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# project1 - .theater xlab="Rating"...
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Sheet1 Page 1 plot(movie.genre\$Action\$rating,movie.genre\$Action\$gross, xlab="Rating", ylab="gross", pch=1 ) points(movie.genre\$Comedy\$rating,movie.genre\$Comedy\$gross, pch=2, col=2 ) points(movie.genre\$Drama\$rating,movie.genre\$Drama\$gross, pch=3, col=3 ) points(movie.genre\$Horror\$rating,movie.genre\$Horror\$gross, pch=4, col=4 ) points(movie.genre\$Kids\$rating,movie.genre\$Kids\$gross, pch=5, col=5 ) legend(locator(1),legend=unique(Genre),pch=1:5, col=1:5) plot(movie.genre\$Action\$rating,movie.genre\$Action\$opening, xlab="Rating", ylab="opening", pch=1 ) points(movie.genre\$Comedy\$rating,movie.genre\$Comedy\$opening, pch=2, col=2 ) points(movie.genre\$Drama\$rating,movie.genre\$Drama\$opening, pch=3, col=3 ) points(movie.genre\$Horror\$rating,movie.genre\$Horror\$opening, pch=4, col=4 ) points(movie.genre\$Kids\$rating,movie.genre\$Kids\$opening, pch=5, col=5 ) legend(locator(1),legend=unique(Genre),pch=1:5, col=1:5) plot(movie.genre\$Action\$rating,movie.genre\$Action\$Open.
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Unformatted text preview: .theater, xlab="Rating", ylab="Open. .theater", pch=1 ) points(movie.genre\$Comedy\$rating,movie.genre\$Comedy\$Open. .theater, pch=2, col=2 ) points(movie.genre\$Drama\$rating,movie.genre\$Drama\$Open. .theater, pch=3, col=3 ) points(movie.genre\$Horror\$rating,movie.genre\$Horror\$Open. .theater, pch=4, col=4 ) points(movie.genre\$Kids\$rating,movie.genre\$Kids\$Open. .theater, pch=5, col=5 ) legend(locator(1),legend=unique(Genre),pch=1:5, col=1:5) plot(movie.genre\$Action\$rating,movie.genre\$Action\$percent, xlab="Rating", ylab="Percent", pch=1 ) points(movie.genre\$Comedy\$rating,movie.genre\$Comedy\$percent, pch=2, col=2 ) points(movie.genre\$Drama\$rating,movie.genre\$Drama\$percent, pch=3, col=3 ) points(movie.genre\$Horror\$rating,movie.genre\$Horror\$percent, pch=4, col=4 ) points(movie.genre\$Kids\$rating,movie.genre\$Kids\$percent, pch=5, col=5 ) legend(locator(1),legend=unique(Genre),pch=1:5, col=1:5)...
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Ask a homework question - tutors are online | 732 | 2,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.406807 |
http://lessonplanspage.com/mathgraphadd1-htm/ | 1,516,780,067,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893530.89/warc/CC-MAIN-20180124070239-20180124090239-00465.warc.gz | 191,014,526 | 28,207 | # Here’s a math lesson on graphing and addition
Subject:
Math
1
``` First Grade
by Vickie
Materials:
```
Fish (various colors) I use magnetic tape and put them up on a chalk board. I also have access to an Ellison machine to die-cut all the fish the same size. I usually have about 5 colors for first grade.
Activity 1: I arrange various #’s of fish on the board.
Then I give the children word problems as described below.
Example… ___red fish + ___blue fish = ______fish in all
They count, record and calculate their answers in groups. I also let them use small chalkboards or white boards to work their problems. They also like to write out their problem and draw the fish to represent the #.
Activity 2:
You can also have the children group the fish by color and make a graph. I let small groups of children place the fish on the white board which is also magnetic. They form the fish in a horizontal or vertical graph and record their totals. You can use any shape. I chose fish as an extension of my ocean unit.
Activity 3:
I also let the children use the fish to tell stories. “One day 2 blue fish went swimming. 3 red fish decided to join them.” The children tell the story, make the problem and conclude their story. “Now there are 5 fish swimming in the sea” They also write the problem for the other students to see.
``` 2+3=5 2
+3
5
```
E-Mail Vicki ! | 333 | 1,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-05 | longest | en | 0.970245 |
https://www.unitsconverters.com/en/Btu(It)Persecondperfoot2-To-Calorie(It)Perminutepercentimeter2/Unittounit-4389-4369 | 1,657,213,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00304.warc.gz | 1,079,727,809 | 38,470 | Formula Used
1 Watt per Meter² = 8.80547457023708E-05 Btu (IT) per Second per Foot²
1 Watt per Meter² = 0.001433075379765 Calorie (IT) per Minute per Centimeter²
1 Btu (IT) per Second per Foot² = 16.2747587878138 Calorie (IT) per Minute per Centimeter²
## Btu (IT) per Second per Foot²s to Calorie (IT) per Minute per Centimeter²s Conversion
BTU/s*ft² stands for btu (it) per second per foot²s and kcal(IT)/min*cm² stands for calorie (it) per minute per centimeter²s. The formula used in btu (it) per second per foot²s to calorie (it) per minute per centimeter²s conversion is 1 Btu (IT) per Second per Foot² = 16.2747587878138 Calorie (IT) per Minute per Centimeter². In other words, 1 btu (it) per second per foot² is 17 times bigger than a calorie (it) per minute per centimeter². To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter²
How to convert btu (it) per second per foot² to calorie (it) per minute per centimeter²? In the heat flux density measurement, first choose btu (it) per second per foot² from the left dropdown and calorie (it) per minute per centimeter² from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from calorie (it) per minute per centimeter² to btu (it) per second per foot²? You can check our calorie (it) per minute per centimeter² to btu (it) per second per foot² converter.
How to convert Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter²?
The formula to convert Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter² is 1 Btu (IT) per Second per Foot² = 16.2747587878138 Calorie (IT) per Minute per Centimeter². Btu (IT) per Second per Foot² is 16.2748 times Bigger than Calorie (IT) per Minute per Centimeter². Enter the value of Btu (IT) per Second per Foot² and hit Convert to get value in Calorie (IT) per Minute per Centimeter². Check our Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter² converter. Need a reverse calculation from Calorie (IT) per Minute per Centimeter² to Btu (IT) per Second per Foot²? You can check our Calorie (IT) per Minute per Centimeter² to Btu (IT) per Second per Foot² Converter.
How many Watt per Meter² is 1 Btu (IT) per Second per Foot²?
1 Btu (IT) per Second per Foot² is equal to 16.2748 Watt per Meter². 1 Btu (IT) per Second per Foot² is 16.2748 times Bigger than 1 Watt per Meter².
How many Kilowatt per Meter² is 1 Btu (IT) per Second per Foot²?
1 Btu (IT) per Second per Foot² is equal to 16.2748 Kilowatt per Meter². 1 Btu (IT) per Second per Foot² is 16.2748 times Bigger than 1 Kilowatt per Meter².
How many Watt per Centimeter² is 1 Btu (IT) per Second per Foot²?
1 Btu (IT) per Second per Foot² is equal to 16.2748 Watt per Centimeter². 1 Btu (IT) per Second per Foot² is 16.2748 times Bigger than 1 Watt per Centimeter².
How many Watt per Inch² is 1 Btu (IT) per Second per Foot²?
1 Btu (IT) per Second per Foot² is equal to 16.2748 Watt per Inch². 1 Btu (IT) per Second per Foot² is 16.2748 times Bigger than 1 Watt per Inch².
## Btu (IT) per Second per Foot²s to Calorie (IT) per Minute per Centimeter²s Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like heat flux density finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like BTU/s*ft² to kcal(IT)/min*cm² through multiplicative conversion factors. When you are converting heat flux density, you need a Btu (IT) per Second per Foot²s to Calorie (IT) per Minute per Centimeter²s converter that is elaborate and still easy to use. Converting Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter² is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Btu (IT) per Second per Foot² to Calorie (IT) per Minute per Centimeter² conversion along with a table representing the entire conversion.
Let Others Know | 1,251 | 4,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.422904 |
http://www.traditionaloven.com/metal/precious-metals/gold/convert-tonne-metric-t-of-gold-to-carat-ct-of-gold.html | 1,518,904,059,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891807825.38/warc/CC-MAIN-20180217204928-20180217224928-00727.warc.gz | 576,894,980 | 14,060 | Gold tonne (Metric) to carats (Metric / mass) of gold converter
# gold conversion
## Amount: tonne (Metric) (t) of gold mass Equals: 5,000,000.00 carats (Metric / mass) (ct) in gold mass
Calculate carats (Metric / mass) of gold per tonne (Metric) unit. The gold converter.
TOGGLE : from carats (Metric / mass) into Metric tonnes in the other way around.
### Enter a New tonne (Metric) Amount of gold to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## gold from tonne (Metric) to carat (Metric / mass) Conversion Results :
Amount : tonne (Metric) (t) of gold
Equals: 5,000,000.00 carats (Metric / mass) (ct) in gold
Fractions: 5,000,000.00 carats (Metric / mass) (ct) in gold
CONVERT : between other gold measuring units - complete list.
## Solid Pure 24k Gold Amounts
This calculator tool is based on the pure 24K gold, with Density: 19.282 g/cm3 calculated (24 karat gold grade, finest quality raw and solid gold volume; from native gold, the type we invest -in commodity markets, by trading in forex platform and in commodity future trading. Both the troy and the avoirdupois ounce units are listed under the gold metal main menu. I advice learning from a commodity trading school first. Then buy and sell.) Gold can be found listed either in table among noble metals or with precious metals.
Is it possible to manage numerous calculations for how heavy are other gold volumes all on one page? Yes, all in one Au multiunit calculator makes it possible managing just that.
Convert gold measuring units between tonne (Metric) (t) and carats (Metric / mass) (ct) of gold but in the other direction from carats (Metric / mass) into Metric tonnes.
conversion result for gold: From Symbol Equals Result To Symbol 1 tonne (Metric) t = 5,000,000.00 carats (Metric / mass) ct
# Precious metals: gold conversion
This online gold from t into ct (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this gold calculator are ...
With the above mentioned units calculating service it provides, this gold converter proved to be useful also as a teaching tool:
1. in practicing Metric tonnes and carats (Metric / mass) ( t vs. ct ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with gold's density values including other physical properties this metal has.
International unit symbols for these two gold measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for tonne (Metric) is: t
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for carat (Metric / mass) is: ct
### One tonne (Metric) of gold converted to carat (Metric / mass) equals to 5,000,000.00 ct
How many carats (Metric / mass) of gold are in 1 tonne (Metric)? The answer is: The change of 1 t ( tonne (Metric) ) unit of a gold amount equals = to 5,000,000.00 ct ( carat (Metric / mass) ) as the equivalent measure for the same gold type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of gold can have a crucial/pivotal role in investments. If there is an exact known measure in t - Metric tonnes for gold amount, the rule is that the tonne (Metric) number gets converted into ct - carats (Metric / mass) or any other unit of gold absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many carats (Metric / mass) ( ct ) of gold are contained in a tonne (Metric) ( 1 t ). Or, how much in carats (Metric / mass) of gold is in 1 tonne (Metric)? To link to this gold - tonne (Metric) to carats (Metric / mass) online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: gold from tonne (Metric) (t) to carats (Metric / mass) (ct) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,183 | 4,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.837128 |
http://www.mathhelp.com/how_to/ratio/introduction_to_ratios.php | 1,406,138,644,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997882928.29/warc/CC-MAIN-20140722025802-00243-ip-10-33-131-23.ec2.internal.warc.gz | 904,666,609 | 5,848 | # PRE-ALGEBRALesson 90 of 171
RATIO
Introduction to Ratios
Students learn that a ratio is a comparison of two numbers or quantities. For example, if there are 2 shaded parts of a picture and 7 unshaded parts, the ratio of shaded parts to unshaded parts is 2 to 7. Note that the ratio 2 to 7 can also be written as 2:7 or 2/7. And since a ratio more...
### Community College College Math
*Also referred to as Elementary Algebra or Beginning Algebra. | 124 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.922068 |
https://www.studypug.com/ca/chem11/percentage-yield-and-atom-economy | 1,534,551,900,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00126.warc.gz | 978,866,456 | 35,958 | # Percentage yield and atom economy
### Percentage yield and atom economy
#### Lessons
In this lesson, we will learn:
• To understand the definitions of atom economy and percentage yield and the difference between them.
• To calculate atom economy and percentage yield from example chemical reactions.
• To explain the importance of atom economy as a chemist when planning chemical reactions and processes.
Notes:
• The atom economy of a reaction is the percentage of atomic mass of useful products in a reaction. It is calculated by: $Atom\;Economy \;$(%)$\;= \frac{Atomic\;mass\;of\;useful\;products}{Total\;atomic\;mass\;of\;products} * 100$
• A high atom economy means that most of what the chemical process makes is useful! In the same way, a low atom economy tells you that a reaction is mostly producing unwanted waste products.
The atom economy is used to show an efficient reaction that makes lots of valuable chemicals and not much waste product that will require disposal.
• A quick tip – reactions with a single product have a 100% atom economy because the only chemical being produced is the desired product.
• The percentage yield of a reaction is the mass of products formed as a percentage of how much could have been formed given the mass of reactants used.
• A low percentage yield means that not much of the reactants you used has become products. A high percentage yield therefore means that a lot of the reactant chemicals you used successfully reacted to make the products.
• Know the difference between the two! Atom economy is about how wasteful the reaction is, and yield is about how much reactant was successfully converted to products!
• Introduction
Why do chemists do chemical reactions?
a)
Things that happen during a chemical reaction.
b)
The definitions of atom economy and percentage yield.
c)
The difference between atom economy and percentage yield.
d)
Environmental and economic chemistry.
• 1.
Find the atom economy and percentage yield of chemical reactions.
Water can be produced by reaction of hydrogen and oxygen gas according to this equation:
2 H$_2 +$O$_2$→2 H$_2$O
a)
What is the atom economy of this reaction?
b)
If the reaction was performed and 150 g was the theoretical yield of water and the actual yield was only 92 g, what is the percentage yield?
• 2.
Find the atom economy and percentage yield of chemical reactions.
The reaction to make iron metal by reduction uses carbon. CO$_2$ is an unwanted side-product. The reaction is as follows:
2 Fe$_2$O$_3 +$3 C→3 CO$_2 +$4 Fe
a)
What is the atom economy of this reaction?
b)
280 g of Fe$_2$O$_3$ was used in this reaction and 104 g of Fe metal was collected. Calculate the theoretical yield and then the percentage yield of this reaction.
c)
This reaction was done again and a percentage yield of 90% was achieved this time. If 250 g of Fe was collected this time, how much Fe$_2$O$_3$ was used up?
• 3.
Compare the viability of two reactions by finding the atom economy and percentage yield.
Magnesium reacts with hydrochloric acid as shown in the following equation:
Mg$+$2 HCl→MgCl$_2 +$H$_2$
a)
If this reaction was performed to make hydrogen gas, what would be the atom economy?
b)
55 g of Mg metal was used in this reaction with HCl in excess. Only 4.21 g of H$_2$ gas was produced from this reaction. What is the percentage yield of this reaction?
c)
Hydrogen gas could also be made by electrolysis of water, in the following reaction:
2 H$_2$O→2 H$_2 +$O$_2$
Find the atom economy of this reaction. Is it a better or worse way to make hydrogen than reacting magnesium with hydrochloric acid? | 835 | 3,625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.941053 |
https://businessyield.com/marketing/total-addressable-market/ | 1,713,732,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00755.warc.gz | 124,024,414 | 41,442 | # TOTAL ADDRESSABLE MARKET: Definition, Calculation, Examples & All You Need
The Total Addressable Market (TAM) of a market is a very important factor for every business because it plays such an important role in determining the growth of the business. TAM, which is also known as the total available market, refers to the overall revenue opportunity that is available to a product or service if that product or service achieves 100% market share. Another name for the TAM is the total available market. The total available market is often referred to as the total prospective market. Let’s have a look at the Total Addressable Market Calculation and an applicable example.
The TAM will illustrate the theoretical total market that could exist for a product under the assumption that access to the product is not restricted in any way, and that the supplier has an unlimited supply of operational resources. In other words, the TAM will assume that access to the product is not restricted in any way.
## What is Total Addressable Market
A product or service’s total addressable market, sometimes known as its total available market (TAM), is a term that is used to characterize the potential for that product or service to generate money.
To be straightforward TAM is a term for annual expected sales within an observable market, TAM is another term for annual projected sales within an observable market.
TAM is a useful tool for prioritizing business possibilities since it provides a quick metric of the underlying potential of a particular opportunity.
Total Addressable Market, or TAM, refers to the total revenue opportunity that is open to a product or service if that product or service gets a share of the market equal to one hundred percent. It is helpful in assessing how much effort and money a person or corporation should invest into a new business line and helps determine whether or not they should do so.
When it comes to contributing to a company’s overall success, the value of TAM simply cannot be overemphasized. In efforts to build a strong company, we have gained knowledge about the many benefits that TAM can bring to a company from seasoned business professionals
Total Addressable Market (TAM) sizing exercises have increased in popularity in recent years as a method for evaluating the potential of a business. On the other hand, TAM is commonly miscalculated and misinterpreted because of its complexity.
When computing TAM, there are three basic calculation methods, as well as one secondary way, which are as follows:
• Top-down takes a macro view of assessing factors right at the very top of an economy.
• Bottom-up looks at a subset of a localized situation and then extrapolates the results to the wider market opportunity.
• Value theory focuses on the positive externalities derived from an offering versus incumbent options.
• The secondary collection method uses verified third-party research (also known as external research).
### #1. Top Down Approach
The top-down total addressable market calculation method is based on the idea that you should begin at the very top of a macro data set and work your way down to identify a market subset. Starting with a population, it makes sense to use geographic, economic, and demographic hypotheses to get rid of subsets that don’t add anything.
The benefit of this approach is that macroeconomic data may be found with reliable and accessible statistics. UN, OECD, The World Bank, and CIA World Factbook are a few examples of sources.
### #2. Bottom-Up Method
The objective of this method is to identify intermediate steps that can be extended to the entire population. Because it is built on a tested data point that can be enlarged to reveal the entire TAM population, it is preferred for being more accurate. Secondary research (news reports, corporate filings) or primary gathering techniques (such as a poll in a local market) are both options.
Since so many broad assumptions are extracted from a small subset, the bottom-up technique may produce an erroneous TAM. This is especially true for a worldwide TAM calculation because of how dramatically different countries’ population densities, economies, and consumer preferences are.
### #3. Value Theory
Top-down and bottom-up strategies assume a new offering fits into established paradigms. The value hypothesis, total addressable market calculation may be best for items or services that can evolve a market or bring value to different consumer groups.
Value theory tries to figure out how much a product’s price can show about its value. A corporation calculates its value and why it should be priced accordingly. When offering new items or cross-selling to existing clients, TAM is calculated using value theory.
Estimating Uber’s addressable value using value theory. Users can drive themselves, take public transportation, or use Uber taxis. Since people are willing to forgo all these options and choose an Uber cab, the firm can measure the value these users obtain from Uber taxis and determine how to price them.
### #4. Secondary Method
External Research
Referencing professional data that has already been collected is a simple and speedy method that can be used to identify a TAM. Although private research studies are pricey, the headline data contained inside them is sometimes made available to the general public through press releases. The benefits of using these numbers are more practical. They are easy to get and, according to the law of the division of labor, were put together by a skilled professional who can be trusted. These are some of the advantages of using these figures.
The disadvantage of this method is that it is a black box in terms of how the number was calculated. If an investor asks you to defend the rationale behind the figures, the only defense will be, “Well, it was published by X, so it’s valid.”
First, multiply your average sales price by your number of current customers. This will yield your annual contract value. Then, multiply your ACV by the total number of customers. This will yield your total addressable market.
TAM= Potential Market Size x Competitive Position
Potential Market Size= Number of potential customers in the market
Competitive Position = Your company’s share of the total number of possible customers.
And, of course, examples make things clearer and more understandable. Let’s take a quick look at this total addressable market example below.
The number of people who buy a bicycle in a single year represents your potential market size, also known as your total addressable market. Let’s say 80 million people in the United States do so.
If a company holds 50% of that share, then that company has the potential to sell this product or service to 40 million customers. If there are two other companies, and each of them holds 20% of that share, then each of those companies has the potential to sell to 20 million customers.
### How is TAM market size calculated?
First, multiply your average sales price by the number of current customers. This will yield your annual contract value. Then, multiply your ACV by the total number of customers. This will yield your total addressable market.
### What Is the Difference Between Tam and Sam?
TAM is a product’s entire market demand. It’s the maximum revenue a business can make in a given market.
Total addressable market helps companies estimate a market’s growth potential.
Whereas
Your business model’s limitations (such as expertise or geography) may prevent you from serving your whole potential market.
Serviceable addressable market helps firms assess the market share they can achieve to set goals.
### How are TAM SAM and SOM calculated?
By conducting a bottom-up examination of the industry, the total addressable market can be determined most accurately. In a bottom-up analysis, the total number of consumers in a market is calculated by dividing it by the average yearly income of each consumer in this market.
## Calculation of the Serviceable Addressable Market (SAM)
Calculating with the SAM formula
## Calculating the Serviceable Obtainable Market (SOM)
Calculating with the SOM formula
Subtract your revenue from the prior year from the serviceable addressable market for your sector. Your market share from the prior year is represented by this %. After that, multiply your market share from the previous year by the serviceable addressable market for your sector from this year.
Always remember that these numbers are primarily estimations.
## Conclusion
Total addressable market is beneficial since it helps to improve one’s awareness of the progress that a firm is making, which, in turn, raises the possibility that one may identify issue areas and make improvements in those areas. However, it is essential to be aware that capturing one hundred percent of a TAM is virtually impossible. This is something that should be kept in mind at all times.
Investors place a high level of importance on the total addressable market since it is a good indicator of the company’s potential for growth. And it should go without saying that prospective investors are on board with the idea of recognizing the TAM of a firm prior to making an investment in that company.
I believe the calculation and example given here for the total addressable market were quite helpful.
## What are the two approaches for calculating your potential TAM?
The two common approaches are
• Top-Down Market Size.
• Bottom-Up Market Size.
## How Is Tam Market Size Calculated?
First, multiply your average sales price by the number of current customers. This will yield your annual contract value. Then, multiply your ACV by the total number of customers. This will yield your total addressable market.
No, it isn’t.
\$100M.
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## CUSTOMER JOURNEY MAP: Meaning, How to Create it, Funnel & Plan
Table of Contents Hide What is a Customer Journey Map?The Benefits of Customer Journey MappingCustomer Journey Map Stages#1.… | 2,005 | 10,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.951428 |
https://calculationcalculator.com/kanal-to-cent | 1,721,321,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00183.warc.gz | 134,397,139 | 22,116 | # Kanal to Cent Conversion
## 1 Kanal is equal to how many Cent?
### 12.5 Cent
##### On This Page:
Kanal and Cent both are the Land measurement unit. Compare values between unit Kanal with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page.
Kanal to Cent conversion allows you to convert value between Kanal to Cent easily. Just enter the Kanal value into the input box, the system will automatically calculate Cent value. 1 Kanal in Cent? In mathematical terms, 1 Kanal = 12.5 Cent.
To conversion value between Kanal to Cent, just multiply the value by the conversion ratio. One Kanal is equal to 12.5 Cent, so use this simple formula to convert -
The value in Cent is equal to the value of Kanal multiplied by 12.5.
Cent = Kanal * 12.5;
For calculation, here's how to convert 10 Kanal to Cent using the formula above -
10 Kanal = (10 * 12.5) = 125 Cent
Kanal Cent Conversion
0.1 1.25 0.1 Kanal = 1.25 Cent
0.2 2.5 0.2 Kanal = 2.5 Cent
0.3 3.75 0.3 Kanal = 3.75 Cent
0.4 5 0.4 Kanal = 5 Cent
0.5 6.25 0.5 Kanal = 6.25 Cent
0.6 7.5 0.6 Kanal = 7.5 Cent
0.7 8.75 0.7 Kanal = 8.75 Cent
0.8 10 0.8 Kanal = 10 Cent
0.9 11.25 0.9 Kanal = 11.25 Cent
1 12.5 1 Kanal = 12.5 Cent
2 25 2 Kanal = 25 Cent
3 37.5 3 Kanal = 37.5 Cent
4 50 4 Kanal = 50 Cent
5 62.5 5 Kanal = 62.5 Cent
6 75 6 Kanal = 75 Cent
7 87.5 7 Kanal = 87.5 Cent
8 100 8 Kanal = 100 Cent
9 112.5 9 Kanal = 112.5 Cent
10 125 10 Kanal = 125 Cent
11 137.5 11 Kanal = 137.5 Cent
12 150 12 Kanal = 150 Cent
13 162.5 13 Kanal = 162.5 Cent
14 175 14 Kanal = 175 Cent
15 187.5 15 Kanal = 187.5 Cent
16 200 16 Kanal = 200 Cent
17 212.5 17 Kanal = 212.5 Cent
18 225 18 Kanal = 225 Cent
19 237.5 19 Kanal = 237.5 Cent
20 250 20 Kanal = 250 Cent
21 262.5 21 Kanal = 262.5 Cent
22 275 22 Kanal = 275 Cent
23 287.5 23 Kanal = 287.5 Cent
24 300 24 Kanal = 300 Cent
25 312.5 25 Kanal = 312.5 Cent
26 325 26 Kanal = 325 Cent
27 337.5 27 Kanal = 337.5 Cent
28 350 28 Kanal = 350 Cent
29 362.5 29 Kanal = 362.5 Cent
30 375 30 Kanal = 375 Cent
31 387.5 31 Kanal = 387.5 Cent
32 400 32 Kanal = 400 Cent
33 412.5 33 Kanal = 412.5 Cent
34 425 34 Kanal = 425 Cent
35 437.5 35 Kanal = 437.5 Cent
36 450 36 Kanal = 450 Cent
37 462.5 37 Kanal = 462.5 Cent
38 475 38 Kanal = 475 Cent
39 487.5 39 Kanal = 487.5 Cent
40 500 40 Kanal = 500 Cent
41 512.5 41 Kanal = 512.5 Cent
42 525 42 Kanal = 525 Cent
43 537.5 43 Kanal = 537.5 Cent
44 550 44 Kanal = 550 Cent
45 562.5 45 Kanal = 562.5 Cent
46 575 46 Kanal = 575 Cent
47 587.5 47 Kanal = 587.5 Cent
48 600 48 Kanal = 600 Cent
49 612.5 49 Kanal = 612.5 Cent
50 625 50 Kanal = 625 Cent
51 637.5 51 Kanal = 637.5 Cent
52 650 52 Kanal = 650 Cent
53 662.5 53 Kanal = 662.5 Cent
54 675 54 Kanal = 675 Cent
55 687.5 55 Kanal = 687.5 Cent
56 700 56 Kanal = 700 Cent
57 712.5 57 Kanal = 712.5 Cent
58 725 58 Kanal = 725 Cent
59 737.5 59 Kanal = 737.5 Cent
60 750 60 Kanal = 750 Cent
61 762.5 61 Kanal = 762.5 Cent
62 775 62 Kanal = 775 Cent
63 787.5 63 Kanal = 787.5 Cent
64 800 64 Kanal = 800 Cent
65 812.5 65 Kanal = 812.5 Cent
66 825 66 Kanal = 825 Cent
67 837.5 67 Kanal = 837.5 Cent
68 850 68 Kanal = 850 Cent
69 862.5 69 Kanal = 862.5 Cent
70 875 70 Kanal = 875 Cent
71 887.5 71 Kanal = 887.5 Cent
72 900 72 Kanal = 900 Cent
73 912.5 73 Kanal = 912.5 Cent
74 925 74 Kanal = 925 Cent
75 937.5 75 Kanal = 937.5 Cent
76 950 76 Kanal = 950 Cent
77 962.5 77 Kanal = 962.5 Cent
78 975 78 Kanal = 975 Cent
79 987.5 79 Kanal = 987.5 Cent
80 1000 80 Kanal = 1000 Cent
81 1012.5 81 Kanal = 1012.5 Cent
82 1025 82 Kanal = 1025 Cent
83 1037.5 83 Kanal = 1037.5 Cent
84 1050 84 Kanal = 1050 Cent
85 1062.5 85 Kanal = 1062.5 Cent
86 1075 86 Kanal = 1075 Cent
87 1087.5 87 Kanal = 1087.5 Cent
88 1100 88 Kanal = 1100 Cent
89 1112.5 89 Kanal = 1112.5 Cent
90 1125 90 Kanal = 1125 Cent
91 1137.5 91 Kanal = 1137.5 Cent
92 1150 92 Kanal = 1150 Cent
93 1162.5 93 Kanal = 1162.5 Cent
94 1175 94 Kanal = 1175 Cent
95 1187.5 95 Kanal = 1187.5 Cent
96 1200 96 Kanal = 1200 Cent
97 1212.5 97 Kanal = 1212.5 Cent
98 1225 98 Kanal = 1225 Cent
99 1237.5 99 Kanal = 1237.5 Cent
100 1250 100 Kanal = 1250 Cent
Kanal is land measurement units commonly used in Northern India states like Uttarakhand, Himachal Pradesh and Jammu and Kashmir. 1 Kanal is equal to 5445 Square Feet (sq ft).
A Cent is a unit of land area measurement which is still used in some parts of South Indian states. 1 Cent is equal to 1/100 Acre. It is still used in many news reports and in the real estate business.
The value in Cent is equal to the value of Kanal multiplied by 12.5.
Cent = Kanal * 12.5;
1 Kanal is equal to 12.5 Cent.
1 Kanal = 12.5 Cent.
On this page, we tried to solve all your queries. Here are some of your query those are resolved
• 1 kanal = cent
• kanal into cent
• kanals to cents
• convert kanal to cent
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→ | 2,081 | 4,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.712992 |
https://www.stockpricetrends.com/company/AVNT | 1,723,180,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00157.warc.gz | 774,840,587 | 40,769 | # Avient Corp (AVNT)
Avient Corporation provides specialty polymer materials, services, and solutions in the United States, Canada, Mexico, Europe, South America, and Asia. The company is headquartered in Avon Lake, Ohio.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 11.51% per annum.
• AVNT price at the close of August 7, 2024 was \$43.43 and was higher than the top border of the primary price channel by \$3.30 (8.22%). This indicates a possible reversal in the primary trend direction.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 53.90% per annum.
• AVNT price at the close of August 7, 2024 was inside the secondary price channel.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of AVNT price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### February 19, 2021 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### February 23, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 11.51% per annum.
• AVNT price at the close of August 7, 2024 was \$43.43 and was higher than the top border of the primary price channel by \$3.30 (8.22%). This indicates a possible reversal in the primary trend direction.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### September 22, 2023 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### August 1, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 53.90% per annum.
• AVNT price at the close of August 7, 2024 was inside the secondary price channel. | 1,080 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.832798 |
https://freetutsdownload.net/physics-100-mastering-kinematics-free-download/ | 1,642,771,922,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00634.warc.gz | 303,733,744 | 44,744 | # Physics 100 – Mastering Kinematics
Note: We have purchased this course/tutorial from Course Site and we’re sharing the download link with you for absolutely FREE. So you can learn & be your own master if you can’t afford to buy this course. But if you have money we strongly suggest you to buy [page_title] course/tutorial. So, the course’s author can help you if you can’t understand something or if you want to learn something spectacular.
## What you’ll learn
• They will learn how to solve any one and two dimensional kinematics problem
• Feel comfortable using kinematics equations and solving graphical problems
• Develop a strategy to solve kinematic problems
• Learn how to solve relative motion problems such as river crossing problems
## Requirements
• Basic Algebra and some basic calculus
## Description
This course covers all of one and two dimensional kinematics and sets the foundation for other topics. The class is perfect for high school (honors and AP students), college and university students taking an introductory physics class. Students will learn to analyze the motion of an object using equations and graphs. I will teach you how to approach problems and help you develop your own problem solving strategies to become a better problem solver.
1) Introduction
• Scientific Notation
• Converting Units
• Problem solving using dimensional analysis
• Prefixes
• Example problems for each topic.
2) One-Dimensional (1D) Kinematics
• Understanding the difference between displacement and distance traveled
• Average velocity vs instantaneous velocity
• Average acceleration vs instantaneous acceleration
• Motion with constant acceleration: The BIG 5 kinematic equations
• Solving problems using kinematic equations
• Using graphs to solve problems
• Free Fall
• Understanding gravity
• Interpreting graphs (position vs time, velocity vs time and acceleration vs time)
• Calculating velocity from position vs time graphs
• Calculating acceleration from velocity vs time graphs
• Calculating displacement from velocity graphs
• Calculating velocity from acceleration graphs.
• +30 fully explained problems
3) Two Dimensional (2D) Kinematics
• Learn how to apply kinematic the equations to analyze 2-dimensional motion
• Analyze the motion of horizontally launched objects
• Able to analyze the motion of a projectile launched at an arbitrary angle
• Calculating the range of projectile, maximum height, and time of flight
• 10 fully explained problems and solutions
4) Relative Motion in 1D and 2D
• Definition of relative velocity
• Learn how to find the relative velocities of objects relative to different objects
• 1D Relative motion problems: objects moving on airport walkways
• 2D Relative motion problems: problems dealing with river crossing.
If at any point you don’t understand something in my videos please feel free to reach out. I’m always willing to help someone learn. Physics Ninja always has your back!
If you have a topic that is missing from this course just email me and i’ll add it to the curriculum.
Happy Learning
Dr. E., Physics Ninja and Expert Physics and Math Teacher.
## Who this course is for:
• First Year Physics and Engineering Students and AP Physics students
• Anyone who wants to learn how to solve kinematic problems
• Students preparing for finals or MCAT exam.
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https://educationexpert.net/mathematics/10691.html | 1,642,883,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00225.warc.gz | 271,916,748 | 6,974 | 23 January, 07:33
# What is the discriminant of the equation x^2+11x-10=0
0
1. 23 January, 08:32
0
The discriminant is b^2-4ac
ax^2+bx+c=0
x^2+11x-10=0
11^2-4 (1) (-10)
1. Simplify 11^2 to 121
121-4 (1) (-10)
2. Simplify 4*1*-10 to - 40
121 - (-40)
3. Simplify brackets
121+40
4. Simplify
161 | 143 | 305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.625323 |
https://hh-energiesysteme.cz/08-28/9619.html | 1,643,100,520,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304798.1/warc/CC-MAIN-20220125070039-20220125100039-00016.warc.gz | 354,070,081 | 9,478 | # Concrete Mortar Mix Formula
• ### Dry Volume Formula Wet Volume Of Concrete And Mortar
Dry volume is greater than 20 of wet volume = 100/100 20/100 = 1 20 As engineers select an additional 5 to 10 for wastages and weak workmanship so that the quantities of cement and sand are not reduced the mortar ratio is taken as follow = 1 20
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• ### How to Mix Mortar with Pictures wikiHow
Use the right amount of water A bag of mortar should be mixed with about three gallons of clean water to achieve the right consistency The amount of water used can vary drastically depending upon the weather how wet the sand is and the variety of mix you re using so read the instructions carefully before adding water Ambient conditions temperature and humidity will affect the mix
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• ### Concrete Mix Design Calculations
Basic Concrete Mix Design Materials Pounds of material S G Abs Volume 667 3 15 X 62 4 Cement 667 3 15 3 39 Total Cementious 667 Miller Stone 1590 2 6 9 80 Evert Sand 1242 2 65 7 51 Water 300 1 481 1590 2 60 X 62 4 1242 2 65 X 62 4 4 81 Air 5 5 1 485 Total 3799 27 00 w cm 0 45 Unit Wt 140 72 Basic Concrete Mix Design Materials Pounds of
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• ### Concrete Mix Ratios Cement Sand Gravel
Typically used as the compound for joining masonry stone or ceramic units together mortar is made by combining cement lime and sand Mortar typically has a higher water to cement ratio when compared with concrete which allows greater workability and is
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• ### Mortar Mix Ratio Proportioning for Masonry Construction
🕑 Reading time 1 minuteMortar mix ratio i e proportioning of cement sand in mortar provides consistency in the performance and appearance of masonry construction Proper proportioning of mortar ingredients helps in having the following advantages Uniformity of strength Uniform workability Uniform color Uniformity of proportions and yields Mainly the proportioning of the cement and sand
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• ### From M to S Types of Mortar and Mortar Mix Ratios
Making your own type S mortar is fairly straight forward Simply combine the following ingredients 2 parts cement 1 part lime and 8 to 9 parts sand This mortar mix ratio is very similar to type O mortar so be sure to carefully measure your ingredients when making either type Type M The last of the four most common mortar types is type M
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• ### How to make lightweight cement and concrete for Stone
For interior applications the lightweight concrete formula is fine The medium weight concrete formula is much better and will hold up many more years in exterior applications A medium mix was used for the stone on the castle in the photos below
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• ### Proper Concrete Mix Procedures Using Portland Cement for
Basic mix formula for thin stone veneer and concrete tile 3 the batch should mix in the mortar mixer for about ten minutes total At six to eight minutes you can still have some un dissolved color particles or dry aggregate Now when you are ready to make your tile or stone fill the can with concrete mix dump it into your mold and
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• ### Cement Mortar Estimation of Cement Sand Water in
Cement Mortar Cement Mortar is one of the most common and cheapest binding materials used in construction industry Cement mortar is basically a mixture of cement sand water is used in various aspects of civil engineering works such as masonry brickwork plastering flooring etc There are two types dry mortar and wet mortar
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• ### Sand Mix/Topping Bedding Mix Concrete Mortar
Sand Mix Topping Bedding Mix is a pre blended mixture of sand and cementitious materials designed for topping bedding applications A high strength multi purpose product designed for a variety of projects from crack repair to thick mortar beds Sand Mix is often referred to as a high strength mortar in many how to manuals but can also be used in contractor grade concrete repair
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• ### What is the formula for mortar mix FindAnyAnswer
How to Calculate the Amount of Mortar Mix Needed Find the area of your brick structure by multiplying length and width Multiply the square footage by seven to estimate the number of bricks you use in your project Divide the total number of bricks by 30 to estimate how many 60 pound bags of mortar mix you need How do you make mortar from scratch Mixing Procedure Put 2/3 to 3/4 of the water into the mixer
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• ### concrete plaster mortar mixes for builders
The amount of water added to a mix must be enough to make the mix workable and plastic 2 Common cement complying with SANS 50197 may be used for concrete plaster or mortar 3 Stone for concrete should be 19 mm or 26 mm size 4 If you use a wheelbarrow for measuring it should be a builder s wheelbarrow which has a capacity of 65 litres
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• ### Mix ratio for mortar for bricks concrete slab floor
Mix ratio for concrete floor Generally mix ratio for concrete floor is 1 2 4 cement sand aggregate this mix ratio for concrete floor is made of 1 part cement 2 parts sand 4 parts coarse aggregate This is normal standard ideal commonly used and best mix ratio for concrete floor
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• ### Refractory concrete mixing recipe Heat resistant
Refractory concrete mixture ingredients recipe for mixing by hand and concrete in building pizza ovens How to make heat resistant concrete type for buildig wood fired pizza ovens and similar applications where heat and heating is present Heat differences in materials Temperature differences in material heat expansion and contraction shrinkage of edges with cooling downwards
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• ### How to Make Fireproof Mortar Hunker
However there is a formula for mixing mortar that will not only resist fire but it will also resist heat damage as well This mixture of mortar is easy to mix up and is ideal for use around fireplaces and other areas where there is a risk of fire or extreme heat Follow the intended uses and thickness instructions for the mortar for best results
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• ### What Is the Formula for Concrete Mix Reference
The formula for concrete mix is one part cement two parts sand and three parts gravel or crushed stone If hand mixing it s inadvisable to exceed a water to cement ratio of 0 55 which translates to 55 pounds of water for every 100 pounds of cement When mixing concrete it s necessary to ensure that there is an adequate amount of aggregate
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• ### How to Mix Portland Mortar 5 Steps with Pictures wikiHow
2 Mix the dry ingredients Using a shovel smaller bucket or scoop put three parts sand and one part cement into a mixing tub or trough wheelbarrow cement mixer or a 5 gallon 19 L bucket Add any dry additives according to the directions on the package and stir the powdered mixture thoroughly
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• ### How to Choose the Right Mortar Mix N O S or M
Type N mortar mix has a medium compressive strength and it is composed of 1 part Portland cement 1 part lime and 6 parts sand It is considered to be a general purpose mix useful for above grade exterior and interior load bearing installations It is also the preferred mortar mix
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• ### Mixing Mortar QUIKRETE Cement and Concrete Products
Add the dry mix into the mixer and allow the mortar to mix for about a minute then add the remaining water as necessary Step 3 Continue to mix for 3 5 minutes until a uniform workable consistency is achieved Step 4 Let the mortar sit undisturbed for about 3 5 minutes to allow the fine aggregate in the mix to fully saturate
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• ### Conventional grade C20 C25 and C30 concrete mix ratio
The basic requirements for the design of concrete mix ratio are 1 meet the strength grade of concrete design 2 meet the concrete peaceability 3 meet the durability of the use of concrete 4 meet the above conditions to save cement and reduce the cost of concrete The strength of concrete is divided into twelve grades such as C7 5 C10
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• ### How to mix cement to make mortar or concrete Marshalls
How to mix cement to make a mortar or concrete mix Cement mixing is a great DIY skill to master and can be applied to a huge range of outdoor building projects around your home With the correct tools materials safety equipment and a bit of elbow grease you can make your own mortar or concrete mixready to use for your next job
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• ### Mortar Mix Ratio Proportioning for Masonry Construction
Mortar Cements mix Cement Lime Mortar Mix This mortar mix is produced by blending lime sand mortar with ordinary Portland cement This mix will gain a well uniform physical property These mortar mix have high excellent workability high capability of water retention increased setting time and provision of additional strength Masonry Cement Mortar This mortar mix was developed to reduce the mortar
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• ### Concrete Mixing RatiosHow To Make Concrete Cement
3000 psi concrete mix ratio To produce a 3000 psi cubic yard of concrete 27 cubic feet the concrete mixture ratio is 517 pounds of cement or 234kg 1560 pounds of sand or 707kg 1600 pounds of stone or 725kg 3234 gallons of water or 132L This mixing ratio will give you a concrete mix that is strong durable and good for most
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• ### The Sand to Mortar Ratio to Lay Concrete Blocks Hunker
The basic mixture for most concrete block projects is a 4 to 1 or 5 to 1 mixture In essence four parts or five of sand are added to one part of cement and then water is added to that until you achieve the texture you want for your particular block project
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• ### Cement mortar ratio for plastering plastering it s
2 concrete quantity calculation for staircase and its formula Plastering is thin layer of cement mortar adhesive material put over the brick wall for protection from environment smooth surface good finishing good looking increasing the strength of brick wall We know that external part of brick wall is more prone to sever climatic condition its needs more protection from outer side
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• ### Calculate Quantity of Cement mortar in Brickwork With
Concrete Mix Design as per IS CODE easy way by excel sheet In this article you will get full detail information about the mix design of concrete also one excel sheet which include all concrete mix design procedure as per is code concrete mix design as per is 10262 so that easily calculate your mix
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• ### Water cement ratio formula Table Calculation for Mortar
Water cement ratio formulaTable Calculation for Mortar Example In this post you will get water cement ratio formula water cement ratio table and calculation for Mortar The water cement ratio is the ratio between the weight of water to the weight of cement used in the concrete mix or the amount of water that we used in the cement concrete
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• ### Concrete Mix Formulas
Low Strength Concrete10 MPa Domestic Foundations Light Footings Bases for Prefabricated Buildings 33 liter 50 Kg Cement 160 liter 105 liter 9 5mm to 13 2mm Add enough water to produce a workable mix 175 liter
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• ### Conventional grade C20 C25 and C30 concrete mix ratio
The basic requirements for the design of concrete mix ratio are 1 meet the strength grade of concrete design 2 meet the concrete peaceability 3 meet the durability of the use of concrete 4 meet the above conditions to save cement and reduce the cost of concrete The strength of concrete is divided into twelve grades such as C7 5 C10
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• ### Water cement ratio formula Table Calculation for Mortar
Water Cement ratio of different grade of ConcreteTable Normally we used the water cement ratio falls under 0 4 to 0 6 per IS Code 10262 2009 for nominal mix i e M7 5 M10 M15 M20 M25 Here M denotes Mix and Number denotes characteristics compressive strength of concrete of 150 mm cube after 28 days
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• ### Mixing Mortar QUIKRETE Cement and Concrete Products
Measure the recommended water amount for the number of bags to be added to the mixer and
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• ### Cement ChemistryThe Concrete Portal
Figure 6 Polished surface of a C 3 S mortar showing hydrating grains of C 3 S the darker shades are C S H deposits while the lighter shades especially as rims around aggregates are deposits of CH Figure 7 Low magnification image showing the polished surface of PC concrete the porosity of the paste and the extent of unhydrated cement are
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• ### concrete plaster mortar mixes for builders
The amount of water added to a mix must be enough to make the mix workable and plastic 2 Common cement complying with SANS 50197 may be used for concrete plaster or mortar 3 Stone for concrete should be 19 mm or 26 mm size 4 If you use a wheelbarrow for measuring it should be a builder s wheelbarrow which has a capacity of 65 litres
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• ### Water cement ratio formula Table Calculation for Mortar
Water Cement ratio of different grade of ConcreteTable Normally we used the water cement ratio falls under 0 4 to 0 6 per IS Code 10262 2009 for nominal mix i e M7 5 M10 M15 M20 M25 Here M denotes Mix and Number denotes characteristics compressive strength of concrete of 150 mm cube after 28 days
Get Price | 2,831 | 13,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.843238 |
https://www.organised-sound.com/how-to-find-total-resistance-in-series-and-parallel-circuit/ | 1,660,394,614,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00355.warc.gz | 818,921,073 | 18,211 | # How To Find Total Resistance In Series And Parallel Circuit
By | July 27, 2022
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Series circuits parallel networks and learn physics for kids resistors in calculate total resistance seriesparallel r1 100 ohm r2 250 inst equivalent of two basic electronics audio part 2 l4 electrical electronic solving study guide inspirit combination calculator the circuit shown comprising simple formula derivation r4 form a or 3 8 simplified formulas dc engineering | 506 | 2,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | longest | en | 0.691185 |
https://community.fabric.microsoft.com/t5/Desktop/How-to-I-use-a-Matrix-Total-based-on-a-Dynamic-Measure-in/m-p/2228837 | 1,723,485,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00301.warc.gz | 137,185,125 | 55,938 | cancel
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Frequent Visitor
## How to I use a Matrix Total, based on a Dynamic Measure, in another measure.
In the Pick below I'm trying to find a way to reference the "Grand Total" amount to be used in another measure. In this case, Specifically, the " ITD Paid Claims Loss Ratio".
Might look something like this (Net Reserves * ITD Paid Claims Loss Ratio TOTAL= BF NET) The problem is I need this to be dynamic based on the slicers selected.
I tried to make a column, but it's not dynamic, (See Second Pick)
I feel like I'm missing something simple. Let me know if you have any questions.
Thank you!
1 ACCEPTED SOLUTION
Super User
Hi,
Hope this helps.
Regards,
Ashish Mathur
http://www.ashishmathur.com
6 REPLIES 6
Super User
Hi,
I am not clear with your expected result. Plese clearly show the expected result (may be in an Excel file with your formula so that your logic can be understood).
Regards,
Ashish Mathur
http://www.ashishmathur.com
Frequent Visitor
ya, sorry this was a bit hard to explain. I was trying to break this into two large chunks.
First - Use the Loss Ratio Total, or Average shown in the Matrix to Calculate the BF Net Column. (See Pick, could not attach Excel)
Second - Figure out how to only Average the Rows that are 100% earned.
Super User
Regards,
Ashish Mathur
http://www.ashishmathur.com
Frequent Visitor
Here is the PBI File. PBI File Example
Super User
Hi,
Hope this helps.
Regards,
Ashish Mathur
http://www.ashishmathur.com
Frequent Visitor
Column F is the same as Column E but uses the Average Loss Ratio of the business that is 100% Earned. (only looking at rows 2-10). Excel Example
I'll start working on making a version of the PBI file that I can share.
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Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data. | 510 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.910751 |
https://www.scribd.com/document/58515634/CCSPowerPicsMath | 1,521,752,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00553.warc.gz | 880,529,569 | 35,054 | # PowerPic Reference Sheets for k-2 Math Common Core Standards
Based on the following PowerPics by Chris Biffle and Jay Vanderfin Kindergarten PowerPics First Grade Math PowerPics Second Grade Math PowerPics Third Grade Math PowerPics
counting 1 to 5
Common Core Standards K.CC.1. Count to 100 by ones and by tens. Question: What is counting 1 to 5? Answer: Counting 1 to 5 is counting 1, 2, 3, 4, 5. Gesture: Hold up one finger for each number until you are holding up five fingers. Teaching Suggestion: Using various groups of objects and numbers, teach students how to count from 1 to 5. Play Yes/No Way! with questions like the following: 1. Is 1, 2, 3 counting 1 to 5? (Vary the sequence, occasionally using the correct number set.) 2. Is this the counting 1 to 5 picture? (Point at various Power Pix; include the counting 1 to 5 PowerPix occasionally.) 3. Is this the counting 1 to 5 gesture? (Make various gestures; include the counting 1 to 5 gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Counting 1 to 5 is counting 1, 2, 3, 4, 6. 2. Counting 1 to 5 is counting 1, 2, 3, 4, 5. 3. Counting 1 to 5 is counting 2, 3, 4, 5. Critical Thinking: Play Compare/Contrast with counting 1 to 5 and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for counting 1 to 5 and other Power Pix.
counting 1 to 10
Common Core Standards K.CC.1. Count to 100 by ones and by tens. Question: What is counting 1 to 10? Answer: Counting 1 to 10 is counting 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Gesture: Hold up one finger for each number until you are holding up ten fingers. Teaching Suggestion: Using various groups of objects and numbers, teach students how to count from 1 to 10. Play Yes/No Way! with questions like the following: 1. Is 1, 2, 3, 4, 5, 6, 7 counting 1 to 10? (Vary the sequence, occasionally using the correct number set.) 2. Is this the counting 1 to 10 picture? (Point at various Power Pix; include the counting 1 to 10 Power Pix occasionally.) 3. Is this the counting 1 to 10 gesture? (Make various gestures; include the counting 1 to 10 gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Counting 1 to 10 is counting 1, 2, 3, 4, 10. 2. Counting 1 to 10 is counting 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 3. Counting 1 to 10 is counting 2, 3, 4, 5, 7, 8, 9, 10 Critical Thinking: Play Compare/Contrast with counting 1 to 10 and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for counting 1 to 10 and other Power Pix.
pointer counting
Common Core Standards K.CC.1. Count to 100 by ones and by tens. K.CC.2. Count forward beginning from a given number within the known sequence (instead of having to begin at 1). K.CC.3. Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects). K.CC.5. Count to answer ―how many?‖ questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1–20, count out that many objects. Question: What is pointer counting? Answer: Pointer counting is pointing and counting. Gesture: Use the index finger of one hand to point at and count three fingers on the other hand (saying, ―one, two, three‖ as you point). Teaching Suggestion: Use various props and/or groups of objects to teach students the concept of pointer counting, i.e. pointing at objects and counting them. Next, have students repeat after you, as you point at objects in the classroom and count aloud. Play Yes/No Way! with questions like the following: 1. Is this pointer counting? (Point at and count various objects in the classroom. Occasionally, to demonstrate incorrect pointer counting, say the alphabet or nonsense words in place of numbers.) 2. Is this the pointer counting picture? (Point at various Power Pix; include the pointer counting Power Pix occasionally.) 3. Is this the pointer counting gesture? (Make various gestures; include the pointer counting gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Pointer counting is pointing and counting. 2. We start counting with one. 3. Pointer counting is counting 4, 8, 9. Critical Thinking: Play Compare/Contrast with the pointer counting and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the pointer counting and other Power Pix.
equal numbers (with 3 = 3) Common Core Standards K.CC.6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.1 K.CC.7. Compare two numbers between 1 and 10 presented as written numerals. Question: What are equal numbers? Answer: Equal numbers are the same numbers. Gesture: Hold up three fingers on each hand, showing equal numbers. Teaching Suggestion: While your students are involved in individual or group projects, write equal number pairs on the board, using numbers from 1 to 10, for example 1 = 1, 3 = 3, etc. In some examples, make one number taller than the other or a different color, so that students understand equal numbers refers to number value not height or color. Using the number pairs on the board and various groups of objects, teach students the concept of equality. Play Yes/No Way! with questions like the following: 1. Are these equal numbers? (Point at various objects, letters, numbers, including the equal numbers on the board.) 2. Is this the equal numbers picture? (Point at various Power Pix; include the equal numbers Power Pix occasionally.) 3. Is this the equal numbers gesture? (Make various gestures; include the equal numbers gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Four and four are equal numbers. 2. Four and six are equal numbers. 3. Three and three are equal numbers. 4. Equal numbers are the same number. Critical Thinking: Play Compare/Contrast with equal numbers and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for equal numbers and other Power Pix.
equals sign
Common Core Standards K.CC.6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.1 K.CC.7. Compare two numbers between 1 and 10 presented as written numerals.
Question: What is the equals sign? Answer: The equals sign means ―the same as.‖ Gesture: Make an equals sign by holding your forearms parallel to the ground in front of your body. Teaching Suggestion: While your students are involved in individual or group projects, write groups of numbers on the board, some of which include the correct use of the equals sign. Explain the concept of equality and the equals sign. Play Yes/No Way! with questions like the following: 1. Is this an equals sign? (Point at various symbols/numbers on the board, including correct examples of the equals sign.) 2. Is this the equals sign picture? (Point at various Power Pix; include the equals sign Power Pix occasionally.) 3. Is this the equals sign gesture? (Make various gestures; include the equals sign gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. The equals sign means ―one more than.‖ 2. The equals sign means ―the same as.‖ 3. Two plus two equals four. Critical Thinking: Play Compare/Contrast with the equals sign and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the equals sign and other Power Pix.
less than
Common Core Standards K.CC.6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.1 K.CC.7. Compare two numbers between 1 and 10 presented as written numerals. Question: What is less than? Answer: Less than is when one group is less than another group. [Note: Though, in general, we try to avoid ―kid talk‖, it is sometimes useful to describe abstract concepts, like less than, in concrete terms, ―one group is less than another‖, as we have here. This teaching preference in early education for the concrete over the abstract dates back, at least, to Piaget.] Gesture: Hold up one finger on one hand and five fingers on the other hand. Say the answer as follows, ―Less than is when one group (waggle one finger) is less than another group (waggle five fingers).‖ Teaching Suggestion: (Teach more than and less than together.) While your students are involved in individual or group projects, make simple drawings of objects on the board, i.e. two circles and two squares, three houses and two cars, one tree and one dog. Use the drawings and various props and/or groups of objects, to teach students the concept of less than. Use pointer counting to compare the number of objects in two groups. Then, Play Yes/No Way! with questions like the following: 1. Is this group less than this group? (Point at various groups of objects and drawings on the board.) 2. Is this the less than picture? (Point at various Power Pix; include the less than Power Pix occasionally.) 3. Is this the less than gesture? (Make various gestures; include the less than gesture occasionally.) 4. Is two apples less than ten apples? (Insert other numbers and examples.) 5. Am I holding up less than five fingers? (Hold up your fingers in various combinations.) Quick Test: Play Cutie with statements like the following: 1. Two is less than one. 2. Less than is when one group is less than another group. 3. You have less than 30 toes. Critical Thinking: Play Compare/Contrast with the less than and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the less than and other Power Pix.
more than
Common Core Standards K.CC.6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.1 K.CC.7. Compare two numbers between 1 and 10 presented as written numerals. Question: What is more than? Answer: More than is when one group is more than another group. [Note: Though, in general, we try to avoid ―kid talk‖, it is sometimes useful to describe abstract concepts, like more than, in concrete terms, ―one group is more than another‖, as we have here. This teaching preference in early education for the concrete over the abstract dates back, at least, to Piaget.] Gesture: Hold up three fingers on one hand and two fingers on the other hand. Say the answer as follows, ―More than is when one group (waggle three fingers) is more than another group (waggle two fingers).‖ Teaching Suggestion: (Teach more than and less than together.) While your students are involved in individual or group projects, make simple drawings of objects on the board, i.e. two circles and two squares, three houses and two cars, one tree and one dog. Use the drawings and various props and/or groups of objects, to teach students the concept of more than. Use pointer counting to compare the number of objects in two groups. Then, Play Yes/No Way! with questions like the following: 1. Is this group more than this group? (Point at various groups of objects and drawings on the board.) 2. Is this the more than picture? (Point at various Power Pix; include the more than Power Pix occasionally.) 3. Is this the more than gesture? (Make various gestures; include the more than gesture occasionally.) 4. Is ten desks more than four desks? (Insert other numbers and examples.) 5. Am I holding up more than two fingers? (Hold up your fingers in various combinations.) Quick Test: Play Cutie with statements like the following: 1. Two is more than one. 2. More than is when one group is more than another group. 3. You have more fingers than eyes. Critical Thinking: Play Compare/Contrast with more than and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for more than and other Power Pix.
Subtraction
minus sign Common Core Standards K.OA Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. Question: What is a minus sign? Answer: A minus sign shows subtraction. Gesture: Hold one forearm parallel to the ground, making a minus sign. Teaching Suggestion: While your students are involved in individual or group projects, write a list of subtraction and addition problems on the board. Use these problems to show students examples of how a minus sign appears in simple subtraction. Play Yes/No Way! with questions like the following: 1. Is this a minus sign? (Point at numbers, minus signs and plus signs on the board.) 2. Is this the minus sign picture? (Point at various Power Pix; include the minus sign Power Pix occasionally.) 3. Is this the minus sign gesture? (Make various gestures; include the minus sign occasionally.) 4. Is the minus sign used in addition? Quick Test: Play Cutie with statements like the following: 1. A minus sign shows one number is added to another. 2. A minus sign is used in subtraction problems. 3. A minus sign is the same as a number. Critical Thinking: Play Compare/Contrast with minus sign and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for minus sign and other Power Pix.
Counting by 2s Common Core Standards 2.OA.3. Determine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends. Question: What is counting by 2s? Answer: Counting by 2s is 2, 4, 6, 8, 10 and so on. Gesture: Hold up 2, 4, 6, 8, and 10 fingers Teaching Suggestion: While your students are involved in individual or group tasks, write lists of numbers on the board. Some lists are counting by 1s, 2s, 5s, 10s and other lists contain random numbers. Explain the concept of counting by 2s. Play Yes/No Way! with one or more questions like the following: 1. Is this counting by 2s? (Point at various lists.) 2. Is 2, 4, 6, 8 counting by 2s? 3. Is this the counting by 2s gesture? (Make various gestures.) 4. Is 1, 2, 3, 4, counting by 2s? 5. Is this the counting by 2s Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Counting by 2s is 9, 7, 18. 2. Counting by 2s is 10, 9, 8, 7. 3. Counting by 2s is 2, 4, 6, 8. Critical Thinking: Play Together/Apart with counting by 2s and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for counting by 2s and other Power Pix.
1s place Common Core Standards K.NBT.1. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. 1.NBT.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. Question: What is the 1s place? Answer: The first number on the right is the 1s place. Gesture: Hold up four fingers on one hand. With the other hand, grab the first finger on the right and wiggle it. Teaching Suggestion: (Teach 1s place, 10s place, 100s place, and 1000s place together.) While your students are involved in individual or group tasks, write a list of numbers on the board and explain the concept of the 1s place. Play Yes/No Way! with one or more questions like the following: 1. Is this the 1s place? (Point at various numbers on the board.) 2. Is the 1s place the first number on the right? 3. Is this the 1s place gesture? (Make various gestures.) 4. Is the 1s place the first number on the left? 5. Is this the 1s place Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. In the number 21, the number 2 is in the 1s place. 2. In the number 9, the 9 is in the 1s place. 3. The 1s place is the first number on the right. Critical Thinking: Play Together/Apart with the 1s place and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the 1s place and other Power Pix.
10s place
Common Core Standards K.NBT.1. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. 1.NBT.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. Question: What is the 10s place? Answer: The second number from the right is the 10s place. Gesture: Hold up four fingers on one hand. With the other hand, grab the second finger from the right and wiggle it. Teaching Suggestion: (Teach 1s place, 10s place, 100s place, and 1000s place together.) While your students are involved in individual or group tasks, write a list of numbers on the board and explain the concept of the 10s place. Play Yes/No Way! with one or more questions like the following: 1. Is this the 10s place? (Point at various numbers on the board.) 2. Is the 10s place the first number on the right? 3. Is this the 10s place gesture? (Make various gestures.) 4. Is the 10s place the second number from the left? 5. Is this the 10s place Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. In the number 42, the number 4 is in the 10s place. 2. In the number 34, the 4 is in the 10s place. 3. The 10s place is the second number from the right. Critical Thinking: Play Together/Apart with the 10s place and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the 10s place and other Power Pix.
100s place Common Core Standards 1.NBT.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. Question: What is the 100s place? Answer: The third number from the right is the 100s place. Gesture: Hold up four fingers on one hand. With the other hand, grab the third finger from the right and wiggle it. Teaching Suggestion: (Teach 1s place, 10s place, 100s place, and 1000s place together.) While your students are involved in individual or group tasks, write a list of numbers on the board and explain the concept of the 100s place. Play Yes/No Way! with one or more questions like the following: 1. Is this the 100s place? (Point at various numbers on the board.) 2. Is the 100s place the third number from the right? 3. Is this the 100s place gesture? (Make various gestures.) 4. Is the 100s place the first number from the right? 5. Is this the 100s place Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. In the number 321, the number 3 is in the 100s place. 2. In the number 431, the number 3 is in the 100s place. 3. The 100s place is the third number from the right. Critical Thinking: Play Together/Apart with the 100s place and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the 100s place and other Power Pix.
1 less than rule Common Core Standards 1.NBT.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. Question: What is the 1 less than rule? Answer: To subtract 1 from a number, subtract 1 from the 1s place. Gesture: Wave 1 finger and then point down. Teaching Suggestion: (Teach the 1 more than rule, 1 less than rule, 10 more than rule and 10 less than rule together.) While your students are involved in individual or group projects, write numbers on the board. Then use these numbers and a number line, and show students how decreasing the number in the 1s place, subtracts 1 from a number. Play Yes/No Way! with questions like the following: 1. Is 10 one less than nine? 2. Is six one less than seven? 3. Is this the 1 less than rule gesture? (Make various gestures.) 4. To subtract 1 from a number, do we subtract 1 from the 1s place? 5. Is this the 1 less than rule Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Four is one less than five. 2. To subtract 1 from a number, subtract 1 from the 1s place. 3. Three is one less than four. Critical Thinking: Play Together/Apart with the 1 less than rule and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the 1 less than rule and other Power Pix.
10 less than rule Common Core Standards 1.NBT.5. Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used. Question: What is the 10 less than rule? Answer: To subtract 10 from a number, subtract 10 from the 10s place. Gesture: Wave 10 fingers and then point down. Teaching Suggestion: (Teach the 1 more than rule, 1 less than rule, 10 more than rule and 10 less than rule together.) While your students are involved in individual or group projects, Write numbers on the board. Then, use these numbers and a number line to show students how decreasing the number in the 10s place, subtracts 10 from a number. Play Yes/No Way! with one or more questions like the following: *** rephrase questions below? 1. Is five, 10 less than 20? 2. Is five, 10 less than 15? 3. Is this the 10 less than rule gesture? (Make various gestures.) 4. To subtract 10 from a number, do we subtract 10 from the 10s place? 5. Is this the 10 less than rule Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Fifteen is 10 less than five. 2. To subtract 10 from a number, subtract 10 from the 10s place. 3. Twenty is 10 less than 30. Critical Thinking: Play Together/Apart with the 10 less than rule and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the 10 less than rule and other Power Pix.
counting by 5s
Common Core Standards 2.NBT.2. Count within 1000; skip-count by 5s, 10s, and 100s. Question: What is counting by 5s? Answer: Counting by 5s is 5, 10, 15, 20 and so on. Gesture: Hold up five fingers over and over as you say, ―5, 10, 15, 20.‖ Teaching Suggestion: While your students are involved in individual or group tasks, write lists of numbers on the board. Some lists are counting by 1s, 2s, 5s, 10s and other lists contain random numbers. Explain the concept of counting by 5s. Play Yes/No Way! with one or more questions like the following: 1. Is this counting by 5s? (Point at various lists.) 2. Is 2, 4, 6, 8 counting by 5s? 3. Is this the counting by 5s gesture? (Make various gestures.) 4. Is 5, 10, 15, 20 counting by 5s? 5. Is this the counting by 5s Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Counting by 5s is 1, 2, 3, 4. 2. Counting by 5s is 5, 10, 15, 20. 3. Counting by 5s is 3, 6, 9, 12. Critical Thinking: Play Together/Apart with counting by 5s and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for counting by 5s and other Power Pix.
counting by 10s
Common Core Standards K.CC.1. Count to 100 by ones and by tens. 2.NBT.2. Count within 1000; skip-count by 5s, 10s, and 100s. Question: What is counting by 10s? Answer: Counting by 10s is 10, 20, 30, 40 and so on. Gesture: Hold up 10 fingers over and over as you say, ―10, 20, 30, 40.‖ California State First Grade Math Standard: Number Sense 2.4: Count by 2s, 5s, and 10s to 100. First Grade Power Pix Math, copyright 2009, Chris Biffle and Jay Vanderfin 30 Teaching Suggestion: While your students are involved in individual or group tasks, write lists of numbers on the board. Some lists are counting by 1s, 2s, 5s, 10s and other lists contain random numbers. Explain the concept of counting by 10s. Play Yes/No Way! with one or more questions like the following: 1. Is this counting by 10s? (Point at various lists.) 2. Is 10, 20, 30, 40 counting by 10s? 3. Is this the counting by tens gesture? (Make various gestures.) 4. Is 5, 10, 15, 20 counting by 10s? 5. Is this the counting by tens Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Counting by 10s is 1, 2, 3, 4. 2. Counting by 10s is 5, 10, 15, 20. 3. Counting by 10s is 10, 20, 30, 40. Critical Thinking: Play Together/Apart with counting by 10s and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for counting by 10s and other Power Pix.
expanded form
Common Core Standards 1.NBT.2. Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: a.10 can be thought of as a bundle of ten ones — called a ―ten.‖ b. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. c. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). Question: What is expanded form? Answer: Expanded form is writing a number in 1s and 10s. Gesture: Hold up two fingers on one hand. With the other hand wiggle the first finger to the right as you say ―1s‖, then the next finger as you say ―10s.‖ Teaching Suggestion: (Teach expanded form after teaching 1s place, 10s place, 100s place.) While your students are involved in individual or group tasks, write pairs of numbers on the board; one of the two numbers is in expanded form. For example, write 34 and 30 + 4. Explain the concept of expanded form to your students. Play Yes/No Way! with one or more questions like the following: 1. Is this number in expanded form? (Point at various numbers on the board.) 2. Is expanded form writing a number in 1s and 10s? 3. Is this the expanded form gesture? (Make various gestures.) 4. Is 20 + 4 the expanded form of 29? 5. Is this the expanded form Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. The expanded form of 34 is 30 + 4. 2. The expanded form of 22 is 2 + 2. 3. Expanded form is writing a number in 1s and 10s. Critical Thinking: Play Together/Apart with expanded form and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for expanded form and other Power Pix.
less than/more than rule Common Core Standards 1.NBT.3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
Question: What is the less than/more than rule? Answer: The less than/more than rule is: the arrow always points at the smaller number. Gesture: Make a V with two fingers on one hand, like the less than/more than symbol. With one finger on the other hand, indicate the point of the V and say ―less than.‖ Then indicate the opening of the V and say ―more than.‖ For additional clarity, make the same symbol with the other hand, so that the V points the opposite direction. Again, indicate the point of the V and say ―less than‖ and the opening of the V and say ―more than.‖ Students will understand the V can point to the left or the right. Teaching Suggestion: While your students are involved in individual or group projects, write examples on the board of less than/more than relationships (10<15, 18>2.) Explain to students that the point on the V always indicates the smaller number. Play Yes/No Way! with one or more questions like the following: 1. Is this the less than/more than symbol? (Point at various symbols/numbers on the board.) 2. Does this say _____? (Fill in the blank, depending on the numbers you are pointing at. If you’re pointing at, 22>13, you would ask, ―Does this say 22 is greater than 13?‖ You could also ask, ―Does this say 22 is less than 13?‖) 3. Is this the less than/more than rule gesture? (Make various gestures.) 4. Is the less than/more than symbol the same as the equals sign? 5. Is this the less than/more than rule Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Nineteen is less than twenty. 2. The less than/more than arrow always points at the smaller number. 3. The less than/more than arrow always points at the larger number. Critical Thinking: Play Together/Apart with the less than/more than rule and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the less than/more than rule and other Power Pix.
Calendar Not in Common Core Standards but is a Core Concept Question: What is a calendar? Answer: A calendar shows the months of the year and the days of the week. Gesture: Move one hand through the air, left to right and then right to left, as if searching for a date on an imaginary calendar. Teaching Suggestion: Use a calendar to show your students its major features. Explain the difference between days, months and a year. (For kindergartners, this may amount to no more than indicating that months have lots of days and years have lots of months.) Play Yes/No Way! with questions like the following: 1. Is this a calendar? (Point at various objects in the classroom, including the calendar.) 2. Is this the calendar picture? (Point at various Power Pix; include the calendar Power Pix occasionally.) 3. Is this the calendar gesture? (Make various gestures; include the calendar gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. The calendar tells us about hours and minutes. 2. A calendar shows the months of the year and the days of the week. 3. A calendar has a big hand and a little hand. Critical Thinking: Play Compare/Contrast with calendar and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for calendar and other Power Pix.
days of the week Not in Common Core Standards but is a Core Concept Question: What are the days of the week? Answer: The days of the week are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Gesture: Hold up one finger for each day of the week, until you are holding up seven fingers. Teaching Suggestion: Use a calendar and other appropriate materials to teach your students the days of the week. Make a one week calendar on the board. Play Yes/No Way! with questions like the following: 1. Is this ______? (Point at various days, on the one week board calendar; ask ―Is this Monday ... Tuesday ... ― ) 2. Is this the days of the week picture? (Point at various Power Pix; include the days of the week Power Pix occasionally.) 3. Is this the days of the week gesture? (Make various gestures; include the days of the week gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. There are seven days of the week. 2. January is one of the days of the week. 3. Monday is one of the days of the week. Critical Thinking: Play Compare/Contrast with days of the week and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for days of the week and other Power Pix.
Evening Not in Common Core Standards but is a Core Concept Question: What is evening? Answer: Evening is after the sun sets. Gesture: Make a circle with both hands and then bring your hands down, (symbolizing the sun setting). Teaching Suggestion: (Teach morning, noon, afternoon and evening together.) Explain the relationship between sunset and evening. Draw or show pictures that represent the evening (stars and moon out, lights on, etc.) Play Yes/No Way! with questions like the following: 1. Is this the evening? (Point at various drawings or pictures.) 2. Is this the evening picture? (Point at various Power Pix; include the evening Power Pix occasionally.) 3. Is this the evening gesture? (Make various gestures; include the evening gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Evening is after the sun sets. 2. Evening is when the sun rises. 3. Evening is nighttime. Critical Thinking: Play Compare/Contrast with evening and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for evening and other Power Pix.
Morning Not in Common Core Standards but is a Core Concept Question: What is a morning? Answer: Morning is when the sun rises and we wake up. Gesture: Make a stretching motion like you are waking up. Teaching Suggestion: (Teach morning, noon, afternoon and evening together.) Explain the concept of morning to your students. Ask exploratory questions like, ―What are things you do in the morning? ... What do your parents do in the morning? ... What do you do on mornings when you go to school? .... What do you do on mornings when you don’t go to school?‖ Play Yes/No Way! with questions like the following: 1. Do we wake up in the morning? (Include examples of activities that are, and aren’t, associated with the morning.) 2. Is this the morning picture? (Point at various Power Pix; include the morning Power Pix occasionally.) 3. Is this the morning gesture? (Make various gestures; include the morning gesture occasionally.) 4. Does morning start before lunch? Quick Test: Play Cutie with statements like the following: 1. Morning is when the sun rises and we wake up. 2. Morning is when the sun sets and we go to sleep. 3. Morning is before lunch. Critical Thinking: Play Compare/Contrast with morning and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for morning and other Power Pix.
Noon Not in Common Core Standards but is a Core Concept Question: What is noon? Answer: Noon is when both clock hands are on the 12 and we eat lunch. Gesture: Put both hands over your head, like clock hands pointing at 12, and then use an imaginary spoon to put food in your mouth. Teaching Suggestion: (Teach morning, noon, afternoon and evening together.) Explain the concept of noon to your students. Use a clock with moveable hands to show students the position of the hands at noon. (Save the concept of midnight for later in the year!) Ask exploratory questions like, ―What are things you do at noon? ... What do your parents do at noon? ... What do you do at noon when are in school? .... What do you do at noon when you don’t go to school?‖ Play Yes/No Way! with questions like the following: 1. Is this noon? (Move the clock hands to various positions; occasionally put both hands pointing at 12.) 2. Is this the noon picture? (Point at various Power Pix; include the noon Power Pix occasionally.) 3. Is this the noon gesture? (Make various gestures; include the noon gesture occasionally.) 4. Is noon before breakfast? Quick Test: Play Cutie with statements like the following: 1. Noon is when the sun rises and we wake up. 2. Noon is when the sun sets and we go to sleep. 3. Noon is when both clock hands are on the 12 and we eat lunch. Critical Thinking: Play Compare/Contrast with noon and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for noon and other Power Pix.
Tomorrow Not in Common Core Standards but is a Core Concept Question: What is tomorrow? Answer: Tomorrow is the day after today. Gesture: Put your palms together and then point both arms straight ahead (as if making an arrow toward the future, tomorrow.) Teaching Suggestion: (teach today, yesterday and tomorrow together.) Using a calendar, show students the day it is today (Monday, Tuesday, Wednesday, etc.) Show students what day it was yesterday and what day it will be tomorrow. Explain to students that tomorrow comes after today. Play Yes/No Way! with questions like the following: 1. Will tomorrow be Thursday? (Substitute other days of the week. Return several times to the correct day for tomorrow.) 2. Is this the tomorrow picture? (Point at various Power Pix; include the tomorrow Power Pix occasionally.) 3. Is this the tomorrow gesture? (Make various gestures; include the tomorrow gesture occasionally.) 4. Was this morning part of tomorrow? 5. Is tomorrow the day after today? Quick Test: Play Cutie with statements like the following: 1. Tomorrow is the day before today. 2. Tomorrow is the day after today. 3. Tomorrow will be ______. (Insert correct or incorrect day.) Critical Thinking: Play Compare/Contrast with tomorrow and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for tomorrow and other Power Pix.
Today Not in Common Core Standards but is a Core Concept Question: What is a today? Answer: Today is the day we are in right now. Gesture: Clap hands twice. And then point at the ground (symbolizing ―right now‖). Say, ―Today is the day we are in right [clap] now [clap].‖ Teaching Suggestion: (Teach today, yesterday and tomorrow together.) Using a calendar, show students the day it is today (Monday, Tuesday, Wednesday, etc.) Show students what day it was yesterday and what day it will be tomorrow. Play Yes/No Way! with questions like the following: 1. Is today Wednesday? (Substitute other days of the week. Return several times to the correct day for today.) 2. Is this the today picture? (Point at various Power Pix; include the today Power Pix occasionally.) 3. Is this the today gesture? (Make various gestures; include the today gesture occasionally.) 4. Was this morning part of today? 5. Will this evening be part of today? Quick Test: Play Cutie with statements like the following: 1. Today is the day we are in right now. 2. Today is ______. (Insert correct or incorrect day.) 3. Today is yesterday. (Or, tomorrow.) Critical Thinking: Play Compare/Contrast with today and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for today and other Power Pix.
Yesterday Not in Common Core Standards but is a Core Concept Question: What is yesterday? Answer: Yesterday is the day before today. Gesture: Jerk your thumb over your shoulder, indicating yesterday is ―past.‖ Teaching Suggestion: (teach today, yesterday and tomorrow together.) Using a calendar, show students the day it is today (Monday, Tuesday, Wednesday, etc.) Show students what day it was yesterday and what day it will be tomorrow. Explain to students that yesterday comes before today. Play Yes/No Way! with questions like the following: 1. Was yesterday Monday? (Substitute other days of the week. Return several times to the correct day for yesterday.) 2. Is this the yesterday picture? (Point at various Power Pix; include the yesterday Power Pix occasionally.) 3. Is this the yesterday gesture? (Make various gestures; include the yesterday gesture occasionally.) 4. Was this morning part of yesterday? 5. Will this evening be part of yesterday? Quick Test: Play Cutie with statements like the following: 1. Yesterday is the day before today. 2. Yesterday was a school day. 3. Yesterday was ______. (Insert correct or incorrect day of the week.) Critical Thinking: Play Compare/Contrast with yesterday and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for yesterday and other Power Pix.
equal height
Common Core Standards K.MD.2. Directly compare two objects with a measurable attribute in common, to see which object has ―more of‖/―less of‖ the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. Question: What is equal height? Answer: Equal height means the same height. (One person and three people on top of each other.) Gesture: Hold both hands up as high as possible, showing they have equal height. Teaching Suggestion: While your students are involved in individual or group projects, make simple drawings of pairs of objects on the board, i.e. a tree and a house, a square and a rectangle, a boy and a girl. In some pairs, the objects are equal height; in other pairs one object is smaller than the other. Use the drawings and various props and/or groups of stackable objects to teach students the concept of equal height. Next, Play Yes/No Way! with questions like the following: 1. Are these equal height? (Point at pairs of various objects, including drawings on the board that do, and do not, represent equal height.) 2. Is this the equal height picture? (Point at various Power Pix; include the equal height Power Pix occasionally.) 3. Is this the equal height gesture? (Make various gestures; include the equal height gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. A boy and a girl four feet tall have equal height. 2. Equal height means the same height. 3. Two things with equal height must be the same color. Critical Thinking: Play Compare/Contrast with equal height and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for equal height and other Power Pix.
sorting
Common Core Standards K.MD.3. Classify objects into given categories; count the numbers of objects in each category and sort the categories by count. 1 Question: What is sorting? Answer: Sorting is putting things together that are similar. Gesture: Grab four fingers of one hand with the other hand, to symbolize bringing things together that are similar (your fingers). Teaching Suggestion: Show students examples of sorting based on a variety of attributes (all green things, all round things, all plastic things, etc.) Play Yes/No Way! with questions like the following: 1. Are all these ______, ______? (Point at groups of objects in the classroom. Create Yes/No Way! questions. Are all these desks, wood? Are all these pencils, yellow? Are all these crayons, square?) 2. Is this the sorting picture? (Point at various Power Pix; include the sorting Power Pix occasionally.) 3. Is this the sorting gesture? (Make various gestures; include the sorting gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. Sorting is putting things together that are similar. 2. I could sort red blocks together because they are all red. 3. I could sort red blocks together because they are all yellow. Critical Thinking: Play Compare/Contrast with sorting and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for sorting and other Power Pix.
big hand on a clock
Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. Question: What is the big hand on a clock? Answer: The big hand on a clock points at the minutes. Gesture: Point one arm straight down. (This is the ―big hand‖ on a clock.) Teaching Suggestion: (Teach little hand on a clock, big hand on a clock and clock together.) Use a clock with moveable hands to teach your students the concept of telling time. Explain that the big hand points at minutes. Play Yes/No Way! with questions like the following: 1. Is this the big hand on a clock? (Point at various objects in the classroom and parts of the clock, the frame, numbers, big hand and little hand.) 2. Is this the big hand on a clock picture? (Point at various Power Pix; include the big hand on a clock Power Pix occasionally.) 3. Is this the big hand on a clock gesture? (Make various gestures; include the big hand on a clock gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. The big hand is longer than the little hand. 2. The big hand on a clock points at minutes. 3. The big hand on a clock points at hours. Critical Thinking: Play Compare/Contrast with big hand on a clock and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for big hand on a clock and other Power Pix.
clock Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. Question: What is a clock? Answer: A clock is for telling time. Gesture: Bend one arm and put it slightly above your head. (This is the little hand on a clock.) Point your other arm straight down. (This is the big hand on a clock.) Teaching Suggestion: (Teach little hand on a clock, big hand on a clock and clock together.) Use a clock with moveable hands to teach your students the concept of telling time. Explain that the little hand points at hours and the big hand points at minutes. Play Yes/No Way! with questions like the following: 1. Is this a clock? (Point at various objects.) 2. Is this the clock picture? (Point at various Power Pix; include the clock Power Pix occasionally.) 3. Is this the clock gesture? (Make various gestures; include the clock gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. A clock is for telling time. 2. A clock shows us the days of the week. 3. A clock tells us what time it is. Critical Thinking: Play Compare/Contrast with clock and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for clock and other Power Pix.
little hand on a clock
Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. Question: What is the little hand on a clock? Answer: The little hand on a clock points at the hours. Gesture: Bend one arm and put it slightly above your head. (This is the little hand on a clock.) Teaching Suggestion: (Teach little hand, big hand and clock together.) Use a clock with moveable hands to teach your students the concept of telling time. Explain that the little hand points at hours. Play Yes/No Way! with questions like the following: 1. Is this the little hand on a clock? (Point at various objects in the classroom and parts of the clock, the frame, numbers, big hand and little hand.) 2. Is this the little hand on a clock picture? (Point at various Power Pix; include the little hand on a clock Power Pix occasionally.) 3. Is this the little hand on a clock gesture? (Make various gestures; include the little hand on a clock gesture occasionally.) Quick Test: Play Cutie with statements like the following: 1. The little hand on a clock is shorter than the big hand on a clock. 2. The little hand on a clock points at minutes. 3. The little hand on a clock points at hours. Critical Thinking: Play Compare/Contrast with little hand on a clock and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for little hand on a clock and other Power Pix.
half hour
Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. Question: What is a half hour? Answer: A half hour is 30 minutes. Gesture: Hold up 10 fingers three times, count aloud, ―10, 20, 30 minutes is half an hour!‖ Teaching Suggestion: (Teach minute, half hour and hour together) While your students are involved in individual or group tasks, write various times on the board: 6:30, 3:15, 2:45, 9:15, 8:30, etc. Using a clock with moveable hands, teach students the concept of a half hour. Point out that a half hour is 30 minutes. Play Yes/No Way! with one or more questions like the following: 1. Does this number show a half hour? (Point at various numbers on the board.) 2. Is a half hour longer than an hour? 3. Is this the half hour gesture? (Make various gestures.) 4. Is a half hour longer than a minute? 5. Is this the half hour Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A half hour is 30 minutes. 2. A half hour is longer than 10 minutes. 3. A half hour is 15 minutes. Critical Thinking: Play Together/Apart with half hour and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for half hour and other Power Pix.
hour
Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. Question: What is an hour? Answer: An hour is 60 minutes. Gesture: Hold up 10 fingers six times, count aloud, ―10, 20, 30, 40, 50, 60 minutes is an hour.‖ Teaching Suggestion: (Teach minute, half hour and hour together.) While your students are involved in individual or group tasks, write various times on the board, 6:30, 3:15, 2:45, 9:15, 8:30, etc. Using a clock with moveable hands, teach students the concept of an hour. Point out that the first number indicates the hour. Play Yes/No Way! with one or more questions like the following: 1. Does this number show the hour? (Pointing at various numbers on the board.) 2. Is an hour longer than a day? 3. Is this the hour gesture? (Make various gestures.) 4. Is an hour longer than half an hour? 5. Is this the hour Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. An hour has 60 minutes. 2. 60 minutes is the same length of time as an hour. 3. If the time is 6:30, the 30 is the hour. Critical Thinking: Play Together/Apart with hour and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for hour and other Power Pix.
minute
Common Core Standards 1.MD.3. Tell and write time in hours and half-hours using analog and digital clocks. 2.MD.7. Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m. Question: What is a minute? Answer: A minute is 60 seconds. Gesture: Hold up 10 fingers six times, count aloud, ―10, 20, 30, 40, 50, 60 seconds is a minute!‖ Teaching Suggestion: (Teach minute, half hour and hour together.) While your students are involved in individual or group tasks, write various times on the board: 6:30, 3:15, 2:45, 9:15, 8:30, etc. Using a clock with moveable hands, teach students the concept of a minute. Explain that a minute has 60 seconds. Ask them to watch a clock with a minute hand and hold up their hands after 60 seconds. Point out that a half hour is 30 minutes. Play Yes/No Way! with one or more questions like the following: 1. Does this number show the minutes? (Point at various numbers on the board.) 2. Is a second shorter than a minute? 3. Is this the minute gesture? (Make various gestures.) 4. Is a minute longer than a second? 5. Is this the minute Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A minute is 60 seconds. 2. A minute is longer than a second. 3. A second is longer than a minute. Critical Thinking: Play Together/Apart with minute and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for minute and other Power Pix.
Dime Common Core Standards Core Concept in first grade 2.MD.8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? Question: What is a dime? Answer: A dime is 10 pennies. Gesture: Hold up one hand and pinch your thumb and finger together, as if holding a penny. Then, with the other hand, hold up five fingers once and then again (symbolizing that a dime is worth 10 pennies.) Students should count ―five‖ as they hold up five fingers the first time and ―ten‖ as they hold up five fingers the second time. Teaching Suggestion: (Teach penny, nickel, dime, quarter and dollar bill together.) Show students dimes and pictures of dimes. Explain that 5 pennies make a nickel, 10 pennies make a dime and that two nickels make a dime. Play Yes/No Way! with one or more questions like the following: 1. Is a dime worth 10 pennies? 2. Is 10 pennies worth a dime? 3. Is this the dime gesture? (Make various gestures.) 4. Is a nickel worth less than a dime? 5. Is this the dime Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A dime is the smallest value coin. 2. A dime is 10 pennies. 3. Five pennies make one dime. Critical Thinking: Play Together/Apart with dime and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for dime and other Power Pix.
Nickel Common Core Standards Core Concept in first grade 2.MD.8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? Question: What is a nickel? Answer: A nickel is five pennies. Gesture: Hold up one hand and pinch your thumb and finger together, as if holding a penny. Then, with the other hand, hold up five fingers (symbolizing that a nickel is worth 5 pennies). Students should say ―five‖ as they hold up five fingers. Teaching Suggestion: (Teach penny, nickel, dime, quarter and dollar bill together.) Show students nickels and pictures of nickels. Explain that 5 pennies make a nickel and that two nickels make a dime. Play Yes/No Way! with one or more questions like the following: 1. Are there any coins worth more than a nickel? 2. Are there any coins worth less than a nickel? 3. Is this the nickel gesture? (Make various gestures.) 4. Is a nickel five pennies? 5. Is this the nickel Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A nickel is the smallest value coin. 2. Two nickels make one penny. 3. Five pennies make one nickel. Critical Thinking: Play Together/Apart with nickel and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for nickel and other Power Pix.
penny
Common Core Standards Core Concept in first grade 2.MD.8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? Question: What is a penny? Answer: A penny is one cent. Gesture: Hold up one hand and pinch your thumb and finger together, as if holding a penny. Teaching Suggestion: (Teach penny, nickel, dime, quarter and dollar bill together.) Show students pennies and pictures of pennies. Explain that 5 pennies make a nickel. Play Yes/No Way! with one or more questions like the following: 1. Are there any coins worth more than a penny? 2. Are there any coins worth less than a penny? 3. Is this the penny gesture? (Make various gestures.) 4. Is a nickel five pennies? 5. Is this the penny Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A penny is the smallest value coin. 2. Two pennies make one nickel. 3. Five nickels make one penny. Critical Thinking: Play Together/Apart with penny and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for penny and other Power Pix.
Quarter Common Core Standards Core Concept in first grade 2.MD.8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? Question: What is a quarter? Answer: A quarter is 25 pennies. Gesture: 5 Teaching Suggestion: (Teach penny, nickel, dime, quarter and dollar bill together.) Show students quarters and pictures of quarters. Explain that 25 pennies make a quarter, 5 nickels make a quarter and so forth. Play Yes/No Way! with one or more questions like the following: 1. Is a quarter worth 10 pennies? 2. Is a quarter worth 25 pennies? 3. Is this the quarter gesture? (Make various gesture.) 4. Is a quarter worth less than a dime? 5. Is this the quarter Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A quarter is worth 25 nickels. 2. A quarter is worth 25 pennies. 3. Five dimes make one quarter. Critical Thinking: Play Together/Apart with quarter and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for quarter and other Power Pix.
pounds and ounces Common Core Standards Core concept Question: What are pounds and ounces? Answer: A pound is 16 ounces. Gesture: Put one hand on top of the other at belt level and slightly bounce your hands up and down as if carrying something. Teaching Suggestion: Explain the concept of pounds to your students and that lbs is an abbreviation for pounds and oz is an abbreviation for ounces. Use a scale and let students guess how many pounds various objects weigh ... including themselves! Play Yes/No Way! with one or more questions like the following: 1. Does a desk weigh more than 5 pounds? 2. Does a pound weigh more than an ounce? 3. Is this the pounds and ounces gesture? (Make various gestures.) 4. Is a pound 16 ounces? 5. Is this the pounds and ounces Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A basketball weighs less than 100 pounds. 2. You weigh more than one ounce. 3. A pound is 16 ounces Critical Thinking: Play Together/Apart with pounds and ounces and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for the pounds and ounces and other Power Pix.
bar graph Common Core Standards 2.MD.10. Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems1 using information presented in a bar graph. 3.MD.3. Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step ―how many more‖ and ―how many less‖ problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets. Question: What is a bar graph? Answer: A bar graph compares numbers using bars. Gesture: Put your two forearms straight up, one higher than the other, as if they were unequal bars on a bar graph. Teaching Suggestion: Show students examples of bar graphs and explain that the height of the bar represents the size of a number. Play Yes/No Way! with one or more questions like the following: 1. Is this number larger than this one? (Pointing at various bars on a bar graph.) 2. Does this bar mean _____ ? (Point at graphs, and ask students about the amount various bars represent.) 3. Is this the bar graph gesture? (Make various gestures.) 4. Is this amount less than this amount? (Ask students to compare two bars on a bar graph.) 5. Is this the bar graph Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Bars on a bar graph stand for numbers. 2. On a bar graph, a tall bar represents a small number. 3. On a bar graph, a short bar represents a small number. Critical Thinking: Play Together/Apart with bar graph and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for bar graph and other Power Pix.
foot and 12 inches Common Core Standards 2.MD.3. Estimate lengths using units of inches, feet, centimeters, and meters. Question: What is a foot and 12 inches? Answer: A foot is made of 12 inches. Gesture: Hold up 10 fingers and then 2 fingers. Then hold your hands approximately a foot apart. Teaching Suggestion: Show students a foot ruler and explain that it is 12 inches long. Give students foot rulers and ask them to individually, or in teams, measure objects in class. Play Yes/No Way! with one or more questions like the following: 1. Are you more than a foot tall? 2. Are you less than an inch tall? 3. Is this the foot and 12 inches gesture? (Make various gestures.) 4. Is a foot longer than an inch? 5. Is this the foot and 12 inches Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. A foot is made of 20 inches. 2. Twenty inches make a foot. 3. A foot is made of 12 inches. Critical Thinking: Play Together/Apart with the foot and 12 inches and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for foot and 12 inches and other Power Pix
tally marks Common Core Standards 1.MD.4. Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. Question: What are tally marks? Answer: Tally marks are marks in groups of five. Gesture: Hold up four fingers on one hand and then cross them with one finger from the other hand, symbolizing four tally marks crossed by a fifth mark. Teaching Suggestion: (Teach tally marks and counting by 5s together.) Show students examples of tally marks and contrast them with groups of numbers and random collections of lines. Play Yes/No Way! With one or more questions like the following: 1. Are these tally marks? (Pointing at various numbers and marks on the board.) 2. Do these tally marks equal ___? (Point at various groups of tally marks and ask students how many marks are in each group.) 3. Is this the tally mark gesture? (Make various gestures.) 4. Does the crossed line stand for the fifth tally mark? 5. Is this the tally mark Power Pix? (Point at various Power Pix.) Quick Test: Play Cutie with statements like the following: 1. Tally marks are a way of counting by five. 2. Two complete groups of tally marks equal 10. 3. Tally marks are a way of counting by 2s. Critical Thinking: Play Together/Apart with tally marks and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for tally marks and other Power Pix.
circle
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a circle? Answer: A circle is a round shape. Gesture: Using the thumb and forefinger of one hand, make a circle. Teaching Suggestion: Show students examples of circles and explain their similarities (each one is round, has no corners or straight lines, etc.) Play Yes/No Way! with questions like the following: 1. Is this a circle? (Hold up a variety of objects.) 2. Is this the circle picture? (Point at various Power Pix; include the circle Power Pix occasionally.) 3. Is this the circle gesture? (Make various gestures; include the circle gesture occasionally.) 4. Is a square a circle? Quick Test: Play Cutie with statements like the following: 1. A circle is made from four straight lines. 2. A circle is a round shape. 3. A circle has four corners. Critical Thinking: Play Compare/Contrast with circle and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for circle and other Power Pix.
Cone
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a cone? Answer: A cone has a circle on one end and a point on the other. Gesture: Hold an imaginary ice cream cone in one hand and take a big bite out of the ice cream. Teaching Suggestion: Show students examples of cones and explain their similarities (a circle at one end and a point at the other.) Play Yes/No Way! with questions like the following: 1. Is this a cone? (Hold up a variety of objects.) 2. Is this the cone picture? (Point at various Power Pix; include the cone Power Pix occasionally.) 3. Is this the cone gesture? (Make various gestures; include the cone gesture occasionally.) 4. Does a cone have a circle on each end? 5. Is an ice cream cone a cone? Quick Test: Play Cutie with statements like the following: 1. A cone has a circle on one end. 2. A cone has a point on one end. 3. A can of soda is a cone. Critical Thinking: Play Compare/Contrast with cone and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for cone and other Power Pix.
Cube
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a cube? Answer: A cube is made of six squares! [sound excited] Gesture: With two hands, shape a cube in the air, as if you are holding top and bottom, side and side, front and back. Teaching Suggestion: Show students examples of cubes and explain their similarities (all have six sides, every side is a square, no sides are rectangles, etc.) Play Yes/No Way! with questions like the following: 1. Is this a cube? (Hold up a variety of objects.) 2. Is this the cube picture? (Point at various Power Pix; include the cube Power Pix occasionally.) 3. Is this the cube gesture? (Make various gestures; include the cube gesture occasionally.) 4. Is a book a cube? 5. Is an ice cube a cube? Quick Test: Play Cutie with statements like the following: 1. A cube is round. 2. A cube is made from squares. 3. A cube has corners. Critical Thinking: Play Compare/Contrast with cube and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for cube and other Power Pix.
Cylinder
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a cylinder? Answer: A cylinder has a circle on both ends. Gesture: Use the thumb and forefinger on each hand to make circles. Then hold these circles in the air to show both ends of a cylinder. Teaching Suggestion: Show students examples of cylinders and explain their similarities (a circle at both ends). Play Yes/No Way! with questions like the following: 1. Is this a cylinder? (Hold up a variety of objects.) 2. Is this the cylinder picture? (Point at various Power Pix; include the cylinder Power Pix occasionally.) 3. Is this the cylinder gesture? (Make various gestures; include the cylinder gesture occasionally.) 4. Does a cylinder have a circle on each end? 5. Does a cylinder have corners? Quick Test: Play Cutie with statements like the following: 1. A cylinder is made from four straight lines. 2. A cylinder has a circle on both ends. 3. A can of soda is a cylinder. Critical Thinking: Play Compare/Contrast with cylinder and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for cylinder and other Power Pix.
Rectangle
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a rectangle? Answer: A rectangle has four sides. Gesture: With one finger, draw an imaginary rectangle in the air. (Make the top and bottom lines very long and the end lines very short, to distinguish rectangles from squares.) Teaching Suggestion: Show students examples of rectangles and explain their similarities (all have four sides, all have four corners, etc.) Distinguish between rectangles and squares (the latter having four equal sides.) Play Yes/No Way! with questions like the following: 1. Is this a rectangle? (Hold up a variety of objects.) 2. Is this the rectangle picture? (Point at various Power Pix; include the rectangle Power Pix occasionally.) 3. Is this the rectangle gesture? (Make various gestures; include the rectangle gesture occasionally.) 4. Is a triangle a rectangle? 5. Is a ball a rectangle? Quick Test: Play Cutie with statements like the following: 1. A rectangle is made from four straight lines. 2. A rectangle is a square. 3. A rectangle has four sides. Critical Thinking: Play Compare/Contrast with rectangle and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for rectangle and other Power Pix.
Sphere
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a sphere? Answer: A sphere is like a ball. Gesture: With one hand, pretend as if you are bouncing a ball on the floor. Teaching Suggestion: Show students examples of spheres and explain their similarities (all are round, can roll in any direction unlike a cylinder, etc.) Also, point out that some spheres are not balls, for example, a globe. Play Yes/No Way! with questions like the following: 1. Is this a sphere? (Hold up a variety of objects.) 2. Is this the sphere picture? (Point at various Power Pix; include the sphere Power Pix occasionally.) 3. Is this the sphere gesture? (Make various gestures; include the sphere gesture occasionally.) 4. Do spheres have corners? 5. Are all spheres blue? Quick Test: Play Cutie with statements like the following: 1. A sphere is round. 2. A sphere is like a ball. 3. A basketball is a sphere. Critical Thinking: Play Compare/Contrast with sphere and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for sphere and other Power Pix.
Square
Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a square? Answer: A square has four equal sides. (Emphasize equal as you say the answer, to distinguish squares from rectangles.) Gesture: With one finger, draw an imaginary square in the air. Teaching Suggestion: Show students examples of squares and explain their similarities (all have four equal sides, all have four corners, etc.) Play Yes/No Way! with questions like the following: 1. Is this a square? (Hold up a variety of objects.) 2. Is this the square picture? (Point at various Power Pix; include the square Power Pix occasionally.) 3. Is this the square gesture? (Make various gestures; include the square gesture occasionally.) 4. Is a square round? 5. Is a triangle a square? Quick Test: Play Cutie with statements like the following: 1. A square is made from four straight lines. 2. A square has four equal sides. 3. A square has three corners. Critical Thinking: Play Compare/Contrast with square and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for square and other Power Pix.
Triangle Common Core Standards K.G.1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2. Correctly name shapes regardless of their orientations or overall size. K.G.3. Identify shapes as two-dimensional (lying in a plane, ―flat‖) or three-dimensional (―solid‖). K.G.4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/―corners‖) and other attributes (e.g., having sides of equal length). K.G.5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. K.G.6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or threedimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. Question: What is a triangle? Answer: A triangle has three sides. Gesture: Bringing together the thumb and forefinger of both hands, make one triangle. Teaching Suggestion: Show students examples of triangles and explain their similarities (all have three sides, all have three corners, etc.) Play Yes/No Way! with questions like the following: 1. Is this a triangle? (Hold up a variety of objects.) 2. Is this the triangle picture? (Point at various Power Pix; include the triangle Power Pix occasionally.) 3. Is this the triangle gesture? (Make various gestures; include the triangle gesture occasionally.) 4. Is a triangle a circle? 5. Does a triangle have three corners? Quick Test: Play Cutie with statements like the following: 1. A triangle is made from four straight lines. 2. A triangle is made from three straight lines. 3. A triangle has three sides. Critical Thinking: Play Compare/Contrast with triangle and other Power Pix. Review: Ask your students to review with each other the question, answer and gestures for triangle and other Power Pix.
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## #1 2005-08-23 08:40:23
mikau
Guest
### money in bags question someone asked
while the forum was hacked, someone asked this question:
hey can anyone help me out?
problem:
Divide \$71 in whole \$ increments into a number of bags so that I can ask for any amount between \$1 and \$71, and you can give me the proper amount by giving me a certain number of therse bags without opening them. What is the minimum number of bags you will require?
thanks
lets see, well if you have to divide them into a group of equal parts, then you'd have to divde it by 71. Anything more then one each and you would not be able to ask for one.
If you can divide it into groups of unequal parts, each bag containing a certain number of whole dollars, then it gets slightly more interesting. If you put 10 in one bag, and put the remaining 61 into individual bags, then you will be able to make the numbers between 1 and 10, by using the additional ones. Therefore I suppose if you put extra in one bag, you need enough singles to cover the numbers from 1 to that point. Lets say we put 36 in one bag and put the remaining 35 into individual bags. For numbers less then 36 you can use the individual bags, for 36 you can use the big bag, any more and you can add the big bag to any number of small bags to make any number from 37 to 71. Now we reduced 71 to 36 bags, lets see if we can the 35 small bags to less bags. Again lets take the midpoint. If 18 in one bag and the remaining 17 into individual bags, we can make the numbers less then 18 by using the individual small bags, we can make 18 using the big bag, and any number between 18 and 36 by using the big bag and adding the individual small bags. So now we've got one big bag with 36, another big bag with 18, and 17 litte bags. If my reasoning proccess is correct, it looks like we are dividing the number by 2 using integer division, then giving the bigger bag the remainder, and the repeat.
So it should be 36, 18, 9, 4, 2, 1, 1. Hope I worked that out right. Kind of a confusing algorithm. I'm not used to doing integer division. lol. Anyway it looks like the answer is 7 bags.
## #2 2005-08-23 08:45:35
mikau
Super Member
Offline
### Re: money in bags question someone asked
weird. I needed to reregister my name. Thats me. Just so you know.
A logarithm is just a misspelled algorithm.
## #3 2005-08-23 08:56:50
MathsIsFun
Offline
### Re: money in bags question someone asked
I knew you would have to
Only because my last backup was just before you registered.
I am in the process of upgrading to the latest version of the forum software to (hopefully) avoid further hacks.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #4 2005-08-23 09:02:35
mikau
Super Member
Offline
### Re: money in bags question someone asked
I'm just ticked cause I lost my introduction post. :-(
A logarithm is just a misspelled algorithm.
## #5 2005-08-23 09:30:25
MathsIsFun
Offline
### Re: money in bags question someone asked
It was great, too.
And I liked your "nothing is hard, it is just unfamiliar" post too.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #6 2005-08-23 09:47:25
mikau
Super Member
Offline
### Re: money in bags question someone asked
I didn't say nothing is hard. Rocks are hard, skateboarding is hard. But math isn't hard. Its just unfamiliar.
Hey, do you think I got the above problem right?
A logarithm is just a misspelled algorithm.
## #7 2005-08-23 14:23:29
ganesh
Moderator
Online
### Re: money in bags question someone asked
Yes, seven bags is right, but I would have put the Dollars in the bags this way:-
1, 2, 4, 8, 16, 32 and 8. | 1,068 | 3,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2014-10 | longest | en | 0.928703 |
https://brainly.ph/question/7896 | 1,487,802,189,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171053.19/warc/CC-MAIN-20170219104611-00227-ip-10-171-10-108.ec2.internal.warc.gz | 697,756,563 | 10,431 | # Derive the formula for each given quantity in the following equations:a. r from c = 2rb. l from p = 21+ 2wc. h from A = 2h 3Vd. a from = a = c b de. s from c = s² m
1
by Drekkemaus
2014-05-12T21:23:32+08:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
You just need to isolate the variable to be derived.
For #1, c = 2πr
to isolate r, you need to divide the equation by 2π.
Therefore, r = c/2π
Try doing the other items. | 200 | 690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-09 | latest | en | 0.937564 |
https://netlib.org/lapack/explore-html/d7/d03/group__hetrs__aa__2stage_ga6c132b7009b4f8854d00191779bfcd7c.html | 1,719,296,493,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00528.warc.gz | 375,285,507 | 5,943 | LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
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## ◆ zhetrs_aa_2stage()
subroutine zhetrs_aa_2stage ( character uplo, integer n, integer nrhs, complex*16, dimension( lda, * ) a, integer lda, complex*16, dimension( * ) tb, integer ltb, integer, dimension( * ) ipiv, integer, dimension( * ) ipiv2, complex*16, dimension( ldb, * ) b, integer ldb, integer info )
ZHETRS_AA_2STAGE
Download ZHETRS_AA_2STAGE + dependencies [TGZ] [ZIP] [TXT]
Purpose:
``` ZHETRS_AA_2STAGE solves a system of linear equations A*X = B with a
hermitian matrix A using the factorization A = U**H*T*U or
A = L*T*L**H computed by ZHETRF_AA_2STAGE.```
Parameters
[in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the details of the factorization are stored as an upper or lower triangular matrix. = 'U': Upper triangular, form is A = U**H*T*U; = 'L': Lower triangular, form is A = L*T*L**H.``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in] NRHS ``` NRHS is INTEGER The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0.``` [in] A ``` A is COMPLEX*16 array, dimension (LDA,N) Details of factors computed by ZHETRF_AA_2STAGE.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [out] TB ``` TB is COMPLEX*16 array, dimension (LTB) Details of factors computed by ZHETRF_AA_2STAGE.``` [in] LTB ``` LTB is INTEGER The size of the array TB. LTB >= 4*N.``` [in] IPIV ``` IPIV is INTEGER array, dimension (N) Details of the interchanges as computed by ZHETRF_AA_2STAGE.``` [in] IPIV2 ``` IPIV2 is INTEGER array, dimension (N) Details of the interchanges as computed by ZHETRF_AA_2STAGE.``` [in,out] B ``` B is COMPLEX*16 array, dimension (LDB,NRHS) On entry, the right hand side matrix B. On exit, the solution matrix X.``` [in] LDB ``` LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N).``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```
Definition at line 139 of file zhetrs_aa_2stage.f.
141*
142* -- LAPACK computational routine --
143* -- LAPACK is a software package provided by Univ. of Tennessee, --
144* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
145*
146 IMPLICIT NONE
147*
148* .. Scalar Arguments ..
149 CHARACTER UPLO
150 INTEGER N, NRHS, LDA, LTB, LDB, INFO
151* ..
152* .. Array Arguments ..
153 INTEGER IPIV( * ), IPIV2( * )
154 COMPLEX*16 A( LDA, * ), TB( * ), B( LDB, * )
155* ..
156*
157* =====================================================================
158*
159 COMPLEX*16 ONE
160 parameter( one = ( 1.0d+0, 0.0d+0 ) )
161* ..
162* .. Local Scalars ..
163 INTEGER LDTB, NB
164 LOGICAL UPPER
165* ..
166* .. External Functions ..
167 LOGICAL LSAME
168 EXTERNAL lsame
169* ..
170* .. External Subroutines ..
171 EXTERNAL zgbtrs, zlaswp, ztrsm, xerbla
172* ..
173* .. Intrinsic Functions ..
174 INTRINSIC max
175* ..
176* .. Executable Statements ..
177*
178 info = 0
179 upper = lsame( uplo, 'U' )
180 IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
181 info = -1
182 ELSE IF( n.LT.0 ) THEN
183 info = -2
184 ELSE IF( nrhs.LT.0 ) THEN
185 info = -3
186 ELSE IF( lda.LT.max( 1, n ) ) THEN
187 info = -5
188 ELSE IF( ltb.LT.( 4*n ) ) THEN
189 info = -7
190 ELSE IF( ldb.LT.max( 1, n ) ) THEN
191 info = -11
192 END IF
193 IF( info.NE.0 ) THEN
194 CALL xerbla( 'ZHETRS_AA_2STAGE', -info )
195 RETURN
196 END IF
197*
198* Quick return if possible
199*
200 IF( n.EQ.0 .OR. nrhs.EQ.0 )
201 \$ RETURN
202*
203* Read NB and compute LDTB
204*
205 nb = int( tb( 1 ) )
206 ldtb = ltb/n
207*
208 IF( upper ) THEN
209*
210* Solve A*X = B, where A = U**H*T*U.
211*
212 IF( n.GT.nb ) THEN
213*
214* Pivot, P**T * B -> B
215*
216 CALL zlaswp( nrhs, b, ldb, nb+1, n, ipiv, 1 )
217*
218* Compute (U**H \ B) -> B [ (U**H \P**T * B) ]
219*
220 CALL ztrsm( 'L', 'U', 'C', 'U', n-nb, nrhs, one, a(1, nb+1),
221 \$ lda, b(nb+1, 1), ldb)
222*
223 END IF
224*
225* Compute T \ B -> B [ T \ (U**H \P**T * B) ]
226*
227 CALL zgbtrs( 'N', n, nb, nb, nrhs, tb, ldtb, ipiv2, b, ldb,
228 \$ info)
229 IF( n.GT.nb ) THEN
230*
231* Compute (U \ B) -> B [ U \ (T \ (U**H \P**T * B) ) ]
232*
233 CALL ztrsm( 'L', 'U', 'N', 'U', n-nb, nrhs, one, a(1, nb+1),
234 \$ lda, b(nb+1, 1), ldb)
235*
236* Pivot, P * B -> B [ P * (U \ (T \ (U**H \P**T * B) )) ]
237*
238 CALL zlaswp( nrhs, b, ldb, nb+1, n, ipiv, -1 )
239*
240 END IF
241*
242 ELSE
243*
244* Solve A*X = B, where A = L*T*L**H.
245*
246 IF( n.GT.nb ) THEN
247*
248* Pivot, P**T * B -> B
249*
250 CALL zlaswp( nrhs, b, ldb, nb+1, n, ipiv, 1 )
251*
252* Compute (L \ B) -> B [ (L \P**T * B) ]
253*
254 CALL ztrsm( 'L', 'L', 'N', 'U', n-nb, nrhs, one, a(nb+1, 1),
255 \$ lda, b(nb+1, 1), ldb)
256*
257 END IF
258*
259* Compute T \ B -> B [ T \ (L \P**T * B) ]
260*
261 CALL zgbtrs( 'N', n, nb, nb, nrhs, tb, ldtb, ipiv2, b, ldb,
262 \$ info)
263 IF( n.GT.nb ) THEN
264*
265* Compute (L**H \ B) -> B [ L**H \ (T \ (L \P**T * B) ) ]
266*
267 CALL ztrsm( 'L', 'L', 'C', 'U', n-nb, nrhs, one, a(nb+1, 1),
268 \$ lda, b(nb+1, 1), ldb)
269*
270* Pivot, P * B -> B [ P * (L**H \ (T \ (L \P**T * B) )) ]
271*
272 CALL zlaswp( nrhs, b, ldb, nb+1, n, ipiv, -1 )
273*
274 END IF
275 END IF
276*
277 RETURN
278*
279* End of ZHETRS_AA_2STAGE
280*
subroutine xerbla(srname, info)
Definition cblat2.f:3285
subroutine zgbtrs(trans, n, kl, ku, nrhs, ab, ldab, ipiv, b, ldb, info)
ZGBTRS
Definition zgbtrs.f:138
subroutine zlaswp(n, a, lda, k1, k2, ipiv, incx)
ZLASWP performs a series of row interchanges on a general rectangular matrix.
Definition zlaswp.f:115
logical function lsame(ca, cb)
LSAME
Definition lsame.f:48
subroutine ztrsm(side, uplo, transa, diag, m, n, alpha, a, lda, b, ldb)
ZTRSM
Definition ztrsm.f:180
Here is the call graph for this function:
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The sum of the measures of the interior angles of a triangle is 180 degrees. TRUE.
s
Question
Updated 9/26/2015 3:14:44 PM
Edited by jeifunk [9/26/2015 2:58:05 PM]
Rating
8
The sum of the measures of the interior angles of a triangle is 180 degrees. TRUE.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 640 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.87177 |
https://infogalactic.com/info/Regenerative_process | 1,537,938,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163326.85/warc/CC-MAIN-20180926041849-20180926062249-00088.warc.gz | 532,935,627 | 14,241 | # Regenerative process
Regenerative processes have been used to model problems in inventory control. The inventory in a warehouse such as this one decreases via a stochastic process due to sales until it gets replenished by a new order.[1]
In applied probability, a regenerative process is a class of stochastic process with the property that certain portions of the process can be treated as being statistically independent of each other.[2] This property can be used in the derivation of theoretical properties of such processes.
## History
Regenerative processes were first defined by Walter L. Smith in Proceedings of the Royal Society A in 1955.[3][4]
## Definition
A regenerative process is a stochastic process with time points at which, from a probabilistic point of view, the process restarts itself.[2] These time point may themselves be determined by the evolution of the process. That is to say, the process {X(t), t ≥ 0} is a regenerative process if there exist time points 0 ≤ T0 < T1 < T2 < ... such that the post-Tk process {X(Tk + t) : t ≥ 0}
• has the same distribution as the post-T0 process {X(T0 + t) : t ≥ 0}
• is independent of the pre-Tk process {X(t) : 0 ≤ t < Tk}
for k ≥ 1.[5] Intuitively this means a regenerative process can be split into i.i.d. cycles.[6]
When T0 = 0, X(t) is called a nondelayed regenerative process. Else, the process is called a delayed regenerative process.[5]
## Properties
$\lim_{t \to \infty} \frac{1}{t}\int_0^t X(s) ds= \frac{\mathbb{E}[R]}{\mathbb{E}[\tau]}.$
where $\tau$ is the length of the first cycle and $R=\int_0^\tau X(s) ds$ is the value over the first cycle.
• A measurable function of a regenerative process is a regenerative process with the same regeneration time[7]
## References
1. Hurter, A. P.; Kaminsky, F. C. (1967). "An Application of Regenerative Stochastic Processes to a Problem in Inventory Control". Operations Research. 15 (3): 467. JSTOR 168455. doi:10.1287/opre.15.3.467.
2. Ross, S. M. (2010). "Renewal Theory and Its Applications". Introduction to Probability Models. pp. 421–641. ISBN 9780123756862. doi:10.1016/B978-0-12-375686-2.00003-0. Cite error: Invalid <ref> tag; name "Ross" defined multiple times with different content
3. Schellhaas, Helmut (1979). "Semi-Regenerative Processes with Unbounded Rewards". Mathematics of Operations Research. 4: 70–78. JSTOR 3689240. doi:10.1287/moor.4.1.70.
4. Smith, W. L. (1955). "Regenerative Stochastic Processes". Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences. 232 (1188): 6–4. Bibcode:1955RSPSA.232....6S. doi:10.1098/rspa.1955.0198.
5. Haas, Peter J. (2002). "Regenerative Simulation". Stochastic Petri Nets. Springer Series in Operations Research and Financial Engineering. pp. 189–273. ISBN 0-387-95445-7. doi:10.1007/0-387-21552-2_6.
6. Asmussen, Søren (2003). "Regenerative Processes". Applied Probability and Queues. Stochastic Modelling and Applied Probability. 51. pp. 168–185. ISBN 978-0-387-00211-8. doi:10.1007/0-387-21525-5_6.
7. Sigman, Karl (2009) Regenerative Processes, lecture notes | 915 | 3,085 | {"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-39 | latest | en | 0.829235 |
https://superioressaywriters.com/2018/06/13/the-hypothesis-tester/ | 1,550,886,190,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249414450.79/warc/CC-MAIN-20190223001001-20190223023001-00625.warc.gz | 709,247,989 | 13,718 | ## the Hypothesis Tester
In 2010, Playbill Magazine, a regionally based magazine, consulted Boos Allen to determine the mean annual household income of its readers-
Using a list of customers provided by Playbill, Boos Allen randomly sampled 300 Playbill customers- From that sample, Boos Allen is confident that
the average Playbill reader’s household income is \$119,155, with a population sample household income standard deviation of \$30,000-
Recently, two Playbill executives suggested that the mean average household income for Playbill readers has increased, and the magazine price
should be raised- As new marketing manager for Playbill, you convince the chief operating officer to complete a second survey with Boos Allen to
confirm that assertion- Yesterday, the new Boos Allen report appeared on your desk- From a new sample of Playbill customers, taken from a list of
recent customers you e-mailed to Boos Allen, the 2012 Playbill customer profile shows a mean annual household income of \$124,450, with a
population standard deviation of household income unchanged at \$30,000-
You realize that you have enough data to perform a one-sample hypothesis test- In order to choose the correct statistical tool to complete this
assessment, consider the following questions:
What do you know about the population being studied? (For example, do you know the standard deviation of the overall population?)
What do you know about the population sample you chose? (For example, which sample statistical parameters do you have, such as mean,
standard deviation, variance, and so on?)
Directions
Solve the following equations, based on the data from the Hypothesis Tester – Single Sample file- Demonstrate your method and display your results
in table format, using Excel or another appropriate computer application-
Identify the null hypothesis, via both a written explanation and a math equation-
Deterrnine the alternative hypothesis, via both a written explanation and a math equation-
Solve the equation to determine whether to accept or reject the null hypothesis-
Deterrnine whether the p-value indicates acceptance or rejection of the null- Use alpha = -05-
Next, address the following in a report to the executives:
Write a three-sentence paragraph that details your recommendations for a course of action, based on your results-
Report the rejection or acceptance of the null, in terms of the scenario results-
Explain why you can be statistically confident that the mean household Playbill reader’s income has increased, decreased, or remained the same-
State whether you would propose that the cost of the magazine be raised-
Describe what would happen if alpha was -01-
Compile your work and report in a 1-2 page Microsoft Word file:
Paste in the tables you used to make your calculations-
Clearly title your tables, including each row and column-
Highlight the results of your data calculations within each table- | 588 | 2,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.914476 |
https://questions.examside.com/past-years/jee/question/if-yx--cot--1left-sqrt-1--sin-x--s-jee-main-mathematics-trigonometric-functions-and-equations-5n4dcc0zbld9yfii | 1,721,281,129,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00380.warc.gz | 416,115,746 | 46,663 | 1
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :
A
$$- {1 \over 2}$$
B
$$-$$1
C
$${1 \over 2}$$
D
0
2
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
A
$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
B
$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
C
$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
D
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
3
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
The derivative of
$${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$$ with
respect to $${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$$ at x = $${1 \over 2}$$ is :
A
$${{2\sqrt 3 } \over 3}$$
B
$${{2\sqrt 3 } \over 5}$$
C
$${{\sqrt 3 } \over {10}}$$
D
$${{\sqrt 3 } \over {12}}$$
4
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$
where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is :
A
$${{a - 2b} \over {a + 2b}}$$
B
$${{a - b} \over {a + b}}$$
C
$${{a + b} \over {a - b}}$$
D
$${{2a + b} \over {2a - b}}$$
EXAM MAP
Medical
NEET | 779 | 1,493 | {"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} | 3.484375 | 3 | CC-MAIN-2024-30 | latest | en | 0.417165 |
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Grade 2 - Mathematics3.2 Money Combinations Using Pictures
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Grade 2 - Mathematics3.2 Money Combinations Using Pictures
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For unlimited printable worksheets & more, go to http://www.kwizNET.com. | 335 | 1,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-24 | latest | en | 0.799252 |
https://www.cfd-online.com/Forums/openfoam/92608-limiting-turbulent-viscosity-print.html | 1,513,406,523,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948583808.69/warc/CC-MAIN-20171216045655-20171216071655-00049.warc.gz | 716,697,121 | 2,621 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- OpenFOAM (https://www.cfd-online.com/Forums/openfoam/)
- - Limiting turbulent viscosity (https://www.cfd-online.com/Forums/openfoam/92608-limiting-turbulent-viscosity.html)
cm_jubayer September 19, 2011 17:24
Limiting turbulent viscosity
Hi,
I want to limit turbulent viscosity but I don't know how to do that in OpenFOAM. Can someone please help me with this :confused:? Thanks.
Jubayer
cm_jubayer September 21, 2011 14:27
Hi, I am using pisoFoam and my fvSchemes is as follows:
ddtSchemes
{
default Euler;
}
{
default cellMDLimited Gauss linear 0.5;
// grad(U) cellLimited Gauss linear 1;
}
divSchemes
{
default none;
div(phi,U) Gauss limitedLinearV 1;
div(phi,k) Gauss limitedLinear 1;
div(phi,omega) Gauss limitedLinear 1;
}
laplacianSchemes
{
default none;
laplacian(nuEff,U) Gauss linear limited 0.5;
laplacian((1|A(U)),p) Gauss linear limited 0.5;
laplacian(DkEff,k) Gauss linear limited 0.5;
laplacian(DomegaEff,omega) Gauss linear limited 0.5;
}
interpolationSchemes
{
default linear;
interpolate(U) linear;
}
{
default limited 0.5;
}
fluxRequired
{
default no;
p;
}
What should I do if I want to limit my nu to specific values like 0 to 1e6?
Jubayer
vkrastev September 22, 2011 06:20
In order to limit the turbulent viscosity you have to modify the source file related to the turbulence model you are using, by adding a limiter in the eddy (turbulent) viscosity calculation formula. Anyway, I don't know what is your application, but you should be careful in adding an arbitrary limiter in a pre-existing turbulence model.
V.
cm_jubayer September 22, 2011 09:54
Jubayer
cm_jubayer September 29, 2011 09:54
Hi,
To bound nut, I have added this line to the LaunderSharmaKE model,
bound(nut_, dimensionedScalar("0", nut_.dimensions(), 10.0));
After compiling and everything, at each time step it showed me the bounding values. But this is not actually bounding the values as I can see that the nut values are going above 10. Can someone please explain me how this bounding actually works? Thanks.
Jubayer
cheng1988sjtu April 6, 2015 14:18
This depends on how to calculate the eddy viscosity, usually the unexpected large eddy viscosity is caused by the deviding a relative small value, for example, in k-Epsilon model, nut = C*k^2/epsilon, to limit the eddy viscosity, an effective way is to limit the smallest epsilon value (but not unphysically large).
Charlie
All times are GMT -4. The time now is 02:42. | 709 | 2,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-51 | latest | en | 0.74627 |
https://forum.allaboutcircuits.com/threads/audio-pre-amplifier-design-please-can-you-tell-me-if-my-theory-is-correct.104247/ | 1,627,083,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00107.warc.gz | 288,835,608 | 25,069 | # Audio Pre- Amplifier Design - Please can you tell me if my theory is correct?
#### Solar shock
Joined Dec 4, 2014
17
Hey guys, (sad face - I got an 'alert' and I just lost this post when I was 90% done, here we go again )
First off, id just like to say; this is my first post, but been a occasional lurker when on the hunt for the ever elusive electronics knowledge, so a big thank you for all the help you have given me; if a little unknowingly. I am predominately looking to consolidate my knowledge and to help it grow about my circuit; but more about that below.
The task I was given: Audio Pre-amplifier Design
The aim is to design and test a pre-amp circuit, with the following specifications:
1. 300Hz to 3.4kHz bandwidth
2. 'in-band' sensitivity of 2V/mV with a max input of +/- 15mV.
3. 'in-band' input impedance should be well defined at 50kohm and output load min value is 2kohm.
4. power supply is +Vcc and 0.
What I have achieved thus far:
I have a working circuit (well eventually i did ), it achieves close to a gain of 2000, the resistors on the first stage are a little out and I only get 225 instead of 250, so when multiplied by the second stage of 8, I end up with about 1800, but this is ok, as the lecturer likes stuff to be a little wrong, as then he likes to see you explain why it is so. For a 1mV input I get close to 2V out.
It runs of a single supply, however I haven't drawn the 741's supply lines in, they go to Vcc and 0.
What Id like help with: Can you please assess my theory of my circuit that I am about to give and then let me know where I am wrong, or areas that can be improved. I have used blue text for the areas that I feel I am most uncertain about.
My Theory/Knowledge:
• The input to the first stage is acting as my DC offset, due to the potential divider their is a DC component added to my AC signal in that is Vcc/2. This is what allows my 741 to swing within a positive voltage range. It also forms part of my HPF, with a total resistance of 50kohm and the 10nF capacitor I get a HPF that has F1 = 318 Hz.
• Due to the virtual earth created by the op-amp, the negative input will attempt to make the difference between the two inputs 0, therefore the negative input will also be at the DC offset of Vcc/2. so it works like a differential amplifier and amplifies my AC input wave + and a little noise to some extent?
• Circle 3 - I am not 100% on what this is achieving? Is this helping to remove any high frequency AC noise that is appearing on the output? by allowing it to pass to ground when the Xc gets small enough?
• The gain of stage 1 is set by the resistors, but is not affected by the capacitor to ground in circle 3? surely at lower frequencies (for example Xc = 4k at 400Hz) the Xc is adjusting the gain of stage 1; as R2 is now effectively 2.2k + Xc?
• Circle 1 - Acts as DC blocking; removing any DC offset including any DC offset errors introduced by the op-amp. It also affects the gain of the second stage; as its Xc value affects the resistance total of R4. ( i found this out, as originally I used a 0.1uF, and found that my Xc at 1500Hz was about 1k, therefore my R4 was about 2.2k, and my output was about half of what I was expecting) - Is that correct? I believe its right, as i solved the issue by making it the 1uF which gave me a much lower Xc which then had a much lower impact on R4 and thus the gain of the second stage.)
• The DC offset is then reintroducted on the positive input of the second stage, again due to the inputs wanting to be equal the negative input is also then up at Vcc/2.
• Circle 2 - This capacitor is acting purely as de-coupling, its there to remove any AC noise that may have gotten onto the supply rails; as this noise would then be amplified / affect the amplification of my wanted signal?
Any extra questions I have:
I have found that the bandwidth is actually about 600Hz to 2.8kHz. Is this reduction due to the capacitors in my circuit?
Responses:
Hey guys, well if you made it this far then thank you for taking the time to read my rather long post. I would ideally like to discuss the answers rather than be given them, so keywords, or areas I should go look up would be greatly appreciated, as I need to write my report on this circuit. I also have quite a few books at my disposal; operational amplifiers by G clayton, op amps for everyone by B carter. So if there is a topic in there I should consult please let me know.
also any knowledge that im completely missing or questions you have do say!
Kindest Regards
Robert
#### Attachments
• 67.1 KB Views: 34
#### Jony130
Joined Feb 17, 2009
5,241
All your capacitors except capacitor in circle 2 has a effect on the amplifier bandwidth.
Circle 3 capacitor is needed because the op-amp is supply from a single power supply. So we need to bias the op amp somewhere in the middle of his "linear region". This is a job for input voltage divider (Ra, Rb). Without the capacitor the output DC voltage will be equal to Vcc (positive sat voltage). But we have this capacitor in the circuit, so DC voltage gain is reduce to one. And the output voltage is equal to 6V.
Also this capacitor from a filter with R2 resistor. Fc = 0.16/R*C = 0.16/(2.2kΩ * 100nF) = 728Hz
Circle 1 capacitor with R4 also form a high pass filter, Fc = 0.16/R*C = 0.16/(1.2k * 1μ) = 133Hz
Circle 2 has now influence amplifier bandwidth. But you should add 10μF or larger capacitor parallel to 0.1μF = 100nF.
Input capacitor (10nF) also is a high pass filter. Fc = 0.16/(Rb||Ra * Cin ) = 0.16/(50kΩ * 10nF) = 320Hz
And finally 4.7nF capacitor with R3 forms low pass filter, F = 0.16/(10kΩ*4.7nF) = 3.4khz
So if you want 300Hz to 3.4kHz bandwidth set all high pass filter at 30Hz or lower except Cin with Ra and Rb.
This filter set to F = 0.16/(Cin * Ra||Rb) = 300Hz.
Also notice that your opamp is very old and has a gain bandwidth product around 1Mhz. Additional your first stage amp will have a very large gain (1 +R1/R2 = 255V/V) so the bandwidth of this stage is reduce to F = 1Mhz/255 = 3.9kHz.
And this is why your measured upper frequency limit is at 2.8Khz instead of 3.4khz. So to reduce this effect try set 0.16/(10kΩ*4.7nF) at 4.2khz or larger.
#### Attachments
• 14.8 KB Views: 44
#### Solar shock
Joined Dec 4, 2014
17
Thanks for the response, I think I understand most of it, but there are some areas I am not sure of;
Circle 3 capacitor is needed because the op-amp is supply from a single power supply. So we need to bias the op amp somewhere in the middle of his "linear region". This is a job for input voltage divider (Ra, Rb). Without the capacitor the output DC voltage will be equal to Vcc (positive sat voltage). But we have this capacitor in the circuit, so DC voltage gain is reduce to one. And the output voltage is equal to 6V.
I am still unsure what capacitor in circle 3 is doing - As is it not capacitor in circle 1 that is blocking the DC component from the input of the second stage?
Also this capacitor from a filter with R2 resistor. Fc = 0.16/R*C = 0.16/(2.2kΩ * 100nF) = 728Hz.
I thought that to create a filter with the resistor the setup had to be with one of them in parallel with respect to the Vout?
Circle 1 capacitor with R4 also form a high pass filter, Fc = 0.16/R*C = 0.16/(1.2k * 1μ) = 133Hz.
See I thought that it was this capacitor that was doing the blocking of the DC offset in the stage 1 output. The capacitor would act as a high pass filter to some extent due to its reactance at low frequencies (so at low frequencies the reactance attenuates the signal), but at the same time it acts to completely block the DC component of the output of stage 1. This is important then, as it prevents the DC bias of the stage 1 affecting the DC bias of stage 2.
Circle 2 has now influence amplifier bandwidth. But you should add 10μF or larger capacitor parallel to 0.1μF = 100nF.
Ok, why do you suggest I add another capacitor in parallel to the 0.1uF? surely that would just give me a combined capacitance of: 10uF + 0.1uF? So the additonal capacitor would be almost un affected by the original already there? Why would that be any different from simply replacing it with a 10uF?
Input capacitor (10nF) also is a high pass filter. Fc = 0.16/(Rb||Ra * Cin ) = 0.16/(50kΩ * 10nF) = 320Hz
And finally 4.7nF capacitor with R3 forms low pass filter, F = 0.16/(10kΩ*4.7nF) = 3.4khz.[/QUOTE]
Yep, completely get that
So if you want 300Hz to 3.4kHz bandwidth set all high pass filter at 30Hz or lower except Cin with Ra and Rb.
This filter set to F = 0.16/(Cin * Ra||Rb) = 300Hz..
Ah ok, so with this your saying, to ensure that I keep the bandwidth at the levels I want I need to make sure that all of the high pass filters are only filtering below the value of the HPF at the input. I get that, makes logical sense
Also notice that your opamp is very old and has a gain bandwidth product around 1Mhz. Additional your first stage amp will have a very large gain (1 +R1/R2 = 255V/V) so the bandwidth of this stage is reduce to F = 1Mhz/255 = 3.9kHz.
And this is why your measured upper frequency limit is at 2.8Khz instead of 3.4khz. So to reduce this effect try set 0.16/(10kΩ*4.7nF) at 4.2khz or larger.
I understand how you calculated the bandwidth of the stage 1 being 3.9kHz, but why does that mean the upper frequency limit is at 2.8kHz?
Surely if the upper limit is 3.9kHz and I am trying to achieve 3.4 that is ok? Or is it then to do with the tolerances of my op-amp? so that while theoretically I should be able to achieve 3.9kHz, in reality the GBP of the op-amp isn't quite 1MHz, so with my gain remaining at 255, I am then having to sacrifice the bandwidth?
So like you suggested, to deal with that I should reduce the gain of the first stage, thus reducing the fact that I am at the GBP limit of the chip, allowing me my bandwidth?
#### Jony130
Joined Feb 17, 2009
5,241
I am still unsure what capacitor in circle 3 is doing - As is it not capacitor in circle 1 that is blocking the DC component from the input of the second stage?
Cin capacitor remove a DC component from the input signal only. But don't forget that Ra and Rb bias the op amp somewhere in the middle of his "linear region" Vcc/2. And this DC voltage is present at the non-inverting input. So we add C3 (circle 3) into the circuit, because we don't want to amplify any DC voltage present at the input of the amplifier. Without C3, any DC voltage present at the "+" input will also be amplified (1+ R1/R2) times. So by adding C3 we "reduce" DC voltage gain to one.
jony130 said:
Also this capacitor from a filter with R2 resistor. Fc = 0.16/R*C = 0.16/(2.2kΩ * 100nF) = 728Hz.
I thought that to create a filter with the resistor the setup had to be with one of them in parallel with respect to the Vout?
Look at this example
For DC signals and for low frequency Xc ≈ ∞. And this is why DC gain is 1V/V (1 + R1/(R2 + Xc)).
But as frequency rise, Xc decrease his value. And at frequency (F1) when Xc = (R1 + R2) amplifier gain start to rise from 1V/V and will reach 11V/V at frequency (F2) when Xc = R2. In reality the gain at F2 is -3dB lower from 11V/V (0.707*11V/V = 7.7V/V). The gain will reach 11V/V at 10*F2.
Ok, why do you suggest I add another capacitor in parallel to the 0.1uF? surely that would just give me a combined capacitance of: 10uF + 0.1uF? So the additonal capacitor would be almost un affected by the original already there? Why would that be any different from simply replacing it with a 10uF?
Because we want to short any signal present at this input to ground.
For example, without C2 any noise from Vcc will appear at the "+" input. And this noise will also be amplified by the amplifier gain (1 + R3/R4). So we add C2 to short any AC voltage present at the amp input.
As for the remaining part of the question please read this (why 100nF||10μF) | 3,293 | 11,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-31 | latest | en | 0.954492 |
http://en.wikibooks.org/wiki/Chess_Opening_Theory/1._e4/1...e5/2._Nf3/2...Nc6/3._d4/3...exd4/4._Bc4 | 1,433,216,927,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195035316.21/warc/CC-MAIN-20150601214355-00095-ip-10-180-206-219.ec2.internal.warc.gz | 60,590,645 | 12,375 | # Chess Opening Theory/1. e4/1...e5/2. Nf3/2...Nc6/3. d4/3...exd4/4. Bc4
< Chess Opening Theory | 1. e4 | 1...e5 | 2. Nf3 | 2...Nc6 | 3. d4 | 3...exd4
Scotch Gambit
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Position in Forsyth-Edwards Notation(FEN)
r1bqkbnr/pppp1ppp/2n5/8/2BpP3/5N2/PPP2PPP/RNBQK2R
# Scotch Gambit
Black can continue the Scotch with 4...Bc5 5.c3 and now 5...Nf6 will transpose into a safe variation of the Giuoco Piano. Alternatively, he can transpose into the Two Knights Defense with 4...Nf6 and White has now two sound options: the obvious 5. e5 and 5. O-O. In the first case 5. e5 white goes for maintaining a strong center and an open game with chances for both sides. 5. O-O gives away the pawn, and while taking it might seem risky due to 6.. Re1 it's perfectly sound, and white will only get his pawn back.
Black can instead accept the gambit with 5...dxc3 but this is riskier because White will gain a lead in development. A possible continuation is 6.Nxc3 (Grandmaster Sveshnikov has played 6.Bxf7+!? Kxf7 7.Qd5+ followed by 8.Qxc5) 6...d6 7.Qb3 Qd7 8.Nd5 Nge7 9.Qc3 0-0.Göring Gambit'. This gambit is quite similar to the Danish Gambit.
## Theory table
For explanation of theory tables see theory table and for notation see algebraic notation.
1. e4 e5 2. Nf3 Nc6 3. d4 exd4 4. Bc4
4 5
Two Knights Defence
by transposition
Bc4
Nf6
= to 1.e4 e5 2.Nf3 Nc6 3.Bc4 Nf6 4.d4 exd4
Scotch Gambit ...
Bc5
c3
Nf6
=
Hungarian Defence
by transposition
...
Be7
d4
-
= | 606 | 1,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.837639 |
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# Question 3 Prove that the points (a,0), (b,0) and (1,1) are collinear if 1/a+1/b=1.
Asked by 28th February 2013, 10:32 AM
Sometimes there can be printing mistakes in the book of sample papers also. If you will try to solve it using the section method and try to find the ratio, you will find that the ratio would come out to be 0 if you consider the points that you provided, which is not possible for collinear points. So, yeah, there can be mistakes everywhere. Sometimes NCERT books (even Board exam papers) also have mistakes, so, its alright.
If the question as I said would have been "Prove that the points (a,0), (0,b) and (1,1) are collinear if 1/a+1/b=1.", then you can solve it as
Let A(a,0), B(0,b) and C(1,1) be the 3 collinear points
Let B divide the line segment joining AC in the ratio of k:1
Then, by section formula, the coordinates of B would be given by (k+a/k+1 , k/k+1)
However, we already know that the coordinates of B is (0,b)
So, equating, 0 = k+a/k+1 i.e. k = -a
Also, b = k/k+1
Substituting k = -a, we get
b = -a/-a+1
i.e. 1/b = a-1/a
i.e. 1/b = 1-1/a
i.e. 1/a+1/b = 1
Hence, proved
Answered by Expert 28th February 2013, 6:35 PM
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You have rated this answer /10 | 508 | 1,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.933881 |
https://www.studypool.com/discuss/506861/describe-how-the-graph-of-the-function-changes-with-time?free | 1,508,844,813,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00864.warc.gz | 980,079,901 | 14,358 | ##### Describe how the graph of the function changes with time
label Algebra
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the exponential function g(t)=500(1-3-t) describes how long it takes for catfish to migrate from a pond. what is the meaning of g(0) ? what are the restraints on the types of numbers used for g(t)/
Apr 30th, 2015
hi, g(t)=500(1-3-t) is not an exponential function, revise the question and I will help you! thanks
Apr 30th, 2015
ok sorry g(t) = 500 (1-3negative t) I don't know how to do a negative exponent on here but that's what it is?
Apr 30th, 2015
you can use "^" as the exponent symbol
Apr 30th, 2015
thank you
Apr 30th, 2015
ok, so is it g(t)=500(1-3^(-t))???
Apr 30th, 2015
yes
Apr 30th, 2015
describes how long it takes for catfish to migrate from a pond.
this question is too vague to answer, there should be additional information related to it. The answer can be t if you have to write one.
what is the meaning of g(0) ?
g(0) is the initial position of the catfish, plug in 0 for g(t)=500(1-3^(-t)), we get 0.
what are the restraints on the types of numbers used for g(t)
t is the time, so t has to be greater than or equal to 0.
Let me know if you have any questions
Apr 30th, 2015
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Apr 30th, 2015
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Apr 30th, 2015
Oct 24th, 2017
check_circle | 417 | 1,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-43 | latest | en | 0.906525 |
http://www.cseblog.com/2014/07/ | 1,519,559,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00295.warc.gz | 427,998,689 | 23,724 | ## Posts
Showing posts from July, 2014
### 3D Tic Tac Toe Puzzle
Source: Shared by Alok Mittal (Cannan Partners)
Problem: A 3x3 tic tac toe has 8 "winning lines" (3 horizontal, 3 vertical and 2 diagonals). How many "winning lines" does the 3x3x3 3D tictactoe have? There is a brute force solution, and then there is the aha! solution.
Update (23 Oct 2014)
Solution: Posted in comments by Anti, Taz, Javier, Shubham Gupta, Leela. Detailed solution and much more advanced problems in the document http://library.msri.org/books/Book42/files/golomb.pdf
### Mad Robot Puzzle
Source:http://nrich.maths.org/
Problem:
A mad robot sets off towards the North East on a journey from the point (0,0) in a coordinate system. It travels in stages by moving forward and then rotating on the spot. It follows these pseudo-code instructions:
SUB JOURNEY
DISTANCE = 1000
WHILE (DISTANCE > 0.001)
MOVE DISTANCE
STOP
ROTATE(90, DEGREES, CLOCKWISE)
DISTANCE = DISTANCE / 2
END WHILE
EXPLODE
END SUB
Where does the robot explode?
Update (23 Oct 2014):
Solution: Posted by me (Pratik Poddar) in comments!
### Social Network Friendship Paradox
Problem / Observation:
The friendship paradox is the phenomenon first observed by the sociologist Scott L. Feld in 1991 that most people have fewer friends than their friends have, on average. Prove it mathematically.
Update (23 Oct 2014):
Solution: Posted by Mike Earnest and Taz in comments! | 392 | 1,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-09 | longest | en | 0.843938 |
www.jrtsportsfitness.com | 1,394,253,495,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999653106/warc/CC-MAIN-20140305060733-00087-ip-10-183-142-35.ec2.internal.warc.gz | 397,702,795 | 4,334 | Bodyfat Percentage Calculator - Calculate Bodyfat %
General Options: Home Articles Workout Exercises Free Information: Free Weight Loss Plan Top Protein Bars Top Protein Powders Top Weight Gainer Supplements Free Tools: Calorie Burning Calculator Calories Calculator Contact:
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# Bodyfat Percentage Calculator
feet inches
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THE NEXT STEP: Once you know your Bodyfat %, use our Calories Calculator to find out your Basal Metabolic Rate (BMR), Lean Body Mass, and what your Daily Recommended Calorie Intake should be in order to meet your weight loss goals.
REMEMBER: There are two guaranteed ways to lose weight: One is to eat less food than your body needs, causing it to break down fat for energy, the other is to exercise more, so that your body demands even more energy. Of course, the ideal scenario is to do both: eat sensibly, and exercise regularly, then you'll soon be able to get your perfect figure.
IMPORTANT HEALTH NOTICE: This site and its contents are for information only, and are not intended to override the advice of a medical professional. If you are in any doubt about your general health and fitness levels (including your Bodyfat %), you should consult a medical professional before embarking on any extreme form of diet or weight loss exercise programme. The important thing is to look after your body, it's the only one you have! | 725 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | longest | en | 0.882587 |
https://physics.stackexchange.com/questions/515901/laminar-vs-turbulent-flow-in-a-partially-filled-pipe | 1,656,385,666,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00494.warc.gz | 507,743,134 | 65,444 | # Laminar vs. Turbulent flow in a partially filled pipe
I've recently been learning about how to use Reynolds number to estimate if a fluid flow is laminar or turbulent while flowing through a pipe, and it has got me wondering if there is a similar way to tell if a flow is laminar or turbulent in a partially filled pipe.
I'm not sure if this makes a difference, but I'm assuming the pipe is half filled and cylindrical.
Update:
Reynolds number for a full pipe is $$\frac{\rho{V}{d}}{\mu}$$, where $$\rho$$ is density, $$V$$ is velocity of the fluid, $$d$$ is diameter of the pipe, and $$\mu$$ is viscosity. How would this formula change for a half pipe?
• Sure, if there is geometric similarity (pipes of the same cross-section shape, filled to the same fraction of the diameter) then transition to turbulence will occur at the same Reynolds number. Nov 24, 2019 at 21:41
• @MaximUmansky, how would the Reynolds number be calculated for a partially filled pipe? Would the diameter be the diameter of the full pipe, the average diameter of the half pipe, or something else? Nov 24, 2019 at 23:13
• It does not really matter, as long as a consistent definition is used. You can use the full pipe diameter, that's convenient. Nov 25, 2019 at 19:53
Interesting question. I briefly skimmed through this paper which described this phenomenon.
They were able to model the flow by introducing a quantity known as the equivalent diameter, which was the diameter of a tube that would have the same cross sectional area as the "partially filled tube", given by:
$$D_\text{eq} = 2\sqrt{\frac{A}{\pi}}$$
where $$A$$ was the cross sectional area of the flow. Substituting in $$A=\frac{1}{2}\pi R^2$$ gives:
$$D_\text{eq} = \frac{d}{\sqrt{2}}$$
so we can say the Reynolds number is:
$$\text{Re} = \frac{\rho V d}{\sqrt{2}\mu}$$
for a half filled pipe. | 485 | 1,851 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-27 | longest | en | 0.942383 |
https://grandpaperwriters.com/question-s-name-a-team-name-and-a-jersey-number-for/ | 1,696,439,777,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00134.warc.gz | 314,105,357 | 11,944 | # Question & Answer: The Blossom Company sells sports decals that can be personalized with a player's name, a team name, and a jersey number for…..
Really need help with this question. Thanks!
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The Blossom Company sells sports decals that can be personalized with a player’s name, a team name, and a jersey number for \$8.00 each. Blossom buys the decals from a supplier for \$2.60 each and spends an additional \$0.50 in variable operating costs per decal. The results of last month’s operations are as follows: Calculate contribution margin per unit. (Round answer to 2 decimal places, e.g. 0.38.) Contribution margin per unit \$ per unit What is Blossom’s monthly breakeven point in units? In dollars? (Use your answer of breakeven units to calculate the breakeven point in dollars. Round Breakeven units and point in dollar to 0 decimal places, e.g. 25,000.) Breakeven point units Breakeven sales \$ What is Blossom’s margin of safety? (Round answers to 0 decimal places, e.g. 25,000.) Margin of safety units Margin of safety \$
Breakeven point=Fixed cost/contribution margin per unit
=\$3,600-(28000*0.5/8)/\$4.9
=378
Breakeven point in dollars=378*\$8
=\$3,024
Margin of safety units=(28000/8)-378
=3122
Margin of safety =\$28000-\$3024
=\$24,976 | 400 | 1,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-40 | latest | en | 0.795899 |
https://www.physicsforums.com/threads/statistical-thermodynamics-multiple-questions.794492/ | 1,709,508,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00858.warc.gz | 932,987,441 | 16,916 | # Statistical Thermodynamics (multiple questions)
• vitom001
In summary, the conversation discusses the relationship between particle density and external potential in a gravity field. It also explores the characteristic function for a system with an interface and fixed pressure or volume. The final equations show that the characteristic function is related to the energy stored in the interface.
## Homework Statement
1.) For N particles in a gravity field, the Hamiltonian has a contribution of external potential only (-mgh). Show that the particle density follows the barometric height equation (1).
2.) For N particles in a open system at constant pressure p and temperature T, let there be an interface with area A in the system (eg. air-water), that is (μ, p, A, T). N, V and U fluctuate in this sytem. Identify the characteristic function X.(2)
3.) Same as above, only for a system where V instead of p is fixed.
## Homework Equations
1.) ρ(h) = ρ(0)·exp(-mgh/kT), with m mass of particle, g gravitational constant, h barometric height, k boltzmann's constant and T temperature.
2.) X = - kT·lnΔ, where Δ is the partition function for the grand canonical ensemble.
3.) Same as above.
## The Attempt at a Solution
1.) From: <N> = kT ∂lnΞ/∂μ = Lx·Ly·∫ρ(h)dh
I’m supposed to arrive at: ρ(h) = ∫[exp(μ/kT)/Λ3]·exp(-mgh/kT) dh
And finally to: ρ(h) = ρ(0) ·exp(-mgh/kT), where ρ(0) = exp(μ/kT)/Λ3
I understand the setup with the expectation value of N, however the transition to the integral and getting the ρ(0) out of the integral is not quite clear to me.
Also, Ξ is the grand-canonical partition function in this case, while Lx·Ly is neglected since they're both uniform and only the barometric height h is considered.2.) By using the equation X = - kT·lnΔ, with Δ = exp(S/k)·exp(-U/kT)·exp(μN/kT)·exp(-pV/kT), I arrive at
X = -ST + U - μN + pV
X = [ (U + pV) - ST ] - μN
X = (H-TS) -μN = G - μN
However, I know the characteristic function is supposed to be X = γ·A, so the energy stored in the interface. I do not understand how I arrive at that conclusion.
3.) Same procedure as above, only here the volume is fixed instead of the pressure, so the derivation should go through the helmholtz energy. I use the same formula X = - kT·lnΔ, however I'm also supposed to implement dF = -SdT - pdV + μdN + γdA, so to my guess the partition function may look something like this:
Δ = exp(S/k)·exp(-U/kT)·exp(μN/kT)·exp(-pV/kT)·exp(γA/kT)
which then gives:
X = (H-TS) -μN + γA = G - μN + γA
As with the previous question, I don't know how I should conclude from this what the characteristic function is.
Actually I managed to figure it out in the end, so thread can be closed.
## 1. What is statistical thermodynamics?
Statistical thermodynamics is a branch of physics that combines the principles of thermodynamics with statistical mechanics to explain and predict the behavior of thermodynamic systems on a microscopic level.
## 2. How is statistical thermodynamics different from classical thermodynamics?
Classical thermodynamics deals with macroscopic systems and their overall properties, while statistical thermodynamics focuses on the behavior and properties of individual particles within a system.
## 3. What are the key principles of statistical thermodynamics?
The key principles of statistical thermodynamics include the Boltzmann distribution, which describes the distribution of particles in a system, and the concept of entropy, which measures the disorder of a system.
## 4. What are the main applications of statistical thermodynamics?
Statistical thermodynamics has numerous applications in fields such as chemistry, biology, and materials science. It is used to understand and predict the behavior of gases, liquids, and solids, as well as chemical reactions and phase transitions.
## 5. How does statistical thermodynamics relate to quantum mechanics?
Statistical thermodynamics is based on the principles of quantum mechanics, which describes the behavior of particles on a subatomic level. It uses concepts such as energy levels and wave functions to explain the behavior of particles within a system. | 1,030 | 4,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.843823 |
https://examfear.com/notes/Class-11/Maths/Probability/372/Events.htm | 1,571,079,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00074.warc.gz | 487,464,182 | 7,152 | Class 11 Maths Probability Events
Event: It is the set of favorable outcome. Also, Event is a subset of the sample space. E.g. Event of getting odd outcome in a throw of a die
Types of Event
• Impossible and Sure Events
• Simple Event
• Compound Event
Impossible event is denoted by φ, while Sure Event is denoted by S.
E.g. in rolling a die, impossible event is that number is more than 6 & Sure event is the event of getting number less than or equal to 6.
Simple Event has only one sample point of a sample space.
E.g. in rolling a die, Simple event could be the event of getting 4.
Compound Event has more than one sample points of a sample space.
E.g. in rolling a die, Simple event could be the event of getting even number
Algebra of Events
• Complementary Event
• Event ‘A or B’
• Event ‘A and B’
• Event ‘A but not B
Complementary Event
Complementary event to A= ‘not A’
Example: If event A= Event of getting odd number in throw of a die, that is {1, 3, 5}
Complementary event to A = Event of getting even number in throw of a die, that is {2, 4, 6}
Or A’ = S- A (where S is the Sample Space)
Event (A or B)
Union of two sets A and B denoted by A ∪ B contains all those elements which are either in A or in B or in both.
When the sets A and B are two events associated with a sample space, then ‘A ∪ B’ is the event ‘either A or B or both’. This event ‘A ∪ B’ is also called ‘A or B’.
Event ‘A or B’ = A ∪ B = {ω : ω ∈ A or ω ∈ B}.
Event ‘A and B’
Intersection of two sets A ∩ B is the set of those elements which are common to both A and B. i.e., which belong to both ‘A and B’.
If A and B are two events, then the set A ∩ B denotes the event ‘A and B’.
Thus, A ∩ B = {ω : ω ∈ A and ω ∈ B}
Event ‘A but not B’
A–B is the set of all those elements which are in A but not in B. Therefore, the set A–B may denote the event ‘A but not B’. A – B = A ∩ B’
Mutually exclusive events
Events A and B are called mutually exclusive events if occurrence of any one of them excludes occurrence of other event, i.e., if they cannot occur simultaneously.
Example: A die is thrown. Event A = All even outcome & event B = All odd outcome. Then A & B are mutually exclusive events, they cannot occur simultaneously.
Exhaustive events
Lot of events that together forms sample space. Example: A die is thrown. Event A = All even outcome & event B = All odd outcome. Even A & B together forms exhaustive events as it forms Sample Space.
. | 664 | 2,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.958759 |
https://www.jiskha.com/display.cgi?id=1340233258 | 1,516,483,171,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889733.57/warc/CC-MAIN-20180120201828-20180120221828-00440.warc.gz | 989,936,432 | 3,569 | # Algebra
posted by .
6b^2 - 4b + 3b^4 - (7b^3 + b^4 - 4b)
• Algebra -
6b^2 - 4b + 3b^4 - (7b^3 + b^4 - 4b) = 6b^2 - 4b + 3b^4 - 7b^3 - b^4 + 4b
Combine like terms.
## Similar Questions
1. ### Algebra 2 URGENT
Simplify: 1/u^2-2u-1/u^2-4 Please Show All Work!
2. ### Algebra 2 URGENT
Simplify: 2x+5/4x^2+2x-5/10x Please Show All Work!
3. ### Algebra 2 URGENT
Simplify: 2x+5/4x^2+2x-5/10x Please Show All Work!
4. ### Algebra 2 URGENT
Simplify: x+2/x^2+x-2/2x Please Show All Work!
6. ### math
2. Simplify. Please be sure to show all of your work. -3(-9) – |-5 – 3 3. Simplify. Please show all of your work 9c^3+7c-(3c^3-12+c) 4. Solve 4x- 3(5x-8) =23-9(x+2) . Please show all of your work 5. Solve the following equation … | 341 | 734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-05 | latest | en | 0.635413 |
https://mitpress.mit.edu/sites/default/files/titles/content/sicm_edition_2/index.html | 1,563,443,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00469.warc.gz | 450,150,904 | 28,629 | # Index
Any inaccuracies in this index may be explained by the fact that it has been prepared with the help of a computer.
Donald E. Knuth, Fundamental Algorithms
(Volume 1 of The Art of Computer Programming)
Page numbers for Scheme procedure definitions are in italics.
Page numbers followed by n indicate footnotes.
0, for all practical purposes, 20 n. See also Zero-based indexing
∘ (composition), 7 n, 510
Γ[q]
for local tuple, 11
Lagrangian state path, 203
γ (configuration-path function), 7
δ function, 454 (ex. 6.12)
δη (variation operator), 26
λ-calculus, 509
λ-expression, 498499
λ-notation, 498 n
ΠL[q] (Hamiltonian state path), 203
χ (coordinate function), 7
σ (phase-space path), 218
ω matrix, 124
$\stackrel{\to }{\omega }$ (angular velocity), 124
ω (symplectic 2-form), 359
C (local-tuple transformation), 44
CH (canonical phase-space transformation), 337 n
D. See Derivative
Dt (total time derivative), 64
∂. See Partial derivative
$\mathcal{E}$ (Euler–Lagrange operator), 98
(energy state function), 82
F1(t, q, q′), 373
F2(t, q, p′), 373
F3(t, p, q′), 374
F4(t, p, p′), 374
H (Hamiltonian), 199
I (identity operator), 517
I with subscript (selector), 64 n, 513
$\stackrel{˜}{J}$ (shuffle function), 350
J, Jn (symplectic unit), 301, 355
L (Lagrangian), 11
L (Lie derivative), 447
P (momentum selector), 199, 220
$\mathcal{P}$ (momentum state function), 79
Q (coordinate selector), 220
$\stackrel{˙}{Q}$ (velocity selector), 64
q (coordinate path), 7
S (action), 10
Lagrangian, 12
(quote in Scheme), 505
, in tuple, 520
:, names starting with, 21 n
; in tuple, 31 n, 520
# in Scheme, 504
{ } for Poisson brackets, 218
[ ] for down tuples, 512
[ ] for functional arguments, 10 n
( ) for up tuples, 512
() in Scheme, 497, 498 n, 503
Action, 913
computing, 1423
coordinate-independence of, 17
free particle, 1420
generating functions and, 421425
Hamilton–Jacobi equation and, 421425
Lagrangian, 12
minimizing, 1823
parametric, 21
principles (see Principle of stationary action)
S, 10
time evolution and, 423425, 435437
variation of, 28
Action-angle coordinates, 311
Hamiltonian in, 311
Hamilton–Jacobi equation and, 413
Hamilton’s equations in, 311
harmonic oscillator in, 346 (eq. 5.31)
perturbation of Hamiltonian, 316, 458
surfaces of section in, 313
Action principle. See Principle of stationary action
Alphabet, insufficient size of, 15 n
Alternative in conditional, 501
angle-axis->rotation-matrix, 184
Angles, Euler. See Euler angles
Angular momentum. See also Vector angular momentum
conservation of, 43, 80, 86, 142143
equilibrium points for, 149
Euler’s equations and, 151153
in terms of principal moments and angular velocity, 136
kinetic energy in terms of, 148
Lie commutation relations for, 452 (ex. 6.10)
as Lie generator of rotations, 440
of free rigid body, 146150, 151153
of rigid body, 135137
sphere of, 148
z component of, 85
Angular velocity vector ($\stackrel{\to }{\omega }$), 124, 139
Euler’s equations for, 151153
kinetic energy in terms of, 131, 134
representation of, 123126
Anomaly, true, 171 n
antisymmetric->column-matrix, 126
Antisymmetry of Poisson bracket, 220
Area preservation
by maps, 278
Liouville’s theorem and, 272
Poincaré–Cartan integral invariant and, 434435
of surfaces of section, 272, 434435
active vs. passive in Legendre transformation, 208
in Scheme, 497
Arithmetic
generic, 16 n, 509
on functions, 18 n, 511
on operators, 34 n, 517
on procedures, 19 n
on symbolic values, 511
on tuples, 509, 513516
Arnold, V. I., xiii, xv n, 113. See also Kolmogorov–Arnold–Moser theorem
Assignment in Scheme, 506508
Associativity and non-associativity of tuple multiplication, 515, 516
Asteroids, rotational alignment of, 151
Astronomy. See Celestial objects
Asymptotic trajectories, 223, 287, 302
Atomic scale, 8 n
Attractor, 274
Autonomous systems, 82. See also Extended phase space
surfaces of section for, 248263
Awake top, 231
Axes, principal, 133
of this dense book, 135 (ex. 2.7), 150
Axisymmetric potential of galaxy, 250
Axisymmetric top
awake, 231
behavior of, 161165, 231232
conserved quantities for, 160
degrees of freedom of, 5 (ex. 1.1)
Euler angles for, 159
Hamiltonian treatment of, 228233
kinetic energy of, 159
Lagrangian treatment of, 157165
nutation of, 162 (fig. 2.5), 164 (ex. 2.15)
potential energy of, 160
precession of, 119, 162 (fig. 2.6), 164 (ex. 2.16)
rotation of, 119
sleeping, 231
symmetries of, 228
Baker, Henry. See Baker–Campbell–Hausdorff formula
Baker–Campbell–Hausdorff formula, 453 (ex. 6.11)
Banana. See Book
Barrow-Green, June, 457
Basin of attraction, 274
Bicycle wheel, 156 (ex. 2.13)
Birkhoff, George David. See Poincaré–Birkhoff theorem
bisect (bisection search), 321, 326
Body components of vector, 134
Boltzmann, Ludwig, 12 n, 203 n, 274 n
Book
banana-like behavior of, 128
rotation of, 119, 150
for down tuples, 512
for functional arguments, 10 n
bulirsch-stoer, 145
Bulirsch–Stoer integration method, 74 n
Butterfly effect, 241 n
C (local-tuple transformation), 44
CH (canonical phase-space transformation), 337 n
Campbell, John. See Baker–Campbell–Hausdorff formula
canonical?, 344
Canonical-H?, 348
Canonical-K?, 348
canonical-transform?, 351
Canonical condition, 342352
Poisson brackets and, 352353
Canonical equations. See Hamilton’s equations
Canonical heliocentric coordinates, 409 (ex. 5.21)
Canonical perturbation theory. See Perturbation theory
Canonical plane, 362 n
Canonical transformations, 335336. See also Generating functions; Symplectic transformations
composition of, 346 (ex. 5.4), 381, 393 (ex. 5.12)
conditions for, 342357
for driven pendulum, 392
general, 342357
group properties of, 346 (ex. 5.4)
for harmonic oscillator, 344
invariance of antisymmetric bilinear form under, 359362
invariance of phase volume under, 358359
invariance of Poisson brackets under, 358
invariants of, 357364 (see also Integral invariants)
as Lie series, 448
Lie transforms (see Lie transforms)
point transformations (see Point transformations)
polar-canonical (see Polar-canonical transformation)
to rotating coordinates, 348349, 377378
time evolution as, 426437
total time derivative and, 390393
Cantorus, cantori, 244 n, 330
car, 503
Cartan, Élie. See Poincaré–Cartan integral invariant
Cauchy, Augustin Louis, 39 n
cdr, 503
Celestial objects. See also Asteroids; Comets; Earth; Galaxy; Hyperion; Jupiter; Mercury; Moon; Phobos; Planets
rotation of, 151, 165, 170171
Center of mass, 121
in two-body problem, 381
Jacobi coordinates and, 409 (ex. 5.21)
kinetic energy and, 121
vector angular momentum and, 135
Central force
collapsing orbits, 389 (ex. 5.11)
epicyclic motion, 381389
gravitational, 31
in 2 dimensions, 40, 227228, 381389
in 3 dimensions, 47 (ex. 1.16), 84
Lie series for motion in, 450
orbits, 78 (ex. 1.30)
reduced phase space for motion in, 405407
Central potential. See Central force
Centrifugal force, 47, 49
Chain rule
for derivatives, 517, 523 (ex. 9.1)
for partial derivatives, 519, 523 (ex. 9.1)
for total time derivatives, 64 (ex. 1.26)
in traditional notation, xiv n
for variations, 27 (eq. 1.26)
Chaotic motion, 241. See also Exponential divergence
homoclinic tangle and, 307
in Hénon–Heiles problem, 259
in restricted three-body problem, 283 (ex. 3.16)
in spin-orbit coupling, 282 (ex. 3.15), 496 (ex. 7.5)
near separatrices, 290, 484, 486
of Hyperion, 151, 170176
of non-axisymmetric top, 263
of periodically driven pendulum, 76, 243
overlapping resonances and, 488
Characteristic exponent, 293
Characteristic multiplier, 296
Chirikov, Boris V., 278 n
Chirikov–Taylor map, 278 n
Church, Alonzo, 498 n
Colon, names starting with, 21 n
Comets, rotation of, 151
Comma in tuple, 520
islands and, 309
of pendulum period with drive, 289, 290
periodic orbits and, 309, 316
rational rotation number and, 316
small denominators and, 475
of some tuple multiplication, 515
of variation (δ) with differentiation and integration, 27
Commutator, 451
of angular-momentum Lie operators, 452 (ex. 6.10)
Jacobi identity for, 451
of Lie derivative, 452 (ex. 6.10)
Poisson brackets and, 452 (ex. 6.10)
compatible-shape, 351 n
Compatible shape, 351 n
component, 15 n, 514
compose, 500
Composition
of canonical transformations, 346 (ex. 5.4), 381, 393 (ex. 5.12)
of functions, 7 n, 510, 523 (ex. 9.2)
of Lie transforms, 451
of linear transformations, 516
of operators, 517
of rotations, 123, 187
Compound data in Scheme, 502504
cond, 500
Conditionals in Scheme, 500501
Configuration, 4
Configuration manifold, 7 n
Configuration path. See Path
Configuration space, 45
Conjugate momentum, 79
non-uniqueness of, 239
cons, 503
Consequent in conditional, 500
Conserved quantities, 78, 195. See also Hénon–Heiles problem, integrals of motion
angular momentum, 43, 80, 86, 142143
coordinate choice and, 7981
cyclic coordinates and, 80
energy, 8183, 142, 211
Jacobi constant, 89 n, 383, 400
Lyapunov exponents and, 267
momentum, 7981
Noether’s theorem, 9091
phase space reduction and, 224226
phase volume (see Phase-volume conservation)
Poisson brackets of, 221
symmetry and, 79, 90
for top, 160
Constant of motion (integral of motion), 78. See also Conserved quantities; Hénon–Heiles problem
Constraint(s)
augmented Lagrangian and, 102, 109
configuration space and, 4
as coordinate transformations, 5963
explicit, 99103
in extended bodies, 4
holonomic, 4 n, 109
integrable, 4 n, 109
linear in velocities, 112
nonholonomic (non-integrable), 112
on coordinates, 101
rigid, 4963
as subsystem couplers, 105
total time derivative and, 108
velocity-dependent, 108
velocity-independent, 101
Constraint force, 104
Constructors in Scheme, 503
Contact transformation. See Canonical transformations
Continuation procedure, 247
Continued-fraction
approximation of irrational number, 325
Contraction of tuples, 514
coordinate, 15 n
action-angle (see Action-angle coordinates)
conserved quantities and choice of, 7981
constraints on, 101
cyclic, 80, 224 n
heliocentric, 409 (ex. 5.21)
ignorable (cyclic), 80
Jacobi, 409 (ex. 5.21)
polar (see Polar coordinates)
redundant, and initial conditions, 69 n
rotating (see Rotating coordinates)
spherical, 84
Coordinate function (χ), 7
Coordinate-independence
of action, 17
of Lagrange equations, 30, 43 (ex. 1.14)
of variational formulation, 3, 39
Coordinate path (q), 7. See also Local tuple
Coordinate selector (Q), 220
Coordinate singularity, 144
Coordinate transformations, 4447
constraints as, 5963
Coriolis force, 47, 49
Correction fluid, 150
Cotangent space, bundle, 203 n
Coupling, spin-orbit. See Spin-orbit coupling
Coupling systems, 105106
Curves, invariant. See Invariant curves
Cyclic coordinate, 80, 224 n
D. See Derivative
D (Scheme procedure for derivative), 16 n, 516
D-as-matrix, 355 n
D-phase-space, 347
∂. See Partial derivative
Dt (total time derivative), 64
d’Alembert–Lagrange principle (Jean leRond d’Alembert), 113
Damped harmonic oscillator, 274
define, 499
definite-integral, 17
Definite integral, 10 n
Definitions in Scheme, 499500
Degrees of freedom, 45
Delta function, 454 (ex. 6.12)
Derivative, 8 n, 516521. See also Total time derivative
as operator, 517
as Poisson bracket, 446
chain rule, 517, 523 (ex. 9.1)
in Scheme programs: D, 16 n, 516
notation: D, 8 n, 516
of function of multiple arguments, 29 n, 518521
of function with structured arguments, 24 n
of function with structured inputs and outputs, 522
of state, 71
partial (see Partial derivative)
precedence of, 8 n, 516
with respect to a tuple, 29 n
determinant, 144
Differentiable manifold, 7 n
Dimension of configuration space, 45
Dirac, Paul Adrien Maurice, 12 n
Dissipation of energy
in free-body rotation, 150
tidal friction, 170
Dissipative system, phase-volume conservation, 274
Dissolution of invariant curves, 329330, 486
Distribution functions, 276
Divided phase space, 244, 258, 286290
Dot notation, 32 n
Double pendulum. See Pendulum, double
down, 15 n, 513
Down tuples, 512
Driven harmonic oscillator, 430 (ex. 6.6)
Driven pendulum. See Pendulum (driven)
Driven rotor, 317, 321
Dt (total time derivative), 97
Dynamical state. See State
$\mathcal{E}$ (Euler–Lagrange operator), 98
(energy state function), 82
Earth
precession of, 176 (ex. 2.18)
rotational alignment of, 151
Effective Hamiltonian, 230
Effects in Scheme, 505508
Eigenvalues and eigenvectors
for equilibria, 293
for fixed points, 296
for Hamiltonian systems, 298
of inertia tensor, 132
for unstable fixed point, 303
Einstein, Albert, 1
Einstein summation convention, 367 n
else, 500
Empty list, 503
Energy, 81
as sum of kinetic and potential energies, 82
conservation of, 8183, 142, 211
dissipation of (see Dissipation of energy)
Energy state function (), 82
Hamiltonian and, 200
Epicyclic motion, 381389
eq?, 505
Equilibria, 222223, 291295. See also Fixed points
for angular momentum, 149
inverted, for pendulum, 246, 282 (ex. 3.14), 491494, 496 (ex. 7.4)
linear stability of, 291295
relative, 149
stable and unstable, 287
Equinox, precession of, 176 (ex. 2.18)
Ergodic motion, 312 n
Ergodic theorem, 251
Euler, Leonhard, 13 n
Euler->M, 139
Euler-state->omega-body, 140
Euler angles, 137141
for axisymmetric top, 159
kinetic energy in terms of, 141
singularities and, 143, 154
Euler–Lagrange equations. See Lagrange equations
Euler-Lagrange-operator ($\mathcal{E}$), 98
Euler–Lagrange operator ($\mathcal{E}$), 98
Euler’s equations, 151157
singularities in, 154
Euler’s theorem on homogeneous functions, 83 n
Euler’s theorem on rotations, 123
Euler angles and, 182
Evolution. See Time evolution of state
evolve, 75, 145, 238
explore-map, 248
Exponential(s)
of differential operator, 443
of Lie derivative, 447 (eq. 6.147)
of noncommuting operators, 451453
Exponential divergence, 241, 243, 263267. See also Chaotic motion; Lyapunov exponent
homoclinic tangle and, 307
Expressions in Scheme, 497
Extended phase space, 394402
generating functions in, 407
F1(t, q, q′), 373
F2(t, q, p′), 373
F3(t, p, q′), 374
F4(t, p, p′), 374
F->C, 46, 96
F->CH, 339
F->K, 340
Fermat, Pierre, 13 (ex. 1.3)
Fermat’s principle (optics), 13 (ex. 1.3), 13 n
Fermi, Enrico, 251
Feynman, Richard P., 12 n
find-path, 21
First amendment. See Degrees of freedom
First integral, 78
elliptic, 299, 320
equilibria or periodic motion and, 290, 295
for Hamiltonian systems, 298
hyperbolic, 299, 320
linear stability of, 295297
manifolds for, 303
parabolic, 299
Poincaré–Birkhoff fixed points, 320
Poincaré–Birkhoff theorem, 316321
rational rotation number and, 316
Floating-point numbers in Scheme, 18 n
Floquet multiplier, 296 n
Flow, defined by vector field, 447 n
Force
central (see Central force)
exerted by constraint, 104
Forced libration of the Moon, 175
Forced rigid body. See Rigid body, forced
Formal parameters
of a function, 14 n
of a procedure, 499
Foucault pendulum, 62 (ex. 1.25), 78 (ex. 1.31)
frame, 76 n
Free libration of the Moon, 175
Free particle
action, 1420
Lagrange equations for, 33
Lagrangian for, 1415
Free rigid body. See Rigid body (free)
Freudenthal, Hans, xiv n
Friction
internal, 150
tidal, 170
Function(s), 510511
arithmetic operations on, 18 n, 511
composition of, 7 n, 510, 523 (ex. 9.2)
homogeneous, 83 n
operator vs., 448 n, 517
orthogonal, tuple-valued, 101 n
parallel, tuple-valued, 101 n
selector (see Selector function)
tuple of, 7 n, 521
vs. value when applied, 509, 510
with multiple arguments, 518, 519, 523 (ex. 9.2)
with structured arguments, 24 n, 519, 523 (ex. 9.2)
with structured output, 521, 523 (ex. 9.2)
Functional arguments, 10 n
Functional mathematical notation, xiv, 509
Function definition, 14 n
Fundamental Poisson brackets, 352
Γ[q]
for local tuple, 11
Lagrangian state path, 203
Galaxy, 248252
axisymmetric potential of, 250
Galilean invariance, 68 (ex. 1.29), 341 (ex. 5.1)
Gamma (Scheme procedure for Γ), 16
optional argument, 36 (ex. 1.13)
Gamma-bar, 95
Gas in corner of room, 273
Generalized coordinates, 68, 39. See also Coordinate(s)
Euler angles as, 138 (see also Euler angles)
Generalized momentum, 79
transformation of, 337 (eq. 5.5)
Generalized velocity, 8
transformation of, 45
Generating functions, 364394
in extended phase space, 407
F1F4, 373374
F1, 364368
F2, 371373
F2 and point transformations, 375376
F2 for polar coordinate transformation, 376377
F2 for rotating coordinates, 377378
integral invariants and, 368373
Lagrangian action and, 421425
Legendre transformation between F1 and F2, 373
mixed-variable, 374
Generic arithmetic, 16 n, 509
Gibbs, Josiah Willard, 12 n, 203 n
Golden number, 325
Golden ratio, a most irrational number, 325
Golden rotation number, 328
Goldstein, Herbert, 119
Goldstein’s hoop, 110
Golf ball, tiny, 108 (ex. 1.41)
Grand Old Duke of York. See neither up nor down
Graphing, 23 (ex. 1.5), 75, 248
Gravitational potential
central, 31
of galaxy, 250
multipole expansion of, 165169
rigid-body, 166
Group properties
of canonical transformations, 346 (ex. 5.4)
of rotations, 187 (see also Euler’s theorem on rotations)
H (Hamiltonian), 199
H-central, 339
H-harmonic, 448
H-pend-sysder, 237
Hamilton, Sir William Rowan, 39 n, 183
Hamiltonian, 199
in action-angle coordinates, 311
computing (see H-…)
cyclic in coordinate, 224 n
energy state function and, 200
for axisymmetric potential, 250
for central potential, 227, 339, 381, 382
for damped harmonic oscillator, 275
for driven pendulum, 392
for driven rotor, 317
for harmonic oscillator, 344
for harmonic oscillator, in action-angle coordinates, 346 (eq. 5.31)
for Kepler problem, 418
for pendulum, 460
for periodically driven pendulum, 236, 476
for restricted three-body problem, 399, 400
for spin-orbit coupling, 496 (ex. 7.5)
for top, 230
for two-body problem, 378
Hénon–Heiles, 252, 455 (ex. 6.12)
Lagrangian and, 200 (eq. 3.19), 210
perturbation of action-angle, 316, 458
time-dependent, and dissipation, 276
Hamiltonian->Lagrangian, 213
Hamiltonian->state-derivative, 204
Hamiltonian flow, 447 n
Hamiltonian formulation, 195
Lagrangian formulation and, 217
Hamiltonian state, 202203
Hamiltonian state derivative, 202, 204
Hamiltonian state path ΠL[q], 203
Hamilton–Jacobi equation, 411413
action-angle coordinates and, 413
action at endpoints and, 425
for harmonic oscillator, 413417
for Kepler problem, 417421
separation in spherical coordinates, 418421
time-independent, 413
Hamilton-equations, 203
Hamilton’s equations, 197200
in action-angle coordinates, 311
computation of, 203205
dynamical, 217
for central potential, 227
for damped harmonic oscillator, 275
for harmonic oscillator, 344
from action principle, 215217
from Legendre transformation, 210211
numerical integration of, 236
Poisson bracket form, 220
Hamilton’s principle, 38
for systems with rigid constraints, 4950
Harmonic oscillator
coupled, 105
damped, 274
decoupling via Lie transform, 442
driven, 430 (ex. 6.6)
first-order equations for, 72
Hamiltonian for, 344
Hamiltonian in action-angle coordinates, 346 (eq. 5.31)
Hamilton’s equations for, 344
Lagrange equations for, 30, 72
Lagrangian for, 21
Lie series for, 448
solution of, 34, 344
solution via canonical transformation, 344
solution via Hamilton–Jacobi, 413417
Hausdorff, Felix. See Baker–Campbell–Hausdorff formula
Heiles, Carl, 241, 248. See also Hénon
Heisenberg, Werner, 12 n, 203 n
Heliocentric coordinates, 409 (ex. 5.21)
Hénon, Michel, 195, 241, 248
Hénon–Heiles problem, 248263
computing surfaces of section, 261263
Hamiltonian for, 252
history of, 248252
integrals of motion, 251, 254, 256260
interpretation of model, 256260
model of, 252254
potential energy, 253
surface of section, 254263
Hénon’s quadratic map, 280 (ex. 3.13)
Heteroclinic intersection, 305
Higher-order perturbation theory, 468473, 489494
History
Hénon–Heiles problem, 248252
variational principles, 10 n, 13 n, 39 n
Holonomic system, 4 n, 109
Homoclinic intersection, 304
Homoclinic tangle, 302309
chaotic regions and, 307
computing, 307309
exponential divergence and, 307
Homogeneous function, Euler’s theorem, 83 n
Huygens, Christiaan, 10 n
Hyperion, chaotic tumbling of, 151, 170176
I (identity operator), 517
I with subscript (selector), 64 n, 513
if, 501
Ignorable coordinate. See Cyclic coordinate
Indexing, zero-based. See Zero-based indexing
Inertia, moments of. See Moment(s) of inertia
Inertia matrix, 128. See also Inertia tensor
Inertia tensor, 127
diagonalization of, 132133
kinetic energy in terms of, 131
principal axes of, 133
transformation of, 130132
Initial conditions. See Sensitivity to initial conditions; State
Inner product of tuples, 515
free-body rotation, 149151
Integers in Scheme, 18 n
Integrable constraints, 4 n, 109
Integrable systems, 285, 309316
periodic orbits of near-integrable systems, 316
perturbation of, 316, 322, 457
surfaces of section for, 313316
Integral, definite, 10 n
Integral invariant
generating functions and, 368373
Poincaré, 362364
Poincaré–Cartan, 402, 431434
Integral of motion, 78. See also Conserved quantities; Hénon–Heiles problem
Integration. See Numerical integration
Invariant curves, 243, 322330
dissolution of, 329330, 486
finding (computing), 326329
finding (strategy), 322325
irrational rotation number and, 322
Kolmogorov–Arnold–Moser theorem, 322
Invariants of canonical transformations, 357364. See also Integral invariants
Irrational number, continued-fraction approximation, 325
Islands in surfaces of section. See also Resonance
for Hénon–Heiles problem, 259
for periodically driven pendulum, 244246, 289290, 483486
for standard map, 279
perturbative vs. actual, 483486
in Poincaré–Birkhoff construction, 321
Poisson series and, 488
secondary, 260, 290
size of, 322, 488
small denominators and, 322, 488
iterated-map, 308 n
Iteration in Scheme, 502
$\stackrel{˜}{J}$ (shuffle function), 350
J, Jn (symplectic unit), 301, 355
J-func, 351
J-matrix, 353
Jac (Jacobian of map), 270
Jacobi, Carl Gustav Jacob, 39 n. See also Hamilton–Jacobi equation
Jacobian, 270
Jacobi constant, 89 n, 383, 400
Jacobi coordinates, 409 (ex. 5.21)
Jacobi identity
for commutators, 451
for Poisson brackets, 221
Jeans, Sir James, “theorem” of, 251
Jupiter, 129 (ex. 2.4)
KAM theorem. See Kolmogorov–Arnold–Moser theorem
Kepler, Johannes. See Kepler…
Kepler problem, 31, 35 (ex. 1.11)
in reduced phase space, 406
reduction to, 378381
solution via Hamilton–Jacobi equation, 417421
Kepler’s third law, 35 (ex. 1.11), 173
Kinematics of rotation, 122126
Kinetic energy
ellipsoid of, 148
in Lagrangian, 3839
as Lagrangian for free body, 122, 141
of axisymmetric top, 159
of free rigid body, 148150
of rigid body, 120122 (see also Rigid body, kinetic energy…)
rotational and translational, 122
in spherical coordinates, 84
Knuth, Donald E., 531
Kolmogorov, A. N.. See Kolmogorov–Arnold–Moser theorem
Kolmogorov–Arnold–Moser theorem, 302, 322
L (Lagrangian), 11
L (Lie derivative), 447
L-axisymmetric-top, 229
L-body, 137
L-body-Euler, 141
L-central-polar, 43, 47
L-central-rectangular, 41
L-free-particle, 14
L-harmonic, 22
L-pend, 52
L-periodically-driven-pendulum, 74
L-rectangular, 213
L-space, 137
L-space-Euler, 141
L-uniform-acceleration, 40, 61
Lagrange, Joseph Louis, 13 n, 39 n
Lagrange-equations, 33
Lagrange equations, 2325
at a moment, 97
computing, 3336
coordinate-independence of, 30, 43 (ex. 1.14)
derivation of, 2530
as first-order system, 72
for central potential (polar), 43
for central potential (rectangular), 41
for damped harmonic oscillator, 275
for driven pendulum, 52
for gravitational potential, 32
for harmonic oscillator, 30, 72
for periodically driven pendulum, 74
for spin-orbit coupling, 173
from Newton’s equations, 3638, 5458
vs. Newton’s equations, 39
numerical integration of, 73
off the beaten path, 97
singularities in, 143
traditional notation for, xiv, 24
uniqueness of solution, 69
Lagrange-interpolation-function, 20 n
Lagrange interpolation polynomial, 20
Lagrange multiplier. See Lagrangian, augmented
Lagrangian, 12
adding total time derivatives to, 65
augmented, 102, 109
coordinate transformations of, 44
cyclic in coordinate, 80
energy and, 12
for axisymmetric top, 159
for central potential (polar), 4243, 227
for central potential (rectangular), 41
for central potential (spherical), 84
for constant acceleration, 40
for damped harmonic oscillator, 275
for driven pendulum, 51, 66
for gravitational potential, 31
for harmonic oscillator, 21
for spin-orbit coupling, 173
for systems with rigid constraints, 49
generating functions and, 421423
Hamiltonian and, 200 (eq. 3.19), 210
kinetic energy as, 14, 122, 141
kinetic minus potential energy as, 3839 (see also Hamilton’s principle)
non-uniqueness of, 6366
parameter names in, 14 n
rotational and translational, 141
symmetry of, 90
Lagrangian-action, 17
Lagrangian->energy, 82
Lagrangian->Hamiltonian, 213
Lagrangian->state-derivative, 71
Lagrangian action, 12
Lagrangian formulation, 195
Hamiltonian formulation and, 217
Lagrangian reduction, 233236
Lagrangian state. See State tuple
Lagrangian state derivative, 71
Lagrangian state path Γ[q], 203
lambda, 498
Lambda calculus, 509
Lambda expression, 498499
Lanczos, Cornelius, 335
Least action, principle of. See Principle of stationary action
Legendre, Adrien Marie. See Legendre…
Legendre polynomials, 167
Legendre-transform, 212
Legendre transformation, 205212
active arguments in, 208
passive arguments in, 208209
of quadratic functions, 211
Leibniz, Gottfried, 10 n
let, 501
let*, 502
Libration of the Moon, 174, 175
Lie, Sophus. See Lie…
Lie-derivative, 448, 448 n
Lie derivative, 447 n
commutator for, 452 (ex. 6.10)
Lie transform and, 447 (eq. 6.147)
operator LH, 447
Lie series, 443451
computing, 448451
for central field, 450
for harmonic oscillator, 448
in perturbation theory, 458460
Lie-transform, 448
Lie transforms, 437443
composition of, 451
computing, 448
exponential identities, 451453
for finding normal modes, 442
Lie derivative and, 447 (eq. 6.147)
in perturbation theory, 458
Lindstedt, A., 471
linear-interpolants, 20 n
Linear momentum, 80
Linear separation of regular trajectories, 263
Linear stability, 290
equilibria and fixed points, 297302
nonlinear stability and, 302
of equilibria, 291295
of fixed points, 295297
of inverted equilibrium of pendulum, 492, 496 (ex. 7.4)
Linear transformations
as tuples, 515
composition of, 516
Liouville, Joseph. See Liouville…
Liouville equation, 276
Liouville’s theorem, 268272
from canonical transformation, 428
Lipschitz condition (Rudolf Lipschitz), 69 n
Lisp, 503 n
list, 503
list-ref, 503
Lists in Scheme, 502504
literal-function, 15, 512, 521
Literal symbol in Scheme, 504505
Local names in Scheme, 501502
Local state tuple, 71
Local tuple, 11
component names, 14 n
functions of, 14 n
in Scheme programs, 15 n
transformation of (C), 44
Log, falling off, 84 (ex. 1.33)
Loops in Scheme, 502
Lorentz, Hendrik Antoon. See Lorentz transformations
Lorentz transformations as point transformations, 399 (ex. 5.18)
Lorenz, Edward, 241 n
Lyapunov, Alexey M.. See Lyapunov exponent
Lyapunov exponent, 267. See also Chaotic motion
conserved quantities and, 267
exponential divergence and, 267
Hamiltonian constraints, 302
linear stability and, 297
M-of-q->omega-body-of-t, 126
M-of-q->omega-of-t, 126
M->omega, 126
M->omega-body, 126, 185
MacCullagh’s formula, 168 n
make-path, 20, 20 n
Manifold
differentiable, 7 n
stable and unstable, 303309
Map
area-preserving, 278
Chirikov–Taylor, 278 n
fixed points of, 295297 (see also Fixed points)
Hénon’s quadratic, 280 (ex. 3.13)
Poincaré, 242
representation in programs, 247
standard, 277280
symplectic, 301
twist, 315
Mars. See Phobos
Mass point. See Point mass
Mathematical notation. See Notation
Mather, John N. (discoverer of sets named cantori by Ian Percival), 244 n
Matrix
orthogonal, 124, 130 n
symplectic, 301, 355, 356 (ex. 5.6)
as tuple, 515
Maupertuis, Pierre-Louis Moreau de, 13 n
Mean motion, 175 n
Mechanics, 1496
Newtonian vs. variational formulation, 3, 39
Mercury, resonant rotation of, 171, 193 (ex. 2.21)
Minimization
of action, 1823
in Scmutils, 19 n, 21 n
minimize, 19 n
Mixed-variable generating functions, 374
Moment(s) of inertia, 126130
about a line, 128
about a pivot point, 159
principal, 132135
of top, 159
conjugate to coordinate (see Conjugate momentum)
conservation of, 7981
generalized (see Generalized momentum)
variation of, 216 n
momentum, 204
Momentum path, 80
Momentum selector (P), 199, 220
Momentum state function ($\mathcal{P}$), 79
Moon
history of, 9 n
libration of, 174, 175
rotation of, 119, 151, 170176, 496 (ex. 7.5)
Moser, Jürgen. See Kolmogorov–Arnold–Moser theorem
Motion
atomic-scale, 8 n
chaotic (see Chaotic motion)
dense, on torus, 312 n
deterministic, 9
epicyclic, 381389
ergodic, 312 n
periodic (see Periodic motion)
quasiperiodic, 243, 312
realizable vs. conceivable, 2
regular vs. chaotic, 241 (see also Regular motion)
smoothness of, 8
tumbling (see Chaotic motion, of Hyperion; Rotation(s), (in)stability of)
multidimensional-minimize, 21, 21 n
Multiplication of operators as composition, 517
Multiplication of tuples, 514516
as composition, 516
as contraction, 514
Multiply periodic functions, Poisson series for, 474
Multipole expansion of potential energy, 165169
n-body problem, 408 (ex. 5.21). See also Three-body problem, restricted; Two-body problem
Nelder–Mead minimization method, 21 n
Newton, Sir Isaac, 3
Newtonian formulation of mechanics, 3, 39
Newton’s equations
as Lagrange equations, 3638, 5458
vs. Lagrange equations, 39
Noether, Emmy, 81 n
Noether’s integral, 91
Noether’s theorem, 9091
angular momentum and, 143
Non-associativity and associativity of tuple multiplication, 515, 516
Non-axisymmetric top, 263
exponential(s) of noncommuting operators, 451453
of some partial derivatives, 427 n, 520
of some tuple multiplication, 516
Nonholonomic system, 112
Nonsingular structure, 368 n
Notation, 509523. See also Subscripts; Superscripts; Tuples
{ } for Poisson brackets, 218
( ) for up tuples, 512
[ ] for down tuples, 512
[ ] for functional arguments, 10 n
ambiguous, xivxv
composition of functions, 7 n
definite integral, 10 n
derivative, partial: ∂, xv, 24, 520
derivative: D, 8 n, 516
functional, xiv, 509
functional arguments, 10 n
function of local tuple, 14 n
selector function: I with subscript, 64 n, 513
total time derivative: Dt, 64
traditional, xivxv, 24, 200 n, 218 n, 509
Numbers in Scheme, 18 n
Numerical integration
of Hamilton’s equations, 236
of Lagrange equations, 73
in Scmutils, 17 n, 74 n, 145
symplectic, 453 (ex. 6.12)
Numerical minimization in Scmutils, 19 n, 21 n
Nutation of top, 162 (fig. 2.5), 164 (ex. 2.15)
Oblateness, 170
omega (symplectic 2-form), 361
omega-cross, 126
Operator, 517
arithmetic operations on, 34 n, 517
composition of, 517
exponential identities, 451453
function vs., 448 n, 517
generic, 16 n
Operators
derivative (D) (see Derivative)
Euler–Lagrange ($\mathcal{E}$), 98
Lie derivative (LH), 447
Lie transform (${E}_{\mathit{ϵ},W}^{\prime }$), 439
partial derivative (∂) (see Partial derivative)
variation (δη), 26
Optical libration of the Moon, 174
Optics
Fermat, 13 (ex. 1.3)
Snell’s law, 13 n
Orbit. See Orbital motion; Phase-space trajectory
Orbital elements, 421
Orbital motion. See also Epicyclic motion; Kepler problem
in a central potential, 78 (ex. 1.30)
Lagrange equations for, 3132
retrodiction of, 9 n
Euler’s equations and, 153154
nonsingular coordinates for, 181191
specified by Euler angles, 138
specified by rotations, 123
Orientation vector, 182
Orthogonal matrix, 124, 130 n
Orthogonal transformation. See Orthogonal matrix
Orthogonal tuple-valued functions, 101 n
Oscillator. See Harmonic oscillator
osculating-path, 96
Osculation of paths, 94
Ostrogradsky, M. V., 39 n
Out-of-roundness parameter, 173
P (momentum selector), 199, 220
$\mathcal{P}$ (momentum state function), 79
p->r (polar-to-rectangular), 46
pair?, 504
Pairs in Scheme, 503
Parallel tuple-valued functions, 101 n
Parameters, formal. See Formal parameters
Parametric path, 20
parametric-path-action, 21
with graph, 23 (ex. 1.5)
Parentheses
in Scheme, 497, 498 n
for up tuples, 512
partial, 33 n
Partial derivative, 24, 518519, 520
chain rule, 519, 523 (ex. 9.1)
notation: ∂, xv, 24, 520
Particle, free. See Free particle
Path, 2
coordinate path (q), 7 (see also Local tuple)
finding, 2023
momentum path, 80
osculation of, 94
parametric, 20
realizable (see Realizable path)
variation of, 12, 18, 26
Path-distinguishing function, 2, 8. See also Action
Path functions, abstraction of, 94
Peak, 222
Pendulum. See also Pendulum (driven); Periodically driven pendulum
behavior of, 223, 286287
constraints and, 103
degrees of freedom of, 5 (ex. 1.1)
double (planar), 6, 117 (ex. 1.44)
double (spherical), 5 (ex. 1.1)
equilibria, stable and unstable, 287
Foucault, 62 (ex. 1.25), 78 (ex. 1.31)
Hamiltonian for, 460
Lagrangian for, 32 (ex. 1.9)
periodically driven pendulum vs., 244
as perturbed rotor, 460473
phase plane of, 223, 286
phase-volume conservation for, 268
spherical, 5 (ex. 1.1), 86 (ex. 1.34)
width of oscillation region, 466
Pendulum (driven), 5052. See also Pendulum; Periodically driven pendulum
drive as modification of gravity, 66
Hamiltonian for, 392
Lagrange equations for, 52
Lagrangian for, 51, 66
Pericenter, 171 n
Period doubling, 245
Periodically driven pendulum. See also Pendulum (driven); Pendulum
behavior of, 196, 244246
chaotic behavior of, 76, 243
emergence of divided phase space, 286290
Hamiltonian for, 236, 476
inverted equilibrium, 246, 282 (ex. 3.14), 491494, 496 (ex. 7.4)
islands in sections for, 244246, 289290, 483486
Lagrange equations for, 74
linear stability analysis, 492, 496 (ex. 7.4)
as perturbed rotor, 476478
phase-space descriptions for, 239
phase space evolution of, 236
resonances for, 481491
spin-orbit coupling and, 173
surface of section for, 242248, 282 (ex. 3.14), 287290, 483494
undriven pendulum vs., 244
with zero-amplitude drive, 286289
Periodically driven systems, surfaces of section, 241248
Periodic motion, 312
fixed points and, 295
integrable systems and, 309, 316
Periodic points, 295
Poincaré–Birkhoff theorem, 316321
rational rotation number and, 316
resonance islands and, 290
Perturbation of action-angle Hamiltonian, 316, 458
Perturbation theory, 457
for many degrees of freedom, 473478
for pendulum, 466468
for periodically driven pendulum, 491494
for spin-orbit coupling, 496 (ex. 7.5)
higher-order, 468473, 489494
Lie series in, 458460
nonlinear resonance, 478494
secular-term elimination, 471473
secular terms in, 470
small denominators in, 475, 476
Phase portrait, 231, 248 (ex. 3.10)
Phase space, 203. See also Surface of section
chaotic regions, 241
divided, 244, 258, 286290
evolution in, 236238 (see also Time evolution of state)
extended, 394402
non-uniqueness, 238239
of pendulum, 223, 286
qualitative features, 242246, 258260, 285286
reduced, 402407
regular regions, 241
two-dimensional, 222
volume (see Phase-volume conservation)
Phase space reduction, 224226
conserved quantities and, 224226
Lagrangian, 233236
Phase-space state function, 519
in Scheme, 521
Phase-space trajectory (orbit)
asymptotic, 223, 287, 302
chaotic, 243, 259
periodic, 309, 312, 316
quasiperiodic, 243, 312
regular, 243, 258
regular vs. chaotic, 241
Phase-volume conservation, 268, 428
for damped harmonic oscillator, 274
for pendulum, 268
under canonical transformations, 358359
Phobos, rotation of, 171
Pit, 222
moment of inertia of, 129 (ex. 2.4)
rotational alignment of, 151
rotation of, 165
plot-parametric-fill, 308
plot-point, 76 n
Plotting, 23 (ex. 1.5), 75, 248
Poe, Edgar Allan. See Pit; Pendulum
Poincaré, Henri, 239 n, 251, 285, 302, 471
Poincaré–Birkhoff theorem, 316321
computing fixed points, 321322
recursive nature of, 321
Poincaré–Cartan integral invariant, 402
time evolution and, 431434
Poincaré integral invariant, 362364
generating functions and, 368373
Poincaré map, 242
Poincaré recurrence, 272
Poincaré section. See Surface of section
Point mass, 4 n. See also Golf ball, tiny
Point transformations, 336341. See also Canonical transformations
computing, 339341
general canonical transformations vs., 357
generating functions for, 375376
polar-rectangular conversion, 339, 376377
to rotating coordinates, 348349, 377378
time-independent, 338
Poisson, Siméon Denis, 33 (ex. 1.10)
Poisson brackets, 218222
canonical condition and, 352353
commutator and, 452 (ex. 6.10)
of conserved quantities, 221
as derivations, 446
fundamental, 352
Hamilton’s equations in terms of, 220
in terms of $\stackrel{˜}{J}$, 352
in terms of symplectic 2-form, ω, 360
invariance under canonical transformations, 358
Jacobi identity for, 221
Lie derivative and, 447
Poisson series
for multiply periodic function, 474
resonance islands and, 488
polar-canonical, 345
Polar-canonical transformation, 344
generating function for, 365
harmonic oscillator and, 346
Polar coordinates
Lagrangian in, 4243
point transformation to rectangular, 339, 376377
transformation to rectangular, 46
Potential. See Central force; Gravitational potential
Potential energy
of axisymmetric top, 160
Hénon–Heiles, 253
in Lagrangian, 3839
multipole expansion of, 165169
Precession
of equinox, 176 (ex. 2.18)
of top, 119, 162 (fig. 2.6), 164 (ex. 2.16)
Predicate in conditional, 500
Predicting the past, 9 n
principal-value, 76 n
Principal axes, 133
of this dense book, 135 (ex. 2.7), 150
Principal moments of inertia, 132135
kinetic energy in terms of, 134, 141, 148
Principle of d’Alembert–Lagrange, 113
Principle of least action. See Principle of stationary action
Principle of stationary action (action principle), 813
Hamilton’s equations and, 215217
principle of least action, 10 n, 13 n, 39 n
statement of, 12
used to find paths, 20
print-expression, 444, 511 n
Probability density in phase space, 276
Procedure calls, 497498
Procedures
arithmetic operations on, 19 n
generic, 16 n
Products of inertia, 128
Q (coordinate selector), 220
$\stackrel{˙}{Q}$ (velocity selector), 64
q (coordinate path), 7
qcrk4 (quality-controlled Runge–Kutta), 145
Quadratic functions, Legendre transformation of, 211
integrable systems and, 311
reduction to, 224 n
Quartet, 300 (fig. 4.5)
Quasiperiodic motion, 243, 312
quaternion->angle-axis, 184
quaternion->RM, 184
quaternion->rotation-matrix, 185
quaternion-state->omega-body, 186
Quaternions, 181191
Hamilton’s discovery of, 39 n
quaternion units, 188
Quotation in Scheme, 504505
qw-state->L-space, 190
qw-sysder, 189
Reaction force. See Constraint force
Realizable path, 9
conserved quantities and, 78
as solution of Hamilton’s equations, 202
as solution of Lagrange equations, 23
stationary action and, 913
uniqueness, 12
Recurrence theorem of Poincaré, 272
Recursive procedures, 501
Reduced mass, 35 (ex. 1.11), 380
Reduced phase space, 402407
Reduction
Lagrangian, 233236
of phase space (see Phase space reduction)
to quadrature, 224 n
ref, 15 n, 514
Regular motion, 241, 243, 258
linear separation of trajectories, 263
Renormalization, 267 n
center, 480
islands (see Islands in surfaces of section)
nonlinear, 478494
of Mercury’s rotation, 171, 193 (ex. 2.21)
overlap criterion, 488489, 496 (ex. 7.5)
for periodically driven pendulum, 481491
spin-orbit, 177181
width, 483 (ex. 7.2), 488
Restricted three-body problem. See Three-body problem, restricted
Rigid body, 120
forced, 154157 (see also Spin-orbit coupling; Top)
free (see Rigid body (free))
kinetic energy, 120122
kinetic energy in terms of inertia tensor and angular velocity, 126129, 131
kinetic energy in terms of principal moments and angular momentum, 148
kinetic energy in terms of principal moments and angular velocity, 134
kinetic energy in terms of principal moments and Euler angles, 141
vector angular momentum, 135137
Rigid body (free), 141
angular momentum, 151153
angular momentum and kinetic energy, 146150
computing motion of, 143145
Euler’s equations and, 151154
(in)stability, 149151
orientation, 153154
Rigid constraints, 4963
as coordinate transformations, 5963
Rotating coordinates
in extended phase space, 400402
generating function for, 377378
point transformation for, 348349, 377378
active, 130
composition of, 123, 187
computing, 93
group property of, 187
(in)stability of, 149151
kinematics of, 122126
kinetic energy of (see Rigid body, kinetic energy…)
Lie generator for, 440
matrices for, 138
of celestial objects, 151, 165, 170171
of Hyperion, 170176
of Mercury, 171, 193 (ex. 2.21)
of Moon, 119, 151, 170176, 496 (ex. 7.5)
of Phobos, 171
of top, book, and Moon, 119
orientation as, 123
orientation vector and, 182
passive, 130
as tuples, 515
Rotation number, 315
golden, 328
irrational, and invariant curves, 322
rational, and commensurability, 316
rational, and fixed and periodic points, 316
Rotor
driven, 317, 321
pendulum as perturbation of, 460473
periodically driven pendulum as perturbation of, 476478
Routh, Edward John
Routhian, 234
Routhian equations, 236 (ex. 3.9)
Runge–Kutta integration method, 74 n
qcrk4, 145
Rx, 63 (ex. 1.25), 93 n
Rx-matrix, 139
Ry, 63 (ex. 1.25), 93 n
Rz, 63 (ex. 1.25), 93 n
Rz-matrix, 139
S (action), 10
Lagrangian, 12
s->m (structure to matrix), 353
s->r (spherical-to-rectangular), 85
Salam, Abdus, 509
Saturn. See Hyperion
Scheme, xvi, 497508, 509. See also Scmutils
for Gnu/Linux, where to get it, xvi
Schrödinger, Erwin, 12 n, 203 n
generic arithmetic, 16 n, 509
minimization, 19 n, 21 n
numerical integration, 17 n, 74 n, 145
operations on operators, 34 n
simplification of expressions, 511
where to get it, xvi
Second law of thermodynamics, 274
Section, surface of. See Surface of section
Secular terms in perturbation theory, 470
elimination of, 471473
Selector function, 64 n, 513
coordinate selector (Q), 220
momentum selector (P), 199, 220
velocity selector ($\stackrel{˙}{Q}$), 64
Selectors in Scheme, 503
Semicolon in tuple, 31 n, 520
Sensitivity to initial conditions, 241 n, 243, 263. See also Chaotic motion
Separatrix, 147, 222. See also Asymptotic trajectories
chaos near, 290, 484, 486
motion near, 302
series, 462
series:for-each, 444
series:sum, 463
set-ode-integration-method!, 145
show-expression, 16, 46 n
Shuffle function $\stackrel{˜}{J}$, 350
Simplification of expressions, 511
Singularities, 202 n
Euler angles and, 143, 154
in Euler’s equations, 154
quaternions, 181191
Sleeping top, 231
Small denominators
for periodically driven pendulum, 477
in perturbation theory, 475, 476
resonance islands and, 322, 488
Small divisors. See Small denominators
Snell’s law, 13 n
Solvable systems. See Integrable systems
solve-linear-left, 71 n
solve-linear-right, 339 n
Spherical coordinates
kinetic energy in, 84
Lagrangian in, 84
Spin-orbit coupling, 165181
chaotic motion, 282 (ex. 3.15), 496 (ex. 7.5)
Hamiltonian for, 496 (ex. 7.5)
Lagrange equations for, 173
Lagrangian for, 173
periodically driven pendulum and, 173
perturbation theory for, 496 (ex. 7.5)
resonances, 177181
surface of section for, 282 (ex. 3.15)
Spivak, Michael, xiv n, 509
Spring–mass system. See Harmonic oscillator
square, 21 n, 499
for tuples, 40 n, 499 n
Stability. See Equilibria; Instability; Linear stability
Stable manifold, 303309
computing, 307309
standard-map, 278
Standard map, 277280
Stars. See Galaxy
State, 6871
evolution of (see Time evolution of state)
Hamiltonian vs. Lagrangian, 202203
in terms of coordinates and momenta (Hamiltonian), 196
in terms of coordinates and velocities (Lagrangian), 69
State derivative
Hamiltonian, 204
Hamiltonian vs. Lagrangian, 202
Lagrangian, 71
State path
Hamiltonian, 203
Lagrangian, 203
State tuple, 71
Stationarity condition, 28
Stationary action. See Principle of stationary action
Stationary point, 2 n
Steiner’s theorem, 129 (ex. 2.2)
String theory, 119 n, 150. See also Quartet
Stroboscopic surface of section, 241248. See also Surface of section
computing, 246
Subscripts
down and, 15 n
for down-tuple components, 513
for momentum components, 79 n, 338 n
for selectors, 513
Summation convention, 367 n
Superscripts
for coordinate components, 7 n, 15 n, 79 n
for up-tuple components, 513
for velocity components, 15 n, 338 n
up and, 15 n
Surface of section, 239248
in action-angle coordinates, 313
area preservation of, 272, 434435
computing (Hénon–Heiles), 261263
computing (stroboscopic), 246
fixed points (see Fixed points)
for autonomous systems, 248263
for Hénon–Heiles problem, 254263
for integrable system, 313316
for non-axisymmetric top, 263
for periodically driven pendulum, 242248, 282 (ex. 3.14), 287290, 483494
for restricted three-body problem, 283 (ex. 3.16)
for spin-orbit coupling, 282 (ex. 3.15)
for standard map, 277280
invariant curves (see Invariant curves)
islands (see Islands in surfaces of section)
stroboscopic, 241248
Symbolic values, 511512
Symbols in Scheme, 504505
Symmetry
conserved quantities and, 79, 90
continuous, 195
of Lagrangian, 90
of top, 228
symplectic-matrix?, 355
symplectic-transform?, 355
symplectic-unit, 355
Symplectic bilinear form (2-form), 359362
invariance under canonical transformations, 359
Symplectic condition. See Symplectic transformations
Symplectic integration, 453 (ex. 6.12)
Symplectic map, 301
Symplectic matrix, 301, 356 (ex. 5.6), 353357
Symplectic transformations, 355. See also Canonical transformations
antisymmetric bilinear form and, 359362
Symplectic unit J, Jn, 301, 355
Syntactic sugar, 499
System derivative. See State derivative
T-body, 134
T-body-Euler, 141
T-func, 347
Taylor, J. B., 278 n
Tensor. See Inertia tensor
Tensor arithmetic
notation and, 79 n, 338 n
summation convention, 367 n
tuple arithmetic vs., 509, 513
Theology and principle of least action, 13 n
Thermodynamics, second law, 274
Three-body problem, restricted, 8690, 283 (ex. 3.16), 399402
chaotic motion, 283 (ex. 3.16)
surface of section for, 283 (ex. 3.16)
Tidal friction, 170
time, 15 n
Time-dependent transformations, 347349
Time evolution of state, 6878
action and, 423425, 435437
as canonical transformation, 426437
in phase space, 236238
Poincaré–Cartan integral invariant and, 431434
energy conservation and, 81
Top
axisymmetric (see Axisymmetric top)
non-axisymmetric, 263
Top banana. See Non-axisymmetric top
Torque, 165 (ex. 2.16)
in Euler’s equations, 154
in spin-orbit coupling, 173
Total time derivative, 6365
adding to Lagrangians, 65
affecting conjugate momentum, 239
canonical transformation and, 390393
commutativity of, 91 n
computing, 97
constraints and, 108
identifying, 68 (ex. 1.28)
notation: Dt, 64
properties, 67
Trajectory. See Path; Phase-space trajectory
Transformation
canonical (see Canonical transformations)
coordinate (see Coordinate transformations)
Legendre (see Legendre transformation)
Lie (see Lie transforms)
orthogonal (see Orthogonal matrix)
point (see Point transformations)
symplectic (see Symplectic transformations)
time-dependent, 347349
Transpose, 351 n
transpose, 126, 351 n
True anomaly, 171 n
Tumbling. See Chaotic motion, of Hyperion; Rotation(s), (in)stability of
Tuples, 512516
arithmetic on, 509, 513516
commas and semicolons in, 520
component selector: I with subscript, 64 n, 513
composition and, 516
contraction, 514
of coordinates, 7
down and up, 512
of functions, 7 n, 521
inner product, 515
linear transformations as, 515
local (see Local tuple)
matrices as, 515
multiplication of, 514516
rotations as, 515
semicolons and commas in, 520
squaring, 499 n, 514
state tuple, 71
up and down, 512
Twist map, 315
Two-body problem, 378381
Two-trajectory method, 265
Undriven pendulum. See Pendulum
Uniform circle map, 326
Uniqueness
of Lagrangian—not!, 6366
of phase-space description—not!, 238239
of realizable path, 12
of solution to Lagrange equations, 69
unstable-manifold, 308
Unstable manifold, 303309
computing, 307309
up, 15 n, 513
Up tuples, 512
Vakonomic mechanics, 114 n
Variation
chain rule, 27 (eq. 1.26)
of action, 28
of a function, 26
of a path, 12, 18, 26
operator: δη, 26
Variational equations, 266
Variational formulation of mechanics, 23, 39
Variational principle. See Principle of stationary action
Vector
body components of, 134
in Scheme, 502504
square of, 18 n, 21 n
vector, 504
vector?, 504
vector-ref, 504
Vector angular momentum, 135137. See also Angular momentum
center-of-mass decomposition, 135
in terms of angular velocity and inertia tensor, 136
in terms of principal moments and Euler angles, 141
Vector space of tuples, 514
Vector torque. See Torque
Velocity. See Angular velocity vector; Generalized velocity
velocity, 15 n
Velocity dispersion in galaxy, 248
Velocity selector ($\stackrel{˙}{Q}$), 64
Web site for this book, xvi
Wheel, 156 (ex. 2.13)
Whittaker transform (Sir Edmund Whittaker), 357 (ex. 5.9)
Width of oscillation region, 466 n
write-line, 505 n
Zero-amplitude drive for pendulum, 286289
Zero-based indexing, 7 n, 15 n, 503, 513, 514, 531 | 16,100 | 49,904 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.658799 |
https://www.beatthegmat.com/a-museum-has-259-paintings-and-11-sculptures-hoa-many-photographs-does-the-museum-have-t333244.html?sid=81113d27781a7194c3d42058f20a7684 | 1,716,490,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00873.warc.gz | 571,617,257 | 208,368 | ## A museum has $$259$$ paintings and $$11$$ sculptures. Hoa many photographs does the museum have?
##### This topic has expert replies
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### A museum has $$259$$ paintings and $$11$$ sculptures. Hoa many photographs does the museum have?
by AAPL » Sun Dec 10, 2023 2:46 pm
00:00
A
B
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E
## Global Stats
A museum has $$259$$ paintings and $$11$$ sculptures. How many photographs does the museum have?
1) If the museum acquires $$6$$ more paintings, then the number of paintings and sculptures combined will be $$6$$ times the number of photographs
2) If the museum acquires $$4$$ more paintings and $$2$$ more sculptures, then the number of photographs will be $$1/6$$ the number of paintings and sculptures combined
OA D
• Page 1 of 1 | 222 | 827 | {"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} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.905203 |
https://www.experts-exchange.com/questions/22982328/Complex-Numbers-in-C.html | 1,518,924,532,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811352.60/warc/CC-MAIN-20180218023321-20180218043321-00325.warc.gz | 879,131,685 | 19,460 | # Complex Numbers in C
Could someone give me sone help on working out complex root using C?
Here is my code so far:
#include <stdio.h>
#include <math.h>
double t1=0,t2=0,t3=0;
double quad1(double t1,double t2, double t3);
double quad2(double t1,double t2, double t3);
main()
{
printf("Enter the X square coeff: ");
scanf("%lf",&t1);
printf("Enter the X coeff: ");
scanf("%lf",&t2);
printf("Enter the last term: ");
scanf("%lf",&t3);
}
double quad1(double t1,double t2, double t3)
{
double ans1=0,det1=0;
det1= pow(t2,2) - (4*t1*t3); // b^2 - 4ac
ans1= (((-1)*t2) + pow(det1,0.5))/(2*t1);
return(ans1);
}
double quad2(double t1,double t2, double t3)
{
double ans2=0,det2=0;
det2= pow(t2,2) - (4*t1*t3); // b^2 - 4ac
ans2= (((-1)*t2) - pow(det2,0.5))/(2*t1);
return(ans2);
}
As you can see, it only works out real roots
Cheers
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Data Warehouse Architect / DBACommented:
Hi jonnytabpni,
The C math routines that you're using (pow) doesn't deal with imaginary numbers. To get the results you seek, you'll need to determine if the result will be real or imaginary, set a flag accordingly, and if imaginary, pass the absolute value of the data item.
The result is that the function will always return the root value, and you'll indicate in a separate field whether the result is real or imaginary.
Good Luck,
Kent
0
Author Commented:
ok fair enough but im confused how to actually work out the complex roots once iv passed the data onto a seperate function.
cheers
0
Data Warehouse Architect / DBACommented:
Hi jonnytabpni,
There are several ways to do this, though they all require a bit of effort.
C functions return one value. Period. So you can't get the function to return the real root and a real/imaginary flag. The two most common ways around this are to return one or both values via function parameters, or to build a 'complex' structure that contains a the two critical components and pass it to the function.
typedef struct
{
int Imaginary;
rel Value;
) compex_t;
Good Luck,
Kent
0
Author Commented:
ok i've got it:
#include <stdio.h>
#include <math.h>
double a,b,c;
int find_roots(double,double,double,double*,double*);
main()
{
double x1,x2;
int type;
printf("Enter the 3 quadratic coeffs: ");
scanf("%lf %lf %lf",&a,&b,&c);
type = find_roots(a,b,c, &x1,&x2);
if(type==0)
{
printf("The 2 solutions of your equation are: %lf and %lf",x1,x2);
if((((-1)*b)/a) == (x1+x2)) printf("\nCheck Confirmed!"); //Check to see if real roots are correct
}
else
{
printf("The 2 complex roots of your equations are %lf + j%lf and %lf - j%lf",x1,fabs(x2),x1,fabs(x2));
}
}
int find_roots(double a,double b, double c,double *x1,double *x2)
{
double ans1=0,det=0;
det= pow(b,2) - (4*a*c); // b^2 - 4ac
if(det>0)
{
det = sqrt(det);
*x1 = ((-1)*(b))/(2*a) + (det/(2*a));
*x2 = ((-1)*(b))/(2*a) - (det/(2*a));
return(0);
}
else
{
*x1 = ((-1)*(b))/(2*a);
*x2 = det/(2*a);
return(1);
}
}
the only thing i need now is a way to check to see if the comple roots are correct
0
Data Warehouse Architect / DBACommented:
Hi jonnytabpni,
That should be pretty easy to do. Build several test cases of known values.
4 ^ 2
9 ^ 2
4 ^ 1/2
9 ^ 1/2
If the values are correct (and they should be) rerun the test with negative coefficients.
Then, just to make sure:
27 ^ 1/3
-27 ^ 1/3
Good Luck,
Kent
0
Commented:
Have you Looked into Complex.h.
0 | 1,086 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-09 | latest | en | 0.718945 |
http://math.stackexchange.com/questions/392250/prove-or-disprove-statements-about-the-greatest-common-divisor/392260 | 1,469,514,567,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00015-ip-10-185-27-174.ec2.internal.warc.gz | 160,464,075 | 19,871 | # Prove or disprove statements about the greatest common divisor
Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.
(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$
(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$
-
Try some simple examples first of all to see which you want to prove or disprove. – anon May 15 '13 at 8:43
Unfortunately, I offer a clue without having actually tried it (this specific problem). Try seeing if they work with small integers, experiment with primes, 2,3,5, and composites, 4,6, etc. If they seem to hold, you may by now have developed a feeling for why you think it might be true, so run with that (i.e., try to prove them true). With further analysis, you may begin to feel that in fact, one is not correct, so you go back to construct a counterexample with this new perspective. And so on. – Brady Trainor May 15 '13 at 8:45
A good general strategy with this sort of problem is to consider an arbitrary prime number $p$ and see how many times it divides both sides. If the prime always divides each side the same number of times, then the two sides are equal. Otherwise, they arent. – 6005 May 15 '13 at 8:46
Can you find a title which is more descriptive of the subject of the question? The current one essentially gives no information about your question! – Mariano Suárez-Alvarez May 15 '13 at 8:50
Apologies, Mariano, I did what I could to make it a little bit more descriptive. – Miguel May 15 '13 at 8:52
Hint for (a): Note that $c|ac$ and $c|bc$.
Hint for (b): Look at special cases, $a=b$, $a=c$ etc
-
I find it helpful to look at the prime factor decomposition of the numbers $a,b$ and $c$. For two numbers $x,y$ the $\gcd(x,y)$ is just the maximal set of common prime factors.
So for (a) you it is clear that the maximal sets of prime factors of $ac$ and $bc$ contain both the prime factors of $c$. So if you take the prime factors of $c$ out, both sets remain maximal.
For (b) I would suggest to play around with some examples.
-
If the highest power of prime $p$ in $a,b,c$ are $r_a,r_b,r_c$ respectively,
the highest power of $p$ in gcd$(a,b)=$min $(r_a,r_b)$
the highest power of $p$ in $c\cdot$ gcd$(a,b)=r_c+$min $(r_a,r_b)$
the highest power of $p$ in gcd$(a\cdot c,b \cdot c)=$min $(r_a+r_c,r_b+r_c)=r_c$+min $(r_a,r_b)$
the highest power of $p$ in gcd$(a\cdot b,c)=$min $(r_a+r_b,r_c)$
the highest power of $p$ in gcd$(a,b)\cdot$gcd $(b,c)=$min $(r_a,r_b)$ + min$(r_b+r_c)$
Can you prove min $(r_a+r_b,r_c)=$min $(r_a,r_b)$ + min$(r_b+r_c)?$
-
Concerning (b), perhaps it's worth mentioning that $\gcd$ and $\operatorname{lcm}$ distribute each with respect to the other, so you have \begin{align} &\gcd(\operatorname{lcm}(a, b), c) = \operatorname{lcm}(\gcd(a, c), \gcd(b, c)), \\& \operatorname{lcm}(\gcd(a, b), c) = \gcd(\operatorname{lcm}(a, c), \operatorname{lcm}(b, c)). \end{align} The first of these equalities is the correct form of (b).
- | 949 | 3,119 | {"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} | 3.734375 | 4 | CC-MAIN-2016-30 | latest | en | 0.912833 |
https://www.tutorsonnet.com/aggregate-demand-multiplier-homework-help.php | 1,579,571,415,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601241.42/warc/CC-MAIN-20200121014531-20200121043531-00382.warc.gz | 1,142,760,116 | 9,193 | # Aggregate Demand Multipler
Introduction
A change in the equilibrium income or output is the result of a shift in the aggregate demand function or the C + I curve. The aggregate demand curve can either shift upwards or downwards. The amount of the change in the income will be a multiple of the amount of the shift in the aggregate demand curve. The multiplier is the amount by which there is a change in equilibrium income or output when autonomous aggregate expenditure increases by one unit. It is this condition necessary for the multiplier to work.
Shifts in the Aggregate Demand and the Multiplier
In a two sector economy, the aggregate demand is a sum of consumption and investment expenditures. It is generally agreed that though both consumption and investment functions undergo a change from one period to another, the consumption function is relatively more stable than the investment function. Thus, the initial changes in income occur more due to the shifts in the investment function.
A few illustrations could explain us clearly the shifts in the aggregate demand as well as the multiplier.
Illustration 16
In an economy, the basic equations are as follows: the consumption function is C = 300 + 0.8Y and investment is Ī = \$ 360 millions. You are required to ascertain the following
1. The equilibrium level of income
2. The equilibrium level of income when planned investment increases from 360 to 400 millions, a total increases of 40 millions
3. The multiplier effect of the 40 millions increases in planned investment.
Solution
The equilibrium condition is given as Y = C + I
Y = 300 + 0.8Y + 360
Y – 0.8Y = 300 + 360
0.2Y = 660
Y = 3,300
1. The equilibrium level of income is Y = 3,300
The equilibrium condition is given as Y = C + I
Thus,
Y = 300 + 0.8Y + 400
Y – 0.8Y = 300 + 400
0.2Y = 700
Y = 700 / 0.2
2. Hence, the equilibrium level of income is 3,500
The equilibrium level of income increases from 3,300 to 3,500crores when planned investment increases from 360 to 400 millions. There is an increase in income by 200 millions. Hence the multiplier effect is
M = 1
1 – b
= 1
1 – 0.8
= 1
0.2
3. The multiplier effect is m is 5
Illustration 17
Presume in an economy the marginal propensity to consume is 0.75 and the level of autonomous investment decreases by 40 millions. Find,
1. The change in the equilibrium level of income
2. The change in consumption expenditures
Solution
Δ Y = m
Δ I
Also,
m = 1
1 – mpc
Thus, Δ Y = Δ I
m = Δ I 1
1 – b
= -40 x 1
1 – 0.75
= -160
(1) Thus, the decrease in autonomous investment causes a decrease in the equilibrium level of income by 160 millions. This effect occurs due to the reverse multiplier.
Y = C + I
Therefore, ΔY = Δ C + Δ S
-160 = Δ C – 40
The autonomous investment decreases by 40 millions, the saving will also decrease by 40 millions
Δ C = - 160 + 40
Δ C = - 120
(2) The consumption expenditure decreases by 120 millions
Illustration 18
Compute the value of the investment multiplier when the marginal propensity to continue is
(1) 0.80, (2) 0.65, (3) 0.40 and (4) 0.25
Find the effect of a decrease in the equilibrium income when autonomous investment decreases by 60 millions when the marginal propensity to consume is (1) 0.80, (2) 0.65, (3) 0.40 and (4) 0.25
Solution
The value of m, the investment multiplier is
m = 1
1 – mpc
Hence,
(1) m = 1
1 – 0.8
= 1 / 0.2 = m = 5
(2) m = 1
1 – 0.65
= 1 / 0.35 = m = 2.87
(3) m = 1
1 – 0.4
= 1 / 0.2 = m = 1.67
(4) m = 1
1 – 0.25
= 1 / 0.75 = m = 1.33
Thus, the decrease in the equilibrium income when autonomous investment decreases by 60 millions is
Δ Y = Δ Im = 60 x 5 = 300
= 60 x 2.87 = 172.2
= 60 x 1.67 = 100.2
= 60 x 1.33 = 79.8
Illustration 19
In an economy the marginal propensity to consume is 0.50. the level of autonomous investment decreases by 60 millions. Find the following
1. The change in the equilibrium level of income
2. The change in autonomous demand
3. The induced change in the consumption expenditure
Solution
Δ Y = m
Δ Ȳ
But, m is the investment multiplier
Also,
m = 1
1 – mpc
Thus,
Δ Y = ΔIm = ΔI 1
1 – mpc
= - 60 x 1
1 – 0.50
= - 120
1. Hence, the decrease in autonomous investment causes a decrease in the equilibrium level of income by 120 millions.
2. The decrease in investment by 60 millions is the change in the level of autonomous demand.
3. Y = C + I
Therefore,
Δ Y = Δ C + ΔS
-120 = ΔC – 60
Δ C = -120 + 60
4. The Consumption expenditure falls by 60 millions
Illustration 20
Presume that in two sector economy, the income is \$ 1000 million while the marginal propensity to consume is 0.40.
Suppose the government wants to increase the income to \$ 1600 million, by an amount of \$ 600 million.
By how much should the autonomous investment be increased?
Solution
The income level = \$1000 millions
The planned income level is \$1600 millions
Change in income = Δ Y = 1600 – 1000
= \$600 millions
But
Δ Y = m = m ΔI 1
Δ I 1 – b
Δ Y = Δ I 1
1 – b
600 = Δ I 1
1 – b
= Δ I 1 / 1-0.4
= Δ I 1 / 0.6
600 = Δ I 1.67
Δ I = 600 / 1.67
Thus, the autonomous investment should be increased by \$360 millions for the income to increase to \$ 1600 millions
An increase in income by \$ 600 millions
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Other topics under National Income and Its Determination: | 2,003 | 8,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-05 | latest | en | 0.87158 |
http://www.jiskha.com/10th_grade/?page=18 | 1,495,586,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607726.24/warc/CC-MAIN-20170524001106-20170524021106-00150.warc.gz | 557,741,015 | 6,922 | fractions
Mr.shafier has been teaching school for 10 years.For 6 of those years he taught 8th grade.what fraction represents the time Mr.shafier has not been teaching 8th grade?I knoe the answere is 2 5ths. but how do i show work?HELP!!!!!!!!!!!!!!!
Solve 3r + 18 ¡Ü 6r
if the ratio is 3:2 and length is 5 inches.what is the widith
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English
1. Mom had to check the timer every 10 minutes. 2. Mom had to check the timer every 10th minute. (Can we use both expressions?) 3. How about going at 6:00 at the art center. 4. How about coming at 6:00 at the art center. 5. How about arriving at 6:00 at the art center. 6. How ...
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What is the mode of these # 19,20,32,26
Art History
7. Compare the naturalism of Chinese art of the 10th to 12th centuries with the Western Romanesque art of the same time period. Use the Guanyin Bodhisattva, the handscroll, Ladies Preparing Newly Woven Silk, and the landscape paintings for specific references (see pages 346-49...
2(2.5b-9) + 6b= -7 what is b
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find the next three numbers in the pattern 2/3, 1 5/12, 2 1/6, 2 11/12 _, _, _
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6 1/8 - x = 1 7/8 ? would it be 4 1/8 = x
In doing alphabetical order which of these words com first?its or it's
CHIEF OR RULER OF A NARRAGANSETT TERRITORY
i am a number between 1 and 5. you can make exactly 3 arrays with me. who am i?
how many miles does the earth travles a minute or second ????
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We are learning about potential and kinetic energy. There are two ramps. One is short and the other is long. The long one is labeled A, and the short one is labeled B. On which one would a marble go the farthest after touching the ground?
Draw a triangle with a perimeter of 50 ft.
Ashford SAP for FINANCIAL AID
I have read in the handbook about SAP.I pass 2 of 3 courses but my GPA IS 2.0 AS REQUIRED.My grade A- C+ and F which is 62.24%,since when has a grade of 62 is a F???? I know it is a D.If I am mistaken about what in the handbook for SAP so be it,but I am going to withdraw ...
Having trouble with division word problems
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what is -8(700-3)= (using distributive property)
what is -8(700-3)= (using distributive property)
Repost perimeter is 30 and area 36
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statistics
In one history class, 34 students took an exam. The mean of the exam was 52 with a standard deviation of 13. If the instructor is going to assume data are normally distributed and assign grades on a curve, then she will assign an A to the top 10% of the class, B's to the next ...
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math
did you know when you try to find the least commen multiple of 3 numbers you have to divide by two after you get your answer my class (ms.lott in texas) told us that for some reason its doubling the number so always divide by two once you have crossed out the commen numbers ...
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write down five" th"words/
6 - m = 2 and 3/10
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algebra
Mixture Problems Template – Price Problems Copy the problem from the lesson: 1. A store manager wishes to reduce the price on her fresh ground coffee by mixing two grades. If she has 50 pounds of coffee which sells for \$10 per pound, how much coffee worth \$6 per pound must ...
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The trem hierarchy refers to
What is 16 - 10 - 2 - 3 = ? What is 8 - 2 - 2 - 2 = ? What is 10 - 3 - 3 - 4 = ? What is 8 - 5 - 5 - 5 = ? What is 7 - 7 - 7 - 7 = ? What is 12 - 15 - 10 - 8 = ? What is 3 - 7 - 4 - 2 = ? What is 9 - 3 - 6 - 5 = ? What is 11 - 9 - 6 - 6 = ? What is 6 - 5 - 4 - 3 = ? What is 13...
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Math
About 3/4 of the track team are girls. About 3/4 of these girls are in 8 th grade. What fraction of th e students are grade 8 girls? 1/3?
1+9a-8=8a
Problem: How many 3/4ths are in 1? Is the calculation correct? 1 / 3/4 1 x 4/3 = 4/3
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7z - z
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Solve for x. 6 - x = -10 + 7 my answer: x = 9
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What is a parent function? | 1,970 | 6,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.925344 |
https://www.transum.org/Maths/Palindromes/Products.asp | 1,723,044,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00833.warc.gz | 799,797,777 | 11,514 | # Palindrome Products
## Use long multiplication and pattern spotting to fill in the table with palindromic products.
##### Menu Level 1Level 2 Help More
This is level 1: Use long multiplication on paper to complete this 3 by 3 table.
For each cell multiply the nomber at the end of the row (blue) by the number at the top of the column (red).
×111111111
11
111
1111
Check
This is Palindrome Products level 1. You can also try:
Level 2 More
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 14 September 'Starter of the Day' page by Trish Bailey, Kingstone School:
"This is a great memory aid which could be used for formulae or key facts etc - in any subject area. The PICTURE is such an aid to remembering where each number or group of numbers is - my pupils love it!
Thanks"
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"Find the starters wonderful; students enjoy them and often want to use the idea generated by the starter in other parts of the lesson. Keep up the good work"
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#### Snooker Investigation
Given the width and height of a snooker table can you predict which pocket the ball will end up in and how many times will it bounce off one of the sides?
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## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
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## Description of Levels
Close
Level 1 - Use long multiplication on paper to complete this 3 by 3 table
Level 2 - Use the patterns you have seen in Level 1 to complete this 5 by 5 table
More Palindromes - Resources to guide your exploration of this fascinating concept.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Example
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.
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Close | 1,189 | 5,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.929013 |
http://kruel.co/tag/introduction/ | 1,638,067,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00039.warc.gz | 42,760,322 | 9,331 | # introduction
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## A Guide to Bayes’ Theorem – A few links
### miscellaneous
A law of probability that describes the proper way to incorporate new evidence into prior probabilities to form an updated probability estimate. Bayesian rationality takes its name from this theorem, as it is regarded as the foundation of consistent rational reasoning under uncertainty. A.k.a. “Bayes’s Theorem” or “Bayes’s Rule”.
Eliezer Yudkowsky is on bloggingheads.tv with the statistician Andrew Gelman.
Several different points of fascination about Bayes…
When looking further, there is however a whole crowd on the blogs that seems to see more in Bayes’s theorem than a mere probability inversion…
Bayesian statistics is a system for describing epistemological uncertainty using the mathematical language of probability.
Bayesian probability is one of the most popular interpretations of the concept of probability.
Edwin T. Jaynes was one of the first people to realize that probability theory, as originated by Laplace, is a generalization of Aristotelian logic that reduces to deductive logic in the special case that our hypotheses are either true or false. This web site has been established to help promote this interpretation of probability theory by distributing articles, books and related material. As Ed Jaynes originated this interpretation of probability theory we have a large selection of his articles, as well as articles by a number of other people who use probability theory in this way…
Bayesian statistics is so closely linked with induction that one often hears it called “Bayesian induction.” What could be more inductive than taking a prior, gathering data, updating the prior with Bayes Law, and limiting to the true distribution of some parameter?
Gelman (of the popular statistics blog) and Shalizi point that, in practice, Bayesian statistics should actually be seen as Popper-style hypothesis-based deduction. The problem is intricately linked to the “taking a prior” above.
Or, how to recognize Bayes’ theorem when you meet one making small talk at a cocktail party.
Still, I’m sure Blogger won’t mind me using their resources instead. The basic idea is that there’s a distinction between true values x and measured values y. You start off with a prior probability distribution over the true values. You then have a likelihood function, which gives you the probability P(y|x) of measuring any value y given a hypothetical true value x.
In other words, What is so special about starting with a human-generated hypothesis? Bayesian methods suggest what I think is the right answer: To get from probabilistic evidence to the probability of something requires combining the evidence with a prior expectation, a “prior probability”, and human hypothesis generation enables this requirement to be ignored with considerable practical success.
Andrew Gelman recently responded to a commenter on the Yudkowsky/Gelman diavlog; the commenter complained that Bayesian statistics were too subjective and lacked rigor. I shall explain why this is unbelievably ironic…
Maybe this kind of Bayesian method for “proving the null” could be used to achieve a better balance.
Bayesian brain is a term that is used to refer to the ability of the nervous system to operate in situations of uncertainty in a fashion that is close to the optimal prescribed by Bayesian statistics.
—————————————
P.S.
Expect this link collection to be permanently updated. | 688 | 3,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.951919 |
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## 数学代写|matlab代写|Vibration of floating bodies
Consider a solid cylinder of radius $a$ that is partially submerged in a bath of pure water as shown in Figure 2.2.2. Let us find the motion of this cylinder in the vertical direction assuming that it remains in an upright position.
If the displacement of the cylinder from its static equilibrium position is $x$, the weight of water displaced equals $A g \rho_{w} x$, where $\rho_{w}$ is the density of the water and $g$ is the gravitational acceleration. This is the restoring force according to the Archimedes principle. The mass of the cylinder is Ah $\rho$, where $\rho$ is the density of cylinder. From Newton’s second law, the equation of motion is
$$\rho A h x^{\prime \prime}+A g \rho_{w} x=0,$$
or
$$x^{\prime \prime}+\frac{\rho_{w} g}{\rho h} x=0 .$$
From Equation 2.2.20 we see that the cylinder will oscillate about its static equilibrium position $x=0$ with a frequency of
$$\omega=\left(\frac{\rho_{w} g}{\rho h}\right)^{1 / 2} .$$
When both $A$ and $B$ are nonzero, it is often useful to rewrite the homogeneous solution, Equation 2.2.5, as
$$x(t)=C \sin (\omega t+\varphi)$$
to highlight the amplitude and phase of the oscillation. Upon employing the trigonometric angle-sum formula, Equation $2.2 .22$ can be rewritten
$$x(t)=C \sin (\omega t) \cos (\varphi)+C \cos (\omega t) \sin (\varphi)=A \cos (\omega t)+B \sin (\omega t) .$$
From Equation 2.2.23, we see that $A=C \sin (\varphi)$ and $B=C \cos (\varphi)$. Therefore,
$$A^{2}+B^{2}=C^{2} \sin ^{2}(\varphi)+C^{2} \cos ^{2}(\varphi)=C^{2},$$
and $C=\sqrt{A^{2}+B^{2}}$. Similarly, $\tan (\varphi)=A / B$. Because the tangent is positive in both the first and third quadrants and negative in both the second and fourth quadrants, there are two possible choices for $\varphi$. The proper value of $\varphi$ satisfies the equations $A=C \sin (\varphi)$ and $B=C \cos (\varphi)$
## 数学代写|matlab代写|DAMPED HARMONIC MOTION
Free harmonic motion is unrealistic because there are always frictional forces that act to retard motion. In mechanics, the drag is often modeled as a resistance that is proportional to the instantaneous velocity. Adopting this resistance law, it follows from Newton’s second law that the harmonic oscillator is governed by
$$m \frac{d^{2} x}{d t^{2}}=-k x-\beta \frac{d x}{d t},$$
where $\beta$ is a positive damping constant. The negative sign is necessary since this resistance acts in a direction opposite to the motion.
Dividing Equation $2.3 .1$ by the mass $m$, we obtain the differential equation of free damped motion,
$$\frac{d^{2} x}{d t^{2}}+\frac{\beta}{m} \frac{d x}{d t}+\frac{k}{m} x=0,$$ or
$$\frac{d^{2} x}{d t^{2}}+2 \lambda \frac{d x}{d t}+\omega^{2} x=0 .$$
We have written $2 \lambda$ rather than just $\lambda$ because it simplifies future computations. The auxiliary equation is $m^{2}+2 \lambda m+\omega^{2}=0$, which has the roots
$$m_{1}=-\lambda+\sqrt{\lambda^{2}-\omega^{2}}, \quad \text { and } \quad m_{2}=-\lambda-\sqrt{\lambda^{2}-\omega^{2}}$$
From Equation 2.3.4 we see that there are three possible cases which depend on the algebraic sign of $\lambda^{2}-\omega^{2}$. Because all of the solutions contain the damping factor $e^{-\lambda t}$, $\lambda>0, x(t)$ vanishes as $t \rightarrow \infty$.
# matlab代写
## 数学代写|matlab代写|Vibration of floating bodies
$$\rho A h x^{\prime \prime}+A g \rho_{w} x=0,$$
$$x^{\prime \prime}+\frac{\rho_{w} g}{\rho h} x=0 .$$
$$\omega=\left(\frac{\rho_{w} g}{\rho h}\right)^{1 / 2} .$$
$$x(t)=C \sin (\omega t+\varphi)$$
$$x(t)=C \sin (\omega t) \cos (\varphi)+C \cos (\omega t) \sin (\varphi)=A \cos (\omega t)+B \sin (\omega t) .$$
$$A^{2}+B^{2}=C^{2} \sin ^{2}(\varphi)+C^{2} \cos ^{2}(\varphi)=C^{2},$$
## 数学代写|matlab代写|DAMPED HARMONIC MOTION
$$m \frac{d^{2} x}{d t^{2}}=-k x-\beta \frac{d x}{d t},$$
$$\frac{d^{2} x}{d t^{2}}+\frac{\beta}{m} \frac{d x}{d t}+\frac{k}{m} x=0,$$
$$\frac{d^{2} x}{d t^{2}}+2 \lambda \frac{d x}{d t}+\omega^{2} x=0 .$$
$$m_{1}=-\lambda+\sqrt{\lambda^{2}-\omega^{2}}, \quad \text { and } \quad m_{2}=-\lambda-\sqrt{\lambda^{2}-\omega^{2}}$$
## 有限元方法代写
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## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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assignmentutor™您的专属作业导师 | 2,195 | 5,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.60836 |
https://www.jiskha.com/questions/1009193/a-wave-travels-a-distance-of-20m-in-3s-the-distance-btw-successive-crests-of-the-wave-is | 1,575,714,724,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00003.warc.gz | 764,175,907 | 4,397 | # physics
A wave travels a distance of 20m in 3s. The distance btw successive crests of the wave is 4cm. What is the frequency of the wave?
1. 👍 0
2. 👎 0
3. 👁 123
1. freq*wavelength=speed=20/3
frequency= 20/(3*.04) hz
1. 👍 0
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## Similar Questions
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More Similar Questions | 784 | 2,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-51 | latest | en | 0.940864 |
https://lafashionjudge.com/cup-stacking-challenge-worksheet/ | 1,653,215,903,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00567.warc.gz | 399,477,094 | 11,630 | # Cup Stacking Challenge Worksheet
Cup Stacking Challenge Worksheet. Your students will have a blast stacking cups while recording data in a scatterplot graph. K 2 practice sport stacking :
Get interactive with this cup stacking challenge pe powerpoint! Betacup coffee cup challenge winners: The united states customary cup holds 8 fluid ounces.
### Mathnook Featured Cool Math Games, Fun Games, Teaching Tools:
Get interactive with this cup stacking challenge pe powerpoint! Provide each team with six cups, one rubber band and four pieces of string that are each two feet long. Take them through these 24 fun, interactive cup stacking challenges as they put their brains and muscles to work at the same time!
### Teams Will Return The Cups To The Original Location Without Talking.
Cup stack challenge collaboration activity instructions divide students into teams of four. Cup stacking challenge worksheet worksheet from novenalunasolitaria.blogspot.com this activity can also be used for steam activities, maker spaces, tinkering labs, o it its a mega Put the cups in your students' hands and let the clock begin.
### Teams Will Stack The Cups In A Tower Formation.
What better way to teach scatterplot graphs by incorporating a stem design challenge! To complete the activity, they will need to identify numerical patterns and make use of certain geometrical Students must work as a team to stack all of the cups in a pyramid.
### • 6 Paper Cups • 1 Rubber Band • 5 Pieces Of String 2.
This is just a small sample of one of my many stem challenges. They cannot touch the cups directly. The united states customary cup holds 8 fluid ounces.
### In Decimals, 1/3 Of A Cup Is.33 Cups, So.33 Cups Plus.33 Cups Equals.66 Cups.
Using their teamwork skills, students will work together to stack their cups into a pyramid with 3 cups on bottom, 2 cups in the middle, and 1 cup on top. If you double 3/4 of a cup, you'll get 6/4 cups, which can be simplified as 3/2 cups or 1 1/2 cups. They cannot touch the cups directly. | 445 | 2,030 | {"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} | 3.703125 | 4 | CC-MAIN-2022-21 | latest | en | 0.887565 |
https://socialsci.libretexts.org/Bookshelves/Sociology/Introduction_to_Sociology/Book%3A_Sociology_(Boundless)/13%3A_Education/13.05%3A_The_Conflict_Perspective_on_Education/13.5E%3A_Tilting_the_Tests-_Discrimination_by_IQ | 1,674,879,707,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00263.warc.gz | 555,889,440 | 30,038 | # 13.5E: Tilting the Tests- Discrimination by IQ
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
IQ is meant to measure intelligence but its validity as a measure of intelligence has been debated.
Learning Objectives
• Discuss the various explanations for the IQ gap, ranging from genetic to environmental factors
## Key Points
• The concept of intelligence itself may be culturally variable.
• Although some find evidence of a race -based IQ gap, others argue that race is not a causal variable and that race-based IQ differences are in fact caused by other differences correlated with race, such as health, wealth, and educational disparities.
• Socioeconomic status can affect many aspects of life, and therefore seems like a likely environmental influence on intelligence.
• Systemically disadvantaged minorities, such as the blacks in the United States, generally perform worse in the educational system and in intelligence tests than the majority groups or less disadvantaged minorities.
• Several studies have proposed that a large part of the IQ gap can be attributed to differences in quality of education.
• Peer groups and family can influence behavior and values.
• Peer groups and family can influence behavior and values.
## Key Terms
• IQ gap: The gap in average IQ scores between populations, usually measured along racial lines, though with much disagreement.
• Environmental factors: Factors that come from one’s environment, upbringing, or social situation, rather than biology.
• intelligence: Capacity of mind, especially to understand principles, truths, facts or meanings, acquire knowledge, and apply it to practice; the ability to learn and comprehend.
Intelligence is commonly measured using intelligence quotient (IQ) tests, which are meant to be a general measure of intelligence. However, IQ tests only measure a narrow band of the broad spectrum of intelligence, excluding factors such as creativity or emotional intelligence. Some researchers have raised more serious questions about the validity of IQ tests for measuring intelligence, especially across cultures. For example, IQ tests may be inappropriate for measuring intelligence in non-industrialized communities, because they focus on modern, rational-style thinking, a type of reasoning that is common in the modern industrial West but may be alien to other cultures. Although some find evidence of a race-based IQ gap, others argue that race is not a causal variable and that race-based IQ differences are in fact caused by other differences such as health, wealth, and educational disparities.
## The IQ Gap
In the United States, IQ tests have consistently demonstrated a significant degree of variation between different racial groups. On average, IQ scores are highest among Asian Americans, lower among whites, and lowest among blacks. Yet these IQ gaps are only observed in average scores and say very little about individuals. Plus, IQ scores show considerable overlap between these group scores, and individuals of each group can be found at all points on the IQ spectrum. Thus, the implications of the IQ gaps are unclear. And while the existence of racial IQ gaps is well-documented, researchers have not reached a consensus as to their cause. In general, explanations fall into one of two camps: genetic explanations and environmental explanations.
## Genetic Explanations
Explanations of ethnically innate intelligence were fairly common early in the twentieth century, with the rise of the American eugenics movement. But after World War II, they quickly fell out of favor over fear of being associated with Nazism. Today, even those who believe intelligence may have some genetic component tend to acknowledge the importance of environmental effects as well.
Many researchers are reluctant to adopt genetic explanations of the IQ gap because of their historical and political implications. The connection between race and intelligence has been a subject of debate in both popular science and academic research since the inception of intelligence testing in the early twentieth century. But even before IQ tests were invented, claims of race-based intelligence gaps were used to justify colonialism, slavery, and racial eugenics. In the late nineteenth and early twentieth centuries, much of the “scientific” evidence for racial intelligence gaps came from measurements such as brain size or reaction times.
The first IQ test was created between 1905 and 1908 and revised in 1916, during a time when Americans were quite concerned about an influx of new immigrants. Different nationalities were sometimes thought to comprise different races, especially nationalities newer to the United States, such as the Irish, the Slavs, and the Italians. Alfred Binet, the developer of these tests, warned that they should not be used to measure innate intelligence or to label individuals. Despite his warnings, the tests were used to evaluate draftees for World War I, and researchers found that people of southern and eastern European backgrounds scored lower than native-born Americans. At the time, such data was used to construct an ethnically based social hierarchy, one in which immigrants were rejected as unfit for service and mentally defective. It was not until later that researchers realized that lower language skills by new English speakers affected their scores on the tests.
## Environmental Explanations
Researchers have suggested a wide array of environmental factors that might influence intelligence. In general, these factors are not mutually exclusive with one another: more than one of them may come into play at the same time. In fact, some may even directly contribute to others. Furthermore, the relationship between genetics and environmental factors is likely complicated. For example, the differences in socioeconomic environment for a child may be due to differences in genetic IQ for the parents, and the differences in average brain size between races could be the result of nutritional factors.
## Socioeconomic Environment
Socioeconomic status can affect many aspects of life, and therefore seems like a likely environmental influence on intelligence. People who grow up in a community with lower socioeconomic status may have fewer enrichment opportunities (like going to museums) or a less stimulating home environment, as well as unequal access to health care, nutritious food, and quality education. But research suggests that differences in socioeconomic status cannot entirely explain the IQ gap. In part, this is because the effects of socioeconomic status are hard to isolate and measure, and are probably not independent of intelligence itself.
13.5E: Tilting the Tests- Discrimination by IQ is shared under a CC BY-SA license and was authored, remixed, and/or curated by LibreTexts. | 1,631 | 7,813 | {"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} | 2.875 | 3 | CC-MAIN-2023-06 | latest | en | 0.73581 |
https://allnurses.com/general-nursing-student/iv-dilution-problem-659471.html | 1,529,765,395,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00395.warc.gz | 556,634,523 | 12,514 | # IV Dilution Problem Help!
1. Hey everyone! I need some help with this problem. I want to know how to work it out, not just the answer please! Any help would be very much appreciated!
A 10 year old child is to receive Zosyn 750 mg IVPB q12h. The drug book states: dilute 10-20 mg/ml and administer over 30 minutes. The pharmacy has sent up Zosyn 800 mg in 15 ml. How should the nurse dilute and administer this medication?
A. Add medication to a 25 ml bag of fluid and administer at 80 ml/hr with a VTBI of 40.
B. Add medication to a 50 ml bag of fluid and administer at 100 ml/hr with a VTBI of 50.
C. Add medication to a 100 ml bag of fluid and administer at 230 ml/hr with a VTBI of 115.
•
Joined: Jan '12; Posts: 1
3. Quote from Crunktastic
Hey everyone! I need some help with this problem. I want to know how to work it out, not just the answer please! Any help would be very much appreciated!
A 10 year old child is to receive Zosyn 750 mg IVPB q12h. The drug book states: dilute 10-20 mg/ml and administer over 30 minutes. The pharmacy has sent up Zosyn 800 mg in 15 ml. How should the nurse dilute and administer this medication?
A. Add medication to a 25 ml bag of fluid and administer at 80 ml/hr with a VTBI of 40.
B. Add medication to a 50 ml bag of fluid and administer at 100 ml/hr with a VTBI of 50.
C. Add medication to a 100 ml bag of fluid and administer at 230 ml/hr with a VTBI of 115.
This is a lousy problem because none of the choices is actually correct... However, answer {C} is arguably the best of the three.
For simplicity, I'll just brute-force it and solve each choice...
{A}
Find the concentration
800 mg
----------------
15 mL + 25 mL
800 mg/40 mL = 20 mg/mL (acceptable, at the high end of the range)
Calculate the quantity (mass, actually) of medicine given:
20 mg/mL x 40 mL = 800 mg (too much... by 7%)
40 mL
-----------
80 mL/hr
40/80 = 0.5 hours or 30 mins (acceptable)
{B}
Following same approach:
Concentration = 800 / (50 + 15) = 12.3 mg/mL (acceptable)
Quantity = 12.3 mg/mL x 50 mL = 615 mg (too low by 18%)
50/100 = 30 min (acceptable)
{C}
Following same approach:
Concentration = 800 / (100 + 15) = 7 mg/mL (too low... though not really problematic)
Quantity = 7 mg/mL x 115 mL = 805 mg (too high by 7%)
230/115 = 30 min (acceptable)
I pick choice {C} because the lower concentration is easier on the veins but still delivers the requisite amount of medication in the ordered time. One could argue in favor of {A} by stating that the smaller volume is preferable in a child but a 10-year-old probably weighs about 35 kg or so and a volume of 3 mL/kg is pretty minimal and not likely to create any issues in a kid without renal or cardiac problems.
I wouldn't be cool giving a dose 7% higher than the one ordered, though... would have to seek MD clarification before I gave it...
Or better yet, just add the 15 mL dose to a 50 mL bag and give 61 mL at 122 mL/hr... and say "pshaww" to stupid multiple-choice questions....
4. Quote from crunktastic
a 10 year old child is to receive zosyn 750 mg ivpb q12h. the drug book states: dilute 10-20 mg/ml and administer over 30 minutes. the pharmacy has sent up zosyn 800 mg in 15 ml. how should the nurse dilute and administer this medication?
a. add medication to a 25 ml bag of fluid and administer at 80 ml/hr with a vtbi of 40.
b. add medication to a 50 ml bag of fluid and administer at 100 ml/hr with a vtbi of 50.
c. add medication to a 100 ml bag of fluid and administer at 230 ml/hr with a vtbi of 115.
not sure what is meant by the highlighted text, except as a primary dilutent (which the pharmacy does). official fda information is:
"reconstituted zosyn solution should be further diluted (recommended volume per dose of 50 ml to 150 ml) in a compatible intravenous solution listed below. administer by infusion over a period of at least 30 minutes. during the infusion it is desirable to discontinue the primary infusion solution."
[color=#333333]
that information instantly eliminates one answer. of the other two choices, the rates would would infuse one in exactly 30 minutes, while the other would infuse in at a little more than 1/2 hours.
is this question purely math,or does it involve critical thinking? i.e. how much & how fast do you want to infuse into a 10 year old?
Last edit by MrChicagoRN on Jan 9, '12
close | 1,171 | 4,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.889974 |
https://www.yummymath.com/post/how-many-pennies-did-otha-save/ | 1,716,234,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00452.warc.gz | 958,185,236 | 11,459 | ### How Many Pennies did Otha Save?
2023-11-20 09:44:32
### How Many Pennies did Otha Save?
Otha Anders used picking up pennies as a way to remember to pray. In this 3-act activity we try to estimate how much he actually collected and stored in 5 gallon water jugs. This activity address modeling with math, estimating, and can serve as a great way for students to reason, problem solve and communicate mathematical thinking while practicing division and multiplication with decimals.
Option 1:
Kick off the activity with the article with vital information grayed out (number of pennies, value and number of jugs). Read the article together as a class. Ask kids how many pennies they think Otha saved over 45 years.
You might see students estimate the number of pennies saved on average per day or per week and then consider how many days or weeks have gone by in 45 years. Students might have many different ways to reason through the problem, so let them go with their intuitive reasoning.
Option 2:
You could also start the activity by sharing the number of five gallon jugs Otha had filled by sharing this version: article with number of jugs showing but the other information grayed out. Or give this info after kids have estimated the number of pennies he saved over 45 years using their own intuitive strategies. There is a video about this story here but be careful, the answer is given in the headline of the story below the video.
This activity could span a number of grade levels. Kids are estimating, reasoning and working with rational numbers. Finally, this is a great activity for high school students to engage in modeling and practice basic operations with rational numbers. It is also very likely that students will engage in the following math practices: MP1, MP2, MP3 and MP4.
Special thanks to Sheri Flecca telling us about this news story and suggesting it as a possible activity.
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Act 1 - Give students time to view the gam... | 605 | 2,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | longest | en | 0.94186 |
https://www.unix.com/man-page/centos/3/dlaic1.f | 1,716,592,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00145.warc.gz | 930,389,976 | 9,177 | # dlaic1.f(3) [centos man page]
```dlaic1.f(3) LAPACK dlaic1.f(3)
NAME
dlaic1.f -
SYNOPSIS
Functions/Subroutines
subroutine dlaic1 (JOB, J, X, SEST, W, GAMMA, SESTPR, S, C)
DLAIC1 applies one step of incremental condition estimation.
Function/Subroutine Documentation
subroutine dlaic1 (integerJOB, integerJ, double precision, dimension( j )X, double precisionSEST, double precision, dimension( j )W, double
precisionGAMMA, double precisionSESTPR, double precisionS, double precisionC)
DLAIC1 applies one step of incremental condition estimation.
Purpose:
DLAIC1 applies one step of incremental condition estimation in
its simplest version:
Let x, twonorm(x) = 1, be an approximate singular vector of an j-by-j
lower triangular matrix L, such that
twonorm(L*x) = sest
Then DLAIC1 computes sestpr, s, c such that
the vector
[ s*x ]
xhat = [ c ]
is an approximate singular vector of
[ L 0 ]
Lhat = [ w**T gamma ]
in the sense that
twonorm(Lhat*xhat) = sestpr.
Depending on JOB, an estimate for the largest or smallest singular
value is computed.
Note that [s c]**T and sestpr**2 is an eigenpair of the system
diag(sest*sest, 0) + [alpha gamma] * [ alpha ]
[ gamma ]
where alpha = x**T*w.
Parameters:
JOB
JOB is INTEGER
= 1: an estimate for the largest singular value is computed.
= 2: an estimate for the smallest singular value is computed.
J
J is INTEGER
Length of X and W
X
X is DOUBLE PRECISION array, dimension (J)
The j-vector x.
SEST
SEST is DOUBLE PRECISION
Estimated singular value of j by j matrix L
W
W is DOUBLE PRECISION array, dimension (J)
The j-vector w.
GAMMA
GAMMA is DOUBLE PRECISION
The diagonal element gamma.
SESTPR
SESTPR is DOUBLE PRECISION
Estimated singular value of (j+1) by (j+1) matrix Lhat.
S
S is DOUBLE PRECISION
Sine needed in forming xhat.
C
C is DOUBLE PRECISION
Cosine needed in forming xhat.
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
September 2012
Definition at line 135 of file dlaic1.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 dlaic1.f(3)```
## Check Out this Related Man Page
```zlaic1.f(3) LAPACK zlaic1.f(3)
NAME
zlaic1.f -
SYNOPSIS
Functions/Subroutines
subroutine zlaic1 (JOB, J, X, SEST, W, GAMMA, SESTPR, S, C)
ZLAIC1 applies one step of incremental condition estimation.
Function/Subroutine Documentation
subroutine zlaic1 (integerJOB, integerJ, complex*16, dimension( j )X, double precisionSEST, complex*16, dimension( j )W, complex*16GAMMA,
double precisionSESTPR, complex*16S, complex*16C)
ZLAIC1 applies one step of incremental condition estimation.
Purpose:
ZLAIC1 applies one step of incremental condition estimation in
its simplest version:
Let x, twonorm(x) = 1, be an approximate singular vector of an j-by-j
lower triangular matrix L, such that
twonorm(L*x) = sest
Then ZLAIC1 computes sestpr, s, c such that
the vector
[ s*x ]
xhat = [ c ]
is an approximate singular vector of
[ L 0 ]
Lhat = [ w**H gamma ]
in the sense that
twonorm(Lhat*xhat) = sestpr.
Depending on JOB, an estimate for the largest or smallest singular
value is computed.
Note that [s c]**H and sestpr**2 is an eigenpair of the system
diag(sest*sest, 0) + [alpha gamma] * [ conjg(alpha) ]
[ conjg(gamma) ]
where alpha = x**H * w.
Parameters:
JOB
JOB is INTEGER
= 1: an estimate for the largest singular value is computed.
= 2: an estimate for the smallest singular value is computed.
J
J is INTEGER
Length of X and W
X
X is COMPLEX*16 array, dimension (J)
The j-vector x.
SEST
SEST is DOUBLE PRECISION
Estimated singular value of j by j matrix L
W
W is COMPLEX*16 array, dimension (J)
The j-vector w.
GAMMA
GAMMA is COMPLEX*16
The diagonal element gamma.
SESTPR
SESTPR is DOUBLE PRECISION
Estimated singular value of (j+1) by (j+1) matrix Lhat.
S
S is COMPLEX*16
Sine needed in forming xhat.
C
C is COMPLEX*16
Cosine needed in forming xhat.
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
September 2012
Definition at line 136 of file zlaic1.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 zlaic1.f(3)```
Man Page | 1,297 | 4,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.757902 |
http://www.parent24.com/Family/Travel/6-classic-car-games-to-keep-your-kids-happy-20140530 | 1,505,821,807,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685129.23/warc/CC-MAIN-20170919112242-20170919132242-00709.warc.gz | 541,402,880 | 19,121 | Matric past exam papers Download old matric exam papers here
6 classic car games to keep your kids happy
If you're tired of hearing "Are we there yet?" from the back seat, try playing these games with your kids.
(iStockphoto.com)
It’s time for your long awaited family road trip. The car is packed, your playlist is ready and the kids are strapped safely in the backseat. But when the first “Are we there yet?” comes from behind you, your patience is tested. Make your road trip fun and bearable for the whole family with these fun games to play in the car.
The Animal Name Game
The game starts with one person naming an animal. Then the next person has to name another animal starting with the last letter of the previous animal. There are no winners in the game, but it encourages kids to think creatively. You can play the game with older children by naming TV shows, cities or countries.
I spy with my little eye
One person thinks of something in the car stating only the letter the word begins with. The other players then ask yes and no questions, like “Can I eat it?” or “Is it soft?” The first person to get the answer right is the winner.
The Alphabet Game
One person keeps an eye on the left-hand side of the road and someone else the right. Each player looks for letters of the alphabet that appear on signs, buildings or license plates on their side. The object of the game is to point out all the letters of the alphabet from A-Z and the first one to spot the entire alphabet is the winner.
The Memory Game
The first player starts by saying “A is for…” and fills the gap with a word starting with A. The next person has to repeat what the first player said and then also add a word starting with the next letter in the alphabet. As they move down the alphabet, the list of words to remember gets longer. By the time you reach the letter Z, that player will have to recite the whole alphabet and all the corresponding words.
Secret Place Race
One person looks at a map and chooses a town, city or river and announces it to the other players. A second player then gets one minute to look at the map and point out where the secret place is.
Humming Game
One person hums the tune to a TV show and the other players have to guess from which show it is. The first person to correctly name the show gets to hum next.
What do you do on family road trips when the little ones get restless? Share your tips by emailing chatback@parent24.com and we may publish your comments.
Source: Parents.com
More on
24.com publishes all comments posted on articles provided that they adhere to our Comments Policy. Should you wish to report a comment for editorial review, please do so by clicking the 'Report Comment' button to the right of each comment.
Comment on this story
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2017-09-19 12:14
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HousesR 1 190 000 | 812 | 3,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | longest | en | 0.953157 |
http://cstheory.stackexchange.com/questions/17914/does-a-noisy-version-of-conways-game-of-life-support-universal-computation | 1,469,483,941,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00216-ip-10-185-27-174.ec2.internal.warc.gz | 58,475,582 | 24,993 | # Does a noisy version of Conway's game of life support universal computation?
Quoting Wikipedia, "[Conway's Game of Life] has the power of a universal Turing machine: that is, anything that can be computed algorithmically can be computed within Conway's Game of Life."
Do such results extend to noisy versions of Conway's Game of Life? The simplest version is that after every round every live cell dies with a small probability $t$ and every dead cell becomes alive with a small probability $s$ (independently).
Another possibility is to consider the following probabilistic variant of the rule of the game itself.
• Any live cell with fewer than two live neighbours dies with probability $1-t$.
• Any live cell with two or three live neighbours lives with probability $1-t$ on to the next generation.
• Any live cell with more than three live neighbours dies with probability $1-t$.
• Any dead cell with exactly three live neighbours becomes a live cell with probability $1-t$.
Question: Do these noisy versions of the Game of Life still support universal computations? If not, what can be said about their "computational power"?
Related information on the computational power of cellular automata and noisy versions of cellular automata will be also much appreciated.
(This question developed from this question on MathOverflow. Vincent Beffara's answer on MO gave interesting references for related results on computational aspects of noisy cellular automata.)
-
@vzn 1) no, this is not the "real question," it's an altogether different question; Gil's question is about robustness of a simple computational model to noise, not about the power of randomness; 2) TMs with a random tape are no more powerful than deterministic TMs, see this answer: cstheory.stackexchange.com/a/1415/4896 – Sasho Nikolov Jun 7 '13 at 6:26
The real question here is if the stochastic/noisy versions of the "Game of Life" still support computation. (If these version support computations in P then their power may go all the way to BPP.) It is possible that the computational power of these stochastic versions of the game of life is much lower. – Gil Kalai Jun 8 '13 at 18:25
Perhaps I'm stating the obvious, but you can just duplicate a configuration enough times so as to guarantee with high probability that a version of the configuration doesn't even have one cell flipped. My personal belief is that we can do much, much better, but at least it's a simple lower bound. – user834 Jun 12 '13 at 16:33
I'm not sure the question is well-defined. Suppose $t=10^{−9}$. It seems to me that you might be able to find a computer that deals with all one-bit errors in the "Game of Life", giving you fault-tolerant computation unless you spontaneously get a large block of errors all at once. But I don't think anything can be robust against all errors. For example, suppose the errors spontaneously create a malevolent adversary determined to disrupt the computation. You might be able to show your computation succeeds with probability $>1−10^{−9}$ but fails with probability $>10^{−10000}$. Does this count? – Peter Shor Jun 13 '13 at 13:20
Peter, if your computation succeed with probability 2/3, I am happy. – Gil Kalai Jun 15 '13 at 11:57
Here are some "nearby best" references, for what it's worth. It would seem the way to go on this question is to reduce it to a question on "noisy Turing machines", which have been studied (somewhat recently), and which are apparently the nearest relevant area of the literature. The basic/general/reasonable answer seems to be that if the TM can resist/correct for noise (as is demonstrated in these references), it's quite likely the CA can also, within some boundaries/thresholds.
The question of how to reduce a "noisy CA" to a "noisy TM" (and vice versa) is more open. It may not be hard but there does not appear to be published research in the area. Another issue is that the noisy TM is a new model and therefore there may be multiple (natural?) ways to represent a noisy TM. For example, the following papers look at disruptions in the state transition function, but another natural model is disruptions in the tape symbols (the latter being more connected to noisy CAs?). There may be some relation between the two.
• Fault-tolerant Turing Machine by Ilir Capuni, 2012 (PhD thesis)
The Turing machine is the most studied universal model of computation. This thesis studies the question if there is a Turing machine that can compute reliably even when violations of its transition function occur independently of each other with some small probability.
In this thesis, we prove the existence of a Turing machine that with a polynomial overheadcan simulate any other Turing machine, even when it is subject to faults of the above type, thereby answering the question that was open for 25 years.
• A Turing Machine Resisting Isolated Bursts Of Faults by Ilir Capuni and Peter Gacs, 2012
• Noisy Turing Machines by Eugene Asarin and Pieter Collins, 2005
(Another question: could there be some connection between noisy TMs and probabilistic Turing Machines?)
-
Gil is asking if the GL is forgetting everything about its initial configuration in time independent of the size, when each cell "disobeys" the transition function independently of other cells with some small probability.
To the best of my knowledge, this is not known for the GL. It is a very interesting question though. If it can withstand the noise, then it should preserve its universality.
A quick overview of the state of the art is as follows.
1. Toom's rule can save one bit forever faults that occur independently of each other with some small probability.
2. It was widely believed (the positive rates conjecture) that all 1 dim CA are ergodic until P. Gacs constructed his multi-scale CA that can simulate any other CA with moderate overhead even when subjected to the aforementioned noise.
3. The question if G(acs)K(urdiumov)L(evin) rule can save one bit forever in the presence of the above noise is still open. Kihong Park -- a student of Gacs --- showed that it wont, when the noise is biased.
4. When the work in 2 was published, M. Blum asked if a TM can carry on its computation if at each step, the transition is not done according to the transition function with some small probability independently of other steps, assuming that the information stored on the tape far from the head does not decay. A positive answer was given by I. Capuni (another student of Gacs) in 2012.
-
"If it is not ergodic, then it will preserve its universality" ... do you have any evidence for this statement? Is this a theorem? Where is it proved? I believe that Gacs's work shows that this is true in at least one case, but I don't see how that proves it holds for Conway's game of Life. – Peter Shor Aug 4 '13 at 12:15
Thanks for pointing out. It is not a theorem but an interesting open question. Not being ergodic seems too little to ask for such a strong statement. – user8719 Aug 4 '13 at 13:45
For starters, keep in mind that research in Conway's Game of Life is still ongoing and future developments may present a far less complicated solution.
Now then. Interestingly enough, this is a topic that is actually as much in line with biology and quantum physics as with traditional computer science. The question at the root of the matter is if any device can effectively resist random alterations to its state. The plain and simple answer is that it is impossible to make a such a machine that is perfectly resistant to such random changes. Of course, this is true in much the same way that quantum mechanics could cause seemingly impossible events. What prevents these events from occurring (leading most people to declare them strictly impossible) is the stupendously small probability such an event has of happening. A probability made so small by the large scale difference between the quantum level and the human level. It is similarly possible to make a state machine that is resistant to small degrees of random change by simply making it so large and redundant that any "change" noticed is effectively zero, but the assumption is that this is not the goal. Assuming that, this can be accomplished in the same way that animals and plants are resistant to radiation or physical damage.
The question then may not be how to prevent low-level disturbances from doing too much damage, but rather how to recover from as much damage as possible. This is where biology becomes relevant. Animals and plants actually have this very ability at the cellular level.(Please note: I am speaking of cells in the biological sense in this answer) Now, in Conway's game of life the notion of building a computing device at the scale of single cells is appealing (it does, after all, make such creations much smaller and more efficient), but while we can build self-reproducing computers (see Gemini), This ignores the fact that the constructor object itself may become damaged by disturbances.
Another, more resilient, way I can see to solve this is to build computers out of self-reproducing redundant parts (think biological cells) that perform their operations, reproduce, and are replaced.
At this point we can see another interesting real-world parallel. These low-level disturbances are akin to the effects of radiation. This is most appreciable when you consider the type of damage that can be done to your cellular automata. It is easy to trigger the cascade failure or "death" of a cell in Conway's Game of Life, much the same as what happens to many cells exposed to radiation. But there exists the worst-case possibility of mutation, creating a "cancerous" cell that continues to reproduce faulty copies of itself that do not aid in the computational process, or produce incorrect results.
As I've said, its impossible to build a system that is entirely foolproof, you can only make it less and less likely for a fault to compromise the entire system. Of course, the fundamental question here is really "are probabilistic simulations themselves Turing complete" which has already been decided to be true. I would have answered that fundamental question initially, save that it wasn't what you asked.
-
Wow! Thanks for the drive-by-downvote! At any rate, I've revised my post, adding some information and sources. Sorry I didn't have the time to do that when I first posted this. I could modify this answer even further to fit community standards, if it wasn't for the fact that no reason was given for the downvote. – Hawkwing Jul 5 '13 at 20:23
As a non-voter, I don't see how this answers Gil's question. You address the question of whether "any device can effectively resist random alterations to its state", which is not what Gil asked. – András Salamon Jul 6 '13 at 11:39
Thanks (non-sarcastically this time) for the comment, András Salamon. I'd vote it useful myself, but I'm still a new user on this overflow site. Anyways, I'm sorry my answer seems off-topic. I did perhaps address the question more loosely than I'd intended, but I feel my answer does respond to the original question by answering a similar question and then drawing parallels between the two. Is this perhaps too roundabout a way of answering? – Hawkwing Jul 8 '13 at 12:16
I am reminded of xkcd 505: A Bunch of Rocks.
Any real-world computer is subject to some level of noise. A simulation of a universal computer in the ideal infinite Conway's Life universe will have a mean time between failures dependent on the engineering details of its design. It will compute reliably for a probabilistically quantifiable period, unreliably for a period of accumulating errors, and then not at all.
I would expect a fuzzy logic or quantum superposition model to demonstrate clearly what reliability should be expected of a particular construction. One may want to simulate the expected outputs of various components, rather than iterating over all of their cells, to whatever degree they can be isolated from each other. One might be able to quantify expected interference from failing components. A genetic algorithm should be the best way to develop fault-{tolerating,resisting,correcting} components with MTBFs as large as desired for a given noise distribution.
-
(mysterious voting here) A quantitative answer would be very speculative. There can't be a more precise answer than "yes, conditionally" without extensive experimentation on some chosen implementation of a UTM. A normal computer in a high-radiation environment is still practically a UTM, if only briefly. – user130144 Mar 4 '14 at 1:50 | 2,710 | 12,598 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-30 | latest | en | 0.917259 |
https://www.teacherspayteachers.com/Product/Spiral-Math-and-Language-Review-February-2nd-Grade-3613797 | 1,545,068,744,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00260.warc.gz | 1,075,161,014 | 20,981 | # Spiral Math and Language Review February | 2nd Grade
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1. February Morning Work 2nd Grade February Daily Math Looking to add rigor to your daily math routine? Is so, then this February Daily Math will do just that. I have been using this in my own 2nd grade classroom for 4 years now. It is amazing to see the growth each student makes by the end of the yea
2. February Morning Work 2nd Grade February Daily Language Looking to add rigor to your daily language routine? Is so, then this February Daily Language will do just that. I have been using this in my own 2nd grade classroom for 4 years now. It is amazing to see the growth each student makes by the en
Bundle Description
February Morning Work for Second Grade
Looking to add rigor to your morning routine? Is so, then this February Daily Math and Language will do just that. I have been using this in my own 2nd grade classroom for 4 years now. It is amazing to see the growth each student makes by the end of the year.
This set includes 20 pages of daily math and 20 pages of daily language work. You can use this many different ways: morning work, homework, seat work, or early finishers.
I chose skills that second grade students would be working on this time time of year. The pages gradually get harder and skills that are covered in the earlier pages are revisited towards the end as well.
February Daily Math
►Expanded Form
►Order Numbers, Median
Perimeter
►True / False Equations
►Word Problems
►Place Value
►Number Sense
►Arrays
►Fractions
►Money
Capacity
►Fact Families
►Part, Part Total
►Forms of Numbers
►Geometry
►10 More, 10 Less, 1 More, 1 Less
►Measurement
►2 Digit Subtraction (regrouping)
►100 More, 100 Less
►Comparing Numbers
►Telling Time
February Daily Language
Skills that are covered:
►Suffixes (-ful)
►Prefixes
►Cause & Effect
►Base Words
►Writing a Question
►Spelling
►Syllables
►Sight Words
►Contractions
►Guide Words
►Fact or Opinion
►Writing a Statement
►Synonyms
►Author's Purpose
►Compound Words
►Long Vowels
►Digraphs
►Reflexive Pronouns
►Kinds of Sentences
►Plural Nouns
►Homophones
►Compare & Contrast
►ABC Order
►Word Families
►Fix the Sentence
►Analogies
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**Higher 1st grade and lower 3rd grade students would benefit from this pack as well.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 763 | 3,055 | {"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} | 3.015625 | 3 | CC-MAIN-2018-51 | latest | en | 0.906117 |
http://blog.vmoroz.com/posts/2016-09-18-parametric-functions-in-scala.html | 1,686,163,640,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654012.67/warc/CC-MAIN-20230607175304-20230607205304-00480.warc.gz | 9,139,556 | 2,843 | ### Parametric polymorphism for functions in Scala
Posted on September 18, 2016
Let’s consider a simple typeclass:
``````trait Addable[T] {
def plus(x: T, y: T): T
}
def plus(x: Int, y: Int) = x + y
}
def plus(x: List[Int], y: List[Int]) = x ++ y
}
def plus[T](x: T, y: T)(implicit ev: Addable[T]) = ev.plus(x, y)
List(1, 2, 3).foldLeft(0)(plus) //> Int = 6
List(List(1), List(2), List(3)).foldLeft(List[Int]())(plus) //> List[Int] = List(1, 2, 3)``````
What happens here is called eta-expansion, and what `foldLeft` is getting is this function (e.g. for the former case):
``````new Function2[Int, Int, Int] {
def apply(x: Int, y: Int): Int = plus[Int](x, y)
}``````
What’s wrong here? We wrap `plus` invocation in anonymous function and effectively transfer control twice with required stack reshuffle, and it happens for every element in a collection (details here). Can we do better?
Let’s change just one line:
``def plus[T](implicit ev: Addable[T]) = (x: T, y: T) => ev.plus(x, y)``
It still works, only now we instantiate `Function2` by calling method `plus` once, and this function is passed to `foldLeft`. What is lost is the ability to call `plus` as a method (i.e. `plus(1,2)`). | 380 | 1,201 | {"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} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.766257 |
http://databasefaq.com/index.php/tag/kernel-density | 1,542,744,051,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746639.67/warc/CC-MAIN-20181120191321-20181120213321-00514.warc.gz | 81,634,908 | 3,124 | FAQ Database Discussion Community
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While trying to port some code from Matlab to R I have run into a problem. The gist of the code is to produce a 2D kernel density estimate and then do some simple calculations using the estimate. In Matlab the KDE calculation was done using the function ksdensity2d.m. In...
## Overlay density plot excludes histogram values
r,ggplot2,frequency,kernel-density
I want to overlay a density curve to a frequency histogram I have constructed. For the frequency histogram I used aes(y=..counts../40) because 40 is my total sample number. I used aes(y=..density..*0.1) to force the density to be somewhere between 0 and 1 since my binwidth is 0.1. However, density curve...
## KDE in python with different mu, sigma / mapping a function to an array
python,arrays,kernel-density
I have a 2-dimensional array of values that I would like to perform a Gaussian KDE on, with a catch: the points are assumed to have different variances. For that, I have a second 2-dimensional array (with the same shape) that is the variance of the Gaussian to be used...
## Custom histogram density evaluation in MatLab
matlab,histogram,kernel-density
Does MatLab have any built in function to evaluate the density of a random variable from a custom histogram? (I suspect there are probably lots of ways to do this, I am just looking to see if there is already any builtin MatLab functionality). Thanks.
## How to compare the distributions of two vectors in R?
python,r,distribution,kernel-density
Here is a screenshot of my dataset: Here's what it's about: Imagine that you work in a delivery company and, for some reason, the package fails to be delivered to the client. The distribution of the number of packages returned changes according to the monetary value of the package, which...
## Inconsistency between gaussian_kde and density integral sum
python,numpy,kernel-density
Can one explain why after estimation of kernel density d = gaussian_kde(g[:,1]) And calculation of integral sum of it: x = np.linspace(0, g[:,1].max(), 1500) integral = np.trapz(d(x), x) I got resulting integral sum completely different to 1: print integral Out: 0.55618 ... | 611 | 2,638 | {"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} | 2.984375 | 3 | CC-MAIN-2018-47 | longest | en | 0.874984 |
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- NUMECA (https://www.cfd-online.com/Forums/numeca/)
- - How to best set boundary conditions? (https://www.cfd-online.com/Forums/numeca/108389-how-best-set-boundary-conditions.html)
DonQuijote October 22, 2012 11:38
How to best set boundary conditions?
Hello,
I'm working in FINE/turbo with a fan blade geometry, but I'm encountering a lot of problems in finding a converged solution.
I've reviewed all properties but I'm unabled to find a correct convergence, so I hope someone could help me. I think the problem should be related with boundary conditions.
I imposed a total pressure condition at inlet (since a don't have any other value), and a average static pressure outlet. Both of them are constant values since I don't have any kind of profile information (in fact, I don't know the static pressure outlet value). How important is to begin the computation with profiles??
Furthermore, thinking about the physics, if the fan has x rotational speed, it is supposed to obtain an y outlet pressure depending of it, so how can I pre-set this unknow initial condition?(So: known p. inlet, known rotational speed, unk. outlet)
I hope someone could enlighten to me...:confused:
Hamidzoka October 23, 2012 00:35
Dear Sergio;
it is not neccessary to begin the computation with profiles and it has no negative effect on convergence. Profiles are just applied to find more realistic results specially when a validation with experimental data is required.
I think you should check the boundary conditions. They should be physically meaningful.
When flow passes thnrough the fan its total pressure increases.
So, you should consider this when setting the boundary conditions.
other possibility is the direction of rotation. check if you have correctly set the direction of rotation. It has a big effect on convergence.
what are your boundary conditions values? what is the level of RPM?
regards
xyz33uu October 23, 2012 04:29
Hello,
Have U ever checked the message box that might show you the problems why U couldn't make a conerged solution. And show me, then maybe I could help.
Best Regards
DonQuijote October 23, 2012 09:37
Thank you Hamidzoka and xyz33uu!
I'll explain better my case to show my main purpose and my problem. I'm studyng a specific fan blade (tunnel fans, not aircraft fan), so I have modelled it in autogrid, and now I want to validate the NUMECA results with information I have. After that, I want to change this blade and optimize it for another design point.
So, I have information about massflow, rpm and thrust. Then I set inlet condition as massflow imposed, and the outlet as averaged static pressure. With this configuration, the solution converges, but the axial thrust doesn't reach the real thrust data. The only value I can change is the outlet static pressure (as I unknow).
Thinking that, I was wondering how this parameter is caculated. Because I thought that axial thrust= (difference between inlet-outlet pressures)* A (Ainlet=Aoutlet). With this formula, I need a pressure difference of about 1500 Pa, but I'm unable to reach the mentioned thrust.
So any ideas that could help me?
Thank you very much for your interest!
Regards
xyz33uu October 23, 2012 22:05
Maybe we have different understanding about thrust. In my opinion, thrust=H（head）*density*gravity*A=(（difference between inlet-outlet pressure）+density*（vout^2-vin^2）/2)*A
DonQuijote October 24, 2012 03:22
:O I completely missed this part!!! Thank you john!!.
Then , if the thrust is T=massflow/g*V2+A*(P2-P1) (since the fan is static, V1=0), and I know all variables except P2 (P1=Patm), I can solve it, but when I use this parameter in Numeca, the resulting axial thrust is three times lower than the expected... Even when I consider a higher pressure outlet, the axial thrust is half the 'theoretical' one...(but, since the outlet pressure imposed in Numeca is static pressure, it has to be as Pamb.. is that correct?)
What I'm missing???
xyz33uu October 24, 2012 06:18
Let's think about the unit of (massflow/g*v2),which is (Kg/s)/(m/s^2)*(m/s) that is equal to Kg. And that is different from the unit of thrust. So I think you have got a wrong equation, right?
DonQuijote October 24, 2012 07:04
Independently of the ecuation (I use to obtain T in kg, changing for that the pressure units from N/m^2 to kg/m^2), I have a given 'theoretical' thrust (from technical data), and comparing it with the Numeca axial thrust, they are not corresponding (600 N to 1000 N).
But the problem converges quite well. I am sure about the rpm, and I've checked massflow, density, rpm in blade and hub area under the blade...
xyz33uu October 24, 2012 08:40
I'm so sorry that I havn't notice you want to calculate the axial thrust. And thrust and axial thrust are different. Axial thrust is a component of thrust.
DonQuijote October 24, 2012 09:10
But if the fan axis is aligned with the entire fan, it is not suposed to be the same? How would you calculate the component?
xyz33uu October 26, 2012 06:31
Have you checked your private message box?
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# IOPC14A ; small doubt
0 given a=n! and b as stated in the question http://www.codechef.com/problems/IOPC14A .Isnt it enough to just find (n!/b)%2 and this can be written as (n!)%2 * (b^(2-2))%2 =(n!)%2 am i wrong in assuming something because i m using fermet's little theorm asked 28 Jun '14, 16:18 1 accept rate: 0%
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# Surveyors Formula for Area of a Polygon
In this post, a math whiz and Python developer demonstrates the scientific power of Python and the NumPy library by using them to calculate the area of a polygon.
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If you know the vertices of a polygon, how do you compute its area? This seems like this could be complicated, with special cases for whether the polygon is convex or maybe other considerations. But as long as the polygon is "simple," i.e. the sides meet at vertices but otherwise do not intersect each other, then there is a general formula for the area.
## The Formula
But what does that mean? The notation is meant to be suggestive of a determinant. It's not literally a determinant because the matrix isn't square. But you evaluate it in a way analogous to 2 by 2 determinants: add the terms going down and to the right, and subtract the terms going up and to the right. That is:
This formula is sometimes called the shoelace formula because the pattern of multiplications resembles lacing a shoe. It's also called the surveyor's formula because it's obviously useful in surveying.
## Numerical Implementation
As someone pointed out in the comments, in practice you might want to subtract the minimum x value from all the x coordinates and the minimum y value from all the y coordinates before using the formula. Why's that?
If you add a constant amount to each vertex, you move your polygon but you don't change the area. So, in theory, it makes no difference whether you translate the polygon before computing its area. But in floating point, it can make a difference.
The cardinal rule of floating point arithmetic is to avoid subtracting nearly equal numbers. If you subtract two numbers that agree to k significant figures, you can lose up to k figures of precision. We'll illustrate this by taking a right triangle with base 2 and height π. The area should remain π as we translate the triangle away from the origin, we lose precision the further out we translate it.
Here's a Python implementation of the shoelace formula.
`````` def area(x, y):
n = len(x)
s = 0.0
for i in range(-1, n-1):
s += x[i]*y[i+1] - y[i]*x[i+1]
return 0.5*s``````
If you're not familiar with Python, a couple things require explanation. First, `range(n-1)` is a list of integers starting at 0 but less than n-1. Second, the -1 index returns the last element of the array.
Now, watch how the precision of the area degrades as we shift the triangle by powers of 10.
`````` import numpy as np
x = np.array([0.0, 0.0, 2.0])
y = np.array([np.pi, 0.0, 0.0])
for n in range(0, 10):
t = 10**n
print( area(x+t, y+t) )``````
This produces:
`````` 3.141592653589793
3.1415926535897825
3.1415926535901235
3.1415926535846666
3.141592651605606
3.1415929794311523
3.1416015625
3.140625
3.0
0.0``````
Shifting by small amounts doesn't make a noticeable difference, but we lose between one and two significant figures each time we increase t by a multiple of 10. We only have between 15 and 16 significant figures to start within a standard floating point number, and so eventually we completely run out of significant figures.
When implementing the shoelace formula, we want to do the opposite of this example: instead of shifting coordinates so that they're similar in size, we want to shift them toward the origin so that they're not similar in size.
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{{ parent.urlSource.name }} | 1,009 | 4,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-13 | latest | en | 0.919217 |
https://forum.arduino.cc/t/binary-counting-using-12-leds/150181/63 | 1,618,564,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00149.warc.gz | 365,667,666 | 8,316 | # Binary counting using 12 LEDs
Can we review what you have got working and what still needs to be done ?
Can you output the current number to the Serial monitor ? Forget the LEDs for now. Can you add to or subtract from the number using the keypad ?
If you have got that far, then converting the number to binary for display on the LEDs should be relatively easy, but let's solve one problem at a time.
AWOL: So you're saying instead of five LEDs lighting (unary), you want to see 101? (where 1 is a lit LED)
If that's all you want, then rip out this bit
``````for(counter=0; counter< sum; counter++){
//digitalWrite(ledPin[counter-1], HIGH);
digitalWrite(ledPin[counter], HIGH);
//delay(1000);
``````
}
``````
and substitute code based on "bitRead", "digitalWrite" and "sum"
``````
Yes, that's exactly what I want
UKHeliBob: Can we review what you have got working and what still needs to be done ?
Can you output the current number to the Serial monitor ? Forget the LEDs for now. Can you add to or subtract from the number using the keypad ?
If you have got that far, then converting the number to binary for display on the LEDs should be relatively easy, but let's solve one problem at a time.
OK, I pressed some combinations of keys and this is the result: 1+ 2 12 1+ 9 19 2+ 7 27 1+ 1 11 11 1+2 12 12 1+0 10 10 0+0 0 0+5 5 5 0+2 2 2 1 0 This is what I got from my serial monitor. The numbers on the left are the combinations of two keys. On the right is the sum Single numbers on the left is what I see on the LEDs (the LEDs turned on) If I press a big number (bigger than 12) all the LEDs will turn on then, by pressing '*' many times I can subtract from the sum and consequently turn LEDs off, one at the time. If I start with: 0+1 then I can press '#' many times and turn LEDs on, one at the time Hope this explains everything
Does adding or subtracting the number entered on the keypad change the number to be displayed OK ?
Yes, that's right. I'm using the code posted by HazardsMind since the last code I posted is not able to turn on the LEDs.
substitute code based on "bitRead", "digitalWrite" and "sum"
for(counter = 0; counter < 12; counter++) { byte My_bits = bitRead(sum, counter); digitalWrite(counter,My_bits); // some delay }
Thanks, I changed the code following the last suggestion. This is the code:
``````#include
int ledPin[12]={
13,12,11,10,9,8,7,6,5,4,3,2};
int number[]={
0,0};
int count = 0;
int keyPressed=0;
int counter=0;
int sum=0;
char firstKey;
char secondKey;
const byte ROWS = 4; //four rows
const byte COLS = 3; //three columns
char keys[ROWS][COLS] = {
{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}
};
byte rowPins[ROWS] = {31, 33, 35, 37}; //connect to the row pinouts of the keypad
byte colPins[COLS] = {39, 41, 43}; //connect to the column pinouts of the keypad
void setup(){
Serial.begin(9600);
for(counter=0; counter<12; counter++)
{
pinMode(ledPin[counter], OUTPUT); // sets the digital pins as output
}
}
void loop(){
/*if (key) {
Serial.println(key);
}*/
if (key) { // blocking from anything not needed
if (key != '*' && key !='#'){
Serial.print(key);
Serial.print(" ");
number[count] = key - '0';
if(count == 1) {
count = 0;
sum=(number[0] *10)+number[1];
Serial.print('\t');
Serial.println(sum);
}
else {
count++;
}
}
}
// end of else if
//Serial.println(number);
if (key == '#') // increment binary counter
{
Serial.println(sum++);
//digitalWrite(13, HIGH);
//delay (1000);
for(counter=0; counter<= 12; counter++)
{
digitalWrite(counter, My_bits);
//delay(1000);
} // end of increment counter
} // end of '#'
if (key == '*') // decrement binary counter
{
Serial.println(sum--);
for(counter; counter > sum; counter-- )
{
if (counter < 0) counter = 0;
digitalWrite(counter, My_bits);
//delay(1000);
} // end of decrement counter
} // end of '*'
}//end of loop
]
``````
It shows some LEDs on but it's not in binary. At least I know how to count in binary. I guess I am very close to my solution.
It shows some LEDs on but it's not in binary.
But you're not going to describe any of it.
OK, good luck with the rest of your project.
It shows some LEDs on but it's not in binary.
Ok so then what is it outputting?
Oh, you need to add +2 to the counter, because your leds start at pin 2. So its, digitalWrite(counter + 2, My_bits);
AWOL:
It shows some LEDs on but it's not in binary.
But you're not going to describe any of it.
OK, good luck with the rest of your project.
Of course I'm going to let you know about my progress after so much trouble. HazardsMind is right, so I added 2 to the counter. Now I can insert any number in binary up to 99 then I can count up by one, by pressing '#' repeatedly. However, I cannot count down, or subtract in binary. Addition only works when adding 1, one at the time That's it.
This is wrong
for(counter; counter > sum; counter-- )
you still need to count up to 12, because your reading the position of the bits in the sum. Check out this link. http://wiring.org.co/reference/bitRead_.html
You are right. Thanks What about the two remaining problems? Counting up by any number and Counting down In other words I want to be able to make binary addition and subtraction Regards
The Arduino is quite capable of performing addition and subtraction.
If you've got the display of any given binary number between 0 and 4095 solved, then isolating your remaining problems should be trivial.
Counting up by any number and Counting down
Ok, so this would mean you need to save the old number and just add or subtract it from the new number. Once you have the sum [u]or[/u] difference of the two, THEN you turn your LEDs on/off.
I suggest you make this, its own function and just pass in the sum or difference. You do know how to make functions that allow data to be inserted, right?
``````for(counter=0; counter < 12; counter++)
{
digitalWrite(counter, My_bits);
//delay(1000);
}
``````
`````` for(counter=0; counter<= 12; counter++)
``````
Probably doing no harm, but how many LEDs have you got, 12 or 13 ?
`````` for(counter; counter > sum; counter-- )
``````
What is the value of counter and sum when this loop executes ?
You could turn the display of the binary representation of your sum variable into a function and call it when necessary. Get it working once and use it many times.
for(counter=0; counter<= 12; counter++) Probably doing no harm, but how many LEDs have you got, 12 or 13 ?
I caught that and fixed the code I gave him. Now did he see the fix, I dont know, most likely not.
Hi guys, Thank you for the last replies. Just wondering what is missing now To get a number displayed on the LEDs I usually get that number plus 1 So I modified sum=(number[0] *10)+number[1]; to sum=(number[0] *10)+number[1]-1; Now I get the right answer on LEDs but the wrong one on the serial monitor(since new sum is sum minus 1) I tried a different combination of: counter+2 or counter +1 but it only works properly when I use counter+2. When using counter+1 I usually get sum minus half
but it only works properly when I use counter+2.
If, instead of "counter", you used the word "outputPinIndex" or "binaryPlace", would that make it easier to understand?
If you do
``````sum = aNumber;
``````
then Serial.print the number and output it to the LEDs in binary and get a different number displayed on each, then obviously one or other is wrong and I know which one my bet would be on. Do I remember correctly that your bit to pin conversion needed an offset of 2 because your LED pin numbers start at 2 ?
Have you tried
``````Serial.println(sum, BIN);
``````
to get a binary output to the serial monitor ?
Have you written a function to input the number in binary to the LEDs as I suggested in an earlier post ? That way you only have one place to look for the problem and put it right and you can test the function in a very simple program before you commit it to the main one,
I tried now what you said but the results are the same: the correct answer on LEDs and sum-1 on serial monitor. This is the code:
``````#include
int ledPin[12]={
13,12,11,10,9,8,7,6,5,4,3,2};
int number[]={0,0};
int count = 0;
int keyPressed=0;
int counter=0;
int sum=0;
char firstKey;
char secondKey;
const byte ROWS = 4; //four rows
const byte COLS = 3; //three columns
char keys[ROWS][COLS] = {
{'1','2','3'},
{'4','5','6'},
{'7','8','9'},
{'*','0','#'}
};
byte rowPins[ROWS] = {31, 33, 35, 37}; //connect to the row pinouts of the keypad
byte colPins[COLS] = {39, 41, 43}; //connect to the column pinouts of the keypad
void setup(){
Serial.begin(9600);
for(counter=0; counter<12; counter++)
{
pinMode(ledPin[counter], OUTPUT); // sets the digital pins as output
}
}
void loop(){
/*if (key) {
Serial.println(key);
}*/
if (key) { // blocking from anything not needed
if (key != '*' && key !='#'){
Serial.print(key);
Serial.print(" ");
number[count] = key - '0';
if(count == 1) {
count = 0;
sum=(number[0] *10)+number[1]-1;
Serial.print('\t');
Serial.println(sum,BIN);
}
else {
count++;
}
}
}
// end of else if
//Serial.println(number);
if (key == '#') // increment binary counter
{
Serial.println(sum++);
for(counter=0; counter<= 12; counter++)
{
digitalWrite(counter+2, My_bits);
//delay(1000);
} // end of increment counter
} // end of '#'
if (key == '*') // decrement binary counter
{
Serial.println(sum--);
for(counter; counter > sum; counter-- )
{
if (counter < 0) counter = 0; | 2,584 | 9,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.914341 |
https://www.physicsforums.com/threads/pchem-maxwell-boltzmann-molecule-mean-speed.769407/ | 1,604,025,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00487.warc.gz | 842,373,577 | 16,630 | # [PChem] Maxwell-Boltzmann Molecule Mean Speed
Hi,
I am having trouble with this concept...
"A mean speed (c) is calculated by multiplying each speed by the fraction of molecules that have that speed, and then adding all the products together. When the speed varies over a continuous range, the sum is replaced by an integral. To employ this approach here, we note that the fraction of molecules with a speed in the range v to v + dv is f(v)dv, so the product of this fraction and the speed is vf(v)dv. The mean speed is obtained by evaluating the integral
$$c=\int vf(v)dv$$."
This passage is dealing with the M-B Speed Distribution. So integrating the function would give you the probability of finding a molecule with the speed between the lower and upper bounds, correct? So if you just took the integral of the function with no bounds it should be equal to one. I am having trouble seeing why integrating the product of the function and the speed leads to an average...
...Could anyone try and clear this up for me?
Thanks
Related Atomic and Condensed Matter News on Phys.org
The integral over the function "f(v)" would equal one. What you integrate here is the function "v*f(v)".
The integral over the function "f(v)" would equal one. What you integrate here is the function "v*f(v)".
Yes sorry, I see that my wording was ambiguous. I am having trouble seeing why integrating v*f(v) gives you an average.
Dr Transport
Gold Member
it is a weighted average
Yes sorry, I see that my wording was ambiguous. I am having trouble seeing why integrating v*f(v) gives you an average.
If f(x) is a probability distribution then the integral over x*f(x) pretty much is the definition of the average, mean, or expectancy value of the property x. Not sure what I can do to help you understanding it. Maybe an example that does not explicity use an integral: Rolling a dice has an equal probability 1/6 for each side of the dice. The expectancy value is <x> = 1*1/6 + 2*1/6 +3*1/6 + 4*1/6 +5*1/6 + 6*1/6 = $$\sum_{i=1}^6 i * p(i)$$, with p(i) being the probability for event/value i.
I am not sure that emphasizing "weighted average" really helps understanding (but this thread has a chance to prove me wrong). Technically, the integral over x*f(x) is a weighted average of x with f(x) being the weights. But in my experience the term "weighted average" is only used when a sensible "unweighted average" would exist, which is not the case here (there is no point in or sensible meaning of "the average real number").
1 person
Thanks for following up :). Yes, I was confused at where the integral c=∫vf(v)dv was coming from (I don't like using random equations which I don't know where they came from) but I figured out it was pretty much just a definition. | 664 | 2,762 | {"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} | 3.59375 | 4 | CC-MAIN-2020-45 | latest | en | 0.923801 |
http://mathhelpforum.com/calculus/105034-equation-normal.html | 1,480,906,605,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541518.17/warc/CC-MAIN-20161202170901-00115-ip-10-31-129-80.ec2.internal.warc.gz | 172,553,270 | 11,410 | # Thread: Equation of the Normal
1. ## Equation of the Normal
Find the equation of the normal to the curve y=3x^2-2x-1 which is parallel to the line y=x-3
2. Originally Posted by creatively12
Find the equation of the normal to the curve y=3x^2-2x-1 which is parallel to the line y=x-3
You want to find the point $(x_0,y(x_0))$ at which $y'(x_0)=-1$.
3. can you please solve the question for me, i couldn't understand what u are saying
12y=12x-17
4. Originally Posted by creatively12
can you please solve the question for me, i couldn't understand what u are saying
No I will not. I don't do that. Sorry.
Now if you what to try to understand how it works, then I will help.
But I will not just hand you the answer.
5. do you know how to find the equation of a line when you are given a point?
6. yes i want to know why we have to calculate the coordinates with the gradient of the tangent and not with the gradient of the normal in this particular question, as i've solved this problem, but after converting the gradient 1 to -1, please tell me that
and thanks for the reply and procedure
7. Originally Posted by creatively12
yes i want to know why we have to calculate the coordinates with the gradient of the tangent and not with the gradient of the normal in this particular question, as i've solved this problem, but after converting the gradient 1 to -1, please tell me that
The slope of the normal line at $(x_0,f(x_0))$ is $\frac{-1}{f'(x_0)}$.
You see the normal is perpendicular to the tangent.
8. Originally Posted by creatively12
can you please solve the question for me, i couldn't understand what u are saying
12y=12x-17
Are you sure this is the answer?
9. Originally Posted by Arturo_026
Are you sure this is the answer?
Yes $12y=12x-17$ is the correct answer.
10. Originally Posted by Arturo_026
Are you sure this is the answer?
yes
11. but Plato, if we have to find the equation of a normal to a curve that is perpendicular to a line, not parallel, then from which gradient (tangent or normal) would we find the coordinates ??
12. Originally Posted by creatively12
but Plato, if we have to find the equation of a normal to a curve that is perpendicular to a line, not parallel, then from which gradient (tangent or normal) would we find the coordinates ??
No you did not read it very closely.
The normal is perpendicular to the tangent not to the given line.
So that case you are finding where the tangent to the curve is perpendicular to the given line.
Recall from basic geometry, two lines perpendicular to the same line, in this case the tangent, are themselves parallel. | 653 | 2,607 | {"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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-50 | longest | en | 0.952798 |
https://www.scirra.com/forum/how-do-i-use-the-tilt-control_t101216 | 1,524,546,846,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00567.warc.gz | 839,651,017 | 9,959 | # How do I use the tilt control
Get help using Construct 2
### » Mon Apr 07, 2014 10:49 am
Hey guys & girls ,
How can I use the alpha, beta, gamma features?
I don't understand how to tell the system to move into the direction the device is tilted, but as in a 8-way direction.
Regards,
Andrei
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### » Mon Apr 07, 2014 1:34 pm
I haven't tried this out yet, but I came across this old blog post here: http://ericterpstra.com/2012/11/construct-2-html5-mobile-game-turkey-trot-of-doom/
On that page, the author has a link for CAPX with accelerometer controls. You could look at that for ideas. Again, I'm not sure if that code is out-dated.
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### » Mon Apr 07, 2014 2:24 pm
Touch conditions
Compare acceleration
Compare the current device's motion as its acceleration on each axis in m/s^2 (meters per second per second). The effect of gravity can be included or excluded, but note that some devices only support accelerometer values including the effect of gravity and will always return 0 for acceleration excluding gravity.
Compare orientation
Compare the device's current orientation, if the device has a supported inclinometer. Alpha is the compass direction in degrees. Beta is the device front-to-back tilt in degrees (i.e. tilting forwards away from you if holding in front of you). A positive value indicates front tilt and a negative value indicates back tilt. Gamma is the device left-to-right tilt in degrees (i.e. twisting if holding in front of you). A positive value indicates right tilt and a negative value indicates left tilt.
What I know would fill a Book , what I don't know would fill a Library
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### » Mon Apr 07, 2014 8:14 pm
Thanks... granpa I know how to read. The problem is that I don't really understand how to set it up...
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### » Fri Apr 11, 2014 6:50 am
I also encounter this problem too! When i tilt my smartphone, the Player doesn't move to the left and right smoothly, it seems like you tap ur left key twice, not as smooth as Doodle Jump or Mega Jump controls. Anyone pls teach us what to put in the Event
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### » Fri Apr 18, 2014 3:32 pm
hello?
great support, super friendly forums... can you help?
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Reputation: 567 | 658 | 2,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-17 | latest | en | 0.892973 |
https://codereview.stackexchange.com/questions/278357/frequency-counter-pattern-can-we-get-other-better-algorithm-having-like-on | 1,718,609,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00866.warc.gz | 155,108,661 | 42,533 | # "Frequency Counter Pattern", Can we get other better algorithm having like, O(n) or O(n log n)
Problem Statement :
Write a function called matchSquare, which accept two arrays (arr1, arr2)
Criteria:
1. The function should return true, if every value in the arr1, has its corresponding values squared in the second array arr2
2. The frequency of the values must be the same.
Ex:
matchSquare([1,2,3], [4,1,9]) //true
matchSquare([1,2,3], [1,9]) //false, frequency and length issue
matchSquare([1,2,1], [4,4,1]) //false (Must have the same frequency),
//because in the arry2, there is 2 4's, which does not have a matching another 2 in the arr1,
My solution (brute force):
function matchSquare(arr1, arr2){
if(arr1.length !== arr2.length){
return false;
}
for(let i = 0; i < arr1.length; i++){ // This is O(n)
let correctIndex = arr2.indexOf(arr1[i] ** 2) // This is O(n)
if(correctIndex === -1) {
return false;
}
arr2.splice(correctIndex,1) //Here again after splice, the re-index of the arr2,O(n)
}
return true;
}
Together it is O(n^2),
Can we get any better approach here?
An explanation if possible would be helpful.
• (The point about frequency is tricky: [-1, 1], [1, 1].) Commented Jul 26, 2022 at 12:45
• Did you yourself write the code presented for review? Commented Jul 26, 2022 at 14:15
• O(n^2) is too high for your function. Looks like O(n log(n)). Each number checked either reduces the number of seaches by one (if matched) for the nexr value or returns false (no match).. Note that javascripts Set is considered O(n) for searches thus const matchSquare = (a, b, s = new Set(b)) => a.every(v => s.has(v * v))); will solve in O(n) time Commented Jul 26, 2022 at 15:21
• @Blindman67, I doubt you understood my question, with your code above the testcase matchSquare([1,2,1], [4,4,1]) should return false. but your code will return true, you are not considering the frequency thing, every item (it may have duplicate items also) from the arr1, should have the same value doubled in arr2,
– RONE
Commented Jul 26, 2022 at 18:11
• @RONE You can use a Map (or second Set) and create it manualy adding a repeat count. Then count repeats down as you find matches Commented Jul 26, 2022 at 18:34
One idea I had. If you copy array1 and square all the values with a map, then sort both arrays, you can use every() to test if they have all the same values at the same indexes. This might be easier to read, but would have to run some tests to compare performance.
function matchSquare( array1, array2 ) {
if ( array1.length != array2.length ) return false;
const testArray1 = [...array1].map( item => item ** 2 ).sort();
const testArray2 = [...array2].sort();
return testArray1.every( ( item, index ) => item == testArray2[index] );
}
console.log( matchSquare( [1,2,3], [4,1,9] ) );
console.log( matchSquare( [1,2,3], [1,9] ) );
console.log( matchSquare( [1,2,1], [4,1,4] ) );
You could also try using toString() and == on each array, instead of every() for the final comparison to see which is faster.
function matchSquare( array1, array2 ) {
if ( array1.length != array2.length ) return false;
const testArray1 = [...array1].map( item => item ** 2 ).sort();
const testArray2 = [...array2].sort();
return testArray1.toString() == testArray2.toString();
}
console.log( matchSquare( [1,2,3], [4,1,9] ) );
console.log( matchSquare( [1,2,3], [1,9] ) );
console.log( matchSquare( [1,2,1], [4,1,4] ) );
"Frequency Counter Pattern", Can we get other better algorithm having like, O(n) or O(n log n)? Yes, one example is the answer by dqhendricks using an algorithm of O(n log n) complexity due to the sort of the arrays. Your approach having a quadratic complexity can be reduced to a linear complexity as suggested by blindman67 by the use of Map where the specification requires maps to be implemented "that, on average, provide access times that are sublinear on the number of elements in the collection". So you can take your arr1 array, square the elements and create your map storing the occurrences of the elements:
let map = arr1.map(elem => elem ** 2)
.reduce((acc, elem) => acc.set(elem, (acc.get(elem) || 0) + 1)
, new Map());
Then you can use the arr2 elements decrementing the occurrences and checking if all occurrences are equal to 0 (so it is falsy) with the every function:
function matchSquareWithMap(arr1, arr2) {
let map = arr1.map(elem => elem ** 2)
.reduce((acc, elem) => acc.set(elem, (acc.get(elem) || 0) + 1)
, new Map());
arr2.forEach(elem => {
let noccurrences = map.get(elem);
if (!noccurrences) { return false; }
map.set(elem, --noccurrences);
});
return [...map.values()].every(elem => !elem);
}
console.log(matchSquareWithMap([1,2,3], [4,1,9]) === true ? 'pass' : 'fail');
console.log(matchSquareWithMap([1,2,3], [1,9]) === false ? 'pass' : 'fail');
console.log(matchSquareWithMap([1,2,1], [4,4,1]) === false ? 'pass' : 'fail');
So you can reach a linear complexity with the use of an additional storing struct like the Map that provide linear access times to its elements. | 1,428 | 5,035 | {"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} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.785941 |
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# Grenoble Ecole De Management
Author Message
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Joined: 18 Jan 2012
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### Show Tags
05 Jun 2012, 06:50
Hi,
Would you rate Grenoble Ecole De Management as a nearly Elite School?
Thx
Manager
Joined: 26 Feb 2012
Posts: 206
Concentration: Finance, Technology
GPA: 3.3
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jun 2012, 07:11
nimc2012 wrote:
Hi,
Would you rate Grenoble Ecole De Management as a nearly Elite School?
Thx
never heard of this school.. what's the average gmat here?
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jun 2012, 07:26
Hi,
The GMAT score is around 550 and its in Grenoble, near Lyon France. It is 63rd in the Economist times.
thx
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jun 2012, 07:29
and which colleges do you consider Near Elite?
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jun 2012, 07:32
Though it is ranked 63rd on the Economist Times...those who have attended the school say it is like the Ivey League in France.
Grenoble is considered the technology hub of Europe.
I would like to find out from anyone who has been a student here or plans to go this year.
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jun 2012, 07:32
that is why I posted in Near Elite and not the other three sections...
thanks
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Re: Grenoble Ecole De Management [#permalink]
### Show Tags
05 Jan 2016, 06:48
Hi guys,
i was trying to gather info about Grenoble MBA program.
Could anyone please highlight how much important is international exposure for the ad-com?
How are the job prospects at Grenoble?
Re: Grenoble Ecole De Management [#permalink] 05 Jan 2016, 06:48
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Cranfield new MSC Global Product Development and Management 0 08 Jun 2010, 01:19
Display posts from previous: Sort by | 957 | 3,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-04 | latest | en | 0.826563 |
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View Full Version : Shadow Mapping and PolygonOffset
Nakoruru
10-11-2002, 06:42 AM
I have never seen a paper or presentation on how to choose a polygon offset when doing shadow mapping.
It seems like there should be a better way than just adjusting it until it looks good.
pocketmoon
10-11-2002, 06:51 AM
Originally posted by Nakoruru:
It seems like there should be a better way than just adjusting it until it looks good.
Apparently that's what Pixar have to do http://www.opengl.org/discussion_boards/ubb/smile.gif
Nakoruru
10-11-2002, 07:06 AM
It must be luxurious to know exactly where all your objects and lights are going to be ^_^
I said I had never read a presentation which covered this, but actually the shadow mapping presentation from nVidia does give some insight into the error range and absolute error that is possible. I just didn't pay much attention to it because before now there was no way to really use it because it involved derivatives in screen space.
Maybe one could use the ddy and ddx functions to calculate a polygon offset per pixel in a fragment program, instead of using a constant one for the whole scene. It would increase the offset in areas likely to have precision problems.
dorbie
10-13-2002, 10:11 PM
Polygon offset already does this. That's why there are two parameters. One is a constant offset the other is a multiplier for derivative of z (measured in units per pixel span).
[This message has been edited by dorbie (edited 10-14-2002).]
SirKnight
10-14-2002, 03:47 AM
The RedBook shows the equation used in that function and also tells you how to choose the best values depending on what you are doing or want.
-SirKnight
Nakoruru
10-14-2002, 05:31 AM
I figured all this out after I actually started playing with it. The problem I have now is that polygon offset uses an implementation dependent value which is guaranteed to make a difference (for the unit parameter). I am wondering if it is possible to determine such a value for variable in a fragment program.
In other words, is there a way to bump a floating point number and guarantee that it will actually be incremented. Perhaps such a function is needed in glslang or Cg (if it does not already exist).
Actually, dorbie, 'already does that' implies that all I wanted to do was implement PolygonOffset in a fragment program. But, I also wanted to find a way to determine which parameters would be best for PolygonOffset --without guessing--, and do that per-pixel if possible. This is because I doubt there is a single good value you can set for an entire scene, and I hate the idea that shadow maps require tweaking.
davepermen
10-14-2002, 06:21 AM
one thing that stops tweaking is don't use a simple comparison, but a smooth curve..
instead of if(length(light - point) > shadowmap[point]) shadowed() else unShadowed(), do a lerp between some range. the range can be determined by the distance to the light, as you can calculate how big the smallest stepsize on the depthmap there is. or you can use even more values to estimate that bether..
but that could kill the artefacts (including some offset of course) quite much, as it would blur them. no thin lines, but some tinywinyshading..
dorbie
10-14-2002, 03:07 PM
It already does this. You don't need to know the value. It does exactly what you need. Not for offsetting in some fragment program but for offsetting for the purpose of artifact 'free' depth map shadows.
When rendering the *depth map* offset by the values guaranteed to displace away from the viewer to avoid self occlusion.
This is *exactly* what is required for perfect depth buffer offset, even with a low resolution depth map and imprecise depth buffer.
The only possible issue is that implementations have been attrocious at correctly assessing glPolygonOffset, thanks in part to complicity of ARB members deliberately not enforcing a useful quality test because they know they'd fail it.
[This message has been edited by dorbie (edited 10-14-2002).]
Nakoruru
10-15-2002, 07:34 AM
"It already does this..."
I know.
Now, lets proceed with the assumption that I am trying to do something for which PolygonOffset cannot be used for whatever reason. For instance, I want to offset something which is not the Z depth value, or a value which I have calculated myself. That is what I was talking about.
I was just saying it would be nice to have a function which incremented or decremented a value by the smallest quantum. This is because foo++ or foo-- on a floating point value is not always guaranteed to actually change the value (at least in C, maybe it will in glslang or Cg).
daveperman,
I tried to use the smoothstep function, but for some reason it did not help at all. All the shadow acne was still present, it was just smoother ^_^
It boggles my mind as to why I could not get it to work. Maybe I made a mistake. Have you tried it yourself? If so then what did your code look like?
EDIT: Dorbie, are you saying that PolygonOffset work perfectly, but that it doesn't because no body is actually complying with the standards? If so, then what values should be passed to PolygonOffset to make it work perectly for shadow maps?
[This message has been edited by Nakoruru (edited 10-15-2002).]
gaby
10-15-2002, 08:31 AM
pocketmoon,
I think that Pixar user needn't to adjust this all time : it might been a generic solution because under Maya, you needn't adjust the offset. The option is present, but I never use it. In this type of rendering engine (Maya, Renderman, Mentalray...), some advanced filtering processing are applied to smooth, blend and discare all bias problems : such filterings are not avaible on actual chips.
Gaby
davepermen
10-15-2002, 09:06 AM
Originally posted by Nakoruru:
I tried to use the smoothstep function, but for some reason it did not help at all. All the shadow acne was still present, it was just smoother ^_^
It boggles my mind as to why I could not get it to work. Maybe I made a mistake. Have you tried it yourself? If so then what did your code look like?
have you ever seen something from me? http://www.opengl.org/discussion_boards/ubb/biggrin.gif i'm a brainy, but i'm terrible bad in some topics to implement that stuff (basically matrices, and they are, uhm, quite important for shadowmapping to get it to work http://www.opengl.org/discussion_boards/ubb/biggrin.gif)
no, i never did it, but i worked with the nvidia shadowing demo, the one with emulated 16bit shadowing and all that fancy thing. and i know like that the artefacts that occur..
basically i suggest to bias them as before, by some constant value (wich _should_ be determinable. somehow. and thats what the topic is about.. http://www.opengl.org/discussion_boards/ubb/biggrin.gif), and then use the smoothstep (very smooth http://www.opengl.org/discussion_boards/ubb/biggrin.gif) to filter away the resting possible artefacts (even while there shouldn't be any http://www.opengl.org/discussion_boards/ubb/biggrin.gif) | 1,677 | 6,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-34 | latest | en | 0.936518 |
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# Thread: consecutive numbers (with wildcards between) detection
1. ## consecutive numbers (with wildcards between) detection
I'm coding an online game and the cards have values from 1-12 and there can be wild cards for any given number... how can i make a function that will detect a "Run of #" including wilds without manually making an if/else statement for each combination that is possible??
• this is what i have tried so far and it ALWAYS comes back with FAILED
PHP Code:
``` function isRun(\$numbervalues,\$num_wilds){ sort(\$numbervalues,SORT_NUMERIC); \$nextnumber = \$numbervalues[0]+1; \$newnumberset=array(); for(\$i=1;\$i<count(\$numbervalues);\$i++){ array_push(\$newnumberset,\$numbervalues[\$i]); } if(in_array(\$nextnumber,\$numbervalues)){ //we're advancing! //start the function again with "next number" and same number of wild cards if(isRun(\$newnumberset,\$num_wilds)){ return true; }else{ return false; } }elseif(\$num_wilds >0){ \$new_wildcount=\$num_wilds-1; //put a wild in for "next number" //start the function again with "next number" and one less wild card if(isRun(\$newnumberset,\$new_wildcount)){ return true; }else{ return false; } }else{ return false; } } \$thenumbers = array(2,3,6); //this array is missing 4 and 5 to form the complete run \$wilds=2; //these 2 wilds should take the place of 4 and 5 in this run if(!isRun(\$thenumbers,\$wilds)){ echo "Failed"; }else{ echo "Success!"; } ```
• Your function was a bit bloated. Here's my version (with a bit of help from a friend):
PHP Code:
``` function isRun2(\$number_values,\$num_wilds) { sort(\$number_values,SORT_NUMERIC); for(\$i=\$number_values[0];\$i<end(\$number_values);\$i++) { if(!in_array(\$i,\$number_values)) { if((\$num_wilds--)==0) { return false; } } } return true; } \$numbers=array(2,4,5,7); \$wilds=2; if(isRun2(\$numbers,\$wilds)) { echo "Success!"; } else { echo "Failed!"; } ```
• sweet! thanks. (and yea, i figured mine was bloated because i just was adding/removing stuff from it trying to get it to work). yours worked. thanks again!
•
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Topic: Answer to Dik T. Winter
Replies: 441 Last Post: Feb 5, 2013 6:25 AM
Messages: [ Previous | Next ]
Virgil Posts: 311 Registered: 5/26/09
Re: Answer to Dik T. Winter
Posted: Jun 17, 2009 5:27 PM
In article
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 17 Jun., 20:19, William Hughes <wpihug...@hotmail.com> wrote:
> > Your claim is that "no possibility exists to construct or to
> > distinguish by one or many or infinitely many nodes
> > of the tree another path."
> >
> > A: actually infinite paths exist,
> > B: the infinite tree contains a path p that can be
> > distinguished from every path of P.
> >
> > You agree
> >
> > A ==> B
> >
> > So if A is true then B is true
> > and your claim is false.
>
> So it is.
> >
> > > What do you understand by "all nodes"?
> >
> > "all nodes of t are in the tree"
> > There is no node that is in t but is not in the tree
>
> That is correct.
> >
> > "all nodes of t are in P"
> > There is no node that is in t but is not in an element of P
>
> Also correct. Every node of t and all predecessor nodes are in one
> path of P.
But unless every chain of successor nodes to any node in t is in P,
there are paths not in P.
> >
> > WM: the list of paths P is the same as the tree
Nope: A set of sets of nodes is not a set of nodes.
> >
> > Nope. You have agreed that the tree contains a subset of
> > nodes that is not contained in one element of the list of paths P.
>
> All subsets of nodes that are in the tree also are in the list of
> paths.
It is not even true that all totally-ordered-by-'ancestor of' sets of
nodes are in P, at least if P is countable. And no set that is not thus
totally ordered can possibly be in P.
--
Virgil
Date Subject Author
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Dik T. Winter
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 Virgil
5/28/09 mueckenh@rz.fh-augsburg.de
5/28/09 Virgil
5/28/09 Dik T. Winter
5/28/09 G. Frege
5/28/09 Jesse F. Hughes
5/29/09 Dik T. Winter
5/29/09 G. Frege
5/29/09 Dik T. Winter
5/30/09 G. Frege
5/30/09 Jesse F. Hughes
6/2/09 Dik T. Winter
5/30/09 Jesse F. Hughes
6/1/09 mueckenh@rz.fh-augsburg.de
6/1/09 Jesse F. Hughes
6/1/09 Virgil
6/2/09 george
6/2/09 Denis Feldmann
6/2/09 Dik T. Winter
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 Guest
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
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6/12/09 Virgil
6/12/09 William Hughes
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6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/14/09 Owen Jacobson
6/14/09 Virgil
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
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6/14/09 William Hughes
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6/14/09 Virgil
6/14/09 William Hughes
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6/14/09 Virgil
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6/14/09 Virgil
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6/14/09 William Hughes
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6/14/09 William Hughes
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6/15/09 William Hughes
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6/15/09 William Hughes
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6/15/09 Virgil
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6/15/09 William Hughes
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6/16/09 Virgil
2/5/13
6/16/09 William Hughes
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6/16/09 Virgil
6/16/09 William Hughes
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6/16/09 William Hughes
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6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
7/17/09 scriber77@yahoo.com
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
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6/17/09 Virgil
6/17/09 William Hughes
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6/18/09 Virgil
6/18/09 William Hughes
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6/18/09 Virgil
6/18/09 William Hughes
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6/19/09 Virgil
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 William Hughes
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
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6/20/09 george
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6/3/09 george
6/3/09 mueckenh@rz.fh-augsburg.de
6/3/09 george
6/3/09 Virgil
6/3/09 Guest
6/3/09 Marshall
6/3/09 Jack Markan
6/3/09 Dik T. Winter
6/3/09 Guest
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
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6/9/09 Virgil
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6/9/09 Virgil
6/10/09 mueckenh@rz.fh-augsburg.de
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6/10/09 mueckenh@rz.fh-augsburg.de
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6/11/09 Virgil
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Rainer Rosenthal
6/11/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 Dik T. Winter
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6/11/09 mueckenh@rz.fh-augsburg.de
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6/11/09 Dik T. Winter
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7/1/09 Guest
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6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Rainer Rosenthal
6/12/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Guest
6/12/09 Rainer Rosenthal
6/12/09 Virgil
6/3/09 george
6/3/09 Dik T. Winter
5/28/09 William Hughes
5/29/09 mueckenh@rz.fh-augsburg.de
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6/2/09 Virgil
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6/30/09 MeAmI.org
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6/18/09 Virgil
6/20/09 Guest | 5,404 | 11,889 | {"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} | 3.609375 | 4 | CC-MAIN-2018-17 | longest | en | 0.908924 |
https://git.xonotic.org/?p=xonotic/xonotic.git;a=commitdiff;h=019698126f64ddf9fdec9a602a94a1331274c606 | 1,643,109,078,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00319.warc.gz | 330,735,217 | 2,752 | author Rudolf Polzer Mon, 21 Jun 2010 13:09:03 +0000 (15:09 +0200) committer Rudolf Polzer Mon, 21 Jun 2010 13:09:03 +0000 (15:09 +0200)
index f9ba64a92215a935e890c8ef0168bc956ac8b58e..6c88ccbea21d4a16c1d6685faf36ee2dbc8e8528 100755 (executable)
@@ -3,19 +3,14 @@
set -e
: \${qual:=95}
+: \${qual_alpha:=99}
for X in "\$@"; do
case "\$X" in
*.jpg)
- if [ -n "\$scaledown" ]; then
- mogrify -geometry "\$scaledown" -quality 100 "\$X"
- fi
jpegoptim --strip-all -m\$qual "\$X"
;;
*.png|*.tga)
- if [ -n "\$scaledown" ]; then
- mogrify -geometry "\$scaledown" -quality 100 "\$X"
- fi
if convert "\$X" -depth 16 RGBA:- | perl -e 'while(read STDIN, \$_, 8) { substr(\$_, 6, 2) eq "\xFF\xFF" or exit 1; ++\$pix; } exit not \$pix;'; then
echo "\$X has no alpha, converting"
convert "\$X" -quality 100 "\${X%.*}.jpg"
@@ -25,7 +20,8 @@ for X in "\$@"; do
echo "\$X has alpha, converting twice"
convert "\$X" -alpha extract -quality 100 "\${X%.*}.jpg"
convert "\$X" -alpha off -quality 100 "\${X%.*}_alpha.jpg"
- jpegoptim --strip-all -m\$qual "\${X%.*}_alpha.jpg"
+ jpegoptim --strip-all -m\$qual "\${X%.*}.jpg"
+ jpegoptim --strip-all -m\$qual_alpha "\${X%.*}_alpha.jpg"
rm -f "\$X"
fi
;; | 589 | 1,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.101478 |
https://www.jiskha.com/display.cgi?id=1285804927 | 1,516,335,122,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00727.warc.gz | 892,142,491 | 4,034 | # Math
posted by .
x=0 2 4 6
y=8 18 28 38
If x=8 , Y=?
pls show step by step
• Math -
48? It appears that the x's are counting by 2's, and the y's are counting by 10's (starting at 8).
• Math -
correct answer but can you put it in equation like y=?? ??
• Math -
I'm not sure, sorry.
• Math -
thx Jen
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More Similar Questions | 758 | 2,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-05 | latest | en | 0.883334 |
https://scicomp.stackexchange.com/questions/31345/whittaker-shannon-interpolation-accuracy-dies-with-speedup-can-it-be-fixed/31351 | 1,701,930,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00772.warc.gz | 561,960,713 | 43,793 | # Whittaker-Shannon interpolation: Accuracy dies with speedup; can it be fixed?
With a truncated Whitaker-Shannon series (cardinal series) $$f(t) = \sum_{j = 0}^{n-1} y_{j} \frac{\sin\left(\pi( \frac{t-t_0}{h} -j)\right)}{\pi\left(\frac{t-t_0}{h}-j\right)}$$ we can naively evaluate the sum by repeated calls to sinc routines, such as the following code does:
Real naive_sum(Real t) const {
using boost::math::constants::pi;
Real f = 0;
for (size_t i = 0; i < m_y.size(); ++i) {
Real arg = pi<Real>()*( (t-m_t0)/m_h - i);
f += m_y[i]*boost::math::sinc_pi(arg);
}
return y;
}
However, repeated calls to sinc are expensive, so we can use the identity $$\sin(\theta - j\pi) = (-1)^{j}\sin(\theta)$$ to write this as $$f(t) = \frac{\sin(\pi(t-t_0)/h)}{\pi} \sum_{j=0}^{n-1} (-1)^{j}\frac{y_{j}}{(t-t_0)/h -j}$$
which can be implemented as follows:
Real operator()(Real t) const {
using boost::math::constants::pi;
using std::sin;
Real y = 0;
Real x = (t - m_t0)/m_h;
for (size_t i = 0; i < m_y.size(); ++i)
{
Real denom = (x - i);
if (denom == 0) {
return m_y[i];
}
if (i & 1) {
y -= m_y[i]/denom;
}
else {
y += m_y[i]/denom;
}
}
return y*sin(pi<Real>()*x)/pi<Real>();
}
However, I have observed a vast decrease in accuracy using the fast method over the slow method. Can the speed of the fast method be preserved without a massive decrease in accuracy?
Working code, for those that care:
#ifndef BOOST_MATH_INTERPOLATORS_WHITAKKER_SHANNON_HPP
#define BOOST_MATH_INTERPOLATORS_WHITAKKER_SHANNON_HPP
#include <boost/math/special_functions/sinc.hpp>
#include <boost/math/constants/constants.hpp>
namespace boost::math::interpolators {
template<class RandomAccessContainer>
class whittaker_shannon {
public:
using Real = typename RandomAccessContainer::value_type;
whittaker_shannon(RandomAccessContainer&& y, Real t0, Real h) : m_y{std::move(y)}, m_t0{t0}, m_h{h}
{
}
Real operator()(Real t) const {
using boost::math::constants::pi;
using std::sin;
Real y = 0;
Real x = (t - m_t0)/m_h;
for (size_t i = 0; i < m_y.size(); ++i)
{
Real denom = (x - i);
if (denom == 0) {
return m_y[i];
}
if (i & 1) {
y -= m_y[i]/denom;
}
else {
y += m_y[i]/denom;
}
}
return y*sin(pi<Real>()*x)/pi<Real>();
}
Real naive_sum(Real t) const {
using boost::math::constants::pi;
Real y = 0;
Real s = pi<Real>()*(t-m_t0)/m_h;
for (size_t i = 0; i < m_y.size(); ++i) {
Real arg = pi<Real>()*( (t-m_t0)/m_h - i);
y += m_y[i]*sinc_pi(arg);
}
return y;
}
Real operator[](size_t i) const {
return m_y[i];
}
private:
RandomAccessContainer m_y;
Real m_t0;
Real m_h;
};
}
#endif
Here's a test that reproduces the phenomenon:
template<class Real>
void test_bump()
{
using std::exp;
using std::abs;
auto bump = [](Real x) { if (abs(x) >= 1) { return Real(0); } return exp(-Real(1)/(Real(1)-x*x)); };
Real t0 = -1;
size_t n = 2049;
Real h = Real(2)/Real(n-1);
std::vector<Real> v(n);
for(size_t i = 0; i < n; ++i) {
Real t = t0 + i*h;
v[i] = bump(t);
}
auto ws = whittaker_shannon(std::move(v), t0, h);
std::mt19937 gen(323723);
std::uniform_real_distribution<long double> dis(-0.95, 0.95);
size_t i = 0;
while (i++ < 1000)
{
Real t = static_cast<Real>(dis(gen));
Real expected = bump(t);
if(!CHECK_MOLLIFIED_CLOSE(expected, ws(t), 50*std::numeric_limits<Real>::epsilon())) {
std::cerr << " Problem occured at abscissa " << t << "\n";
}
}
}
• Have you tried (1) Accumulating positive and negative terms separately and subtracting at the end (2) Compensated addition (Kahan summation)? Apr 2, 2019 at 18:20
• I believe (1) is discussed by Higham where he demonstrates it doesn't work; see doi.org/10.1137/0914050, I just tried (2), and it doesn't help. The summation condition number is not large at the abscissas where the unit tests fail (I computed it around to be ~5 at the problem abscissas.) There are abscissas where the condition number is ~3000, at those points, the unit tests pass. Apr 2, 2019 at 19:48
• (1) Make sure you are compiling with strictest adherence to IEEE-754 (for my Intel compiler that is /fp:strict for example). (2) Try explicitly using of fma() as extensively as possible (convert divisions to multiplication if need be) to guard against subtractive cancellation when combining products. (3) Be skeptical of your reference function, as it may have numerical issues, too. I would suggest using triple the precision of the actual computation you are trying to implement. Apr 2, 2019 at 20:38
• The compilation is with g++-8 -O3 -march=native; none of these flags break IEEE compliance. I feel like my reference function has no need for skepticism, esp. since I draw the test values away from it's problematic range as |x| -> 1. Apr 2, 2019 at 23:16
• I have repro and I think I have identified the culprit. sin of large arguments polluted by rounding error. Suggest switching sin to sin_pi: return y*sin(pi<Real>()*x)/pi<Real>(); --> return y*boost::math::sin_pi(x)/pi<Real>(); Note: I used my own sinpi from this answer instead of Boost. Apr 3, 2019 at 3:55
I was able to reproduce the behavior reported in the question, and traced the observed inaccuracies to the following line:
return y*sin(pi<Real>()*x)/pi<Real>();
The explicit multiplication with a floating-point approximation of π introduces a small error into the argument to sin, which comprises the representational error in the constant and the rounding error added by the multiply and is on the order of 1 ulp. This small error in the argument represents a phase error in sin which grows with the magnitude of x. In this case |x| is on the order of 1000, resulting in quite a bit of error magnification.
Because this is a relatively frequent scenario, various platforms offer a sinpi function that computes sin (π x) with the multiplication by π happening inside sinpi after argument reduction, minimizing the phase error. In the specific case of Boost the desired function is boost::math::sin_pi. Often the sinpi function is also more efficient than the regular sin function, since its internal argument reduction is simpler.
Replacing the original source line with the line below should fix the observed accuracy issue completely:
return y*boost::math::sin_pi(x)/pi<Real>();
• Not only did this solve the problem, the resulting accuracy increases by a factor of 5 over the naive sum. I'll need to keep sin_pi in mind. Apr 3, 2019 at 13:05 | 1,840 | 6,335 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | longest | en | 0.519853 |
https://www.manualslib.com/manual/25891/Casio-Fx-570es-Plus.html?page=20 | 1,537,480,628,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156622.36/warc/CC-MAIN-20180920214659-20180920235059-00080.warc.gz | 802,117,578 | 24,767 | # Casio fx-570ES PLUS User Manual: Using Calc
Casio calculator user's guide.
To obtain the conjugate complex number of 2 + 3
number format:
To obtain the absolute value and argument of 1 +
Absolute Value:
Argument:
Using a Command to Specify the Calculation Result
Format
Either of two special commands (
of a calculation to specify the display format of the calculation results. The
command overrides the calculator's complex number format setting.
' 2 + ' 2
i
2
2
## Using CALC
CALC lets you save calculation expressions that contain variables, which you
can then recall and execute in the COMP Mode (
Mode (
save with CALC.
• Expressions: 2X + 3Y, 2AX + 3BY + C, A + B
• Multi-statements: X + Y : X (X + Y)
• Equalities with a single variable on the left and an expression including
variables on the right: A = B + C, Y = X
(Use
To store 3A + B and then substitute the following values to perform
the calculation: (A, B) = (5, 10), (7, 20)
a
bi
+
)
(CMPLX)
(CMPLX)
= 2
45, 2
i
2
(
)
( ) 45
). The following describes the types of expressions you can
(=) to input the equals sign of the equality.)
3
Prompts for input of a value for A
(Conjg) 2
3
(Abs) 1
(arg)1
r θ
or
45 = ' 2 + ' 2
i
(CMPLX)
(CMPLX)
2
+ X + 3
(A)
5
10
(or
E-19
i
(Complex
i
(
)
i
i
(
)
i
(
)
a
bi
+
) can be input at the end
r θ
(
)
a
bi
(
+
)
) and the CMPLX
i
(B)
Current value of A
)
i
2–3
' 2
45
2 45
' 2 +' 2
i
Math
Math
Fx-991es plus | 487 | 1,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.669961 |
https://brainytermpapers.com/decision-tree-and-value-of-information/ | 1,632,379,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00467.warc.gz | 186,498,090 | 17,225 | # Decision Tree and Value of Information
Scenario: You are deciding among three investments, as you do for Case 4. You have heard of an expert who has a highly reliable “track record” in the correct identification of favorable vs. unfavorable market conditions. You are now considering whether to consult this “expert.” Therefore, you need to determine whether it would be worth paying the expert’s fee to get his prediction. You recognize that you need to do further analysis to determine the value of the information that the expert might provide.
In order to simplify the analysis, you have decided to look at two possible outcomes for each alternative (instead of three). You are interested in whether the market will be Favorable or Unfavorable, so you have collapsed the Medium and Low outcomes. Here are the three alternatives with their respective payoffs and probabilities.
Option A: Real estate development. This is a risky opportunity with the possibility of a high payoff, but also with no payoff at all. You have reviewed all of the possible data for the outcomes in the next 10 years and these are your estimates of the Net
Present Value (NPV) of the payoffs and probabilities:
High/Favorable NPV: \$7.5 million, Pr = 0.5
Unfavorable NPV: \$2.0 million, Pr = 0.5
Option B: Retail franchise for Just Hats, a boutique-type store selling fashion hats for men and women. This also is a risky opportunity but less so than Option A. It has the potential for less risk of failure, but also a lower payoff. You have reviewed all of the possible data for the outcomes in the next 10 years and these are your estimates of the NPV of the payoffs and probabilities.
High/Favorable NPV: \$4.5 million, Pr = 0.75
Unfavorable NPV: \$2.5 million, Pr = 0.25
Option C: High Yield Municipal Bonds. This option has low risk and is assumed to be a Certainty. So there is only one outcome with probability of 1.0:
NPV: \$2.25 million, Pr = 1.0
You have contacted the expert and received a letter stating his track record which you have checked out using several resources. Here is his stated track record:
True State of the Market
Expert Prediction Favorable Unfavorable
Predicts “Favorable” .9 .3
Predicts “Unfavorable” .1 .7
You realize that this situation is a bit complicated since it requires the expert to analyze and predict the state of two different markets: the real estate market and the retail hat market. You think through the issues of probabilities and how to calculate the joint probabilities of both markets going up, both going down, or one up and the other down. Based on your original estimates of success, here are your calculations of the single probabilities and joint probabilities of the markets.
Probabilities Favorable Unfavorable
A: Real Estate 0.50 0.50
B: Just Hats 0.75 0.25
Joint Probabilities
A Fav, B Fav (A+, B+) 0.375
A Unf, B Unf (A-, B-) 0.125
A Fav, B Unf (A+, B-) 0.125
A Unf, B Fav (A-, B+) 0.375
Finally, after a great deal of analysis and calculation, you have determined the Posterior probabilities of Favorable and Unfavorable Markets for the Real Estate business and the boutique hat business.
Real Estate Just Hats
Favorable (F)
Unfavorable (U)
0.45 says “F/F”
Real Estate(F)0.75 (U)0.25
Just Hats(F)0.90 (U)0.10
0.15 says “F/U”
Real Estate(F)0.75 (U)0.25
Just Hats(F)0.30 (U)0.70
0.30 says “U/F”
Real Estate(F)0.125 (U)0.875
Just Hats(F)0.90 (U)0.10
0.10 says “U/U”
Real Estate(F)0.125 (U)0.875
Just Hats(F)0.30 (U)0.70
For example, this table says that there is 45% chance that the expert will predict Favorable for both markets (F/F), and when he makes this prediction, there is a 75% chance that the Real Estate market will be favorable and 25% chance that it won’t, and also a 90% chance that the Hat market will be Favorable and 10% chance it won’t.
You have developed a Decision Tree showing the original collapsed solution and also showing an expanded Decision Tree for evaluating the value of the expert’s information. You need to enter the probabilities into this tree to see if the expert’s information will increase the overall expected value of your decision. Download the Excel file with the incomplete Decision Tree: (See attached file: SLP Assignment_incomplete Decision Tree Excel.xlsx)
Assignment
Complete the information in the Decision Tree in the Excel file. Determine the Expected NPV of the decision if you were to consult the Expert. Does use of the Expert increase the value of your analysis? If so, by how much?
Develop a PowerPoint presentation to your private investment company and explain your analysis and your recommendation. Provide clear rationale/ justification for your decision. Use audio/video feature in PowerPoint to present each slide. Be sure to check the Oral Communication Rubric (See attached file: Oral rubric) to understand the requirements for the PowerPoint presentation.
SLP Assignment Expectations
Analysis: Conduct accurate and complete analysis in Excel. Check the following link on PowerPoint presentation:
https://support.office.com/en-US/article/PowerPoint-training-40e8c930-cb0b-40d8-
82c4-bd53d3398787
Required:
Meet Length requirements: 10-15 slides (not including Cover and Reference pages).
Provide a brief introduction to/background of the problem.
Show analysis that supports Excel analysis and provides thorough discussion of
assumptions, rationale, and logic used.
Offer meaningful and accurate recommendation(s).
Oral presentation of each slide should use video/audio feature in PowerPoint. | 1,316 | 5,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.952332 |
https://www.weegy.com/?ConversationId=64TXICJ0&Link=i | 1,721,349,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00784.warc.gz | 896,106,788 | 9,783 | A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s. What is the magnitude of the final velocity of the red marble?
A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s. 31 cm/s is the magnitude of the final velocity of the red marble. Solution 1.2 g × v_f = 56.7 g·cm/s - 19.25 g·cm/s v_f = (56.7 g·cm/s - 19.25 g·cm/s) / 1.2 g v_f = 31 cm/s
f
Question
Asked 13 days ago|7/5/2024 10:51:17 PM
Updated 8 days ago|7/10/2024 1:33:54 PM
Rating
3
A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s. 31 cm/s is the magnitude of the final velocity of the red marble.
Solution
1.2 g × v_f = 56.7 g·cm/s - 19.25 g·cm/s
v_f = (56.7 g·cm/s - 19.25 g·cm/s) / 1.2 g
v_f = 31 cm/s
Added 13 days ago|7/5/2024 11:09:53 PM
1
m1=3.5 g=0.0035 kg
2=1.2g=0.0012 kg
m2=1.2 g=0.0012 kg
1 =15cm/s=0.15 m/s
v1i=15 cm/s=0.15 m/s
1 =5.5 cm/s=0.055m/s
v1f=5.5 cm/s=0.055 m/s
2i=3.5 cm/s=0.035 m/s
v2i=3.5 cm/s=0.035 m/s
Now, apply conservation of momentum:
1 1 + 2 2 = 1 1f+ 2 2
Substitute the known values:
0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012× 2
0.0035×0.15+0.0012×0.035=0.0035×0.055+0.0012×v2f
Calculate the left-hand side:
0.000525+0.000042=0.0001925+0.0012 2
0.000525+0.000042=0.0001925+0.0012v 2f
Simplify and solve for v2f:
0.0003745=0.0012v2f
v2f =0.0003745/0.0012
2 0.312 m/s
Added 8 days ago|7/10/2024 1:33:54 PM
Deleted by kanand [7/10/2024 1:34:12 PM], Undeleted by kanand [7/10/2024 1:34:15 PM], Rated good by kanand
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,172 | 2,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.717119 |
https://csposts.com/dsa/maximum-subarray-sum-with-unique-values/ | 1,721,426,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00825.warc.gz | 151,955,508 | 61,980 | # Find the Maximum Subarray Sum with Unique Values
You are given an array of positive integers. Find the maximum subarray sum with unique values.
An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).
Example 1
Input nums = [1, 2, 3, 5, 3, 1]
Output 10
Explanation The subarray is [2, 3, 5] having maximum sum and unique values
Example 2
Input nums = [1, 2, 3, 5, 3, 1, 4, 5, 1]
Output 10
Explanation The subarray is [2, 3, 5] or [4, 5, 1] having maximum sum and unique values
## Approach: Two Pointers
Like many subarray related problems, this can also be solved using two pointers.
### Intuition
Let’s keep two pointers (left and right) both pointing to the start of the array. We will also be maintaining a hashmap to keep track of the unique elements considered in our solution.
As we iterate through our sliding window formed by the pointers left and right, we will increment the right pointer if it is not already in the solution subarray (sliding window). If it is present in the sliding window, we will increment the left pointer as long as nums[right] is present. The check of presence is computed using a hashmap which gives an $O(1)$ look-up time complexity.
The value of sum is also tracked and whenever we are incrementing the right pointer, we are adding nums[right] to the sum and whenever we are incrementing the left pointer, we are subtracting nums[left] from sum.
### Implementation
In C++:
int maximumUniqueSubarray(vector<int>& nums) {
unordered_set<int> unique;
int left = 0, right = 0, sum = 0, ans = 0;
while(right < nums.size()) {
// if nums[right] is already in our solution subarray,
// remove the nums[left] item and increment the left
// pointer. Essentially, as long as nums[right] is there
// in our solution subarray we will keep incrementing
// the left pointer
if (unique.count(nums[right])) {
sum -= nums[left];
unique.erase(nums[left]);
++left;
// nums[right] is not in our solution subarray, include it
} else {
sum += nums[right];
unique.insert(nums[right]);
ans = max(ans, sum);
++right;
}
}
return ans;
}
In Python:
def maximumUniqueSubarray(nums):
unique = set()
left, right, sum, ans = 0, 0, 0, 0
while right < len(nums):
# if nums[right] is already in our solution subarray,
# remove the nums[left] item and increment the left
# pointer. Essentially, as long as nums[right] is there
# in our solution subarray we will keep incrementing
# the left pointer
if nums[right] in unique:
sum -= nums[left]
unique.remove(nums[left])
left += 1
# nums[right] is not in our solution subarray, include it
else:
sum += nums[right]
return ans
• Time Complexity: $O(N)$ where $N =$ size of the array
• Space Complexity: $O(N)$, in the worst case all the elements are unique, and we may have to store all of them in the hashmap | 757 | 2,869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "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} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.792708 |
https://onlinecourseaide.com/1824-2/ | 1,618,538,099,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00524.warc.gz | 426,203,451 | 19,787 | # Chapter 1 Practice Problems
Chapter 1 Practice Problems
Practice Problem 12, 15, 19, 20, 21, & 22
Due Week 2 Day 6 (Sunday)
Follow the instructions below to submit your answers for Chapter 1 Practice Problem 12, 15, 19, 20, 21, & 22.
1. Save Chapter 1 Instructions to your computer.
2. Type your answers into the shaded boxes below. The boxes will expand as you type your answers.
3. Resave this form to your computer with your answers filled-in.
4. Attach the saved form to your reply when you turn-in your work in the Assignments section of the Classroom tab. Note: Each question in the assignments section will be listed separately; however, you only need to submit this form one time to turn-in your answers.
Read each question in your textbook and then type your answers for Chapter 1 Practice Problem 12, 15, 19, 20, 21, & 22 in the corresponding spaces below.
12a. equal-interval –
12b. rank-order –
12c. nominal –
12d. ratio scale –
12e. continuous –
15. Type your answers to Practice Problem 15a in the shaded boxes below. The “X” represents the speed; “f” represents the frequency; and “%” represents the relative percentage for each score. Round to the nearest whole percentage. For example, if your remainder is .5 or greater, round up (i.e. 2.5 = 3). Note: Fill-in each box. Please do not delete any scores.
X
f
%
52
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19. Give an example, in words, of something having the distribution shapes listed below.
19a. bimodal –
19b. approximately rectangular –
19c. positively skewed –
20. Describe an example of a misleading graph: Note: Either copy and paste a graph in the first shaded box below, or fully describe the graph if it will not paste. You may also cut and paste a misleading graph onto a Word document and send as an attachment. Please do not include a link to a graph in lieu of pasting a graph onto a Word document. I will not navigate to a link to view your chosen graph.
In the second shaded box, discuss specific characteristics that make the graph misleading.
21a. Explain the idea of frequency table 1-10 below:
21b. Explain the meaning of the pattern of results below:
22a. Explain the idea of frequency table 1-11 below:
22b. Explain the meaning of the pattern of results below:
## Calculate the price of your order
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee | 653 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.858853 |
http://quizlet.com/21395803/macroeconomics-flash-cards/ | 1,427,696,343,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131299114.73/warc/CC-MAIN-20150323172139-00079-ip-10-168-14-71.ec2.internal.warc.gz | 222,435,061 | 19,076 | # Macroeconomics
### 20 terms by Oeltjen
#### Study only
Flashcards Flashcards
Scatter Scatter
Scatter Scatter
## Create a new folder
### Fiscal Policy
When the Federal government uses taxation and spending actions to stimulate the economy it is conducting
### Expansionary fiscal policy
If Congress passes legislation to increase government spending to counter the effects of a recession, then this would be an example of a(n)
### Contractionary fiscal policy
If the U.S. Congress passes legislation to raise taxes to control demand-pull inflation, then this would be an example of a(n)
### Budget deficit
The economy starts out with a balanced Federal budget. If the government then implements expansionary fiscal policy, then there will be a
### Decrease government spending and increase taxes
You are given the following information about aggregate demand at the existing price level for an economy: (1) consumption = \$500 billion; (2) investment = \$50 billion; (3) government purchases = \$100 billion; and (4) net export = \$20 billion. If the full-employment level of GDP for this economy is \$620 billion, then what combination of actions would be most consistent with closing the GDP-gap here?
### Increase government spending and decrease taxes
You are given the following information about aggregate demand at the existing price level for an economy: (1) consumption = \$400 billion; (2) investment = \$40 billion; (3) government purchases = \$90 billion; and (4) net export = \$25 billion. If the full-employment level of GDP for this economy is \$600 billion, then what combination of actions would be most consistent with closing the GDP-gap here?
### \$20 billion
In an economy, the government wants to increase aggregate demand by \$50 billion at each price level to increase real GDP and reduce unemployment. If the MPS is 0.4, then it could increase government spending by:
### Increase taxes by \$16 billion
In an economy, the government wants to decrease aggregate demand by \$48 billion at each price level to decrease real GDP and control demand-pull inflation. If the MPS is 0.25, then it could:
### Serves as an automatic stabilizer for the economy
As the economy declines into recession, the collection of personal income tax revenues automatically falls. This relationship best describes how the progressive income tax system:
### Transfer payments
The so-called "negative taxes" are better known as:
### Is not subject to the timing problems of discretionary policy
One advantage of automatic fiscal policy over discretionary fiscal policy is that automatic fiscal policy:
### Contractionary and worsen the effects of the recession
Assume that the economy is in a recession and there is a budget deficit. A strict balanced-budget amendment that would require the Federal government to balance its budget during a recession would be:
### Discretionary expansionary fiscal policy
The American Recovery and Reinvestment Act of 2009 is a clear example of:
### Increases in government spending and decreases in taxes
The American Recovery and Reinvestment Act of 2009 included mostly:
### Start of the recession and the time it takes to recognize that the recession has started
One timing problem with fiscal policy to counter a recession is a "recognition lag" that occurs between the:
### Time fiscal action is taken and the time that the action has its effect on the economy
One timing problem with fiscal policy to counter a recession is an "operational lag" that occurs between the:
### Increases in government spending may reduce private investment
The crowding-out effect suggests that:
### Government borrows in the money market, thus causing an increase in interest rates
The crowding-out effect arises when:
### Decrease the effectiveness of expansionary fiscal policy
The crowding-out effect works through interest rates to:
### Increased by less than \$100 billion
Assume that if there was no crowding-out, an increase in government spending would increase GDP by \$100 billion. If there had been partial crowding-out, however, then GDP would have:
Example: | 835 | 4,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-14 | latest | en | 0.893041 |
https://www.metaphysical-concepts.com/numerology/what-r-lucky-numbers/ | 1,539,965,297,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512411.13/warc/CC-MAIN-20181019145850-20181019171350-00247.warc.gz | 1,014,525,341 | 9,317 | # What R Lucky Numbers
Be here on Friday, January 26 for your chance to win big! Match all 6 numbers from the lottery ball machine to the numbers on your ticket and you could win up to \$500000 cash!
## What R Lucky Numbers
Nov 11, 2011. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. Let F(X) equals to the number of lucky digits in decimal representation of X. Chef wants to know the number of such integers X, that L ≤ X ≤ R and F(X) is a lucky number.
### What R Lucky Numbers
He’s accused of telling a customer that his lucky number is 4 and he would try. Although many dealers claim to be able to hit a specific group of numbers, it’s impossible to do with an honest wheel, say Charles R. Mousseau of Total.
What R Lucky Numbers
How to Find Your Lucky Numbers in Numerology. Mathematicians tell us that numbers are the language of the universe. Numerologists go so far as to say that the numbers.
What R Lucky Numbers
Alphabets & Numbers. In numerology, when you want to calculate the Name No. for a name, you have to add up the values of all the alphabets in that name.
What R Lucky Numbers
Reading – "We’ve had a lot of large families," said Fiore, "but seven boys, my goodness. It’s been quite an experience having them on our team." Main photo back row, l-r: Anthony, Giovanni, James, and Giuseppe; front row, l-r: Michael, Peter, and David.
Lucky Bamboo plant is not really a bamboo plant. Known as Dracaena sanderiana, it is a member of the lily family that grows in the dark, tropical rain forests of.
Generally people think that they know how to count their own vehicle number. People usually count the numeric values in the registration number of your vehicle and.
Numerology Quran I read some 'miracles' of the qur'an recently. These included many things such as the 3 stages of embryos, the orbits of planets, etc etc. However, one of them claimed that the word yawm (day) is mentioned 365 times in the qur'an, and the word qam. They day your were born determines your life path.
Symbolism and Meaning of number 18. Lots of Fun Facts. Discover why number 18 is lucky in China. Number 18 in the Bible.
Lottery numbers generator – Search your lucky numbers – Good Luck!
Numerology Guidance Card Completion Do you see repeating numbers like 111, 222, 333, and 444? They indicate that your Spirit Guides are trying to get your attention. Find out why. Daily Angel Oracle Card: Completion, from the Numerology Guidance Oracle Card deck, by Michelle Buchanan. Completion: “This card indicates a time of necessary endings and completion in preparation for
R 3 digit numbers 4 digit numbers RAFI 916, 610, 558, 293 9160, 5194, 8923, 6434 RALPH 169, 656, 343, 679 7923. 4959, 2042. 9365 RALSTON 913, 122, 650, 353 5563, 4227, 2145, 2545 RAMSES 914, 595, 802, 282 5585, 0973, 8485 , 1134 RANDALL 912, 759, 488, 234 5898, 2738, 6444, 1435 RANDOLPH.
Lucky number – Wikipedia – The lucky number which removes n from the list of lucky numbers is: (0 if n is a lucky number) 0, 2, 0, 2, 3, 2, 0, 2. A "lucky prime" is a lucky number that is.
11: Of all of the lucky numbers, the number 11 is probably the luckiest. games that Joe DiMaggio recorded a base hit, and 56 is also the number of times Michael Stipe from R.E.M says Yeah in the band’s hit song Man on the Moon.
It’s funny and well put together and while it does occasionally out-clever itself, it’s still very good. “Lucky Number Slevin” is rated R for strong violence, sexuality, and language.
“We got really lucky,” Lupetin added. for example), and also because the group.
One lucky number for today! | Lottery Post – Quote: Originally posted by faith117 on October 26, 2017. Here is one lucky num ber to try! 902 Sun Smiley. All states try this number. Good luck from Faith Sun Smiley. Wisconsin 290 evening 10/29/17. Congrats to all winners! Hyper Banana Hurray! The only time you run out of chances is if you don't take.
May 24, 2017. Some people use family birthdays, others use lucky numbers while there are those that swear by a fail-safe “system". Dr John Haigh, an emeritus professor of mathematics at the University of Sussex who has written a paper on the statistics underlying the National Lottery, said: “Just because a number.
Your Lucky Number according to your name, find lucky number from name. Your lucky numbers come from two sources: your date of birth, and your name (what you go by).
For this nonbeliever, deciphering "Lucky Number Slevin" is not worth spending an extra five minutes. "Lucky Number Slevin" is rated R (Under 17 requires accompanying parent or adult guardian). It has gory violence, with some profanity.
YOUR LUCKY NUMBER OF THE DAY is: 24 (range from 1 to 100). This result is " casted" by the timing and your magnetic field and JUST FOR YOU. You can keep testing your picks until you feel good.
What R Lucky Numbers
What R Lucky Numbers | 1,273 | 4,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-43 | longest | en | 0.944656 |
https://blog.hkis.edu.hk/wp/200673/2016/09/01/ap-computer-science-principles-unit-one-blog/ | 1,675,016,383,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00554.warc.gz | 150,260,151 | 8,592 | # AP Computer Science Principles | Unit One Blog
You will need to keep track of what you have learned in each lesson on a blog so when it comes to your PT (S) tasks you will be able to remember the activities that relate towards them.
Write a piece of reflection of things learned from each lesson. See sections below from unit one, things we have covered so far.
L2. Building your communication device and sending basic messages through the device.
I learned how to communicate without having to use the complicated words and sentences that we use in the English language. What I did was have a selection of different colored Lego bricks, which each represented a different answer. We would point to the one that represented our answer. If we had a green brick be “yes” and a red brick be “no” and I asked “Are you a human?” you would point to the green brick (probably).
L3. Sending 2 bit binary messages to each other using the internet simulator.
In this lesson I basically learned what binary is. I had no previous knowledge of what it was, and now I learned how to use two bits to convey a message with someone else. It was a slow process, but we managed to send simple messages to each other using just ones and zeroes.
L4. Number systems – creating number systems with symbols such as a square, triangle and circle to come up with as many permutations as possible. Look at the number systems reflection discussion.
This activity opened my eyes to the vast possibilities of different number systems. It showed me what happens when you throw more than two bits into a system and how to pinpoint the exact amount of possibilities using just the number of shapes or bits.
We also looked at Nibble, bit, byte, KB, MB, TB. define these.
Nibble: 1/2 a bit.
Bit: 1 bit
Byte: 8 bits.
Kilobyte: 1,000 bytes.
Megabyte: 1,000 kilobytes.
Gigabyte: 1,000 megabytes.
Terabyte: 1,000 gigabytes.
L5. Binary numbers – You made a Flippy do (take a picture of your flippy do) explain why this is useful to learning binary.
This “Flippy Do” is incredibly useful for learning binary. It’s kind of like a binary cheat-sheet of sorts. You just take the binary code, put it in the bottom part (using either ones or zeroes) and add up the numbers above all of the ones. The answer you get is the binary code converted to decimal.
We looked at binary addition and subtraction.
L6. Developing a number sending protocol.
Reflect on the process of sending numbers to your partner to create the shapes you have designed. See folder L6 discussions and homework. Add these to your blog.
We colored in squares to make a shape, then we sent each row as ones and zeroes to each other. Ones represented spots that were filled in, and zeroes represented spots that were empty. As we went through the rows, we eventually passed on the picture to each other.
Reflect on your learning in this lesson. In the space below, enter:
3 recollections (things you remember),
3) I remember converting a complicated sentence into html, then converting the html into ASCII.
2) I remember learning what ASCII even is.
1) I remember sending messages in ASCII between myself and my partner.
2 observations (things you noticed), and
2) I noticed that converting things into ASCII is very difficult and time consuming (at least by hand).
1) I noticed that as we go through more lessons, we are slowly learning more efficient ways of communicating messages.
1 insight (something you fully understand the…
1) | 763 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.945459 |
https://www.physicsforums.com/threads/magnetic-field-direction.944950/ | 1,544,846,977,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826715.45/warc/CC-MAIN-20181215035757-20181215061757-00025.warc.gz | 1,001,333,080 | 17,146 | # I Magnetic Field Direction
1. Apr 16, 2018
### Schfra
When can we use the right hand rule to find the direction of a magnetic field from current?
I know that it works for an infinite wire. Does it work for a finite wire? Where does it work and not work?
2. Apr 16, 2018
### Mister T
It works for both a finite and an infinite wire.
3. Apr 16, 2018
### Schfra
Does it apply just whenever we have a current?
And if it does work on an infinite wire does that mean that there is no magnetic field past either end of a finite wire with current running through it?
4. Apr 16, 2018
### ZapperZ
Staff Emeritus
Please note that the term "finite wire" needs to be taken with a grain of salt. You can't have current (at least, not steady current that are usually used in intro physics classes) in a finite wire. It needs to be closed somehow for there to be current flow.
The RHR is useful and can be easily used for a straight wire segment and also for a loop wire. You line up your thumb with something that you know is flowing straight, OR, you curl your fingers around something that you know is "curling".
This means that for a straight wire segment, current is flowing along the wire, so you line your thumb along it, and the curl of your fingers shows the direction of the magnetic field. For a loop of wire, you curl your fingers in the direction of the current flow in the loop of wire, and your thumb shows the direction of the magnetic field.
RHR is useful if you know how to use it properly.
Zz.
5. Apr 16, 2018
### Schfra
So would it be correct to say whenever you have current through a wire, you can use the RHR to find the direction of the magnetic field around any given point on that wire? A wire being any circular conductor in this case. I suspect that we couldn’t apply the RHR to a wire with a non-symmetrical shape like the in the attached image (assuming the current is flowing out of the image).
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6. Apr 16, 2018
### ZapperZ
Staff Emeritus
If the wire is twisted and in some weird geometry, then you can't simply use the RHR, because this is no longer a straight wire.
Zz.
7. Apr 16, 2018
### Schfra
But what classifies as a wire? Does it have to have circular cross sections, or does it have to be thin?
Also, how can I determine whether or not a wire is a loop or has weird geometry?
8. Apr 16, 2018
### ZapperZ
Staff Emeritus
Now c'mon. Are you telling me that you can't tell the difference between a simple loop and a weird geometry? Again, is this going to be relevant in your class?
If you intend to continue on studying physics, you will have plenty of opportunities to consider and look at many of these complicated situations. Unless you think you have already mastered the basic principle of this concept, I suggest you sharpen your knowledge and skills at solving and addressing the type of problems that are relevant at your level first.
Zz.
9. Apr 16, 2018
### Schfra
It’s not that these specific questions come up in class, but there are questions I’ve seen in class where I’m not sure if the RHR applies or not. I feel uncomfortable applying things like this if I don’t have at least a decent understanding of when they can be applied.
I’m not sure if simple loop means only a circular loop or if it’s more general than that.
10. Apr 16, 2018
### ZapperZ
Staff Emeritus
Then attempt them, see if you got it right, and if you don't understand them, post your question in the HW/Coursework forum. Don't start making things up beyond the scope of what your level of education will cover.
I don't mind going beyond what you need to know, but you must already establish that you have the knowledge for what you should already know. This is not the case here.
Zz. | 941 | 3,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.929898 |
https://prepinsta.com/leetcode-top-100-liked-questions-with-solution/kth-largest-element-in-an-array/ | 1,708,481,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00583.warc.gz | 497,807,183 | 32,778 | # 215. Kth Largest Element in an Array Leetcode Solution
## Kth Largest Element in an Array Leetcode Problem :
Given an integer array nums and an integer k, return the kth largest element in the array.
Note that it is the kth largest element in the sorted order, not the kth distinct element.
Can you solve it without sorting?
## Kth Largest Element in an Array Leetcode Solution :
### Constraints :
• 1 <= k <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4
### Example 1:
• Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
• Output: 4
Intuition :
The intuition for arriving at this solution is grounded in the properties of a max heap (priority queue) and the goal of efficiently finding the kth largest element in an array.
Approach :
1. Priority Queue (Max Heap): The priority_queue in C++ is implemented as a max heap, which means that the element with the highest value will always be at the top of the heap. This is useful for finding the kth largest element because we want to keep track of the largest elements while discarding the smaller ones.
2. Initialization: The priority queue is initialized with the elements from the nums vector. This will automatically build the max heap, with the largest element at the top.
3. Adjusting k: Since we’re interested in finding the kth largest element, we need to adjust the value of k by subtracting 1. This is because the top element of the max heap is the 1st largest element, the next top element will be the 2nd largest, and so on.
4. Loop to Pop Elements: The loop iterates k times, and in each iteration, it removes the top element (the largest element) from the priority queue using pq.pop(). This effectively removes the k-1 largest elements from the heap.
5. Returning Result: After the loop, the priority queue will only contain the kth largest element (since k-1 larger elements have been removed), and this element is returned as the result.
In summary, this code leverages the properties of a max heap (implemented as a priority queue) to find the kth largest element efficiently. By removing the top element (the largest) k-1 times, the code isolates and returns the kth largest element remaining in the heap. This approach ensures that we don’t need to sort the entire array, making the solution more efficient, especially for large arrays.
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https://www.physicsforums.com/threads/spring-problem-perpendicular-force.272752/ | 1,539,945,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512395.23/warc/CC-MAIN-20181019103957-20181019125457-00308.warc.gz | 1,035,672,683 | 14,490 | # Homework Help: Spring problem-perpendicular force
1. Nov 17, 2008
### racast5
okay so i need help with the problem because my teacher gave it to us today and he doesnt teach the class anything.
it regards a body suspended in between two vertical springs, each in equilibrium, with length L.
he asked us to calculate the amount of required force to pull the body horizontally as if along the x axis, with regards to the angle (theta) with would be created by the initial vertical position of the spring and the new position. THEN he asked us to use a taylor series to show him, something, if anyone has any ideas what that series could prove..help? thanks
i just dont know where to start because he never explained the relations when more than one spring is involved, nor pulling them horizontally with a taylor series proof that that matter. the only equation i know is hooks equation f=-kx
2. Nov 18, 2008
### tiny-tim
Welcome to PF!
Hi racast5! Welcome to PF!
You can still use Hooke's law …
the force will still be kx, where the x is the new length of the spring minus its original length …
in other words, you have a thin right-angled triangle, and x is the difference in length between the hyptoneuse and the side.
Since this will be √(1 + something), you can use a Taylor series to approximate it.
3. Nov 18, 2008
### devon cook
I still can't get a reply from anyone. Who is out there?????
Dev
4. Nov 18, 2008
### tiny-tim
the mother-ship has left us …
There is no-one out there …
we are all alone!
5. Nov 18, 2008
### devon cook
You may be right TT. You may be right. Anyway, it's good to make some sort of contact at last. I've been playing around with the maths of a frictionless bead sliding down a parabola y=0.5(x-1)^2 . I got a messy result for its velocity but can't get the same using Hamiltonian mechanics. Anyone done this?
Dev
6. Nov 19, 2008
### tiny-tim
Welcome to PF!
Hi Dev! Welcome to PF!
I'm no good at Hamiltonian mechanics.
You'd better start a new thread (press the NEW TOPIC button on the sub-forum index page), and put the question there (title it "Hamiltonian mechanics").
(that's because hardly anyone new will look at a thread once it's had several replies … which, incidentally, is a good reason not to "bump" your threads)
7. Nov 20, 2008
### devon cook
Thanks TT,
But I can't find the "sub-forum index page" anywhere.
Dev
8. Nov 20, 2008
click!
Hi Dev! | 624 | 2,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-43 | latest | en | 0.94025 |
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584 views | 1,467 | 5,705 | {"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} | 2.671875 | 3 | CC-MAIN-2019-04 | latest | en | 0.936133 |
https://www.quizzes.cc/metric/percentof.php?percent=8480&of=1050 | 1,590,792,981,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00425.warc.gz | 883,379,763 | 3,216 | #### What is 8480 percent of 1,050?
How much is 8480 percent of 1050? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 8480% of 1050 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 8480% of 1,050 = 89040
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
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divided by
Calculating eight thousand, four hundred and eighty of one thousand and fifty How to calculate 8480% of 1050? Simply divide the percent by 100 and multiply by the number. For example, 8480 /100 x 1050 = 89040 or 84.8 x 1050 = 89040
#### How much is 8480 percent of the following numbers?
8480 percent of 1050.01 = 8904084.8 8480 percent of 1050.02 = 8904169.6 8480 percent of 1050.03 = 8904254.4 8480 percent of 1050.04 = 8904339.2 8480 percent of 1050.05 = 8904424 8480 percent of 1050.06 = 8904508.8 8480 percent of 1050.07 = 8904593.6 8480 percent of 1050.08 = 8904678.4 8480 percent of 1050.09 = 8904763.2 8480 percent of 1050.1 = 8904848 8480 percent of 1050.11 = 8904932.8 8480 percent of 1050.12 = 8905017.6 8480 percent of 1050.13 = 8905102.4 8480 percent of 1050.14 = 8905187.2 8480 percent of 1050.15 = 8905272 8480 percent of 1050.16 = 8905356.8 8480 percent of 1050.17 = 8905441.6 8480 percent of 1050.18 = 8905526.4 8480 percent of 1050.19 = 8905611.2 8480 percent of 1050.2 = 8905696 8480 percent of 1050.21 = 8905780.8 8480 percent of 1050.22 = 8905865.6 8480 percent of 1050.23 = 8905950.4 8480 percent of 1050.24 = 8906035.2 8480 percent of 1050.25 = 8906120
8480 percent of 1050.26 = 8906204.8 8480 percent of 1050.27 = 8906289.6 8480 percent of 1050.28 = 8906374.4 8480 percent of 1050.29 = 8906459.2 8480 percent of 1050.3 = 8906544 8480 percent of 1050.31 = 8906628.8 8480 percent of 1050.32 = 8906713.6 8480 percent of 1050.33 = 8906798.4 8480 percent of 1050.34 = 8906883.2 8480 percent of 1050.35 = 8906968 8480 percent of 1050.36 = 8907052.8 8480 percent of 1050.37 = 8907137.6 8480 percent of 1050.38 = 8907222.4 8480 percent of 1050.39 = 8907307.2 8480 percent of 1050.4 = 8907392 8480 percent of 1050.41 = 8907476.8 8480 percent of 1050.42 = 8907561.6 8480 percent of 1050.43 = 8907646.4 8480 percent of 1050.44 = 8907731.2 8480 percent of 1050.45 = 8907816 8480 percent of 1050.46 = 8907900.8 8480 percent of 1050.47 = 8907985.6 8480 percent of 1050.48 = 8908070.4 8480 percent of 1050.49 = 8908155.2 8480 percent of 1050.5 = 8908240
8480 percent of 1050.51 = 8908324.8 8480 percent of 1050.52 = 8908409.6 8480 percent of 1050.53 = 8908494.4 8480 percent of 1050.54 = 8908579.2 8480 percent of 1050.55 = 8908664 8480 percent of 1050.56 = 8908748.8 8480 percent of 1050.57 = 8908833.6 8480 percent of 1050.58 = 8908918.4 8480 percent of 1050.59 = 8909003.2 8480 percent of 1050.6 = 8909088 8480 percent of 1050.61 = 8909172.8 8480 percent of 1050.62 = 8909257.6 8480 percent of 1050.63 = 8909342.4 8480 percent of 1050.64 = 8909427.2 8480 percent of 1050.65 = 8909512 8480 percent of 1050.66 = 8909596.8 8480 percent of 1050.67 = 8909681.6 8480 percent of 1050.68 = 8909766.4 8480 percent of 1050.69 = 8909851.2 8480 percent of 1050.7 = 8909936 8480 percent of 1050.71 = 8910020.8 8480 percent of 1050.72 = 8910105.6 8480 percent of 1050.73 = 8910190.4 8480 percent of 1050.74 = 8910275.2 8480 percent of 1050.75 = 8910360
8480 percent of 1050.76 = 8910444.8 8480 percent of 1050.77 = 8910529.6 8480 percent of 1050.78 = 8910614.4 8480 percent of 1050.79 = 8910699.2 8480 percent of 1050.8 = 8910784 8480 percent of 1050.81 = 8910868.8 8480 percent of 1050.82 = 8910953.6 8480 percent of 1050.83 = 8911038.4 8480 percent of 1050.84 = 8911123.2 8480 percent of 1050.85 = 8911208 8480 percent of 1050.86 = 8911292.8 8480 percent of 1050.87 = 8911377.6 8480 percent of 1050.88 = 8911462.4 8480 percent of 1050.89 = 8911547.2 8480 percent of 1050.9 = 8911632 8480 percent of 1050.91 = 8911716.8 8480 percent of 1050.92 = 8911801.6 8480 percent of 1050.93 = 8911886.4 8480 percent of 1050.94 = 8911971.2 8480 percent of 1050.95 = 8912056 8480 percent of 1050.96 = 8912140.8 8480 percent of 1050.97 = 8912225.6 8480 percent of 1050.98 = 8912310.4 8480 percent of 1050.99 = 8912395.2 8480 percent of 1051 = 8912480 | 1,869 | 4,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2020-24 | latest | en | 0.824517 |
https://stackoverflow.com/questions/19502378/find-the-first-instance-of-a-nonzero-number-in-a-list-in-python/44655598 | 1,726,724,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00823.warc.gz | 500,389,793 | 47,830 | Find the first instance of a nonzero number in a list in Python [duplicate]
I have a list like this:
``````myList = [0.0, 0.0, 0.0, 2.0, 2.0]
``````
I would like to find the location of the first number in the list that is not equal to zero.
``````myList.index(2.0)
``````
It works in this example, but sometimes the first nonzero number will be 1 or 3.
Is there a fast way of doing this?
• just be cautious about comparison with floating point numbers
– Ant
Commented Oct 21, 2013 at 18:57
• I suspect you're getting a lot of down votes because your question does not demonstrate that you've met the "what have you tried?" requirement. Just a helpful tip for future questions. And also a helpful tip for down-voters, let the user know why you're down voting so that they can make corrections. Commented Oct 21, 2013 at 19:04
• Commented May 5, 2017 at 9:16
Use `next` with `enumerate`:
``````>>> myList = [0.0 , 0.0, 0.0, 2.0, 2.0]
>>> next((i for i, x in enumerate(myList) if x), None) # x!= 0 for strict match
3
``````
• What would be the equivalent for numpy arrays ? Commented Dec 15, 2020 at 18:58
• It begs questions about comparison of floating numbers. E.g., what would happen if `42.8/21.4` was used instead of `2.0`? Can you address comparison of floating numbers in your answer (even if it is somehow automatically handled (rounding, etc.))? (But without "Edit:", "Update:", or similar - the answer should appear as if it was written today. And without reference to this comment.) Commented Feb 4, 2022 at 12:58
Use filter
Python 2:
``````myList = [0.0, 0.0, 0.0, 2.0, 2.0]
myList2 = [0.0, 0.0]
myList.index(filter(lambda x: x!=0, myList)[0]) # 3
myList2.index(filter(lambda x: x!=0, myList2)[0]) # IndexError
``````
Python 3: (Thanks for Matthias's comment):
``````myList.index(next(filter(lambda x: x!=0, myList))) # 3
myList2.index(next(filter(lambda x: x!=0, myList2))) # StopIteration
``````
``````# from Ashwini Chaudhary's answer
next((i for i, x in enumerate(myList) if x), None) # 3
next((i for i, x in enumerate(myList2) if x), None) # None
``````
You have to handle special case.
• Great alternative too, main difference from `next` is that `next` provides a default value whereas this would raise an exception. Commented Mar 30, 2015 at 12:18
• for python 3 you can use `myList.index(next(filter(lambda x: x!=0, myList)))` Commented Sep 29, 2016 at 13:48
• I like this answer best, but with the Matthias mod to use "next". Note that in either case, depending on the situation you should consider a test of any(myList) first, to make sure there is at least one nonzero value. It may help with program flow and avoiding exceptions. Commented Apr 9, 2019 at 21:01
You can use numpy.nonzero:
``````myList = [0.0, 0.0, 0.0, 2.0, 2.0]
I = np.nonzero(myList)
# The first index is necessary because the vector is within a tuple
first_non_zero_index = I[0][0]
# 3
``````
Here's a one liner to do it:
``````val = next((index for index,value in enumerate(myList) if value != 0), None)
``````
Basically, it uses next() to find the first value, or return `None` if there isn't one. enumerate() is used to make an iterator that iterates over index,value tuples so that we know the index that we're at.
Use this:
``````[i for i, x in enumerate(myList) if x][0]
``````
Using `next` with `enumerate` is excellent when the array is large. For smaller arrays, I would use `argmax` from NumPy so that you won't need a loop:
``````import numpy as np
myList = [0.0, 0.0, 0.0, 2.0, 2.0]
myArray = np.array(myList)
np.argmax(myArray > 0)
3
``````
• This will not find the first nonzero element, e.g. for `[0, 0, 1, 2]`. Commented Oct 8, 2018 at 13:39
• Yes it will, but it's not very intuitive. All elements in `myArray > 0` are either `False` or `True`, so the maximum of that array is `True`. According to the documentation for numpy.argmax, "In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned." So with your example, if `myArray` is `[0, 0, 1, 2]`, then `myArray > 0` is `[False, False, True, True]` and the `argmax` of that is 2. Commented Feb 13, 2023 at 14:23
What about using enumerate? Check the enumerate documentation.
``````def first_non_zero(mylist):
for index, number in enumerate(mylist):
if number != 0: # or 'if number:'
return index
``````
Do the following:
``````print (np.nonzero(np.array(myList))[0][0])
``````
This is more convenient, because along with finding nonzero values, this can also help to apply logic function directly. For example:
``````print (np.nonzero(np.array(myList)>1))
``````
• Perhaps elaborate a little bit in your answer how floating point comparisons are handled? Presumably it is not exact comparison. (But without "Edit:", "Update:", or similar - the answer should appear as if it was written today.) Commented Feb 4, 2022 at 13:15
Simply use a list comprehension:
``````myDict = {x: index for index, x in enumerate(myList) if x}
``````
The indices of the nonzero elements are `myDict[element]`.
• This doesn't give the index, it just filters out the non-zero elements from the list. Commented Oct 21, 2013 at 18:59 | 1,558 | 5,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.879943 |
http://www.vias.org/comp_geometry/geom_3d_cyl_generic.html | 1,542,663,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00319.warc.gz | 519,922,290 | 2,253 | The Compendium Geometry is an eBook providing facts, formulas and explanations about geometry.
## Generalized Cylinder
A generalized cylinder is generated by moving a straight line g along an arbitrary curve. A generalized cylinder need not to be closed and is sometimes called cylinder surface. A cylinder is called a right cylinder if its cross sections lie directly on top of each other; otherwise, the cylinder is said to be oblique.
The most common special case of a generalized cylinder is a right circular cylinder (or cylinder, for short). Prisms also are a special case of a closed generalized cylinder.
The volume V of the generalized cylinder is given by
V = Abh
with Ab being the area of the basis of the cylinder, and h being the height. The surface S of the (closed) cylinder is given by
S = 2Ab + mpcs
with pcs being the perimeter of the perpendicular cross section of the cylinder.
Last Update: 2010-12-06 | 198 | 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} | 3.125 | 3 | CC-MAIN-2018-47 | latest | en | 0.945809 |
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Lazy Tennis (Posted on 2006-03-27)
In a game of tennis, the player who puts in the most effort in a match, and wins the majority of points, does not necessarily win the match as a whole.
Imagine two tennis players compete in a 5-set match, with each set following the scoring system of tennis, and a first to 7 point tie-break takes place if the score in a set is 6 games each. Let the total number of points won by the person who wins the match be represented by W, and let the total number of points won by the person who loses the match be represented by L.
If by the end of the match L-W is equal to a POSITIVE integer, then what is the maximum value this integer can be? Furthermore, develop an equation to determine the integer formed from L-W for a match of x number of sets.
Note: Enough information regarding the scoring system in tennis required to solve the problem, can be found at http://tennis.about.com/cs/beginners/a/beginnerscore.htm
No Solution Yet Submitted by Chris, PhD Rating: 4.3333 (6 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): Solution, Standing on the shoulders of Tomarken. | Comment 5 of 24 |
(In reply to re(2): Solution, Standing on the shoulders of Hugo. by tomarken)
That seems to be a more correct solution Tomarken, I made an error: I didn't gave the three points to L when the game was lost.
I am not sure if the tie break has to be won 8 to 6 in this problem, as it says a first to 7 point tie-break. And the rules in the given link don't clarify it
Posted by Hugo on 2006-03-27 10:49:01
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https://www.autoitscript.com/forum/topic/165917-using-a-function-inside-of-a-loop/ | 1,722,863,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00618.warc.gz | 515,083,582 | 28,446 | # Using a function inside of a loop
## Recommended Posts
Hey guys, total newbie here, hoping you guys could lend me a hand. I've done some Googling, but I think I am having trouble phrasing what I am trying to do.
Basically, I have a task that involves copying data from a spreadsheet and copying it into another sheet on a daily basis. The two sheets are structured exactly the same every day, so I figured it would be a good candidate for an AutoIT script. Essentially, I am trying to create a loop that will go through each column and copy out the data that I need. While creating the loop, I found that there were several sections where 4-10 of the steps were exactly the same, before going to a new line and having to select different cells/areas. What I would like to do is create functions for these repetitive steps, call them, then continue on with the loop. Example pasted below:
Function:
```Func Excel()
\$i=0
while \$i = 0
Send("^c")
\$i=\$i+1
sleep(300)
WEnd
while \$i = 1
MouseClick("left", 450, 1060, 1)
\$i=\$i+1
sleep(300)
WEnd
while \$i = 2
Send("^v")
\$i=\$i+1
sleep(300)
WEnd
while \$i = 3
Send("{RIGHT}")
\$i=\$i+1
sleep(300)
WEnd
while \$i = 4
MouseClick("left", 450, 1060, 1)
\$i=\$i+1
sleep(300)
WEnd
EndFunc```
Then, after writing the function, I am trying to call it in a loop as seen below:
```while \$i = 1
MouseClick("left", 450, 1060, 1)
\$i=\$i+1
sleep(300)
WEnd
while \$i = 2
Send("^v")
\$i=\$i+1
sleep(300)
WEnd
while \$i = 3
Send("{RIGHT}")
\$i=\$i+1
sleep(300)
WEnd
while \$i = 4
call("excel")
WEnd
EndFunc
while \$i = 5
MouseClick("left", 450, 1060, 1)
\$i=\$i+1
sleep(300)
WEnd```
ETC...
Essentially, the issue I am having, is that the function is successfully called and executes properly, but the loop stops at the function and will not continue on to the next step (\$i = 5 in this case).
That is the gist of it, hopefully I explained it adequately. I am just starting out on this, so I'm sure the code could be tightened up, and I appreciate any time and responses I receive.
Thanks,
##### Share on other sites
@Madrocks - When you post code, please use the provided code quotes, which is to the immediate left of the normal quotes button in the post editor. This makes it far easier for us to read. Don't highlight your code and then use the button, as you need to tell the code editor, it is AutoIt code ... use copy and paste.
That said, I don't see where you are calling a function in a loop, just a lot of steps within a function.
Admittedly I have only taken a quick look at your code, and not taken it apart.
You should use MsgBox's throughout your code, to check on variables etc, to test your logic flow.
Here is a very simple representation of using a Loop and a function, where the row number for instance keeps increasing by one.
```Global \$i, \$row
For \$i = 1 To 5
\$row = \$i
MyFunction()
Next
Exit
Func MyFunction()
do this to \$row
do this to \$row
do this to \$row
etc
EndFunc```
P.S. Welcome to the forum.
Edited by TheSaint
Make sure brain is in gear before opening mouth!
Remember, what is not said, can be just as important as what is said.
Spoiler
What is the Secret Key? Life is like a Donut
If I put effort into communication, I expect you to read properly & fully, or just not comment.
Ignoring those who try to divert conversation with irrelevancies.
If I'm intent on insulting you or being rude, I will be obvious, not ambiguous about it.
I'm only big and bad, to those who have an over-active imagination.
I may have the Artistic Liesense to disagree with you. TheSaint's Toolbox (be advised many downloads are not working due to ISP screwup with my storage)
##### Share on other sites
I added the code quotes and separated them a little bit better above. I see what you are saying with the counting row numbers. Let me see if I can iterate it a little bit better below.
I essentially am trying to go row by row to perform this task, but there are certain blocks where it is the same steps, something like this:
1
2
3
A
B
C
D
E
F
4
5
6
A
B
C
D
E
F
7
8
9
etc
What I am trying to do is essentially create a function for A-F, then write it as
1
2
3
Call function
4
5
6
Call function
etc
in order to save time writing out all of that duplicate code.
Please let me know if that helps.
Thanks,
##### Share on other sites
I'm still not seeing the representation of your function within a loop, just loops within a function.
I added an Exit etc to try and make the logic a little clearer.
When the process gets to Exit your script/program quits.
Row number always remains the same number during the function stage of the process, incrementing each time the function is called during the For...Next loop.
Edited by TheSaint
Make sure brain is in gear before opening mouth!
Remember, what is not said, can be just as important as what is said.
Spoiler
What is the Secret Key? Life is like a Donut
If I put effort into communication, I expect you to read properly & fully, or just not comment.
Ignoring those who try to divert conversation with irrelevancies.
If I'm intent on insulting you or being rude, I will be obvious, not ambiguous about it.
I'm only big and bad, to those who have an over-active imagination.
I may have the Artistic Liesense to disagree with you. TheSaint's Toolbox (be advised many downloads are not working due to ISP screwup with my storage)
##### Share on other sites
```while \$i = 4
call("excel")
WEnd
EndFunc```
Line 35 of the second box above, during the while loop I try to call the function "Excel" that I created in the first box. Perhaps I am not using the right word, I am just trying to nest several commands into a single line so that I don't have to type all of the duplicate code.
Thanks,
##### Share on other sites
If that was there all along, then sorry, I missed it.
I never use Call (well not in recent history anyway), so perhaps it just didn't register when I went zipping through.
You can just use - excel()
Saves on typing.
Make sure brain is in gear before opening mouth!
Remember, what is not said, can be just as important as what is said.
Spoiler
What is the Secret Key? Life is like a Donut
If I put effort into communication, I expect you to read properly & fully, or just not comment.
Ignoring those who try to divert conversation with irrelevancies.
If I'm intent on insulting you or being rude, I will be obvious, not ambiguous about it.
I'm only big and bad, to those who have an over-active imagination.
I may have the Artistic Liesense to disagree with you. TheSaint's Toolbox (be advised many downloads are not working due to ISP screwup with my storage)
##### Share on other sites
Are you copying data from one Excel spreadsheet to another?
If so, then I'm left wondering why you are using AutoIt for most of this, and not just running a VBA macro within Excel?
If need be, you can use AutoIt, to run that macro, easily enough.
Far better to do things within Excel, using its own VBA commands, especially as you can avoid using the Clipboard and mouse clicks, which can be prone to errors.
Edited by TheSaint
Make sure brain is in gear before opening mouth!
Remember, what is not said, can be just as important as what is said.
Spoiler
What is the Secret Key? Life is like a Donut
If I put effort into communication, I expect you to read properly & fully, or just not comment.
Ignoring those who try to divert conversation with irrelevancies.
If I'm intent on insulting you or being rude, I will be obvious, not ambiguous about it.
I'm only big and bad, to those who have an over-active imagination.
I may have the Artistic Liesense to disagree with you. TheSaint's Toolbox (be advised many downloads are not working due to ISP screwup with my storage)
##### Share on other sites
@Madrocks - When you post code, please use the provided code quotes, which is to the immediate left of the normal quotes button in the post editor. This makes it far easier for us to read. Don't highlight your code and then use the button, as you need to tell the code editor, it is AutoIt code ... use copy and paste.
That said, I don't see where you are calling a function in a loop, just a lot of steps within a function.
Admittedly I have only taken a quick look at your code, and not taken it apart.
You should use MsgBox's throughout your code, to check on variables etc, to test your logic flow.
Here is a very simple representation of using a Loop and a function, where the row number for instance keeps increasing by one.
```Global \$i, \$row
For \$i = 1 To 5
\$row = \$i
MyFunction()
Next
Exit
Func MyFunction()
do this to \$row
do this to \$row
do this to \$row
etc
EndFunc```
P.S. Welcome to the forum.
very very nice... I love it..
ill get to that... i still need to learn and understand a lot of codes
Correct answer, learn to walk before you take on that marathon.
##### Share on other sites
It's not actually Excel, its a proprietary application that I can't really get too much into the details of. This particular task is pretty much just copy and pasting, and using AutoIT to automate all the clicks will take a ~1.5 hour task down to about ~10 minutes. I actually have the script mostly written and working, I just really want to trim some lines of code out of the script before I present it to anyone.
So the format you're saying would be the following?
```while \$i = 4
- excel()
WEnd```
##### Share on other sites
Yes that is correct ... obviously without that dash in front of excell()
(I just use 5 spaces etc to give indents to code in the code browser.)
As I did in my example above, I had some of the necessary work, like the row number, done in the main loop that calls the function you want repeated. Any differences between rows (lines, etc) should be done in the main loop, before the function is called. If you make you variables Global, as I did, probably at the beginning of your script, then they are passed around very effectively. The function should only really work on one row (line) at a time.
I know you are using While...Wend, but it may be more effective to use For...Next as I did. But that depends on the rest of your code. The former can problematic and if you make a coding error, and \$i never equals 4, then you can have the process stuck in a loop. That said, I use both methods all the time.
If you are troubleshooting, and using a MsgBox is an issue, as it can be, and using the SciTE console is not an option, then I find, that writing values from variables to a text or INI file, at key moments, can be very useful.
Where a MsgBox can be an issue, is where your active window then loses focus, and is no longer active.
Most code using Clipboard and Mouse functions, often require you to re-establish the active window from time to time, using WinActivate etc. A lot of issues can be traced to that.
Edited by TheSaint
Make sure brain is in gear before opening mouth!
Remember, what is not said, can be just as important as what is said.
Spoiler
What is the Secret Key? Life is like a Donut
If I put effort into communication, I expect you to read properly & fully, or just not comment.
Ignoring those who try to divert conversation with irrelevancies.
If I'm intent on insulting you or being rude, I will be obvious, not ambiguous about it.
I'm only big and bad, to those who have an over-active imagination.
I may have the Artistic Liesense to disagree with you. TheSaint's Toolbox (be advised many downloads are not working due to ISP screwup with my storage)
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1. 60
2. 90
3. 75
4. 45
4
90
Explanation :
No Explanation available for this question
In an organization, 40% of the employees are matriculates, 50% of the remaining are graduates and remaining 180 are post-graduates. How many employees are graduates
1. 360
2. 240
3. 180
4. 300
4
180
Explanation :
No Explanation available for this question
A 130m long train crosses a bridge in 30 seconds at 45 kmph. The length of the bridge is
1. 200m
2. 225m
3. 245m
4. 250m
4
245m
Explanation :
No Explanation available for this question
By selling an article at some price a person gains 10%. If the article is sold at twice of the price, the gain percent will be
1. 20%
2. 60%
3. 100%
4. 120%
4
120\%
Explanation :
No Explanation available for this question
The ratio of the radius and height of a cone is 5:12, respectively. Its volume is 23147cc. Find its slant height.
1. 13cm
2. 14cm
3. 17cm
4. 26cm
4
13cm
Explanation :
No Explanation available for this question
27 students took part in a debate of a college. What is the probability that at least 3 of them have their birth days in the same month
1. 3/27
2. 3/12
3. 12
4. 1
4
12
Explanation :
No Explanation available for this question
A number consists of two digits whose sum is 8. If 8 is subtracted from the number, the digits interchange their places. The number is:
1. 44
2. 35
3. 62
4. 33
4
33
Explanation :
No Explanation available for this question
A horse is tied to a peg hammered at one of the corner of a rectangular grass field of 40 m by 24 m by a rope 14 m long. Over how much area of the field can the horse graze
1. 154 m2
2. 308 m2
3. 240 m2
4. 480 m2
4
154 m2
Explanation :
No Explanation available for this question
The sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 30 cm. The length of the greatest side of the triangle in cm is:
1. 6
2. 10
3. 14
4. 16
4
14
Explanation :
No Explanation available for this question
1. 12 sq. cm
2. 15 sq. cm
3. 18 sq. cm
4. 21 sq. cm
4 | 716 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.81186 |
http://www.jiskha.com/display.cgi?id=1359537052 | 1,498,142,076,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319575.19/warc/CC-MAIN-20170622135404-20170622155404-00466.warc.gz | 640,596,248 | 3,809 | # math
posted by on .
A horse is tied to a post by a rope, 8m long at one corner of a rectangular field. What area of the field can the horse graze?
solution pls... tnx...
• math - ,
Draw yourself a picture. The horse can cover a circular area of pi*R^2. But at the corner of a rectangular field (with width longer than the rope), only one-quarter of the circle will be in the field.
• math - ,
The rope of horse =14m long | 110 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.890813 |
https://www.nagwa.com/en/videos/876106785627/ | 1,603,265,222,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876136.24/warc/CC-MAIN-20201021064154-20201021094154-00578.warc.gz | 838,044,731 | 7,011 | # Video: CBSE Class X • Pack 3 • 2016 • Question 12
CBSE Class X • Pack 3 • 2016 • Question 12
05:27
### Video Transcript
In the given figure, a tent is the shape of a cylinder surmounted by a conical top of the same diameter. The height and diameter of the cylindrical part are 2.1 meters and three meters, respectively, and the slant height of the conical part is 2.8 meters. Determine the cost of the canvas needed to make the tent if the canvas is available at a rate of 500 rupees per meter squared. Use 𝜋 is equal to 22 over seven.
Now for this question, we’ve been asked to find the cost of the canvas needed to make the tent. In order to do so, we’ll first need to find the area of the canvas that makes up the tent in meters squared. And we’ll then need to multiply this by 500 to find the cost in rupees. The question tells us that the tent has one part which is a cylinder and one part which is a cone.
Looking at the diagram, we can understand that the canvas is split into two distinct parts. One is the curved surface area of the conical top, and the other is the curved surface area of the cylinder. To answer this question, we can first write out the formula for the curved surface area of a cone, which is given by 𝜋 times 𝑟 times 𝑙. Next, we can write out the formula for the curved surface area of a cylinder, which is given by two 𝜋 times 𝑟 times ℎ.
Let’s now define the terms written in these equations. Firstly, we can see that both equations contain the term 𝑟. This term 𝑟 is the radius of the cone and the cylinder, respectively. Looking back at the question, we’re told that both the cone and the cylinder have the same diameter, and this is three meters. What this really means is the circle that makes up the base of our cone and the end of our cylinder has a diameter of three meters.
Now we also know that the radius of any circle is half its diameter. This allows us to say that the radius of our circle, and therefore the radius of both our cone and our cylinder, is three over two meters. For now, we’re gonna choose to leave this as a fraction as we may be able to cancel out some factors later.
Now that we have defined 𝑟, let us look at the 𝑙 in the formula for the curved surface area of a cone. This 𝑙 represents the slant height of the cone, and our question has given us this value as 2.8 meters. Finally, we look at the ℎ in the formula for the curved surface area of a cylinder. This ℎ represents the height of the cylinder, and our question has also given us this value, which is 2.1 meters.
Now that we’ve defined all our terms, let’s work on finding the total area of the canvas. We can now remind ourselves that the total area of the canvas is equal to the curved surface area of the cone added to the curved surface area of the cylinder.
Before we substitute in our values, let’s first see what happens when we add the two expressions together. Adding the two expressions together gives us 𝜋𝑟𝑙 plus two 𝜋𝑟ℎ. We can see that both of our terms have a factor of 𝜋𝑟. We’re therefore able to factor this out, to obtain 𝜋𝑟 times 𝑙 plus two ℎ.
Let’s now substitute in the values that we know. Firstly, we have that 𝑟, the radius of the cone and the cylinder, is three over two. Next, we have that 𝑙, the slant height of the cone, is 2.8. And lastly, we have that ℎ, the height of the cylinder, is equal to 2.1. We can now work on our second set of brackets, firstly by evaluating that two times 2.1 is equal to 4.2. We can then see that 2.8 plus 4.2 is a nice integer number, which is seven.
Now that we have simplified, we can recall that the question gives us an approximate value for 𝜋, which is 22 divided by seven. We can therefore also substitute this value into our expression. After doing this, we see that we can perform some more canceling, firstly by recognizing that we have a factor of one over seven and seven, which evaluates to one. So we can cancel these two out. We also see that we have a factor of 22 and a half, and 22 divided by two is 11. This leaves us with the expression 11 times three, which of course evaluates to 33. Since this represents the total area of the canvas, we’ll recall that our units should be in meters squared. And we have therefore found that the total area of the canvas is 33 meters squared.
After finding the total area of the canvas, we must now calculate the cost of the canvas. The question tells us that the canvas is available at a rate of 500 rupees per meter squared. This means that we must multiply 500 by each square meter of canvas that we have in the tent. Our calculation is therefore 500 times 33.
Now we might find this calculation easier if we say that 500 is equal to 1000 times a half. We can then evaluate that half of 33 is 16.5, and 1000 times 16.5 is 16500. In completing this step, we have answered the question. And we should remember that our units for cost in this case are rupees. We have therefore found that the cost of the canvas to make this tent is 16500 rupees. | 1,241 | 4,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2020-45 | latest | en | 0.950075 |
https://community.fabric.microsoft.com/t5/Desktop/Average-of-other-measure-is-this-possible/m-p/3544519 | 1,723,081,317,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00113.warc.gz | 142,180,514 | 53,911 | cancel
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Frequent Visitor
## Average of other measure - is this possible?
Dear Memebrs,
sicne couple of days I'm struggling with PBI on how to calucalte the average of another measure?
Case as follows:
I do have a table with following meaures:
1. Sales per rep
2. Rep Count (distinct count)
3. Business Count (distinct count - how many business given rep supports)
4. Average of business supported by a rep
Table layout
Measure are calculated as follows:
OURep
The only thing I want to achieve is somehow re-calcualte avrage on Manager level to show average of all averages OR sum of all distinct counts on OU level divided by sum of all reps - desired view below:
Desired view
I'm totally vulenrable now - thought it's easy-peasy task but stucked with it for days... Help, please anyone!
1 ACCEPTED SOLUTION
Frequent Visitor
Hello,
thank for you help - already managed to find out the solution for my problem - works like a charm now!
``````# OU* =
COUNTX (
FILTER (
SUMMARIZE (
'DB',
'DB'[Scenario],
'DB'[Sales Rep Number],
'DB'[OU Hierarchy],
"Count OU", COUNT ( 'DB'[OU Hierarchy] ),
"Sales Rep", SUM ( 'DB'[USD] )
),
'DB'[Scenario] = "CY_ACT"
&& [Sales Rep] > 0
),
[Count OU]
)``````
3 REPLIES 3
Frequent Visitor
Hello,
thank for you help - already managed to find out the solution for my problem - works like a charm now!
``````# OU* =
COUNTX (
FILTER (
SUMMARIZE (
'DB',
'DB'[Scenario],
'DB'[Sales Rep Number],
'DB'[OU Hierarchy],
"Count OU", COUNT ( 'DB'[OU Hierarchy] ),
"Sales Rep", SUM ( 'DB'[USD] )
),
'DB'[Scenario] = "CY_ACT"
&& [Sales Rep] > 0
),
[Count OU]
)``````
Community Support
You can refer to the following solution.
Sample data
Create a measure
``````Measure =
VAR a =
SUMMARIZE (
ALLSELECTED ( 'Table' ),
[Manager],
[Column1],
"Average", AVERAGE ( 'Table'[Column3] )
)
RETURN
IF (
ISINSCOPE ( 'Table'[Column1] ),
CALCULATE ( AVERAGE ( 'Table'[Column3] ) ),
AVERAGEX ( FILTER ( a, [Manager] IN VALUES ( 'Table'[Manager] ) ), [Average] )
)
``````
Output
Best Regards!
Yolo Zhu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
Please provide sample data (with sensitive information removed) that covers your issue or question completely, in a usable format (not as a screenshot). Leave out anything not related to the issue.
If you are unsure how to do that please refer to https://community.fabric.microsoft.com/t5/Community-Blog/How-to-provide-sample-data-in-the-Power-BI-...
Please show the expected outcome based on the sample data you provided.
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Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data. | 768 | 2,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.893167 |
https://discuss.leetcode.com/topic/63853/most-straightforward-recursive-java-solution | 1,516,320,485,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887660.30/warc/CC-MAIN-20180118230513-20180119010513-00764.warc.gz | 670,158,384 | 8,837 | # Most Straightforward Recursive Java Solution.
• I came up with a more straightforward way to solve this problem, a little bit like the recursive way above. Instead of do it in the original function, I traverse the tree in inOrder sequence but with the tree side information.
``````int sum;
public int sumOfLeftLeaves(TreeNode root) {
sum = 0;
inOrder(root, false); // false means the next node is not in left side.
return sum;
}
private void inOrder(TreeNode root, boolean left) {
if (root == null) {
return;
}
inOrder(root.left, true); // true means the next node is in left side.
if (root.left == null && root.right == null && left) { //only left leaves would be added to sum.
sum += root.val;
}
inOrder(root.right, false);
}``````
• @ZhuEason Now I realized how unnecessarily complicated my recursion is LOL
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 217 | 915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.785494 |
https://math.stackexchange.com/questions/3428203/r2-with-n-points-removed-is-a-bouquet-of-n-circles | 1,718,680,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00419.warc.gz | 345,360,765 | 36,422 | # $R^2$ with $n$ points removed is a bouquet of $n$ circles
It seems almost obvious that $$\mathbb{R}^2$$ with $$n$$ points removed is homotopy equivalent to the wedge sum of $$n$$ circles. Take sufficiently small circles around the removed points, connect each circle to the next one using non-intersecting paths, and take a deformation retraction which radially projects the points inside each circle out from the removed point, and deformation retracts everything outside onto the circles and paths connecting them. Then the paths can be deformation retracted to points and the points connecting circles pairwise can be brought to a single point by deformation retracting arcs of the circles between two such points.
While the visual intuition is clear, I have yet to encounter a rigorous proof. Constructing an explicit deformation retraction seems daunting. A stronger claim which could help is the claim for any finite sets $$F_1,F_2\subset\mathbb{R}^2$$, if $$\vert F_1\vert=\vert F_2\vert$$, then $$\mathbb{R}^2\setminus F_1$$ is homeomorphic to $$\mathbb{R}^2\setminus F_2$$. If I had this stronger claim, then I think I could prove explicitly that $$\mathbb{R}^2$$ with $$n$$ points removed is homotopy equivalent to a wedge sum of $$n$$ circles. However, the stronger claim seems even more daunting. Does anyone know of a rigorous proof?
Let $$F \subset \mathbb R^2$$ be a finite set. We want to construct a homeomorphism $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F$$ to a fixed finite set of size $$n$$. We will adapt the convention that "applying a map $$f$$ to a point $$p$$" means that the letter $$p$$ now denotes what was previously $$f(p)$$. This gives the following procedure an algorithmic feel, which is easier to follow I hope.
First, note that we can ensure that one of the elements of $$F$$ is zero by just applying a translation. Once we are in this situation, let $$p \in F$$. Then applying a rotation (around zero) arranges that $$p$$ gets mapped to the $$x$$-axis. Applying a transformation of the form $$(x,y) \mapsto (x, \lambda y)$$ for big enough $$\lambda$$ ensures that all $$p \in F \setminus x\text{-axis}$$ have modulus strictly greater than all $$q \in x\text{-axis}$$.
Now comes the key step. Let $$R_{\alpha}: \mathbb R^2 \rightarrow \mathbb R^2$$ be a rotation (around zero) of angle $$\alpha$$ mapping a point $$p \in F \setminus x\text{-axis}$$ of minimal modulus to the $$x$$-axis. Then let $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ be given as $$f(s) = \begin{cases} s & |s| \leq \max_{q \in x\text{-axis}}|q| \\ R_{(L(|s|))\alpha}(s) & \max_{q \in x\text{-axis}}|q|< |s|<\min_{q \in F \setminus x\text{-axis}}|q| \\ R_\alpha & |s| \geq \min_{q \in F \setminus x\text{-axis}} |q|, \end{cases}$$ where $$L: \mathbb R \rightarrow R$$ is a linear function with $$L(\max_{q \in x\text{-axis}} |q|)= 0$$ and $$L(\min_{q \in F\setminus x\text{-axis}}|q|) = 1$$. This $$f$$ can be visualised as some kind of rotation twisting only outside of a ball around zero. Applying this function and again some $$(x,y) \mapsto (x, \lambda y)$$ produces the same situation but with strictly more points on the $$x$$-axis.
Proceeding by induction, we obtain that there exists an $$f: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F$$ to the $$x$$-axis. All that is left to show is that there exists a homeomorphism $$h: \mathbb R^2 \rightarrow \mathbb R^2$$ mapping $$F \subset \mathbb R$$ to some fixed set, say $$\{0,1, \dots, |F|-1\}$$. But this is easy, just consider some piecewise linear function $$h$$ that does this, and then apply $$(x,y) \mapsto (h(x), y)$$ to obtain our desired final function mapping $$F$$ to a specific set, namely $$\{(0,0), (0,1), \dots, (0, |F|-1)\}$$.
• This is great! Thank you! I imagine this could be generalized to $\mathbb{R}^m$ inductively by using the same method to reduce it to $\mathbb{R}^{m-1}$. This definitely puts my nerves at ease. Commented Nov 9, 2019 at 21:48 | 1,171 | 3,953 | {"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": 48, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.914018 |
https://mathtuition88.com/tag/puzzle/ | 1,675,010,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00564.warc.gz | 408,394,021 | 26,021 | ## Homeschool Maths Puzzles with Answers: The “Average” Question
The average of 5 numbers is 73. When 2 numbers were removed, the average decreased by 3. What is the average of the 2 numbers that were removed?
This is part of a series on Homeschool Math Challenging Puzzles, suitable for Grades 2-4. (Of course, students of other grades are also welcome to try them out.) The questions are suitable for:
• Preparation for GEP (Gifted Education Programme) screening and selection tests
• Puzzles for kids interested in math but find school work too easy.
This is literally an “average” question. 😛 The word “average” appears 3 times.
The trick is to focus on the total instead.
The total of the 5 numbers is: 73×5=365.
After removing 2 numbers, the average is 73-3=70.
Hence, the total of the 3 remaining numbers is: 70×3=210.
We can then conclude that, the total of the 2 removed numbers is:
365-210=155.
Hence, the average of the 2 removed numbers is: 155/2=77.5 or 75 1/2.
## Slow and Steady Snail (Homeschool Math Favorite Challenging Puzzles)
This is a favorite type of homeschool math challenging puzzle — The Snail question.
Question: An aquarium is 47 cm deep. A snail starts at the bottom of the aquarium. Each day, during the daytime the snail climbs up 8 cm, and during the nighttime the snail slides down 3 cm. How many days will it take for the snail to climb out of the aquarium?
This is part of a series on Homeschool Math Challenging Puzzles, suitable for Grades 2-4. (Of course, students of other grades are also welcome to try them out.) The questions are suitable for:
• Preparation for GEP (Gifted Education Programme) screening and selection tests
• Puzzles for kids interested in math but find school work too easy.
Solution:
A tempting answer would be 10 days. This is a trick! Those who get the answer “10 days” reason like this: each day the snail moves a net distance of 8-3=5 cm.
Hence, 47/5=9 R 2.
At the end of the 9th day, the snail moved 45 cm. Thus, rounding up will give 10 days as the answer. However, there is a tricky part to the question!
The correct answer is 9 days.
We can do a list:
Day 1 — 5 cm
Day 2 — 10 cm
Day 3 — 15 cm
Day 8 — 40 cm
Day 9 day time — 40 +8 = 48cm > 47 cm !!!
The snail is already out of the aquarium on Day 9!
## Stickers Math Question
Abby, Brian, Charles, Dennis and Eason have 50 stickers altogether.
Abby has the most stickers — she has 12 stickers.
In second place (tied) are Brian and Charles.
In third place is Dennis.
In fourth place is Eason, with 6 stickers.
How many stickers does Dennis have?
This is part of a series on Homeschool Math Challenging Puzzles, suitable for Grades 2-4. (Of course, students of other grades are also welcome to try them out.) The questions are suitable for:
• Preparation for GEP (Gifted Education Programme) screening and selection tests
• Puzzles for kids interested in math but find school work too easy.
For this type of questions, the easiest way to do is “trial and error”, also known as “guess and check”.
Firstly, lets check how many stickers B, C, and D have together:
50-12-6=32
B C D Total (B+C+D) Comments 9 9 14 32 Wrong! Since D is more than A 10 10 12 32 Wrong! Since D is tied with A 11 11 10 32 Correct! 12 12 8 32 Wrong! Since B, C is tied with A
For guess and check, the most important thing is to be systematic, rather than guess wildly. For instance, we can systematically increase our guesses for B, C.
We can see that the only logical answer is:
B= C=11
D=10
## 成语数学
An alternative answer to Q1) 20 除 3 is “陆续不断”.
20除以3，因为它的答案接近于6.6666，所以这道题的答案是陆续不断，或者是六六大顺都行，百分之一就是百里挑一，9寸加1寸等于一尺即是得寸进尺，12345609，七零八落，1、3、5、7、9无双数所以叫做举世无双，或者你把它答出天下无双都行，如此小升初的难题您答对了吗？
1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9
1) 20/3= 6.666 六六大顺
2）百中挑一
3）得寸進尺
4）七零八落
5）举世无双
View original post | 1,191 | 3,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.940501 |
https://www.physicsforums.com/threads/calculating-ixy.702448/ | 1,477,511,632,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720972.46/warc/CC-MAIN-20161020183840-00217-ip-10-171-6-4.ec2.internal.warc.gz | 947,641,203 | 15,724 | # Calculating Ixy ?
1. Jul 22, 2013
### phydis
Here as shown in the diagram,
(1)Is Ixy zero for plane ABCD symmetrical along X axis? If yes, how it happens?
(2)How can we calculate Ixy for the plane PQRS ?
Ixy = the product of moment of area.
Thanks!
#### Attached Files:
• ###### Ixy.jpg
File size:
8.2 KB
Views:
283
2. Jul 22, 2013
### SteamKing
Staff Emeritus
For any plane figure with an axis of symmetry, Ixy = 0 for coordinate axes thru the centroid of the figure.
To confirm this, examine the definition of Ixy = Int(x*y)dA For every element of area dA which lies on one side of the axis of symmetry, there will be a corresponding element of area lying on the opposite side which is the same distance from the axis. When you add up all of these elements of area multiplied by x*y, the integral is necessarily equal to zero.
To calculate the Ixy of rectangle PQRS, apply the parallel axis theorem. | 240 | 915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-44 | longest | en | 0.851499 |
https://www.codingbroz.com/python-program-to-find-cube-of-a-number/ | 1,723,278,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00640.warc.gz | 542,364,560 | 65,260 | # Python Program to Find Cube of a Number
In this post, we will learn how to find the cube of a number using Python Programming language.
The number that is obtained by multiplying an integer to itself three times is known as the cube of a number. For example: The cube of 2 is 2 x 2 x 2 = 8.
We will be using the following ways to find the cube of a number.
1. Using Standard Method
2. Using Exponent Method
3. Using Functions
So, without further ado, let’s begin this tutorial.
## Python Program to Find Cube of a Number
```# Python Program to Find Cube of a Number
num = int(input("Enter an integer: "))
# Calculating cube
cube = num * num * num
# Displaying output
print("Cube of {0} is {1}" .format(num, cube))
```
Output
``````Enter an integer: 7
Cube of 7 is 343
``````
## How Does This Program Work ?
```num = int(input("Enter an integer: "))
```
In this program, the user is asked to enter an integer.
```# Calculating cube
cube = num * num * num
```
Then, we calculate the cube of the entered number by multiplying the number itself by three times.
```# Displaying output
print("Cube of {0} is {1}" .format(num, cube))
```
Finally, the cube of the number is displayed on the screen using print() function.
## Python Program to Find Cube of a Number Using Exponent
```# Python Program to Find the Cube of a Number Using Exponent
num = int(input("Enter an integer: "))
# Calculating cube
cube = num ** 3
# Displaying output
print("Cube of {0} is {1}" .format(num, cube))
```
Output
``````Enter an integer: 6
Cube of 6 is 216
``````
## Python Program to Find Cube of a Number Using Functions
```# Python Program to Find Cube of a Number Using Functions
def cube(num):
return num * num * num
num = int(input("Enter an number: "))
# Calling out function
cube_num = cube(num)
# Displaying output
print("Cube of {0} is {1}" .format(num, cube_num))
```
Output
``````Enter an number: 13
Cube of 13 is 2197
``````
## Conclusion
I hope after going through this post, you understand how to find the cube of a number using Python Programming language.
If you have any doubt regarding the program, feel free to contact us in the comment section. We will be delighted to solve your query. | 585 | 2,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.786175 |
https://web2.0calc.com/questions/pls-help_63432 | 1,620,618,063,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00167.warc.gz | 630,342,328 | 5,383 | +0
# Pls help!
+3
115
1
+272
The following are all 24 numbers made of digits 1, 2, 3, 4 exactly once:
1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,41324213,4231,4312,4321
What is the average of these 24 numbers? Please express the result in decimal form.
Jan 27, 2021 | 138 | 323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.813314 |
https://boardgames.stackexchange.com/questions/12630/in-poker-what-happens-if-paying-a-blind-puts-a-player-all-in/13010 | 1,716,950,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00469.warc.gz | 111,544,928 | 38,676 | # In Poker, what happens if paying a blind puts a player all-in?
I have a question regarding increasing blinds in Texas Hold-em poker. I am developing a poker website. During tournaments players are not allowed to buy chips. Each player starts with 1000 chips to play. We always keep all amounts on the table (player pot, player bet, table pot etc.), in multiples of the smallest denomination, i.e. small blind.
But when blinds are increasing automatically, what should I do in the case when a player is left with a chip smaller than the small blind? Should that player be able to continue the game, or are they busted immediately (and lose the tournament)?
Example : In a tournament of four players, all players begin with 1000 chips to play. After some time, one of the four players remains with 10 chips in the pot, and blinds reach to 500/1000 (SM/BB). What should I do with that player?
If a player cannot cover a blind, that player is all-in and the bets are handled just as if that player had gone all-in on an ordinary bet. The main pot gets an amount of money from each player who bets equal to the all-in player's stake. Any further betting goes into a side pot, which the all-in player is ineligible to win.
• Still one confusion in case of split the main pot when two, three etc. player got equal hand then which should be used as smallest denomination chip i.e. small blind or pot value of player who goes all in ? Aug 5, 2013 at 6:41
• @AjayKhunti: Pot value of the player. If small blind is 1000 and the player only has 500, than 500 of big blind's money goes into the main pot and 1500 into a side pot. Aug 5, 2013 at 18:53
• @AjayKhunti: Consider this an inviolable rule of poker: I can never win more money from you than you can win from me. If I only have 500 chips, then 500 chips is the most I am allowed to win from any player. The entire purpose of a side pot is then to allow other players to make further bets against each other. Feb 6, 2014 at 6:38
In most casinos, \$25 is the smallest tournament chip. However, it may not remain so. Periodically chips are "raced off" or "colored up". How it is done is another topic.
However, lets use something similar to your example: 4 players, one with 10 chips (call him A), the rest with 1000, blinds are 50/100. None of the three players raises preflop, so 4 see the floop. There is a total of 340 in the pot. 40 in the main, which A is able to win, 300 in the side, which A is not able to win. Flop comes. One of the players with chip bets, one folds, the other calls. There is now 40 in the main, 500 in the side.
Say the board is AAAAK. All players have 4 aces with King kicker. They all tie. The two players with chips split the side of 500, they get 250 each.
The main pot is interesting as there is 40 in there and three players to split. Each player gets 10 chips, but there is 10 left over. Who gets the extra chip? The player that is first to act in the hand (like the one who was in the small blind).
• While all of this is true, it's mostly unnecessary in a computer implementation; online poker tends to keep all the chips in play because the primary consideration at play (keeping the quantity of physical chips manageable) is completely irrelevant. That said, even online poker doesn't go to fractional chips so you do have to break ties occasionally, using the "first to act" rule as needed. Feb 6, 2014 at 6:50
• In your example, it should be 310 total in the pot preflop -- 40 in the main and 270 in the side. Feb 6, 2014 at 22:03
It is not different than any other all in.
If someone raises you create a side pot.
Say it is the BB that only has 1/2 a BB. The question is could someone call the 1/2 BB? The common approach is the min bet is still 1 BB.
• Down vote what is the problem? Jan 16, 2017 at 2:59
• I would guess that it is because the question does not mention raising. It asks "if I have \$x, and the blind is \$x ..."; why muddy the waters with raises?
– Mawg
Apr 14, 2021 at 8:34
• My guess is that it's the same answer as sitnaltax's one 4 years prior Jun 11, 2021 at 8:14
calculate the odds, seek for on line poker odds calculator, i think it should not pose a problem. But I think it is necessary to include it so the program will not crash. I would suggest the remainder to go into the next pot or a random generator pops up and notifys players of who gets it. Hm maybe it would be fair the remainder to go to BB. | 1,138 | 4,432 | {"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} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.961775 |
https://educationexpert.net/mathematics/2582281.html | 1,709,360,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00550.warc.gz | 218,791,144 | 7,209 | 13 February, 11:15
# What are the solutions of x^2+3x-4=0
+1
1. 13 February, 13:05
0
x = - 4, x = 1
Step-by-step explanation:
x²+3x-4=0
- 4 = 4 * (-1)
3 = 4 + (-1)
(x + 4) (x - 1) = 0
x + 4 = 0, x - 1 = 0
x = - 4, x = 1 | 127 | 228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.861004 |
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