}ar(RBC) = {1 \over 2}\left\{ {ar(ABC) - ar(ARC)} \right\}$
$= {1 \over 2}\left\{ {ar(ABC) - {1 \over 4}ar(ABC)} \right\} = {3 \over 8}ar(ABC)$
(iii) ar(ARC) = ${1 \over 2}$ ar(CAP) …(ii) [CR is median]
And ar(CAP) = ar(CPB) …(iii) [CP is median]
From equations (ii) and (iii), we have
$ar(ARC) = {1 \over 2}ar(CPB)$
Further, ar(PBQ)=${1 \over 2}$ ar(PBC) [PQ is median]
∴ ar(ARC) = ar(PBQ).
8. In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y Show that:
(i) ΔMBC ≅ ΔABD
(ii) ar(BYXD) = 2ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
[Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.]
Sol. (i) Consider triangles MBC and ABD,
∠MBA = ∠CBD [Each 90°]
Adding ∠ABC to both sides, we have
∠MBA + ∠ABC = ∠CBD + ∠ABC
⇒ ∠MBC = ∠ABD
MB = BA [Sides of a square]
and BC = BD [Sides of a square]
∴ ΔMBC ≅ ΔABD. [SAS]
(ii) Triangle ABD and rectangle BDXY are on the same base BD and between the same parallels BD and AX.
∴ ar(ABD) = ${1 \over 2}$ ar(BDXY)
As ΔABD ≅ ΔMBC [Using result of part (i)]
⇒ ar(ΔABD) = ar(ΔMBC)
∴ ar(MBC) = ${1 \over 2}$ ar(BDXY)
⇒ ar(BDXY) = 2 ar(MBC).
(iii) Triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.
∴ 2 ar(MBC) = ar(MBAN)
∴ ar(MBAN) = ar(BDXY). [Using result from part (ii)]
(iv) ∠ACF = ∠BCE [90° each]
Adding ∠ACB to both sides, we get
∠ACF + ∠ACB = ∠BCE + ∠ACB
⇒ ∠BCF = ∠ ACE
Consider triangles BCF and ACE,
∠BCF = ∠ACE [Proved above]
AC = CF [Sides of a square]
and BC = CE [ Sides of a square]
∴ ΔFCB ≅ ΔACE. [ SAS]
(v) Since rectangle CYXE and ΔACE are on the same base CE and between the same parallels CE and AX.
∴ ar(CYXE) = 2ar(ACE)
Now, use the result from part (iv), we get
ar(ACE) = ar(FCB) [Congruent triangles have equal areas]
Hence, ar(CYXE) = 2ar(FCB)
(vi) As triangle FCB and square ACFG are on the same base CF and between the same parallels CF and BG,
∴ ar(ACFG) = 2ar(FCB)
Comparing this result and the result from part (v),
we get
ar(CYXE) ar(ACFG)
(vii) Adding the results from parts (iii) and (vi), we get
ar(BYXD) + ar(CYXE) = ar(ABMN) + ar(ACFG)
⇒ ar(BCED) = ar(ABMN) + ar(ACFG). | 3,474 | 10,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 46, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-34 | longest | en | 0.655094 |
https://brilliant.org/problems/inspired-by-satyajit-mohanty/ | 1,529,557,505,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00481.warc.gz | 580,986,342 | 11,375 | # Inspired by Satyajit Mohanty
Algebra Level 5
${ 3 }^{ x }+{ 3 }^{ x+1 }+\cdots+{ 3 }^{ x+31032001 }={ 27 }^{ x }+{ 27 }^{ x+1 }+\cdots+{ 27 }^{ x+31032001 }$
If the above equation is true for some integer $$x$$ and it can be expressed in the form of:-
$\large{\frac { \log _{ A }{ \left( \frac { B }{ { A }^{ C }+{ A }^{ D }+1 } \right) } }{ 2 }}$
where $$A,B,C$$ and $$D$$ are positive integers, find $$A+B+C+D$$.
× | 176 | 424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-26 | latest | en | 0.58631 |
https://webdocs.cs.ualberta.ca/~games/domineering/ | 1,537,503,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00230.warc.gz | 657,044,019 | 2,297 | # Obsequi's Domineering Page
(Horizontal to play, horizontal wins.)
Domineering was introduced by Göran Andersson around 1973.
Domineering is a two player, perfect information game. The game is usually played on an nxm grid, although it can be played on any size or shape of grid, in which the two players take turns placing 2x1 tiles on the grid, one vertically and one horizontally. The first player who is unable to play loses.
## Previous Work
The game of Domineering has previously been solved for a number of board sizes, including 9x9 domineering, by DOMI. DOMI is a program which was written by Breuker et al., to solve domineering on different sizes of boards. One of the main purposes of this program was to look at what affect different replacement schemes had on the affectiveness of transposition tables. Beside 9x9 domineering, DOMI also solved the game on a number of other board sizes, these results can be found here.
This information has then also been extended to larger board sizes by using the theoretical properties of this game, see here.
## Obsequi
Obsequi is a program with a purpose very similar to DOMI, to solve the game of domineering on various sizes of boards. This program started as a course project and from there progressed into the subject of my Master's thesis.
This program uses a greatly superior evaluation function to what DOMI used and as a result needs to examine far fewer board positions in order to determine who will win on a a particular board. This also enables Obsequi to solve larger sizes of boards than what DOMI was able to solve.
Obsequi builds off the work done by the creators of DOMI. Some of the ideas which we have used from DOMI are in the areas of move ordering and the use of a transposition table.
Besides the ideas which we have borrowed from DOMI, we have also used a number of other enhancements to help reduce the branching factor, reduce the search depth, and to increase the chance of transpositions in Obsequi's search. Some of these enhancements are:
• Improved cut off bounds.
• Pruning safe moves.
• Improved transposition table replacement scheme.
• Symmetrical move ordering.
## New Results
Obsequi has been able to solve a number of board positions which have never been solved before. The table below shows which boards these are along with the game theoretic value of the board and the number of nodes which it took to prove.
Board Size Result Nodes 4x19 H 314,148,901 4x21 H 3,390,074,758 6x14 H 1,864,870,370 8x10 H 4,125,516,739 10x10 1 3,541,685,253,370
## Other Information and Resources
Updated game theoretical values for various board sizes is here.
Here is a presentation I gave describing the research I did for my thesis.
Also, here is my thesis and the obsequi.tgz for Obsequi.
Email me at nathan_kent_bullock -at- yahoo -dot- ca. Last Updated: June 2002 | 677 | 2,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-39 | longest | en | 0.975215 |
https://math.stackexchange.com/questions/2274548/x-z-in-mathbb-c-z-le-1-is-pz-bar-z-px-y-in-mathbb-rx | 1,563,510,877,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00333.warc.gz | 466,011,373 | 35,049 | # $X:=\{z \in \mathbb C : |z|\le 1\}$ ; is $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ dense in $C(X , \mathbb C)$ under sup metric?
Let $X:=\{z \in \mathbb C : |z|\le 1\}$ , is it true that any element in $C(X , \mathbb C)$ can be uniformly approximated by polynomials , in $z, \bar z$ , with real co-efficients ? If we wanted uniform approximation by polynomials in $z, \bar z$ , with complex co-efficients then I know it would be true by Complex version of Stone-Weierstrass theorem . But I have no idea what happens if we want real-coefficients ( I can't see whether the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ is a subalgebra over $\mathbb C$ or not ) . Please help . Thanks in advance
• Unless I'm mistaken, the function $p(z, \bar{z}) = iz$ is a good example to have in mind. – Cameron Williams May 10 '17 at 11:39
• @CameronWilliams : what to do with that polynomial ? – user228168 May 10 '17 at 13:46
• Can you approximate that function? – Cameron Williams May 10 '17 at 14:09
To expand on the counterexample $p(z, \bar z)=iz$ given by Cameron Williams, note that when restricted to a subinterval of real line, such as $[-1,1]$, the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ reduces to $\mathbb{R}[x]$. Clearly, any function one can approximate by the elements of $\mathbb{R}[x]$ must be real-valued on $[-1,1]$, and this does not include $iz$. | 450 | 1,379 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-30 | latest | en | 0.792332 |
https://transpireonline.blog/2023/05/29/home-loan-closing-tricks/ | 1,696,135,924,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00348.warc.gz | 630,184,022 | 17,400 | # How to Close Home loan Smartly
When it comes to home loan prepayment and EMI increasing, it’s important to approach these strategies carefully and consider their implications. Let’s discuss these concepts with an example using Indian Rupees (INR):
Example: Suppose you have a home loan of INR 50 lakhs (5,000,000) for a period of 20 years at an interest rate of 8% per annum. Here are some strategies and considerations for prepayment and EMI increase:
1. Prepayment:
a. Lump Sum Prepayment: If you have surplus funds, consider making a lump sum prepayment towards your home loan. This reduces the principal amount, resulting in a lower outstanding balance and potential interest savings over the loan tenure. However, before making a prepayment, check if your lender allows it and if any charges or conditions apply.
b. Partial Prepayment vs. Full Prepayment: You can choose between partial prepayment (paying a portion of the outstanding principal) or full prepayment (repaying the entire outstanding loan amount). Partial prepayment allows you to retain some liquidity while reducing the interest burden. Full prepayment helps you become debt-free sooner, but it may impact your overall financial liquidity.
Prepayment based loan closing Example
To calculate the impact of prepayment on the closing of a home loan, you need to consider factors such as the loan amount, interest rate, tenure, and the amount of prepayment. Let’s consider an example using Indian Rupees (INR):
Example: You have a home loan of INR 50 lakhs (5,000,000) for a tenure of 20 years at an interest rate of 8% per annum. Let’s assume you decide to make a prepayment of INR 2 lakhs after paying EMIs for 10 years.
Step 1: Calculate the remaining number of EMIs. To determine the remaining number of EMIs, you need to consider the tenure and the number of EMIs already paid.
In this example, you have paid EMIs for 10 years, which is 10 * 12 = 120 EMIs.
Remaining EMIs = Total number of EMIs – EMIs already paid Remaining EMIs = 20 years * 12 – 120 Remaining EMIs = 240 – 120 Remaining EMIs = 120 EMIs
Step 2: Calculate the outstanding loan balance. To calculate the outstanding loan balance after the prepayment, you can subtract the prepayment amount from the remaining loan amount.
Outstanding Loan Balance = Principal loan amount – Prepayment amount Outstanding Loan Balance = 5,000,000 – 200,000 Outstanding Loan Balance = INR 4,800,000
Step 3: Calculate the reduced EMI required for the remaining tenure. To determine the reduced EMI required to close the loan within the remaining tenure, divide the outstanding loan balance by the remaining number of EMIs.
Reduced EMI = Outstanding Loan Balance / Remaining EMIs Reduced EMI = 4,800,000 / 120 Reduced EMI = INR 40,000
Therefore, to close the loan within the remaining 120 EMIs, the reduced EMI would be INR 40,000.
By making the prepayment of INR 2 lakhs, the outstanding loan balance is reduced, and you can continue paying the reduced EMI for the remaining tenure to close the loan.
It’s important to note that prepayment may involve additional charges or penalties as per the terms of your loan agreement. It’s advisable to consult with your lender to understand any applicable charges and assess the impact on the overall loan repayment before making a prepayment.
1. EMI Increase:
a. Opting for Step-up EMIs: Some lenders offer step-up EMIs, where the EMI amount gradually increases over time. This is beneficial if you expect your income to rise in the future. Starting with lower EMIs initially can ease your financial burden in the early years, and as your income grows, you can comfortably handle higher EMIs.
b. Shortening the Loan Tenure: Another approach is to increase your EMI amount to shorten the loan tenure. By paying higher EMIs, you can repay the loan faster, saving on interest costs. However, this strategy requires careful consideration of your monthly budget and financial capabilities.
It’s important to note that prepayment and EMI increase strategies may involve additional costs, such as prepayment penalties or processing fees. Therefore, it’s advisable to review your loan agreement and consult with your lender to understand the specific terms and conditions associated with these strategies.
Before implementing any strategy, consider your overall financial situation, future income projections, and other financial goals. It’s recommended to seek advice from a financial advisor who can provide personalised guidance based on your circumstances and help you make informed decisions.
EMI increasing based loan closing Example
To calculate the impact of increasing the Equated Monthly Installment (EMI) on the closing of a home loan, you need to consider factors such as the loan amount, interest rate, tenure, and the desired increased EMI. Let’s consider an example using Indian Rupees (INR):
Example: You have a home loan of INR 50 lakhs (5,000,000) for a tenure of 20 years at an interest rate of 8% per annum. Let’s assume your current EMI is INR 40,000.
Step 1: Calculate the remaining number of EMIs. To determine the remaining number of EMIs, you need to consider the tenure and the number of EMIs already paid.
In this example, let’s assume you have already paid 10 years’ worth of EMIs, which is 10 * 12 = 120 EMIs.
Remaining EMIs = Total number of EMIs – EMIs already paid Remaining EMIs = 20 years * 12 – 120 Remaining EMIs = 240 – 120 Remaining EMIs = 120 EMIs
Step 2: Calculate the total outstanding loan balance. To calculate the total outstanding loan balance, you can use the formula:
Outstanding Loan Balance = Principal loan amount * [(1 + r)^n – (1 + r)^p] / [(1 + r)^n – 1]
where, P = EMIs already paid n = Total number of EMIs r = Monthly interest rate
In this example: Principal loan amount = INR 50 lakhs (5,000,000) P = 120 EMIs n = 240 EMIs r = 8% per annum / 12 (monthly interest rate)
Plugging in the values:
r = 8% / 12 = 0.00667
Outstanding Loan Balance = 5,000,000 * [(1 + 0.00667)^240 – (1 + 0.00667)^120] / [(1 + 0.00667)^240 – 1] Outstanding Loan Balance ≈ INR 23,84,136
So, the total outstanding loan balance is approximately INR 23,84,136.
Step 3: Calculate the increased EMI required for loan closure. To determine the increased EMI required to close the loan, subtract the outstanding loan balance from the total loan amount and divide it by the remaining number of EMIs.
Increased EMI = (Principal loan amount – Outstanding Loan Balance) / Remaining EMIs Increased EMI = (5,000,000 – 23,84,136) / 120 Increased EMI ≈ INR 18,797
Therefore, to close the loan within the remaining 120 EMIs, you would need to increase the EMI to approximately INR 18,797.
It’s important to note that any increase in EMI may impact your monthly budget, and you should ensure that the increased EMI is manageable within your financial capabilities. Consulting with your lender and assessing the impact on the overall loan repayment is advisable before making any changes to your EMI. | 1,639 | 7,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.935865 |
https://www.airmilescalculator.com/distance/vns-to-ixb/ | 1,709,410,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00517.warc.gz | 631,724,078 | 32,079 | # How far is Siliguri from Varanasi?
The distance between Varanasi (Lal Bahadur Shastri Airport) and Siliguri (Bagdogra Airport) is 350 miles / 564 kilometers / 304 nautical miles.
The driving distance from Varanasi (VNS) to Siliguri (IXB) is 420 miles / 676 kilometers, and travel time by car is about 8 hours 19 minutes.
350
Miles
564
Kilometers
304
Nautical miles
1 h 9 min
## Distance from Varanasi to Siliguri
There are several ways to calculate the distance from Varanasi to Siliguri. Here are two standard methods:
Vincenty's formula (applied above)
• 350.382 miles
• 563.886 kilometers
• 304.474 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 349.876 miles
• 563.070 kilometers
• 304.034 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Varanasi to Siliguri?
The estimated flight time from Lal Bahadur Shastri Airport to Bagdogra Airport is 1 hour and 9 minutes.
## Flight carbon footprint between Lal Bahadur Shastri Airport (VNS) and Bagdogra Airport (IXB)
On average, flying from Varanasi to Siliguri generates about 77 kg of CO2 per passenger, and 77 kilograms equals 169 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Varanasi to Siliguri
See the map of the shortest flight path between Lal Bahadur Shastri Airport (VNS) and Bagdogra Airport (IXB). | 439 | 1,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.823033 |
https://www.thestudentroom.co.uk/showthread.php?t=2306486 | 1,521,833,695,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648431.63/warc/CC-MAIN-20180323180932-20180323200932-00365.warc.gz | 893,095,945 | 40,309 | x Turn on thread page Beta
You are Here: Home
Higher Physics 1996 Q3 watch
1. Yeah, can someone help me with Higher Physics 1996 Q3 Paper 1?
So confused.
2. Lets see it?
3. (Original post by String.)
Lets see it?
http://mrmackenzie.wikispaces.com/fi...96_HigherI.pdf
Not sure if this will work but there.
4. (Original post by EME Chris)
Yeah, can someone help me with Higher Physics 1996 Q3 Paper 1?
So confused.
This was posted from The Student Room's iPhone 4s app.
5. (Original post by Jack--)
This was posted from The Student Room's iPhone 4s app.
Oh, i thought you could only use d=vt for the horizontal. I was trying to use an equation of motion to find out the time for the vertical.
Thanks
6. http://mrmackenzie.wikispaces.com/fi...96_HigherI.pdf
You can use d=vt because the question states that the vertical component velocity is constant.
7. (Original post by Felix Felicis)
You can't use that equation if the question is something like "Find the time taken for the ball to hit the ground".
I suspect you need but it'd be helpful if you posted the question
Yeah sorry, i've posted the link a few times but it's not coming up. If you type in on Google "fizzics higher physics past papers 1992-2010" and then go to Higher Revision page it will be on there.
8. I should really pay more attention to the question.
Thanks.
9. (Original post by EME Chris)
Yeah sorry, i've posted the link a few times but it's not coming up. If you type in on Google "fizzics higher physics past papers 1992-2010" and then go to Higher Revision page it will be on there.
Ah, then you can use distance = speed x time or whatever letters you want to use - the vertical velocity is constant
10. all wondering what the question is it can be found here. http://mrmackenzie.wikispaces.com/fi...96_HigherI.pdf
12. (Original post by EME Chris)
The magnitude of the frictional force is equal to the the magnitude of the horizontal component of the pulling force (force along dotted line). From this we can draw a triangle to represent the components of the pulling force, P:
From this
so
The magnitude of the frictional force is equal to the the magnitude of the horizontal component of the pulling force (force along dotted line). From this we can draw a triangle to represent the components of the pulling force, P:
From this
so
Appreciate the help.
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http://www.britannica.com/topic/triangle-mathematics | 1,461,982,331,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111581.11/warc/CC-MAIN-20160428161511-00032-ip-10-239-7-51.ec2.internal.warc.gz | 395,821,742 | 12,042 | # Triangle
Mathematics
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• ## equivalence to the area of a circle
mathematics: Archimedes
Archimedes’ result bears on the problem of circle quadrature in the light of another theorem he proved: that the area of a circle equals the area of a triangle whose height equals the radius of the circle and whose base equals its circumference. He established analogous results for the sphere showing that the volume of a sphere is equal to that of a cone whose height equals the radius of the...
• ## Euclidean geometry
Euclidean geometry: Congruence of triangles
Two triangles are said to be congruent if one can be exactly superimposed on the other by a rigid motion, and the congruence theorems specify the conditions under which this can occur. The first theorem illustrated in the diagram is the side-angle-side (SAS) theorem: If two sides and the included angle of one triangle are equal to two sides and the included angle of...
triangle inequality
in Euclidean geometry, theorem that the sum of any two sides of a triangle is greater than or equal to the third side; in symbols, a + b ≥ c. In essence, the theorem states that the shortest distance between two points is a straight line.
• ## law of tangents
tangent (of a curve)
The trigonometric law of tangents is a relationship between two sides of a plane triangle and the tangents of the sum and difference of the angles opposite those sides. In any plane triangle ABC, if a, b, and c are the sides opposite angles A, B, and C, respectively, then
• ## significance of number three
number symbolism: 3
...Egyptian sun god: Khepri (rising), Re (midday), and Atum (setting). In Christianity there is the Trinity of God the Father, God the Son, and God the Holy Spirit. Plato saw 3 as being symbolic of the triangle, the simplest spatial shape, and considered the world to have been built from triangles. In German folklore a paper triangle with a cross in each corner and a prayer in the middle was...
• ## trigonometry
trigonometry: Plane trigonometry
In many applications of trigonometry the essential problem is the solution of triangles. If enough sides and angles are known, the remaining sides and angles as well as the area can be calculated, and the triangle is then said to be solved. Triangles can be solved by the law of sines and the law of cosines (see the table). To secure symmetry in the writing of these...
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Draw the triangle with vertices D(-2,1), U(4,9), and E(10,1) On a Graph.
2. Find DU, UE, AND DE. (Distance)
3. Classify Triangle DUE by its sides.
4. Write the standard form of the equation of each line: DU, UE, and DE.Also, find the slode and then the y-intercept. Start with y=mx+b and move into standard form.
5. Draw the altitude from U to DE.Name it UM
6. What is the equation of the line UM?
7. Find the Midpoint of UE. Write the formula.Name the midpoint O on your diagram.
8. Write the equation of DO in standard form
9. Find the midpoint of DU.
10. Write the standard form of the equation of line EN.
11. Find the Measure of angle EUM using trigonomotry. Round to the nearest tenth.
12. If you rotated triangle DUE 270 degrees CLOCKWISE, what would the new cordinates be?
13. Write the standard form of the equation of the perpindicular bisector of DU.
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# How do you write 78215?
Updated: 9/23/2023
Wiki User
8y ago
Seventy-eight thousand, two hundred fifteen.
Wiki User
8y ago
Anonymous
Lvl 1
3y ago
its not fifteen one tens 5 ones
Anonymous
Lvl 1
3y ago
.
Anonymous
Lvl 1
3y ago
.
Anonymous
Lvl 1
3y ago
.
Kayley Stewart
Lvl 2
2y ago
The answer is seventy-eight thousand, two hundred fifteen.
Anonymous
Lvl 1
3y ago
Seventy-eight thousand, two hundred fifteen
Earn +20 pts
Q: How do you write 78215?
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"Write" is a verb and therefore is the same applied to singular and plural, for example:"I write the music." and "They write the music." | 686 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-10 | latest | en | 0.893674 |
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# cs402 current final term papers
Views: 883
### Replies to This Discussion
Kol share karoo ga bcz kol mera paper ha
Kindly share your papers mera wenesday ko paper hai
plz
kise k pas is ka solution he tu plz share kry
Cs402 theory of automata
Date 16/08/2017 time 11:00 am
Q1. Define by kleen star plus oprator Sigma + = {aa}
Q2. Write three condition of PDA to conversion form.
Q3. There are two TG1 and TG2 transition graph. You may draw TG1+TG2.
Q4. Make a CFG define over sigma = {a,b}
Q5. There are TM (turning machine) graph define RE.
Q6. Define the union of two TG's
Q7. Define the description form of RE sigma = { a,b,?}
Q8. Draw a TG of define over sigma =
(aab+abb+bab+bbb)(a+b)*
is question ky answer share kr dy plz
objective part thory se mooaz file se bhe thy and handouts read ky hn tu mcq bht easy thy
subjective part me ek cfg dia tha us ko pda me convert krna tha=5
ek dfa tha us ko bhe pda me convert krna tha =5
regular expression dia tha fa draw kna tha'=5
cfg bnane the=5
ek diagram de the moore machine ke output find krne the input strings de hoe the =3
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Step by step video & image solution for State the rms value of an alternating current? Write the relation between the rms value and peak value of an alternating current that varies with time by Physics experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## The r.m.s. value of an alternating current is
Aless than zero
Bequal to its peak value
Cless than its peak value
Dgreater than its peak value
• Question 1 - Select One
## The ratio of peak value and r.m.s value of an alternating current is
A1
B12
C2
D1/2
• Question 1 - Select One
## For an alternating current I=I0cosωt, what is the rms value and peak value of current-
AI0,I02
BI02,I0
CI0,I02
D2I0,I02
• Question 1 - Select One
## The peak value of Alternating current is 6amp, then r.m.s. value of current will be
A3A
B33A
C32A
D23A
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